Quasi-Subgroups that are not Subgroups

Let $G$ be a finite quantum group. We associate to an idempotent state $\int_S$quasi-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 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 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.