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This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.

The context: I want to show

If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} \rightarrow W^l$ has kernel whose dimension is independent of $l$.

Black box terminology explanation:


Definition: Let $E_i \rightarrow M$ be two vector bundles, $P:\Gamma(M,E_0) \rightarrow \Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as $$ Pf = \sum_{|\alpha| \le k } A^\alpha(y) \frac{\partial^\alpha}{\partial x_\alpha} f(y).$$

Definition 2: Let $E \rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $\Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $\Gamma(M,E)$ with respect to this norm.


Let us suppose $P:W^{k+l} \rightarrow W^k$ is well defined and:

Theorem: Let $Pu=f$, $f \in W^l$, $u \in W^r$ for some integer $r$, then $u \in W^{l+k}$.


It is claimed that then we have the kernel is independent of $l$.


How does this follow?


Reference: Pg 48-49.

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  • $\begingroup$ It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator). $\endgroup$ Commented Jan 4, 2019 at 20:36
  • $\begingroup$ Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition. $\endgroup$ Commented Jan 4, 2019 at 20:37
  • $\begingroup$ Oh yes, thanks a lot. I will add this. $\endgroup$
    – Bryan Shih
    Commented Jan 4, 2019 at 23:12

1 Answer 1

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Suppose $u \in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 \in W^{\ell}$ for each $\ell,$ so the elliptic regularity theorem gives $u \in W^{k+\ell}$ for all $\ell.$ So if $N_r \subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} \rightarrow W^r,$ we get $N_r \subset N_{\ell}$ for each $\ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.

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