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\title{A variant of Baer's theorem}
\author{Pasha Zusmanovich}
\institute{University of Ostrava}
\date{
AAA103, Tartu
\\
June 9, 2023
}
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\begin{frame}{``Baer's theorem'', 1952}
``If endomorphism rings of vector spaces are isomorphic, then the vector
spaces themselves are isomorphic''.
\bigskip
More precisely:
\bigskip
\begin{block}{Theorem}
Let $V$, $W$ be (infinite-dimensional) vector spaces over a division ring $D$.
If $\Phi: \End_D(V) \to \End_D(W)$ is an isomorphism, then there there is an
isomorphism $\alpha: V \to W$ such that
$$
\Phi(f) = \alpha \circ f \circ \alpha^{-1}
$$
for any $f \in \End_D(V)$.
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\begin{frame}{A bit of history}
\begin{itemize}
\item Eidelheit, Mackey, et al. (end of 1930s-1940s): analytic setting
(bounded operators on Banach spaces, continuous operators on normed spaces)
\item Dieudonn\'e, Jacobson (1940s): rings of finitary linear maps
\item Baer (1952): using properties of idempotents
\item Wolfson (1953, PhD thesis under Baer): using Jacobson's density theorem
\item Racine (1998), Balaba (2005): super- and graded cases
\item lot of authors: modules over abelian groups (``Baer-Kaplansky theorem'')
\end{itemize}
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\begin{frame}{Another variant of Baer's theorem}
Instead of $\End_D(V)$, or its subring $\FEnd_D(V)$ of finitary linear maps,
consider $\FEnd_D(V,\Pi)$, the ring generated by all ``infinitesimal
transvections''
$$
t_{v,f}: u \mapsto vf(u)
$$
where $\Pi$ is a subspace of $V^*$, $v \in V$, $f \in \Pi$.
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\begin{frame}{Another variant of Baer's theorem (cont.)}
\begin{block}{Theorem}
Let $V, W$ be right vector spaces over a division ring $D$, $\Pi$ a nonzero
finite-dimensional subspace of $V^*$, $\Gamma$ a finite-dimensional subspace of
$W^*$, and $\Phi: \FEnd_D(V,\Pi) \to \FEnd_D(W,\Gamma)$ an isomorphism of
$D$-algebras. Then there is an isomorphism of $D$-vector spaces
$\alpha: V \to W$ such that
\begin{equation*}
\Phi(f) = \alpha \circ f \circ \alpha^{-1}
\end{equation*}
for any $f \in \FEnd_D(V,\Pi)$.
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\begin{frame}{Idea of the proof}
$\FEnd_D(V,\Pi)$ generally, does not have idempotents, and Jacobson's density
theorem does not hold, so all previous methods will not work.
\bigskip
Instead, write $\FEnd_D(V,\Pi)$ as $V \otimes_D \Pi$ and use elementary linear
algebra:
$$
\Hom_D(V \otimes_D \Pi, W \otimes_D \Gamma) \simeq
\Hom_D(V,W) \otimes_D \Hom_D(\Pi,\Gamma) ,
$$
Hence $\Phi$ belonging to the left-hand side can be written as some element
of the tensor product at the right-hand side of rank $\rk(\Phi)$.
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\begin{frame}{Idea of the proof (cont.)}
\begin{block}{The crucial lemma}
$$
\Tr \Phi(\xi) = \rk(\Phi) \Tr (\xi)
$$
for any $\xi \in V \otimes_D \Pi$.
\end{block}
\bigskip
Apply the crucial lemma for $\Phi$ and for $\Phi^{-1}$ to get $\rk(\Phi) = 1$,
and the rest is trivial.
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\begin{frame}{Open question}
What about Lie and Jordan rings $\FEnd_D(V)^{(\pm)}$, $\FEnd_D(V,\Pi)^{(\pm)}$?
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{\Huge That's all. Thank you.}
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