The m-accretivity of covariant Schrödinger operators with unbounded drift

Abstract

We work in the context of a geodesically complete Riemannian n-manifold M with a Hermitian vector bundle $\mathcal {V}$ over M, equipped with a metric covariant derivative $\nabla$. We study the operator $H:=\nabla ^{\dagger }\nabla +\nabla _{X}+V$, where $\nabla ^{\dagger }$ is the formal adjoint of $\nabla$, the symbol $\nabla _{X}$ stands for the action of $\nabla$ along a smooth vector field X on M, and V is a locally bounded section of the endomorphism bundle ${\text {End}}\mathcal {V}$. We show that under certain conditions on X and V, the closure $\overline{H|_{C_{c}^{\infty }(\mathcal {V})}}$ of $H|_{C_{c}^{\infty }(\mathcal {V})}$ in $L^p(\mathcal {V})$, where $1<p<\infty$, is a maximal accretive operator. We also show that $\overline{H|_{C_{c}^{\infty }(\mathcal {V})}}$ coincides with the “maximal” realization of H in $L^p(\mathcal {V})$.

Appendices

A Accretivity and m-accretivity of operators in Banach spaces

In this section, we briefly review some abstract notions. A linear operator T on a Banach space ${\mathscr {B}}$ is called accretive if

$$\begin{aligned} \Vert (\xi +T)u\Vert _{{\mathscr {B}}}\ge \xi \Vert u\Vert _{{\mathscr {B}}}, \end{aligned}$$

for all $\xi >0$ and all $u\in {\text {Dom}}(T)$. We note that instead of “T is accretive,” some authors say “$-T$ is dissipative.” If T is a densely defined accretive operator, then T is closable and its closure $\overline{T}$ is also accretive; see Proposition II.3.14 in [4]. There is another description of accretivity. Denoting by ${\mathscr {B}}^{*}$ the dual space of ${\mathscr {B}}$ and using Hahn–Banach theorem, for every $u\in {\mathscr {B}}$ there exists $u^*\in {\mathscr {B}}^*$ such that $\langle u,u^*\rangle =\Vert u\Vert _{{\mathscr {B}}}^2=\Vert u^*\Vert _{{\mathscr {B}}^*}^2$. Here, $\langle u,u^*\rangle$ stands for the evaluation of the functional $u^*$ at u. For every $u \in {\mathscr {B}}$, define

$$\begin{aligned} {\mathscr {J}}(u) :=\{u^*\in {\mathscr {B}}^*:\langle u,u^*\rangle =\Vert u\Vert _{{\mathscr {B}}}^2=\Vert u^*\Vert _{{\mathscr {B}}^*}^2\}. \end{aligned}$$

By Proposition II.3.23 of [4], an operator T is accretive if and only if for every $u\in {\text {Dom}}(T)$ there exists $j(u)\in {\mathscr {J}}(u)$ such that

$$\begin{aligned} {\text {Re}}\langle Tu,j(u)\rangle \ge 0. \end{aligned}$$

A (densely defined) operator T on ${\mathscr {B}}$ is called m-accretive if it is accretive and $\xi +T$ is surjective for all $\xi >0$. A (densely defined) operator T on ${\mathscr {B}}$ is called essentiallym-accretive if it is accretive and $\overline{T}$ is m-accretive. As indicated in Theorem II.3.15 of [4], the negative $-T$ of an m-accretive operator T on ${\mathscr {B}}$ generates a strongly continuous contraction semigroup. There is a link between m-accretivity and self-adjointness of operators on Hilbert spaces: T is a self-adjoint and nonnegative operator if and only if T is symmetric, closed, and m-accretive; see Problem V.3.32 in [10].

B Proof of Lemma 2

Let $U\subset M$ be a coordinate chart around $x\in M$ which admits a smooth orthonormal frame $e_1,\dots ,e_m\in C^{\infty }(U;\mathcal {V})$, where $m={\text {rank}}\mathcal {V}$. Denote the local coordinates in U as $x_1,\cdots ,x_n$ and the corresponding basis of tangent vectors as $\partial _i$, $1 \le i \le n$. Then, as seen in section 3.3.1 of [16], there exists a (necessarily unique) matrix of 1-forms

$$\begin{aligned} F=(F^{s}_{t})\in {\mathrm {Mat}}\big (C^{\infty }(U; T^*M); m\times m\big ) \end{aligned}$$

such that (using Einstein summation convention)

$$\begin{aligned} \nabla _{|U} (u^{s}{e_s}) = (\partial _{r} u^{s}) \mathrm{d}x^r\otimes e_s + u^{t}F^{s}_{rt} \mathrm{d}x^{r}\otimes e_s, \end{aligned}$$

(65)

for all $u\in [W^{1,1}_{{\text {loc}}}(U)]^m\cap [L_{{\text {loc}}}^{\infty }(U)]^m$.

Before starting the actual proof of the lemma, we record a preliminary formula. Recalling the definition $\omega _u(Y) := {\text {Re}}\langle \nabla _Yu, u\rangle$, where Y is a smooth vector field, and using (65), we have

$$\begin{aligned}&\omega _u(\partial _i) = {\text {Re}}\langle \nabla _{\partial _i}u, u\rangle = {\text {Re}}\langle \big (u^jF_{ij}^k + \partial _i(u^k)\big )e_k, u^le_l\rangle \\&\quad = {\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^l}\delta _{kl}\bigg )= \sum _{k=1}^{m}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ), \end{aligned}$$

which leads to the following local formula for $\omega _u$:

$$\begin{aligned} \omega _u = \sum _{k=1}^{m}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg )\mathrm{d}x^i. \end{aligned}$$

(66)

We now prove the properties (i) and (ii). These are the only two parts of the lemma where we use the assumption that $\nabla$ is a metric covariant derivative. In this case, the local connection 1-form $F=(F^{s}_{t})$ is skew-adjoint, that is, $\overline{F^{t}_{s}}=-F^{s}_{t}$, and formula (66) simplifies to

$$\begin{aligned} \omega _u(\partial _{i}) = \sum _{k=1}^{m}{\text {Re}}\big (\partial _i(u^k)\overline{u^k}\big ). \end{aligned}$$

(67)

As the component functions $u^s$, $1\le s\le m$, are complex-valued with $v^s:={\text {Re}}u^s$ and $w^s:={\text {Im}}u^s$, we may view the fiberwise norm $|\cdot |$ in $\mathcal {V}$ as

$$\begin{aligned} |u|^2=\displaystyle \sum _{s=1}^{m}((v^s)^2+(w^s)^2). \end{aligned}$$

We can now use Theorem A of [12] to conclude that the composition of $|\cdot |$ and u satisfy $|u|\in W^{1,1}_{{\text {loc}}}(U)$. Again, by Theorem A of [12] we have for all $1\le i\le n$:

$$\begin{aligned} \partial _{i}|u|=\frac{1}{|u|}\sum _{s=1}^{m}(v^s\partial _{i}v^s+w^s\partial _{i}w^s), \end{aligned}$$

(68)

for all x such that $u(x)\ne 0$, and $\partial _{i}|u|=0$ for all x such that $u(x)=0$. Note that the right-hand side of (68) is the same as

$$\begin{aligned} \frac{1}{|u|}\sum _{s=1}^{m}{\text {Re}}\big (\partial _i(u^s)\overline{u^s}\big ), \end{aligned}$$

which together with (67) leads to

$$\begin{aligned} \partial _{i}|u|=\frac{1}{|u|}\omega _u(\partial _{i}) =\frac{1}{|u|}{\text {Re}}\langle \nabla _{\partial _{i}} u, u\rangle , \end{aligned}$$

for all x such that $u(x)\ne 0$, and $\partial _{i}|u|=0$ for all x such that $u(x)=0$. This shows parts (i) and (ii) of the lemma.

For part (iii), see the proof of equation (1.34) in “Appendix C” of [23]. We now turn to part (iv). We start by expanding the right-hand side:

$$\begin{aligned}&\langle \omega _u, \omega _u\rangle \\&\quad = \bigg \langle \sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg )\mathrm{d}x^i , \sum _{l}{\text {Re}}\bigg (\big (u^tF_{rt}^l + \partial _r(u^l)\big )\overline{u^l}\bigg )\mathrm{d}x^r\bigg \rangle \\&\quad = \sum _{k, l}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ) {\text {Re}}\bigg (\big (u^tF_{rt}^l + \partial _r(u^l)\big )\overline{u^l}\bigg )g^{ir}, \end{aligned}$$

where $g^{ir} = \langle \mathrm{d}x^i, \mathrm{d}x^r\rangle$ denotes the metric on $T^*M$.

We now expand the left-hand side:

$$\begin{aligned}&{\text {Re}}\langle \omega _u \otimes u, \nabla u\rangle \\&\quad ={\text {Re}}\bigg \langle \sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg )\mathrm{d}x^i \otimes u^le_l, (\partial _ru^s + u^tF_{rt}^s)\mathrm{d}x^r\otimes e_s\bigg \rangle \\&\quad = {\text {Re}}\bigg (\sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg )u^l \bigg (\overline{\partial _ru^s + u^tF_{rt}^s}\bigg )g^{ir}\delta _{ls}\bigg ) \\&\quad = {\text {Re}}\bigg (\sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg )\sum _lu^l \bigg (\overline{\partial _ru^l + u^tF_{rt}^l}\bigg )g^{ir}\bigg ). \end{aligned}$$

Keeping in mind the properties

$$\begin{aligned} {\text {Re}}(\alpha z_1)=\alpha ({\text {Re}}z_1),\quad {\text {Re}}(z_1\overline{z_2})= {\text {Re}}(\overline{z_1}{z_2}), \end{aligned}$$

which hold for all $\alpha \in {\mathbb {R}}$ and $z_1,z_2\in {\mathbb {C}}$, we obtain

$$\begin{aligned}&{\text {Re}}\bigg (\sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ) \sum _lu^l \bigg (\overline{\partial _ru^l + u^tF_{rt}^l}\bigg )g^{ir}\bigg ) \\&\quad =\sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ) \sum _l{\text {Re}}\bigg (u^l \bigg (\overline{\partial _ru^l + u^tF_{rt}^l}\bigg )g^{ir}\bigg ) \\&\quad = \sum _{k}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ) \sum _l{\text {Re}}\bigg (\overline{u^l} \bigg (\partial _ru^l + u^tF_{rt}^l\bigg )\bigg )g^{ir} \\&\quad = \sum _{k, l}{\text {Re}}\bigg (\big (u^jF_{ij}^k + \partial _i(u^k)\big )\overline{u^k}\bigg ) {\text {Re}}\bigg ( \bigg (\partial _ru^l + u^tF_{rt}^l\bigg )\overline{u^l}\bigg )g^{ir}. \end{aligned}$$

Therefore, $\langle \omega _u, \omega _u\rangle ={\text {Re}}\langle \omega _u \otimes u, \nabla u\rangle$, and this concludes the proof of part (iv). Part (v) is a direct consequence of part (iv) of this lemma. Part (vi) can be shown similarly as part (iv). $\square$