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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)

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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.

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Thanks, nice answer. – michael faber Mar 13 '13 at 18:43

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