Friedrich's extension of the generator of a continuous time markov chaoin

Question

Consider the infinitesimal generator $G$ of a Markov chain with state space $\mathbb{Z}$ such that it is symmetric with respect to a measure $\mu$ on $\mathbb{Z}$. Then, the operator $(G,C_c(\mathbb{Z}))$, ($(C_c(\mathbb{Z})$ is the set of all functions with finite support) is symmetric on $L^2(\mathbb{Z},\mu)$. Is the operator $(G,C_c(\mathbb{Z}))$ essentially self-adjoint on $L^2(\mathbb{Z},\mu)$?

Answer 1

Not necessarily, I think. Here is an example that comes to my mind, but I did not check all details carefully.

Map $\mathbb{Z}$ to $A = \{2^{-n} : n = \in \mathbb{Z}\}$, and consider a symmetric, martingale continuous-time Markov chain $X_t$ on $A$, with counting measure as the reference measure. Thus, the transition rates are (up to a constant factor) $$q_{2^{-n},2^{-n+1}} = q_{2^{-n+1},2^{-n}} = 2^n,$$ and zero otherwise. This process can be seen as the trace left on $A$ by a Wiener process $W_t$. More precisely, $X_t$ changes its state at moments of time given by $W_t$ hitting a point in $A$, namely: $$T_{n+1} = \inf \{t > T_n : W_t \in A \setminus \{W_{T_n}\}\} ,$$ and we set $X_t = W_{T_n}$ for $t \in [T_n, T_{n+1})$.

Now depending on what we do with the Wiener process $W_t$ at zero (say: kill or reflect), we will get different traces $X_t$. In other words: imposing different boundary conditions "at zero" for $X_t$, we get different Markov chains.

Of course, this can be mapped back to $\mathbb{Z}$ rather than $A$.