# The Unapologetic Mathematician

## Group Invariants

Again, my apologies. What with yesterday’s cooking, I forgot to post this yesterday. I’ll have another in the evening.

Let $V$ be a representation of a finite group $G$, with finite dimension $d_V$. We can decompose $V$ into blocks — one for each irreducible representation of $G$:

$\displaystyle V\cong\bigoplus\limits_{i=1}^kV^{(i)}\otimes\hom_G(V^{(i)},V)$

We’re particularly concerned with one of these blocks, which we can construct for any group $G$. Every group has a trivial representation $V^\mathrm{triv}$, and so we can always come up with the space of “invariants” of $G$:

$\displaystyle V^G=V^\mathrm{triv}\otimes\hom_G(V^\mathrm{triv},V)$

We call these invariants, because these are the $v\in V$ so that $gv=v$ for all $g\in G$. Technically, this is a $G$ module — actually a $G$-submodule of $V$ — but the action of $G$ is trivial, so it feels slightly pointless to consider it as a module at all.

On the other hand, any “plain” vector space can be considered as if it were carrying the trivial action of $G$. Indeed, if $W$ has dimension $d_W$, then we can say it’s the direct sum of $d_W$ copies of the trivial representation. Since the trivial character takes the constant value $1$, the character of this representation takes the constant value $d_W$. And so it really does make sense to consider it as the “number” $d_W$, just like we’ve been doing.

We’ve actually already seen this sort of subspace before. Given two left $G$-modules ${}_GV$ and ${}_GW$, we can set up the space of linear maps $\hom(V,W)$ between the underlying vector spaces. In this setup, the two group actions are extraneous, and so we find that they give residual actions on the space of linear maps. That is we have two actions by $G$ on $\hom(V,W)$, one from the left and one from the right.

Now just like we found with inner tensor products, we can combine these two actions of $G$ into one. Now we have one left action of $G$ on linear maps by conjugation: $(g,f)\mapsto g\cdot f$, defined by

$\displaystyle[g\cdot f](v)=gf(g^{-1}v)$

Just in case, we check that

\displaystyle\begin{aligned}\left[g\cdot(h\cdot f)\right](v)&=g\left[h\cdot f\right](g^{-1}v)\\&=g(hf(h^{-1}(g^{-1}v)))\\&=(gh)f((h^{-1}g^{-1})v)\\&=(gh)f((gh)^{-1}v)\\&=\left[(gh)\cdot f\right](v)\end{aligned}

so this is, indeed, a representation. And what are the invariants of this representation? They’re exactly those linear maps $f:V\to W$ such that

$\displaystyle gf(g^{-1}v)=f(v)$

for all $v\in V$ and $g\in G$. Equivalently, the condition is that

$\displaystyle gf(v)=f(gv)$

and so $f$ must be an intertwinor. And so we conclude that

$\displaystyle\hom(V,W)^G=\hom_G(V,W)$

That is: the space of linear maps from $V$ to $W$ that are invariant under the conjugation action of $G$ is exactly the space of $G$-morphisms between the two $G$-modules.

About these ads

November 12, 2010 -

## 3 Comments »

1. I didn’t know about that other blog! Thanks for the link.

Comment by xammer | November 12, 2010 | Reply

2. [...] Onto Invariants Given a -module , we can find the -submodule of -invariant vectors. It’s not just a submodule, but it’s a direct [...]

Pingback by Projecting Onto Invariants « The Unapologetic Mathematician | November 13, 2010 | Reply

3. [...] 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 is a right -module and is a left -module, then is a [...]

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