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Let $(M,g)$ be a closed, smooth Riemannian manifold. Let $\Delta = -div\nabla$ be the Laplace-Beltrami operator. Let $h$ be a smooth function on $M$. Is there a condition on $h$ weaker than non-negativity such that $\Delta + h$ is a positive or non-negative operator?

I'm thinking of something akin to the following: For the conformal Laplacian, non-negativity of the Yamabe constant is sufficient for any $h$ that is a scalar curvature of a metric in the conformal class.

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  • $\begingroup$ Note the problem has been changed from the original (on 6/21/10), because counter-examples to the original conjecture were easy to come by. $\endgroup$
    – Viktor Bundle
    Commented Jun 21, 2011 at 17:35

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This can't be right as stated: if $h$ takes a negative value at some point $p$ of $M$ then $\Delta+ch$ has a negative eigenvalue for sufficiently large $c$.

Proof: let $f: M \rightarrow {\bf R}$ be a nonnegative smooth function that's positive at $p$ and supported on a small enough neighborhood of $p$ that $f(q)=0$ whenever $h(q)>0$. Then $\langle f, hf \rangle < 0$. Therefore if $c$ is large enough then $\langle f, (\Delta+ch) f \rangle < 0$, which would be impossible if every eigenvalue of $\Delta + ch$ were nonnegative.

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    $\begingroup$ More generally, for each $n$ there exists $c_n$ such that $\Delta + ch$ has at least $n$ negative eigenvalues once $c > c_n$. Proof: let $V$ be an $n$-dimensional space of smooth functions supported on the same neighborhood. Then the quadratic form $⟨f,hf⟩$ is negative-definite on $V$, whence $⟨f,(Δ+ch)f⟩$ is also negative-definite for $c$ large enough, etc. $\endgroup$
    – Noam D. Elkies
    Commented Jun 21, 2011 at 4:16
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    $\begingroup$ In the asymptotic limit $c \to \infty$, semiclassical heuristics predict that the number of negative eigenvalues should approximately equal (up to some constant factor involving $2\pi$) the volume of $\{ (x,p) \in T M: |p|^2 + ch < 0 \}$ in the tangent bundle (or cotangent bundle, if one prefers). This should be rigorously provable by semiclassical analysis as soon as $h$ and $M$ are smooth enough. In the other direction, for potentials with sufficiently small (in $L^{n/2}$ norm) negative part, there should be positive semi-definiteness from the Sobolev inequality (at least for $n>2$). $\endgroup$
    – Terry Tao
    Commented Jun 22, 2011 at 1:49
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I believe the answer is no if $n>2$. Let $g$ be a metric with a negative Yamabe constant. There will be a metric $h$ in the conformal class of $g$ such that $\int_M R_h dv_g> 0$. Let $L_h$ be the conformal Laplacian of the metric $h$. It will possess a negative eigenvalue due to the negativity of the Yamabe constant.

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