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This article is cited in 2 scientific papers (total in 2 papers) Short R. O. Stasenkoab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. IntroductionThe classical Tits-Kantor-Köcher construction is known in the theory of Lie algebras. Given a simple Jordan algebra $J$, it allows one to construct a simple Lie algebra $\mathfrak{g}$ of the following form:
$$
\begin{equation}
\mathfrak{g}=\mathfrak{der}(J)\oplus\mathfrak{sl}_2(J).
\end{equation}
\tag{1.1}
$$
In this construction the commutator of elements of the second term is defined as the usual matrix commutator plus a certain addition from the first term. More precisely, if we identify $\mathfrak{sl}_2(J)$ with $\mathfrak{sl}_2(\mathbb{C})\otimes J$, then the commutator of two elements of the second term is calculated by the following formula:
$$
\begin{equation*}
[X\otimes a, Y\otimes b]=2(X, Y)[L_a, L_b]+[X, Y]\otimes ab \qquad\forall\, X, Y\in \mathfrak{sl}_2(\mathbb{C}), \quad a, b\in J,
\end{equation*}
\notag
$$
where $L_c$ denotes the linear operator of multiplication by the element $c\in J$ and
$$
\begin{equation*}
(X, Y)=\operatorname{tr}(XY) \quad\forall\, X, Y\in \mathfrak{sl}_2(\mathbb{C}).
\end{equation*}
\notag
$$
The Tits-Kantor-Köcher construction can be interpreted as a linear representation of the group $\mathrm{SL}_2(\mathbb{C})$ by automorphisms of the algebra $\mathfrak{g}$, which decomposes into irreducible representations of dimension 1 and 3. Equality (1.1) is just an isotypic decomposition of this representation. A natural generalization of this linear representation is the following concept. Definition 1. Let $S$ be a reductive algebraic group. An $S$-structure on the Lie algebra $\mathfrak{g}$ is a homomorphism $\Phi\colon S\to \operatorname{Aut}(\mathfrak{g})$. Using this terminology, the construction due to Tits, Kantor and Köcher will be called a very short $\mathrm{SL}_2$-structure. A complete analysis of this construction can be found in [1], § 3. Some particular cases of $S$-structures were studied in [1] and [2]. We consider another case of $S$-structures, namely, short $\mathrm{SL}_2$-structures. It is most natural to consider short $\mathrm{SL}_2$-structures on semisimple Lie algebras in connection with the remarkable algebraic properties of the latter. However, it is obvious that considering a short $\mathrm{SL}_2$-structure on a semisimple Lie algebra reduces to considering such structures on every simple component of this algebra. Therefore, in what follows, we always assume that the algebra $\mathfrak{g}$ is simple and $S= \mathrm{SL}_2(\mathbb{C})$. The differential of the map $\Phi$ defines a linear representation of the Lie algebra $\mathfrak{sl}_2$ by derivations of $\mathfrak{g}$:
$$
\begin{equation*}
\mathrm{d}\Phi\colon \mathfrak{sl}_2\to\mathfrak{der(g)}.
\end{equation*}
\notag
$$
Since the algebra $\mathfrak{g}$ is simple, it follows that $\mathfrak{der(g)}\simeq\mathfrak{inn(g)}\simeq\mathfrak{g}$. Hence the image of $\mathfrak{sl}_2$ in $\mathfrak{g}$ under the action of $\mathrm{d}\Phi$ consists of the adjoint operators of the algebra $\mathfrak{g}$. The isotypic decomposition of this representation has the form
$$
\begin{equation*}
\mathfrak{g}=\mathfrak{g}_0\oplus \bigoplus_{i=1}^lV_i\otimes J_i.
\end{equation*}
\notag
$$
In this formula $V_i$ denotes the space of $(i+1)$-dimensional irreducible representation of $\mathfrak{sl}_2$, and $J_i$ is the vector space on which $\mathfrak{sl}_2 $ acts trivially and which is responsible for the multiplicity of the occurrence of the irreducible representation $V_i$ in $\mathfrak{g}$ (namely, the dimension of $J_i$ is equal to this multiplicity). The subspace $\mathfrak{g}_0$ is the isotypic component corresponding to the one-dimensional representation of $\mathfrak{sl}_2$. Definition 2. An $\mathrm{SL}_2$-structure is said to be short if the representation $\mathrm{d}\Phi$ decomposes into irreducible representations of dimension $1$, $2$ and $3$. The irreducible representation of an algebra $\mathfrak{sl}_2$ of dimension 2 is its tautological representation, while the irreducible representation of dimension 3 is the adjoint representation. Thus, for short $\mathrm{SL}_2$-structures, the isotypic decomposition has the form
$$
\begin{equation}
\mathfrak{g}=\mathfrak{g}_0\oplus\mathfrak{g}_1\oplus \mathfrak{g}_2, \qquad\mathfrak{g}_1= \mathbb{C}^2\otimes J_1, \quad\mathfrak{g}_2=\mathfrak{sl}_2\otimes J_2.
\end{equation}
\tag{1.2}
$$
It is important to note that the cases $J_1=0$ (which corresponds to a very short $\mathrm{SL}_2$-structure) and $J_1\neq0$ differ significantly. Throughout what follows we assume that $J_1\neq0$. Using the commutation operation on the Lie algebra $\mathfrak{g}$ and also the invariant inner product, we endow the vector space $J_1$ with the structure of a symplectic space and the space $J_2$ with the structure of a Jordan algebra of symmetric operators in $J_1$. Moreover, the algebra $\mathfrak{g}_0$ is a Lie subalgebra of the Lie algebra $\mathfrak{sp}(J_1)$. It turns out that, among the short $\mathrm{SL}_2$-structures with given $J_1$, there is a maximal one, which means that the space $J_2$ and the Lie algebra $\mathfrak{g}_0$ in it are maximal possible with respect to inclusion. Namely, for this short $\mathrm{SL}_2$-structure we have $\mathfrak{g}_0=\mathfrak{sp}(J_1)$, and for the space $J_2$ we have the Jordan algebra of all symmetric operators of the symplectic space $J_1$. We call this Jordan algebra canonical. We describe the maximal short $\mathrm{SL}_2$-structure in § 3 of this paper. In § 4 we construct a correspondence between simple Lie algebras with short $\mathrm{SL}_2$-structure and the so-called simple symplectic Lie-Jordan structures of the form $(J_1; J_2; \mathfrak{g}_0; \delta_0)$, where $J_2$ is a simple Jordan subalgebra of the algebra of all symmetric operators on the space $J_1$, $\mathfrak{g}_0$ is a reductive subalgebra in the algebra of all symplectic operators on $J_1$, and $\delta_0$ is a symmetric bilinear map, which we describe below. This is the main result of this paper, which is formulated in Theorem 8 of § 4.2. In § 5 a complete classification of short $\mathrm{SL}_2$-structures on simple Lie algebras is carried out, which indicates a simple symplectic Lie-Jordan structure corresponding to every $\mathrm{SL}_2$-structure. The author is very grateful to É. B. Vinberg, without whose advice and recommendations this paper would not have seen light, for the formulation of the problem and scientific guidance, to D. A. Timashev, whose clear instructions concerning text editing have let this paper take a harmonious final form, and also to D. I. Panyushev, who gave an exact reference to a result of B. Kostant which was necessary for this text. § 2. Isotypic decompositionIn this section and throughout what follows, all algebras, algebraic groups and linear representations are considered over the field $\mathbb{C}$ of complex numbers. 2.1. Commutation formulaeConsider an arbitrary reductive algebraic group $S$. For an irreducible representation $\rho$ of it denote the space of this representation by $V_{\rho}$. The following lemmas can be found in [1], § 1.2. Lemma 1. Let $\rho$, $\sigma$ and $\tau$ be irreducible representations of the group $S$. Assume that $\rho\otimes\sigma$ contains $\tau$ with multiplicity 1, and let be a fixed nontrivial $S$-equivariant bilinear map (defined up to a scalar factor). Let $U_{\rho}$, $U_{\sigma}$ and $U_{\tau}$ be vector spaces on which $S$ acts trivially. Then every $S$-equivariant bilinear map can be described by a formula where is some bilinear map. Lemma 2. Let $\rho$ and $\tau$ be irreducible representations of the group $S$. Assume that $\operatorname{Sym}^2\rho$ and $\Lambda^2\rho$ contain $\tau$ with multiplicity not exceeding 1, and let Keeping the assumptions and notation of the introduction, we consider a short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$. In this subsection we derive the main commutation formulae arising when we take the commutator of arbitrary two elements of isotypic components of the algebra $\mathfrak{g}$. The commutator on $\mathfrak{g}$ produces $\mathrm{SL}_2$-equivariant bilinear maps
$$
\begin{equation*}
\mathfrak{g}_i\times\mathfrak{g}_j\longrightarrow\mathfrak{g}_i \otimes\mathfrak{g}_j\longrightarrow\mathfrak{g}, \qquad i\neq j,
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathfrak{g}_i\times\mathfrak{g}_i\longrightarrow \bigwedge^2\mathfrak{g}_i\longrightarrow\mathfrak{g}.
\end{equation*}
\notag
$$
According to the Clebsch-Gordan formula for the tensor product of irreducible representations of the algebra $\mathfrak{sl}_2$, we have
$$
\begin{equation}
V_i\otimes V_j\simeq V_{i+j}\oplus V_{i+j-2}\oplus\dots \oplus V_{|i-j|}.
\end{equation}
\tag{2.1}
$$
Therefore, the following commutation relations hold for the commutators of isotypic components of $\mathfrak{g}$:
$$
\begin{equation}
\begin{gathered} \, \lbrack\mathfrak{g}_0, \mathfrak{g}_i\rbrack \subseteq\mathfrak{g}_i, \quad i=0,1,2,\qquad [\mathfrak{g}_1, \mathfrak{g}_1]\subseteq\mathfrak{g}_0\oplus\mathfrak{g}_2, \\ [\mathfrak{g}_2, \mathfrak{g}_2]\subseteq\mathfrak{g}_0\oplus\mathfrak{g}_2, \qquad [\mathfrak{g}_1, \mathfrak{g}_2]\subseteq \mathfrak{g}_1. \end{gathered}
\end{equation}
\tag{2.2}
$$
In particular, $\mathfrak{g}_0$ and $\mathfrak{g}_0\oplus\mathfrak{g}_2$ are subalgebras; moreover, a very short $\mathrm{SL}_2$-structure is induced on the subalgebra $\mathfrak{g}_0\oplus\mathfrak{g}_2$. On the space $\mathbb{C}^2$ we introduce an $\mathrm{SL}_2$-invariant skew-symmetric bilinear form $\langle\,\cdot\,{,}\,\cdot\, \rangle$ by setting
$$
\begin{equation*}
\langle u, v\rangle :=\det(u, v) \quad\forall\, u, v\in \mathbb{C}^2.
\end{equation*}
\notag
$$
We also define an $\mathrm{SL}_2$-invariant symmetric bilinear form on $\mathfrak{sl}_2$ by the rule
$$
\begin{equation*}
(X, Y) :=\operatorname{tr}(XY) \quad\forall\, X, Y\in \mathfrak{sl}_2.
\end{equation*}
\notag
$$
Finally, we define a symmetric bilinear map $S\colon \mathbb{C}^2\times \mathbb{C}^2 \to \mathfrak{sl}_2$ by the formula
$$
\begin{equation*}
S(u, v)w=\langle w, u\rangle v+\langle w, v\rangle u.
\end{equation*}
\notag
$$
For completeness of presentation we prove an auxiliary proposition, which is well known to experts. We need it for further considerations. Consider an arbitrary simple Lie algebra $\mathfrak{g}$ on which an involutive nonidentity automorphism $\theta$ is defined. The eigenvalues of the automorphism $\theta$ are $1$ and $-1$, and thus there is a decomposition where
$$
\begin{equation*}
\mathfrak{h}=\{\xi\in\mathfrak{g}\colon \theta(\xi)=\xi\}\quad\text{and} \quad \mathfrak{v}=\{\eta\in\mathfrak{g}\colon \theta(\eta)=-\eta\}.
\end{equation*}
\notag
$$
Proposition 1. The subspace $\mathfrak{h}$ is a Lie subalgebra, and its adjoint action on the subspace $\mathfrak{v}$ is faithful. Proof. By the definition of an involutive automorphism the following relations hold: From this we see that $\mathfrak{h}$ is a Lie subalgebra with adjoint action on the subspace $\mathfrak{v}$. We claim that this action is faithful.
To prove this consider the inefficiency kernel of this action: We note that, since $\theta\neq\mathrm{id}$, it follows that $\mathfrak{h}\neq \mathfrak{g}$. It is obvious that $\mathfrak{k}\triangleleft\mathfrak{h}$ and, since $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{v}$ and $\mathfrak{k}$ commutes with $\mathfrak{v}$, $\mathfrak{k}$ is an ideal of the Lie algebra $\mathfrak{g}$, from which we conclude that $\mathfrak{k}=0$ or $\mathfrak{k}=\mathfrak{g}$; however, $\mathfrak{k}\subset\mathfrak{h}\neq\mathfrak{g}$ by definition, and so $\mathfrak{k}=0$. Thus, the adjoint action of $\mathfrak{h}$ on $\mathfrak{v}$ is faithful. This completes the proof of the proposition.Now let a short $\mathrm{SL}_2$-structure be defined on $\mathfrak{g}$. Note that the action of the element $-\operatorname{id}$ of $\mathrm{SL}_2$ on the algebra $\mathfrak{g}$ defines on it an involutive automorphism for which the subalgebra $\mathfrak{g}_0\oplus\mathfrak{g}_2$ is the space of fixed points, and $\mathfrak{g}_1$ is the subspace on which this automorphism acts by multiplication by $-1$. It follows from the proposition proved above that the adjoint action of the algebra $\mathfrak{g}_0\oplus\mathfrak{g}_2$ on $\mathfrak{g}_1$ is faithful. Bearing this in mind, we can identify elements of the space $J_2$ and ones of the algebra $\mathfrak{g}_0$ with the corresponding operators on $J_1$. In what follows we agree to denote elements of $J_1$ by lowercase Latin letters $a, b, c,\dots $, and elements of the space $J_2$ and the algebra $\mathfrak{g}_0$ by uppercase Latin letters $A, B, C,\dots$ . Using Lemma 1, as well as the representation of elements of $J_2$ and $\mathfrak{g}_0$ by operators on $J_1$, we can readily derive the following commutation formulae:
$$
\begin{equation*}
\begin{gathered} \, [D, u\otimes a]=u\otimes Da \quad\forall\, D\in \mathfrak{g}_0, \quad u\in \mathbb{C}^2, \quad a\in J_1; \\ [D, X\otimes A]=X\otimes\nu(D, A) \quad\forall\, D\in \mathfrak{g}_0, \quad X\in \mathfrak{sl}_2, \quad A\in J_2; \\ [X\otimes A, v\otimes b]=Xv\otimes Ab \quad\forall\, v\in \mathbb{C}^2, \quad X\in\mathfrak{sl}_2, \quad b\in J_1, \quad A\in J_2. \end{gathered}
\end{equation*}
\notag
$$
Here $\nu\colon \mathfrak{g}_0\times J_2 \to J_2$ is a bilinear map. To calculate the map $\nu$ we must consider the action of the operator $\nu(D, A)\in J_2$ on an arbitrary vector $b\in J_1$. To this end consider Jacobi’s identity for $X\otimes A\in \mathfrak{g}_2$, $v\otimes b\in\mathfrak{g}_1$ and $D\in\mathfrak{g}_0$:
$$
\begin{equation*}
[[D, X\otimes A], v\otimes b]+[[X\otimes A, v\otimes b], D]+[[v\otimes b , D], X\otimes A]=0.
\end{equation*}
\notag
$$
Using the above commutation relations, we readily conclude that
$$
\begin{equation}
\nu(D, A)=[D, A] \quad\forall\, D\in\mathfrak{g}_0, \quad A\in J_2.
\end{equation}
\tag{2.3}
$$
Thus, using formula (2.3) we define an action of the Lie algebra $\mathfrak{g}_0$ on $J_2$. By Lemma 2, for the commutator of two elements of $\mathfrak{g}_2$ we have
$$
\begin{equation*}
[X\otimes A, Y\otimes B]=[X, Y]\otimes(A\circ B)+\frac{1}{2}(X, Y)\Delta(A, B) \quad\forall\, X, Y\in \mathfrak{sl}_2, \quad A, B\in J_2.
\end{equation*}
\notag
$$
Here $\circ$ denotes a commutative binary operation which endows the space $J_2$ with the structure of a commutative algebra, and $\Delta\colon J_2\times J_2 \to \mathfrak{g}_0$ is a skew-symmetric bilinear map. The coefficient $1/2$ is taken for the convenience of further calculations. Thus, we have a vector space $J_1$ on which the Lie algebra $\mathfrak{g}_0$ acts, as well as the commutative algebra $J_2$ (with respect to multiplication $\circ$) of linear operators on $J_1$. We show below that $J_1$ is a symplectic space, $J_2$ is a subalgebra of the Jordan algebra of all symmetric operators on $J_1$, and $\mathfrak{g}_0$ is a subalgebra of the Lie algebra $\mathfrak{sp}(J_1)$. Using all the above agreements, we can write out a complete list of commutation formulae:
$$
\begin{equation}
[D, u\otimes a]=u\otimes Da \quad\forall\, D\in \mathfrak{g}_0, \quad u\in \mathbb{C}^2, \quad a\in J_1;
\end{equation}
\tag{2.4}
$$
$$
\begin{equation}
[D, X\otimes A]=X\otimes [D, A] \quad\forall\, D\in \mathfrak{g}_0, \quad X\in \mathfrak{sl}_2, \quad A\in J_2;
\end{equation}
\tag{2.5}
$$
$$
\begin{equation}
[u\otimes a, v\otimes b]=S(u, v)\otimes\varphi(a, b)+\langle u, v\rangle\delta(a, b) \quad\forall\, u, v\in \mathbb{C}^2, \quad a, b\in J_1;
\end{equation}
\tag{2.6}
$$
$$
\begin{equation}
[X\otimes A, v\otimes b]=Xv\otimes Ab \quad\forall\, v\in \mathbb{C}^2, \quad X\in\mathfrak{sl}_2, \quad A\in J_2, \quad b\in J_1;
\end{equation}
\tag{2.7}
$$
$$
\begin{equation}
[X\otimes A, Y\otimes B]=[X, Y]\otimes(A\circ B)+\frac{1}{2}(X, Y)\Delta(A, B) \quad\forall\, X, Y\in \mathfrak{sl}_2, \quad A, B\in J_2.
\end{equation}
\tag{2.8}
$$
Here $\bullet$ $\varphi\colon J_1\times J_1 \to J_2$ is a skew-symmetric bilinear map; $\bullet$ $\delta\colon J_1\times J_1 \to \mathfrak{g}_0$ is a symmetric bilinear map; $\bullet$ $\Delta\colon J_2\times J_2 \to \mathfrak{g}_0$ is a skew-symmetric bilinear map; $\bullet$ $\circ$ is a commutative bilinear binary operation on $J_2$. 2.2. The properties of isotypic componentsConsider an arbitrary short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$. Let us write out Jacobi’s identities on $\mathfrak{g}$, from which we will derive identities required for further considerations. For any vector $u\in\mathbb{C}^2$ we denote by $u^*$ the element of the space $(\mathbb{C}^{2})^*$ such that $u^*(v)=\langle v, u\rangle$ $\forall\,v\in\mathbb{C}^2$. Then the following formula holds for the map $S$ (where tensors of type $(1,1)$ are interpreted as linear operators):
$$
\begin{equation*}
S(u, v)=u\otimes v^*+v\otimes u^* \quad\forall\, u, v\in\mathbb{C}^2.
\end{equation*}
\notag
$$
We now prove a lemma that simplifies further calculations. Lemma 3. The following equalities hold for any three vectors $u, v, w\in\mathbb{C}^2$ and an operator $X\in\mathfrak{sl}_2$: a) $\langle Xu, v\rangle=-\langle u, Xv\rangle$; b) $u\otimes (Xv)^*=-(u\otimes v^*)\cdot X$ and $Xv\otimes u^*=X\cdot(v\otimes u^*)$; c) $(X, S(u,v))=2\langle Xu, v\rangle$; d) $\langle u, v\rangle w+\langle v, w\rangle u+\langle w, u\rangle v=0$; e) $u\otimes v^*-v\otimes u^*=\langle u, v\rangle\operatorname{id}$; f) $[X, S(u, v)]=2(S(u, Xv)+\langle u, v\rangle X)$; g) $S(Xu, v)-S(u, Xv)=2\langle u, v\rangle X$. Proof. a) Clearly, $\langle Bu, Bv\rangle=\langle u, v\rangle$ for any $ B\in \mathrm{SL}_2$ and $u, v\in\mathbb{C}^2$. Differentiating this equality at the identity element we obtain the required result.
b) Since the relations in question are operator equalities, let us verify them at an arbitrary vector $w\in\mathbb{C}^2$:
$$
\begin{equation*}
\begin{gathered} \, (Xv\otimes u^*)w=\langle w, u\rangle Xv=X(\langle w, u\rangle v) =(X\cdot(v\otimes u^*))w, \\ (u\otimes (Xv)^*)w=\langle w, Xv\rangle u=-\langle Xw, v\rangle u =(-(u\otimes v^*)\cdot X)w. \end{gathered}
\end{equation*}
\notag
$$
c) The following equalities hold:
$$
\begin{equation*}
\begin{aligned} \, (X, S(u,v)) &=\operatorname{tr}(X\cdot(u\otimes v^*+v\otimes u^*)) =\operatorname{tr}((u\otimes v^*)\cdot X)+\operatorname{tr}((v\otimes u^*)\cdot X) \\ &=-\operatorname{tr}(u\otimes (Xv)^*)-\operatorname{tr}(v\otimes (Xu)^*) =\langle Xv, u\rangle+ \langle Xu, v\rangle \\ &=\langle Xu, v\rangle+\langle Xu, v\rangle=2\langle Xu, v\rangle. \end{aligned}
\end{equation*}
\notag
$$
d) The left-hand side of this identity is a trilinear skew-symmetric map on $\mathbb{C}^2$. Hence it is equal to zero. e) Let us prove this operator identity at an arbitrary $w\in\mathbb{C}^2$. We have
$$
\begin{equation*}
(u\otimes v^*-v\otimes u^*)w=\langle w, v\rangle u-\langle w, u\rangle v=-\langle v, w\rangle u- \langle w, u\rangle v=\langle u, v\rangle w.
\end{equation*}
\notag
$$
f) Using the definition of $S$ and the equalities proved above we obtain
$$
\begin{equation*}
\begin{aligned} \, &2S(u, Xv)+2\langle u, v\rangle X=2(u\otimes (Xv)^*+(Xv)\otimes u^*)+2\langle u, v\rangle X \\ &\qquad =-2(u\otimes v^*)\cdot X+2X\cdot(v\otimes u^*) +2(u\otimes v^*)\cdot X-2(v\otimes u^*)\cdot X \\ &\qquad =[X, 2(v\otimes u^*)]=[X, u\otimes v^*+v\otimes u^*+v\otimes u^*-u\otimes v^*] \\ &\qquad =[X, S(u,v)]+[X, v\otimes u^*-u\otimes v^*] =[X, S(u,v)]+[X, \langle v, u\rangle\operatorname{id}] \\ &\qquad=[X, S(u,v)]. \end{aligned}
\end{equation*}
\notag
$$
g) Using the definition of $S$ and the equalities proved above we obtain
$$
\begin{equation*}
\begin{aligned} \, &S(Xu, v)-S(u, Xv)=Xu\otimes v^*+v\otimes (Xu)^*-u\otimes (Xv)^*-Xv\otimes u^* \\ &\qquad =X\cdot(u\otimes v^*-v\otimes u^*)+(u\otimes (v)^*-v\otimes (u)^*)\cdot X \\ &\qquad=\langle u, v\rangle X+\langle u, v\rangle X=2\langle u, v\rangle X. \end{aligned}
\end{equation*}
\notag
$$
This completes the proof of the lemma. Now we turn to deriving some identities needed for our further reasoning. To do this, we need some facts from the theory of Jordan algebras, a brief summary of which is given below. All these facts can be found in [3]. Definition 3. An algebra $J$ in which the identities $ab=ba$ and $(a^2b)a= a^2(ba)$ hold for arbitrary $a, b\in J$ is called a Jordan algebra. Let $A$ be an arbitrary associative algebra. The vector space $A$ with the operation of Jordan multiplication
$$
\begin{equation*}
a\circ b=\frac{1}{2}(ab+ba) \quad \forall\, a, b\in A
\end{equation*}
\notag
$$
forms an algebra $A^{+}$, which is Jordan. A Jordan algebra which can be embedded in $A^{+}$ for some associative algebra $A$ is called a special Jordan algebra. We denote by $L_c$ the operator of multiplication by an element $c\in J$ of some Jordan algebra $J$. Then transformations of the form $[L_a, L_b]$, where $a, b\in J$, are derivations of $J$. Their linear combinations are called internal derivations. They form an ideal in the Lie algebra $\mathfrak{der}(J)$ of all derivations, which is denoted by $\mathfrak{inn}(J)$. Definition 4. A Jordan algebra $J$ is said to be semisimple if a nondegenerate inner product $(\,\cdot\,{,}\,\cdot\,)$ with associativity property is defined on $J$: Definition 5. A Jordan algebra $J$ is said to be simple if it contains no nontrivial ideals (other than $0$ and $J$). A simple Jordan algebra is semisimple. If a Jordan algebra $J$ is semisimple, then, first, it is a direct sum of its simple ideals, second, $J$ contains the identity element and, third, $\mathfrak{der}(J)=\mathfrak{inn}(J)$. The following assertion also holds for semisimple Jordan algebras: Theorem 1. If $J$ is a finite-dimensional semisimple Jordan algebra over the field $\mathbb{C}$, then the algebra $\mathfrak{der}(J)$ is semisimple. Proof. It follows from [4], Ch. VIII, § 4, that for a semisimple Jordan algebra $J$ over $\mathbb{C}$ the derivation algebra $\mathfrak{der}(J)$ is reductive. According to the theory of very short $\mathrm{SL}_2$-structures developed in [1], § 3, the algebra of derivations $\mathfrak{der}(J)$ of a simple Jordan algebra $J$ coincides with the subspace $\mathfrak{g}_0\subset\mathfrak{g}$, where $\mathfrak{g}$ is a simple Lie algebra with a very short $\mathrm{SL}_2$-structure (that is, of the form (1.1)), which corresponds uniquely to the given Jordan algebra $J$. It follows from the classification of all such structures that for a very short $\mathrm{SL}_2$-structure the component $\mathfrak{g}_0$ is semisimple. For a semisimple Jordan algebra all its derivations are inner, and so the Lie algebra $\mathfrak{der}(J)$ is also semisimple. This completes the proof of the theorem. We return to a short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$. It is clear that a very short $\mathrm{SL}_2$-structure is induced on the reductive Lie subalgebra $\mathfrak{g}_0\oplus\mathfrak{g}_2$ of $\mathfrak{g}$. It follows from the theory of very short $\mathrm{SL}_2$-structures, in particular, that the algebra $J_2$ is endowed with a Jordan algebra structure, and the Lie algebra $\mathfrak{g}_0$ acts on $J_2$ by derivations. For convenience, in what follows we call $J_2$ the Jordan algebra of a short $\mathrm{SL}_2$-structure on $\mathfrak{g}$. Finally, we define multiplication on $J_2$ completely by using Jacobi’s identity. Theorem 2. The Jordan algebra $J_2\subset \mathfrak{gl}(J_1)$ of a short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$ is special, with classical multiplication defined by the formula where the map $\Delta\colon J_2\times J_2 \to \mathfrak{g}_0$ satisfies the relation here equality is treated as the equality of linear operators on $J_1$. Proof. Consider Jacobi’s identity for two elements of $\mathfrak{g}_2$ and one element of $\mathfrak{g}_1$ ($w\in\mathbb{C}^2$, $X, Y\in\mathfrak{sl}_2, A, B\in J_2, c\in J_1$):
$$
\begin{equation*}
[[X\otimes A, Y\otimes B], w\otimes c]+[[Y\otimes B, w\otimes c], X\otimes A]+ [[w\otimes c, X\otimes A], Y\otimes B]=0.
\end{equation*}
\notag
$$
Using the commutator relations (2.4)–(2.8), from this equality we obtain
$$
\begin{equation}
YXw\otimes BAc+[X, Y]w\otimes(A\circ B)c+\frac{1}{2}(X, Y)w\otimes\Delta(A, B)c -XYw\otimes ABc=0.
\end{equation}
\tag{2.10}
$$
Let us transpose $A$ and $B$ in (2.10) using the commutativity of multiplication on $J_2$ and the skew-symmetry of $\Delta$. Then we obtain
$$
\begin{equation}
YXw\otimes ABc+[X, Y]w\otimes(A\circ B)c-\frac{1}{2}(X, Y)w\otimes\Delta(A, B)c -XYw\otimes BAc=0.
\end{equation}
\tag{2.11}
$$
Adding identities (2.11) and (2.10) we see that which implies the first assertion of the theorem.
It can readily be seen that This implies the other assertion of the theorem and completes the proof of the theorem. In connection with Theorem 2 just proved, in the following arguments, when a Jordan algebra is used, we assume by default that this algebra is special and endowed with classical multiplication defined by (2.9). For the simplicity of further reasoning, by the image of a bilinear map $\psi$: ${U\times V\to W}$ we mean the image of the corresponding uniquely defined map from $U\otimes V$ to $W$. It follows from the theory of very short $\mathrm{SL}_2$-structures that the image of $\Delta$ acts on the Jordan algebra $J_2$ by inner derivations, and this action is faithful. Then it follows from Theorem 2 and (2.5) that for arbitrary $A, B, C\in J_2$ this action is calculated by the following formula:
$$
\begin{equation*}
\begin{aligned} \, [[A, B], C] &=ABC+CBA-BAC-CAB \\ &=4(A\circ(B\circ C)-B\circ (A\circ C))=4[L_{A}, L_{B}](C). \end{aligned}
\end{equation*}
\notag
$$
Thus, we can assume that
$$
\begin{equation*}
\mathfrak{inn}(J_2)=[J_2, J_2]=\langle[A, B]\colon A,B\in J_2\rangle.
\end{equation*}
\notag
$$
2.3. The invariant inner productOn a simple Lie algebra $\mathfrak{g}$ there is a unique invariant inner product up to proportionality. Fix it; in what follows we denote this product by round brackets $(\,\cdot\,{,}\,\cdot\,)$. It follows from Schur’s lemma that, with respect to this inner product on $\mathfrak{g}$, the vector spaces $\mathfrak{g}_i$, $i\in\{0, 1, 2\}$, are pairwise orthogonal to each other, that is, the following relations hold: Using the fixed inner product on the algebra $\mathfrak{g}$ we introduce an inner product on the space $J_1$ and the algebra $J_2$ as follows:
$$
\begin{equation*}
\begin{gathered} \, (u\otimes a, v\otimes b)=\langle u, v\rangle\alpha(a, b), \\ (X\otimes A, Y\otimes B)=\frac{1}{2}(X, Y)\beta(A, B), \end{gathered}
\end{equation*}
\notag
$$
where $\alpha$ is a skew-symmetric bilinear form on $J_1$ and $\beta$ is a symmetric bilinear form on $J_2$. The coefficient $1/2$ in the last formula is needed for the convenience of further calculations. It is obvious that both forms thus obtained are nondegenerate. We call the form $\alpha$ the skew-inner product on $J_1$, and the form $\beta$ the inner product on $J_2$. For brevity in what follows, we denote the form $\beta$ using round brackets $(\,\cdot\,{,}\,\cdot\,)$, and the form $\alpha$ using angular brackets $\langle\,\cdot\,{,}\,\cdot\,\rangle $. As concerns the inner and skew-inner products introduced, the following assertion holds. Proposition 2. Inner products on the symplectic space $J_1$ and on the algebras $\mathfrak{g}_0$ and $J_2$ have the following properties: 1) $(A, \varphi(a, b))=\langle Aa, b\rangle=\langle a, Ab\rangle$ for all $ a, b\in J_1$ and $A\in J_2$; 2) $(D, \delta(a, b))=\langle Da, b\rangle=-\langle a, Db\rangle$ for all $ a, b\in J_1$ and $D\in\mathfrak{g}_0$; 3) $(D, [A, B])=([D, A], B)=-(A, [D, B])$ for all $ A, B\in J_2$ and $D\in\mathfrak{g}_0$; 4) $(A\circ B, C)=(A, B\circ C)$ for all $A, B, C\in J_2$. Proof. 1) Consider $X\in\mathfrak{sl}_2, u, v\in\mathbb{C}^2, a, b\in J_1$ and $ A\in J_2$. Then it follows from the invariance of the inner product on $\mathfrak{g}$ that
$$
\begin{equation}
(X\otimes A, [u\otimes a, v\otimes b])=([X\otimes A, u\otimes a], v\otimes b).
\end{equation}
\tag{2.13}
$$
Using relations (2.4)–(2.8) we obtain
$$
\begin{equation*}
\begin{gathered} \, (X\otimes A, [u\otimes a, v\otimes b])=(X\otimes A, S(u, v)\otimes\varphi(a, b)), \\ ([X\otimes A, u\otimes a], v\otimes b)=(Xu\otimes Aa, v\otimes b). \end{gathered}
\end{equation*}
\notag
$$
Hence, using the above definition of the inner product on the spaces $J_1$ and $J_2$, we see that
$$
\begin{equation*}
\frac{1}{2}(X, S(u, v))(A, \varphi(a, b))=\langle Xu, v\rangle\langle Aa, b\rangle.
\end{equation*}
\notag
$$
According to Lemma 3, $(X, S(u, v))=2\langle Xu, v\rangle$. This yields the first equality in part 1).
The second equality in part 1) follows readily from the first equality with $a$ and $b$ interchanged and the skew symmetry of the map $\varphi$. 2) Consider $D\in\mathfrak{g}_0, u, v\in\mathbb{C}^2$ and $ a, b\in J_1$. Then it follows from the invariance of the inner product on $\mathfrak{g}$ that
$$
\begin{equation}
(D, [u\otimes a, v\otimes b])=([D, u\otimes a], v\otimes b).
\end{equation}
\tag{2.14}
$$
Using relations (2.4)–(2.8) and (2.12) we obtain
$$
\begin{equation*}
(D, [u\otimes a, v\otimes b])=\langle u, v\rangle(D, \delta(a, b))
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
([D, u\otimes a], v\otimes b)=(u\otimes Da, v\otimes b).
\end{equation*}
\notag
$$
From the above definition of the skew-inner product on $J_1$ we see that
$$
\begin{equation*}
\langle u, v\rangle(D, \delta(a, b))=\langle u, v\rangle\langle Da, b\rangle.
\end{equation*}
\notag
$$
Dividing the resulting equality by $\langle u, v\rangle$ we obtain the first equality in part 2). The second equality in part 2) follows readily from the first with $a$ and $b$ interchanged and the symmetry of $\delta$.
3) Consider $D\in\mathfrak{g}_0$, $X, Y\in\mathfrak{sl}_2$ and $A, B\in J_2$. Then it follows from the invariance of the inner product on $\mathfrak{g}$ that
$$
\begin{equation}
(D, [X\otimes A, Y\otimes B])=([D, X\otimes A], Y\otimes B).
\end{equation}
\tag{2.15}
$$
Using (2.4)–(2.8) and Theorem 2, we see from (2.15) that
$$
\begin{equation*}
\frac{1}{2}(X, Y)(D, [A, B])=(X\otimes[D, A], Y\otimes B)= \frac{1}{2}(X, Y)([D, A], B).
\end{equation*}
\notag
$$
This implies the first equality in part 3). The second equality in part 3) follows readily from the first with $A$ and $B$ interchanged. 4) Consider $X, Y, Z\in\mathfrak{sl}_2$ and $ A, B, C\in J_2$. Then it follows from the invariance of the inner product on $\mathfrak{g}$ that
$$
\begin{equation}
(X\otimes A, [Y\otimes B, Z\otimes C])=([X\otimes A, Y\otimes B], Z\otimes C).
\end{equation}
\tag{2.16}
$$
Using relations (2.4)–(2.8), from (2.16) we obtain Since $(X, [Y, Z])=([X, Y], Z)$, the equality in part 4) follows from this. This completes the proof of the proposition. In particular, it follows from Proposition 2 and Theorem 2 that $J_2$ is a semisimple Jordan algebra and $\mathfrak{der}(J_2)=\mathfrak{inn}(J_2)=[J_2, J_2]\subset\mathfrak{g}_0$. It also follows from Proposition 2 that $J_2$ consists of symplectic operators on the symplectic space $J_1$, while $\mathfrak{g}_0\subset \mathfrak{sp}(J_1)$. 2.4. Short $\mathrm{SL}_2$-structures and $ {\mathbb{Z}}$-gradingsConsider an arbitrary short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$. Using the map $\mathrm{d}\Phi$ we embed the Lie algebra $\mathfrak{sl}_2$ in $\mathfrak{der}(\mathfrak{g})=\mathfrak{inn}(\mathfrak{g})\simeq\mathfrak{g}$. We denote by $e$, $f$ and $h$ basis elements of the algebra $\mathrm{d}\Phi(\mathfrak{sl}_2)$ which satisfy the following relations:
$$
\begin{equation}
[e, f]=h, \qquad [h, e]=2e\quad\text{and} \quad [h, f]=-2f.
\end{equation}
\tag{2.17}
$$
Let
$$
\begin{equation*}
\widetilde{e}=\mathrm{d}\Phi^{-1}(e), \qquad\widetilde{f}= \mathrm{d}\Phi^{-1}(f)\quad\text{and} \quad\widetilde{h}=\mathrm{d}\Phi^{-1}(h).
\end{equation*}
\notag
$$
Consider the operator $\operatorname{ad}(h)$. Since the dimensions of the irreducible representations of $\Phi$ do not exceed 3, the absolute values of eigenvalues of the operator $\operatorname{ad}(h)$ do not exceed 2. Then, with respect to the operator $\operatorname{ad}(h)$, the algebra $\mathfrak{g}$ expands into a direct sum of eigensubspaces:
$$
\begin{equation}
\mathfrak{g}= \mathfrak{g}^{-2}\oplus\mathfrak{g}^{-1}\oplus\mathfrak{g}^{0} \oplus\mathfrak{g}^{1}\oplus\mathfrak{g}^2, \qquad\mathfrak{g}^{k}= \{\xi\in\mathfrak{g}\colon [h,\xi]=k\xi\}.
\end{equation}
\tag{2.18}
$$
Clearly,
$$
\begin{equation*}
\begin{gathered} \, \mathfrak{g}^0=\mathfrak{g}_0\oplus(\widetilde{h}\otimes J_2), \qquad \mathfrak{g}^{-1}=e_{-1}\otimes J_1, \qquad \mathfrak{g}^1=e_{1}\otimes J_1, \\ \mathfrak{g}^{-2}=\widetilde{f}\otimes J_2\quad\text{and} \quad \mathfrak{g}^{2}=\widetilde{e}\otimes J_2, \end{gathered}
\end{equation*}
\notag
$$
where $\{e_{-1}, e_{1}\}$ is the eigenbasis of $\mathbb{C}^2$ with respect to the operator $\widetilde{h}$. The decomposition (2.18) is a $\mathbb{Z}$-grading, that is, $[\mathfrak{g}^{k},\mathfrak{g}^{l}]\subset\mathfrak{g}^{k+l}$, where $\mathfrak{g}^m=0$ for $|m| > 2$. Without loss of generality we can assume that $h$ belongs to a fixed Cartan subalgebra $\mathfrak{t}$ of the algebra $\mathfrak{g}$ and even to a positive Weyl chamber with respect to some choice of positive roots. Thus, if $\Pi=\{\alpha_1,\dots ,\alpha_{n}\}$ is a system of simple roots of $\mathfrak{g}$, then We write $p_i=\alpha_i(h)\in\mathbb{Z_{+}}$. It is clear that the $\mathbb{Z}$-grading (2.18) is completely determined by the set $\{p_1,\dots ,p_n\}$. Each subspace $\mathfrak{g}^k$ is the sum of the root subspaces $\mathfrak{g}_{\alpha}$ corresponding to roots $\alpha=k_1\alpha_1+\dots + k_n\alpha_n$ such that and the Cartan subalgebra $\mathfrak t$ in the case of $k=0$.We denote the highest root of $\mathfrak{g}$ by $\alpha_{\operatorname{\max}}$. Since the largest eigenvalue of the operator $\operatorname{ad}(h)$ is 2, it follows that $\alpha_{\operatorname{\max}}(h)=2$. If $\alpha_{\operatorname{\max}}=l_1\alpha_1+\dots+l_n\alpha_n$, then we see from the condition $\alpha_{\operatorname{\max}}(h)=2$ that Consider the subspace $\mathfrak{g}^0$. Clearly, it is a Lie subalgebra. There is a decomposition
$$
\begin{equation*}
\mathfrak{g}^{0}=\widetilde{\mathfrak{g}}^{0}\oplus\langle h\rangle,
\end{equation*}
\notag
$$
where $\widetilde{\mathfrak{g}}^{0}$ is the orthogonal complement to $h$. Consider the representation of $\widetilde{\mathfrak{g}}^{0}$ on the space $\mathfrak{g}^{2}$ equal to the restriction of the adjoint representation. We write It is clear that $\Pi_0$ is a system of simple roots of the semisimple part of $\mathfrak{\widetilde{g}}^{0}$. Consider the subspace $\mathfrak{g}^k$ for $k\neq 0$. We decompose it into a direct sum of subspaces $\mathfrak{g}^k_{(\nu)}$, where every $\mathfrak{g}^k_{(\nu)}$ is the sum of root subspaces $\mathfrak{g}_{\alpha}$ corresponding to roots with fixed coefficients at the simple roots $\alpha_i\not\in \Pi_0$ (and satisfying (2.19)). Clearly, each subspace $\mathfrak{g}^k_{(\nu)}$ is invariant with respect to $\mathfrak{g}^0$. The following assertion holds; its proof can be found in [5], Ch. 3, § 3.5. This theorem implies immediately that the representation $\widetilde{\mathfrak{g}}^0$ in each $\mathfrak{g}^k_{(\nu)}$ is also irreducible. We note that in our case the representation $\widetilde{\mathfrak{g}}^0\colon \mathfrak{g}^2$ is always irreducible since, generally speaking, only three cases are possible. 1. There is exactly one simple root $\alpha_i$ with nonzero value $p_i=2$ and, in the decomposition of the highest root with respect to the simple roots, this root occurs with coefficient 1. 2. There are exactly two simple roots $\alpha_i$ and $\alpha_j$ with nonzero values $p_i= p_j=1$; in the decomposition of the highest root with respect to the simple roots, the coefficients at these roots are equal to 1. 3. There is exactly one simple root $\alpha_i$ with nonzero value $p_i=1$, and, in the decomposition of the highest root with respect to the simple roots, this simple root occurs with coefficient 2. Case 1 does not realize since then $\mathfrak{g}^1=\mathfrak{g}^{-1}=0$, which is not the case for short $\mathrm{SL}_2$-structures. In case 3 a simple root is unique, and thus $\mathfrak{g}^2= \mathfrak{g}^2_{(\nu)}$; hence the representation is irreducible. In case 2, in the decomposition of the highest root with respect to the simple roots the coefficients at roots with nonzero values at $h$ are equal to $1$, which means that for all positive roots the coefficients at these roots do not exceed $1$, that is, $\mathfrak{g}^2= \mathfrak{g}^2_{(\nu)}$, and the representation is irreducible again. 2.5. The simplicity of the Jordan algebra of a short $\mathrm{SL}_2$-structureTheorem 4. Assume that a short $\mathrm{SL}_2$-structure is defined on a simple Lie algebra $\mathfrak{g}$, and let $J_2$ be the Jordan algebra of this structure. Then $J_2$ is simple. Proof. This can readily be proved for very short $\mathrm{SL}_2$-structures. Indeed, in this case the Lie algebra $\mathfrak{g}$ has the decomposition $\mathfrak{g}=\mathfrak{g}_0\,{\oplus}\,\mathfrak{sl}_2\,{\otimes}\, J_2$, where $J_2$ is a Jordan algebra and $\mathfrak{g}_0=\mathfrak{der}(J_2)=\mathfrak{inn}(J_2)$. If the $J_2$ has a nontrivial ideal $I$, then the subspace $\mathfrak{g}_0\oplus\mathfrak{sl}_2\otimes I$ is a nontrivial ideal in $\mathfrak{g}$, as follows easily from the commutation formulae.
In the case of short $\mathrm{SL}_2$-structures this direct method of proof is inapplicable, since commuting the term $\mathbb{C}^2\otimes J_1$ with itself produces a term from $\mathfrak{sl}_2\otimes J_2$, which obstructs the use of the above-described method for constructing an ideal in $\mathfrak{g}$. However, the case of short $\mathrm{SL}_2$-structures can be reduced to the case of very short $\mathrm{SL}_2$-structures already considered. To this end we note that, when a short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$ is restricted to the Lie subalgebra $\mathfrak{h}=\mathfrak{g}_0\oplus\mathfrak{g}_2$, a very short $\mathrm{SL}_2$-structure on $\mathfrak{h}$ is obtained. This Lie algebra, generally speaking, is not simple, but it is reductive due to the nondegeneracy of the inner product. Since the semisimple element $h$ (as also the whole of $\mathfrak{sl}_2$) acts trivially on the centre of the algebra $\mathfrak{h}$, replacing this algebra by its commutator subalgebra $[\mathfrak{h}, \mathfrak{h}]$ we may assume that $\mathfrak{h}$ is semisimple. For convenience, we choose a system of simple roots for $\mathfrak{h}$ as follows. The root system of the algebra $\mathfrak{h}$ consists of the roots of $\mathfrak{g}$ taking the values $0$, $2$ or $-2$ at $h$. We assume that the positive roots of $\mathfrak{h}$ are the positive roots of $\mathfrak{g}$ that take the value $0$ at $h$, and also the roots of $\mathfrak{g}$ taking the value $-2$ at $h$. Then it is clear that the set of simple roots of the semisimple algebra $\mathfrak{h}$ consists of all the simple roots of the Lie algebra $\mathfrak{g}$ that vanish at $h$ and the lowest root of $\mathfrak{g}$. It is clear that, in general, the Lie algebra $\mathfrak{h}$ contains several simple components; however, exactly one of these contains a subspace $\mathfrak{sl}_2\otimes J_2$ (precisely the one whose simple root is the lowest root of the algebra $\mathfrak{g}$). Therefore, the algebra $\mathfrak{h}$ can be assumed to be simple, which reduces the question fully to the case of very short $\mathrm{SL}_2$-structures, and for it we have already proved that the Jordan algebra $J_2$ is simple. This completes the proof of the theorem. Consider the Lie algebra $\mathfrak{g}_0$. In view of the commutation relations (2.2), the subspace $[J_2, J_2]\subset\mathfrak{g}_0$ is an ideal of the algebra $\mathfrak{g}_0$. We denote by $\mathfrak{i}_0$ the orthogonal complement to $[J_2, J_2]$ in $\mathfrak{g}_0$ with respect to the inner product introduced above. The subspace $\mathfrak{i}_0\subset\mathfrak{g}_0$ is also an ideal of $\mathfrak{g}_0$ and, because the inner product on $\mathfrak{g}_0$ is nondegenerate, the sum of the dimensions of the ideals $\mathfrak{i}_0$ and $[J_2, J_2]$ is equal to the dimension of the Lie algebra $\mathfrak{g}_0$. Consider the action of the ideal $\mathfrak{i}_0\triangleleft\mathfrak{g}_0$ on the Jordan algebra $J_2$. Using Proposition 2 proved above, for any $A, B\in J_2$ and $D\in \mathfrak{i}_0$ we obtain Therefore, $[D, A]=0$ for any $A\in J_2$, that is, $D$ is an element of the kernel of the action of $\mathfrak{g}_0$ on $J_2$. Conversely, if $[D, A]=0$ for each $A\in J_2$, then it follows from (2.20) that $(D, [A, B])=0$ for any $A, B\in J_2$, that is, $D\in\mathfrak{i}_0$. This immediately implies that $\mathfrak{i}_0$ is exactly the kernel of the action of $\mathfrak{g}_0$ on $J_2$. Using the facts that the Lie algebra $\mathfrak{g}_0$ acts on $J_2$ by commutation and that multiplication on $J_2$ is classical, it is easy to deduce that the Lie algebra $\mathfrak{g}_0$ acts on the Jordan algebra $J_2$ by derivations. Then, since $\mathfrak{i}_0$ is the kernel of this action, it follows that $\mathfrak{g}_0/\mathfrak{i}_0\subset\mathfrak{der}(J_2)$. However, it follows from the simplicity of $J_2$ that that is, Then there is a decomposition of the Lie algebra $\mathfrak{g}_0$ into a direct sum of two ideals:In particular, it will follow from the classification of all short $\mathrm{SL}_2$-structures (described in § 5 of this paper) that, generally speaking, $\mathfrak{i}_0\neq 0$, that is, the Lie algebra $\mathfrak{g}_0$ is not fully determined by the Jordan algebra $J_2$, which distinguishes the case of short $\mathrm{SL}_2$-structures from that of very short ones. Moreover, the Lie algebra $\mathfrak{g}_0$ is not fully determined even by the pair $(J_1; J_2)$, as will also be seen from the classification of all short $\mathrm{SL}_2$-structures. We note that the simplicity of the Jordan Lie algebra $J_2$ implies that $J_2$ contains a unity. Moreover, the following assertion holds. Theorem 5. Let a short $\mathrm{SL}_2$-structure be defined on a simple Lie algebra $\mathfrak{g}$, where $J_1$ is a symplectic space, and let $J_2$ be the Jordan algebra of this structure. Denote the identity element of $J_2$ by $\mathbb{I}$. Then $\mathbb{I}=\operatorname{id},$ that is, Proof. First we derive the identities needed in the proof. Consider the Jacobi identity for arbitrary $X\in\mathfrak{sl}_2$, $u,v\in\mathbb{C}^2$, $A\in J_2$, and $a, b\in J_1$:
$$
\begin{equation*}
[[X\otimes A, u\otimes a], v\otimes b]+[[u\otimes a, v\otimes b], X\otimes A]+ [[v\otimes b, X\otimes A], u\otimes a]=0.
\end{equation*}
\notag
$$
Using equalities (2.4)–(2.8) we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &S(Xu, v)\otimes\varphi(Aa, b)+\langle Xu, v\rangle\delta(Aa, b) \\ \notag &\qquad +[S(u, v), X]\otimes(\varphi(a, b)\circ A) +\frac{1}{2}(S(u, v), X)[\varphi(a, b), A] \\ &\qquad +\langle u, v\rangle X\otimes[\delta(a, b), A] -S(Xv, u)\otimes\varphi(Ab, a)- \langle Xv, u\rangle\delta(Ab, a)=0. \end{aligned}
\end{equation}
\tag{2.22}
$$
Hence
$$
\begin{equation*}
\langle Xu, v\rangle\delta(Aa, b)+\frac{1}{2}(S(u, v), X)[\varphi(a, b), A]- \langle Xv, u\rangle\delta(Ab, a)=0.
\end{equation*}
\notag
$$
Applying Lemma 3 to the resulting equality, we obtain the first of the identities we need: Further, (2.22) also implies that
$$
\begin{equation}
\begin{aligned} \, \notag &S(Xu, v)\otimes\varphi(Aa, b)+[S(u, v), X]\otimes(\varphi(a, b)\circ A) \\ &\qquad\qquad+ \langle u, v\rangle X\otimes[\delta(a, b), A] -S(Xv, u)\otimes\varphi(Ab, a)=0. \end{aligned}
\end{equation}
\tag{2.24}
$$
In the resulting equality we transpose $a$ and $b$ and use the skew symmetry of $\varphi$ and the symmetry of $\delta$. Then we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &S(Xu, v)\otimes\varphi(Ab, a)-[S(u, v), X]\otimes(\varphi(a, b)\circ A) \\ &\qquad\qquad+ \langle u, v\rangle X\otimes[\delta(a, b), A] -S(Xv, u)\otimes\varphi(Aa, b)=0. \end{aligned}
\end{equation}
\tag{2.25}
$$
Adding (2.24) and (2.25) we obtain
$$
\begin{equation*}
(S(Xu, v)-S(Xv, u))\otimes (\varphi(Ab, a)+\varphi(Aa, b))+2\langle u, v\rangle X\otimes[\delta(a, b), A]=0.
\end{equation*}
\notag
$$
Lemma 3 implies that $S(Xu, v)-S(Xv, u)=2\langle u, v\rangle X$, from where we obtain another required identity: Consider Jacobi’s identity for arbitrary $D\in\mathfrak{g}_0, u,v\in\mathbb{C}^2$, and $ a, b\in J_1$:
$$
\begin{equation*}
[[D, u\otimes a], v\otimes b]+[[u\otimes a, v\otimes b], D] +[[v\otimes b, D], u\otimes a]=0.
\end{equation*}
\notag
$$
Using (2.4)–(2.8) we obtain
$$
\begin{equation*}
\begin{aligned} \, &S(u, v)\otimes\varphi(Da, u)+\langle u, v\rangle\delta(Da, b) -S(u, v)\otimes [D, \varphi(a, b)]-\langle u, v\rangle[D, \delta(a, b)] \\ &\qquad\qquad+S(u, v)\otimes\varphi(a, Db)+\langle u, v\rangle\delta(a, Db)=0. \end{aligned}
\end{equation*}
\notag
$$
Hence Dividing this by $\langle u, v\rangle$ we see that Now we have everything we need to prove the assertion of the theorem.
Consider the operator $\mathbb{I}\in J_2$. Since $\mathbb{I}^2=\mathbb{I}$, it follows that $J_1=J_1^0\oplus J_1^1$, where
$$
\begin{equation*}
J_1^0=\operatorname{Ker}\mathbb{I}, \qquad J_1^1=\operatorname{Im}\mathbb{I}.
\end{equation*}
\notag
$$
We claim that $J_1^0=0$, from which the assertion of the theorem will follow.
It is clear that $J_2=J_2 \circ \mathbb{I}$, and therefore the whole Jordan algebra $J_2$ acts trivially on $J_1^0$. We claim that $J_1^0$ is an invariant subspace for the whole Lie algebra $\mathfrak{g}_0$. Because the action of the Jordan algebra $J_2$ on $J_1^0$ is trivial, the commutator subalgebra $[J_2, J_2]\subset\mathfrak{g}_0$ also acts trivially on $J_1^0$, and thus, by (2.21), it is sufficient to prove that $J_1^0$ is invariant under the action of the ideal $\mathfrak{i}_0$. For any $D\in\mathfrak{i}_0$ and $a\in J_1^0$ we have
$$
\begin{equation*}
\mathbb{I}(Da)=[\mathbb{I},D]a+D(\mathbb{I}a)=0 \implies Da\in J_1^0.
\end{equation*}
\notag
$$
This proves the invariance of $J_1^0$ under the action of $\mathfrak{g}_0$. We write
$$
\begin{equation*}
\mathfrak{g}_0^1=\langle\delta(a, b)\colon a\in J_1^0, b\in J_1\rangle.
\end{equation*}
\notag
$$
Consider $\mathfrak{i}=\mathfrak{g}_0^1\oplus(\mathbb{C}^2\otimes J_1^0)$. We claim that $\mathfrak{i}\triangleleft\mathfrak{g}$. Then it will follow that $J_1^0=0$ by the simplicity of $\mathfrak{g}$. We note that
$$
\begin{equation*}
(A, \varphi(a, b))=\langle Aa, b\rangle=0 \quad\forall\, A\in J_2, \quad a\in J_1^0, \quad b\in J_1.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation}
\varphi(a, b)=0 \quad\forall\, a\in J_1^0, \quad b\in J_1.
\end{equation}
\tag{2.28}
$$
Using the invariance of $J_1^0$ under the action of $\mathfrak{g}_0$ and equality (2.27), we can readily show that the subspace $\mathfrak{i}$ is invariant under the operation of commutation with elements of $\mathfrak{g}_0$. Also, using equalities (2.28) and (2.26) obtained above and the triviality of the action of $J_2$ on $J_1^0$, one can show that the subspace $\mathfrak{i}$ commutes with $\mathfrak{g}_2$. It remains to prove that $\mathfrak{i}$ is invariant under commutation with elements of $\mathfrak{g}_1$. Since (2.28) holds, it suffices to show that the operators in $\mathfrak{g}_0^1$ take an arbitrary element of $J_1$ into $J_1^0$.
Applying (2.23) to $a\in J_1^0$ and $A=\mathbb{I}$ and taking (2.28) into account one can see that
$$
\begin{equation*}
\delta(a, \mathbb{I}b)=\delta(\mathbb{I}a, b)=0 \quad\forall\, a\in J_1^0, \quad b\in J_1.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation}
\delta(a, b)=0 \quad\forall\, a\in J_1^0, \quad b\in J_1^1.
\end{equation}
\tag{2.29}
$$
Therefore, it suffices to consider the operators $\delta(a,b)$ for $a,b\in J_1^0$.
We claim that $\delta(a,b)c=0$ for any $a,b\in J_1^0$ and $c\in J_1^1$. Then it follows from the invariance of $J_1^0$ under the action of $\mathfrak{g}_0$ that the operators in $\mathfrak{g}_0^1$ take an arbitrary element of $J_1$ into $J_1^0$. Consider Jacobi’s identity for arbitrary $u,v,w\in\mathbb{C}^2$, $a, b\in J_1^0$ and $c\in J_1^1$:
$$
\begin{equation*}
[[u\otimes a, v\otimes b], w\otimes c] +[[v\otimes b, w\otimes c], u\otimes a]+ [[w\otimes c, u\otimes a], v\otimes b]=0.
\end{equation*}
\notag
$$
Using (2.4)–(2.8) in combination with (2.28) and (2.29) we obtain
$$
\begin{equation*}
\begin{gathered} \, [[u\otimes a, v\otimes b], w\otimes c]=\langle u, v\rangle w\otimes \delta(a, b)c, \\ [[v\otimes b, w\otimes c], u\otimes a]=[[w\otimes c, u\otimes a], v\otimes b]=0. \end{gathered}
\end{equation*}
\notag
$$
Thus, it follows from the above Jacobi identity that $\delta(a, b)c=0$ for any $a, b\in J_1^0$, $c\in J_1^1$.
Hence, $\mathfrak{i}\triangleleft\mathfrak{g}$, and thus the assertion of the theorem is proved. It follows from the proof of the above theorem that, for an arbitrary short $\mathrm{SL}_2$-structure, the map $\delta\colon J_1\times J_1\to\mathfrak{g}_0$, as well as $\varphi\colon J_1\times J_1\to J_2$, is equivariant with respect to the action of $\mathfrak{g}_0$ (identity (2.27)), and that identities (2.23) and (2.26) hold. Since we have the decomposition (2.21), it follows that
$$
\begin{equation*}
\delta=\delta_0+\delta_c, \qquad\delta_0=\pi_{0,0}\delta, \qquad\delta_c=\pi_{0,c}\delta,
\end{equation*}
\notag
$$
where $\pi_{0,0}\colon \mathfrak{g}_0\to \mathfrak{i}_0$ and $\pi_{0,c}\colon \mathfrak{g}_0\to [J_2, J_2]$ are the orthogonal projections onto $\mathfrak{i}_0$ and $[J_2, J_2]$, respectively. Since $\delta$ is $\mathfrak{g}_0$-equivariant, the same is true for $\delta_0$. Since identity (2.23) holds, whose left-hand side belongs to $[J_2, J_2]$, it follows that
$$
\begin{equation*}
\delta_0(Aa, b)=\delta_0(a, Ab) \quad\forall\, A\in J_2, \quad a, b\in J_1.
\end{equation*}
\notag
$$
§ 3. Maximal short $\mathrm{SL}_2$-structureThe objective of this section is to construct an example of a short $\mathrm{SL}_2$-structure that is maximal in the following sense: for this short $\mathrm{SL}_2$-structure the components $J_2$ and $\mathfrak{g}_0$ are maximal with respect to inclusion for a fixed component $J_1$. 3.1. The construction of a maximal short $\mathrm{SL}_2$-structureIn this subsection we consider the algebra $\mathfrak{g}=\mathfrak{so}_{4n+1}$ in a basis in which its elements are matrices of size $(4n+1)\times(4n+1)$ that are skew-symmetric with respect to the secondary diagonal. Given an invariant inner product on $\mathfrak{so}_{4n+1}$, we fix the product
$$
\begin{equation*}
(A, B)=\frac{1}{2}\operatorname{tr}(AB) \quad\forall\, A, B\in\mathfrak{so}_{4n+1}.
\end{equation*}
\notag
$$
Consider a short $\mathrm{SL}_2$-structure on $\mathfrak{so}_{4n+1}$ given by the embedding of the Lie algebra $\mathfrak{sl}_2$ in $\mathfrak{so}_{4n+1}$ under which basis elements $e,f$ and $h$ of $\mathfrak{sl}_2\subset\mathfrak{so}_{4n+1}$ satisfying relations (2.17) become the matrices
$$
\begin{equation*}
e=\begin{pmatrix} 0&I_{2n}\\ 0&0 \end{pmatrix}, \qquad f=\begin{pmatrix} 0&0\\ I_{2n}&0 \end{pmatrix}\quad\text{and} \quad h=\operatorname{diag}\{\underbrace{1,\dots,1}_{2n}, 0, \underbrace{-1,\dots,-1}_{2n}\}.
\end{equation*}
\notag
$$
Here $I_{2n}$ denotes the $2n\times 2n$ matrix of the following form:
$$
\begin{equation*}
I_{2n}=\operatorname{diag}\{\underbrace{1,\dots,1}_{n}, \underbrace{-1,\dots,-1}_{n}\}.
\end{equation*}
\notag
$$
We note that
$$
\begin{equation*}
\mathfrak{so}_{4n+1}=\mathfrak{so}_{4n}\oplus\mathbb{C}^{4n}.
\end{equation*}
\notag
$$
This decomposition can readily be obtained by considering an arbitrary element of the algebra $\mathfrak{so}_{4n+1}$ in the above basis; this element has the following form:
$$
\begin{equation*}
\begin{pmatrix} \begin{matrix} C \end{matrix} &\begin{matrix} \overline{a}_1 \\ \overline{a}_2 \end{matrix} &\begin{matrix} A \end{matrix} \\ \begin{matrix}\overline{b}_{2}^{\,\mathsf{s}}&-\overline{b}_{1}^{\,\mathsf{s}} \end{matrix} &0& \begin{matrix} -\overline{a}_{2}^{\,\mathsf{s}}&-\overline{a}_{1}^{\,\mathsf{s}} \end{matrix} \\ \begin{matrix} B \end{matrix} &\begin{matrix} \overline{b}_{1} \\ -\overline{b}_{2} \end{matrix} &\begin{matrix} -C^{\mathsf{s}}& \end{matrix} \end{pmatrix},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\overline{a}_1=\begin{pmatrix} a_{1}\\ \vdots\\ a_{n} \end{pmatrix}, \qquad\overline{a}_2=\begin{pmatrix} a_{n+1}\\ \vdots\\ a_{2n} \end{pmatrix}, \qquad\overline{b}_1=\begin{pmatrix} b_{1}\\ \vdots\\ b_{n} \end{pmatrix}\quad\text{and} \quad\overline{b}_2=\begin{pmatrix} b_{n+1}\\ \vdots\\ b_{2n} \end{pmatrix}.
\end{equation*}
\notag
$$
Here $A$ and $B$ are $2n\times 2n$ matrices which are skew-symmetric with respect to the secondary diagonal, $C$ is a $2n\times 2n$ matrix, and $C^{\mathsf{s}}$ is the transpose of $C$ with respect to the secondary diagonal. The direct summand of the form $\mathfrak{so}_{4n}$ is formed by the matrices in which $a_i=b_i=0$, $i=1,\dots, 2n$. The direct summand of the form $\mathbb{C}^{4n}$ is formed by the matrices with zero blocks $A$, $B$, $C$, and $C^{\mathsf{s}}$, and matrices in the first direct summand act naturally on the second. The following equalities also hold:
$$
\begin{equation*}
\mathfrak{so}_{4n}=\mathfrak{g}_0\oplus\mathfrak{g}_2\quad\text{and} \quad\mathbb{C}^{4n}=\mathfrak{g}_1.
\end{equation*}
\notag
$$
Further, the quadratic space $\mathbb{C}^{4n}$ is decomposed into a tensor product of the symplectic spaces $\mathbb{C}^2$ and $\mathbb{C}^{2n}=J_1$. Namely, corresponding to a vector in $\mathbb{C}^{4n}$ which is a matrix of the form of a ‘cross’ (that is, a matrix of size $(4n+1)\times(4n+1)$ all of whose nonzero elements belong to the central column or central row)
$$
\begin{equation*}
\begin{pmatrix} &&\overline{a}_1&& \\ &&\overline{a}_2&& \\ \overline{b}_{2}^{\,\mathsf{s}}&-\overline{b}_{1}^{\,\mathsf{s}} &0&-\overline{a}_{2}^{\,\mathsf{s}}&-\overline{a}_{1}^{\,\mathsf{s}} \\ &&\overline{b}_{1}&& \\ &&-\overline{b}_{2}&& \end{pmatrix}
\end{equation*}
\notag
$$
is the element of $\mathbb{C}^2\otimes J_1$ of the form
$$
\begin{equation*}
e_1\otimes(a_1,\dots ,a_{2n})+e_{-1}\otimes(b_1,\dots ,b_{2n}).
\end{equation*}
\notag
$$
Considering two arbitrary elements of the space $\mathfrak{g}_1$ and taking their inner product, we can readily derive that the matrix of the skew-inner product on $J_1$ has the form
$$
\begin{equation*}
\Omega= \begin{pmatrix} &\Omega_n \\ -\Omega_n& \end{pmatrix}.
\end{equation*}
\notag
$$
Here $\Omega_n$ denotes the $n\times n$ matrix of the following form:
$$
\begin{equation*}
\Omega_n=\begin{pmatrix} &&1\\ &\cdot^{\displaystyle{\displaystyle\cdot}^{\displaystyle\cdot}}&\\ 1&& \end{pmatrix}.
\end{equation*}
\notag
$$
The space of skew-symmetric operators in $\mathbb{C}^{4n}$ is decomposed into a direct sum of two tensor products: of the space of skew-symmetric operators on $\mathbb{C}^2$ and the space of symmetric operators on $J_1$ and, conversely, of the space of symmetric operators in $\mathbb{C}^2$ and the space of skew-symmetric operators in $J_1$. The first product is $\mathfrak{sl}_2\otimes J_2=\mathfrak{g}_2$ and, since the space of symmetric operators in $\mathbb{C}^2$ is obviously one-dimensional, we can assume that the other summand is $\mathfrak{sp}(J_1)=\mathfrak{g}_0$. We denote the Jordan algebra of symmetric operators on the symplectic space $J_1$ by $\mathfrak{sym}(J_1)$. Thus, $J_2= \mathfrak{sym}(J_1)$. Using the form of the matrix of skew-inner product on $J_1$ it is easy to obtain a matrix description of the algebra $\mathfrak{sym}(J_1)$ with respect to a basis of $J_1$ in which the skew-inner product has the above matrix $\Omega$. Namely,
$$
\begin{equation*}
\mathfrak{sym}(J_1)=J_2=\biggl\{\begin{pmatrix} X&Y\\ Z&X^{\mathsf{s}} \end{pmatrix}\in\mathfrak{gl}_{2n}\colon X, Y,Z\in\mathfrak{gl}_n,\,Y^{\mathsf{s}}=-Y,\,Z^{\mathsf{s}}= -Z\biggr\}.
\end{equation*}
\notag
$$
In a similar way the algebra $\mathfrak{sp}(J_1)$ in the same basis has the following matrix description:
$$
\begin{equation*}
\mathfrak{sp}(J_1)=\mathfrak{g}_0 =\biggl\{\begin{pmatrix} X&Y\\ Z&-X^{\mathsf{s}} \end{pmatrix}\in\mathfrak{gl}_{2n}\colon X, Y, Z\in\mathfrak{gl}_n,\,Y^{\mathsf{s}}=Y,\,Z^{\mathsf{s}}= Z\biggr\}.
\end{equation*}
\notag
$$
Let us show explicitly how the tensor products of elements of the Jordan algebra $J_2=\mathfrak{sym}(J_1)$ by elements of $\mathfrak{sl}_2$ correspond to elements of $\mathfrak{so}_{4n+1}$. Rename the standard basis of the abstract Lie algebra $\mathfrak{sl}_2$ which satisfies relations (2.17) by $e$, $f$, $h$. Corresponding to the matrix $M\in\mathfrak{sym}(J_1)$ of the form
$$
\begin{equation*}
M=\begin{pmatrix} X&Y\\ Z&X^{\mathsf{s}} \end{pmatrix}, \quad \text{where }\ X, Y, Z\in\mathfrak{gl}_n, \ \ Y^{\mathsf{s}}=-Y \ \text{ and } \ Z^{\mathsf{s}}=-Z,
\end{equation*}
\notag
$$
and the basis element $h\in\mathfrak{sl}_2$ is the matrix $\widehat{M}_{h}\in h\otimes J_2\subset\mathfrak{so}_{4n+1}$ of the form
$$
\begin{equation*}
\widehat{M}_{h}=\begin{pmatrix} &&0&&\\ &M&\vdots&&\\ 0&\dots&0&\dots&0\\ &&\vdots&-M^{\mathsf{s}}&\\ &&0&& \end{pmatrix}.
\end{equation*}
\notag
$$
Here and below blank areas in matrices contain zeros. In a similar way, corresponding to the matrix $M$ and the basis element $e\in\mathfrak{sl}_2$ is the matrix $\widehat{M}_{e}\in e\otimes J_2\subset\mathfrak{so}_{4n+1}$ of the form
$$
\begin{equation*}
\widehat{M}_{e}=\begin{pmatrix} &&0&&\\ &&\vdots&M_{c}&\\ 0&\dots&0&\dots&0\\ &&\vdots&&\\ &&0&& \end{pmatrix},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
M_{c}=M \cdot I_{2n}=\begin{pmatrix} X&-Y\\ Z&-X^{\mathsf{s}} \end{pmatrix}.
\end{equation*}
\notag
$$
Finally, corresponding to the matrix $M$ and the basis element $f\in\mathfrak{sl}_2$ is the matrix $\widehat{M}_{f}\in f\otimes J_2\subset\mathfrak{so}_{4n+1}$ of the form
$$
\begin{equation*}
\widehat{M}_{f}=\begin{pmatrix} &&0&&\\ &&\vdots&&\\ 0&\dots&0&\dots&0\\ &M_{r}&\vdots&&\\ &&0&& \end{pmatrix},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
M_{r}=I_{2n} \cdot M=\begin{pmatrix} X&Y\\ -Z&-X^{\mathsf{s}} \end{pmatrix}.
\end{equation*}
\notag
$$
The explicit embedding $\mathfrak{g}_0=\mathfrak{sp}(J_1)\,{\subset}\,\mathfrak{so}_{4n+1}$ is as follows. To the matrix ${N\,{\in}\,\mathfrak{sp}(J_1)}$ of the form
$$
\begin{equation*}
N=\begin{pmatrix} X&Y\\ Z&-Z^{\mathsf{s}} \end{pmatrix}, \quad\text{where }\ X, Y, Z\in\mathfrak{gl}_n, \ \ Y^{\mathsf{s}}=Y, \ \text{ and } \ Z^{\mathsf{s}}=Z,
\end{equation*}
\notag
$$
we assign the matrix $\widehat{N}_{0}\in\mathfrak{so}_{4n+1}$ such that
$$
\begin{equation*}
\widehat{N}_{0}=\begin{pmatrix} &&0&&\\ &N&\vdots&&\\ 0&\dots&0&\dots&0\\ &&\vdots&-N^{\mathsf{s}}&\\ &&0&& \end{pmatrix}.
\end{equation*}
\notag
$$
We denote the inner product on $J_2$ by $(\,\cdot\,{,}\,\cdot\,)_2$. Considering two elements of $\mathfrak{g}_2$ and taking their inner product, we can readily see that the inner product on the algebra $J_2 \simeq\mathfrak{sym}(J_1)$ has the form
$$
\begin{equation*}
(A, B)_2=\operatorname{tr}(AB) \quad\forall\, A, B\in \mathfrak{sym}(J_1).
\end{equation*}
\notag
$$
We denote the inner product on $\mathfrak{g}_0$ by $(\,\cdot\,{,}\,\cdot\,)_0$. Proceeding similarly with elements of the algebra $\mathfrak{g}_0\simeq\mathfrak{sp}(J_1)$, we see that the inner product on it has the form
$$
\begin{equation*}
(D_1, D_2)_0=\operatorname{tr}(D_1D_2) \quad\forall\, D_1, D_2\in \mathfrak{sp}(J_1).
\end{equation*}
\notag
$$
Summing up the above, for the short $\mathrm{SL}_2$-structure under consideration we have
$$
\begin{equation*}
J_1=\mathbb{C}^{2n}, \qquad J_2=\mathfrak{sym}(J_1)\quad\text{and} \quad\mathfrak{g}_0=\mathfrak{sp}(J_1),
\end{equation*}
\notag
$$
and the inner product on $\mathfrak{g}_0$ and $J_2$ is the standard inner product of linear operators. We call the short $\mathrm{SL}_2$-structure thus constructed a maximal short $\mathrm{SL}_2$-structure. Since the inclusions $J_2\subset\mathfrak{sym}(J_1)$ and $\mathfrak{g}_0\subset\mathfrak{sp}(J_1)$ hold for any short $\mathrm{SL}_2$-structure with symplectic space $J_1$, it follows that, for the maximal short $\mathrm{SL}_2$-structure with symplectic space $J_1$, the algebras $J_2$ and $\mathfrak{g}_0$ are the maximal ones with respect to inclusion, which justifies the term. 3.2. An analysis of the maximal short $\mathrm{SL}_2$-structureOn the symplectic space $J_1$ consider the linear operators $R(a, b)$ of rank $1$ $(a, b\in J_1)$ defined by the formula
$$
\begin{equation*}
R(a, b)c=\langle c, a\rangle b \quad\forall\, a, b, c\in J_1.
\end{equation*}
\notag
$$
It is obvious that the operator $R$ thus introduced has the following property: where $R^*(a, b)\colon J_1\to J_1$ denotes the operator adjoint to $R(a, b)$. Now we introduce the following maps on $J_1$. 1. The map $\varphi_m\colon J_1\times J_1\to\mathfrak{gl}(J_1)$ acting by the formula 2. The map $\delta_m\colon J_1\times J_1\to\mathfrak{gl}(J_1)$ acting by the formula It is clear that for arbitrary $a, b\in J_1$ the operator $\delta_m(a, b)$ is skew-symmetric, and the operator $\varphi_m(a, b)$ is symmetric with respect to the skew-inner product on $J_1$, that is,
$$
\begin{equation*}
\varphi_m(a, b)\in\mathfrak{sym}(J_1)\quad\text{and} \quad \delta_m(a, b)\in\mathfrak{sp}(J_1) \quad\forall\, a, b\in J_1.
\end{equation*}
\notag
$$
It follows from what we proved in § 3.1 for a maximal short $\mathrm{SL}_2$-structure that $J_2\,{=}\, \mathfrak{sym}(J_1)$ and $\mathfrak{g}_0=\mathfrak{sp}(J_1)$. It is easy to conclude from the definition of maps $\delta_m$ and $\varphi_m$ that, for arbitrary $a, b \in J_1$, $A\in J_2$ and $D\in\mathfrak{g}_0$, and Therefore, from Proposition 2, the nondegeneracy of the inner product on $J_2$ and $\mathfrak{g}_0$ and the above equalities we see that the maps $\delta_m$ and $\varphi_m$ coincide with the maps $\delta$ and $\varphi$ arising in the commutation formulae for the maximal short $\mathrm{SL}_2$-structure. Consider an arbitrary short $\mathrm{SL}_2$-structure on the algebra $\mathfrak{g}$ for which $\mathfrak{g}_1=\mathbb{C}^2\otimes J_1$. Let $\mathfrak{g}_0$ be the isotypic component of the trivial representation of $\mathfrak{sl}_2$, and let $J_2$ be the Jordan algebra of this $\mathrm{SL}_2$-structure. Due to the simplicity of the Jordan algebra $J_2$, the restriction of the invariant inner product $(\,\cdot\,{,}\,\cdot\,)_2$ of the Jordan algebra $\mathfrak{sym}(J_1)$ to the simple subalgebra $J_2$ is nondegenerate, and thus it is proportional to the invariant inner product on $J_2$ obtained from the inner product on $\mathfrak{g}$. Thus, the inner product on $J_2$ is proportional to the trace of the product of operators on $J_1$. Choosing an appropriate coefficient for the inner product on $\mathfrak{g}$, we can assume that the inner product on $J_2$ has the form
$$
\begin{equation}
(A, B)=(A, B)_2=\operatorname{tr}(AB) \quad\forall\, A, B\in J_2.
\end{equation}
\tag{3.1}
$$
Let us use equality 3) from Proposition 2. For arbitrary $D\in\mathfrak{g}_0$ and $A, B\in J_2$ we have
$$
\begin{equation*}
(D, [A,B])=([D, A], B)=\operatorname{tr}([D,A]B)=\operatorname{tr}(D[A,B]).
\end{equation*}
\notag
$$
Thus, on the subspace $[J_2, J_2]\subset\mathfrak{g}_0$ the inner product is defined by
$$
\begin{equation}
(D_1, D_2)=(D_1, D_2)_0=\operatorname{tr}(D_1D_2) \quad\forall\, D_1, D_2\in [J_2, J_2].
\end{equation}
\tag{3.2}
$$
Note that, as we can see from the decomposition (2.21) and the nondegeneracy of the inner product on the Lie algebra $\mathfrak{g}_0$, the inner product calculated above is nondegenerate on $[J_2, J_2]$. For convenience, in what follows, we assume that, on the simple Jordan algebra $J_2$, and also on the algebra of its derivations $\mathfrak{der}(J_2)=\mathfrak{inn}(J_2 )=[J_2, J_2]$, the inner product is by default given by the formulae (3.1) and (3.2), respectively. Theorem 6. For every short $\mathrm{SL}_2$-structure with symplectic space $J_1$, the maps $\varphi\colon J_1\times J_1\to J_2$ and $\delta\colon J_1\times J_1\to\mathfrak{g}_0$ arising in the commutation formulae are surjective (as linear maps from $J_1\otimes J_1$), and the equalities $\varphi=\pi_{2}\varphi_m$ and ${\delta_c=\pi_{c}\delta_m}$ hold, where $\pi_{2}\colon \mathfrak{sym}(J_1)\to J_2$ and $\pi_{c}\colon \mathfrak{sp}(J_1)\to [J_2, J_2]$ are the orthogonal projections onto the algebras $J_2$ and $[J_2, J_2]$, respectively. For the maximal short $\mathrm{SL}_2$-structure $\varphi=\varphi_m$ and $\delta=\delta_m$. Proof. For the maximal short $\mathrm{SL}_2$-structure, the equalities $\varphi=\varphi_m$ and $\delta=\delta_m$ have already been proved.
Consider an arbitrary short $\mathrm{SL}_2$-structure and the subspace $\mathfrak{g}_1\oplus[\mathfrak{g}_1, \mathfrak{g}_1]$. It follows from the relations (2.2) that $\mathfrak{g}_1\oplus[\mathfrak{g}_1, \mathfrak{g}_1]\lhd \mathfrak{g}$, so that $\mathfrak{g}_1\oplus[\mathfrak{g}_1, \mathfrak{g}_1]\,{=}\,\mathfrak{g}$. Thus, in connection with the commutation formula (2.6), the following relations hold:
$$
\begin{equation*}
\mathfrak{g}_0=\langle\delta(a,b)\colon a, b\in J_1\rangle\quad\text{and} \quad J_2=\langle\varphi(a,b)\colon a, b\in J_1\rangle.
\end{equation*}
\notag
$$
Hence the maps $\delta$ and $\varphi$ are surjective. Acting similarly to the case of the maximal short $\mathrm{SL}_2$-structure, we see that, according to Proposition 2, for an arbitrary short $\mathrm{SL}_2$-structure and any $a, b\,{\in}\, J_1$, $ A, B\,{\in}\, J_2$ we have and In connection with the nondegeneracy of the inner products on $J_2$ and $[J_2,J_2]$, the map $\varphi$ coincides with $\pi_2\varphi_{m}$, and $\delta_c$ coincides with $\pi_{c}\delta_{m}$. This completes the proof of the theorem. § 4. Short $\mathrm{SL}_2$-structures and Lie-Jordan symplectic structures4.1. Short $\mathrm{SL}_2$-structures and the curvature of a symmetric spaceLet us take a short break from short $\mathrm{SL}_2$-structures. Let $\mathfrak{h}$ be a reductive Lie algebra, and let $R\colon \mathfrak{h}\to\mathfrak{gl}(\mathfrak{v})$ be a faithful orthogonal linear representation on a linear space $\mathfrak{v}$. Let us define the commutators of the elements in $\mathfrak{h}$ with elements in $\mathfrak{v}$ using the representation $R$ by the formula
$$
\begin{equation*}
[\xi, x]=R(\xi)x=-[x, \xi] \quad\forall\,\xi\in\mathfrak{h}, \quad\forall\, x\in\mathfrak{v},
\end{equation*}
\notag
$$
and define the commutators of two elements of $\mathfrak{v}$ as elements of $\mathfrak{h}$ satisfying the condition
$$
\begin{equation}
(\xi,[x,y])=([\xi,x],y) \quad\forall\,\xi\in\mathfrak{h}, \quad\forall\, x, y\in\mathfrak{v},
\end{equation}
\tag{4.1}
$$
where parentheses denote the $\mathfrak{h}$-invariant inner products on $\mathfrak{h}$ and $\mathfrak{v}$. On the space $\mathfrak{v}$ we define a 4-linear form
$$
\begin{equation*}
K(x, y, z, u)=([x, y], [z, u]) \quad\forall\, x, y, z, u\in\mathfrak{v}.
\end{equation*}
\notag
$$
The following result goes back to É. Cartan and Kostant (see, for example, [6], § 1, and [7], § 1). Theorem 7. The operation defined above on the space $\mathfrak{h}\oplus\mathfrak{v}$ defines on it the structure of a $\mathbb{Z}_2$-graded Lie algebra if and only if the form $K(x, y, z, u)$ has the symmetry of the curvature tensor, which, in turn, is equivalent to the validity of Bianchi’s identity For the proof it suffices to verify of Jacobi’s identity in the following four cases: 1) for elements $\xi, \eta, \zeta\in\mathfrak{h}$; 2) for elements $\xi, \eta\in\mathfrak{h}$ and $x\in\mathfrak{v}$; 3) for elements $\xi\in\mathfrak{h}$ and $x, y\in\mathfrak{v}$; 4) for the elements $x, y, z\in\mathfrak{v}$. In case 1) Jacobi’s identity holds because $\mathfrak{h}$ is a Lie algebra. In case 2) it follows from the fact that $R$ is a representation of the Lie algebra $\mathfrak{h}$. Case 3) can be reduced to 2) by taking the inner product of the left-hand side of Jacobi’s identity in question by an arbitrary element $\kappa\in\mathfrak{h}$ and using the invariance of the inner product. In the last case Jacobi’s identity is equivalent to (4.2). Bianchi’s identity does not appear by occasion in the above theorem. The point is that, if $\mathfrak{g}= \mathfrak{h}\oplus\mathfrak{v}$ is a $\mathbb{Z}_2$-graded Lie algebra, then the space $G/H$, where $G$ is a complex Lie group such that $\operatorname{Lie} G=\mathfrak{g}$ and $H$ is a Lie subgroup such that $\operatorname{Lie}H=\mathfrak{h}$, is a complex symmetric space on which there is a $G$-invariant metric tensor induced by the inner product on $\mathfrak{v}$ and a symmetric Levi-Civita connection that preserves it. The Riemann tensor of the symmetric space $G/H$ at the basis point $eH$ has the form
$$
\begin{equation*}
\rho(x, y)=-\operatorname{ad}([x, y])|_{\mathfrak{v}} \quad\forall\, x, y\in\mathfrak{v},
\end{equation*}
\notag
$$
where $\mathfrak{v}$ is identified with the tangent space $T_{eH}(G/H)$. Therefore, the 4-linear form $K$ is, up to sign, a covariant curvature tensor on the space $G/H$ at the point $eH$. A more detailed analysis of this construction can be found in [6], § 1. Let us return to the consideration of short $\mathrm{SL}_2$-structures. We write $\mathfrak{h}= \mathfrak{g}_0\oplus\mathfrak{g}_2$, $\mathfrak{v}\,{=}\,\mathfrak{g}_1$ and, as a representation $R$, consider the adjoint representation. As noted above, the adjoint representation of $\mathfrak{h}$ on $\mathfrak{v}$ is faithful and, due to the invariance of the inner product on $\mathfrak{g}$, it is also orthogonal. Also, due to the properties of the invariant inner product on the Lie algebra $\mathfrak{g}$, the commutation operation satisfies condition (4.1). For $\mathfrak{h}$ and $\mathfrak{v}$ consider the 4-linear form $K$. By Theorem 7 the form $K$ satisfies identity (4.2). In connection with equations (1.2), the 4-linear form $K$ can be rewritten in terms of the space $J_1$. Namely, we set $x= u_1\otimes a$, $y=u_2\otimes b$, $z=u_3\otimes c$, $u=u_4\otimes d$, where $u_i\in\mathbb{C}^2$, $ i=1,\dots,4$, for all $a, b, c, d\in J_1$. Then, using (2.4)–(2.8) we obtain
$$
\begin{equation*}
\begin{aligned} \, K(x, y, z, u) &=(S(u_1, u_2)\otimes \varphi(a, b) \\ &\qquad+\langle u_1, u_2\rangle\delta(a, b), S(u_3, u_4)\otimes \varphi(c, d)+\langle u_3, u_4\rangle\delta(c, d)). \end{aligned}
\end{equation*}
\notag
$$
From this, using the definition of the inner product and Lemma 3 we obtain
$$
\begin{equation}
\begin{aligned} \, \notag K(x, y, z, u) &=(\langle u_3, u_1\rangle\langle u_2, u_4\rangle+\langle u_3, u_2\rangle\langle u_1, u_4\rangle)(\varphi(a, b), \varphi(c, d)) \\ &\qquad +\langle u_1, u_2\rangle\langle u_3, u_4\rangle(\delta(a, b), \delta(c, d)). \end{aligned}
\end{equation}
\tag{4.3}
$$
Now equality (4.2) can be rewritten in the form
$$
\begin{equation}
\begin{aligned} \, \notag &\langle u_1, u_2\rangle\langle u_3, u_4\rangle\bigl((\varphi(b, c), \varphi(a, d))+ (\varphi(a, c), \varphi(b, d))+(\delta(a, b), \delta(c, d))\bigr) \\ \notag &\qquad +\langle u_2, u_3\rangle\langle u_1, u_4\rangle\bigl((\varphi(b, a), \varphi(c, d))+ (\varphi(c, a), \varphi(b, d))+(\delta(b, c), \delta(a, d))\bigr) \\ &\qquad +\langle u_3, u_1\rangle\langle u_2, u_4\rangle\bigl((\varphi(a, b), \varphi(c, d))+ (\varphi(c, b), \varphi(a, d))+(\delta(c, a), \delta(b, d))\bigr)=0. \end{aligned}
\end{equation}
\tag{4.4}
$$
We write
$$
\begin{equation*}
\overline{F}(a, b, c, d)=(\delta(a, b), \delta(c, d))+(\varphi(b, c), \varphi(a, d))+(\varphi(a, c), \varphi(b, d)).
\end{equation*}
\notag
$$
Then (4.4) can be written in the form
$$
\begin{equation}
\begin{aligned} \, \notag &\langle u_1, u_2\rangle\langle u_3, u_4\rangle \overline{F}(a, b, c, d) +\langle u_2, u_3\rangle\langle u_1, u_4\rangle \overline{F}(b, c, a, d) \\ &\qquad\qquad +\langle u_3, u_1\rangle\langle u_2, u_4\rangle \overline{F}(c, a, b, d)=0. \end{aligned}
\end{equation}
\tag{4.5}
$$
We claim that identity (4.5) is equivalent to the symmetry of the form $\overline{F}$. To prove this we use the well-known Plücker identity for $u_1, u_2, u_3, u_4\in\mathbb{C}^2$:
$$
\begin{equation*}
\langle u_1, u_2\rangle\langle u_3, u_4\rangle +\langle u_2, u_3\rangle\langle u_1, u_4\rangle+ \langle u_3, u_1\rangle\langle u_2, u_4\rangle=0.
\end{equation*}
\notag
$$
First, (4.5) follows from the symmetry of $\overline{F}$ by Plücker’s identity. Second, the definition of $\overline{F}$ immediately implies the symmetry of $\overline{F}$ under the permutation of the first two arguments, and also under the permutation of the last two arguments. Therefore, to prove the inverse implication, it suffices to verify the symmetry of $\overline{F}$ under cyclic permutations of the first three arguments. To do this, substitute $u_1=u_3=e_1$ and $u_2=u_4=e_{-1}$ into (4.5). Then $\langle u_1, u_2\rangle\langle u_3, u_4\rangle=- \langle u_2, u_3\rangle\langle u_1, u_4\rangle=1$, and $\langle u_3, u_1\rangle\langle u_2, u_4\rangle=0$, from which the symmetry of $\overline{F}$ under cyclic permutations of the first three arguments follows. As an example, let us calculate the form $\overline{F}_m$ for the maximal short $\mathrm{SL}_2$-structure. We have
$$
\begin{equation*}
\begin{aligned} \, \overline{F}_m(a, b, c, d) &=(\delta_m(a, b), \delta_m(c, d))+(\varphi_m(b, c), \varphi_m(a, d))+(\varphi_m(a, c), \varphi_m(b, d)) \\ &=\langle\delta_m(a, b)c+\varphi_m(b, c)a+\varphi_m(a, c)b , d\rangle \\ &=\frac{1}{2} \bigl\langle-(\langle a, c\rangle b+\langle b, c\rangle a) +(\langle b, a\rangle c\,{-}\,\langle c, a\rangle b)\,{+}\,(\langle a, b\rangle c\, {-}\,\langle c, b\rangle a) ,d\bigr\rangle\,{=}\,0. \end{aligned}
\end{equation*}
\notag
$$
In the general case we use Proposition 2:
$$
\begin{equation*}
\begin{aligned} \, \overline{F}(a, b, c, d) &=(\delta(a, b), \delta(c, d))+(\varphi(b, c), \varphi(a, d)) +(\varphi(a, c), \varphi(b, d)) \\ &=(\delta(a, b)c+\varphi(b, c)a+\varphi(a, c)b, d). \end{aligned}
\end{equation*}
\notag
$$
We write
$$
\begin{equation*}
F(a, b, c)=\delta(a, b)c+\varphi(b, c)a+\varphi(a, c)b.
\end{equation*}
\notag
$$
Thus, with the help of the above formula, the following trilinear map is defined: Since, as shown above, the form $\overline{F}$ is symmetric for an arbitrary short $\mathrm{SL}_2$-structure, it follows that the map $F$ is also symmetric. 4.2. The main theoremIt is now time to formulate and prove the main theorem in our paper. For the convenience of our formulation, we introduce the following definitions. Let $J_1$ be a symplectic space, let $J_2\subset\mathfrak{sym}(J_1)$ be a semisimple Jordan subalgebra whose unity is the identity operator in the space $J_1$, and let $\mathfrak{g}_0\subset\mathfrak{sp}(J_1)$ be a reductive Lie subalgebra such that $[J_2, J_2]\subset \mathfrak{g}_0$ and $[\mathfrak{g}_0, J_2]\subset J_2$. Then the following decomposition holds: where $\mathfrak{i}_0$ is the kernel of the adjoint action of $\mathfrak{g}_0$ on $J_2$. Since the Jordan algebra $J_2$ is semisimple, it follows from Theorem 1 that the Lie algebra $[J_2, J_2]$ is also semisimple. We assume that, on the algebras $\mathfrak{sp}(J_1)$ and $\mathfrak{sym}(J_1)$, as well as on their subalgebras $[J_2, J_2]$ and $J_2$, respectively, a non-degenerate invariant inner product is defined which is equal to the trace of the product of operators.Let a bilinear symmetric $\mathfrak{g}_0$-equivariant surjective map $\delta_0\colon J_1\times J_1\to \mathfrak{i}_0$ be given which satisfies the identity
$$
\begin{equation}
\delta_0(Aa, b)=\delta_0(a, Ab) \quad\forall\, A\in J_2, a, b\in J_1.
\end{equation}
\tag{4.6}
$$
We define maps $\delta\colon J_1\times J_1\to\mathfrak{g}_0$ and $\varphi\colon J_1\times J_1\to J_2$ using the formulae
$$
\begin{equation}
\delta=\delta_0+\delta_c, \qquad\delta_c=\pi_{c}\delta_m\quad\text{and} \qquad\varphi=\pi_2\varphi_m,
\end{equation}
\tag{4.7}
$$
where $\pi_{c}\colon \mathfrak{sp}(J_1)\to [J_2, J_2]$ is the orthogonal projection onto $[J_2, J_2]$ and $\pi_2$: $\mathfrak{sym}(J_1)\to J_2$ is the orthogonal projection onto $J_2$. Definition 6. A quadruple $(J_1; J_2; \mathfrak{g}_0; \delta_0)$ is called a Lie-Jordan symplectic structure if the trilinear map $F\colon J_1\times J_1\times J_1\to J_1$ given by the formula is symmetric. Remark 1. It can readily be noted that the maps $\delta_c$ (and therefore $\delta$) and $\varphi$ are $\mathfrak{g}_0$-equivariant. Indeed, $\delta_c=\pi_{c}\delta_m$, $\varphi= \pi_2\varphi_m$, and the maps $\delta_m$ and $\varphi_m$ are $\mathfrak{sp}(J_1)$-equivariant, and the projections $\pi_{c}$ and $\pi_2$ are $\mathfrak{g}_0$-equivariant as orthogonal projections onto $\mathfrak{g}_0$-invariant subspaces with respect to the invariant inner product. The map $\delta_c$ is surjective as the composition of the surjective maps $\pi_c$ and $\delta_m$. Under the adjoint representation of $\mathfrak{g}_0$ on itself, the irreducible summands included in $\mathfrak{i}_0$ and $[J_2, J_2]$ are pairwise nonisomorphic. Therefore, since the maps $\delta_0$ and $\delta_c$ are $\mathfrak{g}_0$-equivariant, it follows that the map $\delta= \delta_0+\delta_c$ is also surjective as the sum of two surjective homomorphisms of $\mathfrak{g}_0$-modules. Definition 7. The symplectic Lie-Jordan structure $(J_1; J_2; \mathfrak{g}_0; \delta_0)$ is said to be simple if the Jordan algebra $J_2$ is simple. Theorem 8. There is a one-to-one correspondence between simple symplectic Lie-Jordan structures and simple Lie algebras with short $\mathrm{SL}_2$-structure: a simple symplectic Lie-Jordan structure $(J_1; J_2; \mathfrak{g}_0; \delta_0)$ corresponds to a simple Lie algebra $\mathfrak{g}$ of the form Proof. To begin with, note that from every simple Lie algebra with short $\mathrm{SL}_2$-structure, as shown above, one can construct a simple symplectic Lie-Jordan structure. It remains to prove that from a simple symplectic Lie-Jordan structure $(J_1; J_2; \mathfrak{g}_0; \delta_0)$ one can construct a simple Lie algebra $\mathfrak{g}$ with short $\mathrm{SL}_2$-structure. We set $\mathfrak{g}_1= \mathbb{C}^2\otimes J_1$ and $\mathfrak{g}_2=\mathfrak{sl}_2\otimes J_2$. Since formulae (4.10) and (4.13) are identical to the corresponding formulae for a very short $\mathrm{SL}_2$-structure, the subspace $\mathfrak{h}=\mathfrak{g}_0\oplus\mathfrak{g}_2$ is endowed with the structure of a Lie algebra.
Consider the space $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{g}_1$. We define the operation of commutation on it with the help of formulae (4.9)–(4.13). We claim that these formulae endow the space $\mathfrak{g}$ with the structure of a Lie algebra. To show this it suffices to prove that Jacobi’s identity is true in the following cases: 1) for elements $\xi, \eta, \zeta\in\mathfrak{h}$; 2) for elements $\xi, \eta\in\mathfrak{h}$ and $x\in\mathfrak{g}_1$; 3) for elements $\xi\in\mathfrak{h}$ and $x, y\in\mathfrak{g}_1$; 4) for elements $x, y, z\in\mathfrak{g}_1$. In case 1) Jacobi’s identity holds because $\mathfrak{h}$ is a Lie algebra. Using (4.9) and (4.12) we can define a linear map $R\colon \mathfrak{h}\to\mathfrak{gl}(\mathfrak{g}_1)$ of the Lie algebra $\mathfrak{h}$ into the space of linear operators on $\mathfrak{g}_1$. This map is a linear representation of $\mathfrak{h}$. Indeed, let us prove the defining relation of a linear representation of a Lie algebra for $R$, that is, the equality
$$
\begin{equation}
R([\xi, \eta])x=[R(\xi), R(\eta)]x=R(\xi)R(\eta)x-R(\eta)R(\xi)x \quad\forall\, \xi,\eta\in\mathfrak{h}, \quad x\in\mathfrak{g}_1.
\end{equation}
\tag{4.14}
$$
We write $x=u\otimes a\in\mathfrak{g}_1$. By linearity it suffices to prove (4.14) in the following three cases:
$$
\begin{equation*}
R([\xi, \eta])x=u\otimes [D_1, D_2]a=u\otimes D_1D_2a-u\otimes D_2D_1a=[R(\xi), R(\eta)]x.
\end{equation*}
\notag
$$
In case 2) we set $\xi=D\in\mathfrak{g}_0$ and $\eta=X\otimes A\in\mathfrak{g}_2$. Then
$$
\begin{equation*}
R([\xi, \eta])x=Xu\otimes [D, A]a=[D, Xu\otimes Aa] -[X\otimes A, u\otimes Da]=[R(\xi), R(\eta)]x.
\end{equation*}
\notag
$$
In case 3) we set $\xi=X\otimes A$ and $\eta=Y\otimes B\in\mathfrak{g}_2$. Then
$$
\begin{equation*}
\begin{aligned} \, R([\xi, \eta])x &=[X, Y]u\otimes (A\circ B)a+\frac{1}{2}(X, Y)u\otimes[A, B]a \\ &=[X, Y]u\otimes (A\circ B)a+\frac{1}{2}(XY+YX)u\otimes[A, B]a \\ &=XYu\otimes \biggl(A\circ B+\frac{1}{2}[A, B]\biggr)a -YXu\otimes \biggl(A\circ B-\frac{1}{2}[A, B]\biggr)a \\ &=XYu\otimes ABa-YXu\otimes BAa \\ &=[X\otimes A, Yu\otimes Ba]-[Y\otimes B, Xu\otimes Aa]=[R(\xi), R(\eta)]x. \end{aligned}
\end{equation*}
\notag
$$
Thus, $R$ is a linear representation of the Lie algebra $\mathfrak{h}$. The validity of Jacobi’s identity in case 2) follows from the defining relation of the representation $R$. It remains to prove Jacobi’s identity in cases 3) and 4).
Consider case 3). It suffices to prove Jacobi’s identity in the following two subcases: 1) for elements $\xi\in\mathfrak{g}_0$ and $x, y\in\mathfrak{g}_1$; 2) for elements $\xi\in\mathfrak{g}_2$ and $x, y\in\mathfrak{g}_1$. In the first subcase of case 3), for $\xi=D\in\mathfrak{g}_0$, $x=u\otimes a$ and $y= v\otimes b\in\mathfrak{g}_1$, Jacobi’s identity looks as
$$
\begin{equation*}
[[D, u\otimes a], v\otimes b]+[[u\otimes a, v\otimes b], D] +[[v\otimes b, D], u\otimes a]=0.
\end{equation*}
\notag
$$
Applying commutation formulae (4.9)–(4.13) we can rewrite the above equality in the form
$$
\begin{equation}
\begin{aligned} \, \notag &S(u, v)\otimes\varphi(Da, b)+\langle u, v\rangle\delta(Da, b) -S(u, v)\otimes [D, \varphi(a, b)]-\langle u, v\rangle[D, \delta(a, b)] \\ &\qquad\qquad+S(u, v)\otimes\varphi(a, Db)+\langle u, v\rangle\delta(a, Db)=0. \end{aligned}
\end{equation}
\tag{4.15}
$$
Using here the $\mathfrak{g}_0$-equivariance of $\delta$ we obtain
$$
\begin{equation*}
S(u, v)\otimes\varphi(Da, b)- S(u, v)\otimes [D, \varphi(a, b)]+S(u, v)\otimes\varphi(a, Db)= 0.
\end{equation*}
\notag
$$
Clearly, the resulting equality is equivalent to the $\mathfrak{g}_0$-equivariance of $\varphi$.
For further considerations we must derive two auxiliary identities. For arbitrary $A, B\in J_2$ and $a, b\in J_1$ we have
$$
\begin{equation*}
\begin{aligned} \, ([A,\delta(a, b)], B) &=([A,\delta_c(a, b)], B) =(\delta_c(a, b), [B, A])=\langle[B, A]a, b\rangle \\ &=\langle BAa,b\rangle-\langle ABa, b\rangle=\langle BAa, b\rangle-\langle Ba, Ab\rangle \\ &= (\varphi(Aa, b)-\varphi(a, Ab), B). \end{aligned}
\end{equation*}
\notag
$$
Since the above equality holds for any $B\in J_2$, it follows from the nondegeneracy of the inner product in $J_2$ that
$$
\begin{equation}
[A, \delta(a, b)]=\varphi(Aa, b)-\varphi(a, Ab) \quad\forall\, A\in J_2, \quad a, b\in J_1.
\end{equation}
\tag{4.16}
$$
This is one of the identities we need. Let us now derive the other identity. For arbitrary $A, B, C\in J_2$ and $a, b\in J_1$ we have
$$
\begin{equation*}
\begin{aligned} \, & ([A,\varphi(a, b)], [B, C])=(\varphi(a, b), [[B, C], A]) =\langle[[B, C], A]a, b\rangle \\ &\qquad =\langle [B, C]Aa,b\rangle-\langle A[B, C]a, b\rangle= \langle [B, C]Aa,b\rangle-\langle[B,C]a, Ab\rangle \\ &\qquad =(\delta_c(Aa, b)-\delta_c(a, Ab), [B, C]). \end{aligned}
\end{equation*}
\notag
$$
Since the above equality holds for all elements generating the Lie algebra $[J_2, J_2]$, it follows that
$$
\begin{equation*}
[A, \varphi(a, b)]=\delta_c(Aa, b)-\delta_c(a, Ab) \quad\forall\, A\in J_2, \quad a, b\in J_1.
\end{equation*}
\notag
$$
Hence, using (4.6) and the definition of $\delta$ we obtain
$$
\begin{equation}
[A, \varphi(a, b)]=\delta(Aa, b)-\delta(a, Ab) \quad\forall\, A\in J_2, \quad a, b\in J_1.
\end{equation}
\tag{4.17}
$$
This is the other identity we need.
Consider Jacobi’s identity in the second subcase of case 3), for $\xi=X\otimes A\in\mathfrak{g}_2$ and $x=u\otimes a$, $y=v\otimes b\in\mathfrak{g}_1$:
$$
\begin{equation*}
[[X\otimes A, u\otimes a], v\otimes b]+[[u\otimes a, v\otimes b], X\otimes A]+ [[v\otimes b, X\otimes A], u\otimes a]=0.
\end{equation*}
\notag
$$
Applying commutation formulae (4.9)–(4.13) we can rewrite the above equality in the form
$$
\begin{equation}
\begin{aligned} \, \notag &S(Xu, v)\otimes\varphi(Aa, b)+\langle Xu, v\rangle\delta(Aa, b) \\ \notag &\qquad +[S(u, v), X]\otimes(\varphi(a, b)\circ A)+\frac{1}{2}(S(u, v), X)[\varphi(a, b), A] \\ &\qquad+\langle u, v\rangle X\otimes[\delta(a, b), A] -S(Xv, u)\otimes\varphi(Ab, a)-\langle Xv, u\rangle\delta(Ab, a)=0. \end{aligned}
\end{equation}
\tag{4.18}
$$
Using identities (4.16) and (4.17) proved above and Lemma 3, we can rewrite (4.18) as
$$
\begin{equation}
[X,S(u,v)]\otimes\biggl(\frac{1}{2}\varphi(Aa,b)+\frac{1}{2}\varphi(a,Ab)- \varphi(a,b)\circ A\biggr)=0.
\end{equation}
\tag{4.19}
$$
To prove (4.19) we must derive another auxiliary identity. Using the properties of the inner product in $J_2$ and the equality $\varphi= \pi_2\varphi_m$, for arbitrary $A, B\in J_2$ and $a, b\in J_1$ we obtain
$$
\begin{equation*}
\begin{aligned} \, & (A\circ\varphi(a, b), B)=(\varphi(a, b), A\circ B) \\ &\qquad=\frac{1}{2}(AB, \varphi(a, b))+ \frac{1}{2}(BA,\varphi(a, b)) =\frac{1}{2}\langle ABa, b\rangle+\frac{1}{2}\langle BAa, b\rangle \\ &\qquad =\frac{1}{2}\langle Ba, Ab\rangle+\frac{1}{2}\langle BAa, b\rangle =\frac{1}{2}(B, \varphi(a, Ab))+\frac{1}{2}(B,\varphi(Aa, b)). \end{aligned}
\end{equation*}
\notag
$$
Hence
$$
\begin{equation}
A\circ\varphi(a, b)=\frac{1}{2}(\varphi(Aa, b)+\varphi(a, Ab)) \quad\forall\, A\in J_2, \quad a, b\in J_1.
\end{equation}
\tag{4.20}
$$
Using equality (4.20) we obtain identity (4.19), and therefore also Jacobi’s identity in case 3).
Consider case 4). For $x=u\otimes a$, $y=v\otimes b$ and $z=w\otimes c\in\mathfrak{g}_1$ Jacobi’s identity looks as follows:
$$
\begin{equation*}
[[u\otimes a, v\otimes b], w\otimes c] +[[v\otimes b, w\otimes c], u\otimes a] + [[w\otimes c, u\otimes a], v\otimes b]=0.
\end{equation*}
\notag
$$
Using commutation formulae (4.9)–(4.13) we can rewrite it in the form
$$
\begin{equation}
\langle u, v\rangle w\otimes F(a, b, c)+\langle v, w\rangle u\otimes F(b, c, a)+\langle w, u\rangle v\otimes F(c, a, b)=0,
\end{equation}
\tag{4.21}
$$
where
$$
\begin{equation*}
F(a, b, c)=\delta(a, b)c+\varphi(a, c)b+\varphi(b, c)a.
\end{equation*}
\notag
$$
Since, by the definition of the Lie-Jordan symplectic structure $(J_1; J_2; \mathfrak{g}_0; \delta_0)$, the map $F\colon J_1\times J_1\times J_1\to J_1$ is symmetric, it follows that (4.21) can be rewritten as
$$
\begin{equation*}
(\langle u, v\rangle w+\langle v, w\rangle u+\langle w, u\rangle v) \otimes F(a, b, c)=0.
\end{equation*}
\notag
$$
By Lemma 3 we have $\langle u, v\rangle w+\langle v, w\rangle u+\langle w, u\rangle v=0$, from which Jacobi’s identity in case 4) follows.
Thus, the vector space $\mathfrak{g}$ is endowed with the structure of a Lie algebra with short $\mathrm{SL}_2$-structure. We prove that this Lie algebra is simple. Let $\mathfrak{j}$ be an ideal of $\mathfrak{g}$. It is obvious that
$$
\begin{equation*}
\mathfrak{j}= \mathfrak{j}_0\oplus\mathfrak{j}_1\oplus\mathfrak{j}_2, \quad \text{where } \mathfrak{j}_i= \mathfrak{g}_i\cap\mathfrak{j}, \quad i=0,1,2.
\end{equation*}
\notag
$$
Consider $\mathfrak{j}_2$. It is clear that $\mathfrak{j}_2=\mathfrak{sl}_2\otimes I_2$ and $I_2\triangleleft J_2$. Since the Jordan algebra $J_2$ is simple, two cases are possible. 1. $I_2=0$. Then $\mathfrak{j}=\mathfrak{j}_0\oplus\mathfrak{j}_1$. We note that $\mathfrak{j}_1= \mathbb{C}^2\otimes I_1$, where $I_1\subset J_1$. Since $\mathfrak{j}$ is invariant under the operation of commutation with $\mathfrak{g}_1$, it follows from (4.11) that
$$
\begin{equation*}
\varphi(a, b)=0 \quad\forall\, a\in I_1, \quad b\in J_1.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\langle Aa, b\rangle=(A, \varphi(a, b))=0 \quad\forall\, A\in J_2, \quad a\in I_1, \quad b\in J_1.
\end{equation*}
\notag
$$
Using the nondegeneracy of the inner product on $J_1$, we can derive from the above identity that $Aa= 0$ for any $A\in J_2$ and $a\in I_1.$ However, then $I_1=0$ in view of the fact that $J_2$ contains the identity operator. Thus, $\mathfrak{j}=\mathfrak{j}_0$.
Since $\mathfrak{j}$ is invariant under the operation of commutation with $\mathfrak{g}_1$, it follows that $Da= 0$ for any $D\in\mathfrak{j}_0$ and $a\in J_1$, which immediately implies that $\mathfrak{j}=0$. 2. $I_2=J_2$. Then $\mathfrak{j}=\mathfrak{j}_0\oplus\mathfrak{j}_1\oplus\mathfrak{g}_2$. Since $\mathfrak{j}$ is invariant under the operation of commutation with $\mathfrak{g}_1$ and $J_2$ contains the identity operator, it follows that $\mathfrak{j}_1= \mathfrak{g}_1$. Since the map $\delta$ is surjective, $\mathfrak{g}_0$ is generated by the operators $\delta (a, b)$, where ${a, b\in J_1}$. Hence, since the ideal $\mathfrak{j}$ contains all such operators, we have $\mathfrak{j}_0= \mathfrak{g}_0$, and therefore $\mathfrak{j}=\mathfrak{g}$. Thus, the Lie algebra $\mathfrak{g}$ constructed above is simple, which proves Theorem 8. § 5. Classification of short $\mathrm{SL}_2$-structures5.1. PreliminariesConsider an arbitrary short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}$ and keep all the notation of the previous sections; again, we denote by $e$, $f$ and $h$ basis elements of the Lie algebra $\mathfrak{sl}_2\subset\mathfrak{g}$ satisfying relations (2.17). We note that, since $\mathfrak{g}_1=\mathbb{C}^2\otimes J_1\neq 0$, the only possible nonzero value of a simple root of the Lie algebra $\mathfrak{g}$ at the element $h$ is $1$. On the Dynkin diagram of $\mathfrak {g}$, we colour black the vertices corresponding to simple roots taking the value $1$ at $h$, and we call them unit vertices. The vertices of the diagram corresponding to roots taking the zero value at $h$ are left uncoloured; we call them zero vertices. We call the result the diagram of the short $\mathrm{SL}_2$-structure on the Lie algebra $\mathfrak{g}$. A simple example of such a diagram is the diagram of the short $\mathrm{SL}_2$-structure on a simple Lie algebra $\mathfrak{g}=G_2$. The highest root of this algebra has the form where $\alpha_1$ and $\alpha_2$ are simple roots. Then, since $\alpha_{\max}(h)=2$, the only possible case is when $\alpha_2(h)=1$ and $\alpha_1(h)=0$. Then the Dynkin diagram of this case has the form Based on the diagram of the short $\mathrm{SL}_2$-structure, it is easy to calculate the dimension of the centre $\mathfrak{z}^0$ of the algebra $\mathfrak{g}^0$. Since the Cartan subalgebra $\mathfrak{t}$ lies in $\mathfrak{g}^0$, the dimension of the centre is equal to the difference between the rank of $\mathfrak{g}$ and the rank of the semisimple part of $\mathfrak{g}^0$, that is, to the number of black vertices. Theorem 8 enables us to claim that, by listing all existing short $\mathrm{SL}_2$-structures on simple Lie algebras and indicating simple symplectic Lie-Jordan structures corresponding to them, we identify all possible simple symplectic Lie-Jordan structures, which, generally speaking, can be considered on their own, without linking them with $\mathrm{SL}_2$-structures. To perform this, we denote by $\widetilde{G}^0$ the algebraic group whose tangent algebra is the Lie algebra $\widetilde{\mathfrak{g}}^0$ and use the following assertion. Theorem 9. For a $\mathbb{Z}$-grading defined by a semisimple element $h\in\mathfrak{g}$, the following conditions are equivalent. 1. The element $h$ is included in an $\mathfrak{sl}_2$-triple. 2. The representation $\widetilde{G}^0\colon \mathfrak{g}^2$ has no open orbits. 3. The subspace $\mathfrak{g}^2$ has a nontrivial polynomial invariant of the action $\widetilde{G}^0\colon \mathfrak{g}^2$. The proof of the equivalence of parts 1 and 2 of Theorem 9 can be found in [1], § 1.4. Part 2 obviously follows from part 3. The proof of the fact that part 1 implies part 3 can be found in [8], § 1, Propositions 1.1 and 1.2. It is important to note that the $\mathfrak{sl}_2$-triple containing a semisimple element $h$ is uniquely defined up to conjugation with $G^0$, where $G^0$ is the connected algebraic group with the tangent algebra $\mathfrak{g}^0$. Namely, as $e$ we can take any element of the open $G^0$-orbit in $\mathfrak{g}^2$, and $f$ is defined uniquely by the elements $h$ and $e$. Now we have everything we need to classify all short $\mathrm{SL}_2$-structures on simple Lie algebras. 5.2. A classification of short $\mathrm{SL}_2$-structures on classical simple Lie algebrasIn the case of a classical simple Lie algebra $h$ is a diagonal matrix and its diagonal entries can be computed from the values of this element at simple roots. Considering the commutators of this element with arbitrary matrices in $\mathfrak{g}$, one can determine the representation space $\mathfrak{g}^2$ and the group $\widetilde{G}^0$. This specifies the action $\widetilde{G}^0\colon \mathfrak{g}^2$ fully and, using Theorem 9, or rather its third part, it is possible to determine whether or not there is a short $\mathrm{SL}_2$-structure in this case and, if it exists, then under what conditions. Throughout what follows we denote the highest root of the algebra $\mathfrak{g}$ by $\alpha_{\max}$ and simple roots by $\alpha_i$. The case $\mathfrak{g}=\mathfrak{sl}_{n}$. The dependence of the highest root on the simple roots is expressed by the formula Since $\alpha_{\max}(h)=2$ and the set of unit vertices of the diagram is nonempty, the only possible case is
$$
\begin{equation*}
\alpha_i(h)=1, \quad \alpha_j(h)=1, \quad\alpha_k(h)=0, \quad k=1,\dots, (n-1), \quad k\neq i,j, \quad i<j.
\end{equation*}
\notag
$$
In this case $h$ is a diagonal matrix of size $n\times n$ of the following form (in what follows, we denote the $i\times i$ unity matrix by $E^{(i)}$):
$$
\begin{equation*}
h=\begin{pmatrix} cE^{(i)} & 0 & 0 \\ 0 & (c-1)E^{(j-i)} & 0 \\ 0 & 0 & (c-2)E^{(n-j)} \end{pmatrix}.
\end{equation*}
\notag
$$
The numerical factor $c\in\mathbb{C}$ in the above formula is uniquely determined from the condition $\operatorname{tr}(h)=0$. The algebra $\mathfrak{g}$ can be represented schematically as
$$
\begin{equation*}
\mathfrak{g}=\begin{pmatrix} \mathfrak{g}^{0}_{(1)} & \mathfrak{g}^{1}_{(1)} & \mathfrak{g}^{2} \\ \mathfrak{g}^{-1}_{(1)} & \mathfrak{g}^{0}_{(2)} & \mathfrak{g}^{1}_{(2)} \\ \mathfrak{g}^{-2} & \mathfrak{g}^{-1}_{(2)} & \mathfrak{g}^{0}_{(3)} \end{pmatrix}.
\end{equation*}
\notag
$$
This notation means that the algebra $\mathfrak{g}$ is represented as a direct sum of the subspaces $\mathfrak{g}^{k}$, $k\neq 0$, which are the spaces of matrices in $\mathfrak{g}$ that can contain nonzero elements only at positions in the blocks $\mathfrak{g}^k_{(m)}$, $m=1, 2$, in the above expression, and the subspace $\mathfrak{g}^0$ of block-diagonal matrices in $\mathfrak{sl}_n$ consisting of blocks $\mathfrak{g}^0_{(m)}$ for $m=1, 2, 3$. Here the block $\mathfrak{g}^{0}_{(1)}$ has size $i\times i$, the block $\mathfrak{g}^{0}_{(2)}$ has size $(j-i)\times(j-i)$, and $\mathfrak{g}^{0}_{(3)}$ has size $(n-j)\times(n-j)$. The superscript on every block indicates the eigenspace to which this block belongs. Subscripts distinguish different blocks belonging to the same eigenspace. Let us draw the diagram of this case:  Note that the semisimple part of $\mathfrak{g}^0$ coincides with $\mathfrak{sl}_i\oplus\mathfrak{sl}_{j-i}\oplus\mathfrak{sl}_{n-j}$, and we also have $\dim\mathfrak{z}^0=2 $, from which we conclude that $\widetilde{G}^{0}=\mathrm{SL}_i\times \mathrm{SL}_{j-i}\times \mathrm{SL}_{n-j}\times Z$, where $Z$ is the one-dimensional centre. It is also easy to conclude that $\mathfrak{g}^2= \mathbb{C}^i\otimes(\mathbb{C}^{n-j})^{*}$ as a $\widetilde{G}^0$-module. Since the action of the one-dimensional centre $Z$ on $\mathfrak{g}^2$ is trivial, it is obvious that an invariant on $\mathfrak{g}^2$ for the group $\widetilde{G}^{0}$ exists if and only if $i=n-j$, and it is equal to the determinant of the matrix in $\mathfrak{g}^2$. Let us define the simple Lie-Jordan symplectic structure that corresponds to this structure. It is easy to understand that for the short $\mathrm{SL}_2$-structure in question the elements $e$, $f$ and $h$ are matrices of size $n\times n$ of the following form:
$$
\begin{equation*}
h=\begin{pmatrix} E^{(i)} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E^{(i)} \end{pmatrix}, \qquad e=\begin{pmatrix} 0 & 0 & E^{(i)} \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\quad\text{and} \quad f=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ E^{(i)} & 0 & 0 \end{pmatrix}.
\end{equation*}
\notag
$$
Clearly,
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{z}_{\mathfrak{g}^0}(\mathfrak{sl}_2), \quad \mathfrak{g}^1= e_1\otimes J_1,\quad\text{and}\quad \quad\mathfrak{g}^2=e\otimes J_2,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\mathfrak{z}_{\mathfrak{g}^0}(\mathfrak{sl}_2)=\{\xi\in\mathfrak{g}^{0}\colon [\xi, \eta]= 0 \ \ \forall\,\eta\in\mathfrak{sl}_2\subset\mathfrak{g}_2\}.
\end{equation*}
\notag
$$
We write
$$
\begin{equation*}
\begin{aligned} \, \mathfrak{c}_{i,n} &=\Bigl\{T\in\mathfrak{gl}_n\colon T=\operatorname{diag}\{\underbrace{c_1,\dots,c_1}_{i}, \underbrace{c_2,\dots,c_2}_{n-2i}, \underbrace{c_1,\dots,c_1}_{i}\}, \\ &\qquad 2c_1i+c_2(n-2i)=0,\,c_1,c_2\in\mathbb{C}\Bigr\}. \end{aligned}
\end{equation*}
\notag
$$
It is easy to conclude from the above equations that.
$$
\begin{equation*}
\begin{gathered} \, J_1=\mathfrak{g}^1_{(1)}\oplus\mathfrak{g}^1_{(2)} \cong (\mathbb{C}^{i}\otimes(\mathbb{C}^{n-2i})^{*}) \oplus(\mathbb{C}^{n-2i}\otimes(\mathbb{C}^{i})^{*}), \\ \mathfrak{g}^0_{(1)}\oplus\mathfrak{g}^0_{(2)} \oplus\mathfrak{g}^0_{(3)}\supset\mathfrak{g}_0= \{(A, B, A),\text{ where } A\in\mathfrak{gl}_{i}, B\in\mathfrak{gl}_{n-2i}, \,2\operatorname{tr}A+ \operatorname{tr}B=0\}, \\ \mathfrak{g}_0\cong\mathfrak{sl}_i\oplus\mathfrak{sl}_{n-2i} \oplus\mathfrak{c}_{i,n} \end{gathered}
\end{equation*}
\notag
$$
and We define the inner product on $\mathfrak{g}=\mathfrak{sl}_n$ by the formula
$$
\begin{equation*}
(A, B)=(n-2i)\operatorname{tr}(AB) \quad\forall\, A,B\in\mathfrak{sl}_n.
\end{equation*}
\notag
$$
The normalization factor $n-2i$ is chosen in such a way that, as a result of calculations, the inner product on the Jordan algebra $J_2\cong\mathfrak{gl}_i$ is equal to the trace of the product of operators on the space $J_1$. Let us show that this is indeed the case. The actions of the Lie algebra $\mathfrak{g}_0$ and the Jordan algebra $J_2=\mathfrak{gl}_i$ on the space $J_1$ are carried out according to the following formulae:
$$
\begin{equation*}
C(a, b)=(Ca, bC) \quad\forall\, C\in\mathfrak{gl}_i, \quad (a, b)\in J_1,
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(D_1, D_2) (a, b)=(D_1a-aD_2, D_2b-bD_1) \quad\forall\, (D_1, D_2)\in\mathfrak{g}_0, \quad (a, b)\in J_1.
\end{equation*}
\notag
$$
Hence the trace of the product of operators from $J_2$ has the form
$$
\begin{equation*}
\operatorname{tr}(\mathcal{A}_1\mathcal{B}_1)=2(n-2i)\operatorname{tr}(AB) \quad\forall\, A,B\in\mathfrak{gl}_i,
\end{equation*}
\notag
$$
where $\mathcal{C}_1$ denotes the linear operator on $J_1$ corresponding to the matrix $C\in\mathfrak{gl}_i$. Considering an arbitrary element $\widetilde{e}\otimes A\in\mathfrak{g}^2=\widetilde{e}\otimes J_2$ and taking its inner product with an arbitrary element $\widetilde{f}\otimes B\in\mathfrak{g}^{-2}= \widetilde{f}\otimes J_2$ we see that
$$
\begin{equation*}
(A, B)_2=2(n-2i)\operatorname{tr}(AB) \quad\forall\, A, B\in\mathfrak{gl}_i,
\end{equation*}
\notag
$$
consistent with the above agreement. For this choice of the normalization factor of the inner product on $\mathfrak{g}=\mathfrak{sl}_n$, the symplectic structure on the space $J_1$ is specified using the skew-inner product of the following form: for all $ a_1, a_2\in\mathbb{C}^{i}\otimes(\mathbb{C}^{n-2i})^{*}$ and $b_1,b_2\in\mathbb{C}^{n-2i}\otimes(\mathbb{C}^{i})^{*}$,
$$
\begin{equation*}
((a_1, b_1),(a_2, b_2))=(n-2i)\operatorname{tr}(b_1a_2-a_1b_2).
\end{equation*}
\notag
$$
The operation of multiplication on the Jordan algebra $J_2$ looks standard, that is,
$$
\begin{equation*}
A\circ B=\frac{1}{2}(AB+BA) \quad\forall\, A, B\in\mathfrak{gl}_i.
\end{equation*}
\notag
$$
The action of the Lie algebra $\mathfrak{g}_0$ on $J_2$ is carried out according to the formula
$$
\begin{equation*}
(D_1, D_2)(A)=[D_1, A] \quad\forall\, (D_1, D_2)\in\mathfrak{g}_0,\quad A\in J_2.
\end{equation*}
\notag
$$
This readily implies that
$$
\begin{equation*}
\mathfrak{i}_0=\mathfrak{sl}_{n-2i}\oplus\mathfrak{c}_{i,n},\qquad [J_2, J_2]=\mathfrak{sl}_i.
\end{equation*}
\notag
$$
In this case the maps $\varphi$ and $\delta$ have the following form: for all $ a_1, a_2\in\mathbb{C}^{i}\otimes(\mathbb{C}^{n-2i})^{*}$ and $b_1,b_2\in\mathbb{C}^{n-2i}\otimes(\mathbb{C}^{i})^{*}$
$$
\begin{equation*}
\varphi((a_1, b_1), (a_2, b_2))=\frac{1}{2}(a_2b_1-a_1b_2)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\delta((a_1, b_1), (a_2, b_2))=\biggl(-\frac{1}{2}(a_1b_2+a_2b_1), b_1a_2+ b_2a_1\biggr).
\end{equation*}
\notag
$$
Here
$$
\begin{equation*}
\delta_0((a_1, b_1), (a_2, b_2)) =\biggl(-\frac{1}{2i}\operatorname{tr}(a_1b_2+ a_2b_1)E^{(i)}, b_1a_2+b_2a_1\biggr).
\end{equation*}
\notag
$$
In the analysis of other cases we do not describe all actions and maps arising in one short $\mathrm{SL}_2$-structure or another in such detail. If necessary, all this can be done by direct calculations. The case $\mathfrak{g}=\mathfrak{so}_{2n+1}$. We consider the algebra $\mathfrak{g}= \mathfrak{so}_{2n+1}$ in a basis in which its elements are matrices of size $(2n+1)\times(2n+1)$ which are skew-symmetric with respect to the secondary diagonal. Then
$$
\begin{equation*}
\alpha_{\operatorname{\max}}=\alpha_1+2\alpha_2+\dots +2\alpha_n.
\end{equation*}
\notag
$$
Since the set of unit vertices in the diagram of a short $\mathrm{SL}_2$-structure is nonempty, the only possible case for this Lie algebra is as follows: Then $h$ is the following matrix of size $(2n+1)\times(2n+1)$:
$$
\begin{equation*}
h=\begin{pmatrix} E^{(i)}&0&0\\ 0&0&0\\ 0&0&-E^{(i)} \end{pmatrix}.
\end{equation*}
\notag
$$
Similarly to $\mathfrak{g}=\mathfrak{sl}_n$, the algebra $\mathfrak{g}=\mathfrak{so}_{2n+1}$ can schematically be represented as
$$
\begin{equation*}
\mathfrak{g}=\begin{pmatrix} \mathfrak{g}^{0}_{(1)} & \mathfrak{g}^{1} & \mathfrak{g}^{2} \\ \mathfrak{g}^{-1} & \mathfrak{g}^{0}_{(2)} & -(\mathfrak{g}^{1})^{\mathsf{s}} \\ \mathfrak{g}^{-2} & -(\mathfrak{g}^{-1})^{\mathsf{s}} & -(\mathfrak{g}^{0}_{(1)})^{\mathsf{s}} \end{pmatrix}.
\end{equation*}
\notag
$$
Here the block $\mathfrak{g}^{0}_{(1)}$ is of size $i\times i$, $\mathfrak{g}^{0}_{(2)}$ is of size $(2(n-i)+1)\times(2(n-i)+1)$, and the other blocks are of appropriate size. The only difference from the previous case is that now the terms of the decomposition of $\mathfrak{g}$ into a direct sum in accordance with the above formula are: the space of block-diagonal matrices consisting of blocks $\mathfrak{g}^0_{(m)}$, the space of matrices containing nonzero entries only in the block $\mathfrak{g}^{1}$ and the block symmetric to it with respect to the secondary diagonal, which contains the matrix antisymmetric to the one in $\mathfrak{g}^1$ with respect to the secondary diagonal, an analogous summand for the block $\mathfrak{g}^{-1}$, and also two spaces of matrices whose nonzero entries lie in the blocks $\mathfrak{g}^{2}$ and $\mathfrak{g}^{-2}$. This clarifies immediately that $\mathfrak{g}^2$ is the space of matrices of size $i\times i$ that are skew-symmetric with respect to the secondary diagonal; that is, we can conventionally write $\mathfrak{g}^2=\bigwedge^{2}\mathbb{C}^{i}$. Let us now find $\widetilde{G}^{0}$. Consider the diagram  On the one hand it shows us that $\dim\mathfrak{z}^0=1$, and so $\mathfrak{z}^0=\langle h\rangle$ and, on the other hand, that the semisimple part of $\mathfrak{g}^0$ is equal to $\mathfrak{sl}_i\oplus\mathfrak{so}_{2(n-i)+1}$. Hence we conclude that $\widetilde{G}^{0}= \mathrm{SL}_{i}\times \mathrm{SO}_{2(n-i)+1}$, and only the first factor acts on $\mathfrak{g}^{2}$. The invariant of this action is the determinant of matrices in $\mathfrak{g}^{2}$, which is not equal to 0 if and only if $i$ is even. Thus, a short $\mathrm{SL}_2$-structure is only possible for even $i$. Let us define the simple Lie-Jordan symplectic structure corresponding to the $\mathrm{SL}_2$-structure in question. We define the inner product on $\mathfrak{g}=\mathfrak{so}_{2n+1}$ by using the formula
$$
\begin{equation*}
(A, B)=\frac{1}{2}(2(n-i)+1)\operatorname{tr}(AB) \quad\forall\, A,B\in\mathfrak{so}_{2n+1}.
\end{equation*}
\notag
$$
Since $i=2k$, the $\mathfrak{sl}_2$-triple $e$, $f$, $h$ for this structure is given by the matrices of size $(2n+1)\times (2n+1)$ of the following form:
$$
\begin{equation*}
h=\begin{pmatrix} E^{(i)} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E^{(i)} \end{pmatrix}, \qquad e=\begin{pmatrix} 0 & 0 & I_{i} \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}, \qquad f=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ I_{i} & 0 & 0 \end{pmatrix},
\end{equation*}
\notag
$$
where $I_{i}$ is the following matrix of size $i\times i$:
$$
\begin{equation*}
I_{i}=\begin{pmatrix} E^{(k)} & 0\\ 0 & -E^{(k)} \end{pmatrix}.
\end{equation*}
\notag
$$
Proceeding similarly to $\mathfrak{g}=\mathfrak{sl}_n$ we can readily see that, in this case,
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sp}_i\oplus\mathfrak{so}_{2(n-i)+1}, \quad J_1= \mathbb{C}^{i}\otimes(\mathbb{C}^{2(n-i)+1})^{*}\quad\text{and} \quad J_2=\mathfrak{sym}^{-}_i, \qquad i=2k.
\end{equation*}
\notag
$$
Here $\mathfrak{sym}^{-}_i$ for $i=2k$ denotes the Jordan algebra of matrices of order $i$ that are symmetric with respect to a nondegenerate bilinear form. We set
$$
\begin{equation*}
a=\begin{pmatrix} a_1\\ a_2 \end{pmatrix}\in J_1\quad\text{and} \quad b=\begin{pmatrix} b_1\\ b_2 \end{pmatrix}\in J_1,
\end{equation*}
\notag
$$
where $a_i$ and $b_i$ are arbitrary matrices of the size $k\times (2(n-i)+1)$. The symplectic structure on the space $J_1$ is given by the skew-inner product of the form
$$
\begin{equation*}
(a,b)=(2(n-i)+1)\operatorname{tr}(a_1b_2^{\mathsf{s}}-a_2b_1^{\mathsf{s}}).
\end{equation*}
\notag
$$
The operation of multiplication on the Jordan algebra $J_2$ looks standard, that is,
$$
\begin{equation*}
A\circ B=\frac{1}{2}(AB+BA) \quad\forall\, A, B\in\mathfrak{sym}^{-}_i.
\end{equation*}
\notag
$$
The action of the Jordan algebra $J_2=\mathfrak{sym}^{-}_i$ on $J_1$ is carried out by left multiplication of matrices in $J_1$ by the corresponding matrix in $\mathfrak{g}^2$. The action of the Lie algebra $\mathfrak{g}_0$ on $J_1$ is carried out according to the formula
$$
\begin{equation*}
(D_1, D_2)a=D_1a-aD_2 \quad\forall\, D_1\in\mathfrak{sp}_i, \quad D_2\in\mathfrak{so}_{2(n-i)+1}, \quad a\in J_1.
\end{equation*}
\notag
$$
The case $\mathfrak{g}=\mathfrak{so}_{2n}$. Consider the algebra $\mathfrak{g}= \mathfrak{so}_{2n}$ in a basis in which its elements are represented by matrices of size $2n\times 2n$ that are skew-symmetric with respect to the secondary diagonal. We define the inner product on $\mathfrak{so}_{2n}$ similarly to the previous case. For this algebra the dependence of the highest root on the simple roots is as follows:
$$
\begin{equation*}
\alpha_{\max}=\alpha_1+2\alpha_2+\dots +2\alpha_{n-2}+ \alpha_{n-1}+\alpha_{n}.
\end{equation*}
\notag
$$
Generally speaking, four cases are possible here: 1) $\alpha_i(h)=1$ and $\alpha_j(h)=0$ for $j=1,\dots, n$ and $i=2,\dots,(n-2)$, $j\neq i$; 2) $\alpha_n(h)=\alpha_{n-1}(h)=1$ and $\alpha_i(h)=0$ for $i=1,\dots,n$, $i\neq n, n-1$; 3) $\alpha_1(h)=\alpha_n(h)=1$ and $\alpha_i(h)=0$ for $i=1,\dots,n$, $i\neq 1, n$; 4) $\alpha_1(h)=\alpha_{n-1}(h)=1$ and $\alpha_i(h)=0$ for $i=1,\dots,n$, $i\neq 1, n-1$. Cases 3) and 4) are transformed into each other by an automorphism of the Dynkin diagram, and therefore it suffices to consider only one of them. Thus, it remains to consider cases 1), 2) and 3). In case 1) the matrix $h$ has the same form as in the case $\mathfrak{g}=\mathfrak{so}_{2n+1}$, with the only difference that now the central zero matrix has size $2(n-i)\times2(n-i)$. Therefore, the answer here is the same: the invariant (equal to the determinant) exists if and only if $i$ is even (here $\mathfrak{g}^2=\bigwedge^{2}\mathbb{C}^i$ and $ \widetilde{G}^0=\mathrm{SL}_i\times \mathrm{SO}_{2(n-i)}$). The diagram for this case looks as follows:  Here the $\mathfrak{sl}_2$-triple $e$, $f$, $h$ is formed by the matrices of size $2n\times 2n$ analogous to the case $\mathfrak{g}=\mathfrak{so}_{2n+1}$. Therefore, proceeding in the same way as for $\mathfrak{so}_{2n+1}$, we obtain
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sp}_i\oplus\mathfrak{so}_{2(n-i)}, \quad J_1= \mathbb{C}^{i}\otimes(\mathbb{C}^{2(n-i)})^{*}\quad\text{and} \quad J_2=\mathfrak{sym}^{-}_i, \qquad i=2k\leqslant n-2.
\end{equation*}
\notag
$$
The skew-symmetric inner product on $J_1$, as well as the action of the Lie algebra $\mathfrak{g}_0$ and the Jordan algebra $J_2$ on $J_1$ are carried out in this case according to formulae similar to the case $\mathfrak{g}=\mathfrak{so}_{2n+1}$. Consider case 2). Then
$$
\begin{equation*}
h=\operatorname{diag}\{1,\dots ,1,0,0,-1,\dots ,-1\}.
\end{equation*}
\notag
$$
The algebra $\mathfrak{g}=\mathfrak{so}_{2n}$, similarly to the previous case, can be represented conditionally in the form
$$
\begin{equation*}
\mathfrak{g}=\begin{pmatrix} \mathfrak{g}^{0}_{(1)} & \mathfrak{g}^{1}_{(1)} & \mathfrak{g}^{2} \\ \mathfrak{g}^{-1}_{(1)} & \mathfrak{g}^{0}_{(2)} & -(\mathfrak{g}^{1}_{(1)})^{\mathsf{s}} \\ \mathfrak{g}^{-2} & -(\mathfrak{g}^{-1}_{(1)})^{\mathsf{s}} & -(\mathfrak{g}^{0}_{(1)})^{\mathsf{s}} \end{pmatrix}.
\end{equation*}
\notag
$$
Here the block $\mathfrak{g}^{0}_{(1)}$ is of size $(n-1)\times (n-1)$, the block $\mathfrak{g}^{0}_{(2)}$ is of size $2\times 2$, and the remaining blocks are of appropriate sizes. It is clear from the above formula that $\mathfrak{g}^2=\bigwedge^2\mathbb{C}^{n-1}$. Consider the diagram of this representation:  We derive from this diagram that the semisimple part of $\mathfrak{g}^0$ is equal to $\mathfrak{sl}_{n-1}$ and $\dim\mathfrak{z}^0= 2$. It can readily be seen that the vector orthogonal to $h$ in $\mathfrak{z}^0$ corresponds to the block $\mathfrak{g}^{0}_{(2)}$ and, in view of the position of the blocks $\mathfrak{g}^{0}_{(2)}$ and $\mathfrak{g}^2$, it acts trivially on $\mathfrak{g}^2$. Hence we can assume that $\widetilde{G}^{0}=\mathrm{SL}_{n-1}$. Thus, it is obvious that this representation has an invariant (equal to the determinant of a matrix in $\mathfrak{g}^2$) if and only if $n$ is odd. That is, in this case there is a short $\mathrm{SL}_2$-structure for odd $n$. Here the matrices $e$, $f$, $h$ look similarly to the previous case. Therefore,
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sp}_{n-1}\oplus\mathfrak{so}_{2}, \quad J_1= \mathbb{C}^{n-1}\otimes(\mathbb{C}^{2})^{*}\quad\text{and}\quad \quad J_2=\mathfrak{sym}^{-}_{n-1}, \qquad n=2k+1.
\end{equation*}
\notag
$$
The skew-symmetric inner product in $J_1$, as well as the action of the Lie algebra $\mathfrak{g}_0$ and the Jordan algebra $J_2$ on $J_1$ are expressed in this case by formulae similar to ${\mathfrak{g}=\mathfrak{so}_{2n+1}}$. We go over to case 3), the last one. In this case $h$ has the form
$$
\begin{equation*}
h= \operatorname{diag}\biggl\{\frac{3}{2},\frac{1}{2},\dots, \frac{1}{2},-\frac{1}{2},\dots,-\frac{1}{2},-\frac{3}{2}\biggr\}.
\end{equation*}
\notag
$$
Similarly to the previous cases, the algebra $\mathfrak{g}=\mathfrak{so}_{2n}$ can schematically be represented in the form
$$
\begin{equation*}
\mathfrak{g}=\begin{pmatrix} \mathfrak{g}^{0}_{(1)} & \mathfrak{g}^{1}_{(1)} & \mathfrak{g}^{2} & 0 \\ \mathfrak{g}^{-1}_{(1)} & \mathfrak{g}^{0}_{(2)} & \mathfrak{g}^{1}_{(2)} & -(\mathfrak{g}^{2})^{\mathsf{s}} \\ \mathfrak{g}^{-2} & \mathfrak{g}^{-1}_{(2)} & -(\mathfrak{g}^{0}_{(2)})^{\mathsf{s}} & -(\mathfrak{g}^{1}_{(1)})^{\mathsf{s}} \\ 0 & -(\mathfrak{g}^{-2})^{\mathsf{s}} & -(\mathfrak{g}^{-1}_{(1)})^{\mathsf{s}} &-(\mathfrak{g}^{0}_{(1)})^{\mathsf{s}} \end{pmatrix}.
\end{equation*}
\notag
$$
Here $\mathfrak{g}^{0}_{(2)}$, $\mathfrak{g}^{1}_{(2)}$, and $\mathfrak{g}^{-1}_{(2)}$ are blocks of size $(n-1)\times(n-1)$, $\mathfrak{g}^{1}_{(1)}$ and $\mathfrak{g}^{2}$ are rows of length $n-1$, and $\mathfrak{g}^{-2}$ and $\mathfrak{g}^{-1}_{(1)}$ are columns of height $n-1$. Hence $\mathfrak{g}^2=\mathbb{C}^{n-1}$. Consider the diagram of this case:  It is clear from this diagram that $\dim\mathfrak{z}^0=2$, and the semisimple part of $\mathfrak{g}^0$ is equal to $\mathfrak{sl}_{n-1}$. Then $\widetilde{G}^0=\mathrm{GL}_{n-1}$, and we obtain a tautological representation of $\mathrm{GL}_{n-1}$ on the $(n-1)$-dimensional vector space, which obviously has no nontrivial invariants, that is, there is no short $\mathrm{SL}_2$-structure in this case. The case $\mathfrak{g}=\mathfrak{sp}_{2n}$. Consider the algebra $\mathfrak{sp}_{2n}$ in a basis in which elements of the algebra are matrices of the form
$$
\begin{equation*}
\begin{pmatrix} X&Y\\ Z&-X^{\mathsf{s}} \end{pmatrix}\in\mathfrak{gl}_{2n}, \qquad X, Y, Z\in\mathfrak{gl}_n, \quad Y^{\mathsf{s}}=Y, \quad Z^{\mathsf{s}}= Z.
\end{equation*}
\notag
$$
We define the inner product on $\mathfrak{sp}_{2n}$ similarly to the previous case. For this algebra the highest root has the form
$$
\begin{equation*}
\alpha_{\max}=2\alpha_1+\dots + 2\alpha_{n-1}+\alpha_{n}.
\end{equation*}
\notag
$$
Thus, the only possible case here is as follows:
$$
\begin{equation*}
\alpha_i(h)=1\quad\text{and} \quad \alpha_j(h)=0\quad\text{for }\ i=1,\dots,(n-1), \ \ j=1,\dots,n, \ \ j\neq i.
\end{equation*}
\notag
$$
The elements $e$, $f$ and $h$ are matrices of size $2n\times 2n$ similar to those in the case $\mathfrak{g}=\mathfrak{sl}_n$. Therefore, when taking commutators, the subspace $\mathfrak{g}^2$ corresponds to the upper right corner block of size $i\times i$; so it is clear that $\mathfrak{g}^2$ is the space of matrices of size $i\times i$ symmetric with respect to the secondary diagonal. Consider the diagram of this case:  We see from this diagram that $\dim\mathfrak{z}^0=1$, and the semisimple part of $\mathfrak{g}^0$ coincides with $\mathfrak{sl}_i\oplus\mathfrak{sp}_{2(n-i)}$. Hence $\widetilde{G}^0=\mathrm{SL}_i\times \mathrm{Sp}_{2(n-i)}$. Then, since $\mathrm{Sp}_{2(n-i)}$ acts trivially on $\mathfrak{g}^2$, we can assume that $\widetilde{G}^0=\mathrm{SL}_i$. Now it is clear that in this case an invariant of the action is the determinant of the matrix in $\mathfrak{g}^2$, that is, a short $\mathrm{SL}_2$-structure exists for any $i=1,\dots,(n-1)$. In this case
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{so}_{i}\oplus \mathfrak{sp}_{2(n-i)}, \quad J_1= \mathbb{C}^{i}\otimes(\mathbb{C}^{2(n-i)})^{*}, \quad J_2=\mathfrak{sym}^{+}_{i}, \qquad i<n.
\end{equation*}
\notag
$$
Here $\mathfrak{sym}^{+}_i$ denotes the Jordan algebra of matrices of order $i$ that are symmetric with respect to a nondegenerate symmetric bilinear form. The skew-symmetric inner product on $J_1$, as well as the action of the Lie algebra $\mathfrak{g}_0$ and the Jordan algebra $J_2$ on $J_1$ are expressed in this case according to formulae similar to those in the case $\mathfrak{g}=\mathfrak{so}_{2n+1}$. 5.3. Classification of short $\mathrm{SL}_2$-structures on special simple Lie algebrasAlthough each special simple Lie algebra has its own matrix model, all these models are quite cumbersome, and so the study of short $\mathrm{SL}_2$-structures using the method described above is too complicated. Consider the irreducible representation of the Lie algebra $\widetilde{\mathfrak{g}}^{0}$ on $\mathfrak{g}^2$. It is uniquely determined by its highest weight, namely, by its decomposition with respect to the fundamental weights. The highest weight of this representation obviously coincides with the highest root of $\mathfrak{g}$. Its decomposition with respect to the fundamental weights is obtained by calculating the Cartan products between the highest root of $\mathfrak{g}$ and the simple roots of $\mathfrak{g}$ corresponding to zero vertices of the diagram of the short $\mathrm{SL}_2$-structure. Therefore, using only the extended Dynkin diagram, we can determine both the group $\widetilde{G}^0$ itself and the representation space $\mathfrak{g}^2$. For the considerations below we need the following notation. We denote by $S^k$ the space of the spinor representation of the algebra $\mathfrak{so}_k$ and by $S^k_{1/2}$ and $S^k_{-1/2}$ the spaces of the semispinor representations of the corresponding algebra for even $k$. When we restrict the algebra $\mathfrak{so}_{2l}$ to the subalgebra $\mathfrak{so}_{2l-1}$, the representation $S_{1/2}^{2l}$ is transformed into $S^{2l-1}$ and, when we restrict $\mathfrak{so}_{2l+1}$ to the subalgebra $\mathfrak{so}_{2l}$, the space $S^{2l+1}$ is transformed into $S^{2l}=S_{1/2}^{2l}\oplus S_{-1/2}^{2l}$. As will follow from the classification below, two different cases are possible for the symplectic structure of the space $J_1$ resulting from a short $\mathrm{SL}_2$-structure on a special Lie algebra. 1. If the representation of the Lie algebra $\mathfrak{g}_0$ on the space $J_1$ is irreducible, then, according to Schur’s lemma, there is a unique (up to a scalar factor) $\mathfrak{g}_0$-invariant nondegenerate skew-symmetric form. This means that the symplectic structure of the space $J_1$ in this case is completely determined by the representation of $\mathfrak{g}_0$ in $J_1$. We call this the irreducible case of the specification of a symplectic structure on $J_1$. 2. If the representation $\mathfrak{g}_0 \colon J_1$ is reducible, then $J_1$ decomposes into a direct sum of mutually dual $\mathfrak{g}_0$-invariant Lagrangian subspaces, and the inner product of elements of these subspaces is defined as pairing. We call this the Lagrangian case of the specification of a symplectic structure on $J_1$. We write $\alpha_0=-\alpha_{\max}$. We mark the lowest root $\alpha_0$ of the Lie algebra $\mathfrak{g}$ on the diagram of a short $\mathrm{SL}_2$-structure by a double circle, to distinguish it from zero vertices. The case $\mathfrak{g}=G_2$. As noted above, for this algebra there is a unique possible case, when $\alpha_2(h)=1$ and $\alpha_1(h)=0$. Consider the extended Dynkin diagram:  It is clear that $\widetilde{G}^0=\mathrm{SL}_2$. Since $\langle\alpha_{\max}|\alpha_1\rangle=0$, it follows that $\dim\mathfrak{g}^2=1$, and the action $\widetilde{G}^0\colon \mathfrak{g}^2$ is trivial; therefore, all polynomials on $\mathfrak{g}^2$ are invariant, that is, a short $\mathrm{SL}_2$-structure exists in this case. Since $\dim\mathfrak{g}^2=1$, it follows that $\widetilde{\mathfrak{g}}^0=\mathfrak{g}_0=\mathfrak{sl}_2$. It is clear from the diagram that the representation of the algebra $\mathfrak{g}_0=\mathfrak{sl}_2$ on the vector space $\mathfrak{g}^1$ is irreducible, and the lowest weight of this representation is $\alpha_2$. This readily implies that
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sl}_2, \quad J_1=\operatorname{Sym}^3\mathbb{C}^2\quad\text{and} \quad J_2=\mathbb{C}.
\end{equation*}
\notag
$$
Here the irreducible case of the specification of the symplectic structure of $J_1$ is realized. The case $\mathfrak{g}=F_4$. In this algebra
$$
\begin{equation*}
\alpha_{\max}=2\alpha_1+ 4\alpha_2+3\alpha_3+2\alpha_4;
\end{equation*}
\notag
$$
therefore, only two cases are possible: 1) $\alpha_1(h)=1$ and $\alpha_2(h)=\alpha_3(h)=\alpha_4(h)=0$; 2) $\alpha_4(h)=1$ and $\alpha_1(h)=\alpha_2(h)=\alpha_3(h)=0$. In case 1) the Dynkin diagram looks as follows:  Here $\dim\mathfrak{z}^0=1$ and $\widetilde{G}^0=\operatorname{Spin}_7$. It is clear from the diagram that $\langle\alpha_{\max}|\alpha_4\rangle=1$ and $\langle\alpha_{\max}|\alpha_2\rangle= \langle\alpha_{\max}|\alpha_3\rangle=0$, and therefore this representation of $\widetilde{G}^0$ is a vector representation of the spinor group, that is, $\mathfrak{g}^2= \mathbb{C}^{7}$. The inner square of the vector is clearly the invariant of the action of this group, so in this case a short $\mathrm{SL}_2$-structure exists. Moreover, the Lie algebra $\widetilde{\mathfrak{g}}^0$ is isomorphic to $\mathfrak{so}_7$. Also, according to the diagram, it is clear that the representation of $\widetilde{\mathfrak{g}}^0$ on the space $\mathfrak{g}_1$ is irreducible and the lowest weight of this representation is $\alpha_1$, from which we conclude that this is the spinor representation of the algebra $\mathfrak{so}_7$, that is, $J_1=S^7$. It follows from the considerations in the previous subsections that
$$
\begin{equation*}
\mathfrak{g}^0=\widetilde{\mathfrak{g}}^0\oplus\langle\widetilde{h} \otimes\mathbb{I}\rangle= \mathfrak{g}_0\oplus(\widetilde{h}\otimes J_2)\quad\text{and} \quad \mathfrak{g}^2=\widetilde{e}\otimes J_2,
\end{equation*}
\notag
$$
where $\mathbb{I}$ denotes the unity of the Jordan algebra $J_2$, and $\widetilde{e}, \widetilde{f}$ and $ \widetilde{h}$ are basis elements of the abstract Lie algebra $\mathfrak{sl}_2$ satisfying relations (2.17). Consider the action $\widetilde{\mathfrak{g}}^0\colon \mathfrak{g}^2$. It is clear that
$$
\begin{equation*}
\mathfrak{g}_0=\{D\in\widetilde{\mathfrak{g}}^0\colon [D,\widetilde{e}\otimes\mathbb{I}]=0\}
\end{equation*}
\notag
$$
is the stabilizer of the element $\widetilde{e}\otimes\mathbb{I}\in\mathfrak{g}^2$ under the action of $\widetilde{\mathfrak{g}}^0$. Since the Lie algebra $\mathfrak{g}_0$ is reductive, the vector $\widetilde{e}\otimes\mathbb{I}$ is nonisotropic, from which it is easy to derive that $\mathfrak{g}_0=\mathfrak{so}_6$. When $\widetilde{\mathfrak{g}}^0=\mathfrak{so}_7$ is restricted to the Lie subalgebra $\mathfrak{g}_0=\mathfrak{so}_6$, we obtain the space $S^6$ in place of $S^7$, and thus $J_1=S^6=S^6_{1/2}\oplus S^6_{-1/2}$. The symplectic structure of $J_1$ corresponds to the Lagrangian case. Similarly, when we restrict the Lie algebra $\widetilde{\mathfrak{g}}^0=\mathfrak{so}_7$ to the Lie subalgebra ${\mathfrak{g}_0=\mathfrak{so}_6}$, the space $J_2=\mathbb{C}^7$ decomposes into the direct sum of two subspaces orthogonal to each other: the one-dimensional space spanned by the unity $\mathbb{I}$ of the algebra $J_2$ on which $\mathfrak{g}_0$ acts trivially, and a six-dimensional space on which $\mathfrak{g}_0$ acts tautologically. Thus, $J_2=\mathbb{C}^6\oplus\mathbb{C}$. We define the Jordan operation on $J_2$. We note that it suffices to define Jordan multiplication between elements of $\mathbb{C}^6$, since its complementary one-dimensional subspace in $J_2$ is spanned by the unity of $J_2$. It is clear that the Jordan operation on $\mathbb{C}^6$ can be interpreted as a symmetric $\mathfrak{g}_0$-equivariant bilinear map whose image lies in $J_2$. Then this map can be factored through
$$
\begin{equation*}
\operatorname{Sym}^2(\mathbb{C}^6)=\operatorname{Sym}^2_0(\mathbb{C}^6)\oplus\mathbb{C},
\end{equation*}
\notag
$$
where the first term is an irreducible representation of the algebra $\mathfrak{so}_6$ corresponding to double the first fundamental weight. Then it is obvious that the result of the application of the Jordan operation to vectors in $\mathbb{C}^6$ belongs to the one-dimensional subspace of $J_2$ spanned by the unity of $J_2$. Thus,
$$
\begin{equation}
A\circ B=\lambda_{A, B}\mathbb{I}, \qquad A, B\in\mathbb{C}^6\subset J_2.
\end{equation}
\tag{5.1}
$$
Here the coefficient $\lambda_{A, B}\in\mathbb{C}$ is uniquely determined by the pair $A$, $B$. Calculating the traces of the left- and right-hand sides of (5.1) we can readily conclude that
$$
\begin{equation*}
\lambda_{A, B}=\frac{(A, B)}{\operatorname{dim} J_1}.
\end{equation*}
\notag
$$
The operation of Jordan multiplication on $J_2$ is defined by (5.1). In all other cases of short $\mathrm{SL}_2$-structures where $J_2\neq\mathbb{C}$, the Jordan structure is defined similarly. Therefore, we do not dwell on it in such detail any further. Summing up, in this case we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{so}_6, \qquad J_1=S^6_{1/2}\oplus S^6_{-1/2}\quad\text{and} \quad J_2= \mathbb{C}^6\oplus\mathbb{C}.
\end{equation*}
\notag
$$
In case 2) the Dynkin diagram has the form  In this case it is obvious that $\widetilde{G}^0=\mathrm{Sp}_6$, and $\dim\mathfrak{g}^2=1$, since all Cartan numbers between the highest root and the simple roots corresponding to the zero vertices are equal to 0. Similarly to the previous reasoning, we see that in this case there exists a short $\mathrm{SL}_2$-structure and we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sp}_6 \qquad J_1=\bigwedge_0^3(\mathbb{C}^6)\quad\text{and} \quad J_2= \mathbb{C}.
\end{equation*}
\notag
$$
Here $\bigwedge_0^3(\mathbb{C}^6)$ denotes the space of skew-symmetric tensors of the third order whose complete convolution with the symplectic form vanishes. The symplectic structure on $J_1$ also corresponds to the irreducible case. The case $\mathfrak{g}=E_6$. The highest root of this algebra has the form
$$
\begin{equation*}
\alpha_{\max}=\alpha_1+2\alpha_2+3\alpha_3+2\alpha_4+\alpha_5+2\alpha_6.
\end{equation*}
\notag
$$
Four cases are possible here: 1) $\alpha_2(h)=1$ and $\alpha_1(h)=\alpha_3(h)=\dots =\alpha_6(h)=0$; 2) $\alpha_4(h)=1$ and $\alpha_1(h)=\dots =\alpha_3(h)=\alpha_5(h)=\dots =\alpha_6(h)=0$; 3) $\alpha_6(h)=1$ and $\alpha_1(h)=\dots =\alpha_5(h)=0$; 4) $\alpha_1(h)=\alpha_5(h)=1$ and $\alpha_2(h)=\dots =\alpha_4(h)=0$. Cases 1) and 2) are also equivalent, and thus it is sufficient to look only at case 1). We consider it. The Dynkin diagram here has the form  We see from the diagram that $\dim\mathfrak{z}^0=1$, and therefore $\widetilde{G}^0=\mathrm{SL}_2\times \mathrm{SL}_5$. Since the highest root $\alpha_{\max}$ has a nonzero Cartan number only with an extreme simple root of $\mathfrak{sl}_5$ and this number is equal to 1, it follows that $\mathfrak{g}^2=\mathbb{C}^5$, which means that $\mathrm{SL}_2$ acts trivially on $\mathfrak{g}^2$, and there are no nontrivial invariants in this case so that there is no short $\mathrm{SL}_2$-structure. Let us turn to case 3) and its Dynkin diagram.  Proceeding similarly to the previous cases we see that $\widetilde{G}^0=\mathrm{SL}_6$ and $\dim\mathfrak{g}^2= 1$, which means that there are invariants here, that is, a short $\mathrm{SL}_2$-structure exists and we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sl}_6,\qquad J_1=\bigwedge^3(\mathbb{C}^6)\quad\text{and}\quad J_2= \mathbb{C}.
\end{equation*}
\notag
$$
Here the symplectic structure on $J_1$ corresponds to the irreducible case. In the last case, we have the following Dynkin diagram:  Clearly, the semisimple part of $\mathfrak{g}^0$ is equal to $\mathfrak{so}_8$; however, $\dim\mathfrak{z}^0=2$ in this case, that is, $\widetilde{\mathfrak{g}}^0\,{=}\,\mathfrak{so}_8\,{\oplus}\,\langle h_1\rangle$, where $h_1\in\mathfrak{z}^0$, $h_1\bot h$. Consider the involutive nonidentity automorphism of the Dynkin diagram, which is the symmetry with respect to the central vertical segment. This automorphism is an automorphism of the Lie algebra $\mathfrak{g}$; it preserves the roots $\alpha_3$ and $ \alpha_6$ and transposes the roots $\alpha_1$ and $\alpha_5$, and also $\alpha_2$ and $\alpha_4$. The elements $h$ and $h_1$ obviously define a two-dimensional space orthogonal to the four-dimensional space spanned by the simple roots corresponding to zero vertices. The automorphism under consideration preserves this four-dimensional space, and therefore preserves its orthogonal complement. Since the marks on all vertices are preserved under the action of the automorphism, this automorphism also preserves the element $h$. This means that, by involutivity, the automorphism acts on $h_1$ by multiplication by $-1$, and thus we can assume that $\alpha_1(h_1)=-\alpha_5(h_1)=1$, and all other roots are equal to 0 at $h_1$. This means that $\alpha_{\max}(h_1)=0$, that is, $h_1$ acts trivially on $\mathfrak{g}^2$, and thus we can assume that $\widetilde{G}^0= \operatorname{Spin}_8$. As can be seen from the diagram, we have $\langle\alpha_{\max}|\alpha_6\rangle=1$ and $\langle\alpha_{\max}|\alpha_2\rangle= \langle\alpha_{\max}|\alpha_3\rangle= \langle\alpha_{\max}|\alpha_4\rangle=0$, that is, $\mathfrak{g}^2=\mathbb{C}^8$, which means that a short $\mathrm{SL}_2$-structure exists in this case. In this case the representation of the algebra $\widetilde{\mathfrak{g}}^0$ on $J_1$ is reducible, and Similarly to the previous cases, we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{so}_7, \quad J_1=S^7\oplus S^7\quad\text{and} \quad J_2=\mathbb{C}^7\oplus\mathbb{C}.
\end{equation*}
\notag
$$
The symplectic structure on $J_1$ corresponds to the Lagrangian case. The case $\mathfrak{g}=E_7$. As is known, for this algebra
$$
\begin{equation*}
\alpha_{\max}= \alpha_1+2\alpha_2+3\alpha_3+4\alpha_4+3\alpha_5+2\alpha_6+2\alpha_7,
\end{equation*}
\notag
$$
and therefore three cases are possible: 1) $\alpha_2(h)=1$ and $\alpha_1(h)=\alpha_3(h)=\dots =\alpha_7(h)=0$; 2) $\alpha_6(h)=1$ and $\alpha_1(h)=\dots =\alpha_5(h)=\alpha_7(h)=0$; 3) $\alpha_7(h)=1$ and $\alpha_1(h)=\dots =\alpha_6(h)=0$. In case 1) the Dynkin diagram has the form  Using it we see that $\widetilde{G}^0=\mathrm{SL}_2\times \operatorname{Spin}_{10}$. Since the highest root has a nonzero Cartan number only with the first simple root of the algebra $\mathfrak{so}_{10}$, it follows that $\mathfrak{g}^2=\mathbb{C}^{10}$, which means that this representation has an invariant, and we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{sl}_2\oplus\mathfrak{so}_9, \qquad J_1=\mathbb{C}^2\otimes S^9\quad\text{and} \quad J_2= \mathbb{C}^9\oplus\mathbb{C}.
\end{equation*}
\notag
$$
The symplectic structure on $J_1$ corresponds to the irreducible case. In case 2) the Dynkin diagram has the form  Here $\widetilde{G}^0=\operatorname{Spin}_{12}$ and $\dim\mathfrak{g}^2=1$, so there exists an invariant here and we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{so}_{12}, \qquad J_1=S^{12}_{1/2}\quad\text{and} \quad J_2=\mathbb{C}.
\end{equation*}
\notag
$$
Now the symplectic structure on $J_1$ corresponds to the irreducible case. In case 3) the Dynkin diagram has the form  Here $\widetilde{G}^0=\mathrm{SL}_7$ and $\mathfrak{g}^2=\mathbb{C}^{7}$; therefore, this action has no nontrivial invariants. The case $\mathfrak{g}=E_8$. The highest root of this algebra has the form
$$
\begin{equation*}
\alpha_{\max}=2\alpha_1+3\alpha_2+4\alpha_3+5\alpha_4+6\alpha_5+4\alpha_6+ 2\alpha_7+3\alpha_8;
\end{equation*}
\notag
$$
therefore, there are only two possible cases here: 1) $\alpha_1(h)=1$ and $\alpha_2(h)=\dots =\alpha_8(h)=0$; 2) $\alpha_7(h)=1$ and $\alpha_1(h)=\dots =\alpha_6(h)=\alpha_7(h)=0$. The first case has the following Dynkin diagram:  In this case $\widetilde{G}^0=E_7$ and $\dim\mathfrak{g}^2=1$, that is, there exists an invariant for this representation and we have
$$
\begin{equation*}
\mathfrak{g}_0=E_7, \qquad J_1=V(\pi_1)\quad\text{and} \quad J_2=\mathbb{C}.
\end{equation*}
\notag
$$
Here $V(\pi_1)$ denotes the representation space of the first fundamental weight of the algebra $E_7$. The symplectic structure on $J_1$ corresponds to the irreducible case. The Dynkin diagram of the other case looks as follows:  Similarly to the previous reasoning, we see that $\widetilde{G}^0=\operatorname{Spin}_{14}$, and $\mathfrak{g}^2$ is its vector representation, that is, this representation has an invariant, and we have
$$
\begin{equation*}
\mathfrak{g}_0=\mathfrak{so}_{13}, \qquad J_1=S^{13}\quad\text{and} \quad J_2=\mathbb{C}^{13}\oplus\mathbb{C}.
\end{equation*}
\notag
$$
The symplectic structure of the space $J_1$ corresponds to the irreducible case. Thus, all existing short $\mathrm{SL}_2$-structures can be listed in Table 1. Table 1
We note that the Lie algebra $\mathfrak{g}_0$ is not fully defined by the pair $(J_1, J_2)$. For example, in the case of a simple Lie algebra $\mathfrak{g}=E_8$, for $J_1=V(\pi_1)$ the Jordan algebra $J_2$ is one-dimensional, as in the case of the algebra $\mathfrak{g}=\mathfrak{sl}_n$ for $i=1$. If we choose $n$ so that $\dim V(\pi_1)=\dim J_1 (\mathfrak{sl}_n)$, where $J_1 (\mathfrak{sl}_n)= (\mathbb{C}^{i}\otimes(\mathbb{C}^{n-2i})^{*}) \oplus(\mathbb{C}^{n-2i}\otimes(\mathbb{C}^{i})^{*})$, then we can see that the algebras $\mathfrak{g}_0$ are distinct in these cases, while the pairs $(J_1, J_2)$ coincide. Thus, given a pair $(J_1, J_2)$, it is impossible to construct a simple symplectic Lie-Jordan structure in a unique way. 
Citation in format AMSBIB
\Bibitem{Sta23} |
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