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Symmetric matrices and maximal Nijenhuis pencils A. Yu. Konyaevab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 8, DOI: https://doi.org/10.4213/sm9862 
Bibliographic databases:
    § 1. IntroductionThe Frölicher-Nijenhuis brackets of two tensor fields of type $(1, 1)$ (operator fields) $L$ and $R$ on a manifold $\mathrm M^n$ is by definition the expression
$$
\begin{equation}
\begin{aligned} \, \notag [[L, R]](\xi, \eta) &=L [\xi, R\eta]+R[L\xi, \eta]+R [\xi, L\eta]+L[R\xi, \eta] \\ &\qquad - [L\xi, R\eta]-[R\xi, L\eta]-LR[\xi, \eta]-RL[\xi, \eta]. \end{aligned}
\end{equation}
\tag{1.1}
$$
Here $\xi$ and $\eta$ are arbitrary vector fields and $[\,\cdot\,{,}\,\cdot\,]$ denotes the standard commutator of vector fields. We can show that the expression (1.1) defines a $(1, 2)$ tensor which is skew-symmetric with respect to the lower indices. An operator field $L$ is called a Nijenhuis operator if its Nijenhuis torsion, defined by $\mathcal N_L=\frac{1}{2} [[L, L]]$, is zero. Nijenhuis operators arise in various problems in physics, geometry and algebra (for instance, see [1]–[3]). In the case of finite-dimensional integrable systems they play the role of recursion operators. In infinite-dimensional integrable systems, if a pair of contravariant flat metrics $g$, $\overline g$ defines compatible Hamiltonian operators of Dubrovin-Novikov type, then $L=\overline g g^{-1}$ is a Nijenhuis operator (see [4]). In applications Nijenhuis operators often arise as families. We call a subspace $\mathcal P$ of the infinite-dimensional linear space of $(1, 1)$ tensor fields on a manifold a Nijenhuis pencil if $[[L, R]]=0$ for all $L, R \in \mathcal P$. By definition, the centralizer of a Nijenhuis pencil $\mathcal P$ is the linear space of (not necessarily Nijenhuis!) operator fields $L$ such that $[[L, R]]=0$ for all $R \in \mathcal P$. We denote it by $C (\mathcal P)$. Pencils can be finite- and infinite-dimensional. The above definition is equivalent to the condition that the linear space $\mathcal P$ consists of Nijenhuis operators. The simplest example of an infinite-dimensional pencil associated with a Nijenhuis operator $L$ is the linear space spanned by $\operatorname{Id}, L, L^2, L^3, \dots$ . It is well known that any combination of powers of a Nijenhuis operator with constant coefficients is also a Nijenhuis operator. Any subspace of a Nijenhuis pencil is obviously a Nijenhuis pencil again. Thus, it makes sense to introduce the notion of a maximal Nijenhuis pencil. We say that a Nijenhuis pencil $\mathcal P$ is maximal if it is not a subpencil of a larger pencil. In other words, any Nijenhuis operator $L$ such that $[[L, R]]=0$ for all $R \in \mathcal P$ must lie in $\mathcal P$. It is easy to see that if the Nijenhuis operator $L$ can be reduced to the diagonal form $\operatorname{diag}\{u^1, \dots, u^n\}$, then the pencil spanned by its powers is maximal. The following Nijenhuis pencil arises in the description of compatible Poisson structures of hydrodynamic type. Fix coordinates $u^1, \dots, u^n$ on $\mathrm M^n= \mathbb R^n$ and consider all operator fields $L$ of the form where $a^i_j=a^j_i$ and $K, c^1, \dots, c^n$ are arbitrary constants. Such operator fields turn out to be Nijenhuis operators; they form a pencil. We denote it by $\mathcal P_1$. Note that in the finite-dimensional case operators of the form (1.2) for $K \neq 0$ are connected with elliptic coordinates and separation of variables (for instance, see [5]).It has been claimed (see Corollary 5.1 in [4]) that for $n > 2$ the pencil $\mathcal P_1$ is maximal and has dimension $(n+1)(n+2)/2$. In particular, this is the unique pencil satisfying the following three conditions.
We see that Nijenhuis pencils arise at the crossroads of different areas of mathematics, so a description of maximal (in particular, finite-dimensional) Nijenhuis pencils is of considerable interest. In the fixed coordinates consider an arbitrary — not necessarily symmetric — constant matrix $A$ with entries $a^i_j$ and fix $n$ constants $c^1, \dots, c^n$. Consider operators $L$ whose matrices in these particular coordinates have the form These operators form a linear space of dimension $n^2+n$. The following result holds.Theorem 1. All operator fields defined by matrices of the form (1.3) in a fixed system of coordinates form a maximal Nijenhuis pencil. We denote this pencil by $\mathcal P_2$. It elements arise in applications in a natural way. Let $A$ in (1.3) be a Jordan block of maximum size with eigenvalue zero. Then for ${c^1=1}$ and $c^2=\dots=c^n=0$ we obtain an operator of the form
$$
\begin{equation*}
L=\begin{pmatrix} u^1 & 1 & 0 & \dots & 0 & 0\\ u^2 & 0 & 1 & \dots & 0 & 0\\ & & & \ddots & & \\ u^{n-2} & 0 & 0 & \dots & 1 & 0\\ u^{n-1} & 0 & 0 & \dots & 0 & 1\\ u^{n} & 0 & 0 & \dots & 0 & 0 \end{pmatrix}.
\end{equation*}
\notag
$$
This is the so-called canonical form of a differentially nondegenerate Nijenhuis operator. Let $\mathcal S$ denote the Nijenhuis pencil of operators whose matrices are constant and symmetric in the system of coordinates under consideration. Theorem 2. Any maximal Nijenhuis pencil containing $\mathcal S$ coincides with $\mathcal P_1$ or $\mathcal P_2$. Note a fairly unexpected phenomenon: given a fixed system of coordinate, taking all operators whose matrices are constant in these coordinates we obtain of course a Nijenhuis pencil. However, as we see from Theorem 1, this pencil is not maximal. § 2. Proof of Theorem 1Consider the operator field with entries $b^i_j=u^i c^j$. Then we have
$$
\begin{equation*}
\begin{aligned} \, \frac{1}{2}[[B, B]](\partial_{u^i}, \partial_{u^j}) &=B [B \partial_{u^i}, \partial_{u^j}]+B [\partial_{u^i}, B\partial_{u^j}]- [B \partial_{u^i}, B\partial_{u^j}] \\ & =B[c^i u^{\alpha} \partial_{u^\alpha}, \partial_{u^j}]+B [\partial_{u^i}, c^j u^{\alpha} \partial_{u^{\alpha}}]-[c^i u^{\alpha} \partial_{u^\alpha}, c^j u^{\alpha} \partial_{u^{\alpha}}] \\ & =- c^i c^j u^{\alpha}\partial_{u^{\alpha}}+c^i c^j u^{\alpha}\partial_{u^{\alpha}}=0. \end{aligned}
\end{equation*}
\notag
$$
That is, $B$ is a Nijenhuis operator. Consider an operator field $A$ which has a constant matrix $a^i_j$ in this system of coordinates. Then for $i \neq j$ we obtain
$$
\begin{equation*}
\begin{aligned} \, & [[A, B]] (\partial_{u^i}, \partial_{u^j}) =A[B\partial_{u^i}, \partial_{u^j}]+A[\partial_{u^i}, B \partial_{u^j}]-[A\partial_{u^i}, B\partial_{u^j}]- [B\partial_{u^i}, A\partial_{u^j}] \\ &\qquad =A [c^i u^{\alpha} \partial_{u^\alpha}, \partial_{u^j}]+ A[\partial_{u^i}, c^j u^{\alpha} \partial_{u^\alpha}]-[a_i^{\alpha} \partial_{u^\alpha}, c^j u^{\alpha} \partial_{u^\alpha}]-[c^i u^{\alpha} \partial_{u^\alpha}, a_j^{\beta} \partial_{u^\beta}] \\ &\qquad =- c^i a_j^{\alpha} \partial_{u^\alpha}+c^j a_i^{\alpha} \partial_{u^\alpha}-c^j a_i^{\alpha} \partial_{u^\alpha}+c^i a_j^{\alpha} \partial_{u^\alpha}=0. \end{aligned}
\end{equation*}
\notag
$$
As $[[A, A]]=0$, the condition that $A+B$ is a Nijenhuis operator has the form Thus, the linear space $\mathcal P_2$ consists indeed of Nijenhuis operators. Applying the same equality to $L, M \in \mathcal P_2$ we obtain $[[L, M]]=0$, so that $\mathcal P_2$ is a Nijenhuis pencil. Let $\mathcal M$ denote the linear space of operators whose matrices are constant in the system of coordinates under consideration. This is clearly a Nijenhuis pencil. Now we describe its centralizer $C(\mathcal M)$. Fix $i$ and $j$ and consider $A=\partial_{u^i} \otimes \mathrm{d} u^j$. By definition For an arbitrary operator field $R$ with matrix components $r^i_j$ we have
$$
\begin{equation*}
\begin{aligned} \, & [[A, R]](\partial_{u^k}, \partial_{u^s}) =A[r^{\alpha}_k \partial_{u^{\alpha}}, \partial_{u^s}]+ A[\partial_{u^k}, r^{\alpha}_s\partial_{u^{\alpha}}]-\delta^k_j [\partial_{u^i}, r^{\alpha}_s \partial_{u^\alpha}]-\delta^s_j [r^{\alpha}_k \partial_{u^\alpha}, \partial_{u^i}] \\ &\qquad =\frac{\partial r^j_s}{\partial u^k} \partial_{u^i}-\frac{\partial r^j_k}{\partial u^s} \partial_{u^i}-\delta^k_j \frac{\partial r^{\alpha}_s}{\partial u^i} \partial_{u^\alpha}+\delta^s_j \frac{\partial r^{\alpha}_k}{\partial u^i} \partial_{u^\alpha}. \end{aligned}
\end{equation*}
\notag
$$
Plugging in $j=k$ we let $R \in C(\mathcal M)$. In this case
$$
\begin{equation}
0=\biggl( \frac{\partial r^j_s}{\partial u^j}-\frac{\partial r^j_j}{\partial u^s}-\frac{\partial r^i_s}{\partial u^i}\biggr)\partial_{u^i}+\sum_{\alpha \neq i} \frac{\partial r^{\alpha}_s}{\partial u^i} \partial_{u^\alpha}.
\end{equation}
\tag{2.1}
$$
In particular, for each $s$ the component $r^\alpha_s$ is independent of $u^i$ for $\alpha \neq i$. In other words, the row with index $\alpha$ of $R$ depends only on $u^\alpha$. Since $k \neq s$, the expression in brackets in (2.1) assumes the form
$$
\begin{equation*}
\frac{\partial r^j_s}{\partial u^j}-\frac{\partial r^i_s}{\partial u^i}=0.
\end{equation*}
\notag
$$
Since $r^j_s$ depends only on $u^j$ and $r^i_s$ depends only on $u^i$, this equality means that both derivatives are constants. In addition, for different entries in the same column we obtain the same constant. Denoting these constants by $c^i$ we obtain precisely (1.3). Thus we have shown that $\mathcal P_2=C(\mathcal M)$. As this is a Nijenhuis pencil, we see that it is maximal. The proof is complete.
§ 3. Proof of Theorem 2We begin with several lemmas. Lemma 1. For operator fields $B$ and $C$ with matrices $b^i_j=b^i u^j+u^i b^j$ and $C^i_j=u^i u^j$ the following relations hold: Proof. Consider the operator $L$ of the form (1.2). Note that it is selfadjoint with respect to a symmetric nondegenerate flat metric $\delta_{ij}$. Raising the indices of the operator we obtain the contravariant differential form The fixed coordinates are flat for $\delta_{ij}$, that is, $\nabla_k=\partial_{u^k}$. We have the equality
$$
\begin{equation*}
\nabla_k l^{ij} =\partial_{u^k} l^{ij}=b^i \delta^j_k+\delta^i_k b^j-K \delta^i_k u^j-K u^i \delta^j_k =(b^i-K u^i) \delta^j_k+\delta^i_k (b^j-K u^j).
\end{equation*}
\notag
$$
Note that $\operatorname{tr} L=\sum_{p=1}^n (2 b^p u^p-K(u^p)^2)$ and
$$
\begin{equation*}
\frac{1}{2} \delta^{i\alpha}\mathrm{d}_\alpha \operatorname{tr} L=b^i-K u^i.
\end{equation*}
\notag
$$
Thus, the operator $L$ satisfies the condition of geodesic compatibility with the metric $\delta_{ij}$ (see [4], formula 13). If $l^{ij}$ is nondegenerate in a neighbourhood of the origin, then the contravariant metrics $\delta^{ij}$ and $l^{ij}$ are compatible in the sense of Poisson (see [4], Theorem 1). Hence $L$ is a Nijenhuis operator (for example, see [6]).
The contravariant metric $l^{ij}$ is nondegenerate in a neighbourhood of the origin for almost all values of the parameters $a^i_j$, $b^i$ and $K$. Hence $L$ is a Nijenhuis operator for almost all values of these parameters and, by continuity, for all of their values. Then we obtain
$$
\begin{equation*}
\begin{aligned} \, 0 &=[[A+B+C, A+B+C]] \\ &=2\underbrace{[[A, B]]}_{\text{degree 0}}+2\underbrace{[[A, C]]}_{\text{degree 1}}+2\underbrace{[[B, C]]}_{\text{degree 2}}+ \underbrace{[[B, B]]}_{\text{degree 1}}+\underbrace{[[C, C]]}_{\text{degree 3}}. \end{aligned}
\end{equation*}
\notag
$$
The components of the tensor field corresponding to each term are homogeneous polynomials of degree indicated below it. Since their sum is zero, we immediately obtain $[[C, C]]=[[B, C]]=0$. Taking the zero matrix as $A$ we obtain $[[B, B]] = 0$. The proof is complete. Lemma 2. Consider two operator fields $A$ and $B$ with matrices $a^i_j=b^i u^j+c^j u^i$ and $b^i_j=u^i u^j$. Then $[[A, B]]=0$ if and only if $b^i=c^i$ for all $i$. Proof. First we express $A^i_j=(b^i-c^i) u^j+c^j u^i+c^i u^j$ and set $C^i_j=p^i u^j$, where $p^i=b^i-c^i$. By Lemma 1 $[[A, B]]=[[C, B]]$. For $i \neq j$ this yields the following relations (summation in parentheses is performed only over the $\alpha$ !):
$$
\begin{equation*}
\begin{aligned} \, & [[C, B]](\partial_{u^i}, \partial_{u^j})=C[B\partial_{u^i}, \partial_{u^j}]+C[\partial_{u^i}, B\partial_{u^j}]+ B[C\partial_{u^i}, \partial_{u^j}]+B[\partial_{u^i}, C\partial_{u^j}] \\ &\quad -[C\partial_{u^i}, B\partial_{u^j}]-[B\partial_{u^i}, C\partial_{u^j}]=C[u^i u^{\alpha} \partial_{u^{\alpha}}, \partial_{u^j}]+C[\partial_{u^i}, u^j u^{\alpha} \partial_{u^{\alpha}}] \\ &\quad +B[u^i p^{\alpha} \partial_{u^{\alpha}}, \partial_{u^j}]+ B[\partial_{u^i}, u^j p^{\alpha} \partial_{u^{\alpha}}]-[u^i p^{\alpha}\partial_{u^{\alpha}}, u^j u^{\beta} \partial_{u^{\beta}}]-[u^i u^{\alpha}\partial_{u^{\alpha}}, u^j p^{\beta} \partial_{u^{\beta}}] \\ &=C (u^i \partial_{u^j}-u^j \partial_{u^i})-\delta^i_{\beta} u^j u^{\beta} p^{\alpha} \partial_{u^{\alpha}}+u^i p^{\alpha} \delta^j_{\alpha} u^{\beta} \partial_{u^{\beta}}+u^i p^{\alpha} u^j \delta^{\beta}_{\alpha} \partial_{u^{\beta}} \\ &\quad -u^j p^{\beta} \delta^i_{\beta} u^{\alpha}\partial_{u^{\alpha}}- u^i u^j p^{\beta} \delta^{\alpha}_{\beta} \partial_{u^{\alpha}}+u^i u^{\alpha} \delta^j_{\alpha} p^{\beta} \partial_{u^{\beta}} \\ &=- u^j u^i p^{\alpha} \partial_{u^{\alpha}}+u^i p^j u^{\alpha} \partial_{u^{\alpha}}+u^i u^j p^{\alpha} \partial_{u^{\alpha}}-u^j p^i u^{\alpha} \partial_{u^{\alpha}}-u^i u^j p^{\alpha} \partial_{u^{\alpha}}+u^i u^j p^{\alpha} \partial_{u^{\alpha}} \\ &=(u^i p^j-u^j p^i) u^{\alpha} \partial_{u^{\alpha}}. \end{aligned}
\end{equation*}
\notag
$$
As the Frölicher-Nijenhuis bracket vanishes, all coefficients vanish and, in particular, $p^i=p^j=0$. This completes the proof. Lemma 3. Consider the operator field $A$ with matrix $a^i_j=b^i u^j+u^i c^j$. Then the condition $[[A, A]]=0$ holds (that is, $A$ is a Nijenhuis operator) if and only if at least one of the following conditions holds: 1) $b^i=c^i$; 2) $b^i=0$ for all $i=1, \dots, n$. Proof. For $i \neq j$ we obtain
$$
\begin{equation*}
\begin{aligned} \, 0 &=[[A, A]]=A [A\partial_{u^i}, \partial_{u^j}]+A [\partial_{u^i}, A \partial_{u^j}]-[A\partial_{u^i}, A\partial_{u^j}] \\ &=A[u^i b^{\alpha} \partial_{u^{\alpha}}+c^i u^{\alpha} \partial_{u^{\alpha}}, \partial_{u^j}]+A [\partial_{u^i}, u^j b^{\alpha} \partial_{u^{\alpha}}+c^j u^{\alpha} \partial_{u^{\alpha}}] \\ &\qquad -[u^i b^{\alpha} \partial_{u^{\alpha}}+c^i u^{\alpha} \partial_{u^{\alpha}}, u^j b^{\beta} \partial_{u^{\beta}}+c^j u^{\beta} \partial_{u^{\beta}}] \\ &=c^i (u^j b^{\alpha} \partial_{u^{\alpha}}+c^j u^{\alpha} \partial_{u^{\alpha}})-c^j (u^i b^{\alpha} \partial_{u^{\alpha}}+ c^i u^{\alpha} \partial_{u^{\alpha}}) \\ &\qquad -[u^i b^{\alpha} \partial_{u^{\alpha}}, u^j b^{\beta} \partial_{u^{\beta}}]-[c^i u^{\alpha} \partial_{u^{\alpha}}, u^j b^{\beta} \partial_{u^{\beta}}]-[u^i b^{\alpha} \partial_{u^{\alpha}}, c^j u^{\beta} \partial_{u^{\beta}}] \\ &\qquad -[c^i u^{\alpha} \partial_{u^{\alpha}}, c^j u^{\beta} \partial_{u^{\beta}}]=c^i u^j b^{\alpha} \partial_{u^{\alpha}}-c^j u^i b^{\alpha} \partial_{u^{\alpha}}-b^i b^{\alpha} \partial_{u^{\alpha}}+b^j b^{\alpha} \partial_{u^{\alpha}} \\ &\qquad -c^i u^j b^{\alpha} \partial_{u^{\alpha}}+c^i u^j b^{\alpha} \partial_{u^{\alpha}}-c^j u^i b^{\alpha} \partial_{u^{\alpha}}+c^j u^i b^{\alpha} \partial_{u^{\alpha}}-c^i c^j [u^{\alpha} \partial_{u^{\alpha}}, u^{\beta} \partial_{u^{\beta}}] \\ &=( (c^i-b^i) u^j-(c^j-b^j)u^i) b^{\alpha} \partial_{u^{\alpha}}. \end{aligned}
\end{equation*}
\notag
$$
If $b^i=0$ for $i=1, \dots, n$, then it is obvious that the last formula produces zero. If not all the $b^i$ vanish, then we choose $\alpha=k$ such that $b^k \neq 0$. Then the homogeneous linear polynomial in parentheses vanishes, so that $b^i-c^i=0$ and $b^j-c^j=0$. Since $i$ and $j$ can be arbitrary, the proof is complete. Lemma 4. Consider an operator field $B$ with matrix $b^i_j=u^i u^j$ and a constant operator field $A$ with matrix $a^i_j$. Then they commute with respect to the Frölicher-Nijenhuis bracket if and only if $a^i_j=a^j_i$. Proof. For $i \neq j$ we obtain
$$
\begin{equation*}
\begin{aligned} \, & [[A, B]] (\partial_{u^i}, \partial_{u^j})=A[B\partial_{u^i}, \partial_{u^j}]+A[\partial_{u^i}, B \partial_{u^j}]- [A\partial_{u^i}, B\partial_{u^j}]-[B\partial_{u^i}, A\partial_{u^j}] \\ &\qquad =A[u^i u^{\alpha}\partial_{u^{\alpha}}, \partial_{u^j}]+A [\partial_{u^i}, u^j u^{\alpha} \partial_{u^{\alpha}}]- [A^{\alpha}_i \partial_{u^{\alpha}}, u^j u^{\beta} \partial_{u^{\beta}}]-[u^i u^{\alpha} \partial_{u^{\alpha}}, A^{\beta}_j \partial_{u^{\beta}}] \\ &\qquad =u^i a_j^{\alpha} \partial_{u^{\alpha}}-u^j a^{\alpha}_i \partial_{u^{\alpha}}+a^j_i u^{\alpha} \partial_{u^{\alpha}}+u^j a_i^{\alpha} \partial_{u^{\alpha}}-a^i_j u^{\alpha} \partial_{u^{\alpha}}-u^i a^{\alpha}_j \partial_{u^{\alpha}} \\ &\qquad =(a^j_i-a^i_j) u^{\alpha} \partial_{u^{\alpha}}. \end{aligned}
\end{equation*}
\notag
$$
Thus, the Frölicher-Nijenhuis bracket vanishes if and only if $a^i_j=a^j_i$. The proof is complete. Lemma 5. Consider a nontrivial operator field $B$ with matrix $b^i_j=b^i u^j+u^i b^j$ and a constant operator field $A$ with matrix $a^i_j$. Then they commute with respect to the Frölicher-Nijenhuis bracket if and only if $a^i_j=a^j_i$. Proof. For $i \neq j$ we obtain
$$
\begin{equation*}
\begin{aligned} \, & [[A, B]] (\partial_{u^i}, \partial_{u^j})=A[B\partial_{u^i}, \partial_{u^j}]+A[\partial_{u^i}, B \partial_{u^j}]- [A\partial_{u^i}, B\partial_{u^j}]-[B\partial_{u^i}, A\partial_{u^j}] \\ &\qquad =A[u^i b^{\alpha} \partial_{u^{\alpha}}+b^i u^{\alpha} \partial_{u^{\alpha}}, \partial_{u^j}]+A[\partial_{u^i}, u^j b^{\alpha} \partial_{u^{\alpha}}+b^j u^{\alpha}\partial_{u^{\alpha}}] \\ &\qquad\qquad -[a_i^{\alpha} \partial_{u^{\alpha}}, u^j b^{\beta} \partial_{u^{\beta}}+b^j u^{\beta}\partial_{u^{\beta}}]-[u^i b^{\alpha} \partial_{u^{\alpha}}+b^i u^{\alpha} \partial_{u^{\alpha}}, a_j^{\beta} \partial_{u^{\beta}}] \\ &\qquad =b^i a_j^{\alpha} \partial_{u^{\alpha}}-b^j a_i^{\alpha} \partial_{u^{\alpha}}+a^j_i b^{\beta} \partial_{u^{\beta}}+b^j a_i^{\alpha} \partial_{u^{\alpha}}-a^i_j b^{\alpha} \partial_{u^{\alpha}}-b^i a_j^{\alpha} \partial_{u^{\alpha}} \\ &\qquad =(a^j_i-a^i_j)b^{\alpha} \partial_{u^{\alpha}}. \end{aligned}
\end{equation*}
\notag
$$
By assumption there exists $\alpha$ such that $b^{\alpha} \neq 0$. Hence $a^i_j=a^j_i$ in this case. Since $i$ and $j$ can be arbitrary, the proof is complete. Lemma 6. Consider a nontrivial operator field $B$ with matrix $b^i_j= u^i c^j$ and a constant operator field $A$ with matrix $a^i_j$. Then $[[A, B]]= 0$. Proof. For $i \neq j$ we obtain
$$
\begin{equation*}
\begin{aligned} \, &[[A, B]] (\partial_{u^i}, \partial_{u^j})=A[B\partial_{u^i}, \partial_{u^j}]+A[\partial_{u^i}, B \partial_{u^j}]- [A\partial_{u^i}, B\partial_{u^j}]-[B\partial_{u^i}, A\partial_{u^j}] \\ &\qquad =A [c^i u^{\alpha} \partial_{u^{\alpha}}, \partial_{u^j}]+ A[\partial_{u^i}, c^j u^{\alpha} \partial_{u^{\alpha}}]- [a_i^{\alpha} \partial_{u^{\alpha}}, c^j u^{\beta} \partial_{u^{\beta}}]-[c^i u^{\alpha} \partial_{u^{\alpha}}, a_j^{\beta} \partial_{u^{\beta}}] \\ &\qquad =c^i a_j^{\alpha} \partial_{u^{\alpha}}-c^j a_i^{\alpha} \partial_{u^{\alpha}}+c^j a_i^{\beta} \partial_{u^{\beta}}-c^i a_j^{\alpha} \partial_{u^{\alpha}}=0. \end{aligned}
\end{equation*}
\notag
$$
This proves the lemma. Lemma 7. In dimension $n \geqslant 3$ the centralizer $C(\mathcal S)$ consists precisely of the operators with matrices of the form where $a^i_j$, $b^i$, $c^i$ and $K$ are constants. Proof. Let $L \in \mathcal S$ be an operator with diagonal matrix and pairwise distinct diagonal entries $\lambda_i$, $i=1, \dots, n$. Then for a fixed pair $i$, $j$, $i \neq j$, we obtain
$$
\begin{equation*}
\begin{aligned} \, [[R, L]](\partial_{u^i}, \partial_{u^j}) &=L [R\partial_{u^i}, \partial_{u^j}]+L [\partial_{u^i}, R\partial_{u^j}]- [R\partial_{u^i}, L\partial_{u^j}]-[L\partial_{u^i}, R \partial_{u^j}] \\ &=\biggl( \lambda_{\alpha}\, \frac{\partial r^{\alpha}_i}{\partial u^j}-\lambda_{\alpha} \, \frac{\partial r^{\alpha}_j}{\partial u^i}- \lambda_j \, \frac{\partial r^{\alpha}_i}{\partial u^j}+\lambda_i \, \frac{\partial r^{\alpha}_j}{\partial u^i} \biggr) \partial_{u^\alpha}=0. \end{aligned}
\end{equation*}
\notag
$$
Relabelling, we obtain the system
$$
\begin{equation}
(\lambda_{k}-\lambda_j)\, \frac{\partial r^{k}_i}{\partial u^j}- (\lambda_{k}-\lambda_i) \, \frac{\partial r^{k}_j}{\partial u^i}=0.
\end{equation}
\tag{3.3}
$$
Taking $k=i$ we see that $r^i_i$ is independent of $u^j$. Since $j \neq i$ are arbitrary, $r^i_i$ depends only on $u^i$. Taking another set of constants $\overline \lambda_i$, for $k \neq i, j$ we obtain the following system of linear equations:
$$
\begin{equation*}
\begin{aligned} \, & (\lambda_{k}-\lambda_j) \, \frac{\partial r^{k}_i}{\partial u^j}- (\lambda_{k}-\lambda_i) \, \frac{\partial r^{k}_j}{\partial u^i}=0, \\ & (\overline \lambda_{k}-\overline \lambda_j) \, \frac{\partial r^{k}_i}{\partial u^j}-(\overline \lambda_{k}-\overline \lambda_i) \, \frac{\partial r^{k}_j}{\partial u^i}=0. \end{aligned}
\end{equation*}
\notag
$$
For each triple $k$, $i$, $j$ we can select constants $\lambda$ and $\overline \lambda$ so that the matrix of the corresponding system is nonsingular. Hence ${\partial r^k_j}/{\partial u^i}=0$ for $i \neq j, k$. In other words, for $i \neq j$, $r^i_j$ depends on two variables, $u^i$ and $u^j$, at most.
Now fix $i$ and $j$ and consider $L \in \mathcal S$ such that
$$
\begin{equation*}
L \partial_{u^i}=\partial_{u^j}, \qquad L\partial_{u^j}=\partial_{u^i}\quad\text{and} \quad L\partial_{u^k}=0, \quad k \neq i, j.
\end{equation*}
\notag
$$
If $n > 3$, then considering a quadruple of pairwise distinct $i$, $j$, $p$, $q$ we obtain
$$
\begin{equation}
\begin{aligned} \, \notag [[R, L]] (\partial_{u^p}, \partial_{u^q}) &=L[R\partial_{u^p}, \partial_{u^q}]+L[\partial_{u^p}, R\partial_{u^q}] \\ &=\frac{\partial r^i_p}{\partial u^q} \, \partial_{u^j}+\frac{\partial r^j_p}{\partial u^q} \, \partial_{u^i} -\frac{\partial r^i_q}{\partial u^p} \, \partial_{u^j}-\frac{\partial r^j_q}{\partial u^p} \, \partial_{u^i}=0. \end{aligned}
\end{equation}
\tag{3.4}
$$
We see that the result already established ensures the equality to zero: the components $r^i_j$ depend at most on the two variables $u^i$ and $u^j$. Thus, condition (3.4) for quadruples gives us nothing new.
Now assume that $n \geqslant 3$ again. Then for pairwise distinct $i$, $j$ and $k$ we have (summation is only performed with respect to $\alpha$, there is no summation with respect to $i$, $j$ or $k$!):
$$
\begin{equation}
\begin{aligned} \, \notag [[R, L]] (\partial_{u^j}, \partial_{u^k}) &=L[R\partial_{u^j}, \partial_{u^k}]+L[\partial_{u^j}, R \partial_{u^k}]-[L \partial_{u^j}, R \partial_{u^k}] \\ \notag &=\frac{\partial r^i_j}{\partial u^k} \, \partial_{u^j}+\frac{\partial r^j_j}{\partial u^k} \, \partial_{u^i}-\frac{\partial r^i_k}{\partial u^j}\, \partial_{u^j}-\frac{\partial r^j_k}{\partial u^j} \, \partial_{u^i}+\frac{\partial r^{\alpha}_k}{\partial u^i}\, \partial_{u^\alpha} \\ &=\biggl( \frac{\partial r^i_k}{\partial u^i}-\frac{\partial r^j_k}{\partial u^j}\biggr) \partial_{u^i}=0. \end{aligned}
\end{equation}
\tag{3.5}
$$
Here we have used that $r^{\alpha}_i$ depends only on $u^i$ and $u^j$. Differentiating with respect to $u^i$ we obtain
$$
\begin{equation}
\frac{\partial^2 r^i_k}{\partial u^i\, \partial u^i}=0, \qquad i \neq k.
\end{equation}
\tag{3.6}
$$
Now consider the following expression (again, summation is only performed with respect to $\alpha$, rather than with respect to $i$ and $j$!):
$$
\begin{equation*}
\begin{aligned} \, [[R, L]](\partial_{u^i}, \partial_{u^j}) &=L [R\partial_{u^i}, \partial_{u^j}]+L [\partial_{u^i}, R\partial_{u^j}]-[R\partial_{u^i}, L\partial_{u^j}]-[L\partial_{u^i}, R \partial_{u^j}] \\ &=L [R\partial_{u^i}, \partial_{u^j}]+L [\partial_{u^i}, R\partial_{u^j}]-[R\partial_{u^i}, \partial_{u^i}]-[\partial_{u^j}, R \partial_{u^j}] \\ &=\frac{\partial r^i_i}{\partial u^j}\, \partial_{u^j}+\frac{\partial r^j_i}{\partial u^j} \, \partial_{u^i}-\frac{\partial r^i_j}{\partial u^i}\, \partial_{u^j}-\frac{\partial r^j_j}{\partial u^i} \, \partial_{u^i}- \frac{\partial r^{\alpha}_i}{\partial u^i} \, \partial_{u^\alpha}+ \frac{\partial r^{\alpha}_j}{\partial u^j} \, \partial_{u^\alpha} \\ &=\biggl( \frac{\partial r^j_i}{\partial u^j}+\frac{\partial r^i_j}{\partial u^j}-\frac{\partial r^i_i}{\partial u^i}\biggr) \partial_{u^i}-\biggl( \frac{\partial r^i_j}{\partial u^i}+ \frac{\partial r^j_i}{\partial u^i}-\frac{\partial r^j_j}{\partial u^j}\biggr)\partial_{u^j} \\ &\qquad +\sum_{k \neq i, j} \biggl( \frac{\partial r^{k}_j}{\partial u^j}- \frac{\partial r^{k}_i}{\partial u^i}\biggr) \partial_{u^k}=0. \end{aligned}
\end{equation*}
\notag
$$
For a fixed triple of pairwise distinct $i$, $j$ and $k$ we obtain
$$
\begin{equation}
\begin{gathered} \, \frac{\partial r^j_i}{\partial u^j}+\frac{\partial r^i_j}{\partial u^j}-\frac{\partial r^i_i}{\partial u^i}=0, \qquad i \neq j, \\ \frac{\partial r^{k}_j}{\partial u^j}-\frac{\partial r^{k}_i}{\partial u^i}=0, \qquad k \neq i, j. \end{gathered}
\end{equation}
\tag{3.7}
$$
Since $r^i_k$ depends only on $u^i$ and $u^k$, differentiating the second equation in (3.7) with respect to $u^j$ and relabelling $k \to i$, $j \to k$ we obtain
$$
\begin{equation}
\frac{\partial^2 r^i_k}{\partial u^k \,\partial u^k}=0, \qquad i \neq k.
\end{equation}
\tag{3.8}
$$
From (3.6) and (3.8) we see that for $i \neq j$ $r^i_k$ depends quadratically on the variables $u^i$ and $u^j$. Differentiating the first equation in (3.7) with respect to $u^j$ we obtain
$$
\begin{equation*}
\frac{\partial^2 r^i_i}{\partial u^i\, \partial u^j}=0, \qquad i \neq j.
\end{equation*}
\notag
$$
Hence $r^i_i$ depends only on $u^i$, and we can see from (3.7) itself that this dependence is also quadratic. In the general form we can express components of the operator as (without summation with respect to repeating indices), where $a^i_j$, $b^i_j$, $c^i_j$ and $K^i_j$ are constants. For pairwise distinct $i$, $j$ and $k$ equations (3.7) and (3.5) yield the following relations between coefficients:
$$
\begin{equation*}
b^i_j=b^k_j, \qquad c^i_j=c^k_j, \qquad K^i_j=K^k_j=K^i_k.
\end{equation*}
\notag
$$
Denoting $b^i_j$ and $c^i_j$ by $b^i$ and $c^i$ we see that $R$ has the form (3.2). On the other hand the expressions (3.3)–(3.5) and (3.7) show that each operator with matrix of the form (3.2) in the given coordinates commutes with $\mathcal S$ in the sense of the Frölicher-Nijenhuis brackets. The proof of Lemma 7 is thus complete. Now we go over directly to the proof of Theorem 2. Consider some $R \in C(\mathcal S)$. By Lemma 7 the matrix of $R$ has the form $A+B+C$, where $A$ is a constant matrix, $b^i_j=b^i u^j+u^i c^j$ and $c^i_j=- K u^i u^j$. Consider the condition
$$
\begin{equation}
0=[[R, R]] =\underbrace{2[[A, B]]}_{\text{degree 0}}+ \underbrace{2[[A, C]]+[[B, B]]}_{\text{degree 1}}+ \underbrace{2[[B, C]]}_{\text{degree 2}}+\underbrace{[[C, C]]}_{\text{degree 3}}.
\end{equation}
\tag{3.9}
$$
Similarly to the proof of Lemma 1 the degree indicated under a bracket means the degree of components of the corresponding term. Thus we obtain three cases. The case $B \neq 0$, $C \neq 0$. The expression (3.9) yields $[[B, C]]=0$. By Lemma 2 we obtain $b^i_j=b^i u^j+u^i b^j$. By Lemma 5, $a^i_j=a^j_i$. Thus, $R$ lies in the pencil $\mathcal P_1$. The case $B=0$, $C \neq 0$. The expression (3.9) yields $[[A, C]]=0$. From Lemma 4 we obtain $a^i_j=a^j_i$, so that $R$ lies in $\mathcal P_1$ again. The case $B \neq 0$, $C=0$. Then the expression (3.9) yields $[[B, B]]=0$. By Lemma 3 two situations are possible. If $b^i_j=b^i u^j+u^i b^j$, then by Lemma 5 we have $a^i_j=a^j_i$, and $R$ lies in the pencil $\mathcal P_1$. If $b^i_j=u^i c_j$, then by Lemma 6 the operator lies in $\mathcal P_2$. Now note that $C(\mathcal S)=\mathcal P_1+\mathcal P_2$ and $\mathcal P_1 \cap \mathcal P_2=\mathcal S$ by construction. In addition, the set of Nijenhuis operators in the centralizer is $\mathcal P_1 \cup \mathcal P_2$. Each maximal Nijenhuis pencil $\mathcal P$ lies in $C(\mathcal S)$ by construction, so that, as it is a linear space, it either lies in $\mathcal P_1$ or in $\mathcal P_2$. Since all these pencils are maximal, $\mathcal P$ coincides with one of them. 
Citation in format AMSBIB
\Bibitem{Kon23} |
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