# The Unapologetic Mathematician

## Invariant Forms

A very useful structure to have on a complex vector space $V$ carrying a representation $\rho$ of a group $G$ is an “invariant form”. To start with, this is a complex inner product $(v,w)\mapsto\langle v,w\rangle$, which we recall means that it is

• linear in the second slot — $\langle u,av+bw\rangle=a\langle u,v\rangle+b\langle u,w\rangle$
• conjugate symmetric — $\langle v,w\rangle=\overline{\langle w,v\rangle}$
• positive definite — $\langle v,v\rangle>0$ for all $v\neq0$

Again as usual these imply conjugate linearity in the first slot, so the form isn’t quite bilinear. Still, people are often sloppy and say “invariant bilinear form”.

Anyhow, now we add a new condition to the form. We demand that it be

• invariant under the action of $G$$\langle gv,gw\rangle=\langle v,w\rangle$

Here I have started to write $gv$ as shorthand for $\rho(g)v$. We will only do this when the representation in question is clear from the context.

The inner product gives us a notion of length and angle. Invariance now tells us that these notions are unaffected by the action of $G$. That is, the vectors $v$ and $gv$ have the same length for all $v\in V$ and $g\in G$. Similarly, the angle between vectors $v$ and $w$ is exactly the same as the angle between $gv$ and $gw$. Another way to say this is that if the form $B$ is invariant for the representation $\rho:G\to GL(V)$, then the image of $\rho$ is actually contained in the orthogonal group [commenter Eric Finster, below, reminds me that since we’ve got a complex inner product we’re using the group of unitary transformations with respect to the inner product $B$: $\rho:G\to U(V,B)$].

More important than any particular invariant form is this: if we have an invariant form on our space $V$, then any reducible representation is decomposable. That is, if $W\subseteq V$ is a submodule, we can find another submodule $U\subseteq V$ so that $V=U\oplus W$ as $G$-modules.

If we just consider them as vector spaces, we already know this: the orthogonal complement $W^\perp=\left\{v\in V\vert\forall w\in W,\langle v,w\rangle=0\right\}$ is exactly the subspace we need, for $V=W\oplus W^\perp$. I say that if $W$ is a $G$-invariant subspace of $V$, then $W^\perp$ is as well, and so they are both submodules. Indeed, if $v\in W^\perp$, then we check that $gv$ is as well:

\displaystyle\begin{aligned}\langle gv,w\rangle&=\langle g^{-1}gv,g^{-1}w\rangle\\&=\langle v,g^{-1}w\\&=0\end{aligned}

where the first equality follows from the $G$-invariance of our form; the second from the representation property; and the third from the fact that $W$ is an invariant subspace, so $g^{-1}w\in W$.

So in the presence of an invariant form, all finite-dimensional representations are “completely reducible”. That is, they can be decomposed as the direct sum of a number of irreducible submodules. If the representation $V$ is irreducible to begin with, we’re done. If not, it must have some submodule $W$. Then the orthogonal complement $W^\perp$ is also a submodule, and we can write $V=W\oplus W^\perp$. Then we can treat both $W$ and $W^\perp$ the same way. The process must eventually bottom out, since each of $W$ and $W^\perp$ have dimension smaller than that of $V$, which was finite to begin with. Each step brings the dimension down further and further, and it must stop by the time it reaches $1$.

This tells us, for instance, that there can be no inner product on $\mathbb{C}^2$ that is invariant under the representation of the group of integers $\mathbb{Z}$ we laid out at the end of last time. Indeed, that was an example of a reducible representation that is not decomposable, but if there were an invariant form it would have to decompose.

September 27, 2010 -

## 7 Comments »

1. Since we have a complex inner product, should the image of the representation not be the unitary group instead of the orthogonal group?

Comment by ericfinster | September 28, 2010 | Reply

• BTW, lest my previous comment sound ungrateful, I should say that I am very much enjoying this series of posts. Thanks for doing it!

Comment by ericfinster | September 28, 2010 | Reply

• You know, you’re right. I’m so used to the high-level view on these things that I keep forgetting the little differences like that. I’ll have to tweak that…

Comment by John Armstrong | September 28, 2010 | Reply

2. […] saw last time that in the presence of an invariant form, any reducible representation is decomposable, and so any representation with an invariant form is […]

Pingback by Maschke’s Theorem « The Unapologetic Mathematician | September 28, 2010 | Reply

3. […] all -modules are completely reducible, but remember what it really tells us that there is some -invariant inner product on (we’ll have to keep straight the two inner products by which vector space […]

Pingback by The Inner Product of Characters « The Unapologetic Mathematician | October 18, 2010 | Reply

4. […] subspace which is also -invariant. And we can find such a complement if we have a -invariant inner product on our space. And, luckily enough, permutation representations admit a very nice invariant inner […]

Pingback by One Complete Character Table (part 2) « The Unapologetic Mathematician | October 27, 2010 | Reply

5. I feel like this is a little misleading. Most of the time when people say “invariant bilinear form” they really do mean “invariant bilinear form” and not “invariant sesquilinear form.”

Comment by mclaury | August 8, 2012 | Reply