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Let $M$ be a compact Riemannian manifold. Let $S \subset M$ be a smooth hypersurface separating $M$ into two components. Let $g_T$ be a family of Riemannian metric obtained by stretching along $S$, i.e., a fixed tubular neighborhood of $S$ is replaced by $[-T, T] \times S$.

My question is: when $T \to \infty$, what is the asymptotic behavior of the smallest positive eigenvalue of the Laplacian (or Hodge Laplacian) associated to $g_T$? Any reference for such kind of results?

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    $\begingroup$ I think the standard reference for this type of question is Cappell-Lee-Miller (Self-Adjoint operators and Manifold decompositions I), where they consider the limiting behaviors of eigenmodes associated to a 1st order elliptic operator $D$. If you take your $D=d+\delta$ to be the de Rham operator, its square will be the Hodge Laplacian, so their analysis should help in your case. Roughly, small eigenvalues come from contributions of the 2 connected components of $M$, and their "interactions of eigenspaces along $S$." There is the subtlety that they impose APS-boundary conditions near $S$. $\endgroup$
    – Hadrian Quan
    Commented Mar 9, 2019 at 16:36
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    $\begingroup$ They crucially use the correspondence between Dirac operators with APS-boundary conditions on a manifold $Z$ with boundary, and the the $L^2$ boundary conditions on the prolongation $\hat{Z}$ of your manifold to one with infinite cylindrical ends. So it makes sense why you would consider this boundary condition, as it relates to the “limiting problem” of a neck with infinite length. $\endgroup$
    – Hadrian Quan
    Commented Mar 9, 2019 at 16:36
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    $\begingroup$ You could start investigating this with Cheeger's inequality. If $h_T$ is the Cheeger constant associated to $(M,g_T)$, then Cheeger-Buser inequalities show that $\lambda_1\sim \sqrt{h}$ and my guess is that for large $T$, you can realize $h_T$ with a (perturbation of a) cross section of $[-T,T]\times S$. $\endgroup$
    – Neal
    Commented Mar 9, 2019 at 19:18

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The first eigenvalue $\lambda_1$ is of order $T^{-2}$. One can get the upper bound by a direct analysis of the Rayleigh quotient. Let $M_\pm$ be the two ends. First suppose $\operatorname{Vol}(M_+)=\operatorname{Vol}(M_-)$; then the function $$ f(x)=\begin{cases}\sin(\tfrac{1}{2}\pi t/T),& \text{for $x=(s,t)$ in the neck part $S\times [-T,T]$}\\\pm 1,& x\in M_\pm \end{cases} $$ is in the function space $H^1(M)$ and $\int_Mf=0$, so $$ \lambda_1\leq \frac{\int_M|df|^2}{\int_Mf^2} =\frac{(\tfrac{\pi}{2T})^2T\operatorname{Vol}(S)}{T\operatorname{Vol}(S)+2\operatorname{Vol}(M_\pm)}\sim\frac{\pi^2}{4T^2}. $$ If instead $\operatorname{Vol}(M_+)>\operatorname{Vol}(M_-)$ then let $c:=[\operatorname{Vol}(M_+)-\operatorname{Vol}(M_-)]/\operatorname{Vol}(S)$, add the neck portion $S\times [-T, -T+c]$ to $M_-$ to equalize the volumes, and run the above argument with $T-\tfrac{1}{2}c$ replacing $T$.

Remark: Edited to add the above argument, which improves by an asymptotic factor of 4 the upper bound in my original answer.

For the lower bound, as noted by @Neal, it seems that the Cheeger constant $h(M)$ should be achieved by a cross-section $S$ in the neck part, so $$h(M)=\frac{\operatorname{Vol}(S)}{\operatorname{Vol}(M)/2} =\frac{\operatorname{Vol}(S)}{T\operatorname{Vol}(S)+c}\sim \frac{1}{T}. $$ This means that $\lambda_1\geq h(M)^2/4\sim\tfrac{1}{4}T^{-2}$.

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  • $\begingroup$ Quick sketch for why $h$ should be achieved by a cross-section in the neck part. Let $h_E = |E|/\min(A,B)$ for separating hypersurfaces $E$, with $A\cup B = M$ and $\partial A = \partial B = E$. Then $h = \inf_E h_E$. Since $h_{S\times 0} \to 0$, then $h$ must be achieved by a subsurface that enters the neck. Then show if a subsurface $E$ crosses between $M$ and the neck, it must have area that grows with a power of $T$, so $h_E$ will go as something like $1/\sqrt{T}$. That leaves subsurfaces contained within the neck, which have $h\sim 1/T$. $\endgroup$
    – Neal
    Commented Mar 15, 2019 at 16:22

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