On the Spectra of a Family of Geometric Operators Evolving with Geometric Flows

Abstract

In this work we generalise various recent results on the evolution and monotonicity of the eigenvalues of certain family of geometric operators under some geometric flows. In an attempt to understand the arising similarities we formulate two conjectures on the monotonicity of the eigenvalues of Schrödinger operators under geometric flows. We also pose three questions which we consider to be of a general interest.

Appendix: An Alternative Proof of the Evolution Formula

As promised earlier in Sect. 3, we shall now give an alternative proof of the evolution formula for the eigenvalues of the operator $-\mathbb {L}$ evolving with the flow (1.1). This proof conceptually differs from the one we have already presented, for it makes use of the notion of the $\eta$-divergence and some of its properties. Given a symmetric (0, r)-tensor S on M we define the $\eta$-divergence to be the $(0,r-1)$-tensor

$$\begin{aligned} \text {div}_{\eta }S=\text {div} S-\text {d}\eta \circ S, \end{aligned}$$

where $\text {div}S$ is the usual divergence of S and $\eta$ is the drifting function. The identity $\text {div}_{\eta }(\nabla f)=L(f)$ is effortlessly checked. The $\eta$-divergence enjoys the property $\text {div}(e^{-\eta }X)=e^{-\eta }\text {div}_{\eta }X$ for any smooth function f and a vector field X on the manifold M. It is by its virtue that we have a natural extension of the divergence theorem for $\text {div}_{\eta }$. Indeed, given a vector field X on M and the weighted measure $\text {d}\mu =e^{-\eta }\text {d}\partial M$ on the boundary $\partial M$, the divergence theorem takes the form

$$\begin{aligned} \int _M\text {div}_{\eta }X \text {d}m=\int _{\partial M}g(X,\nu )\text {d}\mu , \end{aligned}$$

where $\nu$ is the unit outward normal vector. We shall also need in what follows the property

$$\begin{aligned} \text {div}_{\eta }(fX)=f\text {div}_{\eta }X+g(\nabla f, X). \end{aligned}$$

Taking any function $f\in C_{0}^{\infty }(M)$ and working in coordinates one can straightforwardly verify the identity

$$\begin{aligned} \frac{\text {d}}{\text {d}t}\bigg (\mathbb {L}f\bigg )=\mathbb {L}'f+\mathbb {L}f'. \end{aligned}$$

(8.1)

Moreover, the following formula holds true (see [10])

$$\begin{aligned} \mathbb {L}'f=\bigg \langle \frac{1}{2}\text {d}h-\text {div}_{\eta }{\mathcal {H}},\text {d}f\bigg \rangle -\langle {\mathcal {H}},\nabla ^2f\rangle +cR'f. \end{aligned}$$

(8.2)

Now, with the aforementioned preparatory remarks in mind we can give the second proof of the evolution formula (3.1).

Proof

To begin with, observe that by differentiating in t the obvious identity

$$\begin{aligned} -u(t)\mathbb {L}u(t)=\lambda (t)u^2(t), \end{aligned}$$

we easily obtain

$$\begin{aligned} u'\mathbb {L}u-u\mathbb {L}u'=\lambda 'u^{2}+u\mathbb {L}'u, \end{aligned}$$

which reduces to

$$\begin{aligned} u'Lu-uLu'= \lambda ' u^{2}+u\mathbb {L}'u. \end{aligned}$$

By dint of formula (2.8) we readily perceive that the left hand side vanishes after the integration and therefore the following simple formula holds true

$$\begin{aligned} \lambda '=-\int _{M}u\mathbb {L}'u\,\text {d}m. \end{aligned}$$

(8.3)

It is now evident that we only need to compute the integral on the right hand side. One can calculate this integral in two different ways. The obvious one is to use the variation formula for the operator $\mathbb {L}$. Namely, plugging (8.2) in (8.3) and using the properties of $\text {div}_{\eta }$ as well as the divergence theorem one will eventually arrive at the desired formula. Leaving this computation for the reader to check we shall give another way of computing the integral in (8.3). We observe first that the definition of $\mathbb {L}$ and the integration by part formula yield

$$\begin{aligned} \int _{M}u\mathbb {L}u\,\text {d}m=-\int _{M}\langle \text {d}u,\text {d}u\rangle \text {d}m+c\int _{M}Ru^{2}\text {d}m. \end{aligned}$$

(8.4)

Evidently, we only need to differentiate the latter formula and simplify to formula (3.1). Using the identity (8.1) we can rewrite the left hand side integral of (8.4) as

$$\begin{aligned} \dfrac{\text {d}}{\text {d}t}\left( \int _{M}u\mathbb {L}u\,\text {d}m\right) =\int _{M}u'(\mathbb {L}u)\text {d}m+\int _{M}u\mathbb {L}'u\,\text {d}m+\int _{M}u\mathbb {L}u'\,\text {d}m+\frac{1}{2}\int _{M}hu\mathbb {L}u\,\text {d}m. \end{aligned}$$

Differentiation of the right hand side of (8.4) gives

$$\begin{aligned} \dfrac{\text {d}}{\text {d}t}\left( \int _{M}u\mathbb {L}u\,\text {d}m\right)= & {} -2\int _{M}\langle \text {d}u,\text {d}u'\rangle \text {d}m+\int _{M}{\mathcal {H}}(\nabla u,\nabla u)\text {d}m-\frac{1}{2}\int _{M}h|\nabla u|^{2}\text {d}m\\&+\,c\int _{M}R'u^{2}\text {d}m+c\int _{M}R(u^{2})'\text {d}m+\frac{c}{2}\int _{M}hRu^{2}\text {d}m. \end{aligned}$$

It is also easily seen that the integration by parts formula implies

$$\begin{aligned} \int _{M}u'\mathbb {L}u\,\text {d}m+\int _{M}u\mathbb {L}u'\text {d}m= & {} \int _{M}u'Lu\,\text {d}m+\int _{M}uLu'\,\text {d}m+c\int _{M}R(u^{2})'\text {d}m\\= & {} -\,2\int _{M}\langle \text {d}u,\text {d}u'\rangle \text {d}m+c\int _{M}R(u^{2})'\text {d}m. \end{aligned}$$

Thus, the latter three formulae justify the validity of

$$\begin{aligned} \int _{M}u\mathbb {L'}u\,\text {d}m= & {} \int _{M}{\mathcal {H}}(\nabla u,\nabla u)\text {d}m-\frac{1}{2}\int _{M}h\langle \text {d}u,\text {d}u\rangle \text {d}m\nonumber \\&+\,c\int _{M}R'u^{2}\text {d}m+\frac{c}{2}\int _{M}Rhu^{2}\text {d}m-\frac{1}{2}\int _{M}hu\mathbb {L}u\,\text {d}m. \end{aligned}$$

(8.5)

We shall have to make a little detour at this point. Using the properties of $\text {div}_{\eta }$ we readily compute

$$\begin{aligned} \text {div}_{\eta }(hu\nabla u)= & {} huLu+g(\nabla hu,\nabla u) \\= & {} huLu+h\langle \text {d}u, \text {d}u\rangle +u\langle \text {d}h, \text {d}u\rangle \\= & {} hu(\mathbb {L}u-cRu)+h\langle \text {d}u, \text {d}u\rangle +u\langle \text {d}h, \text {d}u\rangle \\= & {} hu\mathbb {L}u-cRhu^{2}+h\langle \text {d}u, \text {d}u\rangle +u\langle \text {d}h, \text {d}u\rangle . \end{aligned}$$

Now, as we are on a closed manifold, $\int _M\text {div}_{\eta }(hu\nabla u) \text {d}m$ vanishes and we obtain

$$\begin{aligned} \int _{M}h\langle \text {d}u, \text {d}u\rangle \text {d}m=-\int _{M}u\langle \text {d}h, \text {d}u\rangle \text {d}m-\int _{M}hu\mathbb {L}u\,\text {d}m+c\int _{M}Rhu^{2}\text {d}m. \end{aligned}$$

It follows that relation (8.5) can be expressed as

$$\begin{aligned} \int _{M}u\mathbb {L}'u\,\text {d}m=\int _{M}{\mathcal {H}}(\nabla u,\nabla u)\text {d}m+\frac{1}{2}\int _{M}u\langle \text {d}h, \text {d}u\rangle \text {d}m+c\int _{M}R'u^{2}\text {d}m. \end{aligned}$$

Recall now that the Witten-Laplacian can be defined as $\text {div}_{\eta }(\nabla f)=L(f)$. This implies the identity

$$\begin{aligned} \frac{u}{2}\langle \text {d}h, \text {d}u\rangle =\frac{1}{4}\text {div}_{\eta }(h\nabla u^{2})-\frac{1}{4}hL(u^{2}), \end{aligned}$$

which, by dint of the divergence theorem, can be rewritten in the following integral form

$$\begin{aligned} \int _M\frac{u}{2}\langle \text {d}h, \text {d}u\rangle \text {d}m=-\int _M\frac{1}{4}hL(u^{2}) \text {d}m. \end{aligned}$$

We can then write

$$\begin{aligned} \int _{M}u\mathbb {L}'u\,\text {d}m=\int _{M}\left( -\frac{h}{4}L(u^{2})+{\mathcal {H}}(\nabla u,\nabla u)+cR'u^{2}\right) \text {d}m, \end{aligned}$$

which completes the proof. $\square$