The Unapologetic Mathematician

Mathematics for the interested outsider

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.

November 15, 2010 Posted by | Algebra, Group theory, Representation Theory | 1 Comment

   

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