Atiyah-Singer index theorem 2

2 10月 201720 12月 2017 hxypqr留下评论

1. rough outline of heat kernel proof of Atiyah-singer index theorem

1.1. proof strategy

Theorem 1 (Mckean-Singer formula.)
$\displaystyle ind(D^+)=Str(e^{-tD^2})=\int\limits_{x \in M} Str(K(x,y)).$

from this we know Fredholm operator deformation invariance,in the same time we need chern-weil theory.
Main challenge:
1. in the expansion on heat kernel ,we need to proof when ${ t \rightarrow 0}$ , the limit exist and find a way to calculate it.
2. indentify the limit as ${t \rightarrow 0}$ .
Proof: Our proof road锛� mckean-singer formula ${\rightarrow}$ local-index thm ${\rightarrow}$ A-S index thm ${\rightarrow}$ Riemann-roch-Hirzebunch theorem. $\Box$

1.2. preliminary work

superbundle: ${E=E^+ \oplus E^-}$ .
on compact manifold ${M}$ , ${D: \Gamma(M,E) \rightarrow \Gamma(M,E)}$ is a self-adjoint operator . ${ D = }$ . ${D^+=D|_E^+,D^-=D|_E^-}$ .
observe that:
${D}$ is symmetric ${\Longrightarrow}$ eigenvalue space of ${D^2}$ is finite dimention ${\Longrightarrow}$ in particular ${Ker D^2}$ is finite dimesion ${\Longrightarrow}$ ${Ker D}$ is finite dimension.
dimention of superspace ${E = E^+ \oplus E^-}$ :

$\displaystyle dim E=dim E^+ - dim E^-.$

$\displaystyle kerD=kerD^+ \oplus ker D^-$

Def: ${ind D^+=dim ker D^+ -dim ker D^-}$ .
Lemma:
1.Let ${D}$ be a self-adjoint Dirac operator on a clifford module ${E}$ over a compact manifold ${M}$ ,then

$\displaystyle \Gamma(M,E^{\pm})=ker D^{\pm} \oplus im D^{\mp}$

in particular,
$\displaystyle ind D^+ = dim ker D^+ -dim coker D^+$

where coker ${D^+ :=\Gamma(M,E^-)/im D^+}$ .
2.Let ${D}$ be a differential operator acting on a ${Z_2}$ -graded vector bundle ${E}$ ,then ${Str[D,K]=0}$ .
the proof of this two lemma is easy,leave sas exercise.
1.3. Mckean-Singer formula

the formula is:

$\displaystyle ind(D^+)=Str(e^{-tD^2})=\int\limits_{x \in M} Str(K(x,y)).$

the expression of heat operator by spectral measure is:
$\displaystyle e^{-tD^2}=\int\limits_{0}^\infty d^{\lambda t}dE_{\lambda}.$

proof 1:
we have first eigenvalue estimate on compact manifold:
$\displaystyle |Str(e^{-tD^2}-P_0)| \leq Cvol(M)e^{-t\lambda}.$

$\displaystyle \Longrightarrow$

$\displaystyle \lim\limits_{t \rightarrow \infty}Str(e^{-tD^2}) = Str p_0 =dim kerD^+ -dim ker D^- =ind D^+.$

on the other hand ,we need to show ${Str e^{-tD^2}}$ is independent with ${t}$ ,in fact:
$\displaystyle \frac{d}{dt} Str (e^{-tD^2})=-Str(D^2 e^{-tD^2}).$

${D}$ odd parity : ${\Longrightarrow \ D^2e^{-tD^2}=[D,D E^{-tD^2}]}$ .
${[\ \ ,\ \ ]}$ supercommunater ${\Longrightarrow \ \frac{d}{dt}Str (e^{-tD^2})= -Str[D,D e^{-tD^2}]=0.}$
q.e.d
proof2:
by spectral decompositon of ${e^{-tD^2}}$ :
$\displaystyle Str (e^{-tD^2})=\sum\limits_{\lambda \geq 0}(n_{\lambda}^+ - n_{\lambda}^-)e^{-t\lambda}$

observe that: ${n_{\lambda}^+=n_{\lambda}^-}$ for ${\lambda \not=0}$ . ${\Longrightarrow ind D=n_0^+ - n_0^-}$ . (detail in [BGV])
q.e.d
Corallary: the index of a smooth on-parameter family of Dirac operator is constant.
what we have proved is:
$\displaystyle ind D = Str (e^{-tD^2})=\int Str(k(x,y)).$

1.4. analytic formula of ind ${D^+}$

from the discuss of heat kernel in section 2,we know following result(section 2 only discuss the case of function but use the similar way we can get similar result on bundle):

$\displaystyle K_t(x,y) \sim (4 \pi t)^{\frac{n}{2}} \sum\limits_{i=0}^{+\infty} t^i K_i(x), K_i \in \Gamma(M,End(E)).$

on the other hand: ${ D=\left[\begin{array}{ccc} 0 & D^- \\ D^+ & 0 \end{array}\right] \Longrightarrow D^2=\left[\begin{array}{ccc} D^-D^+ & 0\\ 0 & D^+D^- \end{array}\right] }$ .
and use the Mckean-Singer formula,we get: ind (D^+)&=&Str(e^{-tD^2})
&=&Tr(e^{-tD^+D^-})-Tr(e^{-tD^=D^-})
&=&\int\limits_M Tr(K_t(x,y,D^-D^+)) – \int\limits_M Tr(k_t(x,y,D^+D^-))
&=&\sum\limits_{i=0}^{\infty} t^{i-\frac{n}{2}}a_i(D^+D^-) – \sum\limits_{i=0}^{\infty} t^{i – \frac{n}{2}}a_i(D^+D^-). where ${a_i}$ is the heat trace invariants.
take ${ t \rightarrow 0}$ ,the only thing make sense is the series of order ${\frac{n}{2}}$ ,and we want to proof:
$\displaystyle ind(D^+) =a_{\frac{n}{2}}(D^-D^+) -a_{\frac{n}{2}}(D^+D^-).$

But the difficult thing is that the high order series is very hard ro calculate….
our strategy is following:

Step1: proof ${Str(K_t(x,y))}$ has a limit as ${t \rightarrow 0}$ i.e ${Str(K_t(x,y))\stackrel{t \rightarrow 0}{\longrightarrow}}$ index density.
step2:use a rescaling of space,time,clifford bundles ,to find a way that make us only need to calculate the leader coefficient.

1.5. From the McKean鈥揝inger formula to the index theorem

Let ${M}$ be a compact oriented Riemannian manifold of even dimension ${n}$ . We will write ${k_t(x, y)}$ for the heat kernel associated to ${D^2}$ . The diagonal ${k_t(x, x)}$ is a section of ${End(E )}$ which is iso- morphic to ${Cl(M) \otimes End_{Cl(M)}(E )}$ . Using this isomorphism, we define a filtration on ${End(E )}$ , induced by the filtration on ${Cl(M)}$ . Elements of ${End_{Cl(M)}(E )}$ are given 0-degree. Denote by ${Cl_i(M)}$ the subbundle of ${Cl(M)}$ consisting of all elements of degree less or equal to ${i}$ .
the following theorem hold:
Theorem 1. The following statements hold:
1. The coefficients ${k_i}$ have degree less or equal to ${2i}$ . In other words, ${k_i \in \Gamma (M, Cl_{2i}(M) \otimes End_{Cl(M)}(E))}$ . 2. If ${ \sigma(k):= \sum \limits_{i=0}^{n/2}\sigma_{2i}(k_i) \in A(M,End_{Cl(M)}(E))}$ ,where ${\sigma_j :Cl_j(M)鈫扐_j(M)}$ denotes the ${i=0}$ restriction of the symbol map, then:

$\displaystyle \sigma (k) = A藛(M) exp(鈭扚E /S ).$

$\displaystyle ind (D^+)=Str(e^{-tD^2})=\sum\limits_{i=0}^{\infty} t^{i-\frac{n}{2}}a_i(D^+D^-) - \sum\limits_{i=0}^{\infty} t^{i - \frac{n}{2}}a_i(D^+D^-).$

the important observation is that the ${Str}$ of order less than n all vanish,in the other word:

lemma2:for any quadratic space ${v}$ of dimension ${n}$ , ${CL_{n-1}(V) =[CL(V),CL(V)]}$ .

Proof: Let ${e_1,...,e_n}$ be a basis of ${V}$ . For any multi-index ${I \subset \{1,...,n\}}$ , denote by ${c_I}$ the Clifford product ${\prod_{i \in I} c}$ .Then the set ${\{c_I\}}$ is a basis for ${Cl(V)}$ .If ${|I|<n}$ ,there is at least one ${j}$ such that ${j \notin I}$ , and we have
$\displaystyle c(e_I)=-\frac{1}{2}[c_j,c_j c_I].$

q.e.d
so take ${t \rightarrow 0}$ , we get:
$\displaystyle ind (D) =(4 \pi)^{\frac{n}{2}} \sum\limits_{i = \frac {n}{2}}^{\infty} t^{i- \frac{n}{2}}Str(k_i(x)).$

now we want to identify the term ${Str(k_{\frac{n}{2}}(x))}$ as a characteristic form on ${M}$ :
Lemma 3. Let ${V}$ be a Euclidean space. There is, up to a constant factor, a unique supertrace on ${Cl(V)}$ , equal to ${T \circ \sigma}$ where ${\sigma}$ denotes the symbol map and ${T}$ is the projection of ${\alpha \in \wedge V}$ onto the coefficient of ${e_1 \wedge, ... ,\wedge e_n}$ if ${e_1,...,e_n}$ form an oriented orthonormal basis of ${V}$ . Furthermore, the supertrace defined above equals:
$\displaystyle Str(a) = (鈭�2i)^{\frac{n}{2}} (T \circ \sigma(a)).$

rmk:(The map ${T}$ is also called the canonical Berezin integral.)
proof:
the dimension of ${Str}$ space is one because of ${CL_{n-1}(V)=[CL(V),CL(V)]}$ and it never be empty because there is a natural defined supertrace on ${CL(V)}$ .so the only thing we need to do is to determine the constant,in face we only need to calculate the supertarce on any non-zero element .for instance the chirality operator ${\Gamma}$ . We have that ${ Str(\Gamma) = dim (\wedge P) = 2^{n/2}}$ and therefore ${Str(a) = (鈭�2i)^{n/2} T \circ \sigma (a)}$ for all ${a \in Cl(M)}$ .
q.e.d
so we know:
${\forall \ }$ section ${a \otimes b \in \Gamma(M \otimes CL(M) \otimes End_{CL(M)}(E)}$ :
$\displaystyle Str_E( a \otimes b)(x) = (-2i)^{\frac{n}{2}} \sigma_n(a(x))Str _{E/S}(b(x)).$

$\displaystyle Str_E(k_{\frac{n}{2}})(x) = (-2i)^{\frac{n}{2}} Str _{E/S}(\sigma_n(k_{\frac{n}{2}})).$

theorem1 implies then the following theorem for the index of a Dirac operator associated to a Clifford connection which is known as the local index theorem.

Theorem 2. (Local index theorem) Let ${M}$ be a compact, oriented even-dimensional manifold and let ${E}$ be a Clifford module with Clifford connection ${\nabla E}$ . Let ${D}$ be the associated Dirac operator. Then ${\lim\limits_{t \rightarrow 0} Str(k_t(x, x))|dx| }$ exists and is obtained by taking the ${n}$ -th form piece of
$\displaystyle (2\pi i)^{鈭抧/2} \widehat A(M)ch(E /S ).$

rmk:This theorem only holds for Dirac operators associated to Clifford connections, which are those which are compatible with the Clifford action. However, since the index of a Dirac operator is independent of the Clifford superconnection used to define it, we get Atiyah鈥揝inger index formula for any Dirac operator.

Theorem 3. (Atiyah鈥揝inger Index Theorem) Let ${M}$ be a compact, oriented, even-dimensional manifold and let ${D}$ be a Dirac operator on a Clifford module ${E}$ . Then the index of ${D}$ is given by :
$\displaystyle ind D = (2 \pi i)^{-\frac{n}{2}}\int\limits_{M} \widehat A(M) ch(E/S) .$

hence theorem 3 is a consequence of theorem 1.
up to now,to prove index theorem ,we only need to prove theorem 1.
1.6. idea of the proof of theorem 1

we give the clear proof in appendix锛宼here we explain the idea:
To prove Theorem 4.11 we mainly follow Chapter 4 of [BGV] but rearrange the different steps in order to make the proof clearer, at least for us. Let us summarize the first part of the proof:

1. The idea of the proof is to work in normal coordinates ${x}$ around a point ${x_0 \in M}$ . Near the diagonal, we use parallel transport to pull back the heat kernel ${k_t(x, x_0)}$ which is a section of the vector bundle ${E_x \otimes E_x^*}$ and define a new kernel ${k(t, x) := \iota(x_0, x)k_t(x, x_0)}$ , a section of ${End(E_{x_0} ) \cong Cl(V^*) \otimes End(W)}$ for some twisting space ${W}$ . Using the symbol isomorphism ${蟽}$ , we can look at ${k(t, x)}$ as a section of ${\wedge(V^*) \otimes E (W)}$ .

2. We use Lichnerowicz鈥� formula to get the explicit form of the operator ${L}$ such that the kernel ${k(t, x)}$ satisfies the heat equation ${(\partial t + L)k(t, x) = 0}$ .

3. In a third step, we define a rescaling of space, time and the Clifford algebra, introduced by Getzler. This rescaling has the effect that the leading coefficient of the asymptotic expansion of the rescaled kernel is exactly the differential form ${\sigma(k)}$ of theorem 1 which leads to a reformulation of Theorem 1.

\appendix

2. Chern-Weil theory

some text in Appendix A

3. Complete proof of theorem 1

some text in Appendix B

Heat Kernel proof of Index Theory 1

25 9月 201720 12月 2017 hxypqr留下评论

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:

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

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:

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

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

1.4. chern class

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

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.
$\displaystyle F^{E \widehat\otimes W/S =F^{E/S} }\otimes 1_W + 1_E \otimes F^W.$

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

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

Whitney product formula.
1.5. Atiyah-singer index theorem

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

2. heat kernel

2.1. basic setting

Assume M is a general Riemann manifold ,assume

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

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.
$\displaystyle \Phi(\delta)=\{ w \in C^\infty 1-form | \int |w|^2 + \int |\delta w|^2 < +\infty \}$

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),
$\displaystyle \Delta=\Delta_D=\Delta{\mathbb N}=\overline \delta \overline d$

.
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
$\displaystyle e^{-\Delta t} = \int \limits_0^\infty d^{\lambda t}dE_{\lambda} , t>0$

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

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:

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

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

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

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:
$\displaystyle \frac{\partial}{\partial t} f_1 = - \int\limits_0^\infty \lambda e^{-\lambda t} dE_{\lambda}(f) = -\Delta( e^{-\lambda t} f)=\Delta f_1$

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

.
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 ${\longrightarrow P(x,y,t),P(x,y,t) \in C^\infty (M \times M \times R^+)}$ ,
$\displaystyle \lim \limits_{t \rightarrow 0} P(x,y,t) = \delta_y$

. 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:
$\displaystyle P(x,y,t) \sim \frac{exp(-d(x,y)^2/4t)}{(4\pi t)^{n/2}} \sum\limits_i a_i(x,y)t^i.$

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 ${\epsilon \rightarrow 0}$ ,
$\displaystyle \lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t} P(x,y,\epsilon) = P(x,y,t)-\int\limits_0^t e^{-\Delta(t-s)} \circ_x P(x,y,s)ds =\limits_{def} H(x,y,t).$

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:
$\displaystyle |e^{-\Delta(t-s)} \circ_x P(x,y,s)| = O(s^N).$

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:
$\displaystyle \int H(x,y,t)f(y)dy=\lim\limits{\epsilon \rightarrow 0} \int\limits_M e^{-\Delta t} P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}\lim\limits_{\epsilon \rightarrow 0} \int \limits_M P(x,y,\epsilon)f(y)dy\\ =e^{-\Delta t}f(x).$

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
$\displaystyle e^{-\Delta t} f(x) = \int H(x,y,t)f(y) , \ \forall f\in L^2(M)........................................(\star\star).$

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:
$\displaystyle H(x,y,t)=\lim\limits_{\epsilon \rightarrow 0} e^{-\Delta t-\epsilon} P(x,y,\epsilon), \ \forall \epsilon >0.$

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:
$\displaystyle H(x,y,t) = \int\limits_M H(x,z,t-s)H(z,y,s)dz..$

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:

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

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 ${r}$ :
$\displaystyle \Delta \psi = \frac{d^2 \psi}{dr^2} + (\frac{d \ log(\sqrt g)}{dr})\frac{d \psi}{dr}.$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${Lemma 4.1}$ :

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

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:
$\displaystyle \varepsilon(r(x,y),t) \leq H(x,y,t).$

(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:
$\displaystyle H(x,y,t)= \sum e^{-\lambda_i t} f_i(x)f_i(y)$