# Abstract Nonsense

## Representations on Real and Quaternionic Vector Spaces (Examples and Basics)

Point of post: In this post we introduce the idea of representations in to vector spaces that aren’t over $\mathbb{C}$ and discuss how the landscape is different.

Motivation

Up until this point we’ve restricted ourselves to complex representations. By this, I mean we’ve been looking at homomorphism $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ where $\mathscr{V}$ is some complex inner product space. It is entirely possible to consider such homomorphisms into the unitary group of a real inner product space, and less so into a left $\mathbb{H}$-module with an inner product. What we shall see in this post is that the transition isn’t necessarily smooth. Granted, some aspects of the theory transfer quite nicely–in particular our ability to write every representation as a direct sum of irreps. But, we shall see that many of the ‘nicer’ theorems fail. For example, as we shall see it is entirely possible to have an irreducible representation of an abelian group of degree greater than one. Or, we shall see that in general all we can say about the sum of the degrees of the irreps squared is that it is greater than or equal to the order of the group. This turns out mostly to be the fault of one tantalizing fact–Schur’s lemma doesn’t necessarily hold for non-complex representations. So, in this post we shall just begin to survey the landscape, define the notation, prove some very basic theorems, but most importantly look at an example or two.

An Example

Consider for a second the following homomorphism: $\rho:\mathbb{Z}_4\to\mathcal{U}\left(\mathbb{C}^2\right)$ (where $\mathbb{C}^2$ has the usual structure) given by

$1\mapsto\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$

Where of course we extend the homomorphism to the rest of the group via the fact that $1$ is a generator for $\mathbb{Z}_4$. A quick check shows then that $\chi_\rho(0)=2$, $\chi_\rho(1)=0$, $\chi_\rho(2)=-2$, and $\chi_\rho(3)=0$. It’s clear then that $\left\langle \chi_\rho,\chi_\rho\right\rangle=2$ and so by our useful characterization of irreducibility that $\rho$ is not irreducible in the classic sense. Even simpler, we know that $\rho$ can’t be irreducible since every irrep of $\mathbb{Z}_4$, since it’s abelian, must be degree one.  But, let’s forgo our fancy theorems and see what in particular fails for $\rho$–namely what non-trivial proper subspace of $\mathbb{C}^2$ is $\rho$-invariant. But, since two of linear transformations are just multiples of the identity and the other two are just multiplies of $\rho(1)$ we may clearly conclude that a subspace of $\mathbb{C}^2$ is $\rho$-invariant if and only if it is invariant under $\rho(1)$. But, it is of course true that $\rho(1)$ has a non-trivial proper subspace–every linear transformation on complex space has an eigenvalue and thus an eigenvector! Indeed, one can calculate clearly that the characteristic polynomial of $\rho(1)$ is $x^2+1$ so that it’s eigenvalues are $i$ and $-i$ with associated eigenvectors $(1,i)^{\top}$ and $(1,-i)^{\top}$ respectively.

The pivotal thing to notice here is that one can easily make $\rho$ into a real representation. Indeed, clearly the same mapping produces a homomorphism $\rho_r:\mathbb{Z}_4\to\mathcal{U}\left(\mathbb{R}^2\right)$ (where $\mathbb{R}^2$ has the usual structure). That said, a similar analysis shows that under these conditions $\mathbb{R}^2$ does not have any $\rho$-invariant non-trivial proper subspaces. Indeed, the same analysis as above shows that a subspace of $\mathbb{R}^2$ being $\rho$-invariant reduces to it being $\rho(1)$ invariant. But, since $\dim_{\mathbb{R}}\mathbb{R}^2=2$ any non-trivial proper subspace invariant under $\rho(1)$ must be $1$ dimensional and thus the span of an eigenvector. But, since the characteristic polynomial of $\rho(1)$ is $x^2+1$ no such eigenvectors exist since $x^2+1$ has no real solutions, and thus $\rho(1)$ no real eigenvalues. It thus follows that $\mathbb{R}^2$ has no non-trivial proper $\rho$-invariant subspaces, and thus we must conclude that $\rho$ is irreducible in the definition we must inevitably define.

Note the other ‘fishy’ thing going on here. We have just proven that $\rho$ is an ‘irrep’ (whatever that means for real representations) but, it’s evident that $\rho(1)$ commutes with $\rho(k)$ for $k=0,1,2,3$. That said, it’s clear that $\rho(1)$ is not a scalar transformation. Thus, it seems doomed to produce any real representation analog of Schur’s lemma (note that in fact, the failings of Schur’s lemma should have been apparent from the fact that our proof that every irrep of an abelian group has degree one only really depended on Schur’s lemma).

Definitions and Basic Examples

Let $G$ be a finite group. Then, a homomorphism $\rho:G\to\mathcal{U}\left(\mathscr{R}\right)$ where $\mathscr{R}$ is a real inner product space is called a $\mathbb{R}$representation of $G$. We define a $\mathbb{R}$-representation of $G$ to be irreducible if there does not exist a $\rho$-invariant subspaces of $\mathscr{R}$.

Similarly, if $G$ is a finite group then a homomorphism $\rho:G\to\mathcal{U}\left(\mathscr{H}\right)$ where $\mathscr{H}$ is a  quaternionic vector space with a distinguished inner product (note that yes, while technically $\mathscr{H}$ is actually a left $\mathbb{H}$-module since $\mathbb{H}$ is not a field, the theory of modules over division rings is so similar to that of regular vector spaces that we ignore the distinction) is called a $\mathbb{H}$representation of $G$. Furthermore,  we say that such a  $\mathbb{H}$-representation is irreducible if it leaves no $\rho$-invariant subspaces of $\mathscr{H}$.

There are two important things we make note of. Namely, since the proof for the $\mathbb{C}$-representation case did not actually depend on the fact that representations were $\mathbb{C}$-representations and can easily be repeated for $\mathbb{R}$-representations and $\mathbb{H}$-representations we have the result that:

Theorem: Let $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ be a $\mathbb{F}$-representation (where $\mathbb{F}$ is either $\mathbb{R},\mathbb{C}$, or $\mathbb{H}$) then if $\mathscr{W}$ is a $\rho$-invariant subspace then so is $\mathscr{W}^{\perp}$.

Consequently, following the same line of reasoning as before we may conclude that:

Theorem: Let $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ be a $\mathbb{F}$-representation. Then, $\rho$ may be written as the direct sum of irreducibles.

References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.