# Heat Kernel proof of Index Theory 1

1. framework of atiyah singer index theory

1.1. A genus form

${(M,g)}$ campact,complete,Riemann manifold without boundary,dim ${ M=2m ,m \in N^* }$. ${\bigtriangledown^g}$ is the Levi-civita connection on ${TM}$,${R=R_g\in \Omega^2(End(TM))}$.
${\widehat A}$ genus form:

$\widehat A(M,g)=det^{\frac{1}{2}}(\frac{\frac{i}{4\pi}R_g}{sinh(\frac{i}{4\pi}R_g)})\in \Omega(M).$

by chern-weil theory,we know:
${ 1. \widehat A }$ is closed.
${ 2. \widehat A_{g1}-\widehat A_{g2} }$ is exact.
so we can def ${\widehat A_g =\widehat A(M) \ \forall g}$ is a metric on M.
1.2. dirac bundle

${(E,\bigtriangledown^E)}$ is a dirac bundle.
${D: C^\infty(E^+) \rightarrow C^\infty(E^-)}$ is dirac operator.
${F^{E/S}\in End_{Cl(M)}(E)}$ the twisting curvature of ${E}$.

1.3. supertrace

${Str^{E / S}: End_{cl(M)} \rightarrow C_M}$. induce map:

$Str^{E/S} : \Omega(End_{cl(M)} (E) )\rightarrow \Omega(M) \otimes C$

determined by:
$Str^{E / S}( w \otimes T) = w \otimes Str^{E / S}(T). \ \forall w \in \Omega (M). \ \forall T \in End_{cl(M)} E.$

1.4. chern class

$ch^{E / S}(E) = Str^{E /S}[exp(\frac{i}{2\pi}F^{E/S})] \in \Omega(M).$

by chern-weil theory,we know:
1. ${ch^{E/S}(E)}$ closed form.
2. ${ch^{E \ S}(E)}$ only depend on the topology of E.
$F^{E \widehat\otimes W/S =F^{E/S} }\otimes 1_W + 1_E \otimes F^W.$

$F^W = F^{W^+} \oplus F^{W^-}.$

$ch^{(\widehat E \otimes W)/S} ,ch(E \widehat\otimes W)=ch^{E/S}(E/S)(ch(F^{W^+})-ch(F^{W^-})).$

Whitney product formula.
1.5. Atiyah-singer index theorem

$Ind D_E=Dim(ker D_E)-Dim(ker D^*_E)=\int\limits_M \widehat A(M,g)ch^{E/S}(E/S).$

2. heat kernel

2.1. basic setting

Assume M is a general Riemann manifold ,assume

$\Phi(d)= \{f\in C^\infty(M) | \|f \| _1< +\infty\}$

Where ${\ \|f\|_1^2 = \int f^2+\int|df|^2}$.
complete ${\Phi(d) }$ with the norm ${\| . \|_1}$ is the sobolev space ${H^1(M) < L^2(M)}$.
The extension of operator ${d}$ in ${H^1(M)}$ is called ${\overline d}$,assume ${d_c}$ is ${d}$ restrict on ${ C_0^\infty(M)}$,${d_c}$ extend to ${H_0^1(M)}$ called ${\overline d_c}$,${H_0^1(M)}$ is ${C_o^\infty(M)}$ complete with the norm ${\|.\|_1}$, of course ${H_0^1(M)\subset H^1(M)}$.
Sobelev theory tell us, if ${M}$ is a complete Riemann manifold,then ${H_0^1(M)=H^1(M)}$.
Operator ${\delta=-*d*}$,satisfied ${<df,w>=<f,\delta w>}$,where * is the Hodge-star operator ,${w}$ is the ${C^\infty}$ 1-form, iff one of ${f,w}$ is compact supp.
$\Phi(\delta)=\{ w \in C^\infty 1-form | \int |w|^2 + \int |\delta w|^2 < +\infty \}$

By a lemma from Gaffney (Ann. of Math , 60,1954,458-466) we have ${ \overline \delta = \overline d_c^*, \overline \delta_c =\overline d^* }$.
the laplace operator(with Direchlet boundary condition or Neumann boundary condition) is ${\Delta_D = \overline \delta \overline d_c,\Delta_N =\overline \delta_C \overline d}$, When M is with smooth boundary,then ${\Delta =\Delta_N =general \Delta oprator}$,assume M is complete (${H_0^1(M)=H^1(M)}$),then because ${\overline d_C =\overline d,\overline \delta_C =\overline \delta}$,Gaffney proved(Ann. of Math , 60 ,1954,140-145),
$\Delta=\Delta_D=\Delta{\mathbb N}=\overline \delta \overline d$

.
As we all known, ${\Delta}$ is self-adjoint operator,so ${e^{-\Delta t}}$ make up a bounded self-adjoint operator semi group,by the self-adjoint operator theory,if ${dE_ {\lambda} }$ is the spectrum measure of ${\Delta}$,then
$e^{-\Delta t} = \int \limits_0^\infty d^{\lambda t}dE_{\lambda} , t>0$

.
And for ${t>0}$,${e^{-\Delta t}:L^2(M) \longrightarrow \bigcap_0^\infty D(\Delta^i) \subset C^\infty(M) }$.when ${f \in L^2(M)}$,
$\Delta^i(e^{-\Delta t}) = \int \limits_0^\infty \lambda^i e^{-\Delta t} dE_{\lambda}(f).$

where ${ \bigcap_{i=1}^\infty D(\Delta^i) \subset C^\infty(M)}$ is basic on Weyl theory.
2.2. existence of heat kernel

Now we proof the basic fact:
${Thm 2.1}$: assume ${M}$ is a complete manifold,then there exist a heat kernel ${H(x,y,t) \in C^\infty(M \times M \times R^+)}$,and ${ (e^{-\Delta t}f)(x) = \int_M H(x,z,t-s)H(z,y,s)dz}$,${\forall f\in L^2(M)}$,satisfied:
${(1) H(x,y,t)=H(y,x,t).\\ (2) \lim\limits_{t \rightarrow 0^+} H(x,y,t)= \delta_x(y).\\ (3) (\Delta – \frac{\partial}{\partial t})H=0.\\ (4) H(x,y,t) = \int H(x,y,t-s) H(z,y,s)dz. }$
Proof:
(A) first proof, ${\forall f \in L^2(M), e^{-\Delta t} f\in C^\infty(M \times R^+)}$.to proof this,we first proof,in weak sense:

$\frac{\partial}{\partial t}(e^{-\Delta t} f) = \int\limits_0^\infty -\lambda e^{-\lambda t} dE_{\lambda}(f).$

in fact,we have:
${ \frac{\partial}{\partial t} (e^{-\Delta t} f) \\=\lim\limits_{\epsilon \rightarrow 0} \frac{1}{\epsilon} [\int \limits_0^\infty e^{-\lambda(t+\epsilon)}dE_{\lambda}(f)-\int\limits_0^\infty e^{-\lambda t} dE_{\lambda}(f)] \\=\lim\limits_{\epsilon \rightarrow 0}[\int\limits_0^A \frac{e^{-\lambda \epsilon}-1}{\epsilon} e^{-\lambda t} dE_{\lambda}(f)+\int\limits_A^\infty \frac{e^{-\lambda\epsilon}-1}{\lambda\epsilon} \lambda e^{-\lambda t}dE_{\lambda}(f)] \\=\int\limits_0^A -\lambda e^{-\lambda t} dE_{\lambda}(f) + O(Ae^{-At} \|f\|) \\ \longrightarrow \int\limits_0^\infty -\lambda e^{-\lambda t} dE_{\lambda}(f) \ as\ {A \ \rightarrow +\infty} \dotfill (\star).}$
rmk: the limit is take in the ${L^2}$ space,every step in the caculate make sence because of the dominating convergence theorem.
so what have we proved ,in fact we proved in the classical sence, ${ \frac{\partial}{\partial t} (e^{-\Delta t} f) }$ exist. and our strategy is to proof any order of weak derivatives of it exist and then use the embedding theorem to prove it is smooth.
in fact to proof ${\frac{\partial}{\partial t}(e^{-\Delta t} f)=\int\limits_0^\infty -\lambda e^{-\lambda}dE_{\lambda}(f)}$ (in weak sence),we need only to proof:${\forall \psi \in C_0^\infty(M \times R^+)}$ we have:
$\int \frac{\partial \psi}{\partial t}(e^{-\Delta t}f) = -\int \psi(\int\limits_0^\infty -\lambda e^{-\lambda t}dE_{\lambda}(f))$

but because of ${(\star)}$,this is obvious,and similar we can proof that (in the weak sence):
$(\Delta+\frac{\partial^2}{\partial t^2})^i(e^{-\lambda t} f)= \int\limits_0^\infty (\lambda+\lambda ^2)^i e^{-\lambda t} dE_{\lambda}(f)$

rmk: there is some thing to explain,why ${\Delta(e^{-\lambda t} f)= \int\limits_0^\infty \lambda e^{-\lambda t} dE_{\lambda}(f)}$ .
anyway,we observe that ${L= \Delta+ \frac{\partial^2}{\partial t^2}}$ is the laplace operator on ${M \times R^+}$.and we have proved ${e^{-\lambda t}f \in \bigcap^\infty \Phi(L^i)}$,and we know that ${ \bigcap^\infty \Phi(L^i) \subset C^\infty }$.so we proved ${ e^{-\lambda t}f \in C^\infty( M \times R^+).}$
and now we easy to observe that,
if ${f_1(x,t)=e^{-\lambda t} f}$,then:
$\frac{\partial}{\partial t} f_1 = – \int\limits_0^\infty \lambda e^{-\lambda t} dE_{\lambda}(f) = -\Delta( e^{-\lambda t} f)=\Delta f_1$

.
so
$( \Delta – \frac{\partial}{\partial t}) f_1(x,t) = 0$

.
Rmk:${f_1(x,t) \in C^\infty}$,so derivatives is in classical sense.
(B) to proof ${ e^{-\lambda t} f =\int\limits_M H(x,y,t)f(y)dy}$.
by the decomposition of unity,we can assume ${f \in C_0^\infty(M)}$,and ${supp f }$ sufficed small.
consider the operator ${ \circ=\Delta +\frac{\partial}{\partial t}}$ and its quasi fundamental solution (paramatrix) ${\longrightarrow}$ see next section for the serious definition ${\longrightarrow P(x,y,t),P(x,y,t) \in C^\infty (M \times M \times R^+)}$,
$\lim \limits_{t \rightarrow 0} P(x,y,t) = \delta_y$

. and ${\forall N > 0, \lim\limits_{t \rightarrow 0} \circ_x P(x,y,t) =O(t^N)}$,and when ${d(x,y)}$ suffice small,${t \rightarrow +0 }$,we have expansion:
$P(x,y,t) \sim \frac{exp(-d(x,y)^2/4t)}{(4\pi t)^{n/2}} \sum\limits_i a_i(x,y)t^i.$

where ${n =dim M, d(x,y)= x,y}$ Riemann distance.${a(x,y) \in C^\infty(M \times M), a_0(x,y) =1}$,for ${0< \epsilon<s<t-s}$,

${ e^{-\Delta\epsilon}P(x,y,t-\epsilon)-e^{-\Delta(t-\epsilon)}P(x,y,\epsilon)\\ = \int\limits_{\epsilon}^{t-\epsilon} \frac{d}{ds}(e^{-\Delta(t-s)} P(x,y,s))ds\\ =\int\limits_{\epsilon}^{t-\epsilon} [ \Delta e^{ -\Delta(t-s)P(x,y,s)} + e^{-\Delta(t-s)} \frac{\partial P}{\partial s}(x,y,s)]ds\\ =\int\limits_{\epsilon}^{t-\epsilon} e^{-\Delta(t-s)} \circ _xP(x,y,s)ds.}$
use ${t}$ instead of ${t-\epsilon}$,and assume ${\epsilon \rightarrow 0}$,
$\lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t} P(x,y,\epsilon) = P(x,y,t)-\int\limits_0^t e^{-\Delta(t-s)} \circ_x P(x,y,s)ds =\limits_{def} H(x,y,t).$

assume ${ F(x,y,s) = \circ _x P(x,y,s)}$,then
${ \Delta^i \int\limits_0^t e^{- \Delta(t-s)} (x,y,s)ds\\ =\int\limits_0^t\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)} dE_{\lambda}(F(x,y,s))ds\\ =\int\limits_{s_0}^t \int\limits_0^\infty \lambda^i e^{-\lambda(t-s)}dE_{\lambda}(F(x,y,s))ds\\ + \int\limits_0^{s_0}\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)} dE_{\lambda}(F(x,y,s))ds\\ =\int\limits_{s_0}^t\int\limits_0^\infty \lambda^i e^{-\lambda(t-s)}dE_{\lambda}(F(x,y,s))ds+\int\limits_0^{s_0}O(s^N)ds. }$
this is because ${F(x,y,s) =O(s^N)}$,when ${s \rightarrow 0}$,so ${H(x,y,t) \in \Phi(\Delta^i)}$,so ${H(x,y,t) \in C^\infty(M \times M \times R^+)}$.because:
$|e^{-\Delta(t-s)} \circ_x P(x,y,s)| = O(s^N).$

so ${H(x,y,t)}$ and ${P(x,y,t)}$ have the same expansion .because${H(x,y,t) = \lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t} P(x,y,\epsilon)}$ so ${\forall f(y)\in C_0^\infty(M)}$,we have:
$\int H(x,y,t)f(y)dy=\lim\limits{\epsilon \rightarrow 0} \int\limits_M e^{-\Delta t} P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}\lim\limits_{\epsilon \rightarrow 0} \int \limits_M P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}f(x).$

the equality arrive is because of the prop of ${P(x,y,t)}$.
on the other hand ,by the definition of ${H(x,y,t)}$,we can check:
when ${y}$ is fix and ${t>0}$,${H(x,y,t) \in L^2(M)}$, so
$e^{-\Delta t} f(x) = \int H(x,y,t)f(y) , \ \forall f\in L^2(M)………………………………….(\star\star).$

we call ${H(x,y,t)}$ is the kernel of ${e^{-\Delta t}}$.

for prop (1),${H(x,y,t)=H(y,x,t)}$ is from the operator ${\Delta}$ is self-adjoint,prop (2) is just ${(\star\star)}$.

now we proof prop (3),by definition:
$H(x,y,t)=\lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t-\epsilon} P(x,y,\epsilon), \ \forall \epsilon >0.$

and we know ${ (\Delta_x – \frac{\partial}{\partial t})(e^{-\Delta t} P(x,y,t)) =0}$ so${ (\Delta_x -\frac{\partial}{\partial t})H(x,y,t) = 0}$.
proof prop (4),by ${e^{-\Delta s}e^{-\Delta(t-s)} = e^{-\Delta t}}$and ${(\star\star)}$ wo know:
$H(x,y,t) = \int\limits_M H(x,z,t-s)H(z,y,s)dz..$

QED.
rmk: for the manifold with bounded,we can also proof the existence of heat kernel with Dirchlet or Neumann boundary condition.(L.Chavel. Eigenvalues in Riemannian Geometry,Academic Press,1984)
2.3. quasi fundamental solution of heat equation

it is well known that for ${R^n}$, the fundamental solution of heat equation ${(\Delta – \frac{\partial}{\partial t})u=0}$ is ${ exp(-\frac{r^2}{4t})/(4\pi t)^{n/2}}$.
for general Riemann manifold ${M}$,we want to find the fundamental solution of heat equation with this form:

$U(x,y,t) \sim (4 \pi t)^{-\frac{n}{2}} e^{\frac{-d(x,y)^2}{4t} }{\sum\limits_{i \geq 0} \phi_i(x,y) t^i}.$

where ${d(x,y)}$ is the Riemann distance of two point of M. take the normal coordinate around point ${x}$, ${ y^i(i= 1,2,….,n),r=d(x,y)}$.(the length of geodesic connect ${x,y}$).
it is well known that there exist functions ${\psi(r) , \phi(r)}$ only depend on ${r}$ :
$\Delta \psi = \frac{d^2 \psi}{dr^2} + (\frac{d \ log(\sqrt g)}{dr})\frac{d \psi}{dr}.$

$\Delta( \phi \psi) = \phi \Delta \psi +\psi \Delta \phi +2 \frac{d \phi}{dr}\frac{d \psi}{dr}.$

take:
$\psi =(4 \pi t) ^{-\frac{n}{2}} e^{-\frac{r^2}{4t}}$

$\phi = \phi_0+\phi_1 t+…+\phi_N t^N,$

and
$u_N =\psi \phi =(4 \pi t) ^{-\frac{n}{2}} e^{-\frac{r^2}{4t}} \sum\limits_{i=0 … N} \phi_i t^i.$

then
$(\Delta – \frac{\partial}{\partial t}) u_N= \phi(\Delta \psi -\frac{\partial}{\partial t} \psi)+\psi(\Delta \phi -\frac{\partial}{\partial t}\phi)+2\frac{d\phi}{dr}\frac{d \psi}{dr}.$

because of
$\Delta \psi – \frac{\partial}{\partial t}\psi =\frac{d\ log\sqrt{g}}{dr}\frac{d \psi}{dr},$

$\frac{d \psi}{dr}=-\frac{r}{2t}\psi.$

we know:
$(\Delta -\frac{\partial}{\partial t})u_N = \frac{\psi}{t}\sum\limits_{k = 0 … N} [ \Delta \phi_{k-1} -(k+\frac{r}{2}\frac{d \ log\sqrt{g}}{dr})\phi_k -r\frac{d \phi_k}{dr}]t^k ,$

problem become to solve the equations:
$r\frac{d\phi_k}{dr} =(k+\frac{r}{2}\frac{d\ log\sqrt{g}}{dr})\phi_k =\Delta \phi_{k-1} , k=0,1…N.$

which is equivalent to:
$\frac{d}{dr}(r^k g^{\frac{1}{4}}\phi_k)=r^kg{\frac{1}{4}}\Delta\phi_{k-1},k\geq 1.$

$\frac{d \phi_0}{dr}+\frac{d \ log\sqrt{g}}{\partial dr}\phi_0=0 (take\ \ \phi_{-1} =0).$

solve it:
$\phi_0= g^{\frac{1}{4}}$

$\phi_k(x,y) =g^{\frac{1}{4}} r^{-k} \int\limits_0^{r(x,y)} r^{k-1}(\Delta \phi_{k-1})g^{\frac{1}{4}}dr.$

so ${\phi_k \in C^\infty(M),\forall k \in N}$. and:
$(\Delta – \frac{\partial}{\partial t})u_N=(4 \pi t)^{- \frac{n}{2}} e^{-\frac{r^2}{4t}} \Delta \phi_N t^N.$

take the cut function${ \theta \in C_0^\infty}$,${s.t}$:
$\theta(r)=1, when |r| \le \frac{1}{2}.$

$\theta(r)=0, when |r| \geq 1 .$

take:
$P_N(x,y,t) =\theta (r(x,y)) u_N(x,y,t),$

of course:
$Pn(x,y,t) \in C^\infty(M \times M \times R^+),and \ \\\ \lim\limits_{\epsilon \rightarrow 0} P(x,y,\epsilon) = \delta_x(y), (\Delta – \frac{\partial}{\partial t})P_N = O(t^N),$

this is to say:
${P_N(x,y,t)}$ is the quasi fundamental solution of heat equation. from the equation:
$(\Delta – \frac{\partial}{\partial t})u =G,$

$u|_{t=0} = 0,$

we know if ${G}$ has suffise high zero ,then so is the solution of heat equation,so ${P_n(x,y,t)}$ can approximate heat equation to any order.
2.4. basic proposition of heat kernel

${Lemma 4.1}$:

$H(x,y,t) >0 , \ \forall \ t > 0.$

proof strategy : begin with expansion of heat kernel and integral on a geodesic sphere and take the radius ${r \rightarrow 0}$,use the stokes formula and maximal value principle to proof ${H(x,y,t)}$ is always positive.

${Lemma 4.2}$: assume ${M}$ is a constant curvature complete riemann manifold (space form),then ${H(x,y,t)}$ only depends on ${r=d(x,y)}$,and${\frac{\partial H(r,t)}{\partial r} <0}$. proof is similar to ${Lemma 4.1}$.

${Thm 4.1}$(heat kernel comparision theorem,Cheeger-Yau).:
assume ${M}$ is a complete riemann manifold ,${Ric \geq (n-1)k}$,${ \forall x \in M,r_0 >0}$, heat kernel of ${ B(x,r_0), H(x,y,t)}$ and heat kernel ${\varepsilon(r(x,y),t)}$ of geodesic ball ${V(k,r_0)}$ in space form satisfied:
$\varepsilon(r(x,y),t) \leq H(x,y,t).$

(bounded condition is Derichlet condition or Neumann condition).
proof strategy: basically we use the formula ${\frac{1}{2}\Delta(|\bigtriangledown u|^2)=\sum\limits_{i , j} u_{ij}^2+\sum\limits_i u_i(\Delta u)_i+Ric (\bigtriangledown u,\bigtriangledown u)}$.and the two lemma. ${Thm 4.2}$: assume ${M}$ is a compact reimann manifold ,${f_i}$ is a orthonormal basis of special function on ${M}$,${ \lambda_i}$ is the corresponding spectrum,then the fundamental solution (heat kernel) has the expansion:
$H(x,y,t)= \sum e^{-\lambda_i t} f_i(x)f_i(y)$

in particular,
$\sum e^{-\lambda_i t} =\int\limits_M H(x,y,t)dx,$

when ${ t \rightarrow +0}$,
$H(x,y,t) \sim (4 \pi t)^{\frac{n}{2}}e^{\frac{d^2(x,y)}{4t}}\sum\limits_{i \geq 0} \phi_i(x,y) t^i .$

rmk:this thm is well known.