A mass for asymptotically complex hyperbolic manifolds

hal-00429306, version 1 - 2 Nov 2009
A MASS FOR ASYMPTOTICALLY COMPLEX HYPERBOLIC MANIFOLDS. 17
The second term is basically computed in [CH]. Indeed, since H is symmetric, we use
the identity [X·,ej·] = 2Xj + 2X · ej· to obtain
Re II = − i
i
(HX · φ,φ) −
2 2
2m
j,l=1
Hjl (X · ej · el · φ,φ)
= − i
i
(HX · φ,φ) + Tr H (X · φ,φ) .
2 2
In the same way, the third term can be written
Lemma 3.7 — Re
III = − 1
(JHX ·
2
˜
φ,φ − 1
2
2m
j=1
2m
j=1
X · ej · JHej · ˜
φ,φ .
X · ej · JHej · ˜
φ,φ ≈ −2 JHX · ˜
φ,φ .
Proof. Let us set M := JH and Mij := (ei,Mej), so that
2m
j=1
X · ej · JHej · ˜
φ,φ
=
j,k,p
MkjXp
ep · ej · ek · ˜
φ,φ .
Lemma 3.4 ensures JH ≈ HJ. Since H is symmetric and J antisymmetric, we deduce
that M is antisymmetric up to a negligible term. In particular, Mkj ≈ 0 when k = j,
hence
2m
j=1
X · ej · JHej · ˜
φ,φ
≈
j=k
p
MkjXp
ep · ej · ek · ˜
φ,φ .
Given three distinct indices j,k,p we consider the expression
ep · ej · ek · ˜
φ,φ = (ep · ej · ek · (φl−1 − φl),(φl−1 + φl)) .
Property (7) reduces it into
ep · ej · ek · ˜
φ,φ = (ep · ej · ek · φl−1,φl) − (ep · ej · ek · φl,φl−1) .
and since the indices are distinct, this is imaginary. So
Re
2m
j=1
≈ Re
j=k
≈ − Re
X · ej · JHej · ˜
φ,φ
MkjXj
j=k
≈ −2Re
j=k
MkjXj
MkjXj
≈ −2Re MX · ˜
φ,φ
≈ −2 MX · ˜
φ,φ .
ej · ej · ek · ˜
φ,φ
ek · ˜
φ,φ
ek · ˜
φ,φ
+ Re
+ Re
j=k
j=k
MkjXk
MkjXk
ek · ej · ek · ˜
φ,φ
ej · ˜
φ,φ