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

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CONNES-CHERN CHARACTER AND <strong>ETA</strong> COCHAINS 31<br />

2.1.2. q = −1. Now let D be ungraded <strong>and</strong> put ˜D, α, E 1 as in Eqs. (2.12), (2.13), (2.14).<br />

Then by Proposition 2.4 we have<br />

b Ch k (D t )(a 0 , . . . , a k ) = √ π b Ch k (˜D t )(a 0 , . . . , a k ), (2.21)<br />

b /ch k (D t , Ḋt)(a 0 , . . . , a k ) = √ π b /ch k (˜D t , ˜D ˙<br />

t )(a 0 , . . . , a k ). (2.22)<br />

In the collar of the boundary, we write as usual D = Γ d + D dx ∂, <strong>and</strong> thus<br />

( ) ( )<br />

0 Γ d 0<br />

˜D =<br />

Γ 0 dx + D∂<br />

. (2.23)<br />

D ∂ 0<br />

} {{ } } {{ }<br />

=: Γ e =: D f ∂<br />

˜D ∂ is 2–graded <strong>with</strong> respect to<br />

( )<br />

0 1<br />

E 1 = , E<br />

−1 0 2 = −˜Γ =<br />

Note that<br />

For even k we have<br />

( )<br />

0 −Γ<br />

. (2.24)<br />

−Γ 0<br />

αE 1 E 2 = −Γ ⊗ I 2 , ˜D∂ = αE 1 (D ∂ ⊗ I 2 ). (2.25)<br />

(<br />

Str 2 a0 e −σ 0D g 2<br />

∂,t<br />

[˜D∂,t , a 1 ] · . . . · [˜D ∂,t , a k ]e −σ kD g 2)<br />

∂,t<br />

= 1<br />

4π Tr( αE 1 E 2 (αE 1 ) k( ) )<br />

a 0 e −σ 0D 2 ∂,t [D∂,t , a 1 ] · . . . · [D ∂,t , a k ]e −σ kD 2 ∂,t ⊗ I2 (2.26)<br />

= − 1<br />

2π Tr( )<br />

Γa 0 e −σ 0D 2 ∂,t [D∂,t , a 1 ] · . . . · [D ∂,t , a k ]e −σ kD 2 ∂,t .<br />

With respect to the grading given by −iΓ we can now write<br />

−1<br />

2π Tr( Γ·) = 1<br />

2πi Str 0 . (2.27)<br />

Together <strong>with</strong> (2.21) <strong>and</strong> (2.22) we have thus proved:<br />

Proposition 2.6. Let M be an odd dimensional compact manifold <strong>with</strong> boundary <strong>with</strong><br />

an exact b-metric <strong>and</strong> let D be an ungraded Dirac operator. Writing D in a collar of<br />

the boundary (in cylindrical coordinates) in the <strong>for</strong>m<br />

D =: Γ d<br />

dx + D ∂, (2.28)<br />

D ∂ is a graded Dirac type operator <strong>with</strong> respect to the grading operator −iΓ. Furthermore,<br />

we have<br />

b b Ch k−1 (D t ) + B b Ch k+1 (D t ) = 1<br />

2 √ πi Chk (D ∂ ) ◦ i ∗ , (2.29)<br />

d b Ch k (D t ) + b b /ch k−1 (D t ,<br />

dt<br />

Ḋt) + B b /ch k+1 (D t , Ḋt)<br />

= − 1<br />

2 √ πi /chk (D ∂ t , Ḋ∂ t ) ◦ i ∗ .<br />

(2.30)

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