# The Unapologetic Mathematician

## Tensors Over the Group Algebra are Invariants

It turns out that we can view the space of tensors over a group algebra as a subspace of invariants of the space of all tensors. That is, if $V_G$ is a right $G$-module and ${}_GW$ is a left $G$-module, then $V\otimes_G W$ is a subspace of $V\otimes W$.

To see this, first we’ll want to turn $V$ into a left $G$-module by defining

$\displaystyle g\cdot v=vg^{-1}$

We can check that this is a left action:

\displaystyle\begin{aligned}g\cdot(h\cdot v)&=g\cdot(vh^{-1})\\&=vh^{-1}g^{-1}\\&=v(gh)^{-1}\\&=(gh)\cdot v\end{aligned}

The trick is that moving from a right to a left action reverses the order of composition, and changing from a group element to its inverse reverses the order again.

So now that we have two left actions by $G$, we can take the outer tensor product, which carries an action by $G\times G$. Then we pass to the inner tensor product, acting on each tensorand by the same group element. To be more explicit:

$g\cdot(v\otimes w)=(vg^{-1})\otimes(gw)$

Now, I say that being invariant under this action of $G$ is equivalent to the new relation that holds for tensors over a group algebra. Indeed, if $(vg)\otimes w$ is invariant, then

$\displaystyle(vg)\otimes w=(vgg^{-1})\otimes(gw)=v\otimes(gw)$

Similarly, if we apply this action to a tensor product over the group algebra we find

$\displaystyle g\cdot(v\otimes w)=(vg^{-1})\otimes(gw)=v\otimes(g^{-1}gw)=v\otimes w$

so this action is trivial.

Now, we’ve been playing it sort of fast and loose here. We originally got the space $V\otimes_GW$ by adding new relations to the space $V\otimes W$, and normally adding new relations to an algebraic object gives a quotient object. But when it comes to vector spaces and modules over finite groups, we’ve seen that quotient objects and subobjects are the same thing.

We can get a more explicit description to verify this equivalence by projecting onto the invariants. Given a tensor $v\otimes w\in V\otimes_GW$, we consider it instead as a tensor in $V\otimes W$. Now, this is far from unique, since many equivalent tensors over the group algebra correspond to different tensors in $V\otimes W$. But next we project to the invariant

$\displaystyle\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}(vg^{-1})\otimes(gw)$

Now I say that any two equivalent tensors in $V\otimes GW$ are sent to the same invariant tensor in $(V\otimes W)^G$. We check the images of $(vg)\otimes w$ and $v\otimes(gw)$:

\displaystyle\begin{aligned}\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}((vg)h^{-1})\otimes(hw)&=\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}(v(gh^{-1}))\otimes((hg^{-1}g)w)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{k\in G}(vk^{-1})\otimes(k(gw))\end{aligned}

To invert this process, we just consider an invariant tensor $v\otimes w$ as a tensor in $V\otimes_GW$. The “fast and loose” proof above will suffice to show that this is a well defined map $(V\otimes W)^G\to V\otimes_GW$. To see it’s an inverse, take the forward image and apply the relation we get from moving it back to $V\otimes_GW$:

\displaystyle\begin{aligned}\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}(vg^{-1})\otimes(gw)&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}v\otimes(g^{-1}gw)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}v\otimes w\\&=v\otimes w\end{aligned}

And so we’ve established the isomorphism $V\otimes_GW\cong(V\otimes W)^G$, as desired.