What are the irreducible corepresentations of the eight-dimensional Kac-Paljutkin Quantum Group?

The trivial corepresentation is given by $\Delta_{|W}$ where $W$ is just the one dimensional subspace

$$W=\{\lambda(1,1,1,1,I_{2\times 2}):\lambda\in\mathbb{C}\}.$$

By my reckoning there should be seven more one dimensional invariant subspaces and hence irreducible corepresentations.

The eight-dimensional Kac-Paljutkin Quantum Group

Here we give the defining relations and the main structure of an eight-dimensional quantum group introduced by Kac and Paljutkin. This is actually the smallest finite quantum groups that is not a group algebra (e.g. $F(\mathbb{Z}_5)$). In other words, it is the non-commutative C*-Hopf algebra of smallest dimension.

Consider the multi-matrix algebra $$A=\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus M_2(\mathbb{C}),$$ with the usual multiplication and involution. We shall use the basis $e_1=(1,0,0,0,0)$ (with $e_2,\,e_3,\,e_4$ defined in the same way) and

$$ a_{11}=0+ 0+ 0+ 0+\left(\begin{array}{cc}1&0\\0&0\end{array}\right),$$

with the other $a_{ij}$ defined in the same way. The algebra $A$ is an eight-dimensional C*-algebra. Its unit is of course $1_A=e_1+e_2+e_3+e_4+a_{11}+a_{22}$. The following defines a comultiplication on $A$,

$$ \Delta(e_1)=e_1\otimes e_1+e_2\otimes e_2+e_3\otimes e_3+e_4\otimes e_4+\frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{12}\otimes a_{12}+\frac{1}{2}a_{21}\otimes a_{21}+\frac{1}{2}a_{22}\otimes a_{22},$$

$$\Delta(e_2)=e_1\otimes e_2+e_2\otimes e_1+e_3\otimes e_4+e_4\otimes e_3+ \frac{1}{2}a_{11}\otimes a_{22}+\frac{1}{2}a_{22}\otimes a_{11}+\frac{i}{2}a_{21}\otimes a_{12}-\frac{i}{2}a_{12}\otimes a_{21},$$

$$ \Delta(e_3)=e_1\otimes e_3+e_3\otimes e_1+e_2\otimes e_4+e_4\otimes e_2+ \frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{22}\otimes a_{11}-\frac{i}{2}a_{21}\otimes a_{12}+\frac{i}{2}a_{12}\otimes a_{21},$$

$$ \Delta(e_4)=e_1\otimes e_4+e_4\otimes e_1+e_2\otimes e_3+e_3\otimes e_2+ \frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{22}\otimes a_{22}-\frac{1}{2}a_{12}\otimes a_{21}-\frac{1}{2}a_{21}\otimes a_{21},$$

$$ \Delta(a_{11})=e_1\otimes a_{11}+a_{11}\otimes e_1+e_2\otimes a_{22}+a_{22}\otimes e_2+e_3\otimes a_{22}+a_{22}\otimes e_3+e_4\otimes a_{11}+a_{11}\otimes e_4,$$

$$ \Delta(a_{12})=e_1\otimes a_{12}+a_{12}\otimes e_1+ie_2\otimes a_{21}-ia_{12}\otimes e_2-ie_3\otimes a_{21}+ia_{21}\otimes e_3-e_4\otimes a_{12}-a_{12}\otimes e_4,$$

$$ \Delta(a_{21})=e_1\otimes a_{21}+a_{21}\otimes e_1-ie_2\otimes a_{12}+ia_{12}\otimes e_2+ ie_3\otimes a_{12}-ia_{12}\otimes e_3-e_4\otimes a_{21}-a_{21}\otimes e_4,$$

$$ \Delta(a_{22})=e_1\otimes a_{22}+a_{22}\otimes e_1+e_2\otimes a_{11}+a_{11}\otimes e_2+e_3\otimes a_{11}+a_{11}\otimes e_3+e_{4}\otimes a_{22}+a_{22}\otimes e_4.$$

The counit is given by (looking at this we can see the relationship between the counit and the comultiplication with respect to the unit and multiplication in a commutative, group algebra, case. To encode $ ge=g=eg$ any time there is a term of the form $ x\otimes e_1$, there must be a term of the form $ e_1\otimes x$ --- to capture the left and right symmetry $ R_\varepsilon=(I_A\otimes\varepsilon)\circ\Delta=(\varepsilon\otimes I_A)\circ \Delta=L_\varepsilon$. This also shows that, in this case, $\Delta(x)$ must contain $ x\otimes e_1$ and $ e_1\otimes x$ to encode $L_\varepsilon=I_A=R\varepsilon$):

$$ \varepsilon\left(x_1+x_2+x_3+x_4+\left(\begin{array}{cc}c_{11} &c_{12}\\ c_{21}&c_{22}\end{array}\right)\right)=x_1.$$

The antipode is the transpose map, i.e.

$$ S(e_i)=e_i\text{, and }S(a_{jk})=a_{ji}.$$


A corepresentation $\chi$ of a finite quantum group $(A,\Delta)$ on a complex vector space $V$ is a linear map $\chi:V\rightarrow V\otimes A$ that satisfies

$$(\chi\otimes I_A)\circ \chi=(I_V\otimes \Delta)\circ\chi\text{ and }$$ $$(I_V\otimes \varepsilon)\circ \chi=I_V$$

where $\varepsilon$ is the counit of $(A,\Delta)$.

The dimension of the vector space is called the dimension of $\chi$ and is denoted by $d_\chi$.

If $W$ is a vector space which has the property that $\chi(W)\subset W\otimes A$ then $W$ is said to be invariant and $\chi_{|W}$ is called a subrepresentation.

It can be shown that $\chi$ is equivalent to a direct sum of irreducible unitary corepresentations.

Two corepresentations $\chi_1$ and $\chi_2$ are equivalent as corepresentations if they admit and invertible intertwiner: an invertible linear map $T:V_1\rightarrow V_2$ such that

$$\chi_2\circ T=(T\otimes I_A)\circ\chi_1.$$

The space of intertwiners from $\chi_1$ to $\chi_2$ is denoted by $\text{Hom}(\chi_1,\chi_2)$.

Further Background

I had been under the impression that the comultiplication plays the role of the regular representation in the representation theory of finite groups so that all of the irreducible corepresentations could be found by finding subspaces $W$ of $A$ invariant under the comultiplication in the sense that $\Delta(W)\subset W\otimes A$... however I am either having a stupid misunderstanding or else this isn't as easy as I thought.

I am trying to apply the philosophy of a use of the representation theory of finite groups to the corepresentation theory of finite quantum groups and perhaps I have confused myself a little in the process!

  • $\begingroup$ Anyone? I suppose we are looking to solve $\Delta x=\lambda x\otimes A$... $\endgroup$ – JP McCarthy Jun 29 '13 at 12:39
  • $\begingroup$ It may be worth trying to get an answer at mathoverflow.net if you're unable to get one here - certainly by the looks of the question it seems to be a research level question, and there are probably many more people able to answer your question on MathOverflow $\endgroup$ – Andrew D Jun 29 '13 at 12:52
  • 1
    $\begingroup$ There are, up to equivalence, four one-dimensional and one two-dimensional corep. See mathoverflow.net/questions/135241/…. The self-duality of the Kac-Paljutkin quantum group implies that there exists a basis $f_1,f_2,f_3,f_4,b_{11},b_{12},b_{21},b_{22}$ such that $\Delta f_k=f_k\otimes f_k$ and $\Delta b_{jk}=\sum_{\ell=1}^2 b_{j\ell}\otimes b_{\ell k}$. $\endgroup$ – UwF Jul 1 '13 at 18:08
  • $\begingroup$ @UFO Thank you for that however, I am looking at the Franz & Gohm paper and can see the isomorphism thank you... I gave you 100 points on your Mathoverflow answer. $\endgroup$ – JP McCarthy Jul 10 '13 at 10:51
  • $\begingroup$ Answer given here by UFO mathoverflow.net/questions/135241/… Thank you UFO! $\endgroup$ – JP McCarthy Jul 10 '13 at 10:53

This has been answered by UwF over on Mathoverflow.

The above comment also links to a more explicit description of the matrix elements.


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