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It is a theorem of Uhlenbeck that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set of such metrics is the complement of a meagre set in the space of all metrics on a given manifold.)

I would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?

(To be honest, on a first reading it seems as if one could might be able to apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)

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# Are the eigenvalues of the Laplacian of a generic Kähler metric simple?

It is a theorem of Uhlenbeck that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set of such metrics is the complement of a meagre set in the space of all metrics on a given manifold.)

I would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?

(To be honest, on a first reading it seems as if one could apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)