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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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. |
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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. |
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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). |
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