256and- HEAT KERNEL FOR STATIC METRICS
Llx ( d^2 ( x, y)) L,1;!=0 Llx<faktk I L,1;!=0 \7 xtPk tk 12Llx log H N = - + N - 2
4t Lk=O ¢ktk ( L,1;!=0 ¢ktk)= _ Llx ( d
2
( X, Y)) + Llx¢o _ I \7 x ¢0 J 2 + O ( t).
4t ¢0 ¢5Note that since, by (23.125) with x = y, we have
1
¢1 (y,y) = (Llx¢o) (x,y)lx=y = 5R(y)
and ¢0 (y, y) = 1, summing the above two equations and evaluating at x = y
yields( ut ~ + Llx) logHN (x, y, t)I x=y = -~ t + ~R 3 (y) +^0 (t)'
where we also used JV x¢ol^2 (x, y) lx=y = 0 and Llx ( d^2 (x, y)) lx=y = 2n.^12
Since(23.128) log (t,¢kt') ~ log¢o + ::t + 0 (t^2 ),
by (23.127) we have
logHN +~log (47rt) = d
2
~' y) + log¢ 0 + :~t + 0 (t^2 ).Thus, in general, for any x, y EM, we have
(
at^8 + Ux A)(l Og H N + 2 n log ( 47rt )) + ----t---logHN+~log(47rt)(23129) -Llx(d
(^2) (x,y))+4log¢o 2¢1+Llx¢0 1Vx¢ol (^2) O()
- 4t + ¢0 - ¢5 + t.
In particular, evaluating (23.129) along the diagonal x = y, we have
(23.130)
[ ( :t + Llx) (log HN + ~log ( 47rt)) + log HN +!log ( 47rt)] lx=y
n 1
= -2t + 2R (y) + 0 (t).(^12) On the other hand,
( :t -~"') logHN (x,y, t)lx=y = 0 (t).