260
then
- HEAT KERNEL FOR STATIC METRICS
det M ( 8) = 1 + 8 · tr (A) + 8^2 (tr ( B) + t tr^2 (A) - t tr (A^2 ))
+ 83 ( tr (A) ( ~ tr^2 (A) - ~ tr ( (A)
2
) + tr ( B)) )
+ ~ tr ( (A)^3 ) - tr (AB) + tr ( C)
+0(8^4 ).
Moreover, if A = 0, then
(23.139) det M (8) = 1+8^2 tr (B) + 83 tr (C)
- 8^4 (ttr^2 (B) -t tr (B^2 ) +tr (D)) + 0 (8^5 ).
PROOF. Recall that if M = M ( 8) is a time-dependent invertible matrix,
then
(23.140) ds d detM = detM ·tr ( M-^1 dM) d
8 ,
so that if Mis given by (23.138), then
: 8 's=O detM =tr ( d:).
Differentiating (23.140), we have
(23.141)
d
2
det M = det M · (tr^2 (M-^1 dM) -tr (M-^1 dM M-^1 dM))
~ ~ ~ ~
(
+ det M · tr M -1d2M) d
82 ,
so that if Mis given by (23.138), then
We compute
(23.142)
:: 2 1s=O det M = (tr
2
(A) - tr (A
2
) + 2 tr (B)).