I am struggeling with the following comment that I read regarding the de Rham complex:
Define $(d + \delta)_e : C^\infty(\Lambda^e(T^*M)) + C^\infty(\Lambda^o(T^*M))$ where
\begin{equation} \Lambda^e(T^*M) = \oplus_{k} \, \Lambda^{2k}(T^*M), \text{ and } \Lambda^o(T^*M) = \oplus_{k} \, \Lambda^{2k+1}(T^*M) \end{equation} denote the differential forms of even and odd degrees.
Here comes the statement that I don't understand:
$(d + \delta)$ is an elliptic operator since the associated Laplacian $\triangle = (d + \delta)_e^*(d + \delta)_e$ is elliptic since dim$(\Lambda^e)$ = dim$(\Lambda^o)$.
Why can I deduce the ellipticity of $\triangle$ from the fact that the dimensions of these two spaces agree ? Many thanks for any hints !