Just some notes on section 1 of this paper. Flags and notes are added but mistakes are mine alone.

Definition

Let C(G) be the algebra of continuous functions on a compact matrix quantum group. Such an object is given by a matrix u=\{u_{ij}\}_{i,j=1}^N which generates C(G) as a C*-algebra. Furthermore, there exists a C*-algebra homomorphism \Delta:C(G)\rightarrow C(G)\otimes C(G) such that

\displaystyle \Delta(u_{ij})=\sum_{k=1}^N u_{ik}\otimes u_{kj},

and both u and u^T are invertible in M_N(C(G)).

Any subgroup G\subset \text{GL}(N,\mathbb{C}) is such an object, with the u_{ij}\in C(G) given by u_{ij}(g)=g_{ij}\in\mathbb{C}. Furthermore

\mathrm{C}_{\text{comm}}\langle u_{ij}\rangle \cong C(G).

We say that \rho=(\rho_{ij})_{i,j=1}^{d_\rho}\in M_{d_{\rho}}(C(G)) is a representation if it is invertible and

\displaystyle \Delta(\rho_{ij})=\sum_{k=1}^{d_\rho}\rho_{ik}\otimes\rho{kj}.

The transpose \rho^T=(\rho_{ji})_{i,j=1}^N\in M_{d_{\rho}}(C(G)) is also invertible and so we have:

Proposition

The C*algebra generated by the \rho_{ij} is also the algebra of continuous functions on a compact matrix quantum group.

Podlés proceeds to give some properties of representations of a compact matrix quantum group and introduces the index set \text{Irr}(G). It is noted that the matrix element of the trivial representation is \rho_{11}^\tau=\mathbf{1}_G. Then the Haar state is presented as a functional \int_G such that

\displaystyle \int_G \mathbf{1}_G=1,

and \int_G\rho^\alpha_{ij}=0 for \alpha\not\equiv \tau. Where \text{Pol}(G) is the unital *-algebra generated by the u_{ij}, the \rho_{ij}^\alpha form a basis for \text{Pol}(G). The Haar state is invariant in that for f\in C(G),

\displaystyle \left(\int_G\otimes I_{C(G)}\right)\circ \Delta(f)=\left(I_{C(G)}\otimes \int_G\right)\circ \Delta(f)=\left(\int_G f\right)\mathbf{1}_G

which implies in particular that for all \omega\in M_p(G):

\displaystyle \omega\star \int_G=\int_G=\int_G\star\, \omega.

To measure the modular theory of the Haar state there exist matrices Q_\alpha such that

\displaystyle \int_G\left(\rho_{ij}^\alpha\right)^* \rho_{k\ell}^\beta=\frac{\delta_{\alpha\beta}\delta_{l\ell}}{\text{Tr }Q_\alpha}(Q_\alpha^{-1})_{ki}.

It appears that Podlés now defines densities a_{\rho^{ij}_\alpha} such that \mathcal{F}(a_{\rho^{ij}_\alpha})=\rho_{\alpha}^{ij} (dual element). It is important that this be done because not all functionals have densities.

We have that

\rho_\alpha^{ij}\star\rho_\beta^{k\ell}=\delta_{jk}\delta_{\alpha\beta}\rho_\beta^{i\ell},

and if we define

T_{ij}^\alpha=(I_{C(G)}\otimes \rho_\alpha^{ij})\Delta,

and \rho_\alpha=\sum_{s=1}^{d_\rho}\rho_\alpha^{ss}\in C(G)' (so that \rho_0=\int_G), \rho_\alpha(\rho^\beta_{ij})=\delta_{\alpha\beta}\delta_{ij} and

T^\alpha_{ij}C(G)\subset \text{span}\left(\rho_{ji}:j=1,\dots,d_\alpha\right).

Perhaps this should merely be T^\alpha_{ij}\text{Pol}(G)? Well anyway, it’s like a column  of (\rho^\alpha)_{i,j=1}^{d\alpha}.

Now Podlés wants to talk about subgroups and here we will make some notes to make sense of what Podlés does.

Consider the algebra of functions of continuous functions on a compact matrix quantum group (C(H),v), and assume that G has fundamental representation u.

Definition

We say that H is a (compact) subgroup of G if \dim u=\dim v, and there exists a C*-homomorphism \pi_H:C(G)\rightarrow C(H) such that \pi_H(u_{ij})=v_{ij}.

Podlés remarks that this must be a surjection. I suppose if H is generated by the v_{ij} we can simply map the appropriate combination of u_{ij} to hit any combination of v_{ij}.

Let us try and reconcile this definition with the more standard definition. Where \imath:H\hookrightarrow G is the inclusion of H into G, \pi_H is what I would call \mathcal{Q}(\imath). The commutative diagram in the category of compact groups that says that H is a subgroup of G is given by:

\imath\circ m_H=m\circ(\imath\times\imath).

The image of this commutative diagram under \mathcal{Q} is the standard:

\Delta_{C(H)}\circ \pi_H=(\pi_H\otimes \pi_H)\circ \Delta.

Suppose that \pi_H(u_{ij})=v_{ij}. It is very easy to see that this \pi_H satisfies the above.

Classically, in the finite \pi_H=\mathcal{Q}(\imath) picture, given a subgroup H\subset G, and an element f\in F(G), one can see:

\displaystyle f=\sum_{t\in G}a_t\delta_t=\underbrace{\sum_{t\in H}a_t\delta_t}_{=\pi_H(f)}+\sum_{t\not\in H}a_t\delta_t,

so it just chops bits off that are not relevant to G outside H. A simple example of what this looks like in the matrix case might be to take G=M_2(\mathbb{Z}_p) and 

\displaystyle H=\displaystyle \left\{\left(\begin{array}{cc}a & 0 \\ 0 & 0\end{array}\right):a\in\mathbb{Z}_p\right\}

Then the u_{ij}\in C(G) are the coordinate functions but v_{ij}=0 if either i,\,j is two, and in this sense:

\displaystyle f=\sum_{i,j=1}^2c_{ij}u_{ij}=\underbrace{c_{11}u_{11}}_{=\pi_H(f)}+\sum_{i,j=2}c_{ij}u_{ij}.

Getting back to Podlés.

Definition

Let C(X) be the algebra of continuous function on a compact quantum space X and G a compact quantum group. A C*-homomorphism \kappa:C(X)\rightarrow C(X)\otimes C(G) is an action of G on X if

  • (\kappa\otimes I_{C(G)})\circ \kappa=(I_{C(X)}\otimes \Delta)\circ\kappa, and

  • \langle(I_{C(X)}\otimes f)\circ \kappa(g):g\in C(X),\,f\in C(X)\rangle=C(X)\otimes C(G).

I must admit I am not sure what the second condition is about. I assume it is something like the group acts transitively, I don’t know. The first condition, which doesn’t seem to include an analogue of x\overset{e}{\mapsto}x, for actions of finite groups on finite sets at least, is the image under the \mathcal{Q} functor of, where \alpha is a right action of G on X:

\alpha(\alpha(x,g),h)=\alpha(x,gh).

Let X be a quantum space and G a quantum group. Fix a C*-homomorphism \kappa. A subspace W\subset C(X) corresponds to a representation \rho of G if there exist basis elements e_1,\dots,e_d\in W, \dim \rho=d and

\displaystyle \kappa(e_j)=\sum_{i=1}^d e_i\otimes \rho_{ij}.

If \kappa is an action on X then C(X) can be decomposed into subspaces corresponding to irreducible representations of G.

Theorem

Let \kappa be an action of a quantum group G on a quantum space X. Denote E^\alpha=(I_{C(X)}\otimes \rho_\alpha)\circ \kappa, W_\alpha=E^\alpha C(X)\subset C(X).

  1. \displaystyle C(X)=\bigoplus_{\alpha\in\text{Irr}(G)}

  2. For each \alpha\in \text{Irr}(G), there exists a set I_\alpha and W_{\alpha i}, i\in I_\alpha such that 

    • \displaystyle W_\alpha=\bigoplus_{i\in I_\alpha}W_{\alpha i}
    • W_{\alpha i} corresponds to \rho^\alpha (\left.\kappa\right|_{W_{\alpha i}})=\rho^\alpha: W_{\alpha i} \rightarrow W_{\alpha i}\otimes C(G). This is probably where the second condition comes in).

  3. Subspaces V\subset C(X) corresponding to \rho^\alpha are contained in W_\alpha.

  4. |I_\alpha| does not depend on the choice (?) of \{W_{\alpha i}\}_{i\in I_\alpha}. Denoted by c_\alpha, called the multiplicity of \rho^\alpha in the spectrum of \kappa.

Proof: Set

E^{\alpha}_{ij}=(I_{C(X)}\otimes \rho_\alpha^{ij})\circ \kappa: C(X)\rightarrow C(X).

Careful use of tensor product isomorphisms shows that that

E^\alpha_{ij}E^{\beta}_{k\ell}=(I_{C(X)}\otimes \rho_\alpha^{ij}\otimes I_{\mathbb{C}})\circ (\kappa\otimes I_{C(G)})\circ (I_{C(X)}\otimes \rho_\beta^{k\ell})\circ \kappa.

=(I_{C(X)}\otimes (\rho_\alpha^{ij}\otimes \rho_\beta^{k\ell}))\circ (\kappa\otimes I_{C(G)})\circ \kappa

=(I_{C(X)}\otimes (\rho_\alpha^{ij}\otimes \rho_\beta^{k\ell}))\circ (I_{C(X)}\otimes\Delta)\circ \kappa

(I_{C(X)}\otimes \delta_{\alpha\beta}\delta){jk}\rho_\beta^{i\ell})\circ \kappa=\delta_{\alpha\beta}\delta_{jk}(I_{C(X)}\otimes \rho_\beta^{i\ell})\circ \kappa=\delta_{\alpha\beta}\delta_{jk}E_{i\ell}^\beta

Podlés claims that the densities of the elements of the dual basis to the \rho_{ij}^\alpha generate C(G). This is certainly true in the Kac case as the densities are equal to d_\alpha \rho^\alpha_{ij} and these certainly span C(G).

I have been struggling greatly with the second condition, the so called density condition. I understand that if one works in Pol(G) one has a counit and the extension of the condition xe=x gives this density condition, that it isn’t connected with a notion of transitivity. I am going to skip the rest of the proof because I am really interested in coset spaces and I might not need all this machinery to understand those… I can always go back. De Commer has a lot on actions that I can take a look at. I have a similar issue with a corollary to the theorem. Now however Podlés introduces the quotient spaces.

Let H\subset G be a subgroup. Podlés identifies:

C(H\backslash G)=\{f\in C(G)\,:\,(\pi_H\otimes I_{C(G)})\circ \Delta(f)=\mathbf{1}_H\otimes f\}.

Let us show that if G is classical, and H a subgroup, that for a fixed g\in G, \mathbf{1}_{Hg}\in C(H\backslash G). Note firstly that

\displaystyle \mathbf{1}_{Hg}=\sum_{h\in H}\delta_{hg}

\displaystyle \Rightarrow \Delta(\mathbf{1}_{Hg})=\sum_{h\in H}\sum_{t\in G}\delta_{hgt^{-1}}\otimes \delta_t

\displaystyle \Rightarrow (\pi_H\otimes I_{C(G)})\Delta(\mathbf{1}_{Hg})=\sum_{h\in H}\sum_{t\in G}\pi_H(\delta_{hgt^{-1}})\otimes \delta_t

If hgt^{-1} is in H, \pi_H will leave it. Anything else will be killed by \pi_H. So it will be left if there exists an h_t\in H such that 

hgt^{-1}=h_t\Rightarrow hg=h_t t\Rightarrow t\in Hg.

Therefore summing over t is the same as summing over \eta g where \eta\in H. We have also hgt^{-1}=hg(\eta g)^{-1}=h\eta^{-1}. This gives:

\displaystyle \Rightarrow (\pi_H\otimes I_{C(G)})\Delta(\mathbf{1}_{Hg})=\sum_{h\in H}\sum_{\eta \in H}\delta_{h\eta^{-1}})\otimes \delta_{\eta g}

It isn’t immediately clear but this does equal \mathbf{1}_H\otimes \mathbf{1}_{Hg}, and so \mathbf{1}_{Hg}\in C(H\backslash G). The condition is linear in f therefore, classically, 

\displaystyle C(H\backslash G)=\left\{\sum_{[g]\in H\backslash G}\alpha_g\mathbf{1}_{Hg}\,:\,\alpha_g\in\mathbb{C}\right\},

that is functions constant on cosets Hg. Note that the indicator functions are minimal projections in this subspace. I am not sure how much further structure we have… is it an algebra (in the quantum case)? I’m not sure. It is if

(\mathbf{1}_H\otimes f)\Delta(g)=(\mathbf{1}_H\otimes f)(\pi_H\otimes I_{C(G)})\Delta(g).

Now Podlés gives a completely bounded projection C(G)\rightarrow C(H\backslash G):

\displaystyle E_{H\backslash G}=\left(\int_H \otimes I_{C(G)}\right)\circ (\pi_H\otimes I_{C(G)})\circ \Delta.

Classically this takes a function on G and replaces it with a function constant on the cosets of H. What values does it take on a coset Hg? The average of the function on Hg

\displaystyle E_{H\backslash G}f(Hg)=\frac{1}{|H|}\sum_{h\in H}f(\delta^{hg}).

Podlés says that this projection has the property that

(E_{H\backslash G}\otimes I_{C(G)})\Delta=\Delta\circ E_{H\backslash G},

but I am not to sure of the relevance of this. Perhaps it allows the following make sense. 

Define now a map

\kappa_{H\backslash G}=\Delta_{\left.\right|_{C(H\backslash G)}}:C(H\backslash G)\rightarrow C(H\backslash G)\otimes C(G).

Podlés isn’t clear but I am fairly sure of what happens next. Take an irreducible representation of G and map it to a representation of H via \pi_H. This representation is, apparently, not irreducible. This means that it can be decomposed into H representations. The trivial representation, \lambda\mapsto \lambda\otimes \mathbf{1}_H, appears n_\alpha times. Choose a basis such that these trivial representations appear ‘first’, in the top left hand corner, kind of (where \varrho represent H representations):

\pi_H(\rho^\alpha)=\left(\varrho^\tau\right)^{\boxplus n_\alpha}\boxplus \tilde{\varrho},

where \tilde{\varrho} contains no non trivial representations.

It might be difficult to move to group-like-projections as they don’t have the representation theory as far as I know.

Well I think Podlés shows that only the trivial bits correspond to functions constant on cosets, so that, in this basis

C(H\backslash G)=\langle\rho_{ij}^\alpha\,:\,\alpha\in \text{Irr}(G),i,j=1,\dots, n_\alpha\rangle

and so the action is defined on these elements only

\displaystyle \Delta_{\left.\right|_{C(G\backslash G)}}(\rho_{ij}^\alpha)=\sum_{k=1}^{n_\alpha}\rho_{ik}^\alpha\otimes \rho_{kj}^{\alpha},

and essentially

\Delta_{\left.\right|_{C(H\backslash G)}}=\Delta\circ E_{C(H\backslash G)}=(E_{C(H\backslash G)}\otimes I_{C(G)})\circ \Delta.

Let us write out in detail what Podlés writes. 

Let \alpha\in\text{Irr}(G). If \rho=(\rho_{ij})_{i,j=1}^{d} is a representation matrix for G then so is \pi_H(\rho)=(\pi_H(\rho_{ij}))_{i,j=1}^{d}.

Let us show this carefully. We are working with Pol(G) and so we will look at compatability and a counit condition. Firstly recall that \Delta_H\circ \pi_H=(\pi_H\otimes \pi_H)\circ \Delta. Apply this to \rho_{ij}^\alpha to get 

\displaystyle \Delta_H(\pi_H(\rho_{ij}^\alpha)=(\pi_H\otimes\pi_H)\Delta(\rho_{ij}^\alpha)=\sum_k \pi_H(\rho_{ik}^\alpha)\otimes \pi_H(\rho_{kj}^\alpha).

Using the counit condition in H that

(\varepsilon_H\otimes I_{C(H)})\circ \Delta_H=I_{C(H)},

apply both to \pi_H(\rho_{ij}^\alpha), together with the subgroup condition on \pi_H to get:

\displaystyle (\varepsilon_H\otimes I_{C(H)})\circ (\pi_H\otimes \pi_H)\sum_k\rho_{ik}^\alpha\otimes \rho_{kj}^\alpha=\pi_H(\rho_{ij}^\alpha)

\Rightarrow \varepsilon(\pi_H(\rho_{ik}^\alpha))=\delta_{i,k},

so that indeed \pi_H(\rho^\alpha) is a representation of H.

Now Podlés decomposes \pi_H(\rho^\alpha) into a direct sum of irreducible representations of H, within which, as mentioned above, the trivial representation occurs with multiplicity n_\alpha. Now put all these trivial representations in the top left hand corner so that:

\pi_H (\rho^\alpha)=\left(\begin{array}{cccc}\mathbf{1}_H & \dots & 0 & 0 \\ 0 & \ddots & 0 &0 \\ 0 & \cdots & \mathbf{1}_H & 0 \\ 0 & \cdots & 0 & \tilde{\varrho}\end{array}\right),

where \tilde{\varrho} is a direct sum of non-trivial representions of H and there are n_\alpha copies of \mathbf{1}_H. Now what about

\displaystyle \int_H\pi_H(\rho_{ij}^\alpha)?

We know that of all the matrix elements of irreducible representations, \int_G \rho_{ij}^\alpha=0 except for \rho_{ij}^\alpha=\rho_{11}^\tau=\mathbf{1}_G. The first n_\alpha of these:

\pi_H(\rho_{11}^\alpha),\dots,\pi_H(\rho_{n_\alpha n_\alpha})

of these are the matrix elements of the trivial representation, and so we get one, but for i>n_\alpha, we get zero.

Consider now, with the same basis,

\displaystyle E_{C(H\backslash G)} (\rho_{ij}^\alpha)=\left(\int_H \otimes I_{C(G)}\right) \circ (\pi_H\otimes I_{C(G)})\circ \Delta(\rho_{ij}^\alpha)=\sum_{k}\int_H \pi_H(\rho^\alpha_{ik})\rho_{kj}^\alpha=0,

for i>n_\alpha. We find that E_{C(H\backslash G)}\rho_{ij}^\alpha=\rho_{ij}^\alpha for i=1,\dots,n_\alpha.

I think this is all I need from Podlés. Now onto Wang… and I will also want to talk about coideals at some stage.

 

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