# A weak Schur's lemma for non-semisimple finite dimensional algebras

Let $$B \subseteq C$$ be an inclusion of finite dimensional (associative) algebras over a field $$k$$. Assume that $$C$$ is a free $$B$$-module. Let $$\bigoplus_i U_i$$ be a decomposition of $$B$$ into indecomposable $$B$$-modules. Let $$\bigoplus_j V_{ij}$$ be a decomposition of the extension $$C \otimes_B U_i$$ into indecomposable $$C$$-modules.

Question: Pick $$u \in U_i$$ nonzero. Is there $$\varphi_u \in Hom_B(U_i,V_{ij})$$ such that $$\varphi_u(u)$$ is nonzero?

Remark: By Frobenius reciprocity, $$Hom_B(U_i,V_{ij}) \simeq Hom_C(C \otimes_B U_i,V_{ij})$$ which is nonzero.
In the semisimple case, Schur's lemma implies the existence of an injective map $$\varphi$$, which solves the problem for all $$u$$ nonzero. In the non-semisimple case, we cannot expect the existence of such an injective map, we are just asking about the existence of a map $$\varphi_u$$ as above with $$u \not \in ker(\varphi_u)$$.

Assuming a positive answer, some conditions could be useless, but it is exactly the case I need.

## 1 Answer

No, not necessarily.

Consider the case $$B=kH$$, $$C=kG$$ of finite group algebras over a field $$k$$, where $$H\leq G$$. I'll write $$\downarrow$$ and $$\uparrow$$ for restriction and induction. This case has a couple of simplifying features. Firstly, induction is both left and right adjoint to restriction. Secondly, $$kH$$ is selfinjective, and so the projective module $$U_i$$ is also injective. If the answer to the question were "yes", then there would be an injection from $$U_i$$ to a direct sum of copies of $$V_{ij}\!\downarrow$$. But since $$U_i$$ is injective, this would split, and so, by Krull-Schmidt, $$U_i$$ would be a direct summand of $$V_{ij}\!\downarrow$$.

Suppose that $$U_i$$ is the projective cover of the simple $$kH$$-module $$S$$ and $$V_{ij}$$ the projective cover of the simple $$kG$$-module $$T$$.

Then $$V_{ij}$$ is a direct summand of $$U_i\!\uparrow$$ if and only if $$\text{Hom}_{kG}(U_i\!\uparrow, T)\cong\text{Hom}_{kH}(U_i,T\!\downarrow)$$ is nonzero, which is the case if and only if $$T\!\downarrow$$ has a composition factor isomorphic to $$S$$.

$$U_i$$ does not occur as a direct summand of $$V_{ij}\!\downarrow$$ if and only if $$\text{Hom}_{kH}(V_{ij}\!\downarrow,S)\cong\text{Hom}_{kG}(V_{ij},S\!\uparrow)$$ is zero, which is the case if and only if $$S\!\uparrow$$ does not have a composition factor isomorphic to $$T$$.

So we just need to find simple modules $$S$$ for $$kH$$ and $$T$$ for $$kG$$ for which $$S$$ is a composition factor of $$T\!\downarrow$$ but $$T$$ is not a composition factor of $$S\!\uparrow$$.

This happens quite commonly, but $$S$$ can't appear in the head or socle of $$T\!\downarrow$$, or else there would be a map in one direction between $$S$$ and $$T\!\downarrow$$, and hence between $$S\!\uparrow$$ and $$T$$, and so $$T$$ would appear in the socle or head of $$S\!\uparrow$$. So $$T$$ has to be reasonably large.

One example that I happen to be familiar with is where $$k$$ is algebraically closed of characteristic two, $$G=A_5$$ and $$H=A_4$$. Then $$kG$$ has a $$4$$-dimensional simple projective module $$T$$ and $$kH$$ has three one-dimensional simple modules: take $$S$$ to be one of the non-trivial ones. Then $$S$$ and $$T$$ have the required properties.

• Do you expect the existence of a counter-example if the field is algebraically closed of characteristic zero? If yes, what about $k=\mathbb{C}$? – Sebastien Palcoux Mar 14 '19 at 12:25
• @SebastienPalcoux Not with group algebras, of course, as they'd be semisimple. But for abstract algebras I wouldn't expect the characteristic of $k$ to matter. It might just be harder to construct the examples because you don't have the groups to play with. – Jeremy Rickard Mar 14 '19 at 12:39
• Why the projective cover of a simple $kH$-module $S$ must be indecomposable and direct summand of $kH$? – Sebastien Palcoux Mar 14 '19 at 17:07
• @SebastienPalcoux For any finite dimensional algebra, there’s a bijection between (isomorphism classes of) indecomposable projective modules and simple modules, given by taking the quotient by the radical in one direction and taking the projective cover in the other. – Jeremy Rickard Mar 14 '19 at 20:35