4
$\begingroup$

I am reading a book on Hodge theory (Ref: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/hodge-smf.pdf) or for english (http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/hodge-ams.pdf). In the french version I was reading about this finiteness theorem of elliptic operators (page 18)

(Finiteness Theorem) Let $E$ and $F$ two hermitian vector bundles on a compact manifold $M$ such that the rank of $E$ and $F$ are the same and equals to $r$. Let $P:C^{\infty}(M,E)\rightarrow C^{\infty}(M,F)$ be an elliptic differential operator of degree $\delta$. Then

1) $\ker P$ is finite dimensional.

In the proof it says: Garding's inequality shows that $||u||_{s+\delta}\leq C_{s}(||u||_{0})$ for all $u\in\ker P$. By Sobolev's lemma, this means $\ker P$ is closed in $W^{0}(M,E)$.

Why should this be so?

$\endgroup$
1
$\begingroup$

My interpretation is as follows:

Suppose $u_n$ is a sequence in $\ker P$ that is Cauchy in $W^0(M,E)$. Then the given inequality shows that $u_n$ is Cauchy in each of the spaces $W^{s+\delta}(M,E)$. As each of these spaces is complete, we get that $u_n$ converges to some section $u$ in each $W^{s+\delta}$. Since the intersection of all of the $W^{s+\delta}$ is $C^\infty(M,E)$ by the Sobolev lemma, we conclude that $u$ is smooth, and $u_n\to u$ in $C^\infty(M,E)$. I suppose from this it's not hard to see that $u\in \ker P$; thus $\ker P$ is closed.

Disclaimer: I'm nowhere near an expert on this topic, so there could very well be a mistake in this answer.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer! I also got this close, but the problem is that $P$ might not be continuous on $C^{\infty}(M,E)$. $P$ being elliptic might play a role but I am nowhere expert in this area either. If $P$ is continuous, then I can conclude that $\ker P$ is closed in $C^{\infty}(M,E)$ which solves all my problem. $\endgroup$ – enoughsaid05 Feb 27 '14 at 21:12
  • 1
    $\begingroup$ @enoughsaid05: Are you sure $P$ doesn't need to be continuous on $C^\infty(M,E)$, even if $P$ weren't elliptic? My naive understanding is $P$ is locally something like $f\mapsto \sum a_\alpha(x) (D^\alpha f)(x)$, say in the case of line bundles, for simplicity. But the point would then be that $\|Pf\|_{C^r} \leq K\|f\|_{C^{r-s}}$ for some $s$. Doesn't this imply continuity on $C^\infty$? Sorry if this is really naive.... $\endgroup$ – froggie Feb 27 '14 at 21:36
0
$\begingroup$

Here's a way to proceed. The operator $P$ goes from $W^{s+\delta}$ to $W^s$ for every $s\geq 0$. If you take $s'>s$, then ker P is a closed subspace of $W^{s+\delta}$ and $W^{s'+\delta}$ because it is continuous for both these norms. But Rellich's lemma assures you that the image of a closed ball in $W^{s'+\delta}$ is compact in $W^{s+\delta}$, but every element of $\ker P$ of 0-norm 1, is in a closed ball in $W^{s'+\delta}$ thanks to Garding's inequality, and thus a ball of $\ker P$ is relatively compact in $W^{s+\delta}$, and thus $\ker P$ is finite dimensionnal, a fortiori it is closed in $W^0$ (but this is not really needed).

Of course you can also prove the closedness directly (as pointed before me by forggie, but let me complete his answer) just by using his argument, and the fact that the injection $W^s\to W^0$ is continuous, and thus if $f_n$ tends towards $f$ in $W^s$ and towards $g$ in $W^0$, then $f=g$. To sum up, as the kernel $\ker P$ is closed in $W^{d}$, and via Garding's inequality, you readily concludes that if $f_n$ tends toward $f$ in $W^0$, then $f_n$ is d-Cauchy, and thus tends to an element of $ker P$ in $W^d$, which is $f$, which is thus in $\ker P$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.