Back before Christmas I felt I was within a week of proving the following:

Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group G is ergodic if and only if \nu is not concentrated on a proper quasi-subgroup, nor the coset of a ?normal ?-subgroup.

The first part of this conjecture says that if \nu is concentrated on a quasi-subgroup, then it stays concentrated there. Furthermore, we can show that if the random walk is reducible that the Césaro limit gives a quasi-subgroup on which \nu is concentrated.

The other side of the ergodicity coin is periodicity. In the classical case, it is easy to show that if the driving probability is concentrated on the coset of a proper normal subgroup N\lhd G, that the convolution powers jump around a cyclic subgroup of G/N.

One would imagine that in the quantum case this might be easy to show but alas this is not proving so easy.

I am however pushing hard against the other side. Namely, that if the random walk is periodic and irreducible, that the driving probability in concentrated on some quasi-normal quasi-subgroup!

The progress I have made depends on work of Fagnola and Pellicer. They show that if the random walk is irreducible and periodic that there exists a partition of unity \{p_0,p_1,\dots,p_{d-1}\} such that \nu^{\star k} is concentrated on p_{k\mod d}.

This cyclic nature suggests that p_0 might be equal to \mathbf{1}_N for some N\lhd G and perhaps:

\Delta(p_i)=\sum_{j=0}^{d-1}p_{i-j}\otimes p_j,

and perhaps there is an isomorphism G/N\cong C_d. Unfortunately I have been unable to progress this.

What is clear is that the ‘supports’ of the p_i behave very much like the cosets of proper normal subgroup N\lhd G.

As the random walk is assumed irreducible, we know that for any projection q\in 2^G, there exists a k_q\in \mathbb{N} such that \nu^{\star k_q}(q)\neq 0.

Playing this game with the Haar element, \eta\in 2^G, note there exists a k_\eta\in\mathbb{N} such that \nu^{k_\eta}(\eta)>0.

Let \overline{\nu}=\nu^{\star k_\eta}. I have proven that if \mu(\eta)>0, then the convolution powers of \mu\in M_p(G) converge. Convergence is to an idempotent. This means that \overline{\nu}^{\star k} converges to an idempotent \overline{\nu}_\infty, and so we have a quasi-subgroup corresponding to it, say \overline{p}.

The question is… does \overline{p} coincide with p_0?

If yes, is there any quotient structure by a quasi-subgroup? Is there a normal quasi-subgroup that allows such a structure?

Is \overline{p} a subgroup? Could it be a normal subgroup?

As nice as it was to invoke the result that if e is in the support of \nu, then the convolution powers of \nu converge, by looking at those papers which cite Fagnola and Pellicer we see a paper that gives the same result without this neat little lemma.

This paper by Carbone and Pautrat contains the following result, which basically achieves the same as above — except not just on \overline{p} — but in each p_iF(G)p_i:

Consider an irreducible random walk with cyclic projections p_0,p_1,\dots,p_{d-1}.

Then the restriction of (T_\nu)^d=T_{\nu^{\star d}} to p_iF(G)p_i is irreducible and aperiodic,  and (\nu^{\star d})^{\star k}\rightarrow dp_i\int_Gp_i

The explicit state is not a state I note… now I note something far more interesting. I don’t believe that the invariant state above is a state at all but what we (will) see is that:

T_\nu(p_iap_j)=p_{i-1}ap_{j-1}.

This is a game changer. This paper, in the language that I am using, refer not to an element of the dual M_p(G) as a state, but rather the density of a state, a_\nu.

There is a paper of Skalski and Franz that basically says that if the cyclic projections are in the centre of F(G), we have something like p_0 is associated with an actual subgroup.

Fagnola and Pellicer (and my PhD supervisor Stephen Wills) put great importance on the fact that T_\nu is completely positive. This allows us to write, for functions L_\ell \in F(G),

T_\nu(a)=\sum_{\ell=1}^mL_\ell^\ast  aL_\ell.

What would be great is if we could decompose:

F(G)=\sum_{i=0}^{d-1}p_iF(G)p_i.

This does not seem automatic in the quantum case. However by pre- and post-multiplying by \mathbf{1}_G=\sum_ip_i we can write

F(G)=\sum_{i,j=0}^{d-1}p_iF(G)p_j,

and the action of T_\nu on these ‘factors’ is straightforward. Fagnola and Pellicer show that p_iL_\ell=L_\ell p_{i-1}. The following calculation is stated without proof by Carbone and Pautrat.

\begin{aligned} T_\nu(p_iap_j)&=\sum_{\ell=1}^mL_\ell^{\ast}p_iap_j L_{\ell}=\sum_{\ell=1}^m(p_iL_\ell)^\ast a L_\ell p_{j-1} \\ & = \sum_{\ell=1}^m\left(L_\ell p_{i-1}\right)^\ast aL_\ell p_{j-1}  =\sum_{\ell=1}^m p_{i-1}L_\ell^\ast a L_\ell p_{j-1} \\ & =p_{i-1}\left(\sum_{\ell=1}^mL_\ell^\ast a L_\ell\right)p_{j-1}=p_{i-1}T_{\nu}(a)p_{j-1}\end{aligned}

We know that T_{\nu^{\star d}} is ergodic for p_0… my issue is that, in my head, this invariant state should be \rho=\mathcal{F}(dp_0)

Let us look in a paper, or rather series of papers, of Franz and Skalski. Perhaps also at the end we might summarise some things that did not work.

An idempotent state \int_S has a support projection \mathbf{1}_S. In the commutative case, S is a subgroup S\leq G, but in the quantum case S need not be a subgroup and we call S a quasi-subgroup.

We can construct on object that looks like the algebra of functions. Let us list some properties of \mathbf{1}_S. It is a projection. It contains the identity and is closed under inverses. It satisfies:

\Delta(\mathbf{1}_S)(\mathbf{1}_G\otimes\mathbf{1}_S)=\mathbf{1}_S\otimes\mathbf{1}_S=\Delta(\mathbf{1}_S)=(\mathbf{1}_S\otimes \mathbf{1}_G).

Consider F(S)=\mathbf{1}_SF(G)\mathbf{1}_S and, for f\in F(S)

 \Delta_S(f)=(\mathbf{1}_S\otimes \mathbf{1}_S)\Delta(f)((\mathbf{1}_S\otimes \mathbf{1}_S)).

Why is this thing not a subgroup? As the nullspace of \int_S need not be self-adjoint. Do I understand the importance of this? Not necessarily! Presumably if we rather look at the support of the state we do not have that it is self-adjoint.

Here is the big result I am looking at:

If \mathbf{1}_S is in the centre of F(G), then S is a quantum subgroup.

If I could show this for p_0 (from above), then possibilities open.

Unfortunately the decomposition F(G)=\sum_{i,j=0}^{d-1}p_iF(G)p_j provides little help.

Franz and Skalski prove the following:

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

  • S\leq G is a subgroup
  • N_{\int_S}=\{a\in F(G):\int_S|a|^2=0\} is a two-sided or self-adjoint or S invariant ideal of F(G)
  • \mathbf{1}_Sa=a\mathbf{1}_S

We are getting nothing for nothing here!

Here is a different thought. How about we use duality. Hear myself out.

We know that the random walk is irreducible. Can we say that the initial “dual” story with the random walk is that is begins at

\varepsilon=\mathcal{F}\left(\frac{\eta}{\int_G \eta}\right),

i.e. (normalised) \eta, say \tilde{\eta}.

Now we know that p_0\tilde{\eta}p_0=\tilde{\eta}\Rightarrow \tilde{\eta}\in p_0F(G)p_0 and so:

T_\nu^k(\tilde{\eta})\in p_{-k}F(G)p_{-k},

and because the random walk is irreducible, every projection q\in 2^G must be in one of the p_{-k}F(G)p_{-k} (well, this is clearly wrong! Maybe we want it never to be in p_iF(G)p_j). We know that, by the by, because of finiteness, F(G) is equal to the span of its projections.

Well anyway, this gives us, in fact, that

F(G)=\sum_{i=0}^{d-1}p_iF(G)p_i,

and now we know that p_0 gives a subgroup because:

p_0a=p_0\left(\sum_{i=0}^{d-1}p_ia_i p_i\right) =p_0a_0p_0=ap_0.

Bingo bango, jobs a good un.

But is this duality argument valid? Note that

\varepsilon T_\nu^k=\nu^{\star k}.

This implies that for all q\in 2^G, there exists k_q such that:

\varepsilon(T_\nu^{k_q}(q))\neq 0.

 

Hmmm… well, no time to go back on the stuff that did not work.