- THE CROSS CURVATURE FLOW
where we used gtμijk = -Cμijk. By (B.29),
μijmμk£nCjR. = 1μijmμjpqμk£nμ£rs EprEqs
Hence= 1 (-gipgmq + giqgmp) (-gkr 9 ns + gksgnr) EprEqs
= Eik Emn _ Ein Emk.
μijmμk£n\7i\7kcje = \li'Vk (EikEmn -EinEmk).
With this, the lemma follows from the identity(^1 2) μiJm .. μ k" m gpq Rijpf,Cqk + C Emn = det E gmn.
497
To see this, we choose a basis where 9ij = 8ij, Eij and Cij are diagonal,
μ^123 = 1, and Rijk£ =f. 0 only if (i,j) = (k,R.) as unordered pairs. We
compute
1μijlμk£lgpqRijp£Cqk = μk£lgpq R23p£Cqk
= R2323 c22 - R2332 c33
= -E^11 (c22 + c33),
so that ~ μijl μk£l gPq Rijp£Cqk + C E^11 = E^11 en = det E, similarly for the
other diagonal components. We leave it as an exercise to check that the
off-diagonal components are zero. D
As we shall show below, part (1) of Proposition B.23 follows from the
following more general computation. Let
LEMMA B.26. For any a E JR.(B.41).!!:__ f (detE)°'dμ=a { (~\Tijk_yjik\
2
-alTil~-i)(detE)°'dμ
dt JM JM 2 E-1+(1-2a) JM(detE)°'Cdμ..
Here 1TilE-1^2 = (E-^1 ) iJ .. TiTj and