What is the dimension of the kernel of the exterior derivative? Let $M$ be a compact Riemannian manifold. Is there is way we can count the dimension of the kernel of the exterior derivative on forms? 
 A: The kernel of the exterior derivative on 0-forms is the set of locally constant functions, i.e. constant on each component. So its dimension is the number of components, possibly infinite. On any $n$-dimensional manifold, the kernel of the exterior derivative on $k$-forms, for $0\le k\le n$, contains all exact forms $d\eta$ for $\eta$ a $(k-1)$-form. So it is infinite dimensional. By definition, all $k$-forms are zero for $k>n$, so the kernel of the exterior derivative on $k$-forms for $k>n$ is zero.
It is easy to check that the symbol $\sigma=\sigma_d$ of the exterior derivative $d$ is $\sigma(\xi)=\xi \wedge {}$ for any 1-form $\xi$, where $\xi \wedge {}$ means the linear transformation $\omega \mapsto \xi \wedge \omega$. Therefore one easily checks that the complex characteristic variety is empty for $d$ acting on $0$-forms, so indeed $d$ is elliptic on $0$-forms. On the other hand, the complex characteristic variety is the complexified cotangent bundle for $d$ acting on $k$-forms for any $k$ with $1 \le k \le n$. Therefore the exterior derivative is never elliptic as an operator on $k$-forms for $1 \le k \le n$. The story for $\bar\partial$ is a nice exercise too.
