3
$\begingroup$

Consider a holomorphic function $u:C\to C^n$, as $|u|^2 $is sub-harmonic, it satifies a maximum principle.

Do some general kind of complex manifold enjoy such property? Say, square of some distance function(or similar function ) will be sub-harmonic?

$\endgroup$
2
  • 1
    $\begingroup$ Stein manifolds sit inside $\mathbb{C}^n$, so they have maximum principles for various functions (distance functions in the ambient $\mathbb{C}^n$) for maps from $\mathbb{C}$. $\endgroup$
    – Ben McKay
    Mar 21, 2016 at 15:41
  • $\begingroup$ Any holomorphic function on a complex manifold. $\endgroup$ Mar 21, 2016 at 20:18

1 Answer 1

5
$\begingroup$

If $f_1$, $f_2$, ..., $f_n$ are holomorhic functions on an open set $U$ in $\mathbb{C}^k$, then $\sum_{i=1}^n |f_i|^2$ has no local maximum. Since complex manifolds are locally $\mathbb{C}^k$, this is also true if $U$ is an open set in a complex $k$-fold.

In fact more is true: The signature of the Hessian of $\sum_{i=1}^n |f_i|^2$, as a quadratic form on $\mathbb{R}^{2k}$, has at most $k$ negative signs. The proof is pretty straightforward: Suppose $V \subset \mathbb{R}^{2k}$ is a $d$-dimensional real vector space on which the Hessian is negative definite; we must show $d \leq k$. If not, then $V \cap i V$ is nonzero. Note that $V \cap i V$ is a complex subspace of $\mathbb{C}^k$. We can restrict the $f_j$ to holomorphic functions on a complex line in this vector space and violate the standard maximum modulus principle.

This comes up in the proof that $k$-dimensional Stein manifolds have the homotopy type of a $(\leq k)$-dimensional CW-complex: For such a Stein manifold $X$, let $(f_1, \ldots, f_n)$ be a homolomorphic embedding $X \to \mathbb{C}^n$. After possibly replacing $f_i$ by $f_i-a_i$ for some generic $(a_1, \ldots, a_n) \in \mathbb{C}^n$, the function $\sum |f_i|^2$ is a morse function, and we have just computed that the signatures of the critical points are $+^{2k-d} -^d$ for $d \leq k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.