The relation between the semi-groups and transition matrix of a Markov chain. For a finite state space Markov chain, let $P$ be the transition matrix of an ergodic Markov chain with finite state space $\Omega$, and $\pi$ is stationary distribution. For all functions $f: \Omega \to R$, define
$$
P_tf(x):=\sum_{y\in \Omega} P_t(x,y)f(y).
$$
I am confused about this one and the notation from semi-groups of a Markov chain. The definition of semi-groups $P_tf(x):=E[f(X_t)|X_0=x]$.
Question: Does it mean the semi-groups $P_t$ is same as the transition matrix $P_t$?
 A: So we generally think of the transition semigroup $\{P_t\}_t$ as a set of operators on the space of functions $\Omega \rightarrow \mathbf{R}$.
I'll call this space $\mathcal{F}(\Omega, \mathbf{R})$.
Enumerating $\Omega = \{x_1, \dotsc, x_n\}$, we have that $\mathcal{F}(\Omega, \mathbf{R})$ is a vector space with basis $\mathcal{B} = \{e_i : e_i(x) = \delta_{xx_i}\}_{i=1}^n$, where $\delta$ is the Kronecker delta.
With respect to this basis, the vector representation of a function $f$ is defined by $[f]_i = f(x_i)$.
Since $P_t$ is a linear transformation on $\mathcal{F}(\Omega, \mathbf{R})$, we can represent it as a matrix with respect to this basis.
In particular,
\begin{align*}
[P_t]_{ij}
= [P_t e_j]_i 
= (P_t e_j)(x_i) 
= \mathbf{E}(e_j(X_t) | X_0 = x_i) 
= \Pr(X_t = x_j | X_0 = x_i).
\end{align*}
This is the transition probability from $x_i$ to $x_j$ in $t$ steps, which is $(\mathbf{P}^t)_{ij}$, where $\mathbf{P}$ is the transition matrix of this Markov chain.
Thus, for a finite-state Markov chain, the transition matrix is the matrix representation of the linear transformation $P_1$ with respect to the basis $\mathcal{B}$,
