I know the kernel of an elliptic operator on a compact manifold has finite dimension. Is the kernel of an elliptic operator on sections of a vector bundle a finite dimensional space?
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1$\begingroup$ yes. see chap. 10 of these notes. www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$– Liviu NicolaescuCommented Feb 27, 2016 at 13:40
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1$\begingroup$ But assuming the vector bundle is over a compact manifold. $\endgroup$– Deane YangCommented Feb 27, 2016 at 14:07
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$\begingroup$ I think that this result is due to Bochner, Tensor fields with finite bases, Ann. Math. 1951 (53) p. 400-411. But Nicolaescu's book is where I would look for a proof. $\endgroup$– Ben McKayCommented Feb 27, 2016 at 16:58
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In general, no. As an example, the spin Dirac operator on even-dimensional complex hyperbolic space has nontrivial $L^2$-sections in its kernel. Because the space is symmetric, in particular homogeneous, one can conclude that the $L^2$-kernel must be infinite-dimensional. The proof uses some results of Harish-Chandra on representations of semi-simple noncompact Lie groups. For more details, see this article or arXiv:math/9905089.
answered Feb 27, 2016 at 16:32
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4$\begingroup$ For a more elementary example of infinite dimensional kernel on noncompact manifolds, you could just look at the kernel of the Cauchy Riemann operator on complex functions of one complex variable. $\endgroup$ Commented Feb 27, 2016 at 16:41
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$\begingroup$ @BenMcKay You are right. I was somehow automatically thinking of $L^2$ sections, even though this was not demanded. $\endgroup$ Commented Feb 27, 2016 at 17:12