# The Unapologetic Mathematician

## Outer Tensor Products

Let’s say we have two finite groups $G$ and $H$, and we have (left) representations of each one: ${}_GV$ and ${}_HW$. It turns out that the tensor product $V\otimes W$ naturally carries a representation of the product group $G\times H$. Equivalently, it carries a representation of each of $G$ and $H$, and these representations commute with each other. In our module notation, we write ${}_{GH}(V\otimes W)$.

The action is simple enough. Any vector in $V\otimes W$ can be written (not always uniquely) as a sum of vectors of the form $v\otimes w$. We let $G$ act on the first “tensorand”, let $H$ act on the second, and extend by linearity. That is:

\displaystyle\begin{aligned}g(v\otimes w)&=(gv\otimes w)\\h(v\otimes w)&=v\otimes(hw)\end{aligned}

and the action of either $g$ or $h$ on the sum of two tensors is the sum of their actions on each of the tensors.

Now, might the way we write the sum make a difference? No, because all the relations look like

• $(v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w$
• $v\otimes(w_1+w_2)=v\otimes w_1+v\otimes w_2$
• $(cv)\otimes w=v\otimes(cw)$

where in the last equation $c$ is a complex constant. Now, we can check that the actions of $g$ and $h$ give equivalent results on either side of each equation. For instance, acting by $g$ in the first equation we see

\displaystyle\begin{aligned}g((v_1+v_2)\otimes w)&=(g(v_1+v_2))\otimes w\\&=(gv_1+gv_2)\otimes w\\&=(gv_1)\otimes w+(gv_2)\otimes w\\&=g(v_1\otimes w)+g(v_2\otimes w)\end{aligned}

just as we want. All the other relations are easy enough to check.

But do the actions of $G$ and $H$ commute with each other? Indeed, we calculate

\displaystyle\begin{aligned}gh(v\otimes w)&=g(v\otimes(hw))\\&=(gv)\otimes(hw)\\&=h((gv)\otimes w)\\&=hg(v\otimes w)\end{aligned}

So we really do have a representation of the product group.

We have similar “outer” tensor products for other combinations of left and right representations:

\displaystyle\begin{aligned}({}_GV)\otimes(W_H)&={}_G(V\otimes W)_H\\(V_G)\otimes({}_HW)&={}_H(V\otimes W)_G\\(V_G)\otimes(W_H)&=(V\otimes W)_{GH}\end{aligned}

Now, let’s try to compute the character of this representation. If we write the representing homomorphisms $\rho:G\to\mathrm{End}(V)$ and $\sigma:H\to\mathrm{End}(W)$, then we get a representing homomorphism $\rho\otimes\sigma:G\times H\to\mathrm{End}(V\otimes W)$. And this is given by

$\displaystyle\rho\otimes\sigma(g,h)=\rho(g)\otimes\sigma(h)$

Indeed, this is exactly the endomorphism of $V\otimes W$ that applies $\rho(g)$ to $V$ and applies $\sigma(h)$ to $W$, just as we want. And we know that when expressed in matrix form, the tensor product of linear maps becomes the Kronecker product of matrices. We write the character of $\rho$ as $\chi$, that of $\sigma$ as $\psi$, and that of their tensor product as $\chi\otimes\psi$, and calculate:

\displaystyle\begin{aligned}\chi\otimes\psi(g,h)&=\mathrm{Tr}\left((\rho\otimes\sigma(g,h)\right)\\&=\mathrm{Tr}\left(\rho(g)\boxtimes\sigma(h)\right)\\&=\sum\limits_{i=1}^m\sum\limits_{j=1}^n\left(\rho(g)\boxtimes\sigma(h)\right)_{ij}^{ij}\\&=\sum\limits_{i=1}^m\sum\limits_{j=1}^n\rho(g)_i^i\sigma(h)_j^j\\&=\sum\limits_{i=1}^m\rho(g)_i^i\sum\limits_{j=1}^n\sigma(h)_j^j\\&=\mathrm{Tr}\left(\rho(g)\right)\mathrm{Tr}\left(\sigma(h)\right)\\&=\chi(g)\psi(h)\end{aligned}

That is, the character of the tensor product representation is the product of the characters of the two representations.

Finally, if both $\rho$ and $\sigma$ are irreducible, then the tensor product $\rho\otimes\sigma$ is as well. We calculate:

\displaystyle\begin{aligned}\langle\chi_1\otimes\psi_1,\chi_2\otimes\psi_2\rangle&=\frac{1}{\lvert G\times H\rvert}\sum\limits_{(g,h)\in G\times H}\overline{\chi_1\otimes\psi_1(g,h)}\chi_2\otimes\psi_2(g,h)\\&=\frac{1}{\lvert G\rvert}\frac{1}{\lvert H\rvert}\sum\limits_{g\in G}\sum\limits_{h\in H}\overline{\chi_1(g)}\overline{\psi_1(h)}\chi_2(g)\psi_2(h)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\overline{\chi_1(g)}\chi_2(g)\frac{1}{\lvert H\rvert}\sum\limits_{h\in H}\overline{\psi_1(h)}\psi_2(h)\\&=\langle\chi_1,\chi_2\rangle\langle\psi_1,\psi_2\rangle\end{aligned}

In particular, we find that $\langle\chi\otimes\psi,\chi\otimes\psi\rangle=\langle\chi,\chi\rangle\langle\psi,\psi\rangle$. If $\chi$ and $\psi$ are both irreducible characters, then our character properties tell us that both of the inner products on the right are $1$, and we conclude that the inner product on the left is as well, which means that $\chi\otimes\psi$ is irreducible.

November 4, 2010 -