- THE CROSS CURVATURE FLOW
Hence[a-DY (g) (() v]ij = gpq ((i(pVqj + (j(pVqi - (i(jVpq)
= 8i1V1j + Oj1V1i - 8il8j1gpqVpq,
assuming (1 = 0 and (2 = (3 = 0. In other words,vn vn - v22 - v33
V12 V12
a-DY (g)(() V13 V13
V22^0
V33^0
Vz3^0
or in matrix form,
1 0 0 -1 -1 0(^0 1 0 0 0 0)
o-DY(g)(()=
(^0 0 1 0 0 0)
(^0 0 0 0 0 0)
(^0 0 0 0 0 0)
(^0 0 0 0 0 0)
Hence
1 0 0 E22 -1 E33 -1 2E23
0 1 0 -E12 0 -E13
o-D (X + Y) (g) (() =^0
0 1 0 -E13 -E12
0 0 0 En 0 0
0 0 0 0 En 0
0 0 0 0 0 En
which has all positive eigenvalues.
Hence the equation
(B.35)
[)
atg = 2c + Cwg
495
is strictly parabolic. Hence for any metric go with negative sectional cur-
vature, a solution g (t) to (B.35) with g (0) = go exists for a short time.
Similarly to the case of the Ricci flow, we solve the following ODE at each
point in M (see (3.35) on p. 81 of Volume One):
(B.36) :t 'Pt = -W*,
<po= id,where W* (t) is the vector field dual to W (t) with respect tog (t). Pulling
back g (t) by the diffeomorphisms 'Pt, we obtain a solution