## Quantum Subgroups

Let $C(G)$ be a the algebra of functions on a finite or perhaps compact quantum group (with comultiplication $\Delta$) and $\nu\in M_p(G)$ a state on $C(G)$. We say that a quantum group $H$ with algebra of function $C(H)$ (with comultiplication $\Delta_H$) is a quantum subgroup of $G$ if there exists a surjective unital *-homomorphism $\pi:C(G)\rightarrow C(H)$ such that:

$\displaystyle \Delta_H\circ \pi=(\pi\circ \pi)\circ \Delta$.

## The Classical Case

In the classical case, where the algebras of functions on $G$ and $H$ are commutative,

$\displaystyle \pi(\delta_g)=\left\{\begin{array}{cc}\delta_g & \text{ if }g\in H \\ 0 & \text{ otherwise}\end{array}\right..$

There is a natural embedding, in the classical case, if $H$ is open (always true for $G$ finite) (thanks UwF) of $\imath: C(H) \xrightarrow\, C(G)$,

$\displaystyle \sum_{h\in H}a_h \delta_h \mapsto \sum_{g\in G} a_g \delta_g$,

with $a_g=a_h$ for $h\in G$, and $a_g=0$ otherwise.

Furthermore, $\pi$ is has the property that

$\pi\circ\imath\circ \pi=\pi$,

which resembles $\pi^2=\pi$.

In the case where $\nu$ is a probability on a classical group $G$, supported on a subgroup $H$, it is very easy to see that convolutions $\nu^{\star k}$ remain supported on $H$. Indeed, $\nu^{\star k}$ is the distribution of the random variable

$\xi_k=\zeta_k\cdots \zeta_2\cdot \zeta_1$,

where the i.i.d. $\zeta_i\sim \nu$. Clearly $\xi_k\in H$ and so $\nu^{\star k}$ is supported on $H$.

We can also prove this using the language of the commutative algebra of functions on $G$, $C(G)$. The state $\nu\in M_p(G)$ being supported on $H$ implies that

$\nu\circ\imath\circ \pi=\nu\imath\pi=\nu$.

Consider now two probabilities on $G$ but supported on $H$, say $\mu,\,\nu$. As they are supported on $H$ we have

$\mu=\mu\imath\pi$ and $\nu=\nu\imath\pi$.

Consider

$(\mu\star \nu)\imath\pi=(\mu\otimes \nu)\circ \Delta\circ \imath\pi$

$=((\mu\imath\pi)\otimes(\nu\imath\pi))\circ \Delta\circ\imath\pi =(\mu\imath\otimes \nu\imath)(\pi\circ \pi)\Delta\circ\imath\pi$

$=(\mu\imath\otimes\nu\imath)(\Delta_H\circ \pi\circ \imath\circ \pi)=(\mu\imath\otimes\nu\imath)(\Delta_H\circ \pi)$

$(\mu\imath\otimes \nu\imath)\circ (\pi\circ \pi)\circ\Delta(\mu\imath\pi\otimes \nu\imath\pi)\circ\Delta$

$=(\mu\otimes\nu)\circ\Delta=\mu\star \nu$,

that is $\mu\star \nu$ is also supported on $H$ and inductively $\nu^{\star k}$.

## Some Questions

Back to quantum groups with non-commutative algebras of functions.

• Can we embed $C(H)$ in $C(G)$ with a map $\imath$ and do we have $\pi\circ \imath\circ \pi=\pi$, giving the projection-like quality to $\pi$?
• Is $\nu\circ\imath\circ \pi=\nu$ a suitable definition for $\nu$ being supported on the subgroup $H$.

If this is the case, the above proof carries through to the quantum case.

• If there is no such embedding, what is the appropriate definition of a $\nu\in M_p(G)$ being supported on a quantum subgroup $H$?
• If $\pi$ does not have the property of $\pi\circ \imath\circ \pi=\pi$, in this or another definition, is it still true that $\nu$ being supported on $H$ implies that $\nu^{\star k}$ is too?

## Edit

UwF has recommended that I look at this paper to improve my understanding of the concepts involved.