\classheader{2012-10-26}

We'll continue talking about tensor products today.

Recall that any abelian group can be considered as a $\Z$-module, and therefore we can tensor them over $\Z$. In particular, for any rings $A$ and $B$, we can form the tensor product $A\otimes_\Z B$, which is a ring in the obvious way:
\[(a\otimes b)(a'\otimes b')=aa'\otimes bb'.\]
We can do the same for $k$-algebras; given $k$-algebras $A,B$, then $A\otimes_k B$ is a $k$-algebra, with the same operation as above.

A good way of thinking about $(A,B)$-bimodules is as left modules over $A\otimes_\Z (B^{\text{op}})$.

Given a left $A$-module $M$ and a left $B$-module $N$, we can form their ``external tensor product'' $M\otimes_\Z N$, sometimes denoted $M\boxtimes N$, which is a left $A\otimes_\Z B$-module.

Given two groups $G$ and $H$, and a $G$-representation $\rho$ and a $H$-representation $\rho'$, then we can form an external tensor product $\rho\boxtimes\rho'$ which is a $(G\times H)$-representation. To describe it explicitly, given $\rho:G\to\GL(V)$ and $\rho':H\to\GL(V')$, then
$\rho\boxtimes\rho':G\times H\to\GL(V\otimes_k V')$ maps $g\times h$ to $\rho(g)\otimes \rho'(h)$.

Note that the $G$-representation $\rho$ is equivalent to the left $kG$-module $V$, and the $H$-represenation $\rho'$ is equivalent to the left $kH$-module $V'$. We have a canonical isomorphism
\[k(G\times H)\cong kG\otimes_k kH,\]
and $\rho\boxtimes \rho'$ is the $(G\times H)$-represenation that corresponds to the $(kG\otimes_k kH)$-module $V\boxtimes V'$.

Given two representations $\rho$ and $\rho'$ of $G$, then we define $\rho\otimes \rho':G\to\GL(V\otimes V')$ by mapping $g$ to $\rho(g)\otimes \rho'(g)$. This is different than $\boxtimes$; for example, if $\rho'$ is the regular representation $G\lcirclearrowright kG$, then $V\otimes_k kG$ won't be the same $V\otimes_{kG}kG=V$ (recall $M\otimes_A A\cong M$).

In other words, given two modules over an algebra, we can tensor them over the base field, but there won't be any canonical action of the algebra on it. 


The special property $kG$ has is that it is a Hopf algebra, i.e. there is an algebra morphism $\delta:kG\to kG\otimes_kkG$ mapping $g$ to $g\otimes g$.

\begin{proposition}
$\text{}$

\begin{enumerate}
\item If $\rho$ is an irrep of a finite group $G$, and $\rho'$ is an irrep of a finite group $H$, then $\rho\boxtimes \rho'$ is an irrep of $G\times H$.
\item Any irrep of $G\times H$ has the form $\rho\boxtimes \rho'$ for irreps $\rho\in\widehat{G},\rho'\in\widehat{H}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Recall that $\chi_\rho(g)=\tr(\rho(g))$ and $\chi_{\rho'}(h)=\tr(\rho'(h))$. Then
\[\chi_{\rho\,\boxtimes\,\rho'}(g,h)=\tr(\rho(g)\otimes\rho'(h))=\tr(\rho(g))\tr(\rho'(h))=\chi_\rho(g)\chi_{\rho'}(h).\]
Thus
\[(\chi_{\rho\,\boxtimes\, \rho'},\chi_{\rho\,\boxtimes\,\rho'})=(\chi_\rho,\chi_\rho)\cdot(\chi_{\rho'},\chi_{\rho'}).\]
By the orthogonality relations, this is equal to $\# G \cdot\# H=\#(G\times H)$, and therefore $\rho\boxtimes\rho'$ is an irrep of $G\times H$.

To see part 2, we use a counting argument. The number of representations of the form $\rho\boxtimes\rho'$ with $\rho\in\widehat{G},\rho'\in\widehat{H}$ is just $\#(\widehat{G}\times\widehat{H})$. 

\textit{A priori}, $\#(\widehat{G\times H})=\#$ of conjugacy classes in $G\times H$, but a conjugacy class in $G\times H$ is just a product of conjugacy classes from $G$ and $H$, so
\begin{align*}
\#(\widehat{G\times H})&=\#\text{ of conjugacy classes in }G\times H\\
&=(\#\text{ of conjugacy classes in }G)\cdot(\#\text{ of conjugacy classes in }H)\\
&=\#\widehat{G}\cdot\#\widehat{H}.\qedhere
\end{align*}
\end{proof}

If $A$ is a ring, and $M$ is an $A$-bimodule, we can form
\[T_A(M)=A\otimes M\oplus (M\otimes_AM)\oplus(M\otimes_A M\otimes_AM)\oplus\cdots\]
which is an algebra where the operation on simple tensors in a single degree is just concatenation,
\[(m_1\otimes \cdots\otimes m_k)\cdot(n_1\otimes\cdots \otimes n_\ell)=m_1\otimes\cdots\otimes m_k\otimes n_1\otimes\cdots\otimes n_\ell.\]
Fix a finite set $X$, and let $A=k\{X\}=\bigoplus_{x\in X}k\cdot\mathbf{1}_x$. An $A$-module $M$ decomposes as $M=\bigoplus_{x\in X}M_x$ where $M_x$ is a $k$-vector space. Take another $A$-module $N$. How can we think about $M\otimes_A N$?

We know that $M\otimes_A N$ will be a quotient of $M\otimes_kN=\bigoplus_{x,y\in X}M_x\otimes_k N_y$. Specifically,
\[M\otimes_A N=\frac{\bigoplus_{x,y}M_x\otimes_k N_y}{\big\langle(m\cdot\mathbf{1}_z)\otimes n=m\otimes (\mathbf{1}_zn)\text{ for all }z\in X\big\rangle}\cong\bigoplus_{x\in X}M_x\otimes_k N_x.\]
More generally, if $X$ and $Y$ are finite sets, then
\[k\{X\}\otimes_kk\{Y\}=k\{X\times Y\},\]
via the map $\mathbf{1}_x\otimes \mathbf{1}_y\mapsto \mathbf{1}_{x,y}$.

Now let $M, N$ be $k\{X\}$-bimodules. Then $M=(M_{x,y})_{(x,y)\in X\times X}$ because a $k\{X\}$-bimodule is just a left module over $k\{X\}\otimes k\{X\}^{\text{op}}=k\{X\}\otimes k\{X\}=k\{X\times X\}$.

For any $m_{x',y'}\in M_{(x',y')}$, we have
\[\mathbf{1}_x\cdot m_{x',y'}\cdot\mathbf{1}_y=\begin{cases}
m_{x,y} & \text{ if }x=x',y=y',\\
0 & \text{ otherwise}.
\end{cases}\]
Thus
\[M\otimes_AN=\bigoplus_{x_1,x_2\in X}\left(\bigoplus_{y\in X}\left(M_{x_1,y}\otimes_k N_{y,x_2}\right)\right).\]

Let $M$ be a bimodule over $A=k\{X\}$. Then
\[T_AM=A\oplus \left(\bigoplus_{x_1,x_2} M_{x_1,x_2}\right)\oplus\left(\bigoplus_{x_1,x_2,x_3}M_{x_1,x_2}\otimes M_{x_2,x_3}\right)\oplus\left(\bigoplus_{x_1,x_2,x_3,x_4}M_{x_1,x_2}\otimes M_{x_2,x_3}\otimes M_{x_3,x_4}\right)\cdots\]

We will now define a construction called the smash product. Let $A$ be a ring, and let $G$ be a group acting on $A$ by automorphisms. For example, we might have $G\subset A^\times$, and $G\lcirclearrowright A$ via $g\cdot a=gag^{-1}$. Let's use the notation ${}^ga$ for $g$ acting on $a$.

A $G$-equivariant $A$-module $M$ is an $A$-module $A\otimes M\to M$ and a $G$-representation $G\times M\to M$ such that $g(am)={}^gag(m)$. In the case of $G\subset A^\times$, then this just says that
\[g(am)=(gag^{-1})(gm)=(ga)m.\]
The smash product $A\# G$ will be an algebra such that modules over $A\# G$ are equivalent to $G$-equivariant $A$-modules.








