Why are diffeomorphism-invariant PDE not elliptic? In reading geometric analysis papers, I frequently encounter a statement of the form, "The PDE in question is diffeomorphism-invariant, and therefore cannot be elliptic."
My vague understanding is that this might have to do with the space of solutions being infinite-dimensional, or maybe it has to do with elliptic regularity failing.  Either way, I would like more precise clarification.
 A: aes' comment is correct but the fact that elliptic operators on a compact space are Fredholm is, properly speaking, a theorem, and the question can be answered more directly.
To recall the definition:


*

*Let $E,F$ be vector bundles over closed $M$ with $L:C^\infty(E)\to C^\infty(F)$ an order $k$ linear differential operator taking a section of $E$ to a section of $F$, meaning that in any local coordinates $x^i$ on $M$ you can write
$$Lu(p)=\sum_{|\alpha|=k}L^\alpha_p\left(\frac{\partial^\alpha u}{\partial x^\alpha}\right)+\sum_{|\alpha|\leq k-1}\Big\{\Big\}$$
where for each multiindex $\alpha$, $L_p^\alpha:E_p\to F_p$ is a linear map.

*For a cotangent vector $\xi\in T^\ast_pM$ the principal symbol $\sigma L_\xi$ is a linear map $E_p\to F_p$ defined by $e\mapsto \xi_{\alpha_1}\cdots \xi_{\alpha_k}L_p^\alpha(e).$

*We say $L$ is elliptic if for each $\xi$ the principal symbol is an isomorphism.
Now it's possible to say directly that the principal symbol of a diffeomorphism invariant differential operator has a nontrivial kernel, and so can't be elliptic.
Let $\varphi_t:M\to M$ be a 1-parameter family of diffeomorphisms with $\varphi_0=\operatorname{id}_M,$ generated by the vector field $X$. The fact of diffeomorphism invariance says $L(\varphi_t^\ast u)=\varphi_t^\ast(Lu)$; differentiating at $t=0$ gives
$$L(\mathcal{L}_Xu)=\mathcal{L}_X(Lu).$$
Now view these as mappings $X\mapsto L(\mathcal{L}_Xu)$ and $X\mapsto \mathcal{L}_X(Lu)$, and note that the first map has order $k+1$ and the second map has order 1. Now take the (order $k+1$) principal symbol of the equality of the mappings to get
$$\sigma L_\xi\circ \sigma f_\xi=0,$$
where I'm letting $f$ denote the map taking $X$ to $\mathcal{L}_Xu$. The zero on the right hand side is from taking the order $k+1$ principal symbol of an order 1 operator. Anyway, now we see the kernel of $\sigma L_\xi$ contains the image of $\sigma f_\xi$. So in a direct way the diffeomorphism invariance causes a nontrivial kernel; the kernel consists of a bunch of Lie derivative terms.
(This used the composition of principal symbols, see e.g. the top of page 55 in http://www.math.uiuc.edu/~palbin/Math524.Spring2012/LectureNotesMay1.pdf)
