In Euclidean space, $\mathbb R^d$, the Langevin diffusion $${\rm d}X_t=b(X_t){\rm d}t+\sigma(X_t){\rm d}W_t\tag1,$$ where $\sigma:\mathbb R^d\to\mathbb R^{d\times k}$, $$b:=\frac{\Sigma+U}2\nabla\ln p+\frac12\nabla\cdot\Sigma,\tag2$$ $\Sigma:=\sigma\sigma^\ast$ and $(W_t)_{t\ge0}$, $U\in\mathbb R^{d\times d}$ is any anti-symmetric matrix is a $\mathbb R^k$-valued standard Wiener process, has an unique invariant measure with density $p$ with respect to the Lebesgue measure on $\mathbb R^d$.

Now, in my application, I'm working on $[0,1)^d$ instead. So I have a given density $p:[0,1)^d\to[0,\infty)$. In my application, it works nicely to consider $[0,1)^d$ with toroidal boundary and the standard MCMC sampling approach is Metropolis-Hastings with proposal kernel given by the wrapped normal distribution.

What I would like to do now is defining a (continuous-time) Markov process, with values in $[0,1)^d$, with unique invariant measure with density $p$ with respect to the restriction of the Lebesgue measure on $\mathbb R^d$ to $[0,1)^d$. Is this possible?

One might think that we can simply use $(1)$ with toroidal wrapping, but as this paper shows, the resulting process is not even Markov anymore.

EDIT: I think the question is trickier than one might think at first glance. It seems like the Markov property of $(1)$ will not be maintained under the transformation by $$\iota:\mathbb R^d\to[0,1)^d\;,\;\;\;x\mapsto x-\lfloor x\rfloor\tag3.$$ A counterxample is given here.

However, we got the following general result: Assume $(E,\mathcal E)$ is a measurable space and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$. Now, given another measurable space $(\tilde E,\tilde{\mathcal E})$ and an $(\mathcal E,\tilde{\mathcal E})$-measurable $\varphi:E\to\tilde E$ satisfying

1. $\varphi(E)=\tilde E$;
2. If $t\ge0$ and $\tilde f:\tilde E\to\mathbb R$ is bounded and $\tilde E$-measurable, then $$\kappa_t(\tilde f\circ\varphi)=\tilde g\circ\varphi\tag4$$ for some bounded $\tilde{\mathcal E}$-measurable $\tilde g:\tilde E\to\mathbb R$,

then $\tilde g$ is uniquely determined and hence $$\tilde\kappa_tf:=\tilde g\tag5$$ is well-defined. If now $(X_t)_{t\ge0}$ is a Markov process, with respect to some filtration $(\mathcal F_t)_{t\ge0}$, with transiton semigroup $(\kappa_t)_{t\ge0}$, then $$\tilde X_t:=\varphi\circ X_t\;\;\;\text{for }t\ge0$$ is a Markov process with respect to the same filtration $(\mathcal F_t)_{t\ge0}$ with transition semigroup $(\tilde\kappa_t)_{t\ge0}$.

We clearly would want to apply this for $E=\mathbb R^d$, $\tilde E=[0,1)^d$, $\varphi=\iota$ and $(X_t)_{t\ge0}$ given by $(1)$. However, with that I end up with $$\operatorname E\left[f\left(\tilde X_{s+t}\right)\mid\tilde X_s\right]=\operatorname E\left[\left(\kappa_t(f\circ\iota)\right)(X_s)\mid\tilde X_s\right]\tag6.$$ I thought we might need to assume $$\kappa_t(x,y+B)=\kappa_t(x-y,B)\;\;\;\text{for all }x,y\in\mathbb R^d\text{ and }B\in\mathcal B(\mathbb R^d)\tag7$$ in which case $$(\tilde\kappa_tf)=\sum_{k\in\mathbb Z^d}(\kappa_tf)(x-k)\tag8;$$ an assumption which is clearly satisfied when $(X_t)_{t\ge0}$ is a Brownian motion. However, even with that I wasn't able to proceed. In general, we only have $$(\kappa_t(x,\;\cdot\;)\circ\iota^{-1})f=(\kappa_t(f\circ\iota))(x)\tag9,$$ which is not enough for $(4)$.

EDIT 2:

A proof (or reference to it) of why $\tilde X:=\iota\circ X$ is still a Markov process, in case $X$ is a Brownian motion (or, more generally, a Lévy process) would be sufficient for me to start with.