Show this is the only $2$-dimensional $\mathbb{C}$-representation of $Q_8$ up to isomorphism.

group-theorylinear algebrarepresentation-theory

I'm very new to representation theory and trying to wrap my head around it. Specifically, I'm finding it hard to prove negative statements i.e. of the form "there does not exist X" because I'm finding the computations not too difficult but struggling with the theory.

By considering $Q_8$ as a subgroup of $\mathbb{H}$ which can be viewed as a $2$-dimensional vector space over $\mathbb{C}$ with basis $\{1, j \}$ I have arrived at the representation

$$A=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix},B=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

of $Q_8$. ($A$ is the matrix representing $i$, $B$ represents $j$)

And then to show this is irreducible follows from the fact that there's no eigenvectors in common between these two matrices.

But how do I prove the negative of there's no more $2$-dimensional irreducible $\mathbb{C}$-representations up to isomorphism? If you can present it in such a way, a method that works in more general circumstances would be very helpful.

Best Answer

Here's a hands-on approach. We will use one representation theory fact: recall for any complex finite-dimensional representation of a finite group $G$, every operator $\rho(g)$ is diagonalizable. (This follows from using Jordan canonical form from linear algebra.) Moreover, if $g$ has order $n$ then $\rho(g)$'s eigenvalues must evidently be $n$th roots of unity.

First let's consider $\rho(-1)$. Its eigenvalues must be $\pm1$. This gives three cases:

  • $\rho(-1)=\big[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\big]$. This would mean all $\rho(g)$s commute (why?) hence must be scalar transformations (why?) so all 1D subspaces are subreps. Contradiction.
  • $\rho(-1)=\big[\begin{smallmatrix}1&\phantom{-}0\\0&-1\end{smallmatrix}\big]$, up to change-of-basis. Since $-1$ commutes with all $g$s, $\rho(-1)$ commutes with all $\rho(g)$s, so this tells us the coordinate axes are subreps (why?). Contradiction.
  • $\rho(-1)=\big[\begin{smallmatrix}-1&\phantom{-}0\\\phantom{-}0&-1\end{smallmatrix}\big]$. (Conclusion.)

Next let's consider $\rho(\mathbf{i})$. Its eigenvalues must be $\pm i$. We again have cases:

  • $\rho(\mathbf{i})=\pm\big[\begin{smallmatrix}i&0\\0&i\end{smallmatrix}\big]$. This is central, so commutes with $\rho(\mathbf{j})$, a contradiction. (How?)
  • $\rho(\mathbf{i})=\big[\begin{smallmatrix}i&\phantom{-}0\\0&-i\end{smallmatrix}\big]$, up to change-of-basis. (Basis change cannot affect $\rho(-1)=\big[\begin{smallmatrix}-1&\phantom{-}0\\\phantom{-}0&-1\end{smallmatrix}\big]$.)

Finally, consider $\rho(\mathbf{j})=\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]$. Apply $\rho$ to $\mathbf{ij}=-\mathbf{ji}$ and solve for $a,b,c,d$ for $\rho(\mathbf{j})=\pm\big[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}\big]$. Use $\rho(\mathbf{i})$ as a change-of-basis matrix if necessary to ensure $\rho(\mathbf{j})=\big[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}\big]$. Since $\mathbf{i},\mathbf{j}$ generate $Q_8$, this determines $\rho$, and we conclude $\rho$ must be the given matrix representation in some choice of coordinates.

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