- THE CROSS CURVATURE FLOW 491
3.1. The cross curvature tensor. Let (M^3 ,g) be a Riemannian 3-
manifold with either negative sectional curvature everywhere or positivesectional curvature everywhere. Recall the cross curvature tensor c is
defined on p. 87 of Volume One by(B.29)(B.30)CiJ .. -.,--_:_ det E (E -1) ij -_ 2μ 1. ipq μ JTS. EprEqs
- (^1) 8μ pqk μ rs£R· i£pq R kJrs,.
w h ere E ij ::::;= · .1 D Lij · -^1 2 R gij is · th e E" ms t em · t ensor, d t e E ::::;= · det det Eij 9 ij , an d μ ijk are
the components of dμ with indices raised.
The equalities in the definition of Cij are a consequence of the following
identities.
LEMMA B .1 7 (Elementary curvature identities).
(a)(B.31)(b)
d e tE (E-1) ij _ - 8μ^1 pqk μ rs£R· i£pq R k1rs·.PROOF. (a) We compute in an oriented orthonormal basis so that μ123 =
μ^123 = 1 and Eij is diagonal. In such a frame, we have Eu = -R2332, etc.
Given f, define a and b so that abf is a cyclic permutation of 123. Thenμ pqk μ rs£R kjrs -^2 μ pqkR kjab -^2 μ pqa Uj KbR abab +^2 μ pqb Uj KaR baab
= 2 (μPqaoj - μpqb8j) Eu
= 2 (obojoj-o~oioJ-ozopj + ojo~o;) Eu
= 2 (oi (o~ - o~o]) - o~ (oJ - ojo])) Eu
= 2Eq£ or: J - 2Ep£ 8~. J(b) By part (a), we have8μ^1 pqk μ rs£R i£pq R kjrs = 4 1 R i£pq Em£ (KP ujum Kq - umuj KP Kq)
= 4 1 ( Eq £ Ri£jq - EP £ Ri£pj )
- -2 lEPrR-. irpJ·
In a frame as in part (a), aij ~ -! EPr Rirpj is diagonal and
au= --E^1 22 R1221 - -E^1 33 R1331
2 2