232 23. HEAT KERNEL FOR STATIC METRICS
STEP 3. H is the fundamental solution. Recall from Proposition 23.12
that we have for any f E c^0 (M)
lim ( PN(x,t;y,u)f(x)dμ(x)=f(y).
t\,u}M
By (23.59), that His the fundamental solution follows from the claim that
(23.60) lim ( (PN f (DxPN )k) (x, t; y, u) f (x) dμ (x) = 0.
t\,u} M k=l
Now we prove (23.60). For simplicity we take u = 0. By (23.69) and
(23.70) below, we have
(23.61)
~I L.J (DxPN) *kl (x, y, t) :::; Ct N -2 n exp (CVol N (M) _ I! tN-~+1) + d^2 (x,y)
1
e--5t-.
k=l 2
Therefore
(23.62)
(PN * t, (DxPN)•k) (x, y, t)
00
PN (x, z, s) L (DzPN)*k (z, y, t - s) dμ (z) ds
k=l
CVol(M)(t-s) N -"2" n+l d2(z,y)
PN (x, z, s) C (t - s)N-~ e N-~+^1 e -^5 (t-s) dμ (z) ds,
which tends to zero as t "';; 0 uniformly in ( x, y) since N > ~. Thus we
obtain (23.60) for u = 0. The theorem now follows from Lemma 26.4. D
2.3. The heat kernel expansion.
We have the following.
THEOREM 23.17 (Heat kernel expansion). Ast-+ 0, we have
(23.63) H(x, y, t) ~ ( 4?rt)-n/^2 exp ( - d2 ~; y)) ~ti u; (x, y).
More precisely, equation (23.63) means that there exists a sequence of func-
tions Uj E C^00 (M x M) such that
(23.64) H (x, y, t)-( 4?rt)-n/Z exp ( - d2~; y)) t, t•uk( x, y) "c WN(x, y, t ),
with
uo(y, y) = 1
and