Let $X\_0, X\_1, \dots, X\_k$ be smooth vector fields over ${\mathbb R}^n$, and let us consider the operator
$$
L = \sum\_{i=1}^k X\_i^2 + X\_0~.
$$
Here, I assume that Hörmander's bracket condition is satisfied, that is to say that the Lie algebra generated by $X\_0, X\_1, \dots, X\_k$ has full rank at every point in ${\mathbb R}^n$. This implies that $L$ is hypoelliptic. Now the question is: what are the conditions (if any) for the formal adjoint $L^\dagger$ of $L$ to be also hypoelliptic?


Judging by the answer to [this question](https://mathoverflow.net/questions/85578/when-the-adjoint-of-a-hypoelliptic-operator-hypoelliptic) it appears that this is trivially true since Hörmander's condition "does not change by taking adjoints". What this means is unclear to me. We have indeed that $$
L^\dagger = \sum\_{i=1}^k X\_i^2 - X\_0 + f~,
$$
where $f$ is a smooth scalar function, right? If there were no $f$, then this would indeed be trivial, since we find again the same Lie algebra. Since all the treatments of Hörmander's theorem that I have seen discuss operators without such an $f$ (except Hormander's [paper](http://link.springer.com/article/10.1007%2FBF02392081) of 1967), I guess there is an easy way to get rid of this $f$. Do you know how? Or do you have another argument to show that $L^\dagger$ is also hypoelliptic?
Thanks a lot.



Hormander's bracket condition for the adjoint of an operator
The hypoellipticity result is more precise:
you have
$$
Lu \in H^s\_{loc}\Longrightarrow u\in H^{s+2-\delta}\_{loc}\quad\text{ for some $\delta\in [0,2)$,}
$$
and that $\delta$ is linked to the number of brackets of the $X\_j$ needed to generate the full tangent space. For instance, in the elliptic case, where $X\_0=0$ and the $(X\_j)\_{1\le j\le n}$ span the tangent space, we have $\delta=0$. Now if $f\in C^\infty$,


$$
u\in H^s\_{loc},\quad Lu+fu \in H^s\_{loc}\Longrightarrow Lu \in H^s\_{loc}\Longrightarrow u\in H^{\epsilon+s}\_{loc}
$$
with $\epsilon=2-\delta>0$.


We assume now $Lu+fu\in C^\infty$ and we want to prove $u\in C^\infty$:
since it is a local result, we may assume for some $s$ that
$
u\in H^s\_{loc}
$.


Since $Lu+fu\in H^s\_{loc}$, we get from above 
$u\in H^{s+\epsilon}\_{loc}$. 


Since $u, Lu+fu\in H^{s+\epsilon}\_{loc}$ we get 


$u\in H^{s+2\epsilon}\_{loc}$ 


and so on: $u$ is smooth.


P.S. The weird typesetting is due to MathJax.


