# Laplacian spectrum asymptotics in neck stretching

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?

• 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 $S$. – Hadrian Quan Mar 9 '19 at 16:36
• 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. – Hadrian Quan Mar 9 '19 at 16:36
• 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 $[-T,T]\times S$. – Neal Mar 9 '19 at 19:18

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 $$T$$.
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 $$\lambda_1\geq h(M)^2/4\sim\tfrac{1}{4}T^{-2}$$.
• 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 $h\sim 1/T$. – Neal Mar 15 '19 at 16:22