Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I asked this question on Math.Stack but have not had any answers.

Question

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

The trivial corepresentation is given by $\Delta_{|\mathbb{C}1_A}$ where $1_A$ is just the unit.

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. 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 $A$,

$ \scriptsize{ \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}}$ $ \scriptsize{ \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}}$ $ \scriptsize{ \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_{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}}$ $ \scriptsize{ \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_{12}-\frac{1}{2}a_{21}\otimes a_{21}}$ $ \scriptsize{ \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}$ $ \scriptsize{ \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}$ $ \scriptsize{ \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}$ $ \scriptsize{ \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.

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

Background

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!

share|improve this question
1  
Hang on, if all your coreps are one dimensional, wouldn't this make your example cocommutative? –  Yemon Choi Jun 29 '13 at 21:21
    
(The point being that the KP example is neither commutative nor cocommutative) –  Yemon Choi Jun 29 '13 at 21:40
    
@YemonChoi This is an issue that my supervisor has raised. As quantum groups, the algebra of functions on an (finite) abelian group is commutative and all of the irreps are one dimensional. However the algebra of functions on a non-abelian group is also commutative! My supervisor asks, what is an appropriate definition of an abelian quantum group? Are you suggesting that cocummutative is the right answer? And does cocummutative indeed imply that all of the coreps are one dimensional? –  Jp McCarthy Jul 16 '13 at 12:21
    
If you want abelian quantum groups to generalize the notion of abelian group I suspect cocommutative is the right thing. But I am not an expert! (This is related to one of my pet peeves, namely that what people call quantum groups are really the co-ordinate rings of quantum groups, just as Cstar algebras are not really NC topological spaces, but the algebras of functions on NC top spaces.) –  Yemon Choi Jul 16 '13 at 16:13
    
@YemonChoi I must do the sum later on... if the algebra of functions on an abelian group is cocommutative and on a non-abelian group non-cocommutative, then it sounds like a good candidate. Interestingly I was thinking about one dimensional coreps as another candidate but I understand that there are infinite dimensional non-abelian groups with one dimensional representations only. –  Jp McCarthy Jul 17 '13 at 11:47

1 Answer 1

up vote 6 down vote accepted
+100

The Kac-Paljutkin Quantum Group $A$ is self-dual, i.e. the dual space $A^*$ (which is of course again finite quantum group = finite-dimesional $C^*$-Hopf algebra, with the multiplication being the dual of the coproduct of $A$ and the coproduct the dual of the multiplication of $A$) is isomorphic to $A$.

This must have been known already to Kac and Paljutkin. You can find the isomorphism, e.g., in Franz and Gohm, Random Walks on Finite Quantum Groups, in "Quantum Independent Increment Processes II", Lecture Notes in Mathematics Volume 1866, 2006, pp 1-32 .

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.