This sandbox is going to take from a variety of sources, mostly Shuzhou Wang.
C*Ideals
Let be a closed (twosided) ideal in a noncommutative unital algebra . Such an ideal is selfadjoint and so a noncommutative algebra . The quotient map is given by , , where is the equivalence class of under the equivalence relation:
.
Where we have the product
,
and the norm is given by:
,
the quotient is a algebra.
Consider now elements and . Consider
.
The tensor product . Now note that
,
by the nature of the Tensor Product (). Therefore .
Definition
A WC*ideal (W for Woronowicz) is a C*ideal such that , where is the quotient map .
Let be the algebra of functions on a classical group . Let . Let be the set of functions which vanish on : this is a C*ideal. The kernal of is .
Let so that . Note that
and so
.
Note that if . It is not possible that both and are in : if they were , but , which is not in by assumption. Therefore one of or is equal to zero and so:
,
and so by linearity, if vanishes on a subgroup ,
.
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 . Let be the ring homomorphism
.
Then , , and so we have the isomorphism of rings, which presumably carries forward to the algebras of functions level…
A WC*subalgebra of is a W Hopf C*algebra together with an injective morphism of WC*algebras . Such a morphism is a C*morphism such that
.
Note this is a different beast to the surjective C*morphism I have previously seen.
Note that () and (with , and ) are trivial WC*subalgebras of .
Note that in the finite classical case, the set of functions vanishing on a proper subgroup has the property that:
, and
That vanishes on implies that . Could ? Of course not — because then would not be closed under inverses. Similarly , and so for any . From above we know that if , that for all , either or is not an element of , that is either or vanishes on . This implies that
.
This makes a Hopf ideal.
Wang remarks that if the algebra of functions is finite, say , and a WC*ideal, that and (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 , then does not hold. Assume therefore that is a proper ideal. For , a linear map between finite vector spaces with kernel :
.
Let . As , either or . We know that , so that, using the antipodal property:
.
If , then , implying . Therefore we have that if is proper.
I am unable to show the stability of under , and have farmed this question out to MO (where it has been answered: using the below we have ).
First Isomorphism Theorem

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.

For every morphism of WC*algebras. the kernel is a WC*ideal. The image of is a WC*algebra isomorphic to (as defined above). Furthermore this image is a WC*subalgebra of .

Let be the morphism from above. If (should this be an (closed) ideal?), then there is a unique morphism of WC*algebras , such that , where is the quotient map .
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 be the set of roots of . Let be two elements in an ideal. Then has roots at . Note further that this cannot be nonempty, for it it is, then is invertible.
Therefore
,
is nonempty and is a subset of such that every element of vanishes on it.
Next question: is a group? The answer is yes. Let and . Using , and , we know that for all , , either or is in . Therefore
,
and either or is equal to zero. Therefore , and so is closed under multiplication. Is ?
Because is finite, every element has finite order, a least number such that . Therefore . We have shown that 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
.
This follows from the map 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 is of the form for … I have spent some time now on this problem and perhaps it is a waste of time. I have shown if is the pullback of a group homomorphism, then
,
and I am certain this is a morphism of quantum groups. On the other hand I have shown that if is a morphism of classical , that with respect to the basis of delta functions, for all
and I am confident this will yield that 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 be the quotient . Define
,
where . This is well defined. For if , and
,
because , and it follows that is well defined.
Wang claims that is generated by .
2. Let . Define a C*isomorphism by
,
that is . Under this isomorphism
identifies with .
Let us make a commutative diagram for all this:
Is a WC*ideal? Well
.
We know that which implies , and so is a WC*ideal.
It turns out that is an isomorphism of WC*algebras from onto , and
,
and so is a WC*subalgebra of under the natural injection
The following took me a little by surprise:
Proposition
Let be a compact group. We have the correspondences:
, subgroups , WC*ideals
with , 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 . I am reasonably confident that this is the case. Writing it down seems to be a little awkward.
Definition
A compact quantum is called a Wang subgroup of if there is a WC*ideal of , , such that:
If there is a surjective morphism , is called embedded.
Let be Wang subgroup of a compact quantum group. This means there is a WC*ideal such that . Let
be the quotient map.
is said to be normal if for every irreducible representation of , with matrix , the multiplicity of the trivial representation of ; , in the representation is either zero or the dimension of .
Hopefully someone here can help me with how this is a classical result.
Let be the right quotient space, defined by Podlés:
,
This is a compact quantum group, the right quotient of by . We also have a right quotient which is, in general, different to the right quotient. The full quantum group is normal in , and if the counit is bounded, so is the trivial group.
Normal Quantum Subgroups
Let be a quantum subgroup of a quantum group with surjection , . is closed but this is not important at this moment.
What follows can also be done for right quotient spaces.
Define the left quotient
.
These are functions constant on left cosets of . Define
.
These are smooth functions constant on left cosets of . Define
.
This map is a projection of norm one (completely positive and completely bounded conditional expectation) from onto the the continuous functions on the right cosets.
Classically it maps a function on to a function on . The value that takes on is the average over .
We have that and this algebra is dense in .
Proposition
Let be a subgroup of . The following are equivalent.
 is an (injective) WC*subalgebra of .
 same for the right quotient
 the right and left quotient spaces are equal
 for every irreducible with matrix has equal to zero or .
Before we tackle the proof, we must look at the Podlés results that the conditional expectation satisfies
(*)
If we can believe this we have for , as :
.
If we write down (*) as applied to the matrix element of an irreducible representation, , 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:
.
Proof: (3) ($) In general
,
and similar for the left quotient. Letting , we have
.
Let be the multiplicity of the trivial representation of in . It is either or zero as shown (more or less) here.
I am going to leave the rest of the proof and move on
Definition
A quantum subgroup of a CQG 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 . Let be a quantum subgroup of with surjection . Then is normal, where is the embedding of in , is discrete, and . Moreover where
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January 16, 2020 at 12:13 pm
Freslon’s Ergodic Theorem for (Finite) Group Algebras  J.P. McCarthy: Math Page
[…] Well… I have an inkling that because dual groups satisfy what I would call the condition of abelianness (under the ‘quantisation’ functor), all (quantum) subgroups are normal… this is probably an obvious thing to write down (although I must search the literature) to ensure it is indeed known (or is untrue?). Edit: Wang had it already, see the last proposition here. […]