Dirac cohomology on manifolds with boundary and spectral lower bounds

Abstract

Along the lines of the classic Hodge–De Rham theory a general decomposition theorem for sections of a Dirac bundle over a compact Riemannian manifold is proved by extending concepts as exterior derivative and coderivative as well as elliptic absolute and relative boundary conditions for both Dirac and Dirac Laplacian operators. Dirac sections are shown to be a direct sum of harmonic, exact and coexact spinors satisfying alternatively absolute and relative boundary conditions. Cheeger’s estimation technique for spectral lower bounds of the Laplacian on differential forms is generalized to the Dirac Laplacian. A general method allowing to estimate Dirac spectral lower bounds for the Dirac spectrum of a compact Riemannian manifold in terms of the Dirac eigenvalues for a cover of 0-codimensional submanifolds is developed. Two applications are provided for the Atiyah–Singer operator. First, we prove the existence on compact connected spin manifolds of Riemannian metrics of unit volume with arbitrarily large first non zero eigenvalue, which is an already known result. Second, we prove that on a degenerating sequence of oriented, hyperbolic, three spin manifolds for any choice of the spin structures the first positive non zero eigenvalue is bounded from below by a positive uniform constant, which improves an already known result.

Appendix: Some results about second order boundary value problems

Proposition 16

Let the function $a=a(u)$ be a non trivial solution of the linear second order boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -a^{\prime \prime }+qa=0 \\ a^{\prime }(m_0)+\alpha a(m_0)=0, \end{array}\right. } \end{aligned}$$

(128)

where $q\in C^\infty ([m_0,m_1])$ is a smooth function satisfying $q>k^2$ for constants $k, \alpha \in \textbf{R}$ such that $k>0$ and $\alpha \le k.$ Then,  for the unique solution v of the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll}-v^{\prime \prime }+k^2v=0\\ v(m_0)=a(m_0)\\ v^{\prime }(m_0)=a^{\prime }(m_0) \end{array}\right. } \end{aligned}$$

(129)

the following inequality holds on $[m_0,m_1]$:

$$\begin{aligned} \frac{a^{\prime }}{a}\ge \frac{v^{\prime }}{v}. \end{aligned}$$

(130)

In particular

$$\begin{aligned} \liminf _{m_1\rightarrow +\infty }\frac{a^\prime (m_1)}{a(m_1)}\ge \frac{1}{2}k. \end{aligned}$$

(131)

Proof

Without loss of generality we can prove the inequality on $[m_0,m_1[$ and choose $m_0:=0$ and $m_1=+\infty .$ We need to distinguish several cases:

case 0::

$a(0)=0$ never occurs. In fact, both cases $a(0)=0$ and $a^{\prime }(0)=0$ are excluded by the assumption on the non triviality of a and by the existence and uniqueness theorem for the solutions of ordinary differential equations.

case 1::

$a(0)>0.$

Since $a^{\prime }(0)=-\alpha a(0)>-ka(0),$ we obtain $v(u)=a(0)\cosh (ku)+\frac{a^{\prime }(0)}{k}\sinh (ku)>0$ $\forall u\in [0,+\infty [.$ With $w:=a^{\prime }v-av^{\prime }$ it follows $w^{\prime }=(q-k^2)av,$ $w(0)=0$ and $w^{\prime }(0)=(q(0)-k^2)v^2(0)>0.$ So, $\varepsilon _1:=\sup \left\{ u\in ]0,+\infty [\;|\;w^{\prime }>0\;\text {on}\;]0,u[\right\}$ must belong to $]0,+\infty ].$ If $\varepsilon _1<+\infty ,$ then by continuity $w^{\prime }(\varepsilon _1)=0.$

Analogously, since $a(0)>0,$ $\varepsilon _2:=\sup \left\{ u\in ]0,+\infty [\;|\;a>0\;\text {on}\;]0,u[\right\}$ must be in $]0,+\infty ].$ If $\varepsilon _2<+\infty ,$ then by continuity $a(\varepsilon _2)=0.$ Set $\varepsilon :=\min \{\varepsilon _1,\varepsilon _2\}.$ On $[0,\varepsilon [$ one has $w\ge 0,$ i.e. $\frac{a^{\prime }}{a}\ge \frac{v^{\prime }}{v},$ being a and v positive. Integrating both sides of this inequality , one gets $a\ge v$ on $[0,\varepsilon [.$ So, on this interval one has $w^{\prime }(u)=(q-k^2)a(u)v(u)\ge (q(u)-k^2)v^2(u)=(q(u)-k^2)a^2(0)\cosh ^2(ku)$ and $a(u)\ge v=a(0)\cosh (ku).$ Assume now that $\varepsilon <\infty .$ There are two possibilities: if $\varepsilon =\varepsilon _1,$ then by continuity $w^{\prime }(\varepsilon _1)=(q(\varepsilon _1)-k^2)a^2(0)\cosh ^2(k\varepsilon _1)>0;$ if $\varepsilon =\varepsilon _2,$ again by continuity $a(\varepsilon _2)\ge a(0)\cosh (k\varepsilon _2)>0.$ In both cases there is a contradiction, so it must be $\varepsilon =\infty .$ We therefore come to the conclusion that $\frac{a^{\prime }}{a}\ge \frac{v^{\prime }}{v}$ on $[0,+\infty [.$

case 2::

$a(0)<0.$

We set $\bar{a}:=-a$ and $\bar{v}:=-v.$ Case 1 leads to $\frac{\bar{a}^{\prime }}{\bar{a}}\ge \frac{\bar{v}^{\prime }}{\bar{v}}$ on $[0,+\infty [,$ which means $\frac{a^{\prime }}{a}\ge \frac{v^{\prime }}{v}$ on the same interval.

By solving the initial value problem for v,  we can determine v and $v^\prime$ explicitly:

$$\begin{aligned} \begin{aligned} v(u)&=a(m_0)\cosh (k(u-m_0))+\frac{a^{\prime }(m_0)}{k}\sinh (k(u-m_0)) \\&\\ v^{\prime }(u)&=ka(m_0)\sinh (k(u-m_0))+a^{\prime }(m_0)\cosh (k(u-m_0)). \\ \end{aligned} \end{aligned}$$

(132)

Since $v(u)\ne 0$ for $u\in [m_0,m_1]$ we can write:

$$\begin{aligned} \frac{v^{\prime }(u)}{v(u)}=\,k\,\frac{a(m_0)\tanh (k(u-m_0))+\frac{1}{k}a^{\prime }(m_0)}{a(m_0) +\frac{1}{k}a^{\prime }(m_0)\tanh (k(u-m_0))}. \end{aligned}$$

(133)

We insert the boundary condition $a^{\prime }(m_0)+\alpha a(m_0)=0$ and simplify by $a(m_0)\ne 0$:

$$\begin{aligned} \frac{v^{\prime }(u)}{v(u)}=\,k\frac{\tanh (k(u-m_0))-\frac{\alpha }{k}}{1-\frac{\alpha }{k}\tanh (k(u-m_0))}. \end{aligned}$$

(134)

Since $k>\alpha ,$ we obtain

$$\begin{aligned} \lim _{u\rightarrow +\infty }\frac{v^{\prime }(u)}{v(u)}=k \end{aligned}$$

(135)

and the inequality (131) follows from the estimate (130).

$\square$

Proposition 17

Let the function $a=a(u)$ be a non trivial solution of the linear second order boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -a^{\prime \prime }+qa=0 \\ a(m_0)=0, \end{array}\right. } \end{aligned}$$

(136)

where $q\in C^\infty ([m_0,m_1])$ is a smooth function satisfying $q>k^2$ for a constant $k>0.$ Then,  for the unique solution v of the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll}-v^{\prime \prime }+k^2v=0\\ v(m_0)=0\\ v^{\prime }(m_0)=a^{\prime }(m_0) \end{array}\right. } \end{aligned}$$

(137)

the following inequality holds on $]m_0,m_1]$:

$$\begin{aligned} \frac{a^{\prime }}{a}\ge \frac{v^{\prime }}{v}. \end{aligned}$$

(138)

There exist $\delta >m_0$ such that

$$\begin{aligned} \begin{aligned} a(u)&\ge a(\delta )e^{\frac{k(u-\delta )}{2}}>0\quad (a^{\prime }(m_0)>0)\\ a(u)&\le a(\delta )e^{\frac{k(u-\delta )}{2}}<0\quad (a^{\prime }(m_0)<0). \end{aligned} \end{aligned}$$

(139)

Proof

Without loss of generality we can prove the inequality on $[m_0,m_1[$ and choose $m_0:=0$ and $m_1=+\infty .$ We need to distinguish several cases:

case 0::

$a^{\prime }(0)=0$ never occurs. Cf. case 0 in the proof of Proposition 16.

case 1::

$a^{\prime }(0)>0.$

There exist a $\delta >0$ small enough such that $a^{\prime }(\delta )>0$ and $a(\delta )>0.$ Note that $\alpha :=-\frac{a^{\prime }(\delta )}{a(\delta )}< k.$ We can continue by applying Proposition 16 and obtain the result stated.

case 2::

$a(0)<0.$

Analogously to case 2 in the proof of Proposition 16.

$\square$