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This question was previously posted on MSE.

Let $f:\mathbb R^3 \to \mathbb R^3$ be a smooth Lipschitz function (bounded if needed), and $W_t$ a $3$-dimentional Brownian motion. Consider the SDE on $\mathbb R^3 \times \mathbb S^2$ (where $\mathbb S^2$ is the sphere).

\begin{cases} \mathrm{d}X_t = f(X_t) \mathrm{d}t + \mathrm{d} W_t\\ \mathrm{d}s_t = (\mathrm{d}f(X_t) s_t - \langle s_t, \mathrm{d}f(X_t) s_t \rangle s_t) \mathrm{d}t, \end{cases} where $\mathrm{d}f(x)$ is the derivative of $f$ at $x$.

I would like to know is there any condition on $f,$ which makes the above SDE, transitive on $B(0,1) \times \mathbb{S}^2$ (where $B(0,1)$ is the ball of radius $1$ centered in $0$), in the following sense: For every $(x,\theta)\in B(0,1)\times \mathbb S^2$ and open set $U\subset B(0,1)\times \mathbb S^2,$ there exists $t= t(x,\theta,U)$ such that $$\mathbb E\left[\mathbb 1_U (X_t,s_t) \mid (X_0,s_0)= (x,\theta)\right] >0$$

So far, I was not able to figure what condition would imply this property. Can anyone help me?

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(Not a solution, but an idea.)

A general strategy is to make $f$ nearly zero everywhere, with high-frequency fluctuations so that $df(x)$ changes "randomly" with minor changes in $x$. Then $X_t$ will essentially be ordinary Brownian motion because the $f(X_t)dt$ term is nearly zero, and $s_t$ will essentially be Brownian motion on the sphere because $df(X_t)s_t - \langle s_t, df(X_t) s_t \rangle s_t$ is the projection of the "random" vector $df(X_t)s_t$ onto the orthogonal complement of $s_t$. Can somebody suggest a way to fill in the details to remove the scare quotes?

I think some kind of spatial noisiness of $df$ is important to get the transitivity you seek. If $df(X_t)$ stays nearly constant for long stretches of $t$, then $s_t$ gets sucked into the largest-eigenvalue eigenspace of $df$. The only way noise makes its way to $s_t$ is through variability in $df(X_t)$.

As an aside, I find this SDE charming. The point at $X_t$ is being blown around by the vector field $f\colon \mathbb R^3 \to \mathbb R^3$ plus some Brownian noise. Gazing in direction $s_t$, it sees a cluster of its immediate neighbors in just the same predicament, and does its best to anticipate where to look to keep its eye on the centroid of that cluster of neighbors. There's nothing it can do to help, but it's still considerate to check in.

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