# Apocryphal Maschke theorem?

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 ($S\left(A\right)=\mathrm{adj}A$ for every matrix $A\in\mathrm{End}V$.

Is this correct, and how can this be proven?

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What is the standard Hopf algebra structure on $\mathrm{End}\;V$? –  Mariano Suárez-Alvarez Jan 12 '10 at 20:13
Ok, fail on my side. But the question about the isomorphism of $k\left[G\times G]$-modules remains... –  darij grinberg Jan 12 '10 at 20:18
As far as I can tell, your statement of Maschke is a combination of Maschke (= complete reducibility) and Artin-Wedderburn (characterization of semisimple rings). –  S. Carnahan Jan 12 '10 at 20:42
There is no Hopf algebra structure on End(V), in particular there's no counit. Doesn't the result about bimodules follow immediately from the result on the level of rings? –  Noah Snyder Jan 12 '10 at 21:15
@darij: this is me with my TeX-evangelists hat on talking: you do not need to use \left and \right for small things like k[G], where you know that what is being bracketed is not going to be taller than normal (and in fact, you do not want in general for the brackets to grow even if the inside is slightly taller than ordinary---think of an exponent) –  Mariano Suárez-Alvarez Jan 12 '10 at 23:18
By Frobenius reciprocity, $Hom_G(V,k[G]) = Hom_k(V,k),$ 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$. Now if by Maschke's semisimplicity theorem, we know that$k[G] = \bigoplus_V V \otimes_k Hom_G(V,k[G])$, where the sum is over all irreducible left$G$-representations. (Indeed, Mashke shows that this is true for any left$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 Hom_k(V,k) = \bigoplus_V 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 := 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$. When we consider$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 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$-module (if we think of endomorphisms acting on the left), and then$Hom\_k(V,W)$becomes a right$A\_V$-module. So in the case of$W$being the group algebra, we have $$k[G] = \bigoplus\_V Hom\_k(V,k)\otimes_{A\_V} V$$ (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. - add comment 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). - Thanks. As for 1., I have now fixed it. – darij grinberg Jan 12 '10 at 21:58 Thanks; I have corrected my answer to reflect this your comment 1. – Emerton Jan 12 '10 at 22:35 Of course, the direct sum statement is true if you take endomorphisms of$V$over the algebra of endomorphisms of the representation (which Matt Emerton denotes$A_V$). – Ben Webster Jan 13 '10 at 1:33 Yes, this way it's right. – Leonid Positselski Jan 13 '10 at 1:55 add comment 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. - Thanks, the first paragraph solved my confusion. I already realized that$\mathrm{End}V\$ is quite a strange object to give a Hopf algebra structure to... Somehow, whenever I am confused about something, I tend to spread the confusion to other stuff as well. –  darij grinberg Jan 12 '10 at 22:03