Nonlinear Equations - UFRJ

[SEC. 3.3: COMPLEX MANIFOLDS AND EQUATIONS 41
Proof of Theorem 3.6. Let y, z be regular values of f. Since M is
connected, they belong to the same component of M. Let h t (x) =
x + t(z − y), t ∈ [0, 1].
Then, f and f ◦ h(1, ·) are smoothly homotopic, and admit y as
a common regular value. Using the chain rule, we deduce that the
degree of f in y is equal to the degree of f in z.
3.3 Complex manifolds and equations
Let M be a complex manifold. In a neighborhood U of some p ∈ M,
pick a bi-holomorphic function f from U to f(U) ⊆ C n . The pullback
of the canonical orientation of C n by f defines an orientation on
T q M for all q ∈ U. This orientation does not depend on the choice
of f. We call this orientation the canonical orientation of M. We
proved:
Theorem 3.8. Complex manifolds are orientable.
Theorem 3.9. Let M be an n-dimensional complex manifold, without
boundary. Let F be a space of holomorphic functions M → C n .
Given f ∈ F and U open in M, let n U (f) = #f −1 (0)∩U be the number
of isolated zeros of f in U, counted without multiplicity. Then,
n U : F → Z ≥0 is lower semi-continuous at all f where n U (f) < ∞.
Proof. In order to prove lower semi-continuity of n U , it suffices to
prove that for any isolated zero ζ of f, for any δ > 0 small enough,
there is ɛ > 0 such that if ‖g − f‖ < ɛ, then g has a root in B(ζ, δ).
Then pick δ such that two isolated roots of f are always at distance
> 2δ.
Because complex manifolds admit a canonical orientation, the
Brouwer degree of f |B(ζ,δ) is a strictly positive integer. Since it is
locally constant, there is ɛ > 0 so that it is constant in B(f, ɛ).