Let $M$ be a (compact) Riemannian manifold. Let $v$ be a smooth vector field on $M$ with flow $\Theta\_t$. Let $L$ be an elliptic second order differential operator on $M$ that generates the Ito process $X\_t$. 


Define the process $Y\_t := \Theta\_t^\* X\_t$. I read that the generator of $Y\_t$ is the time-dependent differential operator
$$L\_s = \Theta\_s^\* L - v$$
I do not really know how to prove this though, and there is no explanation in the article I read.
Is the underlying probability measure of $Y\_t$ absolutely continuous w.r.t. $X\_t$? If yes, what is its density?


I know the Girsanov formula, which states e.g. that the process generated by $L-v$ has the density
$$\exp\left( \int\_0^t v(B\_s) \mathrm{d} B\_s - \int\_0^t |v(B\_s)|^2 \mathrm{d} s\right)$$
However, in this situation, also the second-order part is altered and I don't know how to deal with this.



Density for Translated Process
The distribution of the process $Y$ is locally equivalent to the distribution of $X$ if and only if the flow $\Theta$ preserves the principal symbol of $L$. 


If $\Theta$ does not preserve the principal symbol of $L$, then the two processes $X$ and $Y$ do not have the same quadratic variation and thus their distributions can not be locally equivalent.


If $\Theta$ preserves the principal symbol of $L$, then the distributions are locally equivalent and we can use Girsanov theorem to compute the density.


