Download Heat kernel expansion: user's manual

The most elegant way to calculate the remaining constant α10 is based on
the Gauss-Bonnet theorem (see [?]). We use here a more lengthy way [?] which
however works perfectly on flat manifolds without boundary and of trivial
topology of Rn . Note that Rn is non-compact. To make the heat kernel well
defined we should suppose certain fall-off conditions on the background fields
and on the smearing function f . The basis in the space of the square integrable
functions is given by the plane waves exp(ikx). Therefore, for M = Rn with
flat metric the heat kernel reads:
K(f ; t) = TrL2 (f exp(−tD))
Z
Z
dn k −ikx
n
= d x
e
trV {f (x) exp(−tD)eikx }
n
(2π)
Z
Z
dn k
n
trV {f (x) exp(t((∇µ + ik µ )2 + E))} .
= d x
n
(2π)
(4.21)
The following integrals will be useful:
Z
dn k −tk2
1
e
=
,
n
(2π)
(4tπ)n/2
Z
dn k −tk2 µ ν
1 µν
1
g ,
e
k k =
n
n/2
(2π)
(4tπ) 2t
Z
dn k −tk2 µ ν ρ σ
1 µν ρσ
1
e
k k k k =
(g g + g µρ g νσ + g µσ g νρ). (4.22)
n
n/2
(2π)
(4tπ) 4t2
Now we isolate exp(−tk 2 ) on the right hand side of (4.21) and expand the rest
of the exponent in a power series of t.
Z
K(f ; t) =
n
d x
Z
d4 k −tk2
e
trV
(2π)n
(
t2
f (x) 1 + t(∇ + E) − 4(k∇)
2
2
t2 2 2
(∇ ∇ + ∇2 E + E∇2 + E 2 )
2
4t3
((k∇)2 E + E(k∇)2 + (k∇)E(k∇))
−
6
4t3
−
((k∇)2 ∇2 + ∇2 (k∇)2 + (k∇)∇2 (k∇))
6
!)
16t4
4
+
(k∇) + . . .
24
+
We use the integrals (4.22) to obtain
38
(4.23)