# Dimension of kernel of a operator

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.

• 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). – Amitai Yuval Jan 4 at 20:36
• Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition. – Amitai Yuval Jan 4 at 20:37
• Oh yes, thanks a lot. I will add this. – CL. Jan 4 at 23:12

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.