Connes-Chern Character for Manifolds with Boundary and ETA ...

CONNES-CHERN CHARACTER AND ETA COCHAINS 35
2.3. The transgression formula. To prove the transgression formula we proceed
analogously and start with the supercommutator
k∑
([
(−1)
∫∆ j b Str q Dt , a 0 e −σ 0D 2 t [Dt , a 1 ] . . .
k+1
j=0
. . . [D t , a j ]e −σ jD 2 t Ḋe −σ j+1D 2 t . . . [Dt , a k ]e −σ k+1D 2 t
])
dσ. (2.44)
We compute this supercommutator using Proposition 1.9. Note that I(Ḋt, λ) is
proportional ∫ to iΓλ + D ∂ . The summand iΓλ contributes a term proportional to
∞
−∞ λe−λ2 dλ = 0. The remaining summand gives, since ∫ e −λ2 dλ = √ π,
k∑
(−1) j 〈a 0,∂ , [D ∂ t , a 1,∂ ], . . . , [D ∂ t , a j,∂ ], Ḋ∂ t , . . . , [D ∂ t , a k,∂ ]〉
j=0
= /ch k (D ∂ , ˙ D ∂ )(a 0,∂ , . . . , a k,∂ ).
(2.45)
Let us again emphasize that here we are in the case q + 1, where the grading is the
induced grading on the boundary and E q+1 = −Γ.
Next we expand the commutator (2.44). However, we will confine ourselves to small
k. The calculation is basically the same as in [GBVF01, p. 451]. The only difference is
that on a closed manifold (2.44) is a priori 0 while here it coincides with the transgressed
Chern character on the boundary.
Case 1: k = 0. (2.44) expands to
On the other hand
b 〈[D t , a 0 ], Ḋt]〉 + b 〈a 0 , [D t , Ḋt]〉. (2.46)
B b /ch 1 (D t , Ḋt)(a 0 ) = b /ch 1 (D t , Ḋt)(1, a 0 )
= b 〈1, Ḋt, [D t , a 0 ]〉 − b 〈1, [D t , a 0 ], Ḋt〉
= − b 〈[D t , a 0 ], 1, Ḋt〉 − b 〈[D t , a 0 ], Ḋt, 1〉
= − b 〈[D t , a 0 ], Ḋt]〉,
(2.47)
by (2.32). Moreover using the well–known formula
we have
hence altogether
d
dt e−σD2 t
∫ σ
= − e (σ−s)D2 t [Dt , Ḋt]e −sD2 t ds, (2.48)
0
d b Ch 0 (D t )(a 0 ) = − b 〈a 0 , [D t ,
dt
Ḋt]〉, (2.49)
d b Ch 0 (D t ) + B b /ch 1 (D t ,
dt
Ḋt) = − /ch 0 (D ∂ t , Ḋ∂ t ). (2.50)