1
$\begingroup$

I am reading the book Index Theorem and the Heat Equation written by Peter.B.Gilkey. Here is my question: Let E be a hermitian vector bundle on a compact smooth manifold M. Let $D : \mathcal{C}^{\infty}(E) \rightarrow \mathcal{C}^{\infty}(E)$ be an elliptic operator of positive order. Then it extends to the completion of $\mathcal {C}^{\infty}(E)$ to $L^2(E)$. It has been said in the first paragraph of chapter 3 of the book mentioned above that this is an unbounded operator and admits a sequence of eigenvalues $0 \leq \lambda_n \rightarrow \infty$. Any reference or proof of this will be helpful.

$\endgroup$

closed as off-topic by Chris Gerig, Michael Renardy, Johannes Ebert, Stefan Kohl, Yemon Choi Sep 8 '14 at 15:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Chris Gerig, Michael Renardy, Johannes Ebert, Stefan Kohl
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This follows from the spectral theorem for self adjoint unbounded operators together with elliptic regulatory: $L^2(E)$ decomposes as the direct sum of finite dimensional eigenspaces for $D$. Any book on PDE's will prove this. $\endgroup$ – Paul Siegel Sep 8 '14 at 13:06
  • $\begingroup$ Also could you please point out why the operator will be unbounded. Thanks. $\endgroup$ – ankit Sep 8 '14 at 13:35
  • 1
    $\begingroup$ Do you agree that differential operators on Euclidean space are unbounded? If so, restrict $D$ to a Euclidean neighborhood in $M$. $\endgroup$ – Paul Siegel Sep 8 '14 at 13:53
  • 1
    $\begingroup$ This is proven in full detail in section 1.6 of the book by Gilkey that you mentioned. $\endgroup$ – Johannes Ebert Sep 8 '14 at 14:59
  • $\begingroup$ This question appears to be off-topic because it is answered by reading the text mentioned in the question $\endgroup$ – Yemon Choi Sep 8 '14 at 15:08
2
$\begingroup$

The fact that (non-trivial) elliptic operators are not bounded on $L^2$ is a special case of the fact that differential operators generally are not continuous (i.e., not bounded) on $L^2$. This much has little to do with the compactness of the manifold.

Note that such operators do not extend "to the completion of $C^\infty(E)$ to $L^2(E)$", as in the question. Yes, such operators have extensions to distribution-valued operators, but that's not immediately to the point.

The fact that symmetric elliptic differential operators have self-adjoint extensions is also a fairly general fact, not depending on compactness, since Friedrichs' self-adjoint extensions always exist for semi-bounded operators. (The von Neumann or Krein general classification of self-adjoint extensions of symmetric unbounded operators requires that we check some conditions (equality of dimensions of deficiency indices...) before knowing there's any self-adjoint extension.)

The property of eigenvalues mentioned in the question is much more specific, depending for proof on having compact resolvent, which follows for compact manifolds. That argument basically reduces/compares to the case of flat tori, that is, products of circles, where Fourier series give a straightforward proof of the relevant Rellich compactness assertion.

$\endgroup$

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