A classical theorem of Clifford states that if G is a finite group and K a field, then every irreducible right KG-module is a completely reducible right KN-module, where N is any normal subgroup of G. Is there a Lie theoretic analog of this result? That is, if L is a finite-dimensional Lie algebra, I an ideal of L, and M an irreducible L-module, is M a completely reducible I-module? I expect the answer is negative, but what about a counterexample?
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asked Jun 23, 2011 at 22:06
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1$\begingroup$ You don't need that $G$ is finite, right? You just need that $N$ has finite index. $\endgroup$– Qiaochu YuanCommented Jun 24, 2011 at 19:08
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$\begingroup$ Yes, you are right. Indeed, Clifford's Theorem holds in the weaker hypothesis in which the normal subgroup N has finite index. $\endgroup$– Salvatore SicilianoCommented Jun 24, 2011 at 20:08
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2 Answers
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The positive results seem at most to be tied to finite dimensional representations in positive characteristic:
Let $\frak H$ be the Heisenberg algebra with basis $x,y,$ where $c$ is central and $[y,x]=c$. It acts on polynomials $k[x]$ with $x$ acting by multiplication by $x$ and $y$ acting as $d/dx$ and $c$ acting as the identity. If $k$ is a field of characteristic $0$ this is an irreducible representation. On the other hand, $y$ and $c$ span a normal subalgebra and the restriction of this representation to it is not irreducible, polynomials of degree $\le n$ is a subrepresentation for all $n$.
Similarly, in characteristic $p$ the action passes to $k[x]/(x^p)$ which is a finite dimensional irreducible representation whose restriction to $\langle y,c\rangle$ is not irreducible (for the same reason as in characteristic $0$).
The problem as compared with the group case is that automorphisms must be replaced by derivations. If $\phi$ is an automorphism of a group $N$ and $V$ is a $N$-representation, then we may define a new $N$-action on $V$ given by $n\cdot v:=\phi(n)v$. It is clear that $V$ is irreducible precisely when this new action is and that if $V$ is a subrepresentation of a $G$-representation $W$ and $\phi\in G$, then $\phi(V)$ is isomorphic to this twisted action. The fact that $\phi(V)$ is semi-simple when $V$ is irreducible is the key fact used in order to prove the group case.
In the Lie algebra case we instead have a derivation $D$ of a Lie algebra $\frak n$ and if $V$ as before lies in $W$ we have $xD(v)=[x,D]v+Dxv=D(x)v+Dxv$. Hence, instead of, in the abstract case, defining a twisted action on $V$ we have to define an action on $V\times V$ given by $x(u,v)=(xu+D(x)v,xv)$. This instead is an extension of $V$ by $V$ and may not be even semi-simple when $V$ is irreducible. This is exactly what happens in the two examples above.
Hence, I am not totally convinced that the statement is true even in the finite dimensional characteristic zero case when say the Lie algebra is not algebraic (i.e., not the Lie algebra of an algebraic group). I do not however have a counterexample.
answered Jun 24, 2011 at 17:40
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$\begingroup$ Yes, your fourth paragraph spells out some details omitted from my proof sketch. The Lie correspondence works in characteristic 0 without extra assumptions (and by Ado's theorem, the Lie algebra is linear, hence one can exponentiate), thus I don't think that algebraicity of the Lie algebra is an issue. The correspondence does, however, break down for infinite-dimensional representations, as in your first example. $\endgroup$ Commented Jun 24, 2011 at 18:25
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$\begingroup$ Yes, you're right, from my point of view the representation extends to the algebraic closure (the group generated by the exponentials). $\endgroup$ Commented Jun 24, 2011 at 18:35
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$\begingroup$ Dear Torsten, thank you for your answer which confirm what I suspected. Really, I was mainly interested to the modular case, so your counterexample is just what I needed. $\endgroup$ Commented Jun 24, 2011 at 20:35
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If the module $M$ is finite-dimensional then the answer seems to be affirmative.
Let $Soc_I(M)$ be the maximal semisimple $I$-submodule of $M,$ which is non-zero by the finite-dimensio-nality of $M.$ It is easy to check that it's also $L$-invariant (thinking in Lie groups, this follows from the formula $ngv=g(g^{-1}ng)v, g\in G, n\in N$), hence $Soc_I(M)$ coincides with $M.$ This precisely means that $M$ is completely reducible as an $I$-module.
answered Jun 24, 2011 at 5:58
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$\begingroup$ Dear Victor, your argument sounds OK: thank you. What about the modular case? At this purpose, I remember that all simple modules of a finite-dimensional Lie algebra over a field of positive characteristic are finite-dimensional (by a theorem of Jacobson). $\endgroup$– user15982Commented Jun 24, 2011 at 14:06
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$\begingroup$ The issue is not the finite-dimensionality of simple modules. Rather, it is the existence of a simple submodule of a given module. For instance, if $\mathfrak{g}$ is a Lie algebra of a positive dimension over a field, the universal enveloping algebra $U(\mathfrak{g})$ is a noncommutative domain and hence has no minimal left or right ideals. $\endgroup$ Commented Jun 24, 2011 at 18:29