# Witten Deformation and Its Application toward Morse Inequalities

Fu-Hsuan Ho

###### Abstract

In this undergraduate thesis, we present an analytical proof of the Morse
inequalities for closed smooth $n$-manifolds following Witten’s approach.
Using techniques from PDE theory, the proof is reduced to study the
eigenspaces and eigenvalues of harmonic oscillators on $\mathbb{R}^{n}$.

##  1 Introduction

Morse theory has been a branch of differential topology that provides a direct
way to understand the topology of smooth manifolds by studying the smooth
functions on them. In the introduction of the classical book [Mil63], Milnor
discussed a toy model of this idea: assume $M=T^{2}$ is a torus tangent to a
plane $V$ and $M^{a}$ be the set of points $p\in M$ so that $f(p)\leq a$. and
let $f:M\rightarrow\mathbb{R}$ be the height above $V$. Milnor pointed out
that the homotopy type changes exactly at the critical points of $f$. These
critical points are all “non-degenerate” (see section 2 for definition), and
near each critical point $p$ one can choose proper coordinate $(x,y)$ so that
$f=f(p)\pm x^{2}\pm y^{2}$. Note that the number of minus signs in the
expression for $f$ at each point is the dimension of the cell we must attach
to go from $M^{a}$ to $M^{b}$, where $a<f(p)<b$. This gives us a motivation to
study critical points of smooth functions on manifolds.

Another formula also shows the deep connection of the topology and critical
points of smooth functions on a manifold, that is, the morse inequalities. Let
$M$ be a $n$-dimensional closed and smooth manifold. Given $f\in
C^{\infty}(M)$, we say $f$ is a Morse function if its critical points are non-
degenerate (these terms will be defined in section 2). Morse inequalities give
a series of upper bound related to the Betti number of $M$. The full statement
is given in the following:

######  Main Theorem 1.1 (Morse Inequalities).

Let $M$ be a closed smooth $n$-dimensional manifold. Suppose $f\in
C^{\infty}(M)$ is a Morse function. Then we have

  * •

Weak Morse Inequalities: for each $0\leq q\leq n$,

$\displaystyle b_{q}\leq m_{q}$ (1.1)

  * •

Strong Morse Inequalities: for each $0\leq q\leq n$

$\displaystyle\sum^{q}_{j=0}(-1)^{q-j}b_{j}\leq\sum_{j=0}^{q}(-1)^{q-j}m_{j},$
(1.2)

and the equality holds when $q=n$, i.e.,

$\displaystyle\sum^{n}_{j=0}(-1)^{n-j}b_{j}=\sum_{j=0}^{n}(-1)^{n-j}m_{j},$
(1.3)

where we denote $m_{j}$ as the cardinality of the set of critical points of
order $j$ and $b_{j}$ be the $j$-th Betti number of $M$ (see (2.8)), for any
$j\in\\{0,\ldots,q\\}$.

Note that (1.3) gives a way to derive $\chi(M)$ by calculating the alternating
sum of numbers of critical points up to index $n$, since the definition of
Euler characteristic number $\chi(M)$ of $M$ is exactly the alternating sum of
Betti numbers.

A topological proof may be founded in [Mil63]. However in 1982, Witten [Wit82]
discovered the first analytic approach to the Morse inequalities through
considering the de Rham complex the deformed exterior derivative
$d_{t}=e^{-tf}de^{tf}$, i.e., the so-called Witten deformation. Inspired by
physics, he adopted the semi-classical analysis of the eigenvalues of some
properly chosen Schrödinger operators on $M$ to attack this problem. His idea
was immediately absorbed by many mathematicians and was turned into
mathematically rigor papers, for example, [HS85]. Another reference is
[CFKS87].

In this paper, we try to provide a rather direct and detailed proof that can
be accessed by undergraduate students who have the basic knowledge of
geometry. In section 2, we recall the basic notations and preliminaries that
are needed in this thesis. Then in section 3, we introduce to the Witten
Laplacian and reduce the proof of Morse inequality to studying the dimension
of the eigenspace of Witten Laplacian, and we do this work in section 4. The
strategy is constructing enough linear independent $q$-forms on $M$ (see
subsection 4.2). Section 5 is devoted to calculating the eigenspace of Witten
Laplacian. While the calculation is essential to our proof, the method we
adopt to find eigenvalues is rather standard and somehow tedious, and thus we
put it in the last section.

The last thing to be notice is that although we only consider the case of
closed manifolds, this method can be extended to more general cases, for
example, manifolds with boundary. Also, the idea of constructing linearly
independent elements of operator’s eigen-space can be extended to understand
other operators we are interested in.

##  2 Notations and Preliminaries

From now on, we fix $M$ to be a smooth and closed n-dimensional manifold. Let
$C^{\infty}(M)$ be the set of real-valued smooth functions of $M$. A critical
point of a given smooth function $f$ is a point $p$ so that for any local
coordinate chart $(U,\varphi,x)$ near $p$,

$\displaystyle\frac{\partial f}{\partial x_{1}}(p)=\cdots=\frac{\partial
f}{\partial x_{n}}(p)=0.$ (2.1)

This definition is independent of our choice of coordinate chart near $p$. For
any other local coordinate chart $(V,\phi,y)$ near $p$, using the change of
coordinate formula we get,

$\displaystyle\frac{\partial f}{\partial y_{i}}(p)=\frac{\partial f}{\partial
x_{j}}(p)\frac{\partial(\phi\circ\varphi^{-1})}{\partial
y_{i}}(p)=0,\quad\forall i=1,\ldots,n.$ (2.2)

A critical point $p\in M$ of $f$ is called non-degenerate if and only if the
matrix

$\displaystyle\left(\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}(p)\right)$ (2.3)

is invertible. We quickly check that this definition is independent of the
choice of the coordinate system: let $(U,\varphi,x)$ be a local chart
satisfying (2.3). Assume $(V,\phi,y)$ is any other local chart with non-empty
intersection with $U$. Then

$\displaystyle\frac{\partial f}{\partial x_{i}\partial x_{j}}(p)$
$\displaystyle=\sum_{k,l}\frac{\partial(\phi\circ\varphi)^{-1}_{k}}{\partial
x_{i}}(p)\frac{\partial f}{\partial y_{k}\partial
y_{l}}(p)\frac{\partial(\phi\circ\varphi)^{-1}_{l}}{\partial x_{j}}(p).$ (2.4)

Because $\phi\circ\varphi^{-1}$ is diffeomorphism, the matrix

$\left(\frac{\partial(\phi\circ\varphi)^{-1}_{k}}{\partial x_{i}}(p)\right)$

is invertible. Therefore, from 2.4 we must derive that

$\left(\frac{\partial^{2}f}{\partial y_{k}\partial y_{l}}(p)\right)$

is also invertible. With the terminologies prepared above, we now give the
definition of Morse function:

######  Definition 2.1.

We say smooth function $f\in C^{\infty}(M)$ is a Morse function if its
critical points are all non-degenerate.

If $p\in M$ is a critical point, we define a symmetric bilinear form

$d^{2}f:T_{p}M\times T_{p}M\rightarrow\mathbb{R}$

called the Hessian of $f$ at $p$ on $T_{p}M$, the tangent space of $M$ at $p$.
If $X,Y\in T_{p}(M)$, we may extend them to smooth vector fields
$\tilde{X},\tilde{Y}$ with $\tilde{X}(p)=X$ and $\tilde{Y}(p)=Y$. Define

$d^{2}f(X,Y)=\tilde{X}(\tilde{Y}(f))(p).$

In the following we will show that $d^{2}f$ is a well-defined symmetric
bilinear functional: note $\tilde{X}(p)$ equals to $X$. Indeeds, by the
formula

$\tilde{X}(p)(\tilde{Y}(f))-\tilde{Y}(p)(\tilde{X}(f))=df[\tilde{X},\tilde{Y}](p)=0,$

$d^{2}f$ is symmetric. Also by the formula above, we have

$X(\tilde{Y}(f))=\tilde{X}(p)(\tilde{Y}(f))=\tilde{Y}(p)(\tilde{X}(f))=Y(\tilde{X}(f))$

and thus $d^{2}f$ is independent of how we extend $X$ and $Y$, i.e., it is
well-defined.

Locally on a chart $(U,\varphi,x)$ near $p$, we may write
$X=\sum_{i=1}^{n}a_{i}\frac{\partial}{\partial x_{i}}|_{p}$ and
$Y=\sum_{j=1}^{n}b_{j}\frac{\partial}{\partial x_{j}}|_{p}$. Locally, we may
extend $Y$ by $\tilde{Y}=\sum_{j=1}^{n}b_{j}\frac{\partial}{\partial x_{j}}$.
Then

$\displaystyle
d^{2}f(X,Y)=X(\tilde{Y}f)(p)=X\left(\sum_{j=1}^{n}b_{j}\frac{\partial}{\partial
x_{j}}\right)=\sum_{i=1,j=1}^{n}a_{i}b_{j}\frac{\partial^{2}f}{\partial
x_{i}\partial x_{j}}(p);$ (2.5)

so the matrix

$\displaystyle\left(\frac{\partial^{2}f}{\partial x_{k}\partial
x_{l}}(p)\right)$ (2.6)

represents the bilinear form $d^{2}f$ under the basis
$\\{\frac{\partial}{\partial x_{i}}|_{p}\\}_{i=1}^{n}$.

Now, we mention the index of $d^{2}f$. For any bilinear form $B$ defined on a
vector space $\mathbf{V}$, the index of $B$ is defined to be the maximal
dimension of any subspace $\mathbf{W}$ on which $B$ is negative definite. When
we say the phrase “the index of $f$ at $p$”, it always refers to the index of
$d^{2}f$ on $T_{p}M$. We can check that the index of $f$ at $p$ is the number
of non-negative eigenvalues of the matrix (2.6). The lemma of Morse, which
will be stated in the following (for the proof, see [Mil63, Lemma 2.2]), shows
that the index of $f$ at $p$ characterizes the local behavior of $f$ near $p$:

######  Lemma 2.2 (Morse).

Let $f$ be a smooth function. Assume $p\in M$ is a critical point of $f$ on
$M$. Then we can find a local chart $(U,\varphi,x)$ near $p$ with
$\varphi(p)=0$ so that

$f\circ\varphi^{-1}(x)=-\frac{1}{2}x^{2}_{1}-\cdots-\frac{1}{2}x_{l}^{2}+\frac{1}{2}x_{l+1}^{2}+\cdots+\frac{1}{2}x_{n}^{2},$

where $l$ is the index of $f$ at $p$.

From now on, we always assume $f$ to be a fixed Morse function on $M$. We
introduce the notation

$\text{Crit}(f)$

as the set of critical points of $f$. By Morse lemma, the set $\text{Crit}(f)$
must be discrete, and it turns out to be finite by compactness of $M$. We also
introduce the notation

$\text{Crit}(f;j)$

to represent the set of critical points on which the index of $f$ equals to
$j$, and we define

$m_{j}=\\#\text{Crit}(f;j)$

to be the cardinality of $\text{Crit}(f;j)$.

###  2.1 de Rham Cohomology

In this subsection, we discuss the basic properties of de Rham cohomology. The
readers may consult Bott and Tu’s book [BT82] for more information.

Denote $\Omega^{0}(M)$ be the space of smooth functions. For any
$k=1,\ldots,n$, let $\Lambda^{k}T^{*}M$ be the bundle of $k$ forms:

$\Lambda^{k}T^{*}M=\coprod_{p\in M}\Lambda^{k}T^{*}_{p}M.$

Fix $p\in M$ and let $(U,\varphi,x)$ be any local chart defined near $p$.
Assume $\\{dx_{1},\ldots,dx_{n}\\}$ is the basis for $T^{*}_{p}M$. Then the
fibre of $\Lambda^{k}T^{*}M$ at $p$ is the vector space
$\Lambda^{k}T^{*}_{p}M$ spanned by

$\\{dx_{i_{1}}\wedge\cdots\wedge d_{i_{k}}:1\leq i_{1}<\cdots<i_{k}\leq n\\}.$

A smooth section of $\Lambda^{k}T^{*}M$ is called a smooth $k$-forms. We
denote

$\Omega^{k}(M)=\Gamma\left(\Lambda^{k}T^{*}M\right)$

as the vector space of smooth $k$-forms. On any local chart $(U,\varphi,x)$, a
smooth $k$-form $\omega$ can be written as

$\displaystyle\omega=\sum_{i_{1}<\cdots<i_{k}}\omega_{i_{1}\cdots
i_{k}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{k}},$ (2.7)

where $\omega_{i_{1}\cdots i_{k}}$ is a smooth function defined on $U$. Once
again, the local representation (2.7) is independent of choice of local chart.
Suppose on $(\tilde{U},\tilde{\varphi},\tilde{x})$ with $\tilde{U}\cap
U\neq\emptyset$. Then there exists

The exterior derivative $d$ is defined to be the unique real linear mapping

$d:\Omega^{q}(M)\rightarrow\Omega^{q+1}(M)$

sending smooth $q$-forms to $q+1$-forms, for all $q=0,\ldots,n$, satisfying
the following properties

  1. (i)

$df$ is the differential of smooth function for any $f\in\Omega^{0}(M)$.

  2. (ii)

$d^{2}f=0$ for any $f\in\Omega^{0}(M)$.

  3. (iii)

$d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^{p}\alpha\wedge d\beta$ where
$\alpha$ is any $p$-form and $\beta$ is any $r$-form.

If there is a need to distinguish those $d$s acting on different form, we will
add an upper index $(q)$ and write $d^{(q)}$.

The complex

$0\rightarrow\Omega^{0}(M)\xrightarrow{d}\Omega^{1}(M)\xrightarrow{d}\cdots\xrightarrow{d}\Omega^{n}(M)\rightarrow
0.$

together with the exterior derivative $d$ is called the de Rham complex on
$M$.

Note $d^{2}$ always vanish. We verify this in the following: let
$\omega\in\Omega^{q}$. Then on any local chart $(U,\varphi,x)$, we can write

$\omega=\sum_{i_{1}<\cdots<i_{q}}\omega_{i_{1}\cdots
i_{q}}dx^{i_{1}}\wedge\cdots\wedge dx^{i_{q}}.$

Applying $d$ twice on $\omega$, we get

$\displaystyle
d^{2}\omega=\sum_{i,j=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial^{2}\omega_{i_{1}\cdots
i_{q}}}{\partial x_{i}\partial x_{j}}dx^{i}\wedge dx^{j}\wedge
dx^{i_{1}}\wedge\cdots\wedge dx^{i_{q}}$
$\displaystyle=\sum_{i<j}\sum_{i_{1}<\cdots<i_{q}}\left(\frac{\partial^{2}\omega_{i_{1}\cdots
i_{q}}}{\partial x_{i}\partial x_{j}}-\frac{\partial^{2}\omega_{i_{1}\cdots
i_{q}}}{\partial x_{j}\partial x_{i}}\right)dx^{i}\wedge dx^{j}\wedge
dx^{i_{1}}\wedge\cdots\wedge dx^{i_{q}}=0$

since $\omega_{i_{1}\cdots i_{q}}$ is a smooth function and hence we can
change the order of derivations.

Thus, it is able to define the $j$-th de Rham cohomology

$H^{j}(M)=\frac{Ker(d:\Omega^{j}(M)\rightarrow\Omega^{j+1}(M))}{Im(d:\Omega^{j-1}(M)\rightarrow\Omega^{j}(M))}$

for all $j=0,\cdots,n$. Since our manifold $M$ is has finite “good cover” (in
the sense of Bott and Tu), $H^{j}(M)$ is a finite dimensional (real) vector
space (the readers may consult their book [BT82, Proposition 5.3.1]). As a
result, it is able to define

$\displaystyle b_{j}=\dim H^{j}(M)$ (2.8)

called the $j$-th Betti number for all $j=0,\cdots,n$.

Fix $g$ to be a Riemannian metric on the tangent bundle $TM$. It also induces
a bundle metric pointwisely on $\alpha,\beta\in\Lambda^{r}T_{p}^{*}M$
(independent of basis we chosen):

$g(\alpha,\beta)=g^{i_{1}j_{1}}\cdots g^{i_{k}j_{k}}\alpha_{i_{1}\ldots
i_{k}}\beta_{j_{1}\ldots j_{k}}.$

which we also denoted by $g$. Let $dv_{g}$ be the density on $M$ induced by
$g$ so that for any chart $(U,\phi,x)$, $w\in C_{0}^{\infty}(U)$,

$\int_{M}w\,dv_{g}=\int_{\phi(U)}w\circ\phi^{-1}(x)\sqrt{|\det(\phi_{*}g)|}dx_{1}\wedge\cdots\wedge
dx_{n}.$

We remark that if $M$ is orientable, then it is no need to take absolute value
above. Through the density $dv_{g}$, we are then able to defined a global
$L^{2}$ inner product by

$(\alpha,\beta):=\int_{M}g(\alpha,\beta)\,dv_{g}.$

Under this inner product $(\cdot,\cdot)$, we denote $d^{*(q)}$ to be the
Hilbert space adjoint of $d^{(q)}$ for all $q=0,\ldots,n$. That is,

$(du,v)=(u,d^{*}v)$

for all $u\in\Omega^{q}(M)$ and $v\in\Omega^{q+1}(M)$ (once more, we will drop
the upper index $(q)$ and write $d^{*}$ if there is no need to clarify which
space $d^{*(q)}$ is acting on). We call the operator

$\triangle^{(q)}:=dd^{*}+d^{*}d:\Omega^{q}(M)\rightarrow\Omega^{q}(M)$

the Hodge Laplacian for all $q=0,\ldots,n$. Note that this operator depends on
the Riemannian metric $g$ since $d^{*}$ depends on the inner product
$(\cdot,\cdot)$ and this inner product is determined by $g$.

We end this section with two lemmas about the local form of $d$ and $d^{*}$:

######  Lemma 2.3.

Fix $q\in\\{0,\ldots,n-1\\}$. For any $\omega\in\Omega^{q}(M)$, we have

$\displaystyle
d\omega=\sum_{i=1}^{n}dx_{i}\wedge\nabla_{\frac{\partial}{\partial
x_{i}}}\omega.$ (2.9)

######  Remark.

  1. (i)

The operator
$\nabla:\Gamma(\Lambda^{q}T^{*}M)\rightarrow\Gamma(\Lambda^{q}T^{*}M\otimes
T^{*}M)$ is natural extension of the Levi-civita connection
$\nabla^{TM}:\Gamma(TM)\rightarrow\Gamma(TM\otimes TM)$.

  2. (ii)

The notation $dx_{k}\wedge$ represents the exterior multiplications by
$dx_{k}$, respectively. On any local chart, $dx_{k}\wedge$ sends an given
$r$-form $\alpha$, $r\in\\{0,\ldots,n\\}$ to $dx_{k}\wedge\alpha$. The
interior multiplication acts on differential forms following the rules:

    1. (a)

if $\alpha$ is a $1$-form, then

$dx_{k}\lrcorner\alpha=\left\langle\alpha,\frac{\partial}{\partial
x_{k}}\right\rangle$

where $\langle\cdot,\cdot\rangle$ is the dual pairing;

    2. (b)

if $\beta$ is a smooth $p$-form and $\gamma$ is a smooth $q$-form, then

$dx_{k}\lrcorner(\beta\wedge\gamma)=(dx_{k}\lrcorner\beta)\wedge\gamma+(-1)^{p}\beta\wedge(dx_{k}\lrcorner\gamma).$

###### Proof.

Note that both sides of (2.9) are section of $\Lambda^{q+1}T^{*}M$. So we can
check that they agree with each other on each $p\in M$. Fix $p\in M$ and we
adopt the normal geodesic coordinate around $p$. Say near $p$, we have

$\omega=\sum_{i_{1}<\cdots<i_{q}}\omega_{i_{1}\cdots
i_{q}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}}.$

Then

$\displaystyle\begin{split}d\omega=\sum_{i_{1}<\cdots<i_{q}}\sum_{i=1}^{n}\frac{\partial\omega_{i_{1}\cdots
i_{q}}}{\partial x_{i}}dx_{i}\wedge dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}}\\\
=\sum_{i_{1}<\cdots<i_{q}}\sum_{i=1}^{n}\nabla_{\frac{\partial}{\partial
x_{i}}}\omega_{i_{1}\cdots i_{q}}dx_{i}\wedge dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}}\\\
=\sum_{i_{1}<\cdots<i_{q}}\sum_{i=1}^{n}dx_{i}\wedge\nabla_{\frac{\partial}{\partial
x_{i}}}(\omega_{i_{1}\cdots i_{q}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}})\\\
=\sum_{i=1}^{n}dx_{i}\wedge\nabla_{\frac{\partial}{\partial
x_{i}}}\omega.\end{split}$ (2.10)

∎

######  Lemma 2.4.

Fix $q\in\\{1,\ldots,n\\}$. For any $\omega\in\Omega^{q}(M)$, we have

$\displaystyle
d^{*}\omega=-\sum_{i=1}^{n}dx_{i}\lrcorner\nabla_{\frac{\partial}{\partial
x_{i}}}\omega.$ (2.11)

######  Remark.

The notation $dx_{k}\lrcorner$ represents the interior multiplications by
$dx_{k}$. On any local chart, the interior multiplication acts on differential
forms following the rules:

  1. 1.

if $\alpha$ is a $1$-form, then

$dx_{k}\lrcorner\alpha=\left\langle\alpha,\frac{\partial}{\partial
x_{k}}\right\rangle$

where $\langle\cdot,\cdot\rangle$ is the dual pairing;

  2. 2.

if $\beta$ is a smooth $p$-form and $\gamma$ is a smooth $q$-form, then

$dx_{k}\lrcorner(\beta\wedge\gamma)=(dx_{k}\lrcorner\beta)\wedge\gamma+(-1)^{p}\beta\wedge(dx_{k}\lrcorner\gamma).$

###### Proof.

With the same reason of the proof lemma 2.3, we can restrict our calculation
on a fixed point $p\in M$. Again, use the normal geodesic coordinate around
$p$. Also, write

$\omega=\sum_{i_{1}<\cdots<i_{q}}\omega_{i_{1}\cdots
i_{q}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}}.$

Applying lemma 2.3, we have for any $u\in\Omega^{q-1}(M)$,

$\displaystyle\begin{split}g(\omega,du)=g(\omega,\sum_{i=1}^{n}dx_{i}\wedge\nabla_{\frac{\partial}{\partial
x_{i}}}u)=\sum_{i=1}^{n}g(dx_{i}\lrcorner\omega,\nabla_{\frac{\partial}{\partial
x_{i}}}u)\\\ =\sum_{i=1}^{n}\frac{\partial}{\partial
x_{i}}g(dx_{i}\lrcorner\omega,u)-\sum_{i=1}^{n}g(\nabla_{\frac{\partial}{\partial
x_{i}}}dx_{i}\lrcorner\omega,u)\\\
=\sum_{i=1}^{n}\nabla_{\frac{\partial}{\partial
x_{i}}}g(dx_{i}\lrcorner\omega,u)-\sum_{i=1}^{n}g(\nabla_{\frac{\partial}{\partial
x_{i}}}dx_{i}\lrcorner\omega,u)\end{split}$ (2.12)

Integrating the both sides of formula (2.12) and noticing that
$(d^{*}\omega,u)=(\omega,du)$, we then derive that

$\displaystyle(d^{*}\omega,u)=(\sum_{i=1}^{n}-\nabla_{\frac{\partial}{\partial
x_{i}}}dx_{i}\lrcorner\omega,u)$ (2.13)

for all $u\in\Omega^{q-1}(M)$, i.e.,
$d^{*}\omega=\sum_{i=1}^{n}-\nabla_{\frac{\partial}{\partial
x_{i}}}dx_{i}\lrcorner\omega$. ∎

##  3 Witten Deformation of Laplacian

This section is divided into the following parts: first, we will introduce the
deformed de Rham cohomologies. We will show that they are all isomorphic to
the usual de Rham cohomologies. Next, we will define the Witten Laplacian. We
will prove a similar inequalities like the one of Morse for the deformed de
Rham cohomologies. The only difference is the upper bound of cohomology
dimension are replaced by things related to the dimension of eigenspaces of
Witten Laplacian.

###  3.1 Deformed de Rham Cohomology

Consider the following deformed exterior derivatives

$d_{t}^{(q)}:\Omega^{q}(M)\rightarrow\Omega^{q+1}(M)$

defined by

$d_{t}^{(q)}=e^{-tf}d^{(q)}e^{tf},\quad\forall t\geqslant 0$

for all $q=0,\ldots,n$. If the content is clear, we will drop the upper index
of $d_{t}^{(q)}$ and write $d_{t}$. We then get the deformed complex

$0\rightarrow\Omega^{0}(M)\xrightarrow{d_{t}}\Omega^{1}(M)\xrightarrow{d_{t}}\cdots\xrightarrow{d_{t}}\Omega^{n}(M)\rightarrow
0,\quad\forall t\geqslant 0.$

Since $d^{2}_{t}=e^{-tf}d^{2}e^{tf}=0$, we can also define
$H^{\bullet}_{t}(M)$ the cohomology of this complex. Note that the defomation
does not change the cohomology data. To be precise,

######  Theorem 3.1.

The map of multiplication

$\begin{array}[]{cccc}e^{-tf}:&H^{j}(M)&\rightarrow&H^{j}_{t}(M)\\\
&u+\text{Im}d&\mapsto&e^{-tf}u+\text{Im}d_{t}\end{array}$

is a vector space isomorphism for all $j=0,\cdots,n$.

###### Proof.

Fix $j=0,\cdots,n$. It suffices to show $e^{-tf}$ is a well-defined bijection
because linearity is quite obvious.

  * •

$e^{-tf}$ is a well-defined map: let

$u\in\text{Im}(d:\Omega^{j-1}(M)\rightarrow\Omega^{j}(M)),$

i.e., $u=dv$ for some $v\in\Omega^{j-1}(M)$. Since

$e^{-tf}u=e^{-tf}dv=d_{t}(e^{-tf}v)$

and $e^{-tf}v\in\Omega^{j-1}(M)$,

$e^{-tf}u\in\text{Im}(d_{t}:\Omega^{j-1}(M)\rightarrow\Omega^{j}(M)).$

  * •

$e^{-tf}$ is injective: assume $e^{-tf}u\in\text{Im}d_{t}$, i.e.,
$e^{-tf}u=d_{t}v$ for some $v\in\Omega^{j-1}(M)$. Then

$u=d(e^{tf}v)\in\text{Im}d.$

  * •

$e^{-tf}$ is surjective: let $u+\text{Im}d_{t}\in H^{j}(M)$. Consider
$e^{tf}u+\text{Im}d$. Since

$d(e^{tf}u)=e^{tf}(d_{t}u)=0,$

$e^{tf}u+\text{Im}d\in H^{j}_{t}(M)$. Also, $e^{-tf}$ maps
$e^{\text{tf}}u+\text{Im}d$ to $u+\text{Im}d_{t}$.

∎

In particular, this theorem tells us

$\dim H_{t}^{q}(M)=\dim H^{q}(M)=b_{q}$

for all $0\leq q\leq n$ and $t\geq 0$.

###  3.2 Witten Laplacian

Let

$d_{t}^{(q)*}:\Omega^{q+1}(M)\rightarrow\Omega^{q}(M)$

be the formal adjoint of $d_{t}^{(q)}:\Omega^{q}(M)\rightarrow\Omega^{q+1}(M)$
with respect to $(\cdot,\cdot)$, i.e.,

$(d_{t}^{(q)}u,v)=(u,d_{t}^{(q)*}v)$

for all $u\in\Omega^{q}(M)$ and $v\in\Omega^{q+1}(M)$. We call the operator

$\triangle_{t}^{(q)}:=d_{t}^{(q)}d_{t}^{(q)*}+d_{t}^{(q)*}d_{t}^{(q)}:\Omega^{q}(M)\rightarrow\Omega^{q}(M)$

the Witten Laplacian. Note that this operator also depends on the Riemannian
metric $g$.

It has the local form stated in the following theorem.

######  Theorem 3.2 (Bocher type formula).

Let $p\in M$. Define

$\Lambda^{\bullet}T^{*}_{p}M:=\bigoplus_{k=0}^{n}\Lambda^{k}T^{*}_{p}M$

Then for each $\omega\in\Lambda^{\bullet}T^{*}_{p}M$,

$\displaystyle\triangle_{t}\omega=\triangle\omega+t|df|^{2}\omega+t\sum_{k,l}\text{Hess}_{f}(\frac{\partial}{\partial
x_{l}},\frac{\partial}{\partial x_{k}})[dx_{l}\wedge,dx_{k}\lrcorner]\omega.$
(3.1)

######  Remark.

  1. (i)

We define $|df|^{2}:=g(df,df)$ and

$\displaystyle\text{Hess}_{f}:=\nabla df\in\Gamma(T^{*}M\otimes T^{*}M).$
(3.2)

The latter is said to be the Hessian of $f$. If $p\in M$ is a critical point
of $f$, then $d^{2}f(p)=\text{Hess}_{f}(p)$ (see Definition 2.1 for the
meaning of $d^{2}f$). A justification is given in the following: let
$\\{\frac{\partial}{\partial x_{i}}\\}_{i=1}^{n}$ be a basis for $T_{p}M$.
Then

$\text{Hess}_{f}(\frac{\partial}{\partial x_{j}},\frac{\partial}{\partial
x_{k}})(p)=\frac{\partial^{2}f}{\partial x_{j}\partial
x_{k}}(p)-\frac{\partial f}{\partial
x_{l}}(p)\Gamma_{jk}^{l}(p)=\frac{\partial^{2}f}{\partial x_{j}\partial
x_{k}}=d^{2}f(\frac{\partial}{\partial x_{j}},\frac{\partial}{\partial
x_{k}})(p)$

by the definition of critical point and formula (2.5).
$[dx_{l}\wedge,dx_{k}\lrcorner]$ is simply a shorthand of $dx_{l}\wedge
dx_{k}\lrcorner-dx_{k}\lrcorner dx_{l}\wedge$. As a result, there is no harm
to called both $d^{2}f$ and $\text{Hess}_{f}$ Hessian.

  2. (ii)

By the local formula above and simple calculations, we notice that
$\triangle^{(q)}_{t}$ is an positive elliptic operator with the symbol

$g^{ij}\xi_{i}\xi_{j}.$

It has an self-adjoint extension

$\triangle^{(q)}_{t}:\text{Dom}\triangle_{t}^{(q)}\subseteq
L^{2}(M)\rightarrow L^{2}(M),$

which we use the same notation. The space $\text{Dom}\triangle_{t}^{(q)}$ is
defined by

$\text{Dom}\triangle_{t}^{(q)}=\\{u\in L^{2}_{q}(M):\triangle_{t}^{(q)}u\in
L^{2}_{q}(M)\\}.$

Following by [Shu87, theorem 8.3], there exists a complete orthonormal system
$\\{f_{j}\\}$ consisting of eigenfunctions of $\triangle_{t}$. Here, $f_{j}$
are all smooth, $\triangle_{t}f_{j}=\lambda_{j}f_{j}$ and $\\{\lambda_{j}\\}$
is a non-negative sequence tends to $+\infty$. Also, the spectrum
$\sigma(\triangle_{t})$ coincides with the set of all and eigenvalues. Note
that for any $\lambda\in\mathbb{R}$, $\lambda\in\sigma(\triangle_{t})$ if and
only if the one of the three conditions holds:

    1. (a)

$\triangle_{t}-\lambda I$ is not injective (in this case, we call $\lambda$ an
eigenvalue),

    2. (b)

$\triangle_{t}-\lambda I$ is not surjective,

    3. (c)

$\triangle_{t}-\lambda I$ is bijective, but it’s inverse

$(\triangle_{t}-\lambda I)^{-1}:L^{2}_{q}(M)\rightarrow L^{2}_{q}(M)$

is not continuous (with respect to $\|\cdot\|$).

###### Proof of Bochner Type Formula.

Firstly, note that

$\displaystyle d_{t}=d+tdf\wedge$ (3.3)

since for all $u\in\Omega^{\bullet}(M)$,

$\displaystyle d_{t}u=e^{-tf}d(e^{tf}u)=e^{-tf}e^{tf}tdf\wedge
u+e^{-tf}e^{tf}du=(d+tdf\wedge)u.$ (3.4)

Next, formula (3.3) yields

$\displaystyle d_{t}^{*}=d^{*}+tdf\lrcorner$ (3.5)

because for all $u\in\Omega^{\bullet}(M)$ and $v\in\Omega^{\bullet+1}(M)$,

$\displaystyle\begin{split}(u,(d^{*}+tdf\lrcorner)v)_{g}=(u,d^{*}v)_{g}+(u,tdf\lrcorner
v)=(du,v)+(tdf\wedge u,v)\\\
=((d+tdf\wedge)u,v)=(d_{t}u,v)=(u,d_{t}^{*}v).\end{split}$ (3.6)

From formula (3.3) and (3.5), we may derive that for all
$u\in\Lambda^{\bullet}T_{p}^{*}M$,

$\displaystyle\begin{split}\triangle_{t}u=(dd^{*}+d^{*}d)u+t(df\wedge
d^{*}u+d(df\lrcorner u)+df\lrcorner du+d^{*}(df\wedge u))\\\ +t^{2}(df\wedge
df\lrcorner+df\lrcorner df\wedge)u.\end{split}$ (3.7)

Since the formula (3.1) is tensorial, we may verify the formula pointwise.
Recall the exponential map

$exp_{p}:B_{0}(r)\rightarrow M$

is a diffeomorphism onto its image ($0$ is the origin of $\mathbb{R}^{n}$ and
$r$ is a sufficiently small positive real number). It yields a local chart
$(U,\varphi,x)$ for $M$ around $p$, called the normal geodesic coordinate. In
the normal geodesic coordinate, the components of the Riemannian metric on
$T_{p}M$ $g_{ij}(p)=g(\frac{\partial}{\partial x_{i}},\frac{\partial}{\partial
x_{j}})$ simplify to $\delta_{ij}$. Thus, this implies

$\\{dx_{i_{1}}\wedge\cdots\wedge dx_{i_{k}}:i_{1}<\cdots<i_{k}\text{ and
}k=0,\ldots,n\\}$

is an orthonormal basis for $\Lambda^{\bullet}T^{*}_{p}M$.

Note that the both sides of formula (3.1) are invariant under coordinate
change, it suffices for us to check it is true under normal coordinate.
Furthermore, we only need to check those $\omega$ that are basis for
$\Lambda^{\bullet}T^{*}_{p}M$. Write

$df=\sum_{a=1}^{n}\frac{\partial f}{\partial x_{a}}dx_{a}\quad\text{and}\quad$

where $dx_{I}=\omega=dx_{i_{1}}\wedge\cdots\wedge dx_{i_{k}}$ ($I=\\{1\leq
i_{1}<\cdots<i_{k}\leq n\\}$). Then

$\displaystyle\begin{split}(df\wedge df\lrcorner+df\lrcorner
df\wedge)dx_{I}=\frac{\partial f}{\partial x_{a}}dx_{a}\wedge\frac{\partial
f}{\partial x_{b}}\frac{\partial}{\partial x_{b}}dx_{I}+\frac{\partial
f}{\partial x_{a}}\frac{\partial}{\partial x_{a}}\frac{\partial f}{\partial
x_{a}}dx_{a}\wedge dx_{I}\\\ =\left(\frac{\partial f}{\partial
x_{a}}\right)^{2}\left((-1)^{l}\delta^{a}_{I}dx_{a}\wedge
dx_{I-\\{a\\}}+\frac{\partial}{\partial x_{a}}(1-\delta^{a}_{I})dx_{a}\wedge
dx_{I}\right)\\\ =\left(\frac{\partial f}{\partial
x_{a}}\right)^{2}\left((-1)^{2l}\delta^{a}_{I}dx_{I}+(1-\delta^{a}_{I})dx_{I}\right)=\left(\frac{\partial
f}{\partial x_{a}}\right)^{2}dx_{I}=|df|^{2}dx_{I}.\end{split}$ (3.8)

where $\delta^{a}_{I}=1$ if $a\in I$, $\delta^{a}_{I}=0$ otherwise. If $a\in
I$, then $l$ is the integer so that $i_{l}=a$ and $dx_{I-\\{a\\}}$ equals to
$dx_{i_{1}}\wedge\cdots\wedge\hat{dx_{i_{l}}}\wedge\cdots\wedge dx_{i_{k}}$.

Finally, also in the normal coordinate near $p$, we calculate the
”coefficient” of $t$ in formula (3.1). Assume near $p$, $u$ can be written as

$u=\sum_{i_{1}<\cdots<i_{q}}u_{i_{1}\cdots i_{q}}dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}}.$

Applying lemmas 2.3 and 2.4 we get

$\displaystyle\begin{split}df\wedge d^{*}u=\sum_{i=1}^{n}\frac{\partial
f}{\partial
x_{j}}dx_{j}\wedge(-\sum_{i=1}^{n}dx_{i}\lrcorner\nabla_{\frac{\partial}{\partial
x_{i}}}u)\\\ =-\sum_{i,j=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial
f}{\partial x_{j}}\frac{\partial u_{i_{1}\cdots i_{q}}}{\partial
x_{i}}dx_{j}\wedge dx_{i}\lrcorner dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}},\end{split}$ (3.9) $\displaystyle\begin{split}df\lrcorner
du=\sum_{i=1}^{n}\frac{\partial f}{\partial
x_{j}}dx_{j}\lrcorner(\sum_{i=1}^{n}dx_{i}\wedge\nabla_{\frac{\partial}{\partial
x_{i}}}u)\\\ =\sum_{i,j=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial
f}{\partial x_{j}}\frac{\partial u_{i_{1}\cdots i_{q}}}{\partial
x_{i}}dx_{j}\lrcorner dx_{i}\wedge dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}},\end{split}$ (3.10) $\displaystyle\begin{split}ddf\lrcorner
u=d\sum_{i=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial f}{\partial
x_{i}}u_{i_{1}\cdots i_{q}}dx_{i}\lrcorner dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}}\\\ \sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{j}\partial
x_{i}}dx_{j}\wedge dx_{i}\lrcorner(\sum_{i_{1}<\cdots<i_{q}}u_{i_{1}\cdots
i_{q}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}})\\\
+\sum_{i,j=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial f}{\partial
x_{j}}\frac{\partial u_{i_{1}\cdots i_{q}}}{\partial x_{i}}dx_{j}\wedge
dx_{i}\lrcorner dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}},\end{split}$ (3.11)

and

$\displaystyle\begin{split}d^{*}df\wedge
u=d\sum_{i=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial f}{\partial
x_{i}}u_{i_{1}\cdots i_{q}}dx_{i}\wedge dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}}\\\ -\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{j}\partial
x_{i}}dx_{j}\lrcorner dx_{i}\wedge(\sum_{i_{1}<\cdots<i_{q}}u_{i_{1}\cdots
i_{q}}dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}})\\\
-\sum_{i,j=1}^{n}\sum_{i_{1}<\cdots<i_{q}}\frac{\partial f}{\partial
x_{j}}\frac{\partial u_{i_{1}\cdots i_{q}}}{\partial x_{i}}dx_{j}\lrcorner
dx_{i}\wedge dx_{i_{1}}\wedge\cdots\wedge dx_{i_{q}}.\end{split}$ (3.12)

Summing up formulas (3.9) to (3.12), we get the coefficient of $t$ equals to

$\displaystyle\begin{split}df\wedge d^{*}u+df\lrcorner du+ddf\lrcorner
u+d^{*}df\wedge u\\\ =\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial
x_{i}\partial x_{j}}dx_{i}\wedge dx_{j}\lrcorner
u-\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}dx_{i}\lrcorner dx_{j}\wedge u\\\
=\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}dx_{i}\wedge dx_{j}\lrcorner
u-\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{j}\partial
x_{i}}dx_{i}\lrcorner dx_{j}\wedge u\\\
=\sum_{i,j=1}^{n}\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}[dx_{i}\wedge,dx_{j}\lrcorner]\wedge u\\\ \end{split}$ (3.13)

because $f$ is smooth so that

$\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}=\frac{\partial^{2}f}{\partial x_{j}\partial x_{i}}$

for all $i,j\in\\{1,\ldots,n\\}$. The notation
$[dx_{i}\wedge,dx_{j}\lrcorner]$ equals to $dx_{i}\wedge dx_{j}\lrcorner-
dx_{j}\lrcorner dx_{i}\wedge$ for all $i,j\in\\{1,\ldots,n\\}$.

In the normal coordinate, we have $\Gamma_{ij}^{k}(p)=0$. Thus,

$\text{Hess}_{f}(\frac{\partial}{\partial x_{i}},\frac{\partial}{\partial
x_{j}})=\frac{\partial^{2}f}{\partial x_{i}\partial x_{j}}(p)-\frac{\partial
f}{\partial x_{k}}(p)\Gamma_{ij}^{k}(p)=\frac{\partial f}{\partial
x_{i}\partial x_{j}}$

and so the last line of formula (3.13) equals to

$\displaystyle\begin{split}\sum_{i,j=1}^{n}\text{Hess}_{f}(\frac{\partial}{\partial
x_{i}},\frac{\partial}{\partial
x_{j}})[dx_{i}\wedge,dx_{j}\lrcorner]u.\end{split}$ (3.14)

This expression is invariant under change of coordinates, and so we complete
the derivation of Bochner formula. ∎

Denote

$E^{r}_{\mu,t}(M)=\\{u\in\Omega^{r}(M):\triangle_{t}^{(r)}u=\mu u\\}$

as the eigenspace of $\mu$ with respect to $\triangle_{t}^{(r)}$ in
$\Omega^{r}(M)$. Note that the deformed exterior derivative

$d_{t}:E^{r}_{\mu,t}(M)\rightarrow E_{\mu,t}^{r+1}(M)$

sends $r$-forms from $E^{r}_{\mu,t}(M)$ to $E_{\mu,t}^{r+1}(M)$, for
$r=0,\ldots,n-1$. Indeeds, let $u\in E_{\mu,t}^{r}(M)$, i.e.,
$\triangle_{t}^{(r)}u=\mu u$ for some $\mu\in\mathbb{R}$. Then we have

$\triangle_{t}^{(r)}(d_{t}u)=(d_{t}d^{*}_{t}+d^{*}_{t}d_{t})(d_{t}u)=d_{t}d^{*}_{t}d_{t}u=d_{t}d^{*}_{t}d_{t}u+d_{t}d_{t}d^{*}_{t}u=d_{t}(\triangle_{t}^{(r)}u)=\mu
d_{t}u.$

Let $I\subset\mathbb{R}$ be a bounded set and we denote

$\begin{array}[]{l}E_{I,t}^{q}(M):=\bigoplus_{\mu\in\text{Spec}\triangle^{(q)}_{t}\cap
I}E_{\mu,t}^{q}(M).\end{array}$

######  Theorem 3.3.

The following sequence of vector spaces

$E^{0}_{(0,\lambda],t}(M)\xrightarrow{d^{(0)}_{t}}E^{1}_{(0,\lambda],t}(M)\xrightarrow{d^{(1)}_{t}}E^{2}_{(0,\lambda],t}(M)\xrightarrow{d^{(2)}_{t}}\cdots\xrightarrow{d^{(n-1)}_{t}}E^{n}_{(0,\lambda],t}(M)\xrightarrow{d^{(n)}_{t}}0$

is exact.

###### Proof.

Fix $r=0,\ldots,n$. We need to show $\ker d^{(r)}_{t}=\text{im}d^{(r-1)}_{t}.$

  * $(\supseteq)$:

Let $u\in\text{im}d^{(r-1)}_{t}$. Then $u=d_{t}v$ for some $v\in
E^{r-1}_{(0,\lambda],t}(M)$. Since $d^{2}_{t}=0$, we must have
$d_{t}u=d^{2}_{t}v=0$. This gives $u\in\ker d^{(r)}_{t}$.

  * $(\subseteq)$:

Let $u\in\ker d^{(r)}_{t}$. We may write

$u=u_{1}+\cdots+u_{N}$

where $u_{j}\in E_{\mu_{j},t}^{r}(M)$, $\mu_{j}$ distinct real positive
numbers. Applying $d_{t}$ on both sides of the equation above, we get

$0=d_{t}u=d_{t}u_{1}+\cdots+d_{t}u_{N}.$

Since each $d_{t}u_{j}$ belongs to different eigenspace, $\\{d_{t}u_{j}\\}$ is
linearly independent. This implies $d_{t}u_{j}=0$ for $j=1,\ldots,N$. Fixing
$j$, we have

$u_{j}=\frac{1}{\mu_{j}}\triangle^{(r)}_{t}u_{j}=\frac{1}{\mu_{j}}(d_{t}d^{*}_{t}+d^{*}_{t}d_{t})u_{j}=\frac{1}{\mu_{j}}d_{t}d^{*}_{t}u_{j}=d_{t}\left(\frac{1}{\mu_{j}}d^{*}_{t}u_{j}\right).$

Summing $u_{j}$ together, we get

$u=d_{t}\left(\frac{1}{\mu_{1}}d^{*}_{t}u_{1}+\cdots+\frac{1}{\mu_{N}}d^{*}_{t}u_{N}\right)\in\text{im}d^{(r-1)}_{t}.\qed$

Note that the spaces $E_{(0,\lambda],t}^{q}(M)$ are finite-dimensional for all
$q=0,\ldots,n$ since each of them is a finite direct sum of eigenspaces, and
the eigenspaces of Witten Laplacian is finite-dimensional. Using Theorem 3.3
and dimension theorem for vector spaces, we then have

$\dim\text{im}\,d^{(q)}_{t}=\dim E^{q}_{(0,\lambda],t}(M)-\dim\ker
d^{(q)}_{t}=\dim E^{q}_{(0,\lambda],t}(M)-\dim\text{im}\,d^{(q-1)}_{t}$

for $q=1,\ldots,n$. For the case $q=0$, observe that $\ker d_{t}^{(}0)$ is a
trivial vector space. For any $u\in\ker d_{t}^{(0)}$, we can write

$\alpha=\alpha_{1}+\cdots+\alpha_{N}$

where $\alpha_{j}\in E^{0}_{\mu_{j},t}(M)$, $\mu_{j}$ distinct real positive
numbers. Applying the Laplacian on both sides we get

$\triangle_{t}^{(0)}\alpha=\mu_{1}\alpha_{1}+\cdots+\mu_{N}\alpha_{N}$

But since

$\triangle_{t}^{(0)}\alpha=d_{t}^{*(0)}d_{t}^{(0)}\alpha=0,$

we must have $u_{1}=\cdots=u_{N}=0$ by linear independence of distinct
eigenspaces. Thus, $\alpha=0$, i.e., $\ker d_{t}^{(0)}=0$.

and $\dim\text{im}d^{(0)}_{t}=\dim E_{(0,\lambda],t}^{0}(M)$. Then we may
derive the following alternating sum

$\displaystyle\sum^{q}_{j=0}(-1)^{q-j}\dim
E_{(0,\lambda],t}^{q}(M)=\dim\text{im}d^{(q)}_{t}\geq 0.$ (3.15)

Moreover, the equation holds when $q=n$, since $\text{rank}d^{(q)}$ vanishes
as $q=n$. As a result, we obtain estimations encountering $\dim H^{j}_{t}(M)$,
and these estimations play important rules in the next section of proving
Morse inequalities.

######  Theorem 3.4.

For any positive real number $\lambda>0$ and $q=0,\ldots,n$, we have

$\displaystyle\dim H^{q}_{t}(M)\leq\dim E^{q}_{[0,\lambda],t}(M)$ (3.16)

and

$\displaystyle\sum^{q}_{j=0}(-1)^{q-j}\dim
H^{j}_{t}(M)\leq\sum^{q}_{j=0}(-1)^{q-j}\dim E_{[0,\lambda],t}^{q}(M).$ (3.17)

The equality of (3.17) holds when $q=n$.

###### Proof.

  1. 1.

First, we proof that

$\dim H^{q}_{t}(M)=\dim\ker\triangle_{t}^{(q)}.$

Let {$v_{1}+\text{im}d_{t}^{(q-1)},\ldots,v_{l}+\text{im}d_{t}^{(q-1)}\\}$ be
the basis for $H^{q}_{t}(M)$. Fix $i=1,\ldots,l$. Then we can see

$\displaystyle d_{t}v_{i}$ $\displaystyle=0$ $\displaystyle(v_{i},d_{t}u)$
$\displaystyle=0,\quad\forall u\in\Omega^{q-1}(M).$

As a consequence, for any $v\in\Omega^{q}(M)$,

$(\triangle_{t}v_{i},u)=(d_{t}v_{i},d_{t}u)+(d_{t}^{*}v_{i},d_{t}^{*}u)=0,$

which means $v_{i}\in\ker\triangle_{t}^{(q)}$. Thus, $\dim
H^{q}_{t}(M)\leq\dim\ker\triangle_{t}^{(q)}$.

On the other hand, let $\\{u_{1},\ldots,u_{l^{\prime}}\\}$ be the basis for
$\ker\triangle_{t}^{(q)}$. Fix $i=1,\ldots,l^{\prime}$. For each
$u\in\Omega^{q}(M)$,

$0=(\triangle_{t}v_{i},v_{i})=(d_{t}v_{i},d_{t}v_{i})+(d^{*}_{t}v_{i},d^{*}_{t}v_{i}),$

which implies that $d_{t}v_{i}=0$ and $d^{*}_{t}v_{i}=0$. As a result,
$v_{i}+\text{im}d_{t}^{q-1}\in H^{q}_{t}(M)$, i.e., $\dim
H^{q}_{t}(M)\geq\dim\ker\triangle_{t}^{(q)}$.

  2. 2.

By the first part of proof, we can immediately prove (3.16) since

$\displaystyle\dim E^{q}_{[0,\lambda],t}(M)=\dim(\ker\triangle_{t}^{(q)}\oplus
E^{q}_{(0,\lambda],t}(M))=\dim\ker\triangle_{t}^{(q)}+E^{q}_{(0,\lambda],t}(M)$
$\displaystyle\geq\dim\ker\triangle_{t}^{(q)}=\dim H^{q}_{t}(M).$

  3. 3.

Part 1 of this proof and inequality (3.15) says that

$\dim H^{q}_{t}(M)=\dim\ker\triangle_{t}^{(q)}$

and

$\sum^{q}_{j=0}(-1)^{q-j}\dim E_{(0,\lambda],t}^{q}(M)\geq 0$

respectively. Then

$\displaystyle\sum^{q}_{j=0}(-1)^{q-j}\dim
E_{[0,\lambda],t}^{q}(M)=\sum^{q}_{j=0}(-1)^{q-j}(\dim
E_{(0,\lambda],t}^{q}(M)+\dim\ker\triangle_{t}^{(q)})$
$\displaystyle=\sum^{q}_{j=0}(-1)^{q-j}\dim
E_{(0,\lambda],t}^{q}(M)+\sum_{i=1}^{q}(-1)^{q-j}\dim H^{q}_{t}(M)$
$\displaystyle\geq\sum_{i=1}^{q}(-1)^{q-j}\dim H^{q}_{t}(M),$

ending the proof of formula (3.17).

∎

In the next section, we will prove the following theorem. If the theorem hold,
plug the previous theorem and then we derive the Morse inequalities.

######  Theorem 3.5.

There exist $C>0$ and $t>0$ such that when $t\geq t_{0}$,

$\dim E^{q}_{[0,e^{-Ct}],t}=m_{q}$

for all $q=0,\ldots,n$.

###  3.3 Locally Flat Metric Near Critical Points

Let $\tilde{g}$ be arbitrary Riemannian metric on $M$. Fix
$p\in\text{Crit}(f)$. Let $(U_{p},\phi)$ be the coordinate chart near $p$
given by Morse lemma (lemma 2.2). We may assume $U_{p}\cap
U_{p^{\prime}}=\emptyset$ when $p\neq{p^{\prime}}$. Because $\phi$ is a local
diffeomorphism from $U$ to $\phi(U_{p}):=\tilde{U_{p}}$, we can define a
Riemannian metric $g_{p}$ on $U$ by

$g_{p}(u,v)(\hat{p})=\langle d\phi_{\hat{p}}(u),d\phi_{\hat{p}}(v)\rangle$

for each $\hat{p}\in U$, $u,v\in T_{\hat{p}}M$, and
$\langle\cdot,\cdot\rangle$ denotes the standard Euclidean metric on
$\mathbb{R}^{n}$.

Denote $V=M-\cup_{p\in\text{Crit}(f)}U_{p}$. Then

$\displaystyle S=\\{U_{p}\\}_{p\in\text{Crit(f)}}\cup\\{V\\}$ (3.18)

is an open covering of $M$. Let $\\{\varphi_{p}\\}\cup\\{\psi\\}$ be a
partition unity subordinate on $S$, where

$\displaystyle\text{supp}\varphi_{p}$ $\displaystyle\subseteq U_{p}$
$\displaystyle\text{supp}\psi$ $\displaystyle\subseteq V$

We define a new metric $g$ by

$\displaystyle g=\tilde{g}\psi+\sum_{p\in\text{Crit(f)}}g_{p}\varphi_{p}$
(3.19)

From now on, we fix $g$ to be the metric defined by (3.19).

###  3.4 Witten Laplacian on $\mathbb{R}^{n}$

Let $p\in\text{Crit}(f;r)$. Let $(U,\varphi,x)$ be local chart $p$ given by
Morse lemma 2.2. Let $u\in C^{\infty}_{0}(U)$ and $J=\\{1\leq
j_{1}<\cdots<j_{q}\leq n\\}$. Then by our construction of $g$, the Bochner
type formula of $\triangle_{t}$ on $U$ becomes

$\displaystyle\triangle_{t}(udx_{J})=\left[-\triangle
u(x)+t^{2}|x|^{2}u(x)\right]dx_{J}+tu(x)\sum_{i=1}^{n}\varepsilon_{j}[dx_{j}\wedge,dx_{j}\lrcorner](dx_{J}).$
(3.20)

where $dx_{J}=dx_{j_{1}}\wedge\cdots\wedge dx_{j_{q}}$,

$\triangle=\frac{\partial^{2}}{\partial
x_{1}^{2}}+\cdots+\frac{\partial^{2}}{\partial
x_{n}^{2}}\qquad|x|^{2}=x_{1}^{2}+\cdots+x_{n}^{2}$

and

$\varepsilon_{j}=\begin{cases}-1&\text{for }j\leq r\\\ 1&\text{for }j>r\\\
\end{cases}$

##  4 A study on eigenspaces

In this section, we will study the eigenspaces of $\triangle_{t}^{(q)}$. We
will show that when $t$ is sufficiently large, there is exactly $m_{q}$
eigenvalues that are bounded by $e^{-Ct}$. Once we establish this, the proof
of main theorem is complete by the reduction we performed last section.

###  4.1 Norms on $\Omega^{q}(M)$

For any $\omega\in\Omega^{q}(M)$, define

$\displaystyle\|\omega\|_{\infty}=\sup_{p\in M}|\omega(p)|,$ (4.1)

called the sup norm of $\omega$ (we write $|\cdot|$ for
$\sqrt{g(\cdot,\cdot)}$). Note that there is another way to define the sup
norm of differential form. Let $\\{(U_{i},\varphi_{i},x)\\}_{i=1}^{n}$ be a
set of charts that covers $M$. Assume further that the exterior bundle
$\Lambda^{q}T^{*}M$ trivializes on each $U_{i}$ with the trivialization
(composited with $\varphi_{i}\times id$):

$\eta_{i}:\Lambda^{q}T^{*}M|_{U_{i}}\xrightarrow{\sim}\varphi_{i}^{-1}(U_{i})\times\mathbb{R}^{\binom{n}{q}}.$

We can define the induced map
$\eta_{i*}:\Gamma(U_{i},E|_{U_{i}})\rightarrow\Gamma(\varphi^{-1}(U_{i}),\mathbb{R}^{\binom{n}{q}})$
by

$\eta_{i*}\xi=\eta_{i}\circ\xi\circ\varphi_{i}^{-1}.$

Finally, let $\\{\rho_{i}\\}_{i=1}^{N}$ be a partition of unity subordinated
to $\\{U_{i}\\}$.

With the notations prepared, we can then define the sup norm of $\omega$ by

$\omega=\sum_{i=1}^{\binom{n}{q}}\sum_{i_{1}<\cdots<i_{q}}\|\omega_{i,i_{1},\ldots,i_{q}}\|_{\infty,\mathbb{R}}$

where $\|\cdot\|_{\infty,\mathbb{R}}$ is the sup norm on $\mathbb{R}$.

With a similar fashion, we can define the Sobolev $s$ norm on $\Omega^{q}(M)$
by

$\|\omega\|_{s}=\sum_{i=1}^{\binom{n}{q}}\sum_{i_{1}<\cdots<i_{q}}\|\omega_{i,i_{1},\ldots,i_{q}}\|_{s,\mathbb{R}}$

where

$\eta_{i*}(\omega\varphi_{i})=\sum_{i_{1}<\ldots<i_{q}}\omega_{i,i_{1},\ldots,i_{q}}dx_{i_{1}}\wedge\cdots\wedge
dx_{i_{q}}$

and $\|\cdot\|$ is the Sobolev $s$ norm on $\mathbb{R}$ defined by

$\|f\|_{s,\mathbb{R}}^{2}=\sum_{|\alpha|\leq
2n}\|\partial^{\alpha}h\|^{2}_{L^{2}(\mathbb{R}^{n})}$

where $|\alpha|=\alpha_{1}+\cdots+\alpha_{n}$ and
$\partial^{\alpha}=\frac{\partial^{\alpha_{1}}}{\partial
x_{1}^{\alpha_{1}}}\cdots\frac{\partial^{\alpha_{n}}}{\partial
x_{n}^{\alpha_{n}}}$. is the $s$ norm on $\mathbb{R}$.

###  4.2 Lower bound of $\dim E^{q}_{[0,e^{-Ct}],t}(M)$

Let $p\in\text{Crit}(f,q)$ and $(U_{p},\varphi,x)$ be a local chart such that
$x(p)=0$. Let $\kappa\in C^{\infty}_{0}(\mathbb{R})$ be a smooth function with
compact support so that

  1. 1.

$\text{supp}\kappa\subseteq[-2,2]$

  2. 2.

$\kappa(y)=1$ for all $x\in[-1,1]$

  3. 3.

$0\leq\kappa(y)\leq 1$ for all $x\in\mathbb{R}$.

For any $\varepsilon>0$, set
$\kappa_{\varepsilon}(y)=\kappa(\varepsilon^{-1}y)$. We may choose
$\varepsilon$ small enough such that the cube centered at $0$ is contained in
$\varphi(U_{p})$ ,i.e.,

$C_{\varepsilon}(0):=\\{x\in\mathbb{R}:\sup_{1\leq i\leq n}|x_{i}|\leq
2\varepsilon\\}\subseteq\varphi(U_{p})$

Now, define
$\chi(x)=\kappa_{\varepsilon}(x_{1})\cdots\kappa_{\varepsilon}(x_{n})$ such
that $\chi=1$ near $p$. Let $v\in\Omega_{0}^{q}(U_{p})\subset\Omega^{q}(M)$,
where $\Omega_{0}^{q}(U_{p})$ means the set of smooth $q$-forms with compact
support in $U_{p}$ defined by

$\displaystyle
v\circ\varphi^{-1}(x)=e^{-\frac{t}{2}|x|^{2}}\chi(x)dx^{1}\wedge
dx^{2}\wedge\cdots\wedge dx^{q}.$ (4.2)

The following lemma gives a upper bound of the $L^{2}$ norm of
$(\triangle_{t}^{(q)})^{j}v$

######  Lemma 4.1.

For all $s\in\mathbb{Z}^{+}$, there exist $t_{s}>0$, depending on $s$, such
that when $t\geq t_{s}$, we have

$\|(\triangle_{t}^{(q)})^{s}v\|\leq\exp(\frac{-t\varepsilon^{2}}{4}).$

###### Proof.

By calculation

$\displaystyle\triangle_{t}^{(q)}v=\triangle_{t}^{(q)}(\chi(x)\exp(-\frac{t|x|^{2}}{2}))=\exp(-\frac{t|x|^{2}}{2})\sum_{i=1}^{n}(2tx_{i}\frac{\partial\kappa_{\varepsilon}(x_{i})}{\partial
x_{i}}-\frac{\partial^{2}\kappa_{\varepsilon}(x_{i})}{\partial
x_{i}^{2}})\prod_{j\neq i}\kappa_{\varepsilon}(x_{j}).$ (4.3)

Define

$\tau_{1}(x_{i})=2tx_{i}\frac{\partial\kappa_{\varepsilon}(x_{i})}{\partial
x_{i}}-\frac{\partial^{2}\kappa_{\varepsilon}(x_{i})}{\partial x_{i}^{2}}$

Note that $\tau_{1}$ is non-negative the support of $\tau_{1}$ is contained in
$[-2\varepsilon,-\varepsilon]\cup[\varepsilon,2\varepsilon]$. By an induction
argument, one can show that

$\displaystyle(\triangle_{t}^{(q)})^{k}v=(\triangle_{t}^{(q)})^{k}(\chi(x)\exp(-\frac{t|x|^{2}}{2}))=\exp(-\frac{t|x|^{2}}{2})\sum_{i=1}^{n}\tau_{k}(x_{i})\prod_{j\neq
i}\kappa_{\varepsilon}(x_{j})$ (4.4)

where
$\text{supp}\tau_{k}\subseteq[-2\varepsilon,-\varepsilon]\cup[\varepsilon,2\varepsilon]$.
Now, by self-adjointness of $\triangle^{(q)}_{t}$ and formula (4.4),

$\displaystyle\|(\triangle^{(q)}_{t})^{s}v\|^{2}$
$\displaystyle=((\triangle^{(q)}_{t})^{s}v,(\triangle^{(q)}_{t})^{s}v)=((\triangle^{(q)}_{t})^{2s}v,v)$
$\displaystyle=n\int_{\mathbb{R}}\exp(-ty^{2})\tau_{2s}(y)\kappa_{\varepsilon}(y)dy\left(\int_{\mathbb{R}}\exp(-ty^{2})\kappa_{\varepsilon}(y)^{2}dy\right)^{n-1}$
$\displaystyle\leq
2n\sup_{y\in\mathbb{R}}\tau_{2s}(y)\int_{\varepsilon}^{\infty}\exp(-ty^{2})(y)dy\left(\int_{\mathbb{R}}\exp(-ty^{2})dy\right)^{n-1}$
$\displaystyle\leq
2n\sup_{y\in\mathbb{R}}\tau_{2s}(y)\frac{1}{t\varepsilon}\exp(-\frac{t\varepsilon^{2}}{2})\left(\frac{\pi}{t}\right)^{(n-1)/2}.$

As a result, we have

$\|(\triangle_{t}^{(q)})^{s}v\|\leq\exp(-\frac{t\varepsilon^{2}}{4})$

when $t\geq t_{2}$ where
$t_{s}=(2n\pi^{(n-1)/2}\sup_{y\in\mathbb{R}}\tau_{2s}(y))^{2/(n+1)}$ depending
on $s$. ∎

There is another inequalities which is needed.

######  Lemma 4.2 (Sobolev Type Inequality).

Let $h\in\Omega^{q}(M)$ be a smooth $q$-form on $M$. Then we have the
following estimation

$\displaystyle\|h\|_{\infty}\leq C\|h\|_{2n}$ (4.5)

for some $C>0$.

###### Proof.

The case on $\mathbb{R}^{n}$: Let $h\in\mathscr{S}(\mathbb{R}^{n})$ (Schwartz
class on $\mathbb{R}^{n}$). For all $x\in\mathbb{R}^{n}$,

$\displaystyle|h(x)|$
$\displaystyle=\left|(2\pi)^{-n}\int_{\mathbb{R}^{n}}e^{i\langle
x,\xi\rangle}\hat{h}(\xi)d\xi\right|\leq(2\pi)^{-n}\int|\hat{h}(\xi)|d\xi$
(4.6)
$\displaystyle=(2\pi)^{-n}\int_{\mathbb{R}^{n}}|\hat{h}(\xi)|(1+|\xi|^{2})^{n}(1+|\xi|^{2})^{-n}d\xi$
(4.7)
$\displaystyle\leq(2\pi)^{-n}\left(\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{2n}|\hat{h}(\xi)|^{2}d\xi\right)^{1/2}\left(\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{-2n}d\xi\right)^{1/2}$
(4.8)

The second line is a common trick to derive upper bound envolving $L^{2}$
norm, and the last inequalities comes from Cauchy-Schwartz.

Note that $\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{-2n}d\xi$ is bounded. Thus, it
remains to estimate the integration of $(1+|\xi|)^{2n}\hat{h}(\xi)$:

$\displaystyle\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{2n}|\hat{h}(\xi)|^{2}=\int_{\mathbb{R}^{n}}\sum_{\alpha_{1}+\cdots\alpha_{n}\leq
2n}\xi_{1}^{2\alpha_{1}}\cdots\xi_{n}^{2\alpha_{n}}\hat{h}(\xi)^{2}d\xi$ (4.9)
$\displaystyle=\sum_{\alpha_{1}+\cdots\alpha_{n}\leq
2n}\int_{\mathbb{R}^{n}}\xi_{1}^{2\alpha_{1}}\cdots\xi_{n}^{2\alpha_{n}}\hat{h}(\xi)^{2}d\xi$
(4.10)

Note that the Fourier transform of
$\xi_{1}^{\alpha_{1}}\cdots\xi_{n}^{\alpha_{n}}\hat{h}(\xi)$ is

$\displaystyle(-i)^{\alpha_{1}+\cdots+\alpha_{n}}\frac{\partial^{\alpha_{1}}}{\partial
x_{1}^{\alpha_{1}}}\cdots\frac{\partial^{\alpha_{n}}}{\partial
x_{n}^{\alpha_{n}}}h(x).$ (4.11)

Applying (4.10), (4.11) and Parseval’s formula (cf. [Hör03, Theorem 7.1.6]),
we then have

$\displaystyle\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{2n}|\hat{h}(\xi)|^{2}=(2\pi)^{n}\sum_{\alpha_{1}+\cdots+\alpha_{n}\leq
2n}\int_{\mathbb{R}^{n}}(\frac{\partial^{\alpha_{1}}}{\partial
x_{1}^{\alpha_{1}}}\cdots\frac{\partial^{\alpha_{n}}}{\partial
x_{n}^{\alpha_{n}}}h(x))^{2}dx$ (4.12)
$\displaystyle=(2\pi)^{n}\sum_{|\alpha|\leq
2n}\|\partial^{\alpha}h\|^{2}_{L^{2}(\mathbb{R}^{n})}=(2\pi)^{n}\|h\|_{2n}^{2},$
(4.13)

where $|\alpha|=\alpha_{1}+\cdots+\alpha_{n}$ and
$\partial^{\alpha}=\frac{\partial^{\alpha_{1}}}{\partial
x_{1}^{\alpha_{1}}}\cdots\frac{\partial^{\alpha_{n}}}{\partial
x_{n}^{\alpha_{n}}}$.

Therefore, from (4.8) and (4.13), we derive that for all $x\in\mathbb{R}^{n}$

$\displaystyle|h(x)|\leq C\|h\|_{2n},$ (4.14)

where $C=(2\pi)^{-n/2}(\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{-2n}d\xi)^{1/2}$,
independent on $x$. So by taking supremum over $x\in\mathbb{R}^{n}$ we prove
that

$\displaystyle\|h\|_{\infty}=\sup_{x\in\mathbb{R}^{n}}|h(x)|\leq C\|h\|_{2n}.$
(4.15)

The case on closed manifold: let $\\{(U_{i},\varphi_{i},x)\\}_{i=1}^{N}$ be a
collection of charts covers $M$. By passing to a finer cover, we may assume
$\Omega^{q}(M)$ trivializes on each $U_{i}$ with the corresponding
trivialization (composited with $\varphi_{i}\times id$)

$f_{i}:E|_{U_{i}}\rightarrow\varphi^{-1}(U_{i})\times\mathbb{R}^{\text{rank}\Omega^{q}(M)}$

By the previous case of the estimation on $\mathbb{R}^{n}$,

$\displaystyle\|h\|_{2n}^{\prime}=\sum_{i=1}^{N}\|f_{i*}(\eta_{i}h)\|_{2n,\mathbb{R}^{n}}\leq\sum_{i=1}^{N}C_{i}\|f_{i*}\eta_{i}h\|_{\infty,\mathbb{R}^{n}}\leq\max_{1\leq
i\leq N}\\{C_{i}\\}\sum_{i=1}^{N}\|f_{i*}\eta_{i}h\|_{\infty,\mathbb{R}^{n}}$
$\displaystyle=\max_{1\leq i\leq N}\\{C_{i}\\}\|h\|^{\prime}_{\infty}$

and this completes the proof. ∎

Now fix $0<C<\varepsilon^{2}/4$. Denote $L_{q}^{2}(M)$ be the completion of
$\Omega^{q}(M)$ with respect to $(\cdot,\cdot)$. Let

$P_{\leq e^{-Ct}}^{q}:L_{q}^{2}(M)\rightarrow E_{[0,e^{-Ct}],t}^{q}(M)$

be the orthogonal projection with respect to $(\cdot,\cdot)$. Denote
$\tilde{v}=P_{\leq e^{-Ct}}^{q}v$ be the orthogonal projection of $v$ onto
$E_{[0,e^{-Ct}],t}^{q}(M)$. We give an upper bound of the sup norm of

$\|v-\tilde{v}\|_{\infty}=\sup_{x\in M}|v(x)-\tilde{v}(x)|$

in the following theorem:

######  Theorem 4.3.

We can find $\varepsilon_{0}>0$ and $t_{0}>0$ such that when $t\geqslant
t_{0}$, we have

$\|v-\tilde{v}\|_{\infty}\leq e^{-\varepsilon_{0}t}.$

###### Proof.

Applying the Gårding inequalities [Tay81, theorem 8.1], we get

$\displaystyle\|v-\tilde{v}\|_{2n}\leq\hat{C}t((\triangle_{t}^{(q)})^{2n}(v-\tilde{v}),v-\tilde{v})+\hat{C}\|v-\tilde{v}\|=\hat{C}t\|(\triangle_{t}^{(q)})^{n}(v-\tilde{v})\|+\hat{C}\|v-\tilde{v}\|$
(4.16)

Note that if we write $v=\sum_{i=0}^{\infty}u_{i}$, $u_{i}\in
E_{\lambda_{i},t}(M)$, then

$\displaystyle\|(\triangle_{t}^{(q)})^{n}(v-\tilde{v})\|=\sum_{i=N+1}^{\infty}\lambda^{n}_{i}\|u_{i}\|\leq\sum_{i=0}^{\infty}\lambda^{n}_{i}\|u_{i}\|=\|(\triangle_{t}^{(q)})^{n}v\|\leq\exp(-\frac{t\varepsilon^{2}}{4}),$
(4.17)

for $t\geq t_{n}$ by lemma 4.1. The number $N$ is the largest integer so that
$\lambda_{i}\leq e^{-Ct}$ for all $i=0,\ldots,N$.

On the other hand, also by lemma 4.1, when $t\geq t_{1}$

$\displaystyle\exp(-\frac{t\varepsilon^{2}}{4})\geq\|(\triangle_{t}^{(q)})v\|\geq\sum_{i=N+1}^{\infty}\lambda_{i}\|u_{i}\|\geq
e^{-Ct}\sum_{i=N+1}^{\infty}\|u_{i}\|=e^{-Ct}\|v-\tilde{v}.\|$ (4.18)

Then we have

$\displaystyle\|v-\tilde{v}\|_{2n}\leq\hat{C}t\exp(-\frac{t\varepsilon^{2}}{4})+\hat{C}\exp(-(\frac{\varepsilon^{2}}{4}-C)t)$
(4.19)

for $t\geq\max\\{t_{1},t_{n}\\}$. Take
$\varepsilon_{0}=\min\\{\varepsilon^{2}/4,\varepsilon^{2}/4-C\\}/2$. Then
there exists $t_{0}\geq\max\\{t_{1},t_{n}\\}$ so that when $t\geq t_{0}$,

$\|v-\tilde{v}\|_{2n}\leq\exp(-\varepsilon_{0}t).$

And by lemma 4.2,

$\|v-\tilde{v}\|_{\infty}\leq\|v-\tilde{v}\|_{2n}\leq\exp(-\varepsilon_{0}t)$

when $t\geq t_{0}$. ∎

Write $\text{Crit}(f;q)=\\{p_{1},\ldots,p_{m_{q}}\\}$. By the theorem above we
know that there exists a sequence of sections
$\beta=\\{\tilde{v}_{1},\ldots,\tilde{v}_{m_{q}}\\}$ so that $\tilde{v}_{j}\in
E_{[0,e^{-Ct}],t}^{q}(M)$ and

$\displaystyle\tilde{v}_{j}=v_{j}+R_{j}$ (4.20)

where $v_{j}$ satisfies formula (4.2) and $\|R_{j}\|_{\infty}\leq
e^{-\varepsilon_{0}t}$.

To end this subsection, We claim that $\beta$ is linearly independent when $t$
is sufficiently large. We define $A(t),\tilde{A(t)}\in M_{m_{q}\times
m_{q}}(\mathbb{R})$ by

$A(t)_{ij}=(v_{i},v_{j})$

and

$\tilde{A}(t)_{ij}=(\tilde{v_{i}},\tilde{v_{j}}).$

By our construction, $A(t)$ is always positive definite diagonal matrix for
any $t>0$. We check it in the following: for $i\neq j$, the supports of
$v_{i}$ and $v_{j}$ are disjoint, and thus $(v_{i},v_{j})=0$. On the other
hand, when $i=j$, $(v_{i},v_{i})>0$ since $v_{i}$ is non-trivial.

In particular, $\det A(t)>0$ for any $t>0$. Since

$\det:M_{m_{q}\times m_{q}}(\mathbb{R})\rightarrow\mathbb{R}$

is a real-valued continuous function, there exists $\delta>0$ so that $\det
B>\det A(t)/2$ for any $B\in M_{m_{q}\times m_{q}}(\mathbb{R})$ with
$\sup_{ij}|A(t)_{ij}-B_{ij}|<\delta$. But by Theorem 4.3,

$|A(t)_{ij}-\tilde{A}_{ij}|\leq(\|R_{i}\|_{\infty}^{2}(v_{j},v_{j})+\|R_{j}\|_{\infty}^{2}(v_{i},v_{i})+\|R_{i}\|_{\infty}\|R_{j}\|_{\infty})\int_{M}dv_{g}<\delta$

when $t$ is sufficiently large. Thus, when $t$ is sufficiently large

$(a_{1}\tilde{v}_{1}+\cdots+a_{m_{q}}\tilde{v}_{m_{q}},a_{1}\tilde{v}_{1}+\cdots+a_{m_{q}}\tilde{v}_{m_{q}})=\sum_{i,j=1}^{m_{q}}a_{i}a_{j}(\tilde{v}_{i},\tilde{v}_{j})$

is identically zero if and only $a_{1}=\cdots=a_{m_{q}}=0$. Thus, there exists
at least $m_{q}$ linear independent elements in $E^{q}_{[0,e^{-Ct}],t}(M)$, in
other words,

$\dim E^{q}_{[0,e^{-Ct}],t}(M)\geq m_{q}.$

###  4.3 Upper bound of $\dim E^{q}_{[0,e^{-Ct}],t}(M)$

Conversely, we need to show $\dim E^{q}_{[0,e^{-Ct},t]}(M)\leq m_{q}$. To do
this, it suffices to show $\beta$ we chose in the last section spans all
elements in $E^{q}_{[0,e^{-Ct},t]}(M)$. We claim that

######  Theorem 4.4.

There exists $C>0$ and $s>0$ so that when $t\geq s$, for all
$u\in\Omega^{q}(M)$, $u\in(\text{span}\beta)^{\perp}$, we have

$(\triangle_{t}^{(q)}u,u)\geq Ct\|u\|^{2}.$

Let $S$ be the open covering appeared in (3.18). Let $\\{\varphi_{U}\\}$ be a
partition of unity subordinated on $S$ so that

$\sum_{U\in S}\varphi_{U}^{2}=1$

and

$\varphi_{U_{p}}=1\text{ on }\text{supp}\chi$

for all $p\in\text{Crit}(f)$.

######  Lemma 4.5.

There exists a constant $C>0$, independent of $t$, so that for all $t\geq 0$
and for all $u\in\Omega^{q}(M)$,

$(\triangle^{(q)}_{t}u,u)\geq\sum_{U\in
S}(\triangle_{t}^{(q)}(\varphi_{U}u),\varphi_{U}u)-C\|u\|^{2}.$

###### Proof.

Applying formula (3.1), we have

$(\triangle_{t}^{(q)}u,u)=\int_{M}\left(|du|^{2}+|d^{*}u|^{2}+t^{2}|df|^{2}|u|^{2}+t(\sum_{k,l}\text{Hess}_{f}(\frac{\partial}{\partial
x_{l}},\frac{\partial}{\partial
x_{k}})[dx_{l}\wedge,dx_{k}\lrcorner]u,u)\right)$

Since $\sum\varphi_{U}d\varphi_{U}=0$ and

$\displaystyle\sum|d(\varphi_{U}u)|^{2}$
$\displaystyle=|du^{2}|+\sum|d\varphi_{U}\wedge u|^{2}$
$\displaystyle\sum|d^{*}(\varphi_{U}u)|^{2}$
$\displaystyle=|d^{*}u^{2}|+\sum|d\varphi_{U}\lrcorner u|^{2}$

we obtain for any $u\in\Omega^{q}(M)$,

$\sum_{U\in
S}(\triangle_{t}^{(q)}(\varphi_{U}u),\varphi_{U}u)=\sum\int_{M}|d\varphi_{U}\wedge
u|^{2}+|d\varphi_{U}\lrcorner u|^{2}dv_{M}+(\triangle_{t}^{(q)}u,u)\leq
C\|u\|^{2}+(\triangle_{t}^{(q)}u,u).$

∎

######  Lemma 4.6.

There exists $C>0$ and $s>0$ such that for all $u\in\Omega^{q}_{0}(V)$,

$\displaystyle(\triangle_{t}^{(q)}u,u)\geq Ct\|u\|^{2},$ (4.21)

when $t\geq s$.

###### Proof.

From the Bochner type formula, we know that

$(\triangle_{t}u,u)=(\triangle
u,u)+t^{2}|df|^{2}\|u\|^{2}+t(\sum_{l,k}\text{Hess}_{f}(\frac{\partial}{\partial
x_{l}},\frac{\partial}{\partial x_{k}})[dx^{l}\wedge,dx^{k}\lrcorner]u,u).$

On $V$, $|df|^{2}>c$ for some $c>0$, so there exists $t_{0}>0$ such that

$\displaystyle(\triangle_{t}u,u)\geq\frac{c}{2}t^{2}\|u\|^{2}\geq\frac{ct_{0}}{2}t\|u\|^{2}$
(4.22)

when $t\geq t_{0}$. Take $C=ct_{0}/2$ and $s=t_{0}$ and we complete the proof.
∎

In section 5, we will introduce a operator $\Box_{t}$ which can be viewed as
the Witten Laplacian on $\mathbb{R}$ (see (4.32)), and the spectrum of
$\Box_{t}$ will be fully characterized. This result is essential in the proof
of Lemma 4.7 and 4.8 below.

######  Lemma 4.7.

For any $p\in\text{Crit}(f)\backslash\text{Crit}(f;l)$ there exists $C>0$ such
that

$(\triangle_{t}^{(l)}u,u)\geq Ct\|u\|^{2}$

for any $u\in\Omega^{l}_{0}(U_{p})$.

###### Proof.

Since in $U_{p}$ the metric is flat and $|df|^{2}=|x|^{2}$, we may regard $u$
as a $l$ form on $\mathbb{R}^{n}$ and $\triangle_{t}$ as $\Box_{t}$. By our
study of the spectrum of $\Box_{t}$, $(\triangle_{t}u,u)\geq Ct\|u\|^{2}$ for
some $C>0$. ∎

######  Lemma 4.8.

For any $p\in\text{Crit}(f;q)$, there exist $C>0$, $s>0$ and $\varepsilon>0$
so that when $t\geq s$,

$(\triangle_{t}^{(q)}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)\geq
C\left(t\|\varphi_{U_{p}}u\|^{2}-e^{-\varepsilon t}\|u\|^{2}\right)$

for all $\varphi_{U_{p}}u\in\Omega^{q}(U_{p})$ and
$u\in(\text{span}\beta)^{\perp}$.

###### Proof.

Like the previous proof, we also regard $u$ as a $q$ form on $\mathbb{R}^{n}$
and $\triangle_{t}$ as $\Box_{t}$. By our calculation in the appendix, there
exists a complete orthonormal system $\\{u_{i}\\}_{i=0}^{\infty}$ with

$\displaystyle\begin{cases}u_{0}=\pi^{-n/4}t^{-n/4}\exp(-tx^{2}/2)dx_{1}\wedge\cdots\wedge
dx_{q}\in\ker{\Box_{t}}\\\ u_{i}\in E_{\lambda_{i}}(\Box_{t})\end{cases}$
(4.23)

also, $0<\lambda_{1}\leq\lambda_{2}\leq$ and $\lambda_{i}\sim O(t)$. Write
$\varphi_{U_{p}}u=\sum_{i=0}^{\infty}a_{i}u_{i}$. Then

$\displaystyle(\Box_{t}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)_{\mathbb{R}^{n}}$
$\displaystyle=\sum_{i=1}^{n}\lambda_{i}a_{i}^{2}\geq
Ct\sum_{i=1}^{n}a_{i}^{2}=Ct(\|\varphi_{U_{p}}u\|^{2}-a_{0}^{2})$
$\displaystyle=Ct(\|\varphi_{U_{p}}u\|^{2}-t^{n/2}(\varphi_{U_{p}}u,e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q})^{2}).$

Triangle inequality gives

$\displaystyle|(\varphi_{U_{p}}u,e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q})|$ $\displaystyle\leq|(\varphi_{U_{p}}u,\chi
e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge dx_{q})|$
$\displaystyle+|(\varphi_{U_{p}}u,(1-\chi)e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q})|$

Cauchy-Schwartz inequality tells us

$\displaystyle|(\varphi_{U_{p}}u,(1-\chi)e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q})|\leq\|\varphi_{U_{p}}u\|\|(1-\chi)e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q}\|$ $\displaystyle\leq Cte^{-t\varepsilon^{2}/4}\|\varphi_{U_{p}}u\|$

Also, we have

$\displaystyle|(\varphi_{U_{p}}u,\chi e^{-tx^{2}/2}dx_{1}\wedge\cdots\wedge
dx_{q})|=|(u,\tilde{v}-R_{p})|=|(u,R_{p})|\leq\|u\|e^{-\epsilon_{0}t}$ (4.24)

Therefore,
$(\triangle_{t}^{(q)}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)\geq\tilde{C}t(\|\varphi_{U_{p}}u\|^{2}-e^{\epsilon_{0}t}\|u\|)$
when $t$ is sufficiently large. ∎

###### Proof of Theorem 4.4.

By Lemma 4.5,

$(\triangle_{t}^{(q)}u,u)\geq\sum_{U\in
S}(\triangle_{t}^{(q)}\varphi_{U_{p}}u,\varphi_{U_{p}}u)-C_{1}\|u\|^{2}$

We have

$\sum_{U\in
S}(\triangle_{t}^{(q)}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)=(\triangle_{t}^{(q)}(\varphi_{V}u),\varphi_{V}u)+\sum_{p\in\text{Crit}(f)\backslash\text{Crit}(f;q)}(\triangle_{t}^{(q)}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)+\sum_{p\in\text{Crit}(f;q)}(\triangle_{t}^{(q)}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)$

Take $C>0$ so that it satisfies Lemmas 4.6, 4.7 and 4.8 such that

$\displaystyle\sum_{U\in S}(\triangle_{t}(\varphi_{U_{p}}u),\varphi_{U_{p}}u)$
$\displaystyle\geq
Ct\|\varphi_{V}u\|^{2}+Ct\sum_{p\in\text{Crit}(f)\backslash\text{Crit}(f;q)}\|\varphi_{U_{p}}u\|^{2}+C\sum_{p\in\text{Crit}(f;q)}(t\|\varphi_{U_{p}}u\|^{2}-\varepsilon^{-\epsilon_{0}t}\|u\|^{2})$
$\displaystyle\geq\frac{t}{\tilde{C}}\|u\|^{2}-\tilde{C}(1+e^{-\epsilon_{0}t})\|u\|^{2}$

Thus,

$(\triangle_{t}^{(q)}u,u)\geq\tilde{\tilde{C}}t\|u\|^{2}.$

when $t\gg 1$ for some $\tilde{\tilde{C}}>0$. ∎

We finally arrive at the stage of showing $\dim E^{q}_{[0,e^{-Ct}],t}(M)\leq
m_{q}$. Assume $\dim E^{q}_{[0,e^{-Ct}],t}(M)>m_{q}$, then there exists $u\in
E^{q}_{[0,e^{-Ct}],t}(M)$ which is perpendicular to
$\text{span}\beta^{\perp}$. But this is impossible, by theorem 4.4,

$(\triangle_{t}^{(q)}u,u)\geq\tilde{\tilde{C}}t\|u\|^{2}$

when $t\gg 1$. On the other hand, because $u\in E^{q}_{[0,e^{-Ct}],t}$,

$(\triangle_{t}^{(q)}u,u)\leq e^{-Ct}\|u\|^{2}.$

Thus, $\tilde{\tilde{C}}t<e^{-Ct}$ when $t\gg 1$, a contradiction. Therefore,

$\dim E^{q}_{[0,e^{-Ct}],t}(M)\leq m_{q}.$

###  4.4 Eigenvalues of Harmonic Oscillator

Consider the operator $H=-\frac{\partial}{\partial x}+x^{2}$ on
$L^{2}(\mathbb{R})$, called the Harmonic Oscillator (In Quantum Mechanics,
this operator is the Hamiltonian of a system called the one-dimensional
Harmonic Oscillator (although we need to time $\frac{1}{2}$ in natural units),
and this is the reason why we adopt this name).

The calculation in the continuing paragraphs will derive the eigenvalues of
$H$. The result is

$EV(H)=\\{2n+1:n\in\mathbb{Z}_{\geq 0}\\}$

where $EV(H)$ denotes the set of eigenvalues of $H$.

Let $\\{A_{n}(x)\\}$ be the polynomials determined by the formula

$\sum_{n=0}^{\infty}A_{n}(x)\frac{\alpha^{n}}{n!}=e^{-\alpha^{2}+2\alpha x}.$

Define

$\phi_{n}(x)=(2^{n}n!)^{-1/2}\pi^{-1/4}A_{n}(x)e^{-x^{2}/2}$

for all $n\in\mathbb{Z}_{\geq 0}$. We claim that $\\{\phi_{n}\\}$ contains all
the eigenfunctions of the operator $H$.

To do this, we need to show that it is a complete orthonormal system for
$L^{2}(\mathbb{R})$. We do this by the following steps:

  1. 1.

Explicit formulas of $\phi_{n}$: we first show the equality

$\phi_{n}(x)=(-1)^{n}(2^{n}n!)^{-1/2}\pi^{-1/4}e^{x^{2}/2}\frac{d^{n}}{dx^{n}}e^{-x^{2}}$

holds. By generalized Cauchy integral formula,

$\displaystyle\frac{d^{n}}{d\alpha^{n}}\bigg{\rvert}_{\alpha=0}$
$\displaystyle e^{-\alpha^{2}+2\alpha x}=\frac{n!}{2\pi
i}\int_{\gamma}\frac{\exp(-\alpha^{2}+2\alpha x)}{\alpha^{n+1}}d\alpha$

replacing $\alpha$ with $\zeta-x$ then gives

$\displaystyle=e^{x^{2}}(-1)^{n}\frac{n!}{2\pi
i}\int_{\gamma^{\prime}}\frac{-\zeta^{2}}{(\zeta-x)^{n+1}}d\zeta=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}},$

where $\gamma$ and $\gamma^{\prime}$ are anticlockwise loop around $0$ and
$x$, respectively. Therefore, the proof follows from the Taylor expansion of
$e^{-\alpha^{2}+2\alpha x}$.

  2. 2.

Recursion formulas of $A_{n}$: note that the polynomials $\\{A_{n}\\}$
satisfies the recursion formulas

$\displaystyle\begin{split}A_{0}&=1\\\ A_{n}^{\prime}&=2nA_{n-1}\\\
A_{n+1}&=2xA_{n}-2nA_{n-1}.\end{split}$ (4.25)

for $n\geq 1$. Formulas (4.25) can be obtained directly: from 1, we know that

$A_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}.$

Applying derivation on both sides gives

$\displaystyle\begin{split}A^{\prime}_{n}(x)&=(-1)^{n}\frac{d}{dx}\left(e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\right)\\\
&=(-1)^{n}2xe^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}+(-1)^{n}e^{x^{2}}\sum_{k=0}^{n}\binom{n}{k}\frac{d^{k}}{dx^{k}}(-2x)\frac{d^{n-k}}{dx^{n-k}}e^{-x^{2}}\\\
&=(-1)^{n}2xe^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}-(-1)^{n}2xe^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}+(-1)^{n-1}2ne^{x^{2}}\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\\\
&=2nA_{n-1}(x).\end{split}$ (4.26)

On the other hand,

$\displaystyle A^{\prime}_{n}(x)$
$\displaystyle=(-1)^{n}\frac{d}{dx}(e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}})$
$\displaystyle=(-1)^{n}2xe^{x^{2}}\frac{d^{n}}{dx^{2}}e^{-x^{2}}+(-1)^{n}e^{x^{2}}\frac{d^{n+1}}{dx^{n+1}}e^{-x^{2}}$
$\displaystyle=2xA_{n}(x)-A_{n+1}(x).$

By formula (4.26), we can replace $A_{n}^{\prime}(x)$ by $2nA_{n-1}(x)$, and
this completes our derivation of (4.25).

Now, by the recursion formulas (4.25) and 1,

$\displaystyle H\phi_{n}+(x^{2}-(2n+1))\phi_{n}$ $\displaystyle=$
$\displaystyle(2^{n}n!)^{-1/2}\pi^{-1/4}(-e^{-x^{2}/2}(x^{2}-1)A_{n}+2xe^{-x^{2}/2}A^{\prime}_{n}-e^{-x^{2}/2}A^{\prime\prime}_{n}+(x^{2}-(2n+1))e^{-x^{2}/2})$
$\displaystyle=$ $\displaystyle
e^{-x^{2}/2}(2^{n}n!)^{-1/2}\pi^{-1/4}(4xnA_{n-1}-4n^{2}A_{n-2}-2nA_{n})=\frac{e^{-x^{2}/2}}{2^{n}n!}(2nA_{n}-2nA_{n})=0.$

This gives $2n+1$ an eigenvalue of $H$ with $\phi_{n}$ the corresponding
eigenfunction.

  3. 3.

Orthogonality: for any two non-negative integers $n$ and $m$, define

$I_{n,m}=\int_{\mathbb{R}}e^{-x^{2}}A_{n}(x)A_{m}(x).$

We claim that $I_{n,m}=2^{n}n!\sqrt{\pi}\delta_{n,m}$: from recursion formulas
(4.25), one can derive immediately that for $n\geq 1$,

$\displaystyle A_{n}^{\prime\prime}(x)-2xA_{n}^{\prime}(x)+2nA_{n}(x)=0$
(4.27)

the verification is left to the readers. Formula (4.27) can be also written as

$e^{x^{2}}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{n}(x))+2nA_{n}=0.$

Assume $n\neq m$. We can then deduce the following to equations hold:

$\displaystyle
e^{x^{2}}A_{m}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{n}(x))+2nA_{n}A_{m}=0$
$\displaystyle
e^{x^{2}}A_{n}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{m}(x))+2mA_{n}A_{m}=0$

Subtracting them then gives

$\displaystyle
2(n-m)e^{-x^{2}}A_{n}A_{m}=\left(A_{n}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{m}(x))-A_{m}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{n}(x))\right).$
(4.28)

By product rule of differentiation,

$\displaystyle
A_{n}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{m}(x))=\frac{d}{dx}(A_{n}e^{-x^{2}}A^{\prime}_{m}(x))-A^{\prime}_{n}e^{-x^{2}}A^{\prime}_{m}(x)$
$\displaystyle
A_{m}\frac{d}{dx}(e^{-x^{2}}A^{\prime}_{n}(x))=\frac{d}{dx}(A_{m}e^{-x^{2}}A^{\prime}_{n}(x))-A_{m}^{\prime}e^{-x^{2}}A^{\prime}_{n}(x)$

Substituting them to formula (4.28) and integrating on $\mathbb{R}$ give us
that

$\displaystyle 2(n-m)\int_{\mathbb{R}}e^{-x^{2}}A_{n}(x)A_{m}(x)dx$
$\displaystyle=\int_{\mathbb{R}}\frac{d}{dx}(A_{n}e^{-x^{2}}A^{\prime}_{m}(x)-A_{m}e^{-x^{2}}A^{\prime}_{n}(x))dx$
$\displaystyle=A_{n}e^{-x^{2}}A^{\prime}_{m}(x)-A_{m}e^{-x^{2}}A^{\prime}_{n}(x)\bigg{|}_{-\infty}^{\infty}=0$

and thus $I_{n,m}=0$ when $n\neq m$.

Now, it remains to show $I_{n,n}=2^{n}n!\sqrt{\pi}$ for all $n\geq 0$. Note
that by recursion formulas (4.25)

$\displaystyle 0=I_{n-1,n+1}$
$\displaystyle=\int_{\mathbb{R}}e^{-x^{2}}A_{n-1}(x)A_{n+1}(x)dx$
$\displaystyle=\int_{\mathbb{R}}2xe^{-x^{2}}A_{n-1}(x)A_{n}(x)dx-2nI_{n-1,n-1}$

or

$\displaystyle
2nI_{n-1,n-1}=\int_{\mathbb{R}}2xe^{-x^{2}}A_{n-1}(x)A_{n}(x)dx$ (4.29)

Because $A_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}$, the right
hand side of (4.29) equals to

$\displaystyle-\int_{\mathbb{R}}2xe^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}dx$
(4.30)

By product rule of differentiation,

$2xe^{x^{2}}\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}=\frac{d}{dx}(e^{x^{2}}\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}})-e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}.$

Pluging in (4.30) and observing that
$A_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}$, we then derive

$\displaystyle 2nI_{n-1,n-1}$
$\displaystyle=\int_{\mathbb{R}}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}dx-\int_{\mathbb{R}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\frac{d}{dx}(e^{x^{2}}\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}})dx$
$\displaystyle=\int_{\mathbb{R}}e^{-x^{2}}A_{n}(x)A_{n}(x)dx+e^{-x^{2}}A_{n}(x)A_{n-1}(x)\bigg{|}_{-\infty}^{\infty}+\int_{\mathbb{R}}e^{-x^{2}}A_{n-1}(x)A_{n+1}(x)dx$
$\displaystyle=I_{n,n}$

This recursive result immediately gives us

$I_{n,n}=2^{n}n!I_{0,0}=2^{n}n!\sqrt{\pi}$

due to the fact that

$I_{0,0}=\int_{\mathbb{R}}e^{-x^{2}}dx=\sqrt{\pi}.$

To sum up, we have proven that $I_{n,m}=2^{n}n!\sqrt{\pi}\delta_{n,m}$. As

$\int_{\mathbb{R}}\phi_{n}(x)\phi_{m}(x)=(2^{n}n!)^{-1/2}\pi^{-1/4}(2^{m}m!)^{-1/2}\pi^{-1/4}I_{n,m}=\delta_{n,m},$

we conclude that the system $\\{\phi_{n}\\}$ is orthogonal.

  4. 4.

Completeness: to show completeness of the system $\\{\phi_{n}\\}$, it suffices
to show $f\in L^{2}(\mathbb{R})$ is orthogonal to the system $\\{\phi_{n}\\}$
if and only if $f=0$ almost everywhere.

For any $f\in L^{2}(\mathbb{R})$ such that $(f,\phi_{n})=0$ for all
$n\in\mathbb{Z}_{\geq 0}$, we claim that $f=0$ almost everywhere. Note that
each

$P_{n}(x)=e^{x^{2}}\frac{d^{n}e^{u^{2}}}{du^{n}}(x)$

is a degree $n$ polynomial. Therefore, the set ${P_{n}}$ is a basis for the
linear space of polynomial. This means

$\int_{\mathbb{R}}f(x)\phi_{n}(x)=(-1)^{n}(2^{n}n!)^{-1/2}\pi^{-1/4}\int_{\mathbb{R}}f(x)e^{-x^{2}/2}P_{n}(x)dx=0$

if and only if

$\int_{\mathbb{R}}f(x)e^{-x^{2}/2}x^{n}dx=0.$

As a result, the entire function

$\displaystyle
F(z)=\int_{\mathbb{R}}f(x)e^{-\frac{x^{2}}{2}+xz}dx=\sum_{n=0}^{\infty}\int_{\mathbb{R}}f(x)e^{-x^{2}/2}x^{n}dx=0.$
(4.31)

If we take $z=-it$, then (4.31) says $\widehat{fe^{-x^{2}/2}}=0$. Fourier
inversion formula then gives $f=0$ almost everywhere, which means the system
$\\{\phi_{n}\\}$ is complete.

Denote $L^{2}_{q}(\mathbb{R}^{n})$ as the set of square integrable $q$-form on
$\mathbb{R}^{n}$. Define an operator $\Box_{t}^{(r)}$ acting on
$L^{2}_{q}(\mathbb{R}^{n})$ by

$\displaystyle\Box_{t}udx_{J}=(-\triangle
u+t^{2}|x|^{2}u)dx_{J}-tu\sum_{j=1}^{r}[dx_{j}\wedge,dx_{j}\lrcorner](dx_{J})+tu\sum_{j=r+1}^{n}[dx_{j}\wedge,dx_{j}\lrcorner](dx_{J})$
(4.32)

where $\triangle$ denotes the standard Laplacian on $\mathbb{R}^{n}$ and
$|J|=q$.

Since $[dx_{j}\wedge,dx_{j}\lrcorner](dx_{J})=\varepsilon_{J}^{j}$, where

$\displaystyle\varepsilon_{J}^{j}=\begin{cases}1&\text{if $j\in J$}\\\
-1&\text{if $j\not\in J$}\end{cases}$

the eigenvalues of $\Box_{t}$ on $L^{2}_{q}(\mathbb{R}^{n})$ are

$\displaystyle\left\\{t\sum_{j=1}^{r}(2N_{j}+1-\varepsilon_{J}^{j})+t\sum_{j=r+1}^{n}(2N_{j}+1+\varepsilon_{J}^{j}):(N_{1},\ldots,N_{n})\in\mathbb{Z}_{\geq
0}\text{ and }|J|=q\right\\}$ (4.33)

So we get

$\displaystyle\ker(\Box_{t}^{(r)})=\begin{cases}0&\text{if $r\neq q$}\\\
\mathbb{R}e^{-tx^{2}}dx_{1}\wedge\cdots\wedge dx_{q}&\text{if
$r=q$}\end{cases}$ (4.34)

The eigenvalues other than $0$ are all of the form $Kt$ for some $K>0$.

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