Just some notes on section 1 of this paperFlags 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.