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In a recent preprint, Haonan Zhang shows that if \nu\in M_p(Y_n) (where Y_n is a Sekine Finite Quantum Group), then the convolution powers, \nu^{\star k}, converges if

\nu(e_{(0,0)})>0.

The algebra of functions F(Y_n) is a multimatrix algebra:

F(Y_n)=\left(\bigoplus_{i,j\in\mathbb{Z}_n}\mathbb{C}e_{(i,j)}\right)\oplus M_n(\mathbb{C}).

As it happens, where a=\sum_{i,j\in\mathbb{Z}_n}x_{(i,j)}e_{(i,j)}\oplus A, the counit on F(Y_n) is given by \varepsilon(a)=x_{(0,0)}, that is \varepsilon=e^{(0,0)}, dual to e_{(0,0)}.

To help with intuition, making the incorrect assumption that Y_n is a classical group (so that F(Y_n) is commutative — it’s not), because \varepsilon=e^{(0,0)}, the statement \nu(e_{(0,0)})>0, implies that for a real coefficient x^{(0,0)}>0,

\nu=x^{(0,0)}\varepsilon+\cdots= x^{(0,0)}\delta^e+\cdots,

as for classical groups \varepsilon=\delta^e.

That is the condition \nu(e_{(0,0)})>0 is a quantum analogue of e\in\text{supp}(\nu).

Consider a random walk on a classical (the algebra of functions on G is commutative) finite group G driven by a \nu\in M_p(G).

The following is a very non-algebra-of-functions-y proof that e\in \text{supp}(\nu) implies that the convolution powers of \nu converge.

Proof: Let H be the smallest subgroup of G on which \nu is supported:

\displaystyle H=\bigcap_{\underset{\nu(S_i)=1}{S_i\subset G}}S_i.

We claim that the random walk on H driven by \nu is ergordic (see Theorem 1.3.2).

The driving probability \nu\in M_p(G) is not supported on any proper subgroup of H, by the definition of H.

If \nu is supported on a coset of proper normal subgroup N, say Nx, then because e\in \text{supp}(\nu), this coset must be Ne\cong N, but this also contradicts the definition of H.

Therefore, \nu^{\star k} converges to the uniform distribution on H \bullet

Apart from the big reason — that this proof talks about points galore — this kind of proof is not available in the quantum case because there exist \nu\in M_p(G) that converge, but not to the Haar state on any quantum subgroup. A quick look at the paper of Zhang shows that some such states have the quantum analogue of e\in\text{supp}(\nu).

So we have some questions:

  • Is there a proof of the classical result (above) in the language of the algebra of functions on G, that necessarily bypasses talk of points and of subgroups?
  • And can this proof be adapted to the quantum case?
  • Is the claim perhaps true for all finite quantum groups but not all compact quantum groups?
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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.

Slides of a talk given at the Irish Mathematical Society 2018 Meeting at University College Dublin, August 2018.

Abstract Four generalisations are used to illustrate the topic. The generalisation from finite “classical” groups to finite quantum groups is motivated using the language of functors (“classical” in this context meaning that the algebra of functions on the group is commutative). The generalisation from random walks on finite “classical” groups to random walks on finite quantum groups is given, as is the generalisation of total variation distance to the quantum case. Finally, a central tool in the study of random walks on finite “classical” groups is the Upper Bound Lemma of Diaconis & Shahshahani, and a generalisation of this machinery is used to find convergence rates of random walks on finite quantum groups.

Distances between Probability Measures

Let G be a finite quantum group and M_p(G) be the set of states on the \mathrm{C}^\ast-algebra F(G).

The algebra F(G) has an invariant state \int_G\in\mathbb{C}G=F(G)^\ast, the dual space of F(G).

Define a (bijective) map \mathcal{F}:F(G)\rightarrow \mathbb{C}G, by

\displaystyle \mathcal{F}(a)b=\int_G ba,

for a,b\in F(G).

Then, where \|\cdot\|_1^{F(G)}=\int_G|\cdot| and \|\cdot\|_\infty^{F(G)}=\|\cdot\|_{\text{op}}, define the total variation distance between states \nu,\mu\in M_p(G) by

\displaystyle \|\nu-\mu\|=\frac12 \|\mathcal{F}^{-1}(\nu-\mu)\|_1^{F(G)}.

(Quantum Total Variation Distance (QTVD))

Standard non-commutative \mathcal{L}^p machinary shows that:

\displaystyle \|\nu-\mu\|=\sup_{\phi\in F(G):\|\phi\|_\infty^{F(G)}\leq 1}\frac12|\nu(\phi)-\mu(\phi)|.

(supremum presentation)

In the classical case, using the test function \phi=2\mathbf{1}_S-\mathbf{1}_G, where S=\{\nu\geq \mu\}, we have the probabilists’ preferred definition of total variation distance:

\displaystyle \|\nu-\mu\|_{\text{TV}}=\sup_{S\subset G}|\nu(\mathbf{1}_S)-\mu(\mathbf{1}_S)|=\sup_{S\subset G}|\nu(S)-\mu(S)|.

In the classical case the set of indicator functions on the subsets of the group exhaust the set of projections in F(G), and therefore the classical total variation distance is equal to:

\displaystyle \|\nu-\mu\|_P=\sup_{p\text{ a projection}}|\nu(p)-\mu(p)|.

(Projection Distance)

In all cases the quantum total variation distance and the supremum presentation are equal. In the classical case they are equal also to the projection distance. Therefore, in the classical case, we are free to define the total variation distance by the projection distance.

Quantum Projection Distance \neq Quantum Variation Distance?

Perhaps, however, on truly quantum finite groups the projection distance could differ from the QTVD. In particular, a pair of states on a M_n(\mathbb{C}) factor of F(G) might be different in QTVD vs in projection distance (this cannot occur in the classical case as all the factors are one dimensional).

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Just back from a great workshop at Seoul National University, I am just going to use this piece to outline in a relaxed manner my key goals for my work on random walks on quantum groups for the near future.

In the very short term I want to try and get a much sharper lower bound for my random walk on the Sekine family of quantum groups. I believe the projection onto the ‘middle’ of the M_n(\mathbb{C}) might provide something of use. On mature reflection, recognising that the application of the upper bound lemma is dominated by one set of terms in particular, it should be possible to use cruder but more elegant estimates to get the same upper bound except with lighter calculations (and also a smaller \alpha — see Section 5.7).

I also want to understand how sharp (or otherwise) the order n^n convergence for the random walk on the dual of S_n is — n^n sounds awfully high. Furthermore it should be possible to get a better lower bound that what I have.

It should also be possible to redefine the quantum total variation distance as a supremum over projections \sim subsets via G \supset S\leftrightarrow \mathbf{1}_S. If I can show that for a positive linear functional \rho that |\rho(a)|\leq \rho(|a|) then using these ideas I can. More on this soon hopefully. No, this approach won’t work. (I have since completed this objective with some help: see here).

The next thing I might like to do is look at a random walk on the Sekine quantum groups with an n-dependent driving probability and see if I can detect the cut-off phenomenon (Chapter 4). This will need good lower bounds for k\ll t_n, some cut-off time.

Going back to the start, the classical problem began around 1904 with the question of Markov:

Which card shuffles mix up a deck of cards and cause it to ‘go random’?

For example, the perfect riffle shuffle does not mix up the cards at all while a riffle shuffle done by an amateur will.

In the context of random walks on classical groups this question is answered by the Ergodic Theorem 1.3.2: when the driving probability is not concentrated on a subgroup (irreducibility) nor the coset of a normal subgroup (aperiodicity).

Necessary and sufficient conditions on the driving probability \nu\in M_p(\mathbb{G}) for the random walk on a quantum group to converge to random are required. It is expected that the conditions may be more difficult than the classical case. However, it may be possible to use Diaconis-Van Daele theory to get some results in this direction. It should be possible to completely analyse some examples (such as the Kac-Paljutkin quantum group of order 8).

This will involve a study of subgroups of quantum groups as well as normal quantum subgroups.

It should be straightforward to extend the Upper Bound Lemma (Lemma 5.3.8) to the case of compact Kac algebras. Once that is done I will want to look at quantum generalisations of ‘natural’ random walks and shuffles.

I intend also to put the PhD thesis on the Arxiv. After this I have a number of options as regard to publishing what I have or maybe waiting a little while until I solve the above problems — this will all depend on how my further study progresses.

 

Slides of a talk given at the Topological Quantum Groups and Harmonic Analysis workshop at Seoul National University, May 2017.

Abstract A central tool in the study of ergodic random walks on finite groups is the Upper Bound Lemma of Diaconis & Shahshahani. The Upper Bound Lemma uses the representation theory of the group to generate upper bounds for the distance to random and thus can be used to determine convergence rates for ergodic walks. These ideas are generalised to the case of finite quantum groups.

Taken from Condition Expectation in Quantum Probabilty by Denes Petz.

In quantum probability there are a number of fundamental questions that ask how faithfully can one quantise classical probability. Suppose that (\Omega,\mathcal{S},P) is a (classical) probability space and \mathcal{G}\subset\mathcal{A} a sub-\sigma-algebra. The conditional expectation of some integrable function f (with respect to some L-space) relative to \mathcal{G} is the orthogonal projection onto the closed subspace L(\mathcal{G}):

\mathbb{E}^{\mathcal{G}}:L(\mathcal{A})\rightarrow L(\mathcal{G})f\mapsto \mathbb{E}(f|\mathcal{G}).

Suppose now that (A,\rho) is a quantum probability space and that B is some C*-subalgebra of A. Can we always define a conditional expectation with respect to B? The answer turns out to be not always, although this paper gives sufficient conditions for the existence of such a projection. Briefly, things work the other way around. Distinguished states give rise to quantum conditional expectations — and these conditional expectations define a subalgebra. We can’t necessarily start with a subalgebra and find the state which gives rise to it — Theorem 2 gives necessary conditions in which this approach does work.

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Taken from Real Analysis and Probability by R.M. Dudley.

For a sequence of n repeated, independent trials of an experiment, some probability distributions and variables converge as n tends to infinity. In proving such limit theorems, it is useful to be able to construct a probability space on which a sequence of independent random variables is defined in a natural way; specifically, as coordinates for a countable Cartesian product.

The Cartesian product of finitely many \sigma-finite measure spaces gives a \sigma-finite measure space. For example, Cartesian products of Lesbesgue measure on the line give Lesbesgue measure on finite-dimensional Euclidean spaces. But suppose we take a measure space \{0,1\} with two points each having measure 1\mu(\{0\})=1=\mu(\{1\}), and form a countable Cartesian product of copies of this space, so that the measure of any countable product of sets equals the product of their measures. Then we would get an uncountable space in which all singletons have measure 1, giving the measure usually called counting measure. An uncountable set with counting measure is not a \sigma-finite space, although in this example it was a countable product of finite measure spaces. By contrast, the the countable product of probability measures will again be a probability space. Here are some definitions.

For each n=1,2,\dots let (\Omega_n,S_n,P_n) be a probability space. Let \Omega be the Cartesian product \displaystyle \prod_{n\geq 1}\Omega_n, that is, the set of all sequences \{\omega_n\}_{n\geq 1} with \omega_n\in\Omega_n for all n. Let \pi_n be the natural projection of \Omega onto \Omega_n for each n\pi_n\left(\{\omega_m\}_{m\geq 1}\right)=\omega_n for all n. Let S be the smallest \sigma-algebra of subsets of \Omega such that for all m\pi_m is measurable from (\Omega,S) to (\Omega_m,S_m). In other words, S is the smallest \sigma-algebra containing all sets \pi_n^{-1}(A) for all n and all A\in S_n.

Let \mathcal{R} be the collection of all sets \displaystyle \prod_{n\geq 1}A_n\subset\Omega where A_n\in \mathcal{S}_n for all n and A_m=\Omega_m except for at most finitely many values of n. Elements of \mathcal{R} will be called rectangles. Now recall the notion of semiring. \mathcal{R} has this property.

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