Quantisation

I had been of the understanding that a quantisation looks as follows. There is some process or property P(X) of a space X which we want to examine. Depending on the type of space X, from a suitable algebra of complex functions on X, F(X), we can recover and examine many of the properties of P(X) by instead looking at F(X): we essentially have the identification X\leftrightarrow F(X). When the process/ property P(X) is about a space then it is said to be classical or commutative because for any x\in X and f,\,g\in F(X) we have that fg=gf because fg(x)=f(x)g(x)=gf(x) as the f(x),\,g(x) lie in the commutative algebra \mathbb{C}.

Now from X we know about F(X) and vice versa. Now a suitably chosen F(X) is just a commutative C*-algebra so what about a non-commutative C*-algebra F — can we examine it’s “underlying space” in the same way?

So essentially, this means that I thought you quantised objects, such as Markov chains, by replacing a commutative C*-algebra F(X) with a not-necessarily commutative one. (This roughly follows my interpretation as per this)

Quantising a Markov Chain

So where is the C*-algebra in a Markov Chain? Well let \xi be a Markov chain on a set X=\{x_1,\dots,x_N\} with initial distribution \nu and transition probabilities \textbf{P}[\xi_{k+1}=x_j|\xi_{k}=x_i]=p(x_i,x_j)=p_{ij}.

The probability distribution of this walk, after n steps, is given by \nu P^n. However the probability measures, M_p(X), on X, which lie in the dual of F(X) (M_p(X)\subset \mathbb{R}^{N} is equipped with the 1-norm while F(X) is equipped with the supremum norm). In fact we can go further, the probability measures comprise the states of A as for any \theta\in M_p(X)\subset F(X)^*,

\|\theta\|_1=1 and \theta\geq0.

Actually in this case (X is a finite set) the positivity of the functional \theta has three equivalent definitions (the first is the same as saying \theta(x)\geq 0 for all x\in X):

  1. The dual basis to (\mathbf{1}_{\{x_1\}},\cdots,\mathbf{1}_{\{x_N\}}), is the basis of delta measures (\delta^{x_1},\cdots,\delta^{x_N}). In this basis, the coefficients of \theta are all positive.
  2. With respect to the involution (a_1,\cdots,a_N)^*=(a_1^*,\cdots,a_N^*), \theta is a positive element in the *-algebra F(X)^* — where multiplication is defined pointwise.
  3. \theta(f)\subset \mathbb{R}^+ for all positive functions f\in F(X)^+.

Usually when we talk about functionals on C*-algebras being positive we are talking about definition 3. i.e. a linear map \varphi:A\rightarrow B between C*-algebras is said to be positive if \varphi(A^+)\subset B^+. I’m not sure if \varphi is then necessarily positive in the C*-algebra L(A,B), the linear maps from A to B (note that L(A,B) may not necessarily be a C*-algebra — I think)?

The states of a C*-algebra, A, are given by:

S(A)=\{\varphi\in A^*:\|\varphi\|=1,\,\varphi\geq 0\}.

It is an easy exercise to show that stochastic operators are characterised as being M_p(X)-stable. In this picture — which looks at the deterministic evolution of \{\nu P^k:k=0,1,\dots,N\} — rather than the random variable picture of the \xi_k:(Y,\mu)\rightarrow X (see p.11 here for this construction), we thus have an initial distribution \theta\in S(A), a stochastic operator:

P:F(X)^*\rightarrow F(X)^*, \theta\mapsto \theta P

which is S(F(X))-stable, and we look at the set of distributions:

\{P^k(\theta):k=1,\dots,n\},

to tell us everything we want to know about the Markov chain. For example, if P^k(\theta) is convergent then we know the walk converges and a fixed point of the stochastic operator is a stationary distribution.

The only thing left to do is to put some conditions on a stochastic operator being ‘S(A)‘-stable — how about the stochastic operator being positive and isometric?

The C*-algebra quantisation is then as follows. Let A be a C*-algebra with dual A^*. Choose an element \psi\in S(A) and a positive linear isometry T:A^*\rightarrow A^* — which is automatically S(A)-stable. The distribution of the quantum Markov chain generated by \psi and T after k steps would then be given by T^k(\psi).

In my interpretation, this is what a quantisation of a Markov chain looks like.

The Coalgebra Picture

This whole piece is written because I don’t understand coalgebras and duality in category theory — and I was trying to do it a different way. At times I can see that the coalgebra is encoding a lot (see some remarks here and here), but really I haven’t a clue what’s going on!

Wikipedia says turn all the arrows around — it also seems to say turn Cartesian products into tensor products (??). The best I can do out of this is I can see that, roughly (with k+1 elements in the cartesian product), the classical Markov chain random variables look like:

\xi_k:X\times X\times X\times\cdots\times X\rightarrow X,

while, roughly, their quantum counterparts (with k+1 elements in the tensor product)

j_k:A\rightarrow A\otimes A\otimes\cdots\otimes A.

 (the \{j_k\} are constructed in Franz & Gohm).

Problems

My construction is leaning towards the fact that the evolution of the T^k(\psi) can tell us if the Markov chain is stationary or if the Markov chain has limiting behaviour (well, in a quantum way). However, I’m looking at the “co” approach of Franz & Gohm, and I think they get to this picture another way. In other words they do their quantisation ‘down there’ first and then ‘lift’ — I have ‘lifted’ first and then quantised ‘up there’. Very precise I know! This (terrible attempt at a) diagram should explain what I mean, the right arrows are quantisations; mine on top — their’s on the bottom: 

\begin{array}{ccc}(M_p(X))P^k &\overset{q}{\rightarrow} & T^k (S(A))\\ \uparrow & & \uparrow\\ \{\xi_k\} & \overset{q}{\rightarrow} &\{j_k\}\end{array}

Suppose that G is a group — then it is kind of clear to me that we need the “co” picture to encode the group laws ‘up’ in F(G) — so if we’re going to encode group laws in a not-necessarily commutative algebra A (quantum group), we will need the “co” picture. In otherwords my approach can’t deal with quantum random walks on groups. 

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