TITLE: When is the adjoint of a hypoelliptic operator also hypoelliptic?
QUESTION [8 upvotes]: Suppose that $M$ is a smooth manifold with a measure $\mu$ and let $L^2(M, \mu)$ be a space of all square-integrable functions on $M$. 
Recall that $L$ is a hypoelliptic differential operator if for every $f \in \mathcal{D}(L)$, if $Lf$ is in $C^\infty(M)$ then $f$ is also in $C^\infty(M)$.
Could anyone give a reference regarding the example when $L$ is hypoelliptic but its adjoint w.r.t to $\mu$ is not hypoelliptic? Could this happen to the Hörmander operators, when $L$ is defined as 
$$
L = \sum_i X_i^2 + X_0
$$
and the Lie algebra generated by $\{X_i\}$ spans the entire tangent space? 
I am mostly interested in the case when $\mu$ is induced by the Riemannian metric and $M$ is complete.
Thanks!

REPLY [6 votes]: Hormander's operator $L=X_0+\sum_{1\le j\le k} X_j^2$, where the $X_j$ are real smooth vector   fields with the Lie algebra of $\{(X_j)\}_{0\le j\le k}$ generating the tangent space is hypoelliptic as well as its adjoint since the Lie algebra condition does not change by taking adjoints.
On the other hand, $\frac{\partial}{\partial t}+t\Delta_x$ is hypoelliptic whereas its adjoint $-\frac{\partial}{\partial t}+t\Delta_x$ is not hypoelliptic,
Bazin.