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

56 MATTHIAS LESCH, HENRI MOSCOVICI, AND MARKUS J. PFLAUM
By Eq. (6.11) we have
( )
b ch k t (D) − b k
˜ch t (D), ch k+1
t (D ∂ ) − ch k−1
t (D ∂ )
(
)
= − T/ch k t (D ∂) ◦ i ∗ , −(b + B) T/ch k t (D ∂)
= (˜b + ˜B) ( )
0, T/ch k t (D ∂) ,
(6.15)
hence the two pairs differ only by a coboundary.
Next let us compute ( b ch k+2
t (D), ch k+3
t (D ∂ ) ) − S ( b ch k t (D), ch k+1
t (D ∂ ) ) in the above
relative cochain complex. Using Eq. (6.4) one checks immediately that
b ch k+2
t (D) − b ch k t (D) = b Ch k+2 (tD) + B b T/ch k+3
t
(D) − B T/ch k+1
t
(D)
= − (b + B) T/ch k+1
t
(D) − T/ch k+2
t
(D ∂ ) ◦ i ∗ .
From the second line of Eq. (6.4) (or from [CoMo93, Sec. 2.1])
one thus gets
ch k+3
t (D ∂ ) − ch k+1
t (D ∂ ) = −(b + B) T/ch k+2
t
(D ∂ ),
( b
ch k+2
t (D), ch k+3
t (D ∂ ) ) − S ( b ch k t (D), ch k+1
t (D ∂ ) )
= (˜b + ˜B) ( − b T/ch k+1
t
(D), T/ch k+2
t
(D ∂ ) ) .
(6.16)
Hence, the relative cocycles ( b ch k+2
t (D), ch k+3
t (D ∂ ) ) and S ( b ch k t (D), ch k+1
t (D ∂ ) ) are
cohomologous. Similarly, one gets
b ch k t (D) − ch k τ(D) =
resp.
= ∑ ( b
Ch k−2j (tD) − b Ch k−2j (τD) ) ∫ t
+ B b /ch k+1 (sD, D) ds
j≥0
τ
= − (b + B) ∑ ∫ t
b /ch k−2j−1 (sD, D) ds − ∑ ∫ t
/ch k−2j (sD ∂ , D ∂ ) ds,
j≥0 τ
j≥0 τ
ch k+1
t (D ∂ ) − ch k+1
τ (D ∂ ) = −(b + B) ∑ j≥0
∫ t
τ
/ch k−2j (sD ∂ , D ∂ ) ds,
hence ( b ch k t (D), ch k+1
t (D ∂ ) ) and ( b ch k τ(D), ch k+1
τ (D ∂ ) ) are cohomologous in the total
relative complex as well. Thus, we have proved (1)-(3) of the following result.
(
Theorem 6.1. (1) The pairs of retracted cochains b
ch k t (D), ch k+1
t (D ∂ ) ) ,
( k b ˜ch t (D), ch k−1
t (D ∂ ) ) , t > 0, t > 0, k ≥ m = dim M, k − m ∈ 2Z are
cocycles in the relative total complex Tot • ⊕ BC •,• (C ∞ (M), C ∞ (∂M)).
(2) They represent the same class in HC n (C ∞ (M), C ∞ (∂M)) which is independent
of t > 0.
(3) They represent the same class in HP • (C ∞ (M), C ∞ (∂M)) which is independent
of k.