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The category of vector (over $k$) representations of a group $G$ is isomorphic to the category of left $k[G]$-modules, where $k[G]$ is the group ring. This isn't just an equivalence of categories, but rather an isomorphism, which is stronger. However we define a tensor product of representations as $g(v\otimes w)= gv\otimes gw$, which differs from the tensor product over $k[G]$ of $k[G]$-modules. For various reasons, we have to define it this way. For starters, one cannot tensor two left modules if the base ring is not commutative.

But there's something unsettling about using a different tensor, about forgetting the base ring. Is there some deeper explanation for this discrepancy other than just "it works"?

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up vote 6 down vote accepted

Left modules over an arbitrary noncommutative ring don't have a natural notion of tensor product. What is true generally is the following: if $R, S$ are two $k$-algebras ($k$ an arbitrary commutative ring) and $V, W$ two modules over $R, S$, then the tensor product $V \otimes_k W$ is naturally a module over $R \otimes_k S$ (by the functoriality of the tensor product).

In particular, the tensor product $V \otimes_k W$ of two $k[G]$-modules is naturally a $k[G] \otimes_k k[G]$-module. The way we turn it into a $k[G]$-module is by defining a map $$k[G] \to k[G] \otimes_k k[G]$$

called the comultiplication. It sends $g$ to $g \otimes g$ and naturally gives $k[G]$ the structure of a bialgebra. In fact $k[G]$ is naturally even a Hopf algebra with antipode given by $g \mapsto g^{-1}$, and this lets us define duals of representations also.

The comultiplication is an extremely natural map. In fact, for any set $S$ we have a canonical diagonal map $$\Delta : S \to S \times S$$

sending $s$ to $(s, s)$, and this induces a comultiplication on the vector space $k[S]$ making it a coalgebra. So in some sense it is even more basic a structure than the multiplication on $k[G]$ and so there is no reason not to use it to define the tensor product.

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However, because $k[G] \cong k[G]^\textrm{op}$, every left-$k[G]$ module is automatically a right-$k[G]$ module. This still gives the wrong tensor product, of course. –  Zhen Lin Feb 22 '12 at 23:15
@Zhen: yes, I observed that the other day and it confused me for awhile until I realized that the tensor product you get this way is very boring. –  Qiaochu Yuan Feb 22 '12 at 23:27
Interesting. So two things: it's not correct to say we use a different tensor product for left $R$-modules and left $k[G]$-modules. Rather the former has no useful tensor product (we discard the $k[G]^{\text{op}}$ construction), while the latter does have one. Second, the reason we do have a tensor product for left $k[G]$-modules is that it they more structure than modules over rings; they are modules over bialgebras with a comultiplication. Seems a little mysterious, but obviously the right answer. –  Joe Hannon Feb 23 '12 at 0:34
Dear Qiaochu, I wouldn't say that the tensor product over $k[G]$ is boring, just that it is different: it is the $G$-coinvariants of the $G$-module structure on $V\otimes_kW$. Sometimes that is something of interest (but sometimes not). (Compare with the fact that $Hom_k(V,W)$ is naturally a $G$-rep'n as well, and its subspace of $G$-invariants is equal to the $k$-vector space $Hom_{k[G]}(V,W)$ of $G$-equivariant maps. Again, some times this is of interest, and sometimes not.) More generally, it is often useful to think of tensor product over a larger ring as a quotient of the tensor ... –  Matt E Feb 23 '12 at 2:59
... product over a smaller ring. (Just as one can think of the Homs over a larger ring as being a submodule of the Homs over a smaller ring.) Regards, –  Matt E Feb 23 '12 at 3:01
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