Lie algebras and the Hörmander condition for SDEs

Question

I have seen different versions and formulations of Hörmander's theorem for SDEs, and I'm a bit unclear on the whole subject and am seeking some clarifications.

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Notations, and definition of Lie bracket

Let $M$ be a $C^\infty$-smooth manifold, and let $\mathcal{V}$ be the set of $C^\infty$ vector fields on $M$. For any $\mathcal{S} \subset \mathcal{V}$ and $x \in M$, let $$ \mathcal{S}(x) = \{v(x) : v \in \mathcal{S}\}\text{.} $$ For any chart $\varphi \colon U \to \mathbb{R}^D$ on $M$ and any $v \in \mathcal{V}$, define $v_\varphi$ to be the expression of $v$ in the coordinates given by $\varphi$; that is, \begin{align*} &v_\varphi \colon \varphi(U)\to \mathbb{R}^D \\ &v_\varphi(y) = \left.\mathrm{d}\varphi\right|_{\varphi^{-1}(y)}(v(\varphi^{-1}(y)))\text{.} \end{align*} With this, define the Lie bracket $[\boldsymbol{\cdot},\boldsymbol{\cdot}] \colon \mathcal{V} \times \mathcal{V} \to \mathcal{V}$ by $$ [u,v]_\varphi(y) = v_\varphi'(y)u_\varphi(y) - u_\varphi'(y)v_\varphi(y) $$ where the prime $'$ denotes the Jacobian matrix. A Lie subalgebra of $\mathcal{V}$ is a vector subspace of $\mathcal{V}$ that is closed under the Lie bracket.

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SDEs and Hörmander conditions

Consider on $M$ the SDE $$ dX_t = b(X_t) dt + \sum_{i=1}^k \sigma_i(X_t) \circ dW_t^i $$ where $b,\sigma_1,\ldots,\sigma_k \in \mathcal{V}$, and $(W_t^1,\ldots,W_t^k)$ is a $k$-dimensional standard Brownian motion. (There is not necessarily any relationship between $k$ and the dimension $D$ of $M$.)

Let

• $\mathcal{V}_1$ be the smallest vector subspace of $\mathcal{V}$ with the properties that $\sigma_1,\ldots,\sigma_k \in \mathcal{V}_1$ and for all $v \in \mathcal{V}_1$, $[v,b],[v,\sigma_1],\ldots,[v,\sigma_k] \in \mathcal{V}_1$;
• $\mathcal{V}_2$ be the smallest Lie subalgebra of $\mathcal{V}$ with the properties that $\sigma_1,\ldots,\sigma_k \in \mathcal{V}_2$ and for all $v \in \mathcal{V_2}$, $[v,b] \in \mathcal{V}_2$.

[I have sometimes seen the condition for "Hörmander's theorem" formulated as $\mathcal{V}_1(x)=T_xM$ for all $x \in M$ (the "parabolic Hörmander condition"). But I have also seen "Hörmander's condition on a collection of vector fields": Ignoring $b$ by setting $b=0$, "Hörmander's condition for vector fields $\sigma_1,\ldots,\sigma_k$" is sometimes defined as $\mathcal{V}_1(x)=T_xM$ for all $x$, and sometimes defined as $\mathcal{V}_2(x)=T_xM$ for all $x$ (and sometimes both, as though it is obvious that the two conditions are equivalent).]

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Questions.

When we refer to "the Lebesgue measure on $M$", we can refer to any measure $\lambda$ on $M$ such that for any chart $\varphi \colon U \to \mathbb{R}^D$, the pushforward $\varphi_\ast(\lambda|_{\mathcal{B}(U)})$ has strictly positive $C^\infty$ density with respect to the Lebesgue measure on $\varphi(U)$.

For each $x \in M$ and $T \geq 0$, define $E_{x,T}$ to be the event that a strong solution of the SDE starting at $X_0=x$ exists over $t \in [0,T]$. If $\mathbb{P}(E_{x,T})>0$, then over the restricted probability space $(E_{x,T},\mathbb{P}(\boldsymbol{\cdot}|E_{x,T}))$ define the random variable $X_T^x$ to be the location at time $T$ of the strong solution starting at $X_0=x$.

• It is clear that $\mathcal{V}_1 \subset \mathcal{V}_2$, but is it necessarily the case that $\mathcal{V}_1=\mathcal{V}_2$?
• If not, do we at least have that $(\mathcal{V}_2(x)=T_xM \ \forall x) \Rightarrow (\mathcal{V}_1(x)=T_xM \ \forall x)$?
• If $\mathcal{V}_2(x)=T_xM$ for all $x \in M$, does it follow that for each $x \in M$ and $t > 0$, if $\mathbb{P}(E_{x,t})>0$ then the law of $X_t^x$ has $C^\infty$ density w.r.t. Lebesgue?
• Going further than this: If $\mathcal{V}_2(x)=T_xM$ for all $x \in M$, does it follow that for each $x \in M$ and $t > 0$, we have $\mathbb{P}(E_{x,t})>0$ and provided $M$ is connected, the density of the law of $X_t^x$ is strictly positive everywhere in $M$? (Or at least, strictly positive Lebesgue-almost everywhere in $M$?)
• If the answer to any part of the above two questions is no, then are there any stronger conditions that make the answer yes (e.g. changing $\mathcal{V}_2$ to $\mathcal{V}_1$, and/or adding a boundedness or growth control condition on the vector fields $b,\sigma_1,\ldots,\sigma_k$ in the case that $M$ is non-compact)?
• Or alternatively, can the conclusions be weakened so that the answer is yes (e.g. only requiring that the law of $X_t^x$ is equivalent to the Lebesgue measure, without requiring any smoothness on the density.)
• Do the answers to any of the above change if we restrict to the case that $b=0$?

Remark. To answer the first question, it is probably sufficient just to determine the answer the following. Allowing ourselves (for simplicity) to take $b=0$ and $k=4$:

Is it necessarily the case that $ [[\sigma_1,\sigma_2],[\sigma_3,\sigma_4]] \in \mathcal{V}_1$?

[Sorry for asking several questions in one post. I know that it is not generally preferred practice on the StackExchange network to do so, but in this case it seems obviously natural to put these questions together.]

Answer 1

After e-mailing an expert about the questions listed for the bounty, I now have some answers! Firstly, yes it is true that $[[\sigma_1,\sigma_2],[\sigma_3,\sigma_4]] \in \mathcal{V}_1$, by the Jacobi identity. (I will explain shortly.) Secondly, assuming global existence of strong solutions, having $\mathcal{V}_1(x)=T_xM$ for all $x$ does imply smooth transition densities, but does not automatically imply everywhere positive density: for example, with $M=\mathbb{R}$, taking $k=1$, $b=\cos$ and $\sigma_1=\sin$, every $x$ where $\sigma_1(x)=0$ has $[\sigma_1,b](x)=-1$, and yet no initial condition starting in the interval $(0,\pi)$ can escape that interval.

The argument that $[[\sigma_1,\sigma_2],[\sigma_3,\sigma_4]] \in \mathcal{V}_1$ fairly easily generalises to showing that, in general, $\mathcal{V}_1=\mathcal{V}_2$, which I will now present. The Jacobi identity can be expressed as saying that for any three smooth vector fields $v_1,v_2,v_3$, $$ [v_1,[v_2,v_3]] = [v_1,v_2,v_3] - [v_1,v_3,v_2] $$ where we use the notation $[v_1,\ldots,v_k]=[\ldots[[v_1,v_2],v_3],\ldots v_k]$. (So just setting $v_1=[\sigma_1,\sigma_2]$ solves the above problem of four vector fields $\sigma_1,\sigma_2,\sigma_3,\sigma_4$.)

Hence, it is easy to show by induction the following:

Lemma. For any $n \geq 3$, there exists a set $P_n$ of permutations of $\{2,\ldots,n\}$ and a function $s_n \colon P_n \to \mathbb{Z}$ such that for $n$ smooth vector fields $v_1,\ldots,v_n$, $$ [v_1,[v_2,\ldots,v_n]] = \sum_{\pi \in P_n} s_n(\pi)[v_1,v_{\pi(2)},\ldots,v_{\pi(n)}]. $$

[In fact, it is not hard to see that one can take $|P_n|=2^{n-2}$ and $\mathrm{range}(s_n)=\{-1,1\}$. A more explicit form can be found in Proposition 3 of https://arxiv.org/pdf/1604.05281.pdf]

Now to show that $\mathcal{V}_1=\mathcal{V}_2$, it is sufficient just to show that $\mathcal{V}_1$ is a Lie algebra. Define $$ \tilde{\sigma}_i = \begin{cases} b & i=0 \\ \sigma_i & i \in \{1,\ldots,k\}. \end{cases} $$ Since the Lie bracket is bilinear, it is easy to check that a linear basis of $\mathcal{V}_1$ is the set of all vector fields of the form $$ [\sigma_{i_1},\tilde{\sigma}_{i_2},\ldots,\tilde{\sigma}_{i_n}], \quad n \geq 1, \, i_1 \in \{1,\ldots,n\}, \, i_2,\ldots,i_n \in \{0,\ldots,n\}\text{.} $$ So, again since the Lie bracket is bilinear, to show that $\mathcal{V}_1$ is a Lie algebra it is sufficient just to show that the Lie bracket of any two such basis elements belongs to $\mathcal{V}_1$. But this is immediate from the Lemma (by setting $v_1$ to be the first of the two basis elements).

As I have said, assuming global existence of strong solutions, if $\mathcal{V}_1$ (which we have now established is the same as $\mathcal{V}_2$) covers the whole tangent space at every point, then the law of $X_t^x$ has $C^\infty$ Lebesgue-density for $t>0$. It does not, in general, follow that the law of $X_t^x$ has everywhere positive density. However, as a result of posting a question on MathOverflow, I have learned that if $b=0$ then it does follow; more generally, whether or not $b=0$, we have that if the Lie algebra generated by $\sigma_1,\ldots,\sigma_k$ covers the whole tangent space at every point (or more generally, if every point is accessible from every other point by $H^1$ sample paths of the driver), then the density is strictly positive everywhere; this can be found in Sec. 3.3.6.1 of https://www.i2m.univ-amu.fr/perso/etienne.pardoux/_media/en:michelpardoux_calculmalliavin.pdf For further discussion, see the MathOverflow question https://mathoverflow.net/questions/424273/