# Abstract Nonsense

## Complex Conjugates (Quaternionic Irreps Are of Even Degree)

Point of post: In this post we show that every quaternionic irrep has even degree.

We are now able to give the final theorem regarding quaternionic representations. We will require the knowledge of the quaternion algebra $\mathbb{H}$ (if this is unfamiliar to you, see here). We then can consider $\mathbb{H}^n$ as a left $\mathbb{H}$-module with the usual operations. We claim that every representation space of a quaternionic representation can be give the structure of such a left $\mathbb{H}$-module. But, since $\mathbb{H}$ is a division ring every finitely generated left $\mathbb{H}$-module is free and also have invariant basis number. It follows then that if $\mathscr{V}$ is a representation space of a quaternionic irrep then $\mathscr{V}\cong\mathbb{H}^n\cong\mathbb{C}^{2n}$ where $n=\dim_{\mathbb{H}}\mathscr{V}^{\mathbb{H}}$ (defined below) and thus in particular the degree of every quaternionic irrep is even.

Theorem: Let $\mathscr{V}$ be a finite dimensional vector  space such that there exists some antilinear map $J$ such that $J^2=-\mathbf{1}$. Then, $\mathscr{V}$ can be given the structure of a left $\mathbb{H}$-module.

Proof: It’s clear that for every $\gamma\in\mathbb{H}$ we have that $\gamma$ can be written uniquely as $\gamma=\alpha+\beta j$ for $\alpha,\beta\in\mathbb{C}$. Indeed, uniqueness is clear and existence follows since we can note that $k=ij$ so that for $r_1,r_2,r_3,r_4\in\mathbb{R}$ one has that

$r_1+r_2i+r_3j+r_4k=r_1+r_2i+r_3j+r_4ij=\left(r_1+r_2i\right)+\left(r_3+r_4i\right)j$

So, for $\alpha+\beta j=\gamma\in\mathbb{H}$ with $\alpha,\beta\in\mathbb{C}$ we define $\gamma v$ for $v\in\mathscr{V}$ by $\gamma v=\left(\alpha\mathbf{1}+\beta J\right)(v)$. We claim that this along with the already existing abelian group structure on $\mathscr{V}$ defines a left $\mathbb{H}$-module structure on $\mathscr{V}$, which we shall denote $\mathscr{V}^{\mathbb{H}}$. Indeed, we need to prove that this definition of scalar multiplication satisfies vector distributivity $\mathbf{(1)}$, left scalar distributivity $\mathbf{(2)}$, associativity $\mathbf{(3)}$, and $1_{\mathbb{H}}v=v$ $\mathbf{(4)}$. Indeed, to prove $\mathbf{(1)}$ we note that if $\gamma=\alpha+\beta j$ and $\gamma'=\alpha'+\beta' j$

\begin{aligned}(\gamma+\gamma')(v) &=\left((\alpha+\alpha')+\left(\beta+\beta'\right)j\right)\\ &= \left((\alpha+\alpha')\mathbf{1}+\left(\beta+\beta'\right)J\right)(v)\\ &= (\alpha+\alpha')v+(\beta+\beta')J(v)\\ &=\alpha v+\alpha'v+\beta J(v)+\beta' J(v)\\ &= \left(\alpha\mathbf{1}+\beta J\right)(v)+\left(\alpha'\mathbf{1}+\beta'J\right)(v)\\ &= \gamma v+\gamma' v\end{aligned}

To prove scalar distributivity we need that for any $\gamma=\alpha+\beta j\in\mathbb{H}$ and $v,v'\in\mathscr{V}$ ones that

\displaystyle \begin{aligned}\gamma\left(v+v'\right) &=\left(\alpha\mathbf{1}+\beta J\right)\left(v+v'\right)\\ &= \alpha\left(v+v'\right)+\beta J\left(v+v'\right)\\ &= \alpha v+\alpha v'+\beta J(v)+\beta'J(v')\\ &= \left(\alpha v+\beta J(v)\right)+\left(\alpha v'+\beta J(v')\right)\\ &= \left(\alpha\mathbf{1}+\beta J\right)(v)+\left(\alpha\mathbf{1}+\beta J\right)(v')\\ &= \gamma v+\gamma v'\end{aligned}

To prove $\mathbf{(3)}$ we merely note that if $\gamma=\alpha+\beta j$ and $\gamma'=\alpha'+\beta' j$ then

\displaystyle \begin{aligned}\gamma(\gamma' v) &= \left(\alpha\mathbf{1}+\beta J\right)\left(\left(\alpha'\mathbf{1}+\beta' J\right)(v)\right)\\ &= \left(\alpha\mathbf{1}+\beta J\right)\left(\alpha' v+\beta ' J(v)\right)\\ &= \alpha\left(\alpha'(v)+\beta' J(v)\right)+\beta J\left(\alpha' v+\beta' J(v)\right)\\ &= \alpha\alpha' v+\alpha \beta' J(v)+\beta\left(\overline{\alpha'}J(v)-\overline{\beta'}v\right)\\ &= \alpha\alpha' v+\alpha\beta' J(v)+\beta\overline{\alpha' }J(v)-\beta\overline{\beta' v}\\ &= \left(\alpha\alpha'-\beta\overline{\beta'}\right)(v)+\left(\alpha\beta'+\beta\overline{\alpha'}\right)J(v)\\ &= \left(\left(\alpha\alpha'-\beta\overline{\beta'}\right)\mathbf{1}+\left(\alpha\beta'+\beta\overline{\alpha'}\right)J\right)(v)\\ &= \left(\left(\alpha\alpha'-\beta\overline{\beta'}\right)+\left(\alpha\beta'+\beta\overline{\alpha'}\right)j\right)v\end{aligned}

Note though that

\begin{aligned}\gamma\gamma' &= \left(\alpha+\beta j\right)\left(\alpha'+\beta' j\right)\\ &= \alpha\alpha'+\alpha\beta'j+\beta j\alpha'+\beta j\beta' j\\ &= \alpha\alpha'+\alpha\beta'j+\beta\overline{\alpha'}j+\beta jj\overline{\beta'}\\ &=\alpha\alpha'+\alpha\beta'j+\beta\overline{\alpha'}j-\beta\beta'\\ &= \left(\alpha\alpha'-\beta\beta'\right)+\left(\alpha\beta'+\beta\overline{\alpha'}\right)j\end{aligned}

and thus comparing these shows that $\gamma(\gamma' v)=(\gamma\gamma')v$. Lastly, axiom $\mathbf{(4)}$ is verified trivially. Thus, all the axioms are satisfied and thus the conclusion follows. $\blacksquare$

Remark: The converse is clearly also true. Indeed, if $\mathscr{V}$ is such that it admits a quaternionic vector space structure then the map $v\mapsto jv$ is a antilinear map which squares to $-\mathbf{1}$.

We now claim that if a vector space above satisfies these properties then it must have even complex dimension.

Theorem: Let $\mathscr{V}$ be a vector space such that there exists an antilinear map $J:\mathscr{V}\to\mathscr{V}$ such that $J^2=-\mathbf{1}$. Then, $2\mid\dim_{\mathbb{C}}\mathscr{V}$.

Proof: By our previous theorem we know that $\mathscr{V}$ admits a quaternionic structure $\mathscr{V}^{\mathbb{H}}$ with multiplication $(\alpha+\beta j)v=\left(\alpha\mathbf{1}+\beta J\right)(v)$. We claim then that if $\left\{x_1,\cdots,x_m\right\}$ is a basis for $\mathscr{V}^{\mathbb{H}}$ then $\mathscr{B}=\left\{x_1,\cdots,x_m,J(x_1),\cdots,J(x_m)\right\}$ is a basis for $\mathscr{V}$. Indeed, it’s clear that $\mathscr{B}$ is linearly independent since if $\alpha_1,\cdots,\alpha_m,\beta_1,\cdots,\beta_m\in\mathbb{C}$ are such that

$\displaystyle \sum_{r=1}^{m}\alpha_r x_r+\sum_{r=1}^{m}\beta_r J(x_r)=\bold{0}$

then

$\displaystyle \sum_{r=1}^{m}\left(\alpha_r\mathbf{1}+\beta_r J\right)(x_j)=\sum_{r=1}^{m}\left(\alpha_r+\beta_r\right)x_r=\bold{0}$

but this implies that $\alpha_r+\beta_r j=\bold{0}$ but this is only true if $\alpha_r=\beta_r=0$ from where linear independence follows. Now, to see that $\text{span}_{\mathbb{C}}\mathscr{B}=\mathscr{V}$ we merely note that for any $v\in\mathscr{V}$ there exists $\gamma_k\in\mathbb{H}\;\; k=1,\cdots,m$ such that

$\displaystyle \sum_{k=1}^{m}\gamma_k x_k=v$

But, there exists $\alpha_k,\beta_k\in\mathbb{C}\;\; k=1,\cdots,m$ such that $\gamma_k=\alpha_k+\beta_k j$ and thus

$\displaystyle \sum_{k=1}^{m}\alpha_k x_k+\sum_{k=1}^{m}\beta_k J(x_k)=\sum_{k=1}^{m}\left(\alpha_k\mathbf{1}+\beta_k J\right)(x_k)=\sum_{k=1}^{m}\gamma_k x_k=v$

from where $\mathscr{V}=\text{span}_{\mathbb{C}}\mathscr{B}$. It follows then that $\dim_{\mathbb{C}}\mathscr{V}=2\dim_{\mathbb{H}}\mathscr{V}^{\mathbb{H}}$ from where the conclusion follows. $\blacksquare$

With this and our previous theorem regarding characterization of quaternionic irreps we may conclude that every quaternonic irrep has even degree.

References:

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

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print