By definition of u we have that
s(\) = log{\\\-1\p-l(g(\))\}.
Also, g(X) - p(Xav) = 0(|A|2) as A - 0, and so p~1(g(X)) - Xav = 0(|A|2). Hence
0 limsups(A) = |a|, and so \a\ 1.
Suppose that \a\ = 1, so that 5(A) = 0, by the maximum principle. The first
step in proving that g(X) = f(aX) is to show that
(2.13) gf(X) G Tj(x) for each A ^ 0,
where T\ is the subbundle of the tangent space of D \ {0} which is the image of
under dp.
Because s(X) = 0, we have that u(g(X)) = log|A| is harmonic on A \ {0}.
We'll derive (2.13) from this and the fact that (
d.dJk u(z)) n a s onh o n e
zero eigenvalue (and the rest positive) for each z, but we have to do this slowly
because u is not necessarily C
Fix A0 G A \ {0}. Thus u(g(X0)) = log |A0| -00, and so g(X0) / 0. Let
ui be a sequence of C°° functions defined on a neighborhood of 7(Ao) such that
u\ u and ddu\ —• ddu uniformly. For A near A0 we have
As / —* 00 the right side tends to
uniformly, while the left side tends to 0 in the sense of distributions in A, be-
cause m(g(X)) tends uniformly to u(g(X)) = log|A|. Thus (2.14) vanishes on a
neighborhood of Ao, and hence on all of A \ {0}, since Ao was arbitrary.
Let L(z) denote the linear map associated to the matrix
u{z), z G
D \ {0}. We saw in the proof of Theorem 2.3 that L(z) is nonnegative for each
2, and so ^'(A) lies in the kernel of L(g(X)), because (2.14) vanishes.
The kernel of L(z) consists of the set of vectors w £TzCn = C n such that
i(w)(ddu | J = 0,
where i(w) denotes the interior product by w. This is also the same as T\.
To see this, let z' - p~l{z)_, fix w G TzCny and_let w' G TzCn be such that
dpz,(w') = w. Then i(w)(ddu \g) = 0 iff i(w')(ddu0 \z,) = 0, because of (2.12).
This last occurs iff w' G S], which is equivalent to w G T\, From here (2.13)
follows easily.
Because T\ is always a complex one-dimensional subspace of TzCn, (2.13)
implies that the tangent space of /(A) at g(X) is T\x^ for all A ^ 0. Thus if
Previous Page Next Page