Essential Self-Adjointness of Perturbed Biharmonic Operators via Conformally Transformed Metrics

Abstract

We give sufficient conditions for the essential self-adjointness of perturbed biharmonic operators acting on sections of a Hermitian vector bundle over a Riemannian manifold with additional assumptions, such as lower semi-bounded Ricci curvature or bounded sectional curvature. In the case of lower semi-bounded Ricci curvature, we formulate our results in terms of the completeness of the metric that is conformal to the original one, via a conformal factor that depends on a minorant of the perturbing potential V. In the bounded sectional curvature situation, we are able to relax the growth condition on the minorant of V imposed in an earlier article. In this context, our growth condition on the minorant of V is consistent with the literature on the self-adjointness of perturbed biharmonic operators on $\mathbb {R}^{n}$.

A Appendix

In this section we gather together various results pertaining to connections and conformal changes of metrics.

The first two formulas of the following proposition describe the chain rules for the Laplacian/Hessian. For the first formula, see exercises 3.4 and 3.9 in [12]. We remind the reader that in our paper the Laplace-Beltrami operator Δ_g is non-negative, which explains the sign difference with the corresponding formulas of [12]. The second formula in proposition A.1 is obtained by combining the formula for the differential of a composition (exercise 3.4 of [12]) with the definition of Hessian Hess_g(f) := ∇^{lc, g}df. (Here, ∇^{lc, g} is the covariant derivative on T^∗M induced from the Levi–Civita connection on (M, g) and d is the standard differential.) For the third formula of proposition A.1, we refer the reader to (III.24) in [15].

Proposition A.1

Let (M, g) be a an n-dimensional Riemannian manifold, with Riemannian metric g. Let $v : M \rightarrow {\mathbb{R}}$ be a smooth function, and $f : U \rightarrow {\mathbb{R}}$ a smooth function, where U is an open set of ℝ containing the range of v. We then have:

• (i) ${\Delta }_{g}(f \circ v) = -f^{\prime \prime }(v)|dv|_{g}^{2} +f'(v){\Delta }_{g}v$,

• (ii) $Hess_{g}(f \circ v) = f^{\prime \prime }(v)dv\otimes dv + f'(v)Hess_{g}(v)$,

• (iii) $|{\Delta }_{g}f|\leq \sqrt {n}|Hess_{g}(f)|_{g}$, where the inequality is understood in pointwise sense.

Next we recall product rules for the formal adjoint of a connection and for the Bochner Laplacian.

Proposition A.2

Let (M, g) be a Riemannian manifold and let E be a Hermitian vector bundle over M with a Hermitian connection ∇. Let Δ_B := ∇^‡∇ denote the associated Bochner Laplacian. Let $f \in W^{2,\infty }_{loc}(M)$ and $u \in W^{2,2}_{loc}(E)$. Then

• (i) $\nabla ^{\dagger }(f\nabla u) = f{\Delta }_{B} u - \nabla _{(df)^\#}u$,

• (ii) ${\Delta }_{B}(fu) = f{\Delta }_{B} u - 2\nabla _{(df)^\#}u + u{\Delta }_{g} f$,

where (df)^# stands for the vector field corresponding to df via the metric g.

Proof

Using integration by parts and the product rule for ∇, for all $v\in C_{c}^{\infty }(E)$ we have

$$ \begin{array}{@{}rcl@{}} (\nabla^{\dagger}(f\nabla u),v)&=&(\nabla u,f\nabla v)=(\nabla u, \nabla(fv))-(\nabla u, df\otimes v)\\ &=&(\nabla^{\dagger}\nabla u, fv)-(\nabla_{(df)^\#} u,v)= (f\nabla^{\dagger}\nabla u, v)-(\nabla_{(df)^\#} u,v), \end{array} $$

which gives the formula (i). The formula (ii) will then follow by using the product rule for ∇, the formula (i) of this proposition, and the following formula (see the equation (III.7) of [15]):

$$ \nabla^{\dagger}(\omega\otimes z)= (d^{\dagger}\omega)z-\nabla_{\omega^\#}z, $$

where $z\in W^{1,2}_{loc}(E)$ and ω is a 1-form on M belonging to $W^{1,\infty }_{loc}({\Lambda }^{1}T^{*}M)$. □

Proposition A.3

Let (M, g) be a Riemannian manifold, with Riemannian metric g. Let $\tilde {g} = \lambda ^{\alpha } g$, where $\lambda : M \rightarrow (0, \infty )$ is a smooth function, and α ∈ ℝ. Let $f : M \rightarrow {\mathbb{C}}$ be a C¹ function on M. Suppose we have the bound $|df|_{\tilde {g}} \leq \phi$, where $\phi : M \rightarrow [0, \infty )$. Then we have

$$ |df|_{g} \leq \lambda^{\alpha/2}\phi. $$

Proof

This is a pointwise estimate, so it suffices to work in coordinates. Let (xⁱ) be local coordinates about a point in M. We then compute

$$ \langle df , df\rangle_{g} = g^{ij}\frac{\partial{f}}{\partial{x}^{i}} \frac{\partial{f}}{\partial{x}^{j}} = \lambda^{\alpha}\tilde{g}^{ij}\frac{\partial{f}}{\partial{x}^{i}} \frac{\partial{f}}{\partial{x}^{j}} = \lambda^{\alpha}\langle df, df\rangle_{\tilde{g}}. $$

The result then follows. □

Proposition A.4

Let (M, g) be a Riemannian manifold, with Riemannian metric g. Let $\tilde {g} = \lambda ^{\alpha } g$, where $\lambda : M \rightarrow (0, \infty )$ is a smooth function, and α ∈ ℝ. Let $P : M \rightarrow {\mathbb{C}}$ be a C² function on M. We then have the following formula for Δ_gP in terms of ${\Delta }_{\tilde {g}}P$

$$ {\Delta}_{g}P = \lambda^{\alpha}{\Delta}_{\tilde{g}}P + \bigg{(}\frac{2\alpha -n\alpha}{2}\bigg{)}\lambda^{(\alpha -1)} \langle dP, d\lambda\rangle_{\tilde{g}}. $$

Proof

In local coordinates (xⁱ) we can write ${\Delta }_{g}P = \frac {1}{\rho }\frac {\partial }{\partial {x^{j}}}\bigg {(} \rho g^{jk}\frac {\partial {P}}{\partial {x^{k}}} \bigg {)}$, where $\rho = \sqrt {det(g_{ij})}$ with g_ij being the components of the metric tensor in (xⁱ) coordinates. It is then easy to see that $\tilde {\rho } = \sqrt {det(\tilde {g}_{ij})} = \lambda ^{n\alpha /2}\rho$.

In these local coordinates, we compute

$$ \begin{array}{@{}rcl@{}} {\Delta}_{g}P &=& \frac{1}{\rho}\frac{\partial}{\partial{x^{j}}}\bigg{(} \rho g^{jk}\frac{\partial{P}}{\partial{x^{k}}} \bigg{)} \\ &=& \frac{1}{\lambda^{-n\alpha/2}\tilde{\rho}} \frac{\partial}{\partial{x^{j}}}\bigg{(} \lambda^{-n\alpha/2}\tilde{\rho}\lambda^{\alpha}\tilde{g}^{jk} \frac{\partial{P}}{\partial{x}^{k}}\bigg{)} \\ &=& \lambda^{\alpha}{\Delta}_{\tilde{g}}P + \frac{1}{\lambda^{-n\alpha/2}\tilde{\rho}}\tilde{\rho} \tilde{g}^{jk}\frac{\partial{P}}{\partial{x^{k}}} \frac{\partial}{\partial{x^{j}}}\bigg{(} \lambda^{\frac{-n\alpha + 2\alpha}{2}} \bigg{)} \\ &=& \lambda^{\alpha}{\Delta}_{\tilde{g}}P + \bigg{(} \frac{2\alpha - n\alpha}{2} \bigg{)} \lambda^{(\alpha - 1)}\tilde{g}^{jk}\frac{\partial{P}}{\partial{x}^{k}} \frac{\partial{\lambda}}{\partial{x}^{j}} \\ &=& \lambda^{\alpha}{\Delta}_{\tilde{g}}P + \bigg{(} \frac{2\alpha - n\alpha}{2} \bigg{)}\lambda^{(\alpha - 1)} \langle dP, d\lambda\rangle_{\tilde{g}}. \end{array} $$

□

Corollary A.5

Assume that the hypotheses of proposition A.4 are satisfied. Suppose we have the bound $|dP|_{\tilde {g}} \leq h_{1}$ and $|{\Delta }_{\tilde {g}}P| \leq h_{2}$, where $h_{i} : M \rightarrow [0, \infty )$ for i = 1, 2. Then

$$ |{\Delta}_{g}P|\leq \lambda^{\alpha}h_{2} + \bigg{|}\frac{2\alpha - n\alpha}{2} \bigg{|} \lambda^{\frac{\alpha - 2}{2}}|d\lambda|_{g}h_{1}. $$

Proof

Using the formula $\langle dP, d\lambda \rangle _{\tilde {g}}=\lambda ^{-\alpha }\langle dP, d\lambda \rangle _{g}$, we write proposition A.4 in a slightly different form:

$$ {\Delta}_{g}P = \lambda^{\alpha}{\Delta}_{\tilde{g}}P + \bigg{(}\frac{2\alpha -n\alpha}{2}\bigg{)}\lambda^{-1} \langle dP, d\lambda\rangle_{g}. $$

Using the above formula, we estimate

$$ \begin{array}{@{}rcl@{}} |{\Delta}_{g}P| &\leq& \lambda^{\alpha}|{\Delta}_{\tilde{g}}P| + \bigg{|}\frac{2\alpha - n\alpha}{2} \bigg{|}\lambda^{-1} |dP|_{g}|d\lambda|_{g} \\ &=& \lambda^{\alpha}|{\Delta}_{\tilde{g}}P| + \bigg{|}\frac{2\alpha - n\alpha}{2} \bigg{|}\lambda^{-1} |dP|_{g}|d\lambda|_{g} \\ &\leq& \lambda^{\alpha}h_{2} + \bigg{|}\frac{2\alpha - n\alpha}{2} \bigg{|}\lambda^{-1} |dP|_{g}|d\lambda|_{g}. \end{array} $$

We can then estimate the |dP|_g term using proposition A.3, and the corollary follows. □

We will need a formula that tells us how the Ricci curvature changes under a conformal transformation. For a proof of this proposition, the reader can consult [1].

Proposition A.6

Let (M, g) be a Riemannian manifold of dimension n, with Riemannian metric g. Let $f\in C^{\infty }(M)$ be a real-valued function and let $\tilde {g} = e^{2f}g$. Let Ric_g and $Ric_{\tilde {g}}$ denote Ricci curvature tensors with respect to g and $\tilde {g}$ respectively. We then have the following formula:

$$ Ric_{\tilde{g}} = Ric_{g} - (n-2)(Hess_{g}(f) - df\otimes df) + ({\Delta}_{g}f - (n-2)|df|_{g}^{2})g. $$

We will be using the above proposition in the following way:

Proposition A.7

Let (M, g) be an n-dimensional Riemannian manifold. Let $Q : M \rightarrow [1, \infty )$ be a smooth function satisfying the following bounds:

• (i) |dQ|_g ≤ CQ^1/4 for some constant C ≥ 0,

• (ii) |Hess_g(Q)|_g ≤ CQ^− 1/2 for some constant C ≥ 0.

Let $\tilde {g} = Q^{-3/2}g = e^{2Log(Q^{\frac {-3}{4}})}g$. Assume that Ric_g ≥ 0. Then, $Ric_{\tilde {g}}\geq -K$, where K ≥ 0 is some constant.

Proof

For f = Log(Q^− 3/4), we compute

$$ Hess_{g}(f) - df\otimes df = \frac{3}{4}\frac{1}{Q^{2}}dQ\otimes dQ - \frac{3}{4}\frac{1}{Q}Hess_{g}(Q), $$

where we used proposition A.1(ii). Applying the assumptions (i) and (ii), we obtain

$$ |Hess_{g}(f) - df\otimes df|_{g} \leq CQ^{-3/2}. $$

Appealing to proposition A.1(i), we have

$$ {\Delta}_{g}f = -\frac{3}{4}Q^{-2}|dQ|^{2}_{g} - \frac{3}{4}Q^{-1}{\Delta}_{g}Q. $$

Using the assumptions (i) and (ii) together with proposition A.1(iii), we get

$$ |{\Delta}_{g}f| \leq CQ^{-3/2}. $$

Keeping in mind that $\langle \cdot ,\cdot \rangle _{g}=Q^{3/2}\langle \cdot ,\cdot \rangle _{\tilde {g}}$, the result then follows from proposition A.6 and the assumption that Ric_g ≥ 0. □

Proposition A.8

Let (M, g) be an n-dimensional Riemannian manifold. Let $Q : M \rightarrow [1, \infty )$ be a smooth function satisfying the following bounds:

• (i) |dQ|_g ≤ CQ^3/4 for some constant C ≥ 0,

• (ii) |Hess_g(Q)|_g ≤ CQ^1/2 for some constant C ≥ 0.

Let $\tilde {g} = Q^{-1/2}g = e^{2Log(Q^{\frac {-1}{4}})}g$. Assume that Ric_g ≥ 0. Then, $Ric_{\tilde {g}}\geq -K$, where K ≥ 0 is some constant.

Proof

Using f = Log(Q^− 1/4) and following the same pattern as in the proof of proposition A.7, we get the estimates

$$ |Hess_{g}(f) - df\otimes df|_{g} \leq CQ^{-1/2}, $$

$$ |{\Delta}_{g}f| \leq CQ^{-1/2}. $$

Keeping in mind the rule $\langle \cdot ,\cdot \rangle _{g}=Q^{1/2}\langle \cdot ,\cdot \rangle _{\tilde {g}}$ and the assumption Ric_g ≥ 0, we get the result from proposition A.6. □