# The spectrum of a Markov Operator and Invariant Measures

Suppose I have a discrete-time Markov Chain (in an infinite dimensional state space $\Omega$) with Markov operator $P$, a linear operator on the space of bounded measurable functions on $\Omega$. (Or possibly some Banach space of functions, like Lipschitz functions)

And suppose an invariant measure $\mu$ exists on $\Omega$, i.e., a measure such that for any bounded measurable function $g$ on $\Omega$, we have $$\int_\Omega Pg \; \mathrm d \mu = \int_\Omega g \; \mathrm d \mu.$$

My general question is the following:

What can knowledge about the spectrum of the generator $P$ reveal about the invariant measure $\mu$?

It's well known that information about the spectrum of $P$ can give information about convergence of the chain to $\mu$ (namely via the spectral gap and Markov mixing times), but what about the structure of $\mu$ itself?

For example, if I know that $P$ can be written as a small $\epsilon$ perturbation of another operator, i.e., $P = P(\epsilon) = P_0 + \epsilon P_1$ with invariant measure $\mu(\epsilon)$, and I know that for all $\epsilon$ small enough, I know something about the spectrum of $P(\epsilon)$ (like perhaps that its essential spectrum is bounded away from $1$), can I determine that, for example, the function $\epsilon \mapsto \int g \; \mathrm d \mu(\epsilon)$ is analytic near $0$, for all bounded measurable $g$?

I've been reading Kato's book on Perturbation Theory, but there is no mention of measures. I also realize that you may need quite a bit more knowledge of $P$ to say anything conclusive, perhaps that it is self-adjoint on some Hilbert Space, etc... Just mention whatever assumptions are necessary when answering. I would also appreciate any references that may be helpful.

This is a vague, open-ended question, so perhaps it should be a community wiki. Please comment if you think it should be and I can change it.

-
I expect that knowledge of $\mbox{spec} P$ will reveal very little if any nontrivial information about $\mu$. The reason is that the set of transition kernels sharing a common $\mu$ is quite large. This is spelled out for the case of finite state spaces in my old question mathoverflow.net/questions/14729 – Steve Huntsman Jun 4 '12 at 20:13