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.

3 Comments »

I’m still reading and enjoying every posting in this series. Busy writing nonfiction for pay, and interviewing for teaching jobs. Please keep up this dazzlingly clear and painstaking exposition!

Comment by Jonathan Vos Post | November 5, 2010 | Reply
[…] say we have two left -modules — and — and form their outer tensor product . Note that this carries two distinct actions of the same group , and these two actions commute […]

Pingback by Inner Tensor Products « The Unapologetic Mathematician | November 5, 2010 | Reply
[…] now that we have two left actions by , we can take the outer tensor product, which carries an action by . Then we pass to the inner tensor product, acting on each tensorand by […]

Pingback by Tensors Over the Group Algebra are Invariants « The Unapologetic Mathematician | November 15, 2010 | Reply

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