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Realization of arbitrary Lie algebras by automorphisms of M. A. Stepanova Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 2, DOI: https://doi.org/10.4213/im9469 
Bibliographic databases:
    § 1. IntroductionLie groups and algebras, which are classical objects of analysis, algebra, and geometry, appear from consideration of symmetries of geometrical objects. Today, there is a well developed theory of abstract finite dimensional Lie groups and algebras. However, it is natural to ask where and how they appear in mathematics. In this paper, we discuss the realization of finite-dimensional Lie algebras by two classes of vector fields, which are interesting from the point of view of complex analysis: namely, by holomorphic automorphisms of real submanifolds in complex space and symmetries of analytic differential equations. Symmetric objects attract attention of geometers not only because they carry plenty of non-trivial mathematical structures, but also because they are beautiful in aesthetic terms. Symmetries appears in connection with mappings between manifolds, equivalence and classification problems, and also the search of invariants. Questions related to symmetries can be divided into two classes: local and global. When one considers manifolds locally, all of them are topologically the same. Therefore, in questions related to analysis of germs of manifolds, the topology of manifolds does not matter in essence, in contrast to global problems, for which the topology is of importance. We will focus on local questions. In the analysis of germs, it suffices to consider the Lie algebra of automorphisms and symmetries only because the corresponding local Lie group is uniquely determined. Along with the question of the description of automorphisms for a particular germ of $\mathrm{CR}$ manifold, it is natural to enquire about the richness of the class of Lie algebras of automorphisms for all possible germs. There are only few possibilities for a real analytic hypersurface in $\mathbb{C}^2$, which is a simplest $\mathrm{CR}$ manifold: the dimension of the automorphism algebra of a germ of such manifold is either infinite or not greater than $8$ (see [1]). The analogous result also holds for hypersurfaces in $\mathbb{C}^3$: if the dimension of the automorphism algebra of a real analytic hypersurface is finite, then it is not greater than $15$ (see [2]). However, if there are no restrictions on the dimension of the ambient complex space, it turns out that any Lie algebra can be realized with the help of automorphism algebras of hypersurfaces. More precisely, for any finite dimensional real Lie algebra $\mathfrak{h}$, we construct an explicit example of a germ $\Gamma_0$ of a real analytic hypersurface $\Gamma$ such that its automorphism algebra is isomorphic to $\mathfrak{h}$. Similar results were obtained by other authors. For instance, in [3] it was shown that, for any linear connected Lie group, there exists a bounded strictly pseudoconvex domain whose group of holomorphic automorphisms is isomorphic to the given group. In [4], each connected Lie group was shown as being realizable as the group of holomorphic automorphisms of a hyperbolic Stein manifold. The results of [3] and [4] are global, that is, they are related to the full automorphism group. Our considerations are local and independent of [3] and [4]. Lie groups and algebras appears in mathematics also differently as symmetries of differential equations. In this case, the same questions arise: how rich is the class of symmetries for all kinds of equations and whether it is possible to realize any Lie algebra with the help of them? The symmetries of equations are closely related to the automorphisms of $\mathrm{CR}$ manifolds under certain non-degeneracy conditions (the manifold should be generic and finitely non-degenerate, see the definitions below). Namely, with the help of the construction, going back to Cartan and Segre (see [5] and [6]), we can associate with a real analytic $\mathrm{CR}$ manifold a system of analytic partial differential equations such that the Lie algebra of its infinitesimal symmetries is isomorphic to the complexification of the algebra of holomorphic automorphisms of a given $\mathrm{CR}$ manifold (see [7]). The key tool here is Segre manifolds, which allow one to translate questions from the language of $\mathrm{CR}$ geometry to that of differential equations, and vice versa. It is worth noting that it is also possible to apply the Cartan–Segre construction to degenerate manifolds, but in this case, the equations become singular and their analysis becomes more complicated (see, for instance, [8]). The above hypersurfaces endowed with automorphism algebras are not finitely non-degenerate at the origin, and therefore, to apply the above construction we will give another example of real analytic $\mathrm{CR}$ manifold (of codimension three) which is finitely non-degenerate at the origin. Using this example, we also obtain a realization of the complexification of an arbitrary algebra $\mathfrak{h}$ by symmetries of differential equations. The resulting realizations of Lie algebras by automorphisms of $\mathrm{CR}$ manifolds and symmetries of differential equations can be looked upon as representations of Lie algebras by vector fields together with more familiar linear representations. At the same time, we get an affirmative answer to the following question: Is it possible to realize any real finite dimensional Lie algebra by vector fields on a finite dimensional space? Note that the finite dimensionality of the algebra is important. As natural to expect, not every infinite dimensional Lie algebra can be realized by vector fields in a finite dimensional space (and therefore, it cannot be realized also by the two specified classes of vector fields under consideration); see Remark 7. § 2. Realization by automorphisms of $\mathrm{CR}$ manifoldsLet $\mathfrak{h}$ be an arbitrary real Lie algebra of dimension $m < \infty$. By the Ado theorem, $\mathfrak{h}$ is isomorphic to a subalgebra of the algebra $\mathfrak{gl}(n,\mathbb{R})$ for some $n$, and therefore, we can assume that $\mathfrak{h} \subseteq \mathfrak{gl}(n,\mathbb{R})$. Let $e_1,\dots,e_m$ be a basis for $\mathfrak{h}$. Consider the local Lie pseudogroup $H$ corresponding to the algebra $\mathfrak{h}$. We can introduce local coordinates in $H$ in the following way: to a point $(\tau_1,\dots,\tau_m)$ from a neighbourhood of the origin in $\mathbb{R}^m$ we associate a point $C(\tau)=\exp(\tau_1e_1+\dots+\tau_me_m)$ from a neighbourhood of the unity in $H$. Let $(z_1,\dots,z_n,t_1,\dots,t_m,w=u+iv)$ be coordinates in $\mathbb{C}^{n+m+1}$. Let $z$ be the vector $(z_1,\dots,z_n)$, and let $t$ be the vector $(t_1,\dots,t_m)$. Next, let
$$
\begin{equation*}
\begin{aligned} \, P_1(z,\bar{z}) &=2 \operatorname{Re} \biggl(\sum_{j=1}^n|z_j|^{32n} \bigl(z_j^{4j-2}\bar{z}_j^{4j-1}+z_j^{4j}\bar{z}_j^{4j+1}\bigr)\biggr), \\ P_2(z,\bar{z}) &=2 \operatorname{Re}\biggl(\sum_{j=1}^n |z_j|^{40n} \bigl(z_j^{4j-2}\bar{z}_j^{4j-1}+z_j^{4j}\bar{z}_j^{4j+1}\bigr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Let $P(z,\bar{z},t,\bar{t}\,) =P_1(\overline{C(t)}z,C(t)\bar{z})+P_2(C(t)z,\overline{C(t)z})$, where $C(t)$ is the above matrix in which the real argument $\tau \in \mathbb{R}^m$ is replaced by the complex argument $t\in \mathbb{C}^m$. Consider a germ $\Gamma_0$ at the origin of the hypersurface $\Gamma=\Gamma(\mathfrak{h}) \subset \mathbb{C}^{n+m+1}$ given in a neighbourhood of the origin by the equation
$$
\begin{equation*}
v=P(z,\bar{z},t,\bar{t}\,)+u P^4(z,\bar{z},t,\bar{t}\,).
\end{equation*}
\notag
$$
By $\operatorname{aut} \Gamma_0$ we denote the Lie algebra of infinitesimal holomorphic automorphisms of the germ $\Gamma_0$. This algebra consists of the vector fields, which generate $1$-parameter groups of local biholomorphic automorphisms of the germ $\Gamma_0$. Before proceeding with the proof, we need two remarks. Remark 1. The local pseudogroup of automorphisms of the germ $\Gamma_0$ contains all transformations of the form $\{z\to Az, \, t\to b(t), \, w\to w\}$. Here, $A \in H$ is a non-degenerate matrix with constant real coefficients, and an analytic vector- valued function $b(t)$ is such that $C(t)A=C(b(t))$. At the same time, we can uniquely reconstruct $b(t)$ from the matrix $A$ (see Lemma 6 below). Since $H$ is a local pseudogroup corresponding to $\mathfrak{h}$, we have $\mathfrak{h} \subseteq \operatorname{aut} \Gamma_0$. The polynomials $P_1$ and $P_2$ in the definition of the function $P$ are chosen so that $H$ does not contain other transformations, that is, $\mathfrak{h} = \operatorname{aut} \Gamma_0$. Remark 2. Instead of the polynomials $P_1$ and $P_2$ we can choose two different generic polynomials of sufficiently high degree. The property of being generic is considered with respect to the standard topology of the Euclidean space of polynomials in the variables $z$, $\bar{z}$, $t$, $\bar{t}$ of uniformly bounded degree. Instead of $P_1$ and $P_2$, we can also choose a generic real analytic function, that is, a point from an open dense set with respect to the topology of uniform convergence on compact sets in the space of functions which are real analytic in a neighbourhood of the origin. We give the explicit expressions for the polynomials $P_1$ and $P_2$ to make the examples constructive. Proof of Theorem 1. Let
$$
\begin{equation*}
\begin{aligned} \, &\Xi = 2 \operatorname{Re} \biggl(f_1(z,t,w)\, \frac{\partial}{\partial z_1}+\dots+f_n(z,t,w)\, \frac{\partial}{\partial z_n} \\ &\qquad+r_1(z,t,w)\, \frac{\partial}{\partial t_1}+\dots+r_m(z,t,w)\, \frac{\partial}{\partial t_m}+g(z,t,w)\, \frac{\partial}{\partial w}\biggr), \qquad \Xi \in \operatorname{aut} \Gamma_0. \end{aligned}
\end{equation*}
\notag
$$
The tangency condition has the form
$$
\begin{equation}
\begin{gathered} \, \Xi\biggl(\frac{(w-\bar{w})}{2i}-P(z,\bar{z},t,\bar{t}\,) -\biggl(\frac{w+\bar{w}}{2}\biggr)P^4(z,\bar{z},t,\bar{t}\,)\biggr)=0 \\ \text{for } \quad w=u+i\bigl(P(z,\bar{z},t,\bar{t}\,)+u P^4(z,\bar{z},t,\bar{t}\,)\bigr). \end{gathered}
\end{equation}
\tag{2.1}
$$
The left-hand side of (2.1) can be expanded in a Taylor series in the vector variables $z$, $\bar{z}$, $t$, $\bar{t}$ and in the variable $u$. This fact will be used below. Equality (2.1) is equivalent to vanishing of the coefficients of all the monomials in this expansion. Let $z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}u^{\nu}$ be the monomial $z_1^{\alpha_1}\cdots z_n^{\alpha_n}\bar{z}_1^{\beta_1} \cdots \bar{z}_n^{\beta_n}t_1^{\gamma_1}\cdots t_m^{\gamma_m}\bar{t}_1^{\delta_1} \cdots \bar{t}_m^{\delta_m}u^{\nu}$. For a multidegree $\alpha=(\alpha_1,\dots,\alpha_n)$ (and, similarly, for multidegrees $\beta=(\beta_1,\dots,\beta_m)$, $\gamma=(\gamma_1,\dots,\gamma_n)$ and $\delta=(\delta_1,\dots,\delta_m)$), we set $|\alpha|=\alpha_1+\dots+\alpha_n$. We divide the proof into several lemmas. Proof. Let $g(z,t,w)=\sum_{\nu=0}^{\infty}g_{\nu}(z,t)w^{\nu}$. Substituting $\bar{z}=0$ and $\bar{t}=0$ into (2.1), we obtain whence $g_{\nu}(z,t)=g_{\nu} \in \mathbb{R}$. Here and in what follows, we will not introduce new notation for a function with fewer number of variables, but we will only write its arguments (if there are any).
Let $c_{11}$ be the element of the first row and the first column of the matrix $C(t)$. Then $c_{11}=1+l_1t_1+\dots+l_mt_m+o(1)$, where $o(\nu)$ are the terms of degree greater than $\nu$. We also have
$$
\begin{equation*}
\begin{aligned} \, &P(z,\bar{z},t,\bar{t}\,)=2 \operatorname{Re} \biggl((1+(16n+2)\sum_{j=1}^ml_j\bar{t}_j +(16n+3)\sum_{j=1}^ml_jt_j)z_1^{16n+2}\bar{z}_1^{16n+3}\biggr) \\ &\qquad+2 \operatorname{Re} \biggl((1+(16n+4)\sum_{j=1}^ml_j\bar{t}_j +(16n+5)\sum_{j=1}^ml_jt_j)z_1^{16n+4}\bar{z}_1^{16n+5}\biggr)+\cdots, \end{aligned}
\end{equation*}
\notag
$$
where dots stand for the monomials of the form $z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with multidegrees $\alpha$, $\beta$, $\gamma$, $\delta$ such that at least one of the three conditions holds:
1) $\sum_{j=2}^n(\alpha_j+\beta_j)>0$, 2) $\alpha_1+\beta_1>32n+9$, 3) $|\gamma|+|\delta|>1$. Let
$$
\begin{equation*}
f_1(z,t,w) =z_1\sum_{\nu=0}^{\infty}a_{\nu}w^{\nu}+\widetilde{f}_1(z,t,w),\qquad r_j(z,t,w) =\sum_{\nu=0}^{\infty}b_jw^{\nu}+\widetilde{r}_j(z,t,w),
\end{equation*}
\notag
$$
where $a_{\nu} \in \mathbb{C}$, $b_{\nu} \in \mathbb{C}$, $\frac{\partial}{\partial z_1} (\widetilde{f}_1)(0,0,w) = 0$, $\widetilde{r}_j(0,0,w) = 0$.
In (2.1), we write all expressions of the form $au^{\nu}z_1^{16n+2}\bar{z}_1^{16n+3}$, $au^{\nu}\bar{z}_1^{16n+2}z_1^{16n+3}$ and $au^{\nu}z_1^{16n+4}\bar{z}_1^{16n+5}$, $a=\mathrm{const}$. All these monomials appear in the expression
$$
\begin{equation*}
\operatorname{Im} (g_{\nu+1}w^{\nu+1})- 2 \operatorname{Re} \biggl(a_{\nu}w^{\nu}z_1\,\frac{\partial}{\partial z_1}\biggr)P -2 \operatorname{Re}\biggl(\sum_{j=1}^mb_jw^{\nu}\, \frac{\partial}{\partial t_j}\biggr)P.
\end{equation*}
\notag
$$
We have
$$
\begin{equation}
\begin{aligned} \, &\biggl((\nu+1)g_{\nu+1}-(16n+2)a_{\nu}-(16n+3)\bar{a}_{\nu} \\ &\qquad-(16n+3)\sum_{j=1}^mb_jl_j-(16n+2)\sum_{j=1}^m\bar{b}_jl_j\biggr) u^{\nu}z_1^{16n+2}\bar{z}_1^{16n+3}=0, \\ &\biggl((\nu+1)g_{\nu+1}-(16n+3)a_{\nu}-(16n+2)\bar{a}_{\nu} \\ &\qquad-(16n+2)\sum_{j=1}^mb_jl_j-(16n+3)\sum_{j=1}^m\bar{b}_jl_j\biggr) u^{\nu}\bar{z}_1^{16n+2}z_1^{16n+3}=0, \\ &\biggl((\nu+1)g_{\nu+1}-(16n+4)a_{\nu}-(16n+5)\bar{a}_{\nu} \\ &\qquad-(16n+5)\sum_{j=1}^mb_jl_j-(16n+4)\sum_{j=1}^m\bar{b}_jl_j\biggr) u^{\nu}z_1^{16n+4}\bar{z}_1^{16n+5}=0. \end{aligned}
\end{equation}
\tag{2.2}
$$
Consider (2.2) as a system of linear equations with respect to the three variables $g_{\nu+1}$, $(a_{\nu}+\sum_{j=1}^m\bar{b}_jl_j)$, and $(\bar{a}_{\nu}+\sum_{j=1}^mb_jl_j)$. The determinant of the matrix of this system is $4 \nu + 4$, and hence, for $\nu \geqslant 0$ this system is non-degenerate, therefore, it has only zero solution. Thus, $g_{\nu+1}=0$ for $\nu \geqslant 0$, that is, $g(z,t,w)=g_0 \in \mathbb{R}$. Let us write the monomial of the form $a(z_1^{16n+2}\bar{z}_1^{16n+3})^4$ in (2.1), $a=\mathrm{const}$. This monomial appears in the expression $-g_0P^4$. We have $g_0(z_1^{16n+2}\bar{z}_1^{16n+3})^4=0$, whence $g_0=0$. This proves Lemma 1. Lemma 2. $f_l(z,t,w)=f_l(z,t)$, $r_l(z,t,w)=r_l(z,t)$, that is, the functions $f_l(z,t,w)$ and $r_l(z,t,w)$ do not depend on $w$. Proof. Now let Then $\Xi=2 \operatorname{Re} \sum_{\nu=0}^{\infty}w^{\nu}X_{\nu}$, where We claim that $X_{\nu}=0$ for $\nu>0$.
Writing all expressions in (2.1) of the form $au^{\nu}z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ for $\nu>0$, $a=\mathrm{const}$ with either $|\alpha| \leqslant 24n+1$ or $|\beta| \leqslant 24n+1$, we have Let us write all the expressions of the form $au^{\nu}z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ in (2.1) for $\nu>0$, $a=\mathrm{const}$ for which either $32n+4 \leqslant |\alpha| \leqslant 48n+2$ or $32n+4 \leqslant |\beta| \leqslant 48n+2$ holds. We have
$$
\begin{equation}
i\nu u^{\nu-1}PX_{\nu}P-i\nu u^{\nu-1}P\overline{X}_{\nu}P=0.
\end{equation}
\tag{2.4}
$$
Let $c_{j l}$ be the element of the matrix $C(t)$ at row $j$ and column $l$. For simplicity, in the rest of the proof of Lemma 2, we omit the argument $t$ in the matrix $C$. Let $(C\bar{z})_j$ and $(\bar{C}z)_j$ be, respectively, the $j$th coordinate of the vectors $(C\bar{z})$ and $(\bar{C}z)$. In (2.5), we write the expressions of the form $az^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with $|\alpha| \geqslant 16n+4j-3$, $|\beta|=16n+4j-1$. We have
$$
\begin{equation}
\begin{aligned} \, &(16n+4j-2)\biggl(\sum_{l=1}^nf_{l \nu}\bar{c}_{j l}\biggr) (\bar{C}z)_j^{16n+4j-3}(C\bar{z})_j^{16n+4j-1} \nonumber \\ &\qquad+(16n+4j-1)\biggl(\sum_{l=1}^mr_{l \nu}\frac{\partial}{\partial t_l}(C\bar{z})_j\biggr)(\bar{C}z)_j^{16n+4j-2}(C\bar{z})_j^{16n+4j-2}=0. \end{aligned}
\end{equation}
\tag{2.6}
$$
Next, in (2.5), we write the expressions of the form $az^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with $|\alpha| \geqslant 16n+4j-1$ and $|\beta|=16n+4j+1$. We have
$$
\begin{equation}
\begin{aligned} \, &(16n+4j)\biggl(\sum_{l=1}^nf_{l \nu}\bar{c}_{j l}\biggr) (\bar{C}z)_j^{16n+4j-1}(C\bar{z})_j^{16n+4j+1} \nonumber \\ &\qquad+(16n+4j+1)\biggl(\sum_{l=1}^mr_{l \nu}\frac{\partial}{\partial t_l}(C\bar{z})_j\biggr) (\bar{C}z)_j^{16n+4j}(C\bar{z})_j^{16n+4j}=0. \end{aligned}
\end{equation}
\tag{2.7}
$$
$$
\begin{equation}
\sum_{l=1}^nf_{l \nu}\bar{c}_{j l}=0, \qquad \sum_{l=1}^mr_{l \nu}\, \frac{\partial}{\partial t_l}(C\bar{z})_j=0.
\end{equation}
\tag{2.8}
$$
Since the matrix $C(t)$ is non-degenerate for all sufficiently small $t$, from the first equality in (2.8) we have $f_{l \nu}=0$. The second equality in (2.8) can be written as
$$
\begin{equation*}
\sum_{l=1}^mr_{l \nu}\, \frac{\partial}{\partial t_l}(C\bar{z}) =\sum_{l=1}^mr_{l \nu}\, \frac{\partial}{\partial t_l}(C)\bar{z}=0,
\end{equation*}
\notag
$$
which is equivalent to the equality
$$
\begin{equation*}
\sum_{l=1}^mr_{l \nu}\, \frac{\partial}{\partial t_l}(C)=0.
\end{equation*}
\notag
$$
Since the matrices $\frac{\partial}{\partial t_l}(C)$, $1 \leqslant l \leqslant m$, are linearly independent for all sufficiently small $t$, we have $r_{l \nu}=0$.
The proof of Lemma 2 is complete. Thus, $\Xi=2 \operatorname{Re} X_0$, where
$$
\begin{equation*}
X_0 = f_{1 0}(z,t)\, \frac{\partial}{\partial z_1}+\dots+f_{n 0}(z,t)\, \frac{\partial}{\partial z_n}+r_{1 0}(z,t)\, \frac{\partial}{\partial t_1}+\dots+r_{m 0}(z,t)\, \frac{\partial}{\partial t_m}.
\end{equation*}
\notag
$$
In what follows, for simplicity, the index zero will be omitted:
$$
\begin{equation*}
X = f_1(z,t)\, \frac{\partial}{\partial z_1}+\dots+f_n(z,t)\, \frac{\partial}{\partial z_n}+r_1(z,t)\, \frac{\partial}{\partial t_1}+\dots+r_m(z,t)\, \frac{\partial}{\partial t_m}.
\end{equation*}
\notag
$$
Proof. Let where $z^{\alpha}=z_1^{\alpha_1}\cdots z_n^{\alpha_n}$. We claim that $f_{l \alpha}(t)=0$ for $|\alpha|=\alpha_1+\dots+\alpha_n \neq 1$ and $r_{l \alpha}(t)=0$ for $|\alpha|=\alpha_1+\dots+\alpha_n \neq 0$.
In (2.1), we write the expressions of the form $z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with $|\alpha| \geqslant 16n+4j$ and $|\beta|=16n+4j-2$. We have
$$
\begin{equation}
\begin{aligned} \, &(16n+4j-1)\biggl(\sum_{l=1}^n\biggl(\sum_{|\alpha| \geqslant 2}f_{l \alpha}(t)z^{\alpha}\biggr) \bar{c}_{j l}\biggr) (C\bar{z})_j^{16n+4j-2}(\bar{C}z)_j^{16n+4j-2} \nonumber \\ &+(16n+4j-2)\biggl(\sum_{l=1}^m\biggl(\sum_{|\alpha| \geqslant 1}r_{l \alpha}(t)z^{\alpha}\biggr) \frac{\partial}{\partial t_l}(C\bar{z})_j\biggr) (C\bar{z})_j^{16n+4j-3} (\bar{C}z)_j^{16n+4j-1}=0. \end{aligned}
\end{equation}
\tag{2.9}
$$
In (2.1), we write the expressions of the form $z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with $|\alpha| \geqslant 16n+4j+2$ and $|\beta|=16n+4j$. We have
$$
\begin{equation}
\begin{aligned} \, &(16n+4j+1)\biggl(\sum_{l=1}^n\biggl(\sum_{|\alpha| \geqslant 2}f_{l \alpha}(t)z^{\alpha}\biggr) \bar{c}_{j l}\biggr)(C\bar{z})_j^{16n+4j}(\bar{C}z)_j^{16n+4j} \nonumber \\ &+(16n+4j)\biggl(\sum_{l=1}^m\biggl(\sum_{|\alpha| \geqslant 1}r_{l \alpha}(t)z^{\alpha}\biggr) \frac{\partial}{\partial t_l}(C\bar{z})_j\biggr) (C\bar{z})_j^{16n+4j-1}(\bar{C}z)_j^{16n+4j+1}=0. \end{aligned}
\end{equation}
\tag{2.10}
$$
$$
\begin{equation*}
\sum_{l=1}^n\biggl(\sum_{|\alpha| \geqslant 2}f_{l \alpha}(t)z^{\alpha}\biggr)\bar{c}_{j l} =\sum_{l=1}^m\biggl(\sum_{|\alpha| \geqslant 1}r_{l \alpha}(t)z^{\alpha}\biggr) \frac{\partial}{\partial t_l}(C\bar{z})_j=0,
\end{equation*}
\notag
$$
whence
$$
\begin{equation}
\sum_{l=1}^nf_{l \alpha}(t)\bar{c}_{j l} =0 \quad \text{for} \quad |\alpha| \geqslant 2,
\end{equation}
\tag{2.11}
$$
$$
\begin{equation}
\sum_{l=1}^mr_{l \alpha}(t)\, \frac{\partial}{\partial t_l}(C\bar{z})_j =0 \quad \text{for} \quad |\alpha| \geqslant 1.
\end{equation}
\tag{2.12}
$$
Since the matrix $C(t)$ is non-degenerate for all sufficiently small $t$, from (2.11) we find that $f_{l \alpha}(t)=0$ for $|\alpha|\geqslant 2$. Condition (2.12) can be written as
$$
\begin{equation*}
\sum_{l=1}^mr_{l \alpha}(t)\, \frac{\partial}{\partial t_l}(C\bar{z}) =\sum_{l=1}^mr_{l \alpha}(t)\, \frac{\partial}{\partial t_l}(C)\bar{z}=0,
\end{equation*}
\notag
$$
which is equivalent to the equality
$$
\begin{equation*}
\sum_{l=1}^mr_{l \alpha}(t)\, \frac{\partial}{\partial t_l}(C)=0.
\end{equation*}
\notag
$$
The matrices $\frac{\partial}{\partial t_l}(C)$, $1 \leqslant l \leqslant m$, are linearly independent for all sufficiently small $t$, and so $r_{l \alpha}(t)=0$ for $|\alpha|\geqslant 1$.
In (2.1), we write the expressions of the form $z^{\alpha}\bar{z}^{\beta}t^{\gamma}\bar{t}^{\delta}$ with $|\alpha| = 16n+4j-3$, $|\beta|=16n+4j-1$. We have
$$
\begin{equation*}
\biggl(\sum_{l=1}^nf_{l 0}(t)\bar{c}_{j l}\biggr) (16n+4j-2)(\bar{C}z)_j^{16n+4j-3}(C\bar{z})_j^{16n+4j-1}=0.
\end{equation*}
\notag
$$
Hence $\sum_{l=1}^nf_{l 0}(t)\bar{c}_{j l}=0$, which, as before, implies that $f_{l 0}(t)=0$. This proves Lemma 3. Thus, we have $\Xi=2 \operatorname{Re} X$, where
$$
\begin{equation*}
X = f_1(z,t)\, \frac{\partial}{\partial z_1}+\dots+f_n(z,t)\, \frac{\partial}{\partial z_n}+r_1(t)\, \frac{\partial}{\partial t_1} +\dots+r_m(t)\, \frac{\partial}{\partial t_m},
\end{equation*}
\notag
$$
and $f_j(z,t)$ depends linearly on $z$ (the linearity includes the condition $f_j(0,t)=0$). Thus, we have the following result. Lemma 4. The locally biholomorphic transformation corresponding to the field $\Xi$ has the form $F(\Xi)=\{z\to A(t)z, \, t\to b(t), \, w\to w\}$, where $A(t)$ is a non-degenerate matrix. Proof. To find a 1-parameter subgroup of transformations
$$
\begin{equation*}
(z,t,w)\to \bigl(Z(\tau,z,t,w),T(\tau,z,t,w),W(\tau,z,t,w)\bigr)
\end{equation*}
\notag
$$
with parameter $\tau$ corresponding to the field, it is necessary to solve the system of ordinary differential equations with the initial conditions
$$
\begin{equation*}
Z(0,z,t,w)=z, \qquad T(0,z,t,w)=t, \qquad G(0,z,t,w)=w,
\end{equation*}
\notag
$$
where $Z=Z(z,t,w)$ is an $n$-dimensional vector-valued function, $T=T(z,t,w)$ is an $m$-dimensional vector-valued function, and $G=G(z,t,w)$. From the system we find that $T(\tau,z,t,w)=T(\tau,t)$ and $G(\tau,z,t,w)=w$. Plugging this expression for $T$ in the system $Z'=f(Z,T)$, $Z(0,z,t,w)=z$, we obtain the Cauchy problem for the linear system with parameter $t$, namely Let $\Phi(\tau,t)$ be its fundamental matrix of solutions. Then the general solution of the system has the form $\Phi(\tau,t)c$, where $c=(c_1,\dots,c_n)$ for some constants $c_1,\dots,c_n$. Substituting the initial conditions in the general solution, we find that $c_1,\dots,c_n$ are linearly expressible in terms of $z_1,\dots,z_n$. Substituting $\tau=1$, which corresponds to the exponential map of the tangent vector, we obtain the required form of the transformation.
The proof of Lemma 4 is complete. Proof. The condition that $F$ is a self map of $\Gamma_0$ is as follows: whence
Thus, $C(t)=C(b(t))\overline{A(t)}$. Since the elements of the matrix $C$ are holomorphic, $A(t)=A$ is a matrix with constant complex coefficients. From (2.13) we have
$$
\begin{equation*}
\begin{aligned} \, P_1(\overline{C}z,C\bar{z}) &=P_1\bigl(\overline{C(b(t))}Az,C(b(t))\overline{Az}\, \bigr), \\ P_2(Cz,\overline{Cz}\,) &=P_2\bigl(C(b(t))Az,\overline{C(b(t))Az}\,\bigr), \end{aligned}
\end{equation*}
\notag
$$
whence $C(b(t))\bar{A}=C(b(t))A$. Therefore, $\bar{A}=A$ by non-degeneracy of $C(b(t))$ for small $t$. This means that $z\to Az$ is a real linear mapping, and hence $f_l(z)$ is a real linear function.
The proof of Lemma 5 is complete. Since $C(b(t)) \in H$ and $C(t) \in H$, we have $A = (C(b(t)))^{-1}C(t) \in H$. In addition, we have the following result. Lemma 6. For any matrix $A \in H$, there exists a unique analytic vector-valued function $b(t)$ such that $F(\Xi)=\{z\to Az, \, t\to b(t), \, w\to w\}$ is an automorphism of $\Gamma_0$. Vice versa, for any analytic vector-valued function $b(t)$, there exists a unique matrix $A \in H$ such that $F(\Xi)=\{z\to Az, \, t\to b(t), \, w\to w\}$ is an automorphism of $\Gamma_0$. In particular, for $A=\mathrm{Id}$ we have $b(t)=t$. Conversely, for $b(t)=t$ we have $A=\mathrm{Id}$. Proof of Lemma 6. The vector-valued function $b(t)$ is uniquely defined by the equality $C(b(t))=C(t)A^{-1}$. In fact, to find the coordinates of the vector-valued function $b(t)$, it is necessary to decompose the matrix $\log(C(t)A^{-1})$ with respect to the basis $e_1,\dots,e_m$ of the Lie algebra $\mathfrak{h}$. The coefficients of the decomposition are as required. Here, we choose the matrix logarithm $\log$ such that all elements of the matrix $\log(C(t)A^{-1})$ are in a neighbourhood of the origin, and hence all coordinates $b(t)$ are in a neighbourhood of the origin. For $A=\mathrm{Id}$, we have $C(b(t))=C(t)$, whence $b(t)=t$. On the other hand, the matrix $A$ is uniquely defined by the equality $C(b(t))A=(C(b(t)))^{-1}C(t)$. For $b(t)=t$, we have $C(t)=C(t)A^{-1}$, which implies that $A=\mathrm{Id}$.
The proof of Lemma 6 is complete. Therefore, $F$ is completely defined by the action on the coordinate $z$. But the Lie algebra corresponding to the local group $H$ is $\mathfrak{h}$, and hence the automorphism algebra is isomorphic to $\mathfrak{h}$. The proof of Theorem 1 is complete. Remark 3. The hypersurface $\Gamma$ thus constructed has finite Bloom–Graham type at the origin. But it is also possible to construct a hypersurface of infinite type at the origin for which the automorphism algebra at the origin is isomorphic to $\mathfrak{h}$. It can be defined in a neighbourhood of the origin by the equation The proof of the fact that the automorphism algebra of the germ of this hypersurface is isomorphic to $\mathfrak{h}$ is analogous to the above proof. Remark 4. If we have a realization of an arbitrary algebra as an automorphism algebra of a hypersurface (of codimension one), we can also obtain a realization by the automorphisms of a $\mathrm{CR}$ manifold of an arbitrary codimension $k$. For this purpose it suffices to take the direct product $\Pi$ of the germ $\Gamma_0$ of the hypersurface and $(k-1)$ germs of hypersurfaces in $\mathbb{C}^2$ with zero automorphism algebra. As an example of such hypersurface, we can take The automorphism algebra of the direct product of germs of finite type is isomorphic to the direct sum of the algebras of the factors (see [9]). Thus, the algebra of the germ $\Pi$ is also isomorphic to $\mathfrak{h}$. Remark 5. In the proof of Theorem 1, we also proved that, for any real Lie algebra $\mathfrak{h}$, there exists an analytic function which is invariant under the action of the local pseudogroup $H$ of biholomorphic transformations such that the Lie group corresponding to $H$ is isomorphic to $\mathfrak{h}$. This function is $P(z,\bar{z},t,\bar{t}\,)$, which, in addition, is invariant only under the action of the elements in $H$. § 3. Realization by symmetries of differential equationsUnder certain non-degeneracy conditions for a $\mathrm{CR}$ manifold, we can associate with this manifold a system of analytic partial differential equations such that its symmetry algebra (see Definition 3 below) is isomorphic to the complexification of the automorphism algebra of the germ of the given $\mathrm{CR}$ manifold. We will use the construction, which extends the idea of B. Segre to the case of finitely non-degenerate manifolds (see [7]). Let us give the necessary definitions. Definition 1. Let $y=(y_1,\dots,y_{N})$ be coordinates in $\mathbb{C}^N$. A $\mathrm{CR}$ vector field on a manifold $M$ is a vector field which is tangent to $M$ and has the form where $F_j$ are smooth complex valued functions. Next, let $\{\rho_j=0, \, 1\leqslant j \leqslant k\}$ be a system of defining equations for $M$. Here, we assume that the $\mathbb{R}^{k}$-valued mapping $\rho=(\rho_1,\dots,\rho_{k})$ has maximal real rank. Let us recall that the Levi forms of the manifold $M$ at a point $p$ are the restrictions of the Hermitian forms
$$
\begin{equation*}
\mathcal{L}_{p}^{j}(d y,d \bar{y})=\sum_{\alpha,\beta}\frac{\partial^2 \rho_j}{\partial y_{\alpha}\, \partial \bar{y}_{\beta}}(p,\bar{p})\, dy_{\alpha}\, d\bar{y}_{\beta}
\end{equation*}
\notag
$$
to the complex tangent space. Before proceeding with the definition of finite non-degeneracy, we note that the notion of finite non-degeneracy extends that of the Levi non-degeneracy. There are two different points of view on the Levi non-degeneracy at a point $p$. Under the first approach, the following two conditions are assumed to hold: a) the Levi forms $\mathcal{L}_{p}^{j}(u,v)$, $1\leqslant j \leqslant k$ have no common kernel (that is, if $\mathcal{L}_{p}^{j}(e,v)=0$ for all $j$ and $v$, then $e=0$), b) the Levi forms are linearly independent. Another approach here differs from the first one by the replacement of condition a) by the stronger condition of existence of a linear combination of the Levi forms which is a non-degenerate Hermitian form. In this case, the manifold is said to be strongly Levi non-degenerate. Note that the 1-non-degeneracy is precisely the Levi non-degeneracy in the sense of conditions a) and b) (see [10]). Now, let $L_1,\dots,L_{\nu}$ be a basis for $\mathrm{CR}$ vector fields in a neighbourhood of a point $p$ on $M$,
$$
\begin{equation*}
\frac{\partial \rho_j}{\partial y}=\biggl(\frac{\partial \rho_j}{\partial y_1}, \dots,\frac{\partial \rho_j}{\partial y_{N}}\biggr),\qquad L^{\alpha} =L_1^{\alpha_1}\cdots L_{\nu}^{\alpha_{\nu}},\quad |\alpha|=\alpha_1+\dots+\alpha_{\nu}.
\end{equation*}
\notag
$$
Definition 2. A manifold $M$ is called $d$-non-degenerate at point $p$ if there exists a natural number $d$ such that the complex linear span of the system of the vectors $\{L^{\alpha}(\partial \rho_j/\partial y)(p,\bar{p}), \, |\alpha|\leqslant d, \, 1\leqslant j \leqslant k\}$ is $\mathbb{C}^N$. A manifold $M$ is called finitely non-degenerate at a point $p$ if it is $d$-non-degenerate at $p$ for some $d$. We use the definition of finite non-degeneracy from the book [10]. Note that the hypersurface $\Gamma$ constructed in § 2 is not finitely non-degenerate at the origin. Therefore, to apply the construction from [7], we will write a germ of a different germ (which is finite non-degenerate at the origin) of a $\mathrm{CR}$ manifold $\mathcal{M}$ whose holomorphic automorphisms are capable of realizing the Lie algebra $\mathfrak{h}$. Let $(z_1,\dots,z_n,\, x_1,\dots,x_n,\, t_1,\dots,t_m$, $w_1=u_1+iv_1$, $w_2=u_2+iv_2$, $w_3=u_3+iv_3)$ be coordinates in $\mathbf{C}^{2n+m+3}$. Let $z$ be the vector $(z_1,\dots,z_n)$, let $t$ be the vector $(t_1,\dots,t_m)$, let $x=(x_1,\dots,x_n)$, and let $w=u+iv$ be the vector $(w_1=u_1+iv_1$, $w_2=u_2+iv_2$, $w_3=u_3+iv_3)$. Next, let We also set
$$
\begin{equation*}
\begin{aligned} \, Q(z,\bar{z},t,\bar{t}\,) &=2 \operatorname{Re} \bigl(Q_1\bigl(\overline{C(t)}z,C(t)\bar{z}\bigr)\bigr), \\ R(x,\bar{x},t,\bar{t},u_3) &=P(x+\mathbf{1},\bar{x}+\mathbf{1},t,\bar{t}\,) +u_3P^4(x+\mathbf{1},\bar{x}+\mathbf{1},t,\bar{t}\,) \\ &\qquad-P(x+\mathbf{1},\mathbf{1},t,0)-P(\mathbf{1},\bar{x}+\mathbf{1},0, \bar{t}\,) -u_3P^4(\mathbf{1},\mathbf{1},0,0), \end{aligned}
\end{equation*}
\notag
$$
where $\mathbf{1}=(1,\dots,1)$ is the vector of $n$ ones, and $C(t)$ are the matrices defined in § 2. Note that near the point $(x_0,\bar{x}_0,t_0,\bar{t}_0,u_0)=(\mathbf{1},\mathbf{1},0,0,0) $ the function $R$ equals to the expression $\bigl(P(x,\bar{x},t,\bar{t}\,)+u_3P^4(x,\bar{x},t,\bar{t}\,)\bigr)$ from which the pluriharmonic terms are subtracted. Consider the germ $\mathcal{M}_0$ at the origin of the manifold $\mathcal{M} \subset \mathbb{C}^{2n+m+3}$ given near the origin by the system of equations
$$
\begin{equation*}
\begin{aligned} \, v_1 &=P(z,\bar{z},t,\bar{t}\,)+u_1P^4(z,\bar{z},t,\bar{t}\,), \\ v_2 &=Q(z,\bar{z},t,\bar{t}\,)+u_2Q^4(z,\bar{z},t,\bar{t}\,), \\ v_3 &=R(x,\bar{x},t,\bar{t},u_3). \end{aligned}
\end{equation*}
\notag
$$
The origin in the space $\mathbb{C}^{2n+m+3}$ is denoted by $\mathbf{0}$. Proof. Let
As a basis for $\mathrm{CR}$ vector fields on $\mathcal{M}$ we can take the set (see formula (1.6.2) in [10])
$$
\begin{equation*}
\begin{aligned} \, L_{1 j} &=\frac{\partial}{\partial \bar{z}_j} +\frac{2i}{1+iP^4}\, \frac{\partial \rho_1}{\partial \bar{z}_j}\, \frac{\partial}{\partial \bar{w}_1} +\frac{2i}{1+iQ^4}\, \frac{\partial \rho_2}{\partial \bar{z}_j}\, \frac{\partial}{\partial \bar{w}_2}, \\ L_{2 j} &=\frac{\partial}{\partial \bar{t}_j}+\frac{2i}{1+iP^4}\, \frac{\partial \rho_1}{\partial \bar{t}_j}\, \frac{\partial}{\partial \bar{w}_1} +\frac{2i}{1+iQ^4}\, \frac{\partial \rho_2}{\partial \bar{t}_j}\, \frac{\partial}{\partial \bar{w}_2} +\frac{2i}{1+2i(\partial R/\partial \bar{w}_3)}\, \frac{\partial \rho_3}{\partial \bar{t}_j}\, \frac{\partial}{\partial \bar{w}_3}, \\ L_{3 j} &=\frac{\partial}{\partial \bar{x}_j}+\frac{2i}{1+2i(\partial R/\partial \bar{w}_3)}\, \frac{\partial \rho_3}{\partial \bar{x}_j}\, \frac{\partial}{\partial \bar{w}_3}. \end{aligned}
\end{equation*}
\notag
$$
Let $y=(z;x;t;w)=(z_1,\dots,z_n;x_1,\dots,x_n;t_1,\dots,t_m;w_1,w_2,w_3)$ be a vector coordinate in $\mathbb{C}^{2n+m+3}$. We claim that the linear span of the set
$$
\begin{equation*}
\biggl\{\frac{\partial \rho_{k}}{\partial y},\, L_{1 j}^{16n+j+3} \biggl(\frac{\partial \rho_2}{\partial y}\biggr), \, L_{3 j}L_{3 l}^{16n+4l} \biggl(\frac{\partial \rho_3}{\partial y}\biggr), \, L_{3 j} \biggl(\frac{\partial \rho_3}{\partial y}\biggr)\biggr\}
\end{equation*}
\notag
$$
for all admissible values of $j$, $l$, $k$ is $\mathbb{C}^{2n+m+3}$. This means that $M$ is $d$-non-degenerate at the origin for $d=\max(16n+m+3,20n+1)$.
Let us first show that the system of vectors
$$
\begin{equation*}
\biggl\{\frac{\partial \rho_{k}}{\partial y},\, L_{1 j}^{16n+j+3} \biggl(\frac{\partial \rho_2}{\partial y}\biggr), \, L_{3 j}L_{3 l}^{16n+4l}\biggl(\frac{\partial \rho_3}{\partial y}\biggr)\biggr\}
\end{equation*}
\notag
$$
has rank $n+m+3$. For each of the variables $w$, $z$, $t$, $x$, we have:
1) $(\partial \rho_{k}/\partial w_{k})(\mathbf{0}) \neq 0$, the remaining first-order partial derivatives of the functions $\rho_{k}$ vanishing; 2) the expressions
$$
\begin{equation*}
\frac{\partial^{16n+j+3}}{\partial \bar{z}_j^{16n+j+3}}\,\frac{\partial \rho_2}{\partial z_l}(\mathbf{0})
\end{equation*}
\notag
$$
do not vanish only for $j=l$;
3) writing the expressions of the form $a\bar{x}_l^{16n+4l}\bar{x}^{\beta}t^{\gamma}$ for $|\beta|=|\gamma|=1$ in the equality $\rho_3=0$, we have
$$
\begin{equation*}
\bar{x}_{\nu}^{16n+4l}\biggl(\biggl(\sum_{j=1}^mt_je_j\biggr)\bar{x}\biggr)_{\nu}=0,
\end{equation*}
\notag
$$
where, as above, $e_j$ are the basis elements of the Lie algebra $\mathfrak{h}$, and the lower index $\nu$ at the end of the formula stands for the $\nu$th coordinate of the vector. Consequently,
$$
\begin{equation*}
\frac{\partial}{\partial \bar{x}_{\mu}}\, \frac{\partial^{16n+4l}}{\partial \bar{x}_{\nu}}\, \frac{\partial \rho_3}{\partial t_j}=e_{j \mu \nu},
\end{equation*}
\notag
$$
where $e_{j \mu \nu}$ is the element in the $\mu$th row and $\nu$th column of the matrix $e_j$. But the matrices $e_j$, $1 \leqslant j \leqslant m$, are linearly independent, and hence from the set of the vectors $\{(e_{1 \mu \nu},\dots,e_{m \mu \nu}), 1 \leqslant \mu \leqslant n, \, 1 \leqslant \nu \leqslant n\}$, we can choose $m$ linearly independent vectors.
Now from 1)–3) we find that, in the coordinates $(w_1,\, w_2,\, w_3,\, z_1,\dots,z_n,\, t_1,\dots, t_m,\, x_1,\dots,x_n)$, the linear span of the set
$$
\begin{equation*}
\biggl\{\frac{\partial \rho_{k}}{\partial y},\, L_{1 j}^{16n+j+3} \biggl(\frac{\partial \rho_2}{\partial y}\biggr), \, L_{3 j}L_{3 l}^{16n+4l} \biggl(\frac{\partial \rho_3}{\partial y}\biggr)\biggr\}
\end{equation*}
\notag
$$
for all admissible values of $j$, $l$, $k$ has the form
$$
\begin{equation}
(\,\underbrace{*,\dots,*}_{n+m+3},\underbrace{0,\dots,0}_n\,),
\end{equation}
\tag{3.2}
$$
where $*$ is an arbitrary complex number.
Now let us show that the system $\{L_{3 j}(\partial \rho_3/\partial x), \, 1 \leqslant j \leqslant n\}$ has full rank, which is equal to $n$. This follows from the fact that the function $R(x,\bar{x},t,\bar{t},u_3)$ after the substitution $t=\bar{t}=0$, $u_3=0$ has the form where
$$
\begin{equation}
a_{\nu}=\frac{\partial^2 R}{\partial x_{\nu} \bar{x}_{\nu}}(\mathbf{0}) \neq 0.
\end{equation}
\tag{3.3}
$$
In contrast, these $n$ vectors have the form
$$
\begin{equation}
(\,\underbrace{0,\dots,0}_{n+m+3},\underbrace{*,\dots,*}_n\,).
\end{equation}
\tag{3.4}
$$
It now remains to note that the union of the systems of vectors (3.2) and (3.4) has rank $2n+m+3$. This follows from the presence of the block structure (of the two blocks with zeroes). The proof of Lemma 7 is complete. Theorem 2. The automorphism algebra $\operatorname{aut} \mathcal{M}_0$ is isomorphic to $\mathfrak{h}$. Proof. Let
$$
\begin{equation*}
\begin{aligned} \, \Xi &= 2 \operatorname{Re} \biggl(f_1(z,x,t,w)\, \frac{\partial}{\partial z_1}+\dots+f_n(z,x,t,w)\, \frac{\partial}{\partial z_n} \\ &\qquad\qquad+q_1(z,x,t,w)\, \frac{\partial}{\partial x_1}+\dots+q_n(z,x,t,w)\, \frac{\partial}{\partial x_n} \\ &\qquad\qquad+r_1(z,x,t,w)\, \frac{\partial}{\partial t_1}+\dots+r_m(z,x,t,w)\, \frac{\partial}{\partial t_m} \\ &\qquad\qquad+g_1(z,x,t,w)\, \frac{\partial}{\partial w_1}+g_2(z,x,t,w)\, \frac{\partial}{\partial w_2}+g_3(z,x,t,w)\, \frac{\partial}{\partial w_3}\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $\Xi \in \operatorname{aut} \mathcal{M}_0$. The tangency condition reads as where
$$
\begin{equation*}
\begin{aligned} \, w_1 &=u_1+i\bigl(P(z,\bar{z},t,\bar{t}\,)+u_1P^4(z,\bar{z},t,\bar{t}\,)\bigr), \\ w_2 &=u_2+i\bigl(Q(z,\bar{z},t,\bar{t}\,)+u_2Q^4(z,\bar{z},t,\bar{t}\,)\bigr), \\ w_3 &=u_3+i\bigl(R(x,\bar{x},t,\bar{t},u_3)\bigr), \qquad j=1,2,3. \end{aligned}
\end{equation*}
\notag
$$
The left-hand sides of (3.5) can be expanded in a Taylor series in the vector variables $z$, $\bar{z}$, $x$, $\bar{x}$, $t$, $\bar{t}$, $u$. This fact will be used below. The proof is based on several lemmas. We split the coefficients of the field $\Xi$ into two groups: $\{f_l,r_l,q_l\}$ and $g_j$. The coefficients from these groups will be gradually simplified in the lemmas. Lemmas 8 and 9 deal with the coefficients $g_j$, and Lemmas 10–13 pertain to the coefficients $\{f_l,r_l,q_l\}$; the final simplification is achieved in the remaining lemmas. The notation for the coefficients of a Taylor series introduced in the proof of a lemma is kept only in the proof. We also need the following remark. Remark 6. Let a real analytic manifold $M$ have finite dimensional algebra of infinitesimal holomorphic automorphisms at least at one point. Then $M$ is holomorphically non-degenerate (see [10]). We now proceed with the proof of the lemmas. Lemma 8. $g_j(z,x,t,w)=g_j(w)$ for $j=1,2,3$; that is, the function $g_j(z,x,t,w)$ depends only on $w$. In addition, $g_j(w)$ is a real analytic function. Proof. Let
$$
\begin{equation*}
g_j(z,x,t,w)=\sum_{\lambda, \mu, \nu \geqslant 0}^{\infty} g_{j \lambda \mu \nu}(z,x,t)w_1^{\lambda}w_2^{\mu}w_3^{\nu}.
\end{equation*}
\notag
$$
Substituting $\bar{z}=0$, $\bar{t}=0$ into (3.5), we obtain
$$
\begin{equation*}
\sum_{ \lambda, \mu, \nu \geqslant 0}\bigl(g_{j \lambda \mu \nu}(z,x,t) u_1^{\lambda}u_2^{\mu}u_3^{\nu}-\bar{g}_{j \lambda \mu \nu}(0,0,0) u_1^{\lambda}u_2^{\mu}u_3^{\nu}\bigr)=0,
\end{equation*}
\notag
$$
whence $g_{j \lambda \mu \nu}(z,x,t)=g_{j \lambda \mu \nu} \in \mathbb{R}$, proving Lemma 8. Proof. Let us prove the claim for $g_1,g_2$ and $g_3$ in succession.
1) Writing the expressions of the forms $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1\bar{z}_1^{16n+4}$ and $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} |x_1|^2$, where $a=\mathrm{const}$, in the equality $\Xi(\rho_1)=0$, we have
$$
\begin{equation*}
g_{1 \lambda (\mu+1) \nu}u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1\bar{z}_1^{16n+4}=0, \qquad a_1g_{1 \lambda \mu (\nu+1)}u_1^{\lambda}u_2^{\mu}u_3^{\nu} |x_1|^2=0,
\end{equation*}
\notag
$$
where $a_1$ is given by (3.3) with $\nu=1$. Hence $g_{1 \lambda \mu \nu}=0$ for $\mu, \nu >0$. This means that $g_1$ depends only on $w_1$.
2) Writing the expressions of the forms $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1^{16n+2}\bar{z}_1^{16n+3}$ and $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} |x_1|^2$, where $a=\mathrm{const}$, in the equality $\Xi(\rho_2)=0$, we have
$$
\begin{equation*}
g_{2 (\lambda+1) \mu \nu}u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1^{16n+2}\bar{z}_1^{16n+3}=0, \qquad a_1g_{2 \lambda \mu (\nu+1)}u_1^{\lambda}u_2^{\mu}u_3^{\nu} |x_1|^2=0,
\end{equation*}
\notag
$$
where $a_1$ is given by (3.3) with $\nu=1$.
Hence $g_{2 \lambda \mu \nu}=0$ for $\lambda, \nu >0$. This means that $g_2$ depends only on $w_2$. 3) Writing the expressions of the form $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1^{16n+2}\bar{z}_1^{16n+3}$ and of the form $a u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1\bar{z}_1^{16n+4}$, where $a=\mathrm{const}$, in the equality $\Xi(\rho_3)=0$, we have
$$
\begin{equation*}
g_{3 (\lambda+1) \mu \nu}u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1^{16n+2}\bar{z}_1^{16n+3}=0, \qquad g_{3 \lambda (\mu+1) \nu}u_1^{\lambda}u_2^{\mu}u_3^{\nu} z_1\bar{z}_1^{16n+4}=0,
\end{equation*}
\notag
$$
where $a_1$ is given by (3.3) with $\nu=1$.
Hence $g_{3 \lambda \mu \nu}=0$ for $\lambda, \mu >0$. This means that $g_3$ depends only on $w_3$. The proof of Lemma 9 is complete. Lemma 10. $f_l(z,x,t,w)=f_l(z,t,w)$, $r_j(z,x,t,w)=r_j(z,t,w)$ for $1\leqslant l \leqslant n$, $1\leqslant j \leqslant m$; that is, the functions $f_l(z,x,t,w)$ and $r_j(z,x,t,w)$ do not depend on $x$. Proof. Let
$$
\begin{equation*}
f_l=\sum_{\alpha, \mu, \nu}f_{l \alpha \mu \nu}(z,t,w_1)x^{\alpha}w_2^{\mu}w_3^{\nu}, \qquad r_j=\sum_{\alpha, \mu, \nu}r_{j \alpha \mu \nu}(z,t,w_1)x^{\alpha}w_2^{\mu}w_3^{\nu}.
\end{equation*}
\notag
$$
By $X_{\alpha \mu \nu}$ we denote the holomorphic vector field
$$
\begin{equation*}
\begin{aligned} \, &f_{1 \alpha \mu \nu}(z,t,w_1)\, \frac{\partial}{\partial z_1} +\dots+f_{n \alpha \mu \nu }(z,t,w_1)\, \frac{\partial}{\partial z_n} \\ &\qquad+r_{1 \alpha \mu \nu}(z,t,w_1)\, \frac{\partial}{\partial t_1} +\dots+r_{m \alpha \mu \nu}(z,t,w_1)\, \frac{\partial}{\partial t_m}. \end{aligned}
\end{equation*}
\notag
$$
For $|\alpha|>0$, we write all the summands in the equality $\Xi(\rho_1)=0$ of the form $F(z,\bar{z},t,\bar{t},u_1)x^{\alpha}w_2^{\mu}w_3^{\nu}$. We have
This means that the field $X_{\alpha \mu}$ is tangent to the hypersurface $\rho_1=0$. But from Theorem 1 and Remark 6 it follows that this hypersurface is holomorphically non-degenerate, hence $X_{\alpha \mu \nu}=0$ for all $\alpha$. Thus, $f_l$ and $r_j$ do not depend on $x$. The proof of Lemma 10 is complete. Similarly, considering the equality $\Xi(\rho_3)=0$, we arrive at the following result. Lemma 11. $q_l(z,x,t,w)=q_l(x,t,w)$, $r_j(z,t,w)=r_j(t,w)$ for $1\leqslant l \leqslant n$, $1\leqslant j \leqslant m$; that is, the functions $q_l(z,x,t,w)$ and $r_j(z,t,w)$ do not depend on $z$. Lemma 12. $q_l(x,t,w)=q_l(x,t,w_3)$, $r_j(t,w)=r_j(t,w_3)$ for $1\leqslant l \leqslant n$, $1\leqslant j \leqslant m$; that is, the functions $q_l(x,t,w)$ and $r_j(t,w)$ do not depend on $w_1$ and $w_2$. Proof. Let
$$
\begin{equation*}
q_l=\sum_{\mu, \nu}q_{l \mu \nu}(x,t,w_3)w_1^{\mu}w_2^{\nu}, \qquad r_j=\sum_{\mu, \nu}r_{j \mu \nu}(t,w_3)w_1^{\mu}w_2^{\nu}.
\end{equation*}
\notag
$$
We denote by $X_{\mu \nu}$ the holomorphic vector field For $\nu>0$, writing the expressions in the equality $\Xi(\rho_3)=0$ of the form we have
For $\nu>0$, writing the expressions of the form
$$
\begin{equation*}
F(x,\bar{x},t,\bar{t},u_3)u_1^{\mu}u_2^{\nu}z_1\bar{z}^{16n+4}
\end{equation*}
\notag
$$
in the equality $\Xi(\rho_3)=0$, we have
$$
\begin{equation}
u_1^{\mu}u_2^{\nu-1}z_1\bar{z}^{16n+4}\bigl(i\nu X_{\mu \nu}(R)-i\nu \overline{X}_{\mu \nu}(R)\bigr)=0.
\end{equation}
\tag{3.7}
$$
From (3.6) and (3.7) we obtain $X_{\mu \nu}R=0$ for $\nu>0$. This means that the holomorphic vector field $X_{\mu \nu}$ is tangent to the hypersurface given by the equation $\{v_3=R(x,\bar{x},t,\bar{t},u_3)\}$. This hypersurface is biholomorphically equivalent to $M$. Hence it is holomorphic non-degenerate by Theorem 1 and Remark 6. Therefore, $X_{\mu \nu}=0$ for $\nu>0$, that is, $X_{\mu \nu}$ does not depend on $w_2$. Similarly, for $\mu>0$, writing all expressions of the form $F(x,\bar{x},t,\bar{t},u_3)u_1^{\mu}$ in the equation $\Xi(\rho_3)=0$, we have
$$
\begin{equation}
u_1^{\mu}\bigl(X_{\mu \nu}(R)+\overline{X}_{\mu \nu}(R)\bigr)=0.
\end{equation}
\tag{3.8}
$$
Next, for $\mu>0$, writing all expressions of the form
$$
\begin{equation*}
F(x,\bar{x},t,\bar{t},u_3)u_1^{\mu}z_1^{16n+2}\bar{z}_1^{16n+3}
\end{equation*}
\notag
$$
in $\Xi(\rho_3)=0$, we have
$$
\begin{equation}
u_1^{\mu-1}z_1^{16n+2}\bar{z}_1^{16n+3}\bigl(i\mu X_{\mu \nu}(R)-i\mu \overline{X}_{\mu \nu}(R)\bigr)=0.
\end{equation}
\tag{3.9}
$$
From (3.8) and (3.9) we obtain $X_{\mu \nu}R=0$ for $\mu>0$. Hence, arguing as in the proof of this lemma, we find that $X_{\mu \nu}$ also does not depend on $w_1$. The proof of Lemma 12 is complete. Lemma 13. $f_l(z,t,w)=f_l(z,t,w_1)$, $r_j(t,w_3)=r_j(t)$ for $1\leqslant l \leqslant n$, $1\leqslant j \leqslant m$; that is, the function $f_l(z,t,w)$ does not depend on $w_2$ and $w_3$, and the function $r_j(t,w_3)$ does not depend on $w_3$. Proof. Let We denote by $X_{\mu \nu}$ the holomorphic vector field
Also let
$$
\begin{equation*}
f_{l \mu \nu}=\sum_{\alpha}f_{l \alpha \mu \nu}(t,w_1)z^{\alpha}.
\end{equation*}
\notag
$$
For $\mu>0$, writing the expressions of the form $F(t,u_1)u_2^{\mu}u_3^{\nu}z^{\alpha}z_l^{16n+l+2}\bar{z}_l$ in the equality $\Xi(\rho_2)=0$ with $|\alpha|>1$, we have
$$
\begin{equation*}
(16n+l+3)u_2^{\mu}u_3^{\nu}z_l^{16n+l+2}\bar{z}_l \biggl(\sum_{|\alpha|>1}f_{l \alpha \mu \nu}(t,u_1)z^{\alpha}\biggr)=0,
\end{equation*}
\notag
$$
whence $f_{l \alpha \mu \nu}=0$ for $|\alpha|>1$.
For $\mu>0$, writing the expressions of the form $F(t,u_1)u_2^{\mu}u_3^{\nu}z^{\alpha}z_l^{16n+l+2}\bar{z}_l$ in the equality $\Xi(\rho_2)=0$ with $|\alpha|=1$, we have
$$
\begin{equation}
u_2^{\mu}u_3^{\nu}z_l^{16n+l+2}\biggl((16n+l+3) \sum_{|\alpha|=1}(f_{l \alpha \mu \nu}(t,u_1)z^{\alpha}\bar{z}_l +\bar{f}_{l \alpha \mu \nu}(0,u_1)z_l\bar{z}^{\alpha})\biggr)=0,
\end{equation}
\tag{3.10}
$$
whence $f_{l \alpha \mu \nu}=0$ for $z^{\alpha} \neq z_l$.
For $\mu>0$, writing the expressions of the form $F(t,u_1)u_2^{\mu}u_3^{\nu}\bar{z}_l^{16n+l+3}z_l$ in the equality $\Xi(\rho_2)=0$., we have
$$
\begin{equation}
u_2^{\mu}u_3^{\nu}\bar{z}_l^{16n+l+2}\bigl(f_{l \alpha \mu \nu}(t,u_1)z_l\bar{z}_l +(16n+l+3)\bar{f}_{l \alpha \mu \nu}(0,u_1)z_l\bar{z}_l\bigr)=0,
\end{equation}
\tag{3.11}
$$
where $\alpha$ is such that $z^{\alpha} = z_l$. From (3.10) and (3.11) we obtain that $f_{l \alpha \mu \nu}=0$ for $z^{\alpha} = z_l$, that is, $f_{l \alpha \mu \nu}=0$ for $|\alpha|=1$.
It suffices to consider the case $|\alpha|=0$. For $\mu>0$, writing the expressions of the form $F(t,u_1)u_2^{\mu}u_3^{\nu}z_l^{16n+l+2}\bar{z}_l$ in the equality $\Xi(\rho_2)=0$., we have
$$
\begin{equation*}
(16n+l+3)u_2^{\mu}u_3^{\nu}z_l^{16n+l+2}\bar{z}_lf_{l 0 \mu \nu}(t,u_1)=0,
\end{equation*}
\notag
$$
whence $f_{l 0 \mu \nu}=0$.
Thus, $f_{l \alpha \mu \nu}=0$ for all $\alpha$, that is, $f_l$ does not depend on $w_2$. Next, let
$$
\begin{equation*}
f_l=\sum_{\nu}f_{l \nu}(z,t,w_1)w_3^{\nu}, \qquad r_j=\sum_{\nu}r_{j \nu}(z,t)w_3^{\nu}.
\end{equation*}
\notag
$$
We denote by $X_{\nu}$ the holomorphic vector field
$$
\begin{equation*}
f_{1 \nu}(z,t,w_1)\, \frac{\partial}{\partial z_1}+\dots+f_{n \nu}(z,t,w_1)\, \frac{\partial}{\partial z_n}+r_{1 \nu}(z,t)\, \frac{\partial}{\partial t_1} +\dots+r_{m \nu}(z,t) \, \frac{\partial}{\partial t_m}.
\end{equation*}
\notag
$$
For $\nu>0$, writing the expressions of the form $F(z,\bar{z},t,\bar{t},u_1)u_3^{\nu}$ and of the form $F(z,\bar{z},t,\bar{t},u_1)u_3^{\nu}|x_1|^2$ in the equality $\Xi(\rho_1)=0$, we have
$$
\begin{equation}
u_3^{\nu}(X_{\nu}\rho_1+\overline{X}_{\nu}\rho_1) =0 \quad \text{for} \quad \rho_1 =0,
\end{equation}
\tag{3.12}
$$
$$
\begin{equation}
a_1|x_1|^2u_3^{\nu-1}(i\nu X_{\nu}\rho_1-i\nu \overline{X}_{\nu}\rho_1) =0 \quad \text{for} \quad \rho_1 =0,
\end{equation}
\tag{3.13}
$$
where $a_1$ is given by (3.3) with $\nu=1$.
From (3.12) and (3.13), we have $X_{\nu}\rho_1=0$ for $\nu>0$ and $\rho_1=0$. As a result, the holomorphic vector field $X_{\nu}$ is tangent to the hypersurface $\Gamma$. This hypersurface is holomorphically non-degenerate. Hence $X_{\nu}=0$ for $\nu>0$, that is, $X_{\nu}$ does not depend on $w_3$, and therefore, $f_l$, $r_j$ also do not depend on $w_3$. The proof of Lemma 13 is complete. Lemma 14. $g_1(w_1)=0$, $f_l(z,t,w_1)=f_l(z)$ for $1 \leqslant l \leqslant n$; that is, the function $f_l(z,t,w_1)$ depends only on $z$, and $f_l(z,t,w_1)$ is a real linear function. Proof. In view of the above lemmas, using the condition $\Xi \in \operatorname{aut} \mathcal{M}_0$, we find that the field belongs to $\operatorname{aut} \Gamma_0$.
Thus, we are in the situation of Theorem 1. Now the required result follows from Lemmas 1–5. The proof of Lemma 14 is complete. Proof. We set
Let us show by induction on $\nu$ that $g_{\nu}=0$ for all $\nu > 0$. The induction base. Since $f_l$ and $r_l$ do not depend on $w_2$, the coefficient of the monomial $z_1\bar{z}_1^{16n+4}$ in $\Xi(\rho_2)=0$ is equal to $g_1$, whence $g_1=0$. The induction step. Assume that $g_1 = {\cdots} = g_{\nu} = 0$. We claim that $g_{\nu+1} = 0$. Since $f_l$ and $r_l$ do not depend on $w_2$, the coefficient of the monomial $u_2^{\nu}z_1\bar{z}_1^{16n+4}$ in $\Xi(\rho_2)=0$ is equal to $(\nu+1) g_{\nu+1}$, whence $g_{\nu+1}=0$. Hence $g_2(w_2) = g_0 = \mathrm{const}$. But the coefficient of the monomial $z_1^4\bar{z}_1^{64n+16}$ is $-g_0$, whence $g_2(w_2)=g_0=0$. The proof of Lemma 15 is complete. Proof. Consider the functions
The vector field
$$
\begin{equation*}
\begin{aligned} \, &2 \operatorname{Re} \biggl(f_1(z)\, \frac{\partial}{\partial z_1} +\dots+ f_n(z)\, \frac{\partial}{\partial z_n}+f_1(x-1)\, \frac{\partial}{\partial x_1}+\dots+ f_n(x-1)\, \frac{\partial}{\partial x_n} \\ &\qquad\qquad +r_1(t)\, \frac{\partial}{\partial t_1} +\dots+r_m(t)\, \frac{\partial}{\partial t_m} \biggr) \end{aligned}
\end{equation*}
\notag
$$
belongs to $\operatorname{aut} \mathcal{M}_0$. Hence the vector field
$$
\begin{equation*}
2 \operatorname{Re} \biggl(\hat{q}_1(x,t,w_3)\, \frac{\partial}{\partial x_1}+\dots+ \hat{q}_n(x,t,w_3)\, \frac{\partial}{\partial x_n}+g_3(w_3)\, \frac{\partial}{\partial w_3}\biggr)
\end{equation*}
\notag
$$
also belongs to $\operatorname{aut} \mathcal{M}_0$. We need to prove that $\hat{q}_l(x,t,w_3)=0$, $g_3(w_3)=0$.
Let
$$
\begin{equation*}
\hat{q}_1(x,t,w_3)=\sum_{\mu=0}^{\infty}\sum_{\nu=0}^{\infty}q_{\mu \nu}x_1^{\mu}w_3^{\nu} +\widetilde{q}_1(x,t,w_3), \qquad g_3=\sum_{\nu=0}^{\infty}g_{\nu}w_3^{\nu},
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\frac{\partial^{\mu+\nu} \widetilde{q}_1}{\partial x_1^{\mu}\, \partial w_3^{\nu}}(\mathbf{0})=0
\end{equation*}
\notag
$$
for all $\mu,\nu \geqslant 0$.
Writing the monomials of the form $ax_1$, $ax_1\bar{x}_1$, $ax_1^2\bar{x}_1$, $ax_1\bar{x}_1^2$, $ax_1^3\bar{x}_1$, $ax_1^2\bar{x}_1^2$, $ax_1\bar{x}_1^3$, $ax_1^4\bar{x}_1$, $ax_1^3\bar{x}_1^2$, $ax_1^2\bar{x}_1^3$, $ax_1\bar{x}_1^4$ in the equality $\Xi(\rho_3)=0$, we get the system of linear equations of full rank in the variables $b_1$, $q_{00}$, $q_{10}$, $q_{20}$, $q_{30}$, $q_{40}$, whence $b_1 = q_{00} = q_{10} = q_{20} = q_{30} = q_{40}=0$. Let us give a detailed argument. Let
$$
\begin{equation*}
\begin{alignedat}{5} A_1 &= \frac{\partial^2 \hat{q}_1}{\partial x_1\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_2 &=\frac{\partial^3 \hat{q}_1}{\partial x_1^2\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_3 &=\frac{\partial^4 \hat{q}_1}{\partial x_1^2\, \partial \bar{x}_1^2}(\mathbf{0}), \\ A_4 &= \frac{\partial^4 \hat{q}_1}{\partial x_1^3\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_5 &= \frac{\partial^5 \hat{q}_1}{\partial x_1^4\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_6 &= \frac{\partial^5 \hat{q}_1}{\partial x_1^3\, \partial \bar{x}_1^2}(\mathbf{0}). \end{alignedat}
\end{equation*}
\notag
$$
The values $A_1,\dots,A_6$ are easy to compute, each of them is a polynomial of $n$ with natural coefficients of degree at most five. We do not give the explicit formulas, and note only that all $A_1,\dots,A_6$ are non-zero. Writing the monomials of the form $ax_1$, we obtain the equation whence $q_{00}=0$.Writing the monomials of the form $ax_1^2\bar{x}_1$, $ax_1\bar{x}_1^2$, $ax_1^3\bar{x}_1$, $ax_1^2\bar{x}_1^2$, we get the system of equations
$$
\begin{equation*}
\begin{gathered} \, 2 A_1 (q_{10}-g_1+\bar{q}_{10})=0,\qquad 4 A_1 q_{20}+4 A_2(2 q_{10}-g_1+\bar{q}_{10})=0, \\ 4 A_1 \bar{q}_{20}+4 A_2( 2 \bar{q}_{10}-g_1+q_{10})=0,\quad 8 A_2 (q_{20} +\bar{q}_{20})+8 A_3 (2 q_{10}-g_1+2 \bar{q}_{10})=0, \end{gathered}
\end{equation*}
\notag
$$
whence $g_1=\operatorname{Re} q_{10}=\operatorname{Re} q_{20}=0$.
Next, writing the monomials of the form $ax_1^3\bar{x}_1$, we have
$$
\begin{equation*}
12 A_1q_{30} +24 i A_4 \operatorname{Im} q_{10}+24 i A_2 \operatorname{Im} q_{20}=0,
\end{equation*}
\notag
$$
whence $\operatorname{Re} q_{30}=0$.
Writing the monomials of the form $ax_1^4\bar{x}_1$, we have
$$
\begin{equation*}
48 A_1q_{40}+144 i A_5 \operatorname{Im} q_{10} +144 i A_4 \operatorname{Im} q_{20}+96 i A_2 \operatorname{Im} q_{30}=0,
\end{equation*}
\notag
$$
whence $\operatorname{Re} q_{40}=0$.
Next, writing the monomials of the form $ax_1^2\bar{x}_1$, $ax_1^3\bar{x}_1$, $ax_1^4\bar{x}_1$, $ax_1^3\bar{x}_1^2$, we obtain the system of equations
$$
\begin{equation}
\begin{gathered} \, 4 i A_2 \operatorname{Im} q_{10}+4 i A_1 \operatorname{Im} q_{20}=0, \\ 24 i A_4 \operatorname{Im} q_{10}+24 i A_2 \operatorname{Im} q_{20} +12 i A_1 \operatorname{Im} q_{30}=0, \\ 144 i A_5 \operatorname{Im} q_{10}+144 i A_4 \operatorname{Im} q_{20} +48 i A_1 \operatorname{Im} q_{40}+96 i A_2 \operatorname{Im} q_{30}=0, \\ 24 i A_6 \operatorname{Im} q_{10}+48 i A_3 \operatorname{Im} q_{20} -24 i A_4 \operatorname{Im} q_{20}+24 i A_2 \operatorname{Im} q_{30}=0, \end{gathered}
\end{equation}
\tag{3.14}
$$
whence $\operatorname{Im} q_{10}=\operatorname{Im} q_{20} =\operatorname{Im} q_{30}=\operatorname{Im} q_{40}=0$.
So, $b_1 = q_{00} = q_{10} = q_{20} = q_{30} = q_{40}=0$. Now, let us prove by induction on $\mu$ that $q_{\mu 0}=0$ for all $\mu \geqslant 4$. The induction base was already verified. Let us proceed with the induction step. Let $q_{00} = q_{10} = \dots = q_{(\mu-1)0} = q_{\mu 0}=0$. We claim that $q_{(\mu+1) 0}=0$. To this end, we write the monomials of the form $ax_1^{\mu+1}\bar{x}_1$. We have whence $q_{\mu 0}=0$ for all $\mu$.Next, we prove by induction on $\nu$ that $b_{(\nu+1)}=q_{\mu \nu}=0$ for all $\nu \geqslant 0$ and all $\mu$. The induction base was already proved. Let us now proceed with the induction step. Assume that
$$
\begin{equation*}
q_{1 \nu} = \dots = q_{(\mu-1) \nu} = q_{\mu \nu}=0,\qquad b_1=\dots=b_{\nu}=0.
\end{equation*}
\notag
$$
Let us show that $q_{\mu (\nu+1)}=0$ for all $\mu$. With this aim, we write the monomials of the form $ax_1$, $ax_1\bar{x}_1w_3^{\nu}$, $ax_1^2\bar{x}_1w_3^{\nu}$, $ax_1\bar{x}_1^2w_3^{\nu}$, $ax_1^3\bar{x}_1w_3^{\nu}$, $ax_1^2\bar{x}_1^2w_3^{\nu}$, $ax_1\bar{x}_1^3w_3^{\nu}$, $ax_1^4\bar{x}_1w_3^{\nu}$, $ax_1^3\bar{x}_1^2w_3^{\nu}$, $ax_1^2\bar{x}_1^3w_3^{\nu}$, $ax_1\bar{x}_1^4w_3^{\nu}$ in the equality $\Xi(\rho_3)=0$. As a result, we obtain the system of linear equations of full rank in the variables $b_{\nu+1}$, $q_{0 \nu}$, $q_{1 \nu}$, $q_{2 \nu}$, $q_{3 \nu}$, $q_{4 \nu}$, whence $b_{\nu+1} = q_{0 \nu} = q_{1 \nu} = q_{2 \nu} = q_{3 \nu} = q_{4 \nu}=0$. The proof is precisely similar to that for the case $\nu=0$. Next, writing the monomials of the form $ax_1^{\mu+1}\bar{x}_1w_3^{\nu}$, we obtain $q_{\mu \nu}=0$ for all $\mu$ (where we also proceed as in the case $\nu=0$).
Thus,
$$
\begin{equation*}
\frac{\partial^{\mu+\nu} \hat{q}_1}{\partial x_1^{\mu}w_3^{\nu}}(x,t,w_3)=0\quad \text{for all }\mu, \nu\text{ and }g_3(w_3)=0.
\end{equation*}
\notag
$$
(Note that if $n=1$, then everything is proved.)
In the same way, it can be shown that
$$
\begin{equation*}
\frac{\partial^{\mu+\nu} \hat{q}_l}{\partial x_1^{\mu}w_3^{\nu}}(x,t,w_3)=0\quad \text{for}\quad 2 \leqslant l \leqslant n
\end{equation*}
\notag
$$
(however, the arguments in this case can be simplified if we use the equality $g_3(w_3)=\mathbf{0}$). The only difference is that instead of the numbers $A_1,\dots,A_6$ we need to substitute the numbers
$$
\begin{equation*}
\begin{alignedat}{5} A_1(l) &= \frac{\partial^2 \hat{q}_l}{\partial x_1\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_2(l) &= \frac{\partial^3 \hat{q}_l}{\partial x_1^2\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_3(l) &= \frac{\partial^4 \hat{q}_l}{\partial x_1^2\, \partial \bar{x}_1^2}(\mathbf{0}), \\ A_4(l) &= \frac{\partial^4 \hat{q}_l}{\partial x_1^3\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_5(l) &= \frac{\partial^5 \hat{q}_l}{\partial x_1^4\, \partial \bar{x}_1}(\mathbf{0}), &\qquad A_6(l) &= \frac{\partial^5 \hat{q}_l}{\partial x_1^3\, \partial \bar{x}_1^2}(\mathbf{0}). \end{alignedat}
\end{equation*}
\notag
$$
Now we claim that
$$
\begin{equation*}
\frac{\partial \hat{q}_1}{\partial x_2}(\mathbf{0})=0.
\end{equation*}
\notag
$$
Let
$$
\begin{equation*}
\begin{alignedat}{5} a_1 &= \frac{\partial \hat{q}_1}{\partial x_2}(\mathbf{0}), &\qquad a_2 &=\frac{\partial^2 \hat{q}_1}{\partial x_1 x_2}(\mathbf{0}), &\qquad a_3 &=\frac{\partial^2 \hat{q}_1}{\partial x_2^2}(\mathbf{0}), \\ a_4 &= \frac{\partial \hat{q}_2}{\partial x_1}(\mathbf{0}), &\qquad a_5 &=\frac{\partial^2 \hat{q}_1}{\partial x_1 x_2}(\mathbf{0}), &\qquad a_6 &=\frac{\partial^2 \hat{q}_1}{\partial x_1^2}(\mathbf{0}). \end{alignedat}
\end{equation*}
\notag
$$
Writing the monomials of the form $az_2z_1 \bar{z}_1^2$ in the equality $\Xi(\rho_3)=0$, we have whence $a_2=0$.Writing the monomials of the form $az_2z_1 \bar{z}_2^2$ in the equality $\Xi(\rho_3)=0$, we have whence $a_5=0$.Next, writing the monomials of the form $az_1^2\bar{z}_2$, $az_2^2\bar{z}_1$, $az_1z_2\bar{z}_1$, $az_1z_2\bar{z}_2$ in the equality $\Xi(\rho_3)=0$, we obtain the system
$$
\begin{equation*}
\begin{gathered} \, 4 A_1(2) a_6 + 4 A_2(1) \bar{a}_1 = 0,\qquad 4 A_1(1) a_3 + 4 A_2(2) \bar{a}_4 = 0, \\ 2 A_1(1) a_2 + 4 A_2(1) a_1 = 0,\qquad 2 A_1(2) a_5 + 4 A_2(1) a_4 = 0, \end{gathered}
\end{equation*}
\notag
$$
whence $a_1=a_3=a_4=a_6=0$.
Proceeding similarly and replacing 2 by $l$ in the above arguments, we have
$$
\begin{equation*}
\frac{\partial \hat{q}_1}{\partial x_l}(\mathbf{0})=0.
\end{equation*}
\notag
$$
A similar argument shows that, for all admissible $j$, $l$,
$$
\begin{equation*}
\frac{\partial \hat{q}_j}{\partial x_l}(\mathbf{0})=0.
\end{equation*}
\notag
$$
Now, let us prove by induction on $|\alpha|$ that
$$
\begin{equation*}
\frac{\partial^{\alpha} \hat{q}_j}{\partial x^{\alpha}}(\mathbf{0})=0.
\end{equation*}
\notag
$$
The induction bases was already verified. Let us proceed with the induction step.
Let
$$
\begin{equation*}
\frac{\partial^{\alpha} \hat{q}_j}{\partial x^{\alpha}}(\mathbf{0})=0 \quad\text{for all } \alpha, \ \ |\alpha|< \nu.
\end{equation*}
\notag
$$
We claim that $\frac{\partial^{\alpha} \hat{q}_j}{\partial x^{\alpha}}(\mathbf{0})=0$ for all $\alpha$, $|\alpha| = \nu$. To this end, we write the monomials of the form $x^{\alpha}\bar{x}_j$ in the equality $\Xi(\rho_3)=0$ for $|\alpha| = \nu$. As a result, we have
$$
\begin{equation*}
A_1(j)\, \frac{\partial^{\alpha} \hat{q}_j}{\partial x^{\alpha}} (\mathbf{0})\, x^{\alpha}\bar{x}_j=0,
\end{equation*}
\notag
$$
which gives that
$$
\begin{equation*}
\frac{\partial^{\alpha} \hat{q}_j}{\partial x^{\alpha}}(\mathbf{0})=0 \quad\text{for } |\alpha| = \nu.
\end{equation*}
\notag
$$
A similar argument involving induction on $|\beta|$ shows that
$$
\begin{equation*}
\frac{\partial^{\alpha+\beta} \hat{q}_j}{\partial x^{\alpha}t^{\beta}}(\mathbf{0})=0\quad \text{for all }\alpha.
\end{equation*}
\notag
$$
As a concluding step, using induction on $\nu$, we verify that
$$
\begin{equation*}
\frac{\partial^{\alpha+\beta+\nu} \hat{q}_j}{\partial x^{\alpha}t^{\beta}w_3^{\nu}}(\mathbf{0})=0\quad \text{for all }\alpha\text{ and }\beta.
\end{equation*}
\notag
$$
Consequently,
$$
\begin{equation*}
\frac{\partial \hat{q}_j}{\partial x^{\alpha}t^{\beta}w_3^{\nu}}(\mathbf{0})=0 \quad\text{for all }\alpha, \beta, \nu,
\end{equation*}
\notag
$$
and hence $\hat{q}_j(x,t,w_3)=0$ for all $j$.
The proof of Lemma 16 is complete. Now from Lemma 16 we obtain that an infinitesimal automorphism of the germ $\mathcal{M}_0$ is completely defined by the vector-valued coefficient of $\partial/\partial z$ of the vector field, as in Theorem 1. From this we obtain the required isomorphism of the algebras. The proof of Theorem 2 is complete. Now we will construct the desired system of equations. The manifold $\mathcal{M}$ is finitely non-degenerate at the origin, and hence the construction of [7] applies. As a result, we find a system of partial differential equations whose Lie algebra of symmetries is isomorphic to the complexification of the automorphism algebra of the given $d$-non-degenerate germ. Let us give the definition of the symmetry algebra in the general case. Let $Z=(Z_1,\dots,Z_{N})$ be independent variables in the space $\mathbb{C}^N$, and $f(Z)=(f_1(Z),\dots,f_K(Z))$ be unknown holomorphic functions. Consider the system $\mathcal{S}$ of holomorphic differential equations on the vector-valued function $f(Z)$. Definition 3 (see also [11]). The symmetry group of the system $\mathcal{S}$ is the local Lie group of all complex transformations of a domain in the space $\mathbb{C}_Z^N\times \mathbb{C}_f^K$ of dependent and independent variables which maps a graph of each solution of the system in a graph of a solution of the same system. The Lie symmetry algebra of the system $\mathcal{S}$ is the Lie algebra corresponding to the symmetry group of the system $\mathcal{S}$ (that is, this is the tangent space at the unity of the group). The systems that follow below have a quite specific form, which will be clear from the construction. Namely, their number and order depend on the number $d$, which in our case is $\max(16n+m+3,20n+1)$, as was shown in the proof of Lemma 7. More precisely, the order of the equations is not greater than $d$, and the number of equations is $\bigl(\operatorname{dim} J_d^{2n+m+3} - (2n+m+3)\bigr)$, where $\operatorname{dim} J_d^{2n+m+3}$ is the dimension of the jet space of $(2n+m+3)$ variables of order $\leqslant d$. In addition, each equation is solved with respect to one of the derivatives. Now we turn to the construction of the required system. Let us replace the variables $\bar{z}$, $\bar{x}$, $\bar{t}$, $\bar{w}$ by the new complex variables $\zeta \in \mathbb{C}^n$, $\xi \in \mathbb{C}^n$, $\tau \in \mathbb{C}^m$, $\omega \in \mathbb{C}^3$ respectively. As a result, from $\mathcal{M}$ we can construct its external complexification, that is, the complex analytic manifold in a neighbourhood of the origin in the space $\mathbb{C}^{2(2n+m+3)}$ given by the system of equations
$$
\begin{equation}
\begin{aligned} \, \frac{w_1-\omega_1}{2i} &=P(z,\zeta,t,\tau) +\biggl(\frac{w_1+\omega_1}{2}\biggr)P^4(z,\zeta,t,\tau), \\ \frac{w_2-\omega_2}{2i} &=Q(z,\zeta,t,\tau) +\biggl(\frac{w_2+\omega_2}{2}\biggr)Q^4(z,\zeta,t,\tau), \\ \frac{w_3-\omega_3}{2i} &=R\biggl(x,\xi,t,\tau,\biggl(\frac{w_3+\omega_3}{2}\biggr)\biggr). \end{aligned}
\end{equation}
\tag{3.15}
$$
In equations (3.15), we can consider $w_1$, $w_2$, $w_3$ as dependent variables, the variables $z$, $x$, $t$ as independent variables, and $\zeta$, $\xi$, $\tau$, $\omega$, as parameters. Solving equations (3.15), which are linear with respect to $w_1$, $w_2$, $w_3$, we obtain the following analytic expressions in terms of the dependent variables and parameters:
$$
\begin{equation}
\begin{aligned} \, w_1 &=W_1(z,x,t,\zeta, \xi, \tau, \omega), \\ w_2 &=W_2(z,x,t,\zeta, \xi, \tau, \omega), \\ w_3 &=W_3(z,x,t,\zeta, \xi, \tau, \omega). \end{aligned}
\end{equation}
\tag{3.16}
$$
Let the vector-valued variable $Z$ denote the set $(\zeta, \xi, \tau, \omega)$. Consider all possible differential operators $\partial^{|\alpha|}/\partial Z^{\alpha}$ for $|\alpha|\leqslant d$. From the condition of $d$-non-degeneracy at the origin it follows that there exist multiindices $\alpha_{\nu}$, $1\leqslant \nu \leqslant 2n+m+3$, such that the mapping
$$
\begin{equation}
Z \to \frac{\partial^{|\alpha_{\nu}|}}{\partial Z^{\alpha_{\nu}}}(w_{j(\nu)}-W_{j(\nu)})
\end{equation}
\tag{3.17}
$$
has full rank at the origin. Here, $j(\nu)$ ranges over $1$, $2$ or $3$, and we formally assume that
$$
\begin{equation*}
\frac{\partial^0}{\partial Z^0}(w_{j(\nu)}-W_{j(\nu)})=w_{j(\nu)}-W_{j(\nu)}.
\end{equation*}
\notag
$$
Hence by the implicit mapping theorem, we can express $Z$ as an analytic function of the variables $(z,x,t,w,\frac{\partial^{|\alpha|}}{\partial Z^{\alpha}}w_{j(\nu)})$, that is,
$$
\begin{equation*}
Z=Z\biggl(z,x,t,w,\frac{\partial^{|\alpha|}}{\partial Z^{\alpha}}w_{j(\nu)}\biggr).
\end{equation*}
\notag
$$
The required system reads as
$$
\begin{equation*}
\frac{\partial^{|\alpha|} w_j}{\partial Z^{\alpha}} =\frac{\partial^{|\alpha|}}{\partial Z^{\alpha}}(W_j) \biggl(z,x,t,Z\biggl(z,x,t,w, \frac{\partial^{|\alpha|}}{\partial Z^{\alpha}}w_{j(\nu)}\biggr)\biggr),
\end{equation*}
\notag
$$
where $\alpha$ takes all values with $|\alpha|\leqslant d$, except $\alpha_{\nu}$. Applying the Frobenius theorem to the space of $d$-jets, we find that this system is completely integrable. The family of its solutions is given by the above complexification of the manifold $\mathcal{M}$. Remark 7. The above construction demonstrates that even in the case of a hypersurface (a manifold of codimension one) the corresponding system consists of several equations if the $\mathrm{CR}$-dimension of the hypersurface exceeds one. In addition, the change of the codimension of the initial manifold from one to three only slightly increases the number of equations in the corresponding system. Remark 8. There exist infinite dimensional Lie algebras which cannot be realized as a Lie algebra of vector fields on a finite dimensional space. For example, the Virasoro algebra cannot be realized in vector fields on a finite dimensional manifold (see [12]). In conclusion, we formulate two questions. Question 1. According to the previous remark, not every infinite dimensional Lie algebra can be realized by automorphisms of $\mathrm{CR}$ manifolds. But the list of realizable algebras is fairly wide — it contains, for example, the automorphism algebras of all holomorphically degenerate manifolds. The question is how to describe the infinite dimensional Lie algebras which can be realized as an automorphism algebra of a germ of a $\mathrm{CR}$ manifold? As is known, there is no unified theory of infinite dimensional Lie algebras (see [13]). Consequently, the strategy of studying the infinite dimensional Lie algebras is completely different from that in the finite dimensional case. Namely, one studies here not abstract Lie algebras, but concrete wide classes. There are four such classes: vector fields, matrices over some algebra of functions, operators on a Hilbert or a Banach space, and the Kac–Moody algebras. Infinite-dimensional algebras which can be realized as automorphisms of $\mathrm{CR}$ manifolds can also be of interest as a special subclass of the Lie algebras of vector fields. Question 2. Is it possible to realize an arbitrary real finite dimensional Lie algebra by symmetries of differential equation? (Here, we mean not only its complexification, which was deal with above.)
Citation in format AMSBIB
\Bibitem{Ste24} |
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