Unique 2-dimensional irreducible representation

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.

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.

• That clarifies a lot, thanks! But I think you are indeed missing something in 2: I think the two-dimensional representation you are giving there is not irreducible (at least if $K$ contains $\pm\sqrt{-1}$? $z$ decomposes into the eigenspaces of $\pm\sqrt{-1}$ and they are stable under the action of $x$ and $y$ as they just act as scalars? So that should be a decomposition into subrepresentations? Or am I missing something? Commented Mar 30, 2020 at 12:02
• @mathstackuser Yes that's right, if the field contains square roots of -1 then $z$ would indeed diagonalise there. If the field does not contain square roots of -1, then that 2-dimensional representation is irreducible I think. Commented Mar 30, 2020 at 13:08