5
$\begingroup$

This questions concerns elliptic complexes and is closely related to Green's operator of elliptic differential operator.

Let $T_f:\Gamma(E)\rightarrow\Gamma(F)$ be an elliptic partial differential operator of degree $r$ with coefficients $f \in \Gamma(L(J^r E, F))$, where $J^r E$ is the $r$-th jet bundle of $E$. Given coefficients $f_0$, let $Z^\pm$ be finite-dimensional vector spaces, $T^+: Z^+ \to \Gamma(F)$ and $T^-: \Gamma(E) \to Z^-$ be continuous linear maps such that $$\begin{pmatrix}T_{f_0} & T^+ \\ T^- & 0\end{pmatrix}: \Gamma(E) \times Z^+ \to \Gamma(F) \times Z^- $$ is invertible (for example, we may choose $Z^- = \ker T_{f_0}$ and $Z^+ = \mathrm{coker} \, T_{f_0}$ with $T^\pm$ being the canonical projection/injection). In Theorem II.3.3.3 (p. 157f) of Hamilton's work on the Nash-Moser inverse function theorem it is claimed (and attributed to "standard Fredholm theory") that there exists an open neighborhood $U$ of $f_0$ in $\Gamma(L(J^r E, F))$ such that $$\begin{pmatrix}T_{f} & T^+ \\ T^- & 0\end{pmatrix}$$ is invertible for all $f \in U$.

Side question:

Do you know a reference for the existence of such an open neighborhood $U$?

Real question:

What is the proper generalization of this picture using extended invertible maps to elliptic complexes $\Gamma(E) \overset{T_f}{\to} \Gamma(F) \overset{S_g}{\to} \Gamma(G)$?

The obvious guess is that one goes over to an extended sequence $$\Gamma(E) \times Z^+ \to \Gamma(F) \times H \to \Gamma(G) \times Z^-,$$ which is exact. Is this worked-out somewhere? (I feel like this is a basic statement in elliptic theory but I couldn't find a reference for both questions.)

$\endgroup$
1
  • 1
    $\begingroup$ On the side question: In a Banach space setting, the openness you state follows directly from the openness of isomorphisms (and it is sufficient to consider some suitable Banach space setting, since e.g. ker consists of smooth functions as $T_{f_0}$ is elliptic). I think the Fredholm theory is only invoked to see that the collection of all $f$, for which $\tilde{L}(f)$ is an iso, is open (in the same way as when proving that the subspace of Fredholm operators is open in bounded operators). $\endgroup$
    – user_1789
    Oct 3, 2018 at 21:29

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.