Page 257 - Curvature and Homology

EXERCISES

Put g(Y, JX) = sin 8; then,
g(R(Y, JX)Y, JX) = - K(Y, JX) eoss8.
Hence, since
h 5 K(aX + by, J(aX + by)) 5 1,
h(as + b2)= 5 d K(X, JX) + 2asba [K(X,Y) + 3K(Y, JX) cos'fl
+ P K(Y, JY) 5 (as + be)'
for any a, b E R, and so
2h - 1 5 K(X,Y) + 3K(Y, JX) cos28 5 2 - h .

Similarly, from
5 K(aX + b JY,J(aX + b JY)) 5 1,
we deduce

Consequently,

for any X and Y. In particular, Kt JX,Y) 2 i(3h - 2), and so from (3)

5. Show that for every orthonormal set of vectors {x,Y, Jx, JY}

B. Reduction of a real 2-form of bidegree (1.1)

1. At each point P E M, show that there exists a basis of Tp of the form

(i = 1,3, ..a, 2p - 1 ; k = 2p + 1, ..., n) such that only those components of a
real 2-form a of bidegree (1,l) of the form a,,,, ai+l,(i+l) +, at.i+l = ai+.(i+l)
a,,, may be different from zero.
To see this, observe that Tp may be expressed as the direct sum of the
2-dimensional orthogonal eigenspaces of a. Since a is real and of bidegree (1,1),
,1(X,Y) = a( JX, JY) for any two vectors X and Y (cf. V. C.6). Let V be such a
subspace. Put v = V + JV. In general, JV ;f V; however, JV = v. Tpis a
direct sum of subspaces of the type given by v. Only two cases are possible for v: