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\title{Lie $p$-algebras of cohomological dimension one}
\author{Pasha Zusmanovich}
\institute{University of Ostrava}
\date{7ECM, Berlin \\ July 19, 2016}
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(Based in arXiv:1601.00352)
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\begin{frame}{What is cohomological dimension?}
$A$ -- an object in a category with ``good'' cohomology theory
\\
(Lie algebras, associative algebras, groups, ...)
\bigskip
Cohomological dimension of $A$: the least number $n$ such that
$$
\Homol^{n+1} (A,\>\cdot\>) = 0
$$
\begin{block}{A natural question}
Objects of ``small'' cohomological dimension (say, $\le 1$).
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\begin{block}{Example}
Free objects have cohomological dimension $1$.
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\begin{frame}{Groups of cohomological dimension $\le 1$}
\begin{block}{Elementary fact}
Groups of cohomological dimension $0$ are trivial.
\end{block}
\begin{block}{A classical theorem (Stallings, Swan, late 1960s)}
Groups of cohomological dimension $1$ are free.
\end{block}
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\begin{frame}{Associative algebras of cohomological dimension $\le 1$}
\begin{block}{A classical result \\
(Eilenberg, Hochschild, Rosenberg, Zelinsky, ..., 1940-1950s)}
Associative algebras of cohomological dimension $0$ are finite-dimensional
separable algebras (roughly, semisimple algebras).
\end{block}
\vskip 1cm
Associative algebras of cohomological dimension $1$:
\\
interesting, but messy
(Cuntz--Quillen, 1995).
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\begin{frame}{Lie algebras of cohomological dimension $\le 1$}
\begin{block}{Elementary fact}
Lie algebras of cohomological dimension $0$ are trivial.
\end{block}
\begin{block}{Question (Bourbaki)}
Is it true that Lie algebras of cohomological dimension $1$ are free?
\end{block}
\begin{block}{Theorem (G.L. Feldman, 1983)}
Yes, for $2$-generated algebras.
\end{block}
\begin{block}{Theorem (A.A. Mikhalev--Umirbaev--Zolotykh, 1994)}
No, generally. Example:
$$
\left\langle\> x, y, z \vertbar x + [y,z] + (\ad x)^p z = 0 \>\right\rangle
$$
over a field of characteristic $p>2$.
\end{block}
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\begin{frame}{Lie algebras of cohomological dimension $1$}
\begin{block}{Remaining questions}
What about characteristic $0$? characteristic $2$? Lie $p$-algebras?
\end{block}
\begin{block}{Main fact about cohomological dimension}
Cohomological dimension does not increase when passing to subalgebras.
In particular, the class of Lie algebras of cohomological dimension $1$ is
closed with respect to subalgebras.
\end{block}
\begin{block}{Proof}
Shapiro's lemma about cohomology of coinduced modules.
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\begin{frame}{Lie $p$-algebras of cohomological dimension $1$}
\begin{block}{Example}
Free Lie $p$-algebras have cohomological dimension $\infty$.
\end{block}
\begin{block}{Proof}
The free Lie $p$-algebra of rank $1$,
$\langle x, x^{[p]}, x^{[p]^2}, \dots \rangle$, is infinite-dimensional and
abelian, hence has cohomological dimension $\infty$.
\end{block}
\begin{block}{Theorem}
A Lie $p$-algebra of cohomological dimension $1$ is $1$-dimensional.
\end{block}
\begin{block}{Proof}
The condition
$$
x^{[p]} = \lambda(x) x
$$
+ Feldman's result.
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\begin{frame}{On the condition $x^{[p]} = \lambda(x) x$}
\begin{block}{A famous question (Jacobson, 1960s)}
Is a Lie $p$-algebra satisfying
$$
x^{[p]^{n(x)}} = x ,
$$
abelian?
\end{block}
\begin{block}{Theorem}
A Lie $p$-algebra over an algebraically closed field satisfying
$$
x^{[p]^{n(x)}} = \lambda(x) x ,
$$
with $n(x)$'s bounded, contains a nonzero $p$-nilpotent element.
\end{block}
\begin{block}{Corollary}
An alternative proof of the theorem about Lie $p$-algebras of cohomological
dimension one.
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\begin{frame}{Lie $p$-algebras of \emph{restricted} cohomological dimension $\le 1$}
Category of Lie $p$-algebras:
\\
restricted cohomology $\leadsto$ \emph{restricted} cohomological dimension
($\cd_*$)
\begin{block}{Theorem (Hochschild, 1950s)}
Lie $p$-algebras of restricted cohomological dimension $0$ are
finite-dimensional tori.
\end{block}
\begin{block}{Examples}
Let
$$
0 \to L_1 \to L \to L_2 \to 0 .
$$
Then
$$
\cd_*(L_1) = \genfrac{}{}{0pt}{}{0}{1} \quad\text{and}\quad
\cd_*(L_2) = \genfrac{}{}{0pt}{}{1}{0} \quad\Rightarrow\quad \cd_*(L) = 1 .
$$
(Follows from the Lyndon--Hochschild--Serre spectral sequence).
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\begin{frame}{Lie $p$-algebras of restricted cohomological dimension $1$}
\begin{block}{Conjecture}
Any Lie $p$-algebra of restricted cohomological dimension $1$ is of the form
$$
( \dots ((\mathscr L \bowtie T_1) \bowtie T_2) \dots ) \bowtie T_n ,
$$
where $\mathscr L$ is a free Lie $p$-algebra, $T_i$'s are finite-dimensional
tori, and $\bowtie \>\in \{ \rtimes, \ltimes \}$.
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\begin{block}{Some facts supporting the conjecture}
For a Lie $p$-algebra of restricted cohomological dimension $1$ the following
holds:
\begin{itemize}
\item A $p$-subalgebra is either a torus, or is infinite-dimensional.
\item
An abelian $p$-subalgebra is either a torus, or \\
(the free Lie $p$-algebra of rank $1$) $\oplus$ (torus).
\item It has (ordinary) cohomological dimension $\infty$.
\end{itemize}
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{\Huge
That's all. Thank you.
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