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A quantum group $A$ here is an algebraic compact quantum group --- a Hopf*-algebra with a Haar State. Here $\hat{A}$ is the set of linear functionals $\{\mathcal{F(a)}:a\in A\}$ of the form $\mathcal{F}(a)(b)=h(ba)$.

We have for $f\in A$ a result by Van Daele that

$$f=\hat{\psi}(\hat{S}(\cdot)\mathcal{F}(f)).\qquad(*)$$

Here $\mathcal{F}:A\rightarrow \hat{A}$ is, as before, defined by $\mathcal{F}(f)(a)=h(af)$. The antipode on $\hat{A}$ is given by $\hat{S}:\hat{A}\rightarrow \hat{A}$ and defined by

$$\hat{S}(\mathcal{F}(a))b=\mathcal{F}(a)S(b)=h(S(b)a),$$ where $S:A\rightarrow A$ is the antipode of $A$. The Haar state on $\hat{A}$ is $\hat{\psi}$ and is defined by $$\hat{\psi}(\mathcal{F}(a))=\varepsilon(a),$$ where $\varepsilon\in A'$ is the counit of $A$.

In $(\star)$, $f$ is taken to be in the second hat-dual which is isomorphic to $A$. Hence $(\star)$ is giving us

$$f(\mathcal{F}(x))=\mathcal{F}(x)(f)=\hat{\psi}(\hat{S}(\mathcal{F}(x))\mathcal{F}(f)).$$

The multiplication is that of $A'$ namely the convolution defined by $$\nu\star\mu(a)=(\nu\otimes\mu)\Delta(a).$$

I have been trying to look at $(\star)$ by expanding

$$\mathcal{F}(x)=\sum_i\beta_i\mathcal{F}(\rho_i),$$

where the sum is over the matrix elements, $\rho_i$, of the inequivalent, unitary, one dimensional corepresentations... actually I want $A$ to be cocommutative so that these guys span $A$.

I think $\mathcal{F}(\rho_i)=\delta^{\rho_i^*}$ and when I run the numbers through I end up wanting to calculate
$$\hat{S}(\mathcal{F}(x))\mathcal{F}(\rho_i)\rho_k.$$ I believe these are equal to $$\delta_{i,k}h(\rho_kf).$$

Now what I want to do is expand $f=\sum_i\alpha_i\rho_i$ and using the result $h(\rho_i^*\rho_j)=\delta_{i,j}$, get a nice answer. However I have $\rho_kf$ not $\rho_k^*f$...

OK, that is the context, now onto my question.

Now, following Timmermann's book consider the following.

Let $\kappa:V\rightarrow V\otimes A$ be a corepresentation of $A$ on a vector space $V$. The matrix elements of $\kappa$, $\{\rho_{ij}\}$ are given by $$\kappa(e_j)=\sum_i e_i\otimes\rho_{ij},$$ where $\{e_i\}$ is a basis of $V$. Denote by $\bar{V}$ the conjugate vector space of $V$ and by $v\mapsto \bar{v}$ is the canonical conjugate-linear isomorphism. Then the map $$\overline{\kappa}:\bar{V}\rightarrow \bar{V}\otimes A,\,\,\bar{e_j}\mapsto \sum_i\bar{e_i}\otimes\rho_{ij}^*,$$ is a corepresentation again, called the conjugate of $\kappa$. More details below but now my question

  1. Are $\kappa$ and $\bar{\kappa}$ equivalent or inequivalent?
  2. If inequivalent in general, is it the case that they are equivalent for one dimensional corepresentations?

Corepresentations $\kappa_1:V_1\rightarrow V_1\otimes A$ and $\kappa_2:V_2\rightarrow V_2\otimes$ are equivalent if there is an invertible linear map $T:V_1\rightarrow V_2$ such that

$$\kappa_2\otimes T=(T\otimes I_A)\kappa_1.$$ Maps that satisfy this are called intertwiners.

We have

$$\bar{\kappa}C=(C\otimes\text{Inv})\kappa,$$ where $C:V\rightarrow \bar{V}$ is the conjugate-linear isomorphism and $\text{Inv}(a)=a^*$.

More details by Timmermann:

  • If $T$ intertwines $\kappa_1$ and $\kappa_2$, then $\bar{T}$ intertwines $\overline{\kappa_1}$ and $\overline{\kappa_2}$ where $\bar{T}\bar{v}=\overline{Tv}$.
  • $\bar{\kappa}$ is irreducible if and only if $\kappa$ is (no invariant subspaces $U\subset V$ such that $U\mapsto \kappa(U)\subset U\otimes A$).
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  • $\begingroup$ Perhaps the downvoter might have explained why the downvote. I have asked some questions about quantum groups on math.stackexchange but have always had to put it to the MO for lack of answers. $\endgroup$ – JP McCarthy Nov 6 '14 at 8:30
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Before thinking about complicated quantum groups, it's best to first understand the simplest situation. So let's just look at when G is a finite group and A is the Hopf algebra of functionals on G. In the finite dimensional setting A-comodules is the same thing as $A^*$-modules, and in this case $A^*$ is just the group ring $\mathbb{C}[G]$. Modules over the group ring are exactly the same thing as group representations. This is the simplest example that you should always have in mind when thinking about quantum groups.

So in this specific case your question becomes "Are all finite dimensional unitary representations of finite groups self-dual?" The answer to that is obviously no. (If this answer is not obvious to you, then I'd strongly recommend reading an intro book on group representations before trying to tackle quantum group representations.)

Fancier examples would be to take A to be the ring of functions on SU(3) or to be the quantum group O_q(SU(3)). In either case the defining vector representation won't be self-dual.

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  • $\begingroup$ I have the matrix elements for $A=F(\mathbb{Z}_3)$ are $\rho_1=1_A$, $\rho_2=(1\,\,\,e^{2\pi i/3}\,\,\,e^{4\pi i/3})^T$ and $\rho_3=(1\,\,\,e^{4\pi i/3}\,\,\,e^{2\pi i/3})^T$. But $\kappa_2$ is not equivalent to $\kappa_3$. Is not $\overline{\kappa_1}=\kappa_2$? Is this (en.wikipedia.org/wiki/Frobenius%E2%80%93Schur_indicator) worth looking at? $\endgroup$ – JP McCarthy Nov 5 '14 at 20:18
  • $\begingroup$ i.e. I don't see how this answers my question? $\endgroup$ – JP McCarthy Nov 6 '14 at 8:29
  • $\begingroup$ @JpMcCarthy: Ok, I rewrote it with more detail. $\endgroup$ – Noah Snyder Nov 6 '14 at 15:33
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    $\begingroup$ No problem. For that the easiest example is the trivial corep which is always selfdual. $\endgroup$ – Noah Snyder Nov 10 '14 at 14:34
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    $\begingroup$ Even easier, you can take consider the dual of Noah's finite group example. Then the representation category is equivalent to the symmetric category $\mathrm{Vec}_G$ of $G$-graded vector spaces in which every simple object is self-dual iff $G$ is an elementary abelian 2-group. $\endgroup$ – Marcel Bischoff Dec 28 '16 at 19:00

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