MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that the eigenfunctions of the Laplacian on a compact manifold $M$ form a countable basis of $H^1(M)$ and $L^2(M)$.

If $L$ is a $2k$-order elliptic operator, do the eigenfunctions of $L$ form a basis for $H^k(M)$? References/more detail would be appreciated. Thanks.

(Crossposted from due to lack of replies)

share|cite|improve this question
Another reference which proves this (for self-adjoint, elliptic operators of positive order) is Theorem III.§5.8 in the book "Spin Geometry" by Lawson and Michelsohn. – AlexE Jun 26 '13 at 18:04
up vote 11 down vote accepted

The answer is no, if the Fredholm index of $L$, which is the integer $\mathrm{ind}\, L=\dim\mathrm{ker}\, L- \dim \mathrm{ker} \, L^\ast$, is negative. (Regard $L$ as an unbounded operator on $L^2(M)$ with domain $D(L)=H^{2k}(M)$.) A proof by contradiction goes as follows. If the eigenvectors formed a basis of $H^k(M)$, then their span would also be dense in $L^2(M)$. The elements of $\mathrm{ker}\, L^\ast$, i.e. of the null space of the adjoint operator, are orthogonal to the range of $L$, which contains the eigenvectors with non-zero eigenvalues. The assumption $\mathrm{ind}\, L<0$ implies that there exists a non-zero element of $\mathrm{ker}\, L^\ast$ orthogonal to all eigenvectors. Thus we obtain the contradiction that the span of the eigenvectors is not dense in $L^2(M)$.

I have no explicit example of a scalar elliptic operator of even order with negative index. Because of $\mathrm{ind}\, L^\ast= -\mathrm{ind}\, L$ it would suffice to find an operator with non-zero index. By the Atiyah-Singer index theorem, this is a topological condition on the manifold $M$ and on the principal symbol of the operator $L$.

You received the answer "yes" to your question on SE under the additional assumption that $L$ is self-adjoint and coercive. It was explained there how this follows using the spectral theorem and using, what I am accustomed to call, Gelfand triplets. As a reference for the latter I mention the book: Wloka, Partial Differential Equations. Note that self-adjointness implies that the index is zero.

share|cite|improve this answer
Thanks, nice answer. – michael faber Mar 13 '13 at 18:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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