Quasi-Subgroups that are not Subgroups

Let G be a finite quantum group. We associate to an idempotent state \int_Squasi-subgroup S. This nomenclature must be included in the manuscript under preparation.

As is well known from the GNS representation, positive linear functionals can be associated to closed left ideals:

\displaystyle N_{\rho}:=\left\{ f\in F(G):\rho(|f|^2)=0\right\}.

In the case of a quasi-subgroup, S\subset G, my understanding is that by looking at N_S:=N_{\int_S} we can tell if S is actually a subgroup or not. Franz & Skalski show that:

Let S\subset G be a quasi-subgroup. TFAE

  • S\leq G is a subgroup
  • N_{\int_S} is a two-sided or self-adjoint or S invariant ideal of F(G)
  • \mathbf{1}_Sa=a\mathbf{1}_S

I want to look again at the Kac & Paljutkin quantum group \mathfrak{G}_0 and see how the Pal null-spaces N_{\rho_6} and N_{\rho_7} fail these tests. Both Franz & Gohm and Baraquin should have the necessary left ideals.

The Pal Null-Space N_{\rho_6}

The following is an idempotent probability on the Kac-Paljutkin quantum group:

\displaystyle \rho_6(f)=2\int_{\mathfrak{G}_0}f\cdot (e_1+e_4+E_{11}).

Hence:

N_{\rho_6}=\langle e_1,e_3,E_{12},E_{22}\rangle.

If N_{\rho_6} were two-sided, N_{\rho_6}F(\mathfrak{G}_0)\subset N_{\rho_6}. Consider E_{21}\in F(\mathfrak{G}_0) and

E_{12}E_{21}=E_{11}\not\in N_{\rho_6}.

We see problems also with E_{12} when it comes to the adjoint E_{12}^{\ast}=E_{21}\not\in N_{\rho_6} and also S(E_{12})=E_{21}\not\in N_{\rho_6}. It is not surprise that the adjoint AND the antipode are involved as they are related via:

S(S(f^\ast)^\ast)=f.

In fact, for finite or even Kac quantum groups, S(f^\ast)=S(f)^\ast.

Can we identity the support p? I think we can, it is (from Baraquin)

p_{\rho_6}=e_1+e_4+E_{11}.

This does not commute with F(G):

E_{21}p_{\rho_6}=E_{21}\neq 0=p_{\rho_6}E_{21}.

The other case is similar.

This tells us something. Unfortunately that something is rather ‘finite-specific’, but I guess we have thrown our eggs in that basket at this stage. The algebra of functions on a finite quantum group G is a ‘full’ matrix algebra:

\displaystyle F(G)=\bigoplus_{n_i\in \mathbb{N}} M_{n_i}(\mathbb{C}),

and the only central projections are of the form:

Z(F(G))=\sum_{\{n_j\}\subset \{n_i\}}I_{n_j}.

This probably does not give us much unless we can attack from a representation-theoretic-angle.

More Notes

One thing I have not explicitly done is show that in the case of a periodic random walk is that the projection p_0 is a group-like projection. This is pretty vital to anything that I am doing.

There are two ways to come at this problem. The first is to show that p_0 satisfies:

\Delta(p_0)(p_0\otimes p_0)=p_0\otimes p_0.

It is possibly easier to come at this from the other side: show that p_0 is the support of an invariant state. Franz & Skalski show that if \phi is idempotent, there exists a group-like projection p such that:

\phi=\mathcal{F}\left(\frac{p}{\int_G p}\right).

This includes a normalisation factor. It might be an idea to prove that this p is indeed the support of \phi.

First of all, we do have that:

\phi(pa)=\phi(a).

Note

\displaystyle \phi(p)=\int_Gp\dfrac{1}{\int_G p}p=1.

Let p_\phi be the support projection of \phi, the smallest projection p' such that \phi(p')=1. This implies that p_\phi<p so that p-p_\phi is a projection. Thus:

\phi(|p-p_\phi|^2)=\phi(p)-\phi(p_\phi)=0,

so that p-p_\phi\in N_\phi. Let q_\phi=\mathbf{1}_G-p_\phi. We can show that N_\phi=F(G)q_\phi and bq_\phi=b for all b\in N_\phi. In particular:

(p-p_\phi)q_\phi=p-p_\phi.

It is clear that \phi(pa)=\phi(a) is not enough to show that p is the support projection — for example \phi(\mathbf{1}_Ga)=\phi(a).

We need to use the fact, where \tilde{p}=\frac{p}{\int_G p}, that

\phi(a)=\int_Ga\tilde{p},

and that \int_G is faithful. Suppose that p_\phi<p so that p-p_\phi is a projection.

Consider

\displaystyle \int_G(p-p_\phi)=\int_G(p-p_\phi p)=\int_G p-\int_Gp_\phi p.

Note

\displaystyle \int_Gp_\phi p=\int_Gp\cdot \int_Gp_\phi \tilde{p}=\int_G p\cdot \phi(p_\phi)=\int_Gp\cdot 1 =\int_Gp,

so that \int_G (p-p_\phi)=0, and as \int_G is faithful, p-p_\phi=0, and so the group-like projection of an idempotent state is also its support.

To finish this off, we must show that the idempotent state associated to p_0 is indeed idempotent… it should be by (T_\nu)^d. I have a little confusion here… we have invariant states and we have idempotent states. These are not precisely the same thing… I am fairly sure that this “invariant state” is dp_0, which is not a state. We might have something like:

T_\nu^d(d\,p_0)=dp_0.

The problem is this right eigenvector… does it necessarily give a left eigenvector that is a state… This is a problem. Recall that \int_Gp_0=d. Can we maybe get from an (actual) invariant state to an idempotent?

Perhaps using the convolution in F(G),

p_0\star_{F(G)}p_0=\frac{1}{d}p_0?

This is now next week’s problem.