For geometric purposes, it is often very useful to use the language of vector fields to study diffusion operators.

Let be a non-empty open set. A

smooth vector field on is a smooth map

We will often regard a vector field as a differential operator acting on the space of smooth functions as follows:

By the chain rule, we note that is a derivation, that is an operator on

, linear over , satisfying for ,

Conversely, it easily seen that any derivation on defines a vector field on (Pick , observe that if is a smooth function such that then and then, to compute use a Taylor expansion around ).

With these notations, it is readily checked that if are smooth vector fields on , then the second order differential operator

is a diffusion operator. Here 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 if and only if for every , the vectors form a basis of .

**Proposition: **Let

be a diffusion operator on such that the ‘s and the ‘s are smooth functions. Let us assume that for every , the rank of the matrix is constant equal to . Then, there exist smooth vector fields on such that are linearly independent and

**Proof: **We first assume . Since the matrix is symmetric and positive, it admits a unique symmetric and positive square root . Let us assume for a moment that the ‘s are smooth functions, in that case by denoting

the vector fields are linearly independent and it is readily seen that the differential operator

is actually a first order differential operator and thus a vector field. We therefore are let to prove that the ‘s are smooth functions. Let be a bounded non empty set of and be any contour in the half plane that contains all the eigenvalues of , $x \in \mathcal{O}$. We claim that

Indeed, if is another contour in the half plane whose interior contains , as a straightforward application of the Fubini’s theorem and Cauchy’s formula we have

In the last expression above, we may modify into a circle . Then by choosing big enough (), and expanding in powers of , we see that

As a conclusion,

so that, as we claimed it,

This expression of the square root of clearly shows that the ‘s are smooth functions. By putting things together, we therefore proved the proposition in the case .

Let us now turn to the case . By smoothly choosing an orthonormal basis of which is adapted to the orthogonal decomposition

we get a decomposition

where is an orthogonal matrix with smooth coefficients and where $\mathcal{V}(x)$ is a symmetric and positive matrix with smooth coefficients. We may now apply the first part of the proof to the matrix and easily conclude