Symmetric Feller processes and Dirichlet forms Let $(G, \mathcal D)$ be a densely defined operator on $C_0$ (continuous functions vanishing at infinity on some nice topological space) whose closure $\bar G$ generates a Feller semigroup and let $X$ be a Markov process corresponding to it. 
Assume that $\mathcal D\subset C_K$ (continuous functions with compact support),
that $G(\mathcal D)\subset C_K$ and that
$G$ is symmetric with respect to a Radon measure $m$ (Edit: with full support, but
not necessarily finite), i.e.
$$\int Gf\  g \ dm = \int f \ Gg\ dm \quad\text{for every } f,g\in \mathcal D.$$
I guess that the Dirichlet form $\mathcal E$ of $X$ (defined as in the book of Fukushima/Oshima/Takeda
by using the transition kernel, see (1.4.13) on p.30 in the last edition) 
is given by the closure of
$$\mathcal D\ni f,g \mapsto \int Gf\ g dm.$$
In other terms the Friedrichs extension of $G$ in $L^2(dm)$ should be the generator
of the $L^2$ semigroup induced by $X$. (Edit: by $L^2$ semigroup induced by $X$ I mean 
the semigroup corresponding to the Dirichlet form $\mathcal E$.) 
Is this true? I didn't find a reference nor a simple argument
 for showing this.   
Or is it possible that a selfadjoint extension other than the Friedrichs one generates the $L^2$ semigroup induced by $X$? 
Edit: From the answer of Byron Schmuland it is clear to me that the guess is true 
if the state space is compact. Observe that in this case $G$ is essentially selfadjoint in $L^2$, so the Friedrichs extension is just the closure of $G$ and there are no other selfadjoint extensions. I'm still confused about the case of noncompact state space. I would also appreciate partial answers which work for some concrete example of $G$  (say elliptic partial differential operators, or discrete operators).   
 A: I've decided to post an incomplete preliminary answer. 
I ran into your problem when I was writing [1]. On page
258 you will see my resolution. 
I should point out that in my case, the underlying space
$X$ was compact, and that $m$ was a finite measure with full
support. Thus, $C(X)$ embeds into $L^2(X;m)$ with a continuous, 
 linear injection in the obvious way. This may not hold in the 
locally compact case, and I'm not sure how serious a problem that is.
Translated into your notation,
 and letting $\tilde G$ be the Friedrichs extension we note 
that $\bar G$ and $\tilde G$ agree on $\cal D$
and so the resolvent operators $\bar R_\lambda$ and $\tilde R_\lambda$
  agree on $(\lambda-G)({\cal D})$. We deduce that $\bar R_\lambda= \tilde R_\lambda$
on $C(X)$ and using the Yosida approximation conclude the same 
about the semigroup operators $\bar T_t$ and $\tilde T_t$.
I hope this is of some help. If anything is unclear, let me know.
[1] A result on the infinitely many neutral alleles diffusion model. 
Journal of Applied Probability 28, 253-267 (1991). 
A: I think that the guess is true under the general assumptions I made, by following Byron Schmuland's reasoning. Let me spell it out the way I understood it. 
I denote by $T$ the $L^2(dm)$ semigroup induced by $X$, 
by $\bar T$ the Feller semigroup generated by the closure of $G$ in $C_0$
and by $\tilde T$ the $L^2(dm)$ semigroup generated by the Friedrichs extension of $G$.
The semigroup $T$ is characterized by
$T=\bar T$  on $L^2(dm)\cap C_0$   (I take this as definition of $T$ as in Fukushima et al.)
So it is enough to show that $\tilde T = \bar T$ on $C_K$ (which is both dense in $L^2(dm)$
and $C_0$).
By Yosida approximation it is enough to show that the corresponding resolvents satisfy for $\lambda>0$
$\tilde R_\lambda = \bar R_\lambda$ on $C_K$
By definition of resolvent  $\tilde R_\lambda= \bar R_\lambda$ on $\mathcal F:=(\lambda-G)(\mathcal D)$. The $C_0$-closure of $\mathcal F$ is $C_0$ since $G$ generates, in particular it contains $C_K$. It follows that also the $L^2$ closure of $\mathcal F$ contains $C_K$, so we are done. 
Observe that under the assumptions I made in the question, $G$ is automatically essentially selfadjoint, so there is no other selfadjoint extension other than the Friedrichs one. 
(a criterion for essential selfadjointness is that
$(\lambda -G)(\mathcal D)$ is dense in $L^2$ which we have shown above). So the interesting case I was wondering about actually doesn't happen. 
Let me know if there is a flaw in what I wrote.
