## Lecture 5. Hormander’s type operators

For geometric purposes, it is often very useful to use the language of vector fields to study diffusion operators.
Let $\mathcal{O} \subset \mathbb{R}^n$ be a non-empty open set. A
smooth vector field  $V$ on $\mathcal{O}$ is a smooth map

$\begin{array}{llll} V: & \mathcal{O} & \rightarrow & \mathbb{R}^{n} \\ & x & \rightarrow & (v_{1}(x),...,v_{n}(x)). \end{array}$

We will often regard a vector field $V$ as a differential operator acting on the space of smooth functions $f: \mathcal{O} \rightarrow \mathbb{R}$ as follows:

$(Vf) (x)=\sum_{i=1}^n v_i (x) \frac{\partial f}{\partial x_i}.$
By the chain rule, we note that $V$ is a derivation, that is an operator on
$C^{\infty} (\mathcal{O} )$, linear over $\mathbb{R}$, satisfying for $f,g \in C^{\infty} (\mathcal{O} )$,

$V(fg)=(Vf)g +f (Vg).$
Conversely, it easily seen that any derivation on $C^{\infty} (\mathcal{O} )$ defines a vector field on $\mathcal{O}$ (Pick $x_0 \in \mathcal{O}$, observe that if $g$ is a smooth function such that $g(x_0)=0$ then $V((x-x_0)g)(x_0)=0$ and then, to compute $Vf(x_0)$ use a Taylor expansion around $x_0$).

With these notations, it is readily checked that if $V_0,V_1,\cdots, V_d$ are smooth vector fields on $\mathbb{R}^n$, then the second order differential operator

$L=V_0+\sum_{i=1}^d V_i^2$
is a diffusion operator. Here $V_i^2$ has to be understood as the operator $V_i$ composed with itself. Diffusion operators that may be written under the previous form are called Hormander’s type diffusion operators. It is easily seen that a Hormander’s type diffusion operator is elliptic in $\mathbb{R}^n$ if and only if for every $x \in \mathbb{R}^n$, the vectors $V_1(x),\cdots,V_d(x)$ form a basis of $\mathbb{R}^n$.
Proposition:  Let

$L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},$

be a diffusion operator on $\mathbb{R}^n$ such that the $\sigma_{ij}$‘s and the $b_i$‘s are smooth functions. Let us assume that for every $x \in \mathbb{R}^n$, the rank of the matrix $(\sigma_{ij}(x))_{1 \le i,j \le n}$ is constant equal to $d$. Then, there exist smooth vector fields $V_0,V_1,\cdots, V_d$ on $\mathbb{R}^n$ such that $V_1,\cdots, V_d$ are linearly independent and

$L=V_0+\sum_{i=1}^d V_i^2.$

Proof:  We first assume $d=n$.  Since the matrix $(\sigma_{ij}(x))_{1 \le i,j \le n}$ is symmetric and positive, it admits a unique symmetric and positive square root $v(x)=(v_{ij}(x))_{1 \le i,j \le n}$. Let us assume for a moment that the $v_{ij}$‘s are smooth functions, in that case by denoting

$V_i=\sum_{j=1}^n v_{ij} \frac{\partial}{\partial x_j},$
the vector fields $V_1,\cdots,V_n$ are linearly independent and it is readily seen that the differential operator

$L-\sum_{i=1}^n V_i^2$
is actually a first order differential operator and thus a vector field. We therefore are let to prove that the $v_{ij}$‘s are smooth functions.  Let $\mathcal{O}$ be a bounded non empty set of $\mathbb{R}^n$ and $\Gamma$ be any contour in the half plane $\mathbf{Re} (z) >0$ that contains all the eigenvalues of $\sigma(x)$, $x \in \mathcal{O}$. We claim that

$v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.$
Indeed, if $\Gamma'$ is another contour in the half plane $\mathbf{Re} (z) >0$ whose interior contains $\Gamma$, as a straightforward application of the Fubini’s theorem and Cauchy’s formula we have

$\begin{array}{ll} & \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \times \frac{1}{2 \pi} \int_{\Gamma'} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \\ =&\frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} (\sigma(x)-z\mathbf{I}_n)^{-1}(\sigma(x)-z'\mathbf{I}_n)^{-1} dz dz' \\ = & \frac{1}{4 \pi^2} \int_{\Gamma'}\int_{\Gamma} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}-(\sigma(x)-z'\mathbf{I}_n)^{-1}}{z-z'} dz dz' \\ = & - \frac{1}{4 \pi^2} \int_{\Gamma}\int_{\Gamma'} \sqrt{z z'} \frac{ (\sigma(x)-z\mathbf{I}_n)^{-1}}{z'-z} dz' dz \\ =& - \frac{1}{4 \pi^2} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \left( \int_{\Gamma'}\frac{ \sqrt{z z'} }{z'-z} dz' \right) dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} z dz \\ =& \frac{1}{2 i \pi} \int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} \sigma(x) dz. \end{array}$
In the last expression above, we may modify $\Gamma$ into a circle $\Gamma_R =\{ z, | z|=R \}$. Then by choosing $R$ big enough ($R > \sup_{x \in \mathcal{O}} \sqrt{ \| \sigma (x) \|}$), and expanding $\int_{\Gamma} (\sigma(x)-z\mathbf{I}_n)^{-1} dz$ in powers of $z$, we see that

$\lim_{R \to +\infty} \frac{1}{2 i \pi} \int_{\Gamma_R} (\sigma(x)-z\mathbf{I}_n)^{-1} dz= \mathbf{Id} .$
As a conclusion,

$\left( \frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz \right)^2 = \sigma(x),$
so that, as we claimed it,

$v(x)=\frac{1}{2 \pi} \int_{\Gamma} \sqrt{z} (\sigma(x)-z\mathbf{I}_n)^{-1} dz, \quad x \in \mathcal{O}.$
This expression of the square root of $\sigma$ clearly shows that the $v_{ij}$‘s are smooth functions. By putting things together, we therefore proved the proposition in the case $d=n$.

Let us now turn to the case $d < n$. By smoothly choosing an orthonormal basis of $\mathbb{R}^n$ which is adapted to the orthogonal decomposition

$\mathbb{R}^n=\mathbf{Ker} (\sigma(x) ) \oplus \mathbf{Ker} (\sigma(x) )^\perp,$
we get a decomposition

$\sigma(x)=M(x)\left( \begin{array}{ll} \mathcal{V}(x) & 0 \\ 0 & 0 \end{array} \right)M(x)^{-1},$
where $M(x)$ is an orthogonal matrix with smooth coefficients and where $\mathcal{V}(x)$ is a $d \times d$ symmetric and positive matrix with smooth coefficients. We may now apply the first part of the proof to the matrix $\mathcal{V}(x)$ and easily conclude $\square$

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