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

Posted on 7/21/2010
UQ ID:153
Science
MathOverflow
Original

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 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!)

Keywords & Tags
dg.differential-geometry
complex-geometry
riemannian-geometry
fa.functional-analysis
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K2-Think

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Qwen2.5-32B
Sep 18, 2025, 01:50 PM

The question of whether the eigenvalues of the Laplacian (acting on functions) are simple for a generic Kähler metric within a fixed Kähler class can be addressed by adapting arguments from Uhlenbeck's theorem on generic Riemannian metrics.

Key Observations:

  1. Space of Kähler Metrics: A compact complex manifold with a fixed Kähler class admits a space of Kähler metrics parameterized by ...
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GPT-5

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Sep 2, 2025, 06:56 AM

Yes. The analogue of Uhlenbeck’s result holds if you restrict to Kähler metrics, even within a fixed Kähler class: for a residual (comeagre) set of Kähler metrics in a given class, the scalar Laplacian has simple spectrum.

Why the same argument goes through

  • Parametrization of the space: Fix a background Kähler form ω0 in the given Kähler class. Every other Kähler form in the class can be writte...
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Claude Opus 4

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Anthropic
Aug 1, 2025, 12:00 AM

This is an interesting question about the genericity of simple eigenvalues for the Laplacian on Kähler manifolds. Let me address what is known about this problem.

The answer is yes - for a generic Kähler metric in a fixed Kähler class, the eigenvalues of the Laplacian acting on functions are simple. This follows from adapting Uhlenbeck's arguments to the Kähler setting, though there are some ...

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Claude 3.7 Sonnet

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Anthropic
Aug 1, 2025, 12:00 AM

Eigenvalues of the Laplacian for Generic Kähler Metrics

This is an interesting question about the spectral properties of Laplacians in Kähler geometry. Let me address it in detail.

Background

Uhlenbeck's theorem shows that for generic Riemannian metrics, the Laplacian acting on functions has simple eigenvalues (all eigenspaces are 1-dimensional). The key question here is whether this prope...

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o3 Pro

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Aug 1, 2025, 12:00 AM

Yes – after one restricts the variation of the metric to a fixed Kähler class, the Laplacian on functions still has a simple spectrum for a “generic’’ metric. In other words

For every compact complex manifold (M , J) and every Kähler class [ω0], the set
G = { ϕ ∈ C∞(M)/ ℝ : ωϕ := ω0 + i∂∂̄ϕ is Kähler
and all non–zero eigenvalues of Δϕ are simpl...

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DeepSeek R1

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DeepSeek
deepseek-ai/DeepSeek-R1
Aug 1, 2025, 12:00 AM
<think> Okay, the user is asking whether the eigenvalues of the Laplacian are simple for a generic Kähler metric. Let me start by recalling Uhlenbeck's theorem. She proved that for a generic Riemannian metric, the Laplacian has simple eigenvalues. The question is whether this holds in the Kähler case, specifically for metrics in a fixed Kähler class.

First, I need to understand the setup. The use...

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o3

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OpenAI
Aug 1, 2025, 12:00 AM

Yes – after fixing a complex structure and a Kähler class, the Laplace operator on functions still has a simple spectrum for a generic Kähler metric. Uhlenbeck’s argument can be repeated almost verbatim once one knows that, inside a fixed Kähler class, one can perturb the metric in “enough directions’’ to split any multiple eigen-value. That is indeed the case because variations of a Kähler metr...

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Gemini 2.5 Pro

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Google
Aug 1, 2025, 12:00 AM

This is an excellent question that touches upon a subtle but important point distinguishing the space of all Riemannian metrics from the more rigid space of Kähler metrics within a fixed Kähler class.

The short answer is: No, this is not true in general. The validity of the statement depends crucially on the complex manifold itself, specifically on its group of holomorphic automorphisms.

Her...

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o4 mini

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OpenAI
Aug 1, 2025, 12:00 AM

Yes – nothing essentially new happens in the Kähler‐case, one can literally run Uhlenbeck’s transversality argument in the smaller Banach manifold of Kähler metrics in a fixed class. In fact:

  1. Fix a reference Kähler form ω₀ on M and identify the space of C^ℓ–Kähler metrics in [ω₀] with the open set of real C^ℓ–functions φ (mod constants) for which ωφ:=ω₀+i∂∂̄φ>0.

  2. Consider the “eval...

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