Unique 2-dimensional irreducible representation

Question

Let $K$ be a field whose characteristic is different from $2$ containing a primitive fourth root of unity. $H$ is defined as follows: It is generated by $x,y,z$ that satisfy the relations $x^2=y^2=1, z^2=\tfrac{1}{2}(1+x+y-xy), xy=yx, xz=zy, yz=zx $.

I would like to show the following: $H$ has up to isomorphism a unique $2$-dimensional irreducible representation.

I managed to show that there is an irreducible $2$-dimensional representation $\rho\colon H\to M(2\times2,K)$ given by $\rho(x)=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$, $\rho(y)=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}$, $\rho(z)=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$.

How do I show know that any other irreducible $2$-dimensional representation is isomorphic to this one? Does that somehow come out of the general theory or do I have to construct such an isomorphism explicitly? I have no idea where to start. I already know how $H$ becomes a Hopf algebra, that $H$ is semisimple and that there are $4$ $1$-dimensional representations.

Answer 1

Let $V$ be a finite-dimensional representation of $H$. Since $x$ and $y$ commute and are digonalisable (since they each square to 1), they simultaneously diagonalise $V$ into four possible simultaneous eigenspaces $V_{++}$, $V_{+-}$, $V_{-+}$, $V_{--}$, where for example we have $V_{+-} = \{v \in V \mid xv = v \text{ and } yv = -v\}$. Along these eigenspaces, the relation $z^2 = \frac{1}{2}(1 + x + y - xy)$ says that $z^2$ must act by $1, 1, 1$, and $-1$ respectively.

The relations $xz=zy$ and $yz=zx$ show that multiplication by $z$ maps between these simultaneous eigenspaces. For example, we have for all $v \in V_{+-}$ that $x(zv) = z(yv) = -zv$ and $y(zv) = z(xv) = zv$ and hence multiplication by $z$ takes $V_{+-}$ to $V_{-+}$. Applying $z$ again takes $V_{-+}$ back to $V_{+-}$, since we already determined that $z^2 = 1$ on $V_{++} \oplus V_{+-} \oplus V_{-+}$. Therefore multiplication by $z$ restricts to:

1. An involution on $V_{++}$.
2. An order-4 endomorphism of $V_{--}$.
3. An involution swapping $V_{+-}$ and $V_{-+}$.

This means that all three of $x, y, z$ preserve $V_{++}$, so it is always a direct summand of $V$. Similarly, $V_{--}$ is always a direct summand of $V$, and finally we have $V_{+-} \oplus V_{-+}$ being a direct summand.

Now we can try to classify irreducibles. If $V$ is irreducible then in particular it is indecomposable, so we can assume one of the following cases:

1. $V = V_{++}$, i.e. both $x$ and $y$ act as multiplication by 1. In this case, the only relation left to satisfy is that $z^2 = 1$. Therefore there are two one-dimensional irreps of this form, where $z$ acts by $1$ and $-1$ respectively. There is no higher dimensional irrep of this form, since $z$ would diagonalise into $+1$ and $-1$ eigenspaces and give a subrepresentation.
2. $V = V_{--}$, where both $x$ and $y$ act as multiplication by $-1$. The only relation remaining to satisfy is $z^2 = -1$, and so the one-dimensional irreps are determined by sending $z$ to $\pm \sqrt{-1}$ (depending on whether this element is in your field or not). There is a two-dimensional irrep, by sending $z$ to a matrix like $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, i.e. some matrix whose minimal polynomial is $z^2 + 1$. I think you are missing this one from your classification, unless I'm missing something.
3. $V = V_{+-} \oplus V_{-+}$. The dimension of $V$ must be a multiple of $2$, since $z$ gives an isomorphism between $V_{+-}$ and $V_{-+}$. If $V$ is two-dimensional, then let $v_{+-} \in V_{+-}$ be nonzero and set $v_{-+} = z v_{+-}$, then $(v_{+-}, v_{-+})$ gives a basis of $V$ with matrices exactly as in your question.

If you want to show that those two kinds of two-dimensional irreps are indeed a classification, suppose that you have representations $V$ and $W$ of the same "type" (for example, both of the $V_{+-} \oplus V_{-+}$ kind), pick a basis as done above, and write down a linear map between those bases. Then check that the map is a map of representations.