11

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.

Jesse Madnick
  • 32,819
  • 1
    There's a discussion of this correspondence (for the Ricci curvature) in "The Ricci flow in Riemannian Geometry" by B.Andrews and C. Hopper (p. 83 onward). –  May 12 '15 at 07:28
  • 7
    Elliptic operators are Fredholm, which means the dimension of the kernel and the dimension of the cokernel are finite. If the equation is diffeomorphism invariant then that gives an infinite dimensional subspace of the kernel of its linearization, which is an elliptic operator, a contradiction. – aes May 19 '15 at 07:13
  • 1
    @aes: Thank you. Could you please turn your comment into an answer? – Jesse Madnick May 20 '15 at 08:09

1 Answers1

3

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)

youler
  • 2,608