Apocryphal Maschke theorem?

Question

This may be totally trivial or wrong. I am just posting this because I am sick and tired of trying to understand this myself and I am sure someone out here can just answer it out of his head in 2 minutes.

Let $G$ be a finite group, and $k$ an algebraically closed field whose characteristic is not a divisor of $\left|G\right|$. The Maschke theorem in its standard form states that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as $k$-algebras. This quickly yields that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as left $k\left[G\right]$-modules, where the $k\left[G\right]$-module structure on $\mathrm{End}V$ is given by $\left(gF\right)\left(v\right)=gF\left(v\right)$ for every $g\in k\left[G\right]$, $f\in\mathrm{End}V$ and $v\in V$.

Now, I've overheard that there exists a stronger form of this, stating that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as left $k\left[G\times G\right]$-modules, where the $k\left[G\times G\right]$-module structure on $k\left[G\right]$ is defined by $\left(g,h\right)u=guh^{-1}$ for any $g\in G$, $h\in G$ and $u\in k\left[G\right]$, and the $k\left[G\times G\right]$-module structure on $\mathrm{End}V$ is defined by $\left(g,h\right)F=gFh^{-1}$ for any $g\in G$, $h\in G$ and $F\in \mathrm{End}V$. This would follow from $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as $k$-Hopf algebras, where the Hopf algebra structure on $k\left[G\right]$ is the standard one ($S\left(g\right)=g^{-1}$ for every $g\in G$), and the Hopf algebra structure on $\mathrm{End}V$ is the standard one as well.

[EDIT: as the comments explained, there is no such thing as a standard Hopf algebra structure on $\mathrm{End}V$, and so "$k$-Hopf algebras" should be "$k$-algebras".]

Is this correct, and how can this be proven?

Answer 1

The result about bimodules is true, and standard. Here is one way to see it.

By Frobenius reciprocity, $\operatorname{Hom}_G(V,k[G]) = \operatorname{Hom}_k(V,k)$, since $k[G]$ is the induction (or coinduction, depending on your terminology) of the trivial representation of the trivial subgroup of $G$ to $G$.

Since Frobenius reciprocity is functorial, one easily sees that this canonical isomorphism is an isomorphism of right $G$-representations, where the source has a right $G$-action coming from the right $G$-action on $k[G]$, and the target has a right $G$-action coming as the transpose of the left $G$-action on $V$.

Now by Maschke's semisimplicity theorem, we know that $k[G] = \bigoplus_V V \otimes_k \operatorname{Hom}_G(V,k[G])$, where the sum is over all irreducible left $G$-representations. (Indeed, Maschke shows that this is true for any left $G$-module in place of $k[G]$.) Again, this is a natural isomorphism, and so respects the right $G$-actions on source and target.

Combined with the preceding computation, we find that $k[G] = \bigoplus_V V\otimes_k \operatorname{Hom}_k(V,k) = \bigoplus_V \operatorname{End}(V),$ as both left and right $G$-modules, as required.

[Edit:] Leonid's remark about $k$ being needing to be big enough in his answer below is correct. Each simple $V$ comes equipped with an associated division algebra of $G$-endomorphisms $A_V := \operatorname{End}_G(V)$. The representation $V$ is absolutely irreducible (i.e stays irred. after passing to any extension field) if and only if $A_V = k$. When we consider $\operatorname{Hom}_k(V,W)$ for another left $G$-module $W$, this is naturally an $A_V$-module, and Maschke's theorem will say that $W = \bigoplus_V \operatorname{Hom}_k(V,W)\otimes_{A_V} V$. (I have written the factors in the tensor product in this order because $V$ is naturally a left $A_V$-module (if we think of endomorphisms acting on the left), and then $\operatorname{Hom}_k(V,W)$ becomes a right $A_V$-module.)

So in the case of $W$ being the group algebra, we have $$k[G] = \bigoplus_V \operatorname{Hom}_k(V,k)\otimes_{A_V} V$$ (an isomorphism of $G$-bimodules).

If all the $V$ are absolutely irreducible, e.g. if $k$ is algebraically closed, then all the $A_V$ just equal $k$, and the preceding direct sum reduces to what I wrote above, and what was written in the question.

Answer 2

1. The assertion that k[G] is isomorphic to the direct sum of End(V) as a k-algebra is not true for an arbitrary field k of the characteristic not dividing |G|. Some additional condition must be imposed ensuring that k is big enough, e.g., it would suffice that k be algebraically closed. E.g., for k being the field of real numbers and G being a cyclic group of order greater than 2 this is not true.

2. The assertion about k[G×G]-module isomorphism of course follows from the assertion about k-algebra isomorphism, as isomorphic k-algebras are also isomorphic bimodules over themselves. One also has to remember where this isomorphism of k-algebras comes from, namely, that each projection k[G] → End(V) comes from the action of G in V.

3. As to the Hopf algebra structure on k[G], it corresponds to the tensor category structure on the category of G-modules, i.e., to recover it one needs to know how tensor products of irreducible representations decompose into direct sums of irreducible representations (not only the multiplicities, but the actual decompositions).

Answer 3

The isomorphism in all of the forms you wrote is the obvious one (send an element of kG to the endomorphism it induces on all simples), so you just have to notice that this map is a map is a map of bimodules.

I don't know what you're talking about with respect to Hopf algebras. There is no Hopf algebra structure on $\mathrm{End} V$. You seem to just mean "algebra equipped with an anti-involution" but even that doesn't make any sense, since it only makes sense to think of adjoint as an antiendomorphism of $\mathrm{End} V$ if you've picked an isomorphism of $V$ and its dual.

On a side note: the other answerers are correct that if not all irreducible representations of $G$ are defined over $k$, the map above is not surjective if one interprets $\mathrm{End}(V)$ as $\mathrm{End}_k(V)$. If one instead takes endomorphisms which commute with endomorphisms of the representation (what Matt Emerton denotes $A_V$), the statement can be saved: the map $kG\cong \oplus \mathrm{End}_{A_V}(V)$, since any semi-simple algebra is equal to its own double commutator on any faithful representation.

Answer 4

I would tend to take the perspective that imposing the stated structure of (left) $k[G \times G]$-module on $k[G]$ naturally imposes a $k[G \times G]$-module structure on each ${\rm End}_{k}(V).$ As long as ${\rm char}(k)$ is not a divisor of $|G|$, we know that $k[G]$ is a semisimple $k$-algebra. On the other hand, we know that $k[G]$ is a completely reducible $k[G \times G]$-module with the stated action. The simple summands as $k[G \times G]$ module are mutually annihilating minimal two sided ideals of $k[G]$, so each one is a simple $k$-algebra in its own right. In the considered case that $k$ is algebraically closed, each of these simple algebras has the form ${\rm End}_{k}(V)$ for a simple $k[G]$-module $V.$ In general, (when $k$ is not necessarily algebraically closed),each simple algebra has the form $M_{d}({\rm End}_{kG}(V))$ for some integer $d$ and some simple $kG$-module $V$ (recall that ${\rm End}_{kG}(V)$ is a $k$-division algebra by Schur's Lemma).