# Automorphisms and derivations of algebras of infinite matrices

Oksana Bezushchak  Faculty of Mechanics and Mathematics, Taras Shevchenko
National University of Kyiv, Volodymyrska, 60, Kyiv 01033, Ukraine _Email
address_<EMAIL_ADDRESS>
###### Abstract.

We describe automorphisms and derivations of several important associative and
Lie algebras of infinite matrices over a field.

###### Key words and phrases:

_Key words and phrases._ Algebra of infinite matrices; locally matrix algebra;
derivation; automorphism.

###### 1991 Mathematics Subject Classification:

2020 _Mathematics Subject Classification._ Primary 15B30; Secondary 16W20,
16W25.

00footnotetext: 00footnotetext:

##  1\. Introduction

Let $\mathbb{F}$ be a ground field and let $A$ be an associative
$\mathbb{F}$-algebra. Recall that an associative algebra $A$ gives rise to the
Lie algebra $A^{(-)}=(A,[a,b]=ab-ba).$ For an arbitrary element $a\in A$ the
operator $\text{ad}(a):x\to[a,x]$ is an (_inner_) _derivation_ of the
associative algebra $A$ and of the Lie algebra $A^{(-)}.$ If $J$ is an ideal
of the algebra $A$ then $\text{ad}_{J}(a)$ denotes the restriction of the
derivation $\text{ad}(a)$ to $J.$

An associative algebra $A$ with $1$ gives rise to the group $G(A)$ of
invertible elements of $A.$ For an arbitrary invertible element $a\in G(A)$
the conjugation $x\to a^{-1}xa,$ $x\in A,$ is an (_inner_) _automorphism_ of
the algebra $A$ and, hence, of the algebra $A^{(-)}.$

Let $I$ be an infinite set and let $\mathbb{N},$ $\mathbb{Z}$ denote the set
of positive integers and the set of integers, respectively.

######  Example 1.

Let $M_{\infty}(I,\mathbb{F})$ be the associative algebra of all $(I\times
I)$-matrices over $\mathbb{F}$ having finitely many nonzero entries
$($finitary matrices$).$ The Lie algebras

$\mathfrak{gl}_{\infty}(I,\mathbb{F})=M_{\infty}(I,\mathbb{F})^{(-)},\quad\quad\mathfrak{sl}_{\infty}(I,\mathbb{F})=[\mathfrak{gl}_{\infty}(I,\mathbb{F}),\mathfrak{gl}_{\infty}(I,\mathbb{F})]$

and their representation attracted considerable attention in the literature
$($see, _[12, 17])_.

A linear transformation $*:A\to A$ of an associative algebra $A$ is called an
involution if $(a^{*})^{*}=a,$ $(ab)^{*}=b^{*}a^{*}$ for arbitrary elements
$a,b\in A.$ The subspace of skew-symmetric elements $K(A,^{*})=\\{a\in
A\,|\,a^{*}=-a\\}$ is a subalgebra of the Lie algebra $A^{(-)}.$ The algebra
$M_{\infty}(I,\mathbb{F})$ is equipped with the transpose and the symplectic
involutions that give rise to Lie algebras
$\mathfrak{o}_{\infty}(I,\mathbb{F})$ and
$\mathfrak{sp}_{\infty}(I,\mathbb{F})$ of skew-symmetric elements,
respectively.

######  Example 2.

Let $M_{rcf}(I,\mathbb{F})$ be the associative algebra of all $(I\times
I)$-matrices over $\mathbb{F}$ having finitely many nonzero entries in each
row and in each column. It is easy to see that $M_{\infty}(I,\mathbb{F})$ is
an ideal of the algebra $M_{rcf}(I,\mathbb{F}).$ The algebra
$M_{rcf}(I,\mathbb{F})$ gives rise to the Lie algebra

$\mathfrak{gl}_{rcf}(I,\mathbb{F})=M_{rcf}(I,\mathbb{F})^{(-)}$

and to the group $GL_{rcf}(I,\mathbb{F})=G(M_{rcf}(I,\mathbb{F}).$

The transpose and the symplectic involutions on $M_{\infty}(I,\mathbb{F})$
naturally and uniquely extend to involutions of $M_{rcf}(I,\mathbb{F}).$

######  Example 3.

Let $M(I,\mathbb{F})$ denote the associative algebra of $(I\times I)$-matrices
over the field $\mathbb{F}$ having finitely many nonzero entries in each
column. If $V$ is a vector space over $\mathbb{F}$ of dimension $|I|$ then the
algebra $\text{End}_{\mathbb{F}}(V)$ of all linear transformations of $V$ is
isomorphic to $M(I,\mathbb{F}).$

Let $\mathfrak{gl}(I,\mathbb{F})=M(I,\mathbb{F})^{(-)}.$ The group
$G(M(I,\mathbb{F}))$ is isomorphic to the group $GL(V)$ of all invertible
linear transformations of $V.$

######  Example 4.

Let $MJ(\mathbb{F})$ be the algebra of Jacobi matrices, that is,
$(\mathbb{Z}\times\mathbb{Z})$-matrices having finitely many nonzero
diagonals. In other worlds, a matrix $a=(a_{ij})_{i,j\in\mathbb{Z}}$ lies in
$MJ(\mathbb{F})$ if there exists $n\in\mathbb{N}$ such that $a_{ij}=0$
whenever $|i-j|>n.$

The Lie algebras $\mathfrak{gl}_{J}(\mathbb{F})=MJ(\mathbb{F})^{(-)}$ found
applications in the theory of solitones _([9, 11])_.

######  Example 5.

Let $\text{End}_{fin}(V)$ be the subalgebra of $\text{End}_{\mathbb{F}}(V)$
that consists of all linear transformations of finite range.

N. Jacobson ([14], Chap. 9, Sec. 11, Th. 7) showed that for an arbitrary
associative algebra $A,$ $M_{\infty}(I,\mathbb{F})\subseteq A\subseteq
M(I,\mathbb{F}),$ every automorphism $\varphi$ of $A$ is a conjugation by an
invertible element from $M(I,\mathbb{F}),$ $\varphi(a)=x^{-1}ax,$ $a\in A.$

K.-H. Neeb [16] and N. Stumme [19] described derivations and automorphisms of
Lie algebras $\mathfrak{sl}_{\infty}(I,\mathbb{F}),$
$\mathfrak{o}_{\infty}(I,\mathbb{F}),$ $\mathfrak{sp}_{\infty}(I,\mathbb{F})$
over a field of zero characteristic.

##  2\. Main results

Theorems 2 and 4 below describe all derivations of the associative
$\mathbb{F}$-algebras $M_{\infty}(I,\mathbb{F}),$ $\text{End}_{fin}(V),$
$M_{rcf}(I,\mathbb{F}),$ $MJ(\mathbb{F}),$ $M(I,\mathbb{F})$ (Theorem 2) and
all automorphisms of the associative $\mathbb{F}$-algebras
$M_{\infty}(I,\mathbb{F}),$ $\text{End}_{fin}(V),$ $M_{rcf}(I,\mathbb{F}),$
$M(I,\mathbb{F})$ (Theorem 4). Theorem 5 describes anti-automorphisms of the
algebras $M_{\infty}(I,\mathbb{F}),$ $M_{rcf}(I,\mathbb{F}),$ $MJ(\mathbb{F})$
and shows that the algebras $\text{End}_{fin}(V),$ $M(I,\mathbb{F})$ do not
have any anti-automorphisms.

Finally, using the proof of Herstein‘s conjectures [13] by K.I. Beidar, M.
Brešar, M. Chebotar and W.S. Martindale (see, [2, 3, 4]) we describe
derivations of Lie algebras $\mathfrak{sl}_{\infty}(I,\mathbb{F}),$
$\mathfrak{o}_{\infty}(I,\mathbb{F}),$ $\mathfrak{sp}_{\infty}(I,\mathbb{F}),$
$\mathfrak{gl}_{rcf}(I,\mathbb{F}),$ $\mathfrak{gl}_{J}(\mathbb{F})$ in
Theorem 3 and automorphisms of Lie algebras
$\mathfrak{sl}_{\infty}(I,\mathbb{F}),$ $\mathfrak{o}_{\infty}(I,\mathbb{F}),$
$\mathfrak{sp}_{\infty}(I,\mathbb{F}),$ $\mathfrak{gl}_{rcf}(I,\mathbb{F}),$
$\text{End}_{fin}(V)^{(-)},$ $\mathfrak{gl}(I,\mathbb{F})$ in Theorem 6.

When dealing with Lie algebras of infinite matrices we use the following
theorem which is of some independent interest.

######  Theorem 1.

For an arbitrary infinite set $I$ we have:

  1. _(a)_

$M(I,\mathbb{F})=[M(I,\mathbb{F}),M(I,\mathbb{F})],$

  2. _(b)_

$M_{rcf}(I,\mathbb{F})=[M_{rcf}(I,\mathbb{F}),M_{rcf}(I,\mathbb{F})],$

  3. _(c)_

$MJ(\mathbb{F})=[MJ(\mathbb{F}),MJ(\mathbb{F})],$

  4. _(d)_

$\mathfrak{o}_{\infty}(I,\mathbb{F})=[\mathfrak{o}_{\infty}(I,\mathbb{F}),\mathfrak{o}_{\infty}(I,\mathbb{F})],$
$\mathfrak{sp}_{\infty}(I,\mathbb{F})=[\mathfrak{sp}_{\infty}(I,\mathbb{F}),\mathfrak{sp}_{\infty}(I,\mathbb{F})].$

In particular, the algebras $\mathfrak{gl}(I,\mathbb{F}),$
$\mathfrak{gl}_{rcf}(I,\mathbb{F})$ and $\mathfrak{gl}_{J}(\mathbb{F})$ are
perfect.

######  Theorem 2.

  1. _(a)_

An arbitrary derivation of the algebra $M_{\infty}(I,\mathbb{F})$ is of the
type

$\emph{ad}_{M_{\infty}(I,\mathbb{F})}(a),\quad\text{where}\quad a\in
M_{rcf}(I,\mathbb{F});$

  2. _(b)_

an arbitrary derivation of the algebra $\emph{End}_{fin}(V)$ is of the type

$\emph{ad}_{\emph{End}_{fin}(V)}(a),\quad\text{where}\quad
a\in\emph{End}_{\mathbb{F}}(V);$

  3. _(c)_

all derivations of the algebras $M_{rcf}(I,\mathbb{F}),$ $MJ(\mathbb{F}),$
$M(I,\mathbb{F})$ are inner.

######  Theorem 3.

Let $\mathbb{F}$ be a field of the characteristic not equal to $2.$

  1. _(a)_

An arbitrary derivation of the Lie algebra
$\mathfrak{sl}_{\infty}(I,\mathbb{F})$ is of the type

$\emph{ad}_{\mathfrak{sl}_{\infty}(I,\mathbb{F})}(a),\quad\text{where}\quad
a\in\mathfrak{gl}_{rcf}(I,\mathbb{F});$

  2. _(b)_

an arbitrary derivation of the Lie algebra
$\mathfrak{o}_{\infty}(I,\mathbb{F})$ $($resp.
$\mathfrak{sp}_{\infty}(I,\mathbb{F}))$ is of the type

$\emph{ad}_{\mathfrak{o}_{\infty}(I,\mathbb{F})}(a)\quad(\text{resp.}\quad\emph{ad}_{\mathfrak{sp}_{\infty}(I,\mathbb{F})}(a)),$

where $a\in K(M_{rcf}(I,\mathbb{F}),t)$ $($resp. $K(M_{rcf}(I,\mathbb{F}),s);$

  3. _(c)_

all derivations of Lie algebras $\mathfrak{gl}(I,\mathbb{F}),$
$\mathfrak{gl}_{rcf}(I,\mathbb{F}),$ $\mathfrak{gl}_{J}(\mathbb{F})$ are
inner.

######  Theorem 4.

  1. _(a)_

An arbitrary automorphism $\varphi$ of the algebra
$M_{\infty}(I,\mathbb{F})$ is a conjugation by an element from
$GL_{rcf}(I,\mathbb{F}).$ In other words, there exists an element $x\in
GL_{rcf}(I,\mathbb{F})$ such that $\varphi(a)=x^{-1}ax$ for all $a\in
M_{\infty}(I,\mathbb{F});$

  2. _(b)_

an arbitrary automorphism of the algebra $\emph{End}_{fin}(V)$ is a
conjugation by an element from $GL(V);$

  3. _(c)_

all automorphisms of the algebras $M_{rcf}(I,\mathbb{F}),$ $M(I,\mathbb{F})$
are inner.

Given an algebra $A$ an invertible linear transformation $\varphi:A\to A$ is
called an _anti-automorphism_ if $\varphi(ab)=\varphi(b)\varphi(a)$ for
arbitrary elements $a,b\in A.$ It is easy to see that an anti-automorphism is
an isomorphism $A\to A^{op}$ with the opposite algebra $A^{op}=(A,a*b=ba).$
The transpose transformations $(a_{ij})^{t}=(a_{ji}),$ $a_{ij}\in\mathbb{F},$
are anti-isomorphisms of the algebras $M_{\infty}(I,\mathbb{F}),$
$M_{rcf}(I,\mathbb{F}),$ $MJ(\mathbb{F}).$

######  Theorem 5.

  1. _(a)_

An arbitrary anti-automorphism of one of the algebras
$M_{\infty}(I,\mathbb{F}),$ $M_{rcf}(I,\mathbb{F}),$ $MJ(\mathbb{F})$ is a
composition of the transpose and an automorphism $($see, Theorem
$\ref{Th_IN_M_3_NEW});$

  2. _(b)_

algebras $\emph{End}_{fin}(V)$ and $M(I,\mathbb{F})$ do not have anti-
isomorphisms. In other words, these algebras are not isomorphic to their
opposite algebras.

######  Theorem 6.

Let $\mathbb{F}$ be a field of the characteristic not equal to $2.$

  1. _(a)_

An arbitrary automorphism $\varphi$ of the Lie algebra
$L=\mathfrak{sl}_{\infty}(I,\mathbb{F})$ is of the type $\varphi(a)=x^{-1}ax,$
$a\in L,$ or of the type $\varphi(a)=-x^{-1}a^{t}x,$ $a\in L,$ where $x\in
GL_{rcf}(I,\mathbb{F});$

  2. _(b)_

an arbitrary automorphism $\varphi$ of the Lie algebra
$L=\mathfrak{o}_{\infty}(I,\mathbb{F})$ $($resp.
$\mathfrak{sp}_{\infty}(I,\mathbb{F}))$ is of the type $\varphi(a)=x^{-1}ax,$
$a\in L,$ where $x\in GL_{rcf}(I,\mathbb{F})$ and $xx^{t}\in\mathbb{F}$
$($resp. $xx^{s}\in\mathbb{F});$

  3. _(c)_

an arbitrary automorphism $\varphi$ of the Lie algebra
$\mathfrak{gl}_{rcf}(I,\mathbb{F})$ is of the type $\varphi(a)=x^{-1}ax,$
$a\in\mathfrak{gl}_{rcf}(I,\mathbb{F}),$ or of the type
$\varphi(a)=-x^{-1}a^{t}x,$ $a\in\mathfrak{gl}_{rcf}(I,\mathbb{F}),$ where
$x\in G(M_{rcf}(I,\mathbb{F}));$

  4. _(d)_

an arbitrary automorphism $\varphi$ of the Lie algebra
$\mathfrak{gl}(I,\mathbb{F})$ is of the type $\varphi(a)=x^{-1}ax,$
$a\in\mathfrak{gl}(I,\mathbb{F}),$ where $x\in G(M(I,\mathbb{F}));$

  5. _(e)_

an arbitrary automorphism $\varphi$ of the Lie algebra
$\emph{End}_{fin}(V)^{(-)}$ is of the type $\varphi(a)=x^{-1}ax,$
$a\in\emph{End}_{fin}(V)^{(-)},$ where $x\in G(\emph{End}_{fin}(V)).$

Remark. We don‘t have a description of automorphisms of the algebra
$\mathfrak{gl}_{J}(\mathbb{F}).$

##  3\. Derivations of associative algebras of infinite matrices

Recall the definition of the Tykhonoff topology. Let $X,$ $Y$ be arbitrary
sets. Let $\text{Map}(X,Y)$ be the set of mappings $X\rightarrow Y.$ For
distinct elements $a_{1},\ldots,a_{n}\in X$ and arbitrary elements
$b_{1},\ldots,b_{n}\in Y,$ $n\geq 1,$ consider the subset

$M(a_{1},\ldots,a_{n};b_{1},\ldots,b_{n})=\\{f:X\to Y\,|\,f(a_{i})=b_{i},\
1\leq i\leq n\\}$

of $\text{Map}(X,Y).$ The Tykhonoff topology on $\text{Map}(X,Y)$ is generated
by all open sets of this type. In other words, $\text{Map}(X,Y)$ as a
topological space is homeomorphic to the space $Y^{|X|},$ the Tykhonoff
product of $|X|$ copies of $Y,$ where $Y$ is equipped with the discrete
topology.

An associative $\mathbb{F}$-algebra $A$ is called a _locally matrix algebra_
if for each finite subset of $A$ there exists a subalgebra $B\subset A$
containing this subset and isomorphic to the algebra $M_{n}(\mathbb{F})$ of
$(n\times n)$-matrices over $\mathbb{F}$ for some positive integer $n.$ For
properties and theory of locally matrix algebras, see [1, 5, 6, 7, 8, 15].

If $A$ is an associative $\mathbb{F}$-algebra and $M$ is an $A$-bimodule then
the space $C^{1}(A,M)$ of all _bimodule derivations_ $A\to M$ lies in
$\text{Map}(X,Y)$ and, therefore, is equipped with the Tykhonoff topology. Let
$B^{1}(A,M)$ be the space of all _inner bimodule derivations_ $A\rightarrow
M.$ The factor-space $H^{1}(A,M)=C^{1}(A,M)/B^{1}(A,M)$ is called the first
cohomology space (for details see, [10, 18]).

We will need the following generalization of Theorem 1(1) from [5].

######  Lemma 1.

If $A$ is a locally matrix algebra then for an arbitrary $A$-bimodule $M$ the
subspace $B^{1}(A,M)$ is dense in the space $C^{1}(A,M)$ in the Tykhonoff
topology.

###### Proof.

We need to show that for arbitrary elements $a_{1},\ldots,a_{n}\in A$ and an
arbitrary derivation $d:A\to M$ there exists an element $x\in M$ such that
$[x,a_{i}]=d(a_{i}),$ $1\leq i\leq n.$

There exists a subalgebra $A_{1}\subset A$ such that $a_{1},\ldots,a_{n}\in
A_{1}$ and $A_{1}\cong M_{k}(\mathbb{F}).$ Then

$M_{1}=d(A_{1})+A_{1}d(A_{1})+d(A_{1})A_{1}+A_{1}d(A_{1})A_{1}$

is a finite-dimensional $A_{1}$-submodule of $M$ and the restriction of the
derivation $d$ to $A_{1}$ is a derivation $A_{1}\to M_{1}.$ Since every
bimodule derivation of a matrix algebra over a field is inner it follows that
there exists an element $x\in M_{1}$ such that $d(a)=[x,a]$ for all elements
$a\in A_{1}.$ This completes the proof of the lemma. ∎

Let $(I\times I)(\mathbb{F})$ denote the vector space of all $(I\times
I)$-matrices over $\mathbb{F}.$ Since for arbitrary matrices $a\in(I\times
I)(\mathbb{F}),$ $b\in M_{\infty}(\mathbb{F})$ the products $ab,$ $ba$ are
well defined it follows that the vector space $(I\times I)(\mathbb{F})$ is a
bimodule over $M_{\infty}(\mathbb{F}).$ We will start with the following
proposition.

######  Proposition 1.

$H^{1}(M_{\infty}(I,\mathbb{F}),(I\times I)(\mathbb{F}))=(0).$

###### Proof.

Let $d:M_{\infty}(I,\mathbb{F})\to(I\times I)(\mathbb{F})$ be a bimodule
derivation. By Lemma 1, for an arbitrary nonempty finite subset $J\subset I$
there exists an element $y_{J}\in(I\times I)(\mathbb{F})$ such that
$d(a)=[y_{J},a]$ for all elements $a\in M_{|J|}(\mathbb{F}).$

Divide the matrix $y_{J}$ into blocks

$y_{J}=\begin{pmatrix}y_{J}(11)&y_{J}(12)\\\
y_{J}(21)&y_{J}(22)\end{pmatrix},$

where $y_{J}(11)\in M_{|J|}(\mathbb{F});$ $y_{J}(12)$ is a
$(J\times(I\,\diagdown J))$-matrix over $\mathbb{F};$ $y_{J}(21)$ is a
$((I\,\diagdown J)\times J)$-matrix over $\mathbb{F};$ and $y_{J}(22)$ is a
$((I\,\diagdown J)\times(I\,\diagdown J))$-matrix over $\mathbb{F}.$ For an
arbitrary element $a\in M_{|J|}(\mathbb{F})$ we have

$[y_{J},a]=\begin{pmatrix}[y_{J}(11),a]&-a\,y_{J}(12)\\\
y_{J}(21)\,a&0\end{pmatrix}.$

Let $J_{1}$ be another finite subset of $I,$ $J\subseteq J_{1}.$ Then
$[y_{J},a]=[y_{J_{1}},a]$ for all elements $a\in M_{|J|}(\mathbb{F}).$ Hence
the $(J\times J)$-minor of $y_{J_{1}}$ differs from $y_{J}(11)$ by a scalar
matrix.

We define a $(I\times I)$-matrix $y$ as follows. Fix $i_{0}\in I.$ Let
$y_{i_{0},i_{0}}=0.$ All matrices $y_{J},$ where $J$ runs over finite subsets
of $I,$ $i_{0}\in J,$ can be selected so that $(y_{J})_{i_{0},i_{0}}=0.$

For $i,j\in I$ choose a finite subset $J\subset I$ such that $i,j,i_{0}\in J.$
Define $y_{ij}=(y_{J})_{ij}.$ Let $J_{1}$ be a finite subset of $I,$
$J\subseteq J_{1}.$ Since $y_{J}(11)$ and the $(J\times J)$-minor of
$y_{J_{1}}$ differ by a scalar and
$(y_{J})_{i_{0},i_{0}}=(y_{J_{1}})_{i_{0},i_{0}}=0$ it follows that
$(y_{J})_{ij}=(y_{J_{1}})_{ij}.$ Hence $y_{ij}$ does not depend on a choice of
the subset $J.$

For an arbitrary element $a\in M_{\infty}(I,\mathbb{F})$ the $(J\times
J)$-minors of $(I\times I)$-matrices $d(a)=[y_{J},a]$ and $[y,a]$ coincide as
long as $J\subset I$ is a finite subset, $i_{0}\in J,$ and $a\in
M_{|J|}(\mathbb{F}).$ This implies that $d(a)=[y,a]$ and completes the proof
of the proposition. ∎

######  Lemma 2.

Let $y\in(I\times I)(\mathbb{F}).$

  1. _(a)_

The inclusion $[y,M_{\infty}(I,\mathbb{F})]\subseteq M(I,\mathbb{F})$ implies
$y\in M(I,\mathbb{F}).$

  2. _(b)_

The inclusion $[y,M_{\infty}(I,\mathbb{F})]\subseteq M_{rcf}(I,\mathbb{F})$
implies $y\in M_{rcf}(\mathbb{F}).$

###### Proof.

Let $e_{pq},$ $p,q\in I,$ denote the matrix unit having $1$ at the position
$(p,q)$ and zeros elsewhere. We have

$[y,e_{jj}]=y\,e_{jj}-e_{jj}\,y,\quad[y,e_{jj}]_{ij}=\begin{cases}y_{ij},&i\neq
j,\\\ 0,&i=j.\end{cases}$

If $[y,e_{jj}]\in M(I,\mathbb{F})$ then $y_{ij}\neq 0$ for only finitely many
$i\in I.$ Hence $y\in M(I,\mathbb{F}).$ Similarly,

$[y,e_{ii}]_{ij}=\begin{cases}-y_{ij},&i\neq j,\\\ 0,&i=j.\end{cases}$

If $[y,e_{ii}]\in M_{rcf}(I,\mathbb{F})$ then $y_{ij}\neq 0$ for only finitely
many $j\in I.$ This completes the proof of the lemma. ∎

###### Proof of Theorem 2.

(a) Let $d:M_{\infty}(I,\mathbb{F})\rightarrow M(I,\mathbb{F})$ be a bimodule
derivation. By Proposition 1, there exists an $(I\times I)$-matrix
$y=(y_{ij})_{i,j\in I}$ such that $d(a)=[y,a]$ for an arbitrary element $a\in
M_{\infty}(I,\mathbb{F}).$ By Lemma 2(a), the matrix $y$ lies in
$M(I,\mathbb{F}).$ Hence

(1) $H^{1}(M_{\infty}(I,\mathbb{F}),M(I,\mathbb{F}))=(0).$

Let $d:M_{\infty}(I,\mathbb{F})\rightarrow M_{\infty}(I,\mathbb{F})$ be a
derivation. By the above, there exists an $(I\times I)$-matrix $y$ such that
$d(a)=[y,a]$ for an arbitrary element $a\in M_{\infty}(I,\mathbb{F}).$ By
Lemma 2(b), we have $y\in M_{rcf}(I,\mathbb{F}).$

(b) Choose a basis $v_{i},$ $i\in I,$ in the vector space $V.$ Linear
transformations from $\text{End}_{fin}(V)$ have matrices lying in the
subalgebra $M_{r-fin}(I,\mathbb{F})$ of $M(I,\mathbb{F})$ that consists of
$(I\times I)$-matrices that are column-finite and have finite range. These two
conditions are equivalent to matrices from $M_{r-fin}(I,\mathbb{F})$ having
finitely many nonzero rows.

We have

$M_{\infty}(I,\mathbb{F})\subset M_{r-fin}(I,\mathbb{F})\lhd M(I,\mathbb{F}).$

Let $d:M_{r-fin}(I,\mathbb{F})\rightarrow M_{r-fin}(I,\mathbb{F})$ be a
derivation. The restriction of $d$ to $M_{\infty}(I,\mathbb{F})$ is a bimodule
derivation from $M_{\infty}(I,\mathbb{F})$ to $M(I,\mathbb{F}).$ By (1), there
exists a matrix $y\in M(I,\mathbb{F})$ such that $d(a)=[y,a]$ for an arbitrary
element $a\in M_{\infty}(I,\mathbb{F}).$

Consider the derivation

$d\,^{\prime}=d-\text{ad}_{M_{r-fin}(I,\mathbb{F})}(y)$

of the algebra $M_{r-fin}(I,\mathbb{F}).$ We have
$d\,^{\prime}(M_{\infty}(I,\mathbb{F}))=(0).$ For an arbitrary element $a\in
M_{r-fin}(I,\mathbb{F})$ and arbitrary indices $i,j\in I$ we have

$d\,^{\prime}(e_{ii}\,a\,e_{jj})=e_{ii}\,d\,^{\prime}(a)\,e_{jj}=0.$

Hence, $d\,^{\prime}(a)_{ij}=0.$ We showed that $d\,^{\prime}=0,$ which
completes the proof of the part (b).

(c) Let $d:M_{rcf}(I,\mathbb{F})\rightarrow M_{rcf}(I,\mathbb{F})$ be a
derivation. There exists a matrix $y\in M(I,\mathbb{F})$ such that
$d(a)=[y,a]$ for an arbitrary element $a\in M_{\infty}(I,\mathbb{F}).$ By
Lemma 2, the inclusion

$[y,M_{\infty}(I,\mathbb{F})]\subseteq M_{rcf}(I,\mathbb{F})$

implies $y\in M_{rcf}(I,\mathbb{F}).$ Consider the derivation

$d\,^{\prime}=d-\text{ad}(y)\ \text{of}\ M_{rcf}(I,\mathbb{F}),\quad
d\,^{\prime}(M_{\infty}(I,\mathbb{F}))=(0).$

As in the proof of Theorem 2$(b),$ for arbitrary indices $i,j\in I$ we have

$d\,^{\prime}(e_{ii}\,M_{rcf}(I,\mathbb{F})\,e_{jj})=e_{ii}\,d\,^{\prime}(M_{rcf}(I,\mathbb{F}))\,e_{jj}=0$

which implies $d\,^{\prime}=0,$ $d=\text{ad}(y).$

Let $d:MJ(\mathbb{F})\rightarrow MJ(\mathbb{F})$ be a derivation. Since

$M_{\infty}(\mathbb{Z},\mathbb{F})\subset MJ(\mathbb{F})\subset
M_{rcf}(\mathbb{Z},\mathbb{F}),$
$H^{1}(M_{\infty}(I,\mathbb{F}),M(I,\mathbb{F}))=(0)$

by Lemma 2, there exists a matrix $y\in M_{rcf}(\mathbb{Z},\mathbb{F})$ such
that $d(a)=[y,a]$ for an arbitrary element $a\in
M_{\infty}(\mathbb{Z},\mathbb{F}).$ Consider the bimodule derivation

$d\,^{\prime}=d-\text{ad}(y),\quad d\,^{\prime}:MJ(\mathbb{F})\rightarrow
M_{rcf}(\mathbb{Z},\mathbb{F}),\quad
d\,^{\prime}(M_{\infty}(\mathbb{Z},\mathbb{F}))=(0).$

Let us show that $d\,^{\prime}(MJ(\mathbb{F}))=(0).$ Let $a\in
MJ(\mathbb{F}),$ $a=(a_{ij})_{i,j\in\mathbb{Z}},$ $a_{ij}\in\mathbb{F}.$ There
exists $k\geq 1$ such that $a_{ij}=0$ whenever $|i-j|>k.$

Let $n\geq 1.$ Consider matrices $a\,^{\prime}(n),$ $a\,^{\prime\prime}(n):$

$a\,^{\prime}(n)_{ij}=\begin{cases}a_{ij}&\text{if}\ |i|,|j|\leq n\\\
0&\text{otherwise},\end{cases}\quad\quad
a\,^{\prime\prime}(n)_{ij}=\begin{cases}a_{ij}&\text{if}\ |i|>n\ \text{or}\
|j|>n\\\ 0&\text{otherwise}.\end{cases}$

Clearly, $a=a\,^{\prime}(n)+a\,^{\prime\prime}(n).$ If
$a\,^{\prime\prime}(n+k)_{ij}\neq 0$ then $|i|>n$ and $|j|>n,$ that is,

$a\,^{\prime\prime}(n+k)\in\begin{pmatrix}\ 0_{n\times n}\ &\vline&\ 0\ \\\
\hline\cr\ 0\ &\vline&\ *\ \end{pmatrix},$

where $0_{n\times n}$ denote zero $(n\times n)$-matrix. Let us show that

$d\,^{\prime}(a\,^{\prime\prime}(n+k))\in\begin{pmatrix}\ 0_{n\times n}\
&\vline&\ 0\ \\\ \hline\cr\ 0\ &\vline&\ *\ \end{pmatrix}$

as well. Indeed, if $i\leq n$ or $j\leq n$ then

$d\,^{\prime}(a\,^{\prime\prime}(n+k))_{ij}=d\,^{\prime}(e_{ii}\,a\,^{\prime\prime}(n+k)\,e_{jj})=0$

since $e_{ii}\,a\,^{\prime\prime}(n+k)\,e_{jj}=0.$ We have
$d\,^{\prime}(a\,^{\prime}(n+k))=0.$ Hence

$d\,^{\prime}(a)\in\begin{pmatrix}\ 0_{n\times n}\ &\vline&\ 0\ \\\ \hline\cr\
0\ &\vline&\ *\ \end{pmatrix}$

for any $n\geq 1.$ Hence $d\,^{\prime}(a)=0,$ $d(a)=[y,a]$ for an arbitrary
$a\in MJ(\mathbb{F}).$

Consider the matrix

$E_{1}=\sum_{i\in\mathbb{Z}}e_{i,i+1}\in MJ(\mathbb{F}).$

It is easy to see that if $y\not\in MJ(\mathbb{F})$ then $[y,E_{1}]\not\in
MJ(\mathbb{F})$ as well. This implies that $y\in MJ(\mathbb{F}).$

Finally, let $d:M(I,\mathbb{F})\rightarrow M(I,\mathbb{F})$ be a derivation.
The ideal $M_{r-fin}(I,\mathbb{F})$ is invariant with respect to all
derivations. By Theorem 2$(b),$ there exists an element $y\in M(I,\mathbb{F})$
such that

$d\,|\,_{M_{r-fin}(I,\mathbb{F})}=\text{ad}_{M_{r-fin}(I,\mathbb{F})}(y).$

Let $d\,^{\prime}=d-\text{ad}(y).$ Then
$d\,^{\prime}(M_{r-fin}(I,\mathbb{F}))=(0).$ Arguing as above, we get
$d\,^{\prime}=0.$ This completes the proof of Theorem 2. ∎

##  4\. Lie algebras of infinite matrices

We will start with the proof of Theorem 1.

Let $I$ be an arbitrary infinite set. It follows that $|I|=|\mathbb{N}\times
I|.$ That is why without loss of generality, we will assume that $I$ is a
direct product of $\mathbb{N}$ with another set $J,$ $|J|=|I|,$
$I=\mathbb{N}\times J.$

Therefore, we can view an arbitrary $(I\times I)$-matrix as an
$(\mathbb{N}\times\mathbb{N})$-matrix over $(J\times J)(\mathbb{F}).$ For two
positive integers $i,j\in\mathbb{N}$ and a $(J\times J)$-matrix $x$ let
$e_{ij}(x)$ denote an $(I\times I)$-matrix having the block $x$ at the
position $(i,j)$ and zeros elsewhere.

Consider the matrix

$E=\sum_{i=1}^{\infty}e_{i,i+1}(\text{Id})\in(I\times I)(\mathbb{F}),$

where Id is the identity $(J\times J)$-matrix. The $(I\times I)$-matrix $E$
contains $\leq 1$ nonzero elements in each row and in each column. Therefore,
for an arbitrary $(I\times I)$-matrix $x$ the products $Ex,$ $xE$ make sense.

Let $a=(a_{ij})_{i,j\in\mathbb{N}},$ $a_{ij}\in(J\times J)(\mathbb{F}).$
Consider the $(I\times I)$-matrix

(2)
$\tilde{a}=(\tilde{a}_{ij})_{i,j\in\mathbb{N}},\quad\tilde{a}_{ij}=\sum_{k=0}^{\infty}a_{i-1-k,j-k},$

where we let $a_{ij}=0$ for $i\leq 0$ or $j\leq 0.$ It is easy to see that for
$a\in M(I,\mathbb{F})$ the matrix $\tilde{a}$ also lies in $M(I,\mathbb{F}).$

######  Lemma 3.

111V.V. Sergeichuk (2020), personal communication. $[E,\widetilde{a}]=a.$

###### Proof.

We have

$E\
\widetilde{a}=\Big{(}\sum_{i=1}^{\infty}e_{i,i+1}(1)\Big{)}\Big{(}\sum_{j,k=1}^{\infty}e_{jk}(\widetilde{a}_{jk})\Big{)}=\sum_{i,k=1}^{\infty}e_{ik}(\widetilde{a}_{i+1,k});$
$\widetilde{a}\
E=\Big{(}\sum_{j,k=1}^{\infty}e_{jk}(\widetilde{a}_{jk})\Big{)}\Big{(}\sum_{i=1}^{\infty}e_{i,i+1}(1)\Big{)}=\\\
=\sum_{j,i=1}^{\infty}e_{j,i+1}(\widetilde{a}_{ji})=\sum_{i\geq 1,k\geq
2}e_{ik}(\widetilde{a}_{i,k-1}).$

Finally,

$[E,\widetilde{a}]=\sum_{i,k=1}^{\infty}e_{ik}(\widetilde{a}_{i+1,k}-\widetilde{a}_{i,k-1}),$

where $\widetilde{a}_{i,0}=0.$ It remains to verify that

(3) $\widetilde{a}_{i+1,j}-\widetilde{a}_{i,j-1}=a_{ij}$

for all $i,j\in\mathbb{N}.$

Define $a_{ij}=0$ whenever $i\leq 0$ or $j\leq 0.$ Then

$\widetilde{a}_{ij}=\sum_{k=0}^{\infty}a_{i-1-k,j-k}.$

Now,

$\widetilde{a}_{i+1,j}-\widetilde{a}_{i,j-1}=\sum_{k=0}^{\infty}a_{i-k,j-k}-\sum_{k=0}^{\infty}a_{i-1-k,j-1-k}=a_{ij},$

which completes the proof of the lemma. ∎

###### Proof of Theorem 1.

(a) An arbitrary $(I\times I)$-matrix $a$ over the field $\mathbb{F}$ can be
divided into blocks $a=(a_{ij})_{\mathbb{N}\times\mathbb{N}},$ where each
block $a_{ij}$ is a $(J\times J)$-matrix. If $a$ lies in
$\mathfrak{gl}(I,\mathbb{F})$ then each block $a_{ij}$ lies in
$\mathfrak{gl}(J,\mathbb{F}).$ The reverse statement is not true.

Suppose that $a\in\mathfrak{gl}(I,\mathbb{F}).$ We will show that the matrix
$\widetilde{a}$ (see above) also lies in $\mathfrak{gl}(I,\mathbb{F}).$ Choose
a column indexed by $(n,\alpha)\in\mathbb{N}\times J,$ $n\in\mathbb{N},$
$\alpha\in J.$ We need to verify that the $\alpha$-th column of the $(I\times
J)$-matrix

$\begin{pmatrix}\widetilde{a}_{1n}\\\ \widetilde{a}_{2n}\\\
\vdots\end{pmatrix}$

contains finitely many nonzero entries. Since the $(I\times I)$-matrix $a$ has
finitely many nonzero entries in each column it follows that for an arbitrary
$l\in\mathbb{N}$ there exists a positive integer $N(l,\alpha)$ such that the
$(J\times J)$-matrix $a_{kl}$ has zero $\alpha$-th column for $k>N(l,\alpha).$

Let $i>n+\max(N(1,\alpha)+\cdots+N(n,\alpha))=s.$ Then $i-2\geq n-1.$ By the
equality (2), we have

$\widetilde{a}_{in}=a_{i-1,n}+\cdots+a_{i-n,1}.$

All summands on the right hand side have zero $\alpha$-th column. Hence the
matrix $\widetilde{a}_{in}$ has zero $\alpha$-th column. The matrix

$\begin{pmatrix}\widetilde{a}_{1n}\\\ \vdots\\\
\widetilde{a}_{sn}\end{pmatrix}$

has finitely many nonzero entries in the $\alpha$-th column since the
$(J\times J)$-matrices $\widetilde{a}_{1n},$ $\dots,$ $\widetilde{a}_{sn}$
have finitely many nonzero entries in each column.

We showed that the matrix $\widetilde{a}$ lies in
$\mathfrak{gl}(I,\mathbb{F}).$ As above, let

$E=\sum_{i=1}^{\infty}e_{i,i+1}(\text{Id}_{J}),$

where $\text{Id}_{J}$ is the identity $(J\times J)$-matrix. By Lemma 3, we
have

$a=[E,\widetilde{a}]\in[\mathfrak{gl}(I,\mathbb{F}),\mathfrak{gl}(I,\mathbb{F})].$

Hence, we proved Theorem 1(a).

(b) We need to show that for an arbitrary matrix $a\in M_{rcf}(I,\mathbb{F})$
the matrix $\tilde{a}$ also lies in $M_{rcf}(I,\mathbb{F}).$ Consider indices
$(k,j),(k\,^{\prime},j\,^{\prime})\in\mathbb{N}\times J.$ The entry
$\tilde{a}_{(k,j),(k\,^{\prime},j\,^{\prime})}$ of the matrix $\tilde{a}$ is
the $(j,j\,^{\prime})$-th entry of the block $\tilde{a}_{k,k\,^{\prime}}.$ We
have

$\tilde{a}_{k,k\,^{\prime}}=\sum_{t=0}^{k-2}a_{k-1-t,k\,^{\prime}-t}.$

Hence,

$\tilde{a}_{(k,j),(k\,^{\prime},j\,^{\prime})}=(\tilde{a}_{k,k\,^{\prime}})_{j,j\,^{\prime}}=\sum_{t=0}^{k-2}(a_{k-1-t,k\,^{\prime}-t})_{j,j\,^{\prime}}=\sum_{t=0}^{k-2}a_{(k-1-t,j),(k\,^{\prime}-t,j\,^{\prime})}.$

We showed that the $(k,j)$-th row of the matrix $\tilde{a}$ is a sum of
permuted $(1,j)$-th, $\ldots,$ $(k-1,j)$-th rows of the matrix $a.$ Hence,
every row of the matrix $\tilde{a}$ contains finitely many nonzero elements,
$\tilde{a}\in M_{rcf}(I,\mathbb{F}).$ This proves the part (b) of the theorem.

(c) Consider a $(\mathbb{Z}\times\mathbb{Z})$-matrix

$E_{J}=\sum_{i\in\mathbb{Z}}e_{i,i+1}(1).$

For a $(\mathbb{Z}\times\mathbb{Z})$-matrix $a=(a_{ij})_{i,j\in\mathbb{Z}}$
consider the matrix $\tilde{a}=(\tilde{a}_{ij})_{i,j\in\mathbb{Z}},$ where

$\tilde{a}_{ij}=\begin{cases}a_{i-1,j}+\cdots+a_{1,2-i+j}&\hbox{for $i\geq
2$}\\\ 0&\hbox{for $i=1$}\\\ -a_{i,j+1}-\cdots-a_{0,j-i+1}&\hbox{for $i\leq
0$.}\end{cases}$

We claim that $[E_{J},\tilde{a}]=a.$ Indeed, a straightforward computation
shows that

  1. $(1)$

$[E_{J},\tilde{a}]_{ij}=\tilde{a}_{i+1,j}-\tilde{a}_{i,j+1},$

  2. $(2)$

$\tilde{a}_{i+1,j}-\tilde{a}_{i,j+1}=a_{ij}$

for all $i,j\in\mathbb{Z}.$

It remains to check that for an arbitrary matrix $a\in MJ(\mathbb{F})$ the
matrix $\tilde{a}$ also lies in $MJ(\mathbb{F}).$ Suppose that $a_{ij}=0$
whenever $|i-j|>k.$ The expression for $\tilde{a}_{ij}$ implies that
$\tilde{a}_{ij}=0$ whenever $|i-j|>k+1.$ Hence $\tilde{a}\in MJ(\mathbb{F}).$
This completes the proof of the part (c).

(d) The Lie algebra $\mathfrak{o}_{\infty}(I,\mathbb{F})$ (resp.,
$\mathfrak{sp}_{\infty}(I,\mathbb{F})$) is simple since it has a local system
of simple finite-dimensional subalgebras
$\mathfrak{o}_{\infty}(I_{0},\mathbb{F})$ (resp.,
$\mathfrak{sp}_{\infty}(J_{0},\mathbb{F})$), where $I_{0},$ $J_{0}$ run over
all nonempty finite subsets of $I,$ the order of $J_{0}$ is even. This implies
$\mathfrak{o}_{\infty}(I,\mathbb{F})=[\mathfrak{o}_{\infty}(I,\mathbb{F}),\mathfrak{o}_{\infty}(I,\mathbb{F})]$
and
$\mathfrak{sp}_{\infty}(I,\mathbb{F})=[\mathfrak{sp}_{\infty}(I,\mathbb{F}),\mathfrak{sp}_{\infty}(I,\mathbb{F})]$.
∎

In [13], I.N. Herstein formulated a series of conjectures about links between
derivations and automorphisms of an associative algebra $A$ and derivations
and automorphisms of Lie algebras $[A,A],$ $K(A,*).$ These conjectures were
proved by K. Beidar, M. Brešar, M. Chebotar and W. Martindale in [2, 3, 4].

We will formulate here and in Sec. 5 only particular cases of their results
that are directly related to this work:

Theorem I (K.I. Beidar, M. Brešar, M. Chebotar and W.S. Martindale; see, Cor.
1.4(b), [4]) Let $A$ be a simple associative algebra with the center $Z.$ The
algebra $A$ is not finite-dimensional over $Z.$ Let $d$ be a derivation of the
Lie algebra $[A,A]\diagup[A,A]\cap Z.$ Then there exists a derivation
$\tilde{d}:A\rightarrow A$ of the associative algebra $A$ such that
$d(a)-\tilde{d}(a)\in Z$ for an arbitrary element $a\in[A,A].$

Theorem II (K.I. Beidar, M. Brešar, M. Chebotar and W.S. Martindale; see, Cor.
1.9(b), [4]) Let $A$ be a simple associative algebra with an involution
$*:A\rightarrow A.$ Let $Z$ be the center of the algebra $A.$ The algebra $A$
is not finite-dimensional over $Z.$ Let $d$ be a derivation of the Lie algebra
$[K,K]\diagup[K,K]\cap Z,$ where $K=K(A,*)=\\{a\in A\,|\,a^{*}=-a\\}.$ Then
there exists a derivation $\tilde{d}:A\rightarrow A$ of the associative
algebra $A$ such that $d(a)-\tilde{d}(a)\in Z$ for an arbitrary element
$a\in[K,K].$

###### Proof of Theorem 3.

(a) The algebra $A=M_{\infty}(I,\mathbb{F})$ is simple and has zero center.
Let $d$ be a derivation of the Lie algebra
$\mathfrak{sl}_{\infty}(I,\mathbb{F})=[A,A].$ By Theorem I above, there exists
a derivation $\tilde{d}$ of the algebra $A$ that coincides with $d$ on
$[A,A].$ By Theorem 2(a), the derivation $\tilde{d}$ looks as
$\text{ad}_{\mathfrak{sl}_{\infty}(I,\mathbb{F})}(a),$ where $a\in
M_{rcf}(I,\mathbb{F}).$

(b) Now, let $*$ be the transpose involution or the symplectic involution of
the algebra $A=M_{\infty}(I,\mathbb{F}).$ Then
$\mathfrak{o}_{\infty}(I,\mathbb{F})$ (resp.,
$\mathfrak{sp}_{\infty}(I,\mathbb{F})$) is the Lie algebra of skew-symmetric
elements $K=K(A,*).$ By Theorem II above, there exists a derivation
$\tilde{d}$ of the algebra $A$ that coincides with $d$ on $[K,K].$ By Theorem
1$(d),$ $[K,K]=K.$ Theorem 2(a) implies that the derivation $\tilde{d}$ is the
restriction of a derivation $\text{ad}_{M_{\infty}(I,\mathbb{F})}(a),$ where
$a\in M_{rcf}(I,\mathbb{F}).$

Both the transpose and the symplectic involution on the algebra
$M_{\infty}(I,\mathbb{F})$ extend to $M_{rcf}(I,\mathbb{F}).$ Let

$a=a_{h}+a_{k},\quad\text{where}\quad a_{h}^{*}=a_{h},\quad a_{k}^{*}=-a_{k}.$

For an arbitrary element $b\in K$ we have $[a,b]=[a_{h},b]+[a_{k},b]\in K.$
However,

$[a_{h},b]^{*}=[a_{h},b],\quad[a_{k},b]^{*}=-[a_{k},b].$

Hence $[a_{h},K]=(0),$ $[a,b]=[a_{k},b],$ that is,
$\text{ad}_{K}(a)=\text{ad}_{K}(a_{k}),$ where $a_{k}\in
K(M_{rcf}(I,\mathbb{F}),*).$

(c) Let $d$ be a derivation of the Lie algebra
$\mathfrak{gl}(I,\mathbb{F})=M(I,\mathbb{F})^{(-)}.$ By Theorem I of Beidar-
Brešar-Chebotar-Martindale [4], there exists a derivation $\tilde{d}$ of the
associative algebra $M(I,\mathbb{F})$ such that $d(a)-\tilde{d}(a)$ lies in
the center $Z$ of $M(I,\mathbb{F})$ for all elements
$a\in[M(I,\mathbb{F}),M(I,\mathbb{F})].$ By Theorem 1(a),
$[M(I,\mathbb{F}),M(I,\mathbb{F})]=M(I,\mathbb{F}),$ hence
$d(a)-\tilde{d}(a)\in Z$ for all elements $a\in M(I,\mathbb{F}).$

Now, for arbitrary elements $a,b\in M(I,\mathbb{F})$ we have

$d([a,b])=[d(a),b]+[a,d(b)]=[\tilde{d}(a),b]+[a,\tilde{d}(b)]=\tilde{d}([a,b]).$

Hence, $d=\tilde{d}$ on $[M(I,\mathbb{F}),M(I,\mathbb{F})].$ Again by Theorem
1(a), we get $d=\tilde{d}.$ It remains to refer to Theorem 2(c). This proves
the assertion about derivations of $\mathfrak{gl}(I,\mathbb{F}).$

Let $d$ be a derivation of the Lie algebra
$\mathfrak{gl}_{rcf}(I,\mathbb{F}).$ The subalgebra
$\mathfrak{sl}_{\infty}(I,\mathbb{F})$ is an ideal of
$\mathfrak{gl}_{rcf}(I,\mathbb{F}).$ For arbitrary elements
$a,b\in\mathfrak{sl}_{\infty}(I,\mathbb{F})$ we have

$d([a,b])=[d(a),b]+[a,d(b)]\in[\mathfrak{gl}_{rcf}(I,\mathbb{F}),\mathfrak{sl}_{\infty}(I,\mathbb{F})]\subseteq\mathfrak{sl}_{\infty}(I,\mathbb{F}).$

Hence the ideal $\mathfrak{sl}_{\infty}(I,\mathbb{F})$ is differentially
invariant. By Theorem 3$(a),$ there exists an element
$a\in\mathfrak{gl}_{rcf}(I,\mathbb{F})$ such that $d(x)=[a,x]$ for an
arbitrary element $x\in\mathfrak{sl}_{\infty}(I,\mathbb{F}).$ Consider the
derivation $d\,^{\prime}=d-\text{ad}(a)$ of the algebra
$\mathfrak{gl}_{rcf}(I,\mathbb{F}).$ We have

$d\,^{\prime}(\mathfrak{sl}_{\infty}(I,\mathbb{F}))=(0).$

For arbitrary elements $x\in\mathfrak{sl}_{\infty}(I,\mathbb{F}),$
$y\in\mathfrak{gl}_{rcf}(I,\mathbb{F})$ we have

$0=d\,^{\prime}([x,y])=[x,d\,^{\prime}(y)].$

Hence $d\,^{\prime}(\mathfrak{gl}_{rcf}(I,\mathbb{F}))$ lies in the
centralizer of $\mathfrak{sl}_{\infty}(I,\mathbb{F}).$ It is easy to see that
the centralizer of $\mathfrak{sl}_{\infty}(I,\mathbb{F})$ in the algebra
$\mathfrak{gl}_{rcf}(I,\mathbb{F})$ is the space of scalar matrices
$\alpha\cdot\text{Id},$ where $\alpha\in\mathbb{F}$ and Id is the identity
$(I\times I)$-matrix. Hence

$d\,^{\prime}(\mathfrak{gl}_{rcf}(I,\mathbb{F}))\subseteq\mathbb{F}\cdot\text{Id}.$

For arbitrary elements $a,b\in\mathfrak{gl}_{rcf}(I,\mathbb{F})$ we have
$d\,^{\prime}([a,b])=[d\,^{\prime}(a),b]+[a,d\,^{\prime}(b)]=0.$ Hence,

$d\,^{\prime}([\mathfrak{gl}_{rcf}(I,\mathbb{F}),\mathfrak{gl}_{rcf}(I,\mathbb{F})])=(0).$

By Theorem 1$(a),$

$[\mathfrak{gl}_{rcf}(I,\mathbb{F}),\mathfrak{gl}_{rcf}(I,\mathbb{F})]=\mathfrak{gl}_{rcf}(I,\mathbb{F}),$

which implies $d\,^{\prime}=0,$ $d=\text{ad}(a).$ This completes the proof of
the part concerning $\mathfrak{gl}_{rcf}(I,\mathbb{F}).$

Let $d:\mathfrak{gl}_{J}(\mathbb{F})\rightarrow\mathfrak{gl}_{J}(\mathbb{F})$
be a derivation. As above, we conclude that the ideal
$\mathfrak{sl}_{\infty}(\mathbb{Z},\mathbb{F})$ of the algebra
$\mathfrak{gl}_{J}(\mathbb{F})$ is differentially invariant. By Theorem
3$(a),$ there exists a matrix $a\in M_{rcf}(\mathbb{Z},\mathbb{F})$ such that
$d(x)=[a,x]$ for an arbitrary element
$x\in\mathfrak{sl}_{\infty}(\mathbb{Z},\mathbb{F}).$ Consider the bimodule
derivation

$d\,^{\prime}:\mathfrak{gl}_{J}(\mathbb{F})\rightarrow\mathfrak{gl}_{rcf}(\mathbb{Z},\mathbb{F}),\quad
d\,^{\prime}(x)=d(x)-[a,x],\quad x\in\mathfrak{gl}_{J}(\mathbb{F}).$

We have $d\,^{\prime}(\mathfrak{sl}_{\infty}(\mathbb{Z},\mathbb{F}))=(0).$ As
above, for arbitrary elements
$x\in\mathfrak{sl}_{\infty}(\mathbb{Z},\mathbb{F}),$
$y\in\mathfrak{gl}_{J}(\mathbb{F})$ we have

$0=d\,^{\prime}([x,y])=[x,d\,^{\prime}(y)].$

Hence $d\,^{\prime}(\mathfrak{gl}_{J}(\mathbb{F}))$ lies in the centralizer of
$\mathfrak{sl}_{\infty}(\mathbb{Z},\mathbb{F})$ in the algebra
$\mathfrak{gl}_{rcf}(\mathbb{Z},\mathbb{F}).$ In other words,

$d\,^{\prime}(\mathfrak{gl}_{J}(\mathbb{F}))\subseteq\mathbb{F}\cdot\text{Id},$

where Id is the identity $(\mathbb{Z}\times\mathbb{Z})$-matrix. By Theorem
1(a),
$\mathfrak{gl}_{J}(\mathbb{F})=[\mathfrak{gl}_{J}(\mathbb{F}),\mathfrak{gl}_{J}(\mathbb{F})]$.
So, by Proposition 1(2), we conclude that $d\,^{\prime}=0.$ Hence $d(x)=[a,x]$
for an arbitrary element $x\in\mathfrak{gl}_{J}(\mathbb{F}).$

It was already mentioned in the proof of Theorem 2(c) that if a
$(\mathbb{Z}\times\mathbb{Z})$-matrix $a$ does not lie in
$\mathfrak{gl}_{J}(\mathbb{F})$ then the matrix $[a,E_{1}],$
$E_{1}=\sum_{i\in\mathbb{Z}}e_{i,i+1}(1),$ also does not lie in
$\mathfrak{gl}_{J}(\mathbb{F}).$ Hence $a\in\mathfrak{gl}_{J}(\mathbb{F}),$
which completes the proof of Theorem 3. ∎

##  5\. Automorphisms of algebras of infinite matrices

Let $V$ be an infinite-dimensional vector space and let $V^{*}$ be the dual
space of $V.$ A subspace $W\subseteq V^{*}$ is called total if

$\bigcap_{w\in W}\ker w=(0).$

Recall the action of the algebra $\text{End}_{\mathbb{F}}(V)$ on the dual
space $V^{*}$: for a linear transformation $a\in\text{End}_{\mathbb{F}}(V)$
and a linear functional $f\in V^{*}$ we define $(fa)(v)=f(a(v)),$ $v\in V.$

For a total subspace $W\subseteq V^{*}$ consider the subalgebra
$\text{End}_{\mathbb{F}}(V|W)=\\{a\in\text{End}_{\mathbb{F}}(V)\,|\,Wa\subseteq
W\\}.$ Then the subalgebra

$\text{End}_{fin}(V|W)=\text{End}_{\mathbb{F}}(V|W)\cap\text{End}_{fin}(V)$

is a locally matrix algebra that can be identified with the tensor product
$V\otimes_{\mathbb{F}}W.$ Moreover, $\text{End}_{fin}(V|W)$ is dense in
$\text{End}_{\mathbb{F}}(V)$ in the Tykhonoff topology (see, [14]).

Let $\mathcal{E}=\\{e_{i},i\in I\\}$ be a basis of the vector space $V$ and
let $\mathcal{E}^{*}=\\{e_{i}^{*},i\in I\\}\subset V^{*}$ be the dual basis:
$e_{i}^{*}(e_{j})=\delta_{ij}e_{j}.$ It is easy to see that the subspace
$\text{Span}_{\mathbb{F}}(\mathcal{E}^{*})$ is total. In the basis
$\mathcal{E}$ the algebra
$\text{End}_{\mathbb{F}}(V|\text{Span}_{\mathbb{F}}(\mathcal{E}^{*}))$
corresponds to the matrix algebra $M_{rcf}(I,\mathbb{F}).$ The subalgebra
$\text{End}_{fin}(V|\text{Span}_{\mathbb{F}}(\mathcal{E}^{*}))$ corresponds to
$M_{\infty}(I,\mathbb{F}).$

If $W=V^{*}$ then $\text{End}_{\mathbb{F}}(V|W)=\text{End}_{\mathbb{F}}(V),$
$\text{End}_{fin}(V|W)=\text{End}_{fin}(V).$

N. Jacobson ([14], Chap. 9, Sec. 11, Th. 7) proved that for an arbitrary
algebra $A,$

$\text{End}_{fin}(V|W)\subseteq A\subseteq\text{End}_{\mathbb{F}}(V|W)$

and an arbitrary automorphism $\varphi$ of $A$ there exists an element $x\in
G(\text{End}_{\mathbb{F}}(V|W))$ such that $\varphi(a)=x^{-1}ax,$ $a\in A.$

Substituting $W=\text{Span}_{\mathbb{F}}(\mathcal{E}^{*})$ or $W=V^{*}$ we
obtain Theorem 4.

In the proof of Theorem 6 on automorphisms of Lie algebras, we will again rely
on the work [4] of K.I. Beidar, M. Brešar, M. Chebotar and W.S. Martindale on
Herstein‘s Conjectures.

The results in [4] are quite general. As in Sec. 4, we will use only a small
part of them that is directly related to our proof.

Theorem III (K.I. Beidar, M. Brešar, M. Chebotar and W.S. Martindale; see,
Cor. 1.2 (c), [4]) Let $\emph{char}\,\mathbb{F}\neq 2$ and let $A$ be a simple
associative $\mathbb{F}$-algebra with the center $Z$ and $A$ is not finite-
dimensional over $Z.$ If $\alpha$ is an automorphism of the Lie algebra
$[A,A]$ then there exist mappings $\varphi,\psi:A\rightarrow A$ such that

  1. $(i)$

$\varphi$ is either an automorphism or the negative of an anti-automorphism of
$A,$

  2. $(ii)$

$\psi(A)\subseteq Z,$ $\psi([A,A])=(0),$

and $\alpha(a)=\varphi(a)+\psi(a)$ for all $a\in[A,A].$

Theorem IV (K.I. Beidar, M. Brešar, M. Chebotar and W.S. Martindale; see, Cor.
1.7, [4]) Let $\emph{char}\,\mathbb{F}\neq 2$ and Let $A$ be a simple
associative $\mathbb{F}$-algebra with an involution and let $K$ be the Lie
algebra of skew-symmetric elements. Suppose that the algebra $A$ is not
finite-dimensional over $Z.$ Then for an arbitrary automorphism $\alpha$ of
the Lie algebra $[K,K]$ there exists an automorphism $\varphi$ of the algebra
$A$ such that $\alpha(a)=\varphi(a)$ for all $a\in[K,K].$

Theorem III above implies that in order to describe Lie automorphisms of a
simple associative algebra we need to describe its anti-automorphisms first.

###### Proof of Theorem 5.

(a) The algebras $M_{\infty}(I,\mathbb{F}),$ $M_{rcf}(I,\mathbb{F}),$
$MJ(\mathbb{F})$ are closed with respect to the transpose $t.$ If $*$ is an
anti-automorphism then the composition $\varphi$ of $*$ and $t$ is an
automorphism, $(a^{t})^{*}=\varphi(a)$ for an arbitrary element $a.$ Hence,
$a^{*}=\varphi(a^{t}).$

(b) N. Jacobson ([14], Chap. IX.12, Th. 8) showed that if an algebra $A,$
$\text{End}_{fin}(V|W)\subseteq A\subseteq\text{End}(V|W),$ has an anti-
isomorphism then $\dim_{\mathbb{F}}V=\dim_{\mathbb{F}}W.$ P. Erdos and J.
Kaplansky (see, [14], Chap. IX.5, Th. 2) showed that

$\dim_{\mathbb{F}}V^{*}=|\mathbb{F}|^{\dim_{\mathbb{F}}V}>\dim_{\mathbb{F}}V.$

Hence, the algebras

$\text{End}_{fin}(V)=\text{End}_{fin}(V|V^{*})\quad\text{and}\quad
M(I,\mathbb{F})\cong\text{End}_{\mathbb{F}}(V)=\text{End}(V|V^{*})$

do not have anti-isomorphisms. ∎

###### Proof of Theorem 6.

The part (a) immediately follows from Theorem III above applied to the algebra
$A=M_{\infty}(I,\mathbb{F})$ and Theorem 4(a).

(b) Let ∗ be the transpose or the symplectic involution on the algebra
$A=M_{\infty}(I,\mathbb{F}).$ Then
$K=K(A,^{*})=\mathfrak{o}_{\infty}(I,\mathbb{F})$ or
$\mathfrak{sp}_{\infty}(I,\mathbb{F}).$ By Theorem 1(d), we have $K=[K,K].$ By
Theorem IV above, an arbitrary automorphism $\varphi$ of the Lie algebra $K$
lifts to an automorphism of the algebra $A.$ By Theorem 4$(a),$ there exists
an element $x\in G(M_{rcf}(I,\mathbb{F}))$ such that $\varphi(a)=x^{-1}ax$ for
all elements $a\in K.$ Hence

$(x^{-1}ax)^{*}=-x^{*}a(x^{*})^{-1}=-x^{-1}ax,$

which implies

$(xx^{*})a(xx^{*})^{-1}=a.$

We proved that the element $xx^{*}$ commutes with all elements from $K.$ Since
the algebra $A$ is generated by the subspace $K$ it follows that the element
$xx^{*}$ commutes with all elements from $M_{\infty}(I,F)$ and, therefore, is
a scalar matrix $xx^{*}=\alpha\cdot\text{Id},$ $0\neq\alpha\in\mathbb{F}.$
This completes the proof of the part (b).

The part (c) immediately follows from Theorem III above, Theorem 1, Theorem
4(c), and parts (d),(e) follow from Theorem IV above, Theorem 5(b) and Theorem
4(b),(c). This completes the proof of Theorem 6.∎

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