This sandbox is going to take from a variety of sources, mostly Shuzhou Wang.

## C*-Ideals

Let $J\subset C(X)$ be a closed (two-sided) ideal in a non-commutative unital $C^*$-algebra $C(X)$. Such an ideal is self-adjoint and so a non-commutative $C^*$-algebra $J=C(S)$. The quotient map is given by $\pi:C(X)\rightarrow C(X)/C(S)$, $f\mapsto f+J$, where $f+J$ is the equivalence class of $f$ under the equivalence relation:

$f\sim_{J} g\Rightarrow g-f\in C(S)$.

Where we have the product

$(f+J)(g+J)=fg+J$,

and the norm is given by:

$\displaystyle\|f+J\|=\sup_{j\in C(S)}\|f+j\|$,

the quotient $C(X)/ C(S)$ is a $C^*$-algebra.

Consider now elements $j_1,\,j_2\in C(S)$ and $f_1,\, f_2\in C(X)$. Consider

$j_1\otimes f_1+f_2\otimes j_2\in C(S)\otimes C(X)+C(X)\otimes C(S)$.

The tensor product $\pi\otimes \pi:C(X)\otimes C(X)\rightarrow (C(X)/C(S))\otimes (C(X)/ C(S))$. Now note that

$(\pi\otimes\pi)(j_1\otimes f_1+f_2\otimes j_2)=(0+J)\otimes(f_1+J)+$

$(f_2+J)\otimes(0+J)=0$,

by the nature of the Tensor Product ($0\otimes a=0$). Therefore $C(X)\otimes C(S)+C(S)\otimes C(X)\subset \text{ker}(\pi\otimes\pi)$.

### Definition

A WC*-ideal (W for Woronowicz) is a C*-ideal $J=C(S)$ such that $\Delta(J)\subset \text{ker}(\pi\otimes\pi)$, where $\pi$ is the quotient map $C(G)\rightarrow C(G)/C(S)$.

Let $F(G)$ be the algebra of functions on a classical group $G$. Let $H\subset G$. Let $J$ be the set of functions which vanish on $H$: this is a C*-ideal. The kernal of $\pi:F(G)\rightarrow F(G)/J$ is $J$.

Let $\delta_s\in J$ so that $s\not\in H$. Note that

$\displaystyle\Delta(\delta_s)=\sum_{t\in G}\delta_{st^{-1}}\otimes\delta_t$

and so

$\displaystyle(\pi\otimes \pi)\Delta(\delta_s)=\sum_{t\in G}\pi(\delta_{st^{-1}})\otimes \pi(\delta_t)$.

Note that $\pi(\delta_t)=0+J$ if $t\not\in H$. It is not possible that both $st^{-1}$ and $t$ are in $H$: if they were $st^{-1}\cdot t\in H$, but $st^{-1}\cdot t=s$, which is not in $H$ by assumption. Therefore one of $\pi(\delta_{st^{-1}})$ or $\pi(\delta_t)$ is equal to zero and so:

$(\pi\otimes\pi)\Delta(\delta_s)=0$,

and so by linearity, if $f$ vanishes on a subgroup $H$,

$\Delta(f)\subset \text{ker}(\pi\otimes\pi)$.

In this way, WC*-ideals generalise functions which vanish on distinguished subgroups. In fact, without checking all the details, I imagine that first isomorphism theorem can show that $F(G)/ J=F(H)$. Let $\pi_H:F(G)\rightarrow F(H)$ be the ring homomorphism

$\displaystyle\pi_H\left(\sum_{t\in G}a_t\delta_t\right)=\sum_{t\in H}a_t\delta_t$.

Then $\text{ker}\,\pi_H=J$, $\text{im}\,\pi_H=F(H)$, and so we have the isomorphism of rings, which presumably carries forward to the algebras of functions level…

A WC*-subalgebra of $C(G)$ is a W Hopf C*-algebra $C(H)$ together with an injective morphism of WC*-algebras $C(H)\rightarrow C(G)$. Such a morphism is a C*-morphism $\pi:C(H)\rightarrow C(G)$ such that

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

Note this is a different beast to the $\pi: C(G)\rightarrow C(H)$ surjective C*-morphism I have previously seen.

Note that $C(G)$ ($\pi=I_{C(G)}$) and $F(\bullet)$ (with $\pi=\eta_{C(G)}$, and $\Delta_{\bullet}(\bullet)=\mathbf{1}_\bullet\otimes\mathbf{1}_\bullet$) are trivial WC*-subalgebras of $C(G)$.

Note that in the finite classical case, the set of functions vanishing on a proper subgroup $H\subset G$ has the property that:

$S(J)\subset J$, and $\varepsilon(J)=\{0\}$

That $\delta_s$ vanishes on $H$ implies that $s\not\in H$. Could $s^{-1}\in H$? Of course not — because then $H$ would not be closed under inverses. Similarly $e\in H$, and so $j(e)=\varepsilon(j)=0$ for any $j\in J$. From above we know that if $s\not\in H$, that for all $t\in G$, either $st^{-1}$ or $t$ is not an element of $H$, that is either $\delta_{st^{-1}}$ or $\delta_t$ vanishes on $H$. This implies that

$\Delta(J)\subset J\otimes F(G)+F(G)\otimes J$.

This makes $J$Hopf ideal.

Wang remarks that if the algebra of functions is finite, say $F(G)$, and $J$ a WC*-ideal, that $S(J)\subset J$ and $\varepsilon(J)=\{0\}$ (but makes no claim to the above).

There is a slight error if the ideal is not proper. If the ideal is the whole of $F(G)$, then $\varepsilon(J)=\{0\}$ does not hold. Assume therefore that $J$ is a proper ideal. For $\pi: F(G)\otimes F(G)/J$, a linear map between finite vector spaces with kernel $\text{ker }\pi=J$:

$\text{ker }(\pi\otimes \pi)=J\otimes F(G)+F(G)\otimes J$.

Let $j\in J$. As $\Delta(j)=\sum j_{(1)}\otimes j_{(2)}\in\text{ker }(\pi\otimes \pi)$, either $j_{(1)}$ or $j_{(2)}\in J$. We know that $S(J)\subset J$, so that, using the antipodal property:

$\sum S(j_{(1)})j_{(2)}=\eta(\varepsilon(j))\in J$.

If $\varepsilon(j)\neq 0$, then $\eta(\varepsilon(j)) =\varepsilon(j)\mathbf{1}_G\in J$, implying $J=F(G)$. Therefore we have that $\varepsilon(J)=\{0\}$ if $J$ is proper.

I am unable to show the stability of $J$ under $S$, and have farmed this question out to MO (where it has been answered: using the below we have $\pi\circ S_{F(G)}=S_{C(G)/J}\circ \pi$).

### First Isomorphism Theorem

1. The quotient of a WC*-algebra by a WC*-ideal has a unique WC*-algebra structure such that the quotient map is a morphism of WC*-algebras.

2. For every morphism $\theta :C(G_1)\rightarrow C(G_2)$ of WC*-algebras. the kernel is a WC*-ideal. The image of $\theta$ is a WC*-algebra isomorphic to $C(G_1)/\ker \theta$ (as defined above). Furthermore this image is a WC*-subalgebra of $C(G_2)$.

3. Let $\theta$ be the morphism from above. If $J\subset \ker\theta$ (should this be an (closed) ideal?)then there is a unique morphism of WC*-algebras $\tilde{\theta}:C(G_1)/J\rightarrow C(G_2)$, such that $\tilde{\theta}\circ\pi=\theta$, where $\pi$ is the quotient map $C(G_1)\rightarrow C(G_1)/J$.

What does this look like classically?

1.Well first of all if the ideal is the full ideal then we are talking about the algebra of functions on the trivial group. So let us suppose that the ideal is proper. Ideals cannot contain invertible elements. This means that only functions with roots are in the ideal. Let $R_j$ be the set of roots of $j\in J$. Let $j_1,\,j_2\in J$ be two elements in an ideal. Then $j_1j_2$ has roots at $R_{j_1}\cap R_{j_2}$. Note further that this cannot be non-empty, for it it is, then $j_1j_2\in J$ is invertible.

Therefore

$\displaystyle Z(J)=\bigcap_{j\in J}j^{-1}(\{0\})$,

is non-empty and is a subset of $G$ such that every element of $J$ vanishes on it.

Next question: is $Z(J)$ a group? The answer is yes. Let $g_1,\,g_2\in Z(J)$ and $j\in J$. Using $\Delta(j)(g_1,g_2)=j(g_1g_2)$, and $\Delta j\in\ker(\pi\otimes\pi)$, we know that for all $j\in J$, $\Delta(j)=\sum j_{(1)}\otimes j_{(2)}$, either $j_{(1)}$ or $j_{(2)}$ is in $J$. Therefore

$\Delta(j)(\delta^{g_1}\otimes \delta^{g_2})=\sum j_{(1)}(\delta{g_1})j_{(2)}(\delta^{g_2})$,

and either $j_{(1)}(\delta^{g_1})$ or $j_{(2)}(\delta^{g_2})$ is equal to zero. Therefore $\Delta(j)(g_1,g_2)=j(g_1g_2)=0$, and so $Z(J)$ is closed under multiplication. Is $e\in Z(J)$

Because $G$ is finite, every element $g\in G$ has finite order, a least number $o(g)$ such that $g^{o(g)}=e$. Therefore $g^{o(g)-1}=g^{-1}$. We have shown that $Z(J)$ is closed under multiplication and so the result follows. Unfortunately this very “set of points” argument does not transfer easily to the quantum case.

Now presumably in this commutative case, we have

$F(G)/J\cong F(H)$.

This follows from the map $\pi_H$ above.

2. To show this we probably have to show that a morphism of algebras on finite groups corresponds to a group homomorphism. I would suggest that every morphism $\theta:F(G_1)\rightarrow F(G_2)$ is of the form $\theta(f)=f\circ \varphi$ for $\varphi: G_{2}\rightarrow G_1$… I have spent some time now on this problem and perhaps it is a waste of time. I have shown if $\Gamma:=(\mathbb{C}\varphi)^T$ is the pullback of a group homomorphism, then

$\Gamma(\delta_{g_j})=\sum_{h_i\in G_2}\delta_{g_j,\varphi(h_i)}\delta_{h_i}$,

and I am certain this is a morphism of quantum groups. On the other hand I have shown that if $\theta$ is a morphism of classical $F(G_i)$, that with respect to the basis of delta functions, for all $x,y,g\in G_1$

$\theta(xy,g)=\sum_{t\in G_1}\theta(x,gt^{-1})\theta(y,t)$

and I am confident this will yield that $\theta$ is the pushback of a group homomorphism, but I am perhaps wasting my time. Let us move onto the proof of the quantum result.

Proof: 1. Let $\tilde{C(G_1)}$ be the quotient $C(G)/J$. Define

$\tilde{\Delta}(\tilde{f})=(\pi\otimes\pi)\Delta f$,

where $\tilde{f}=\pi(f)$. This is well defined. For if $\tilde{f_1}=\tilde{f_2}$, $\tilde{f}_2-\tilde{f}_1\in J$ and

$\tilde{\Delta}(\tilde{f}_2-\tilde{f}_1)=(\pi\otimes\pi)\Delta(f_2-f_1)=0$,

because $\tilde{f}_2=\tilde{f}_1\Rightarrow f_2-f_1\in J\Rightarrow \Delta(f_2-f_1)\in\ker(\pi\otimes\pi)$, and it follows that $\tilde{\Delta}$ is well defined.

Wang claims that $C(G)/J$ is generated by $\pi(\rho_{ij}^\alpha)$.

2. Let $J=\ker \pi$. Define a C*-isomorphism $\rho:C(G_1)/J\rightarrow \theta(C(G_1))$ by

$\rho(\tilde{f})=\theta(f)$,

that is $\rho\pi=\theta$. Under this isomorphism

$\pi:C(G_1)/J\rightarrow \tilde{A}$ identifies with $\theta:C(G_1)\rightarrow \theta(C(G_1))$.

Let us make a commutative diagram for all this:

Is $J$ a WC*-ideal? Well

$(\theta\otimes\theta)\circ\Delta_1(f)=\Delta_2\circ \theta(f)$

$\Rightarrow (\pi\otimes\pi)\Delta_1(f)=(\rho^{-1}\otimes\rho^{-1})\Delta_2\theta(f)$.

We know that $c\in \ker\theta\Rightarrow (\pi\otimes\pi)\Delta_1(c)=0$ which implies $c\in\ker(\pi\otimes\pi)$, and so $J$ is a WC*-ideal.

It turns out that $\rho$ is an isomorphism of WC*-algebras from $\tilde{A}$ onto $\theta(C(G_1))$, and

$\tilde{\Delta}=(\rho^{-1}\otimes\rho^{-1})\Delta_2\rho$,

and so $(\theta(C(G_1)),\Delta_{\left.\right|_{\theta(C(G_1))}})$ is a WC*-subalgebra of $C(G_2)$ under the natural injection $\quad\bullet$

The following took me a little by surprise:

### Proposition

Let $G$ be a compact group. We have the correspondences:

$H\subset G$, subgroups $\longleftrightarrow C(G)/J\cong C(H)$, WC*-ideals

$G/N$ with $N\lhd G\longleftrightarrow C(G/N)\subset C(G)$, WC*-subalgebras.

OK… a WC*subalgebra is an injective WC*-morphism. I am fairly sure that the map in question, in the classical case, is simply $\mathbf{1}_{Ng}\mapsto \sum_{n\in N}\delta_{ng}$. I am reasonably confident that this is the case. Writing it down seems to be a little awkward.

### Definition

A compact quantum $H$ is called a Wang subgroup of $G$ if there is a WC*-ideal $G$ of $C(G)$, $J$, such that:

$C(H)\cong C(G)/J$

If there is a surjective morphism $C(G)\rightarrow C(H)$, $H$ is called embedded.

Let $N$ be Wang subgroup of a compact quantum group. This means there is a WC*-ideal $J$ such that $C(N)\cong C(G)/J$. Let

$\theta:C(G)\rightarrow C(G)/J\cong C(N)$

be the quotient map.

$N$ is said to be normal if for every irreducible representation $\kappa$ of $G$, with matrix $(\rho_{ij})$, the multiplicity of the trivial representation of $N$; $\lambda\mapsto \lambda\otimes \mathbf{1}_N$, in the representation $\theta(\rho_{ij})$ is either zero or the dimension of $\kappa$.

Hopefully someone here can help me with how this is a classical result.

Let $N\backslash G$ be the right quotient space, defined by Podlés:

$C(N\backslash G)=\{f\in C(G)\,:\,(\theta\otimes I_{C(G)})\Delta(f)=\mathbf{1}_B\otimes f\}$,

This $N\backslash$ is a compact quantum group, the right quotient of $G$ by $N$. We also have a right quotient $G/N$ which is, in general, different to the right quotient. The full quantum group $G$ is normal in $G$, and if the counit is bounded, so is the trivial group.

## Normal Quantum Subgroups

Let $(N,\pi)$ be a quantum subgroup of a quantum group with surjection $\pi: C(G)\rightarrow C(N)$, $\widehat{\pi}:\text{Pol}(G)\rightarrow \text{Pol}(N)$. $(N,\pi)$ is closed but this is not important at this moment.

What follows can also be done for right quotient spaces.

Define the left quotient

$C(G/N)=\{f\in C(G)\,:\,(I_{C(G)}\otimes\pi)\circ \Delta(f)=f\otimes \mathbf{1}_N\}$.

These are functions constant on left cosets of $N$. Define

$\text{Pol}(G/N)=\text{Pol}(G)\cap C(G/N)$.

These are smooth functions constant on left cosets of $N$. Define

$E_{G/N}=(I_{C(G)}\otimes \int_N\circ \pi)\circ \Delta$.

This map is a projection of norm one (completely positive and completely bounded conditional expectation) from $C(G)$ onto the the continuous functions on the right cosets.

Classically it maps a function on $G$ to a function on $G/N$. The value that $E_{G/N}(f)$ takes on $gN$ is the average over $\{gn\,:\,n\in N\}$.

We have that $\text{Pol}(G/N)=E_{G/N}(\text{Pol}(G))$ and this algebra is dense in $C(G/N)$.

### Proposition

Let $N$ be a subgroup of $G$. The following are equivalent.

1. $C(N/G)$ is an (injective) WC*-subalgebra of $C(G)$.
2. same for the right quotient
3. the right and left quotient spaces are equal
4. for every irreducible $\kappa$ with matrix $\rho=(\rho_{ij})$ has $\int_{N} \pi (\rho)=(\int_N \pi (\rho_{ij}))_{i,j=1}^d$ equal to zero or $I_{d}$.

Before we tackle the proof, we must look at the Podlés results that the conditional expectation satisfies

$(E_{N\backslash G}\otimes I_{C(G)})\circ \Delta =\Delta\circ E_{N\backslash G}$ (*)

If we can believe this we have for $f\in C(N\backslash G)$, as $E_{N\backslash G}(f)=f$:

$\Delta f=(E_{N\backslash G}\otimes I_{C(G)})\circ \Delta f=\sum E_{N\backslash G}(f_{(1)})\otimes f_{(2)}\subset C(N\backslash G)\otimes C(G)$.

If we write down (*) as applied to the matrix element of an irreducible representation, $\rho_{ij}^\alpha$, we find that it is actually rather trivial, and essentially due to the identities doing nothing on certain factors. In fact both are equal to:

$(\int_N\circ \pi)\otimes I_{C(G)\otimes C(G)}\Delta^2(\rho_{ij}^\alpha)$.

Proof: (3) $\Rightarrow$ (\$) In general

$\Delta(\text{Pol}(N\backslash G))\subset \text{Pol}(N\backslash G)\otimes \text{Pol}(G)$,

and similar for the left quotient. Letting $B:=\text{Pol}(N\backslash G)=\text{Pol}(G/N)$, we have

$\Delta(B)\subset B\otimes B$.

Let $n_{\alpha}$ be the multiplicity of the trivial representation of $N$ in $\pi(\kappa)$. It is either $d_\rho$ or zero as shown (more or less) here.

I am going to leave the rest of the proof and move on $\bullet$

### Definition

A quantum subgroup $N$ of a CQG $G$ is said to be normal if it satisfies the any of the equivalent conditions of the preceding proposition.

There is another condition given above about the multiplicity of the trivial representation… it is equivalent to condition (4).

The following seems a generalisation of the fact that every subgroup of an abelian group is normal.

### Proposition

Let $\mathbb{C}\Gamma=:C(\widehat{\Gamma})$. Let $N$ be a quantum subgroup of $G$ with surjection $\pi:C(\widehat{\Gamma})\rightarrow C(N)$. Then $N$ is normal, where $s\hookrightarrow \delta^s$ is the embedding of $\Gamma$ in $\mathbb{C}\Gamma$, $L:=\pi(\Gamma)$ is discrete, and $C(N)=\mathbb{C}L=C(\widehat{L})$. Moreover $C(\widehat{\Gamma}/N)=\mathbb{C}K$ where $K=\ker(\Gamma\rightarrow L)$ $\bullet$