Taken from Random Walks on Finite Quantum Groups by Franz & Gohm

In this section we will show how one can recover a classical Markov chain from a quantum Markov chain. We will apply a folklore theorem that says one gets a classical Markov process, if a quantum Markov process can be restricted to a commutative algebra.

We might like to do more. This result recovers a Markov chain — if the quantum process is in fact a random walk on a finite quantum group can we recover the group, the transition probabilities (yes), the driving probability measure??

## Conjecture

If we restrict a random walk on a finite quantum group to a commutative subalgebra we can recover a random walk on a finite group.

For random walks on quantum groups we have the following result.

## Theorem 6.1

Let $A$ be a finite quantum group $\{j_n\}_{n\geq 0}$ a random walk on a finite dimensional $A$-comodule algebra $B$, and $B_0$ a unital abelian sub-*-algebra of $B$. The algebra $B_0$ is isomorphic to the algebra of functions on a finite set, say $B_0\cong F(X)$ where $X={1,\dots,d}$.

If the transition operator $T_\phi$ of $\{j_n\}_{n\geq 0}$ leaves $B_0$ invariant, then there exists a classical Markov chain $\{\xi_n\}_{n\geq 0}$ with values in $X$, whose probabilities can be computed as time-ordered moments of $\{j_n\}_{n\geq 0}$, i.e.

$P(\xi_0=i_0,\dots,\xi_\ell=i_\ell)=\Psi\left(j_0\left(\mathbf{1}_{\{i_0\}}\right)\cdots j_\ell\left(\mathbf{1}_{\{i_\ell\}}\right)\right)$

for all $\ell\geq 0$ and $i_0,\dots,i_\ell\in X$.

Proof : We use the indicator functions $\mathbf{1}_{\{1\}},\dots,\mathbf{1}_{\{d\}}$,

$\mathbf{1}_{\{i\}}(j)=\delta_{ij}$$1\leq i,\,j\leq d$,

as a basis for $B_0\subset B$. They are positive, therefore

$\lambda_1=\Psi\left(j_0\left(\mathbf{1}_{\{1\}}\right)\right)$, …, $\lambda_d=\Psi\left(j_0\left(\mathbf{1}_{\{d\}}\right)\right)$

are non-negative. Since furthermore

$\lambda_1+\cdots+\lambda_d=\Psi\left(j_0\left(\mathbf{1}_{\{1\}}\right)\right)+\cdots+\Psi\left(j_0\left(\mathbf{1}_{\{d\}}\right)\right)=\Psi\left(j_0\left(\mathbf{1}\right)\right)=\Psi(\mathbf{1})=1$,

these numbers define a probability measure on $X$.

Define now $\{(p(i,j))\}_{1\leq i,j\leq d}$ by

$\displaystyle T_\phi\left(\mathbf{1}_{\{j\}}\right)=\sum_{i=1}^dp(i,j)\mathbf{1}_{\{i\}}$.

Since $T_\phi=(I_{F(X)}\otimes \phi)\circ\Delta$ is positive ($\checkmark$), we have $p(i,j)\geq 0$ for $1\leq i,\,j\leq d$. Furthermore, $T_\phi(\mathbf{1})=\mathbf{1}$ implies

$\displaystyle \mathbf{1}=T_\phi(\mathbf{1})=T_\phi\left(\sum_{j=1}^d\mathbf{1}_{\{j\}}\right)=\sum_{j=1}^d\sum_{i=1}^dp(i,j)\mathbf{1}_{\{i\}}=\sum_{i=1}^d\mathbf{1}_{\{i\}}\underbrace{\sum_{j=1}^dp(i,j)}_{\overset{!}{=}1}$.

Hence $[p(i,j)]_{1\leq i,\,j\leq d}$ is a stochastic operator.

Therefore there exists a unique Markov chain $\{\xi_n\}_{n\geq 0}$ with initial distribution $\{\lambda_{i}\}_{1\leq i,\,j\leq d}$ and transition matrix $\{p(i,j)\}_{1\leq i,\,j\leq d}$.

We show by induction that the equation in the theorem statement holds.

For $\ell=0$ this is clear by the definition of $\lambda_1,\dots,\lambda_d$. Suppose now that $\ell\geq 1$ and $i_0,\dots,i_\ell\in X$. Then we have

$\Psi\left(j_0(\mathbf{1}_{\{i_0\}})\cdots j_\ell(\mathbf{1}_{\{i_\ell\}})\right)$

$=\Psi\left(j_0(\mathbf{1}_{\{i_0\}})\cdots j_{\ell-1}(\mathbf{1}_{\{i_{\ell-1}\}})\left(j_{\ell-1}({\mathbf{1}_{\{i_\ell\}}}_{(1)})\otimes{\mathbf{1}_{\{i_\ell\}}}_{(2)}\right)\right)$

$=\Psi\left(j_0(\mathbf{1}_{i_0})\cdots j_{\ell-1}(\mathbf{1}_{\{i_{\ell-1}\}}{\mathbf{1}_{\{i_\ell\}}}_{(1)})\right)\phi\left({\mathbf{1}_{\{i_\ell\}}}_{(2)}\right)$

$=\Psi\left(j_0(\mathbf{1}_{\{i_0\}})\cdots j_{\ell-1}\left(\mathbf{1}_{\{i_{\ell-1}\}}T_\phi(\mathbf{1}_{\{i_\ell\}})\right)\right)$

$=\Psi\left(j_0(\mathbf{1}_{\{i_0\}})\cdots j_{\ell-1}(\mathbf{1}_{\{i_{\ell-1}\}})\right)p(i_{\ell-1},i_{\ell})$

$=\lambda_{i_0}p(i_0,i_1)\cdots p(i_{\ell-1},i_\ell)$

$=P(\xi_0=i_0,\dots,\xi_\ell=i_\ell)$ $\bullet$

### Remark

If the condition that $T_\phi$ leaves $A_0$ invariant is dropped, then one can still compute the “probabilities” but in general they are no longer positive or even real, and so it is impossible to construct a classical stochastic process $\{\xi_n\}_{n\geq 0}$ from them.

### Example

The comodule algebra $B=\mathbb{C}^4$ that we considered here is abelian, so we can take $B_0=B$. For any pair of states $\psi$ on $B$ and $\phi$ on $A$ (the Kac-Paljutkin quantun group), we get a random walk on $B$ and a corresponding Markov chain on $X={1,2,3,4}$. We identify $F(X)$ with $B$ by $v_i\equiv\mathbf{1}_{\{i\}}$ for $i\in X$.

The initial distribution is given by $\lambda_i=\psi(v_i)$ and the transition matrix is given by (3.1) on page 10.

### Example

Let us now consider random walks on the Kac-Paljutkin quantum group $A$ itself. For the defining relations, the calculation of the dual of $A$ and a parameterisation of all states see here. Let us consider here transition states of the form

$\phi=\mu_1\eta_1+\mu_2\eta_2+\mu_3\eta_3+\mu_4\eta_4$,

with the $\mu_i$ positive real numbers summing to $1$.

The transition operators $T_\phi=(I_A\otimes\phi)\circ\Delta$ of these states leave the abelian subalgebra $A_0=\text{Lin}\{e_1,e_2,e_3,e_4\}\cong \mathbb{C}^4$ invariant. The transition matrix of the associated classical Markov chain on $X=\{1,2,3,4\}$ that arises by identifying $e_i\equiv\mathbf{1}_{\{i\}}$ for $i\in X$ has the form

$\left(\begin{array}{cccc}\mu_1&\mu_2&\mu_3&\mu_4\\\mu_2&\mu_1&\mu_4&\mu_3\\\mu_3&\mu_4&\mu_1&\mu_2\\\mu_4&\mu_3&\mu_2&\mu_1\end{array}\right)$.

This actually the transition matrix of a random walk on the the group $\mathbb{Z}_2\times\mathbb{Z}_2$.

A pertinent point here is that in this case we actually have a random walk on a finite group rather than just an ordinary Markov chain. If we had an ordinary Markov chain there is no distinction between $\mathbb{Z}_2\times\mathbb{Z}_2$ and $\mathbb{Z}_4$. In fact a quick calculation shows that this is the stochastic operator of the random walk driven by the probability $\nu\in M_p(\mathbb{Z}_2\times\mathbb{Z}_2)$

$\nu((0,0))=\mu_1$$\nu((1,0))=\mu_2$$\nu((0,1))=\mu_3$ and $\nu((1,1))=\mu_4$.

Could it be the the stochastic operator of a random walk on $\mathbb{Z}_4$ — well not always because these have to be circulant matrices. The only time when this stochastic operator is circulant is when $\mu_2=\mu_4$

The subalgebra $\text{Lin}\{a_{11},a_{12},a_{21},a_{22}\}\cong M_2(\mathbb{C})$ is also invariant under these states, $T_\phi$ acts on it by

$T_\phi(M)=\mu_1 M\mu_2 V_2^*M V_2+\mu_3 V_3^*MV_3+\mu_4 V^*_4 MV_4$

for

$M=aa_{11}+ba_{12}+ca_{21}+da_{22}\equiv\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$a,\,b,\,c,\,d\in\mathbb{C}$,

with

$V_2=\left(\begin{array}{cc}0&i\\1&0\end{array}\right)$$V-3=\left(\begin{array}{cc}0&-i\\1&0\end{array}\right)$$V_4=\left(\begin{array}{cc}1&0\&-1\end{array}\right)$.

Let $u=(\cos \vartheta,e^{i\delta}\sin\vartheta)$ be a unit vector in $\mathbb{C}^2$ and denote by $P_u$, the orthogonal projection onto $\text{Lin}\{u\}$. The maximum abelian subalgebra $A_u=\text{Lin}\{P_u,\mathbf{1}-P_u\}\subset M_2(\mathbb{C})\subset A$ is in general not invariant under $T_\phi$.

For example, for $u=(1/\sqrt{2},1/\sqrt{2})$ we get the algebra

$A_u=\text{Lin}\left\{\begin{array}{cc}a&b\\b&a\end{array}:a,\,b\in\mathbb{C}\right\}$.

It can be identified with $F(\{1,2\})$ via

$\left(\begin{array}{cc}a&b\\b&a\end{array}\right)\equiv (a+b)\mathbf{1}_{\{1\}}+(a-b)\mathbf{1}_{\{2\}}$.

Specialising to the transition state $\phi=\eta_2$ and starting from the Haar measure $\psi=\eta$, we see that the time-ordered joint moment is negative and thus can not be obtained from a classical Markov chain.

### Example

For states in $\text{Lin}\{\eta_1,\eta_2,\eta_3,\eta_4,\alpha_{11}+\alpha_{22}\}$, the centre $Z(A)=\text{Lin}\{e_1,e_2,e_3,e_4,a_{11},a_{22}\}$ of $A$ is invariant under $T_\phi$. A state on $A$ parameterised by (B.1) belongs to this set if and only if $x=y=z=0$ ($\checkmark$). With respect to the basis $\{e_1,e_2,e_3,e_4,a_{11}+a_{22}\}$ of $Z(A)$ we get

${T_\phi}_{|Z(A)}=\left(\begin{array}{ccccc}\mu_1&\mu_2&\mu_3&\mu_4&\mu_5\\\mu_2&\mu_1&\mu_4&\mu_3&\mu_5\\\mu_3&\mu_4&\mu_1&\mu_2&\mu_5\\\mu_4&\mu_3&\mu2&\mu_1&\mu_5\\[1ex] \mu_5/4&\mu_5/4&\mu_5/4&\mu_5/4&1-\mu_5\end{array}\right)$.

This is not the stochastic operator of a random walk on a finite group — it is not doubly stochastic. Furthermore this puts paid to the conjecture made above (however I am under the impression that there is an isomorphism between abelian finite quantum groups and finite groups — more on this later).

For Lévy processes or random walks on quantum groups there exists another way to prove the existence of a classical version that does not use the Markov property. We will illustrate this with an example.

### Example

We consider restrictions to the centre $Z(A)$ of $A$. If $a\in Z(A)$, then $a\otimes 1_A\in Z(A\otimes A)$ and therefore

$[a\otimes 1_A,\Delta (b)]=0$ for all $a,\,b\in Z(A)$.

This implies that the range of the restriction $\{{j_n}_{|Z(A)}\}_{n\geq 0}$ of any random walk on $A$ to $Z(A)$ is commutative, i.e.

$[j_\ell(a),j_n(b)]$

$=[(j_0\star k_1\star\cdots \star k_\ell)(a),(j_0\star k_1\cdots \star k_n(b))]$

$=[(j_0\star k_1\star\cdots\star k_\ell)(a),(j_0\star k_1\star\cdots\star k_\ell)(b_{(1)})(k_{\ell+1}\star\cdots\star k_n)(b_{(2)})]$

$=\nabla(j_\ell\otimes(k_{\ell+1}\star\cdots\star k_n))([a\otimes 1_A,\Delta (b)])=0$

for all $0\leq\ell\leq n$ and $a,\,b\in Z(A)$ (I must admit I’m not sure of the calculus of these calculations). Therefore the restriction $\{{j_n}_{|Z(A)}\}_{n\geq 0}$ corresponds to a classical process.

Let us now takes states for which $T_\phi$ does not leave the centre of the Kac-Paljutkin quantum group invariant; for example

$\phi_c=\frac{1+c}{2}\alpha_{11}+\frac{1-c}{2}\alpha_{22}$ for $z\in [-1,1]$.

In this particular case we have the invariant commutative subalgebra $A_0=\text{Lin}\{e_1,e_2,e_3,e_4,a_{11},a_{22}\}$ which contains the centre $Z(A)$. If we identify $A_0$ with $F(1,\dots,6)$ via $e_i\equiv\mathbf{1}_{\{i\}}$$a_{11}\equiv\mathbf{1}_{\{5\}}$ and $a_{22}\equiv \mathbf{1}_{\{6\}}$, then the transition matrix is given at the bottom of page 21.

The classical process corresponding to the centre $Z(A)$ arises from this Markov chain by “gluing” the two states $5$ and $6$ into one. More precisely, if $\{\xi_n\}_{n\geq 0}$ is a Markov chain that has the same time-ordered moments as $\{j_n\}_{n\geq 0}$ restricted to $A_0$, and if $g:\{1,\dots,6\}\rightarrow\{1,\dots,5\}$ is the mapping defined by $g(i)=i$ for $i=1,\dots,5$ and $g(6)=5$, then $\{\zeta_n\}_{n\geq 0}$ with $\zeta_n=g\circ \xi_n$, for $n\geq 0$, has the same joint moments as $\{j_n\}_{n\geq 0}$ restricted to the centre $Z(A)$ of $A$. Note that $\{\zeta_n\}_{n\geq 0}$ is not a Markov process.