Rank of a Holomorphism

Question

Let $f \colon M \rightarrow N$ be a holomorphism of complex manifolds. Let $p \in M$. Let $(U,\phi)$ and $(V,\psi)$ be coordinate charts on $M,N$, respectively, satisfying $U \ni p$ and $V \ni f(p)$.

I don't know what you call it, but as $f$ is holomorphic, the function $F \colon \phi(U) \rightarrow \psi(V)$ given by $F = \psi \circ f \circ \phi^{-1}$ is a holomorphism (in the regular $\mathbb{C}^n$ sense).

Define the rank of $f$ at the point $p \in M$, denoted rk$_pf$, as the rank of the Jacobi matrix of $F$ at the point $\phi(p)$. (I'm guessing readers already know this.) Here is my question:

Is there an open neighbourhood of $p$ in which the rank of $f$ is constant, or at least $\geq$ or $\leq$ rk$_pf$?

A map between complex manifolds is a holomorphism if, locally, its corresponding function in $\Bbb C$ (here I call it $F$) is differentiable, or however else you want to classify it. (There are numerous equivalent ways.) — Open Season, Sep 17 '14 at 01:47

score 1 · Accepted Answer · answered Sep 17 '14 at 01:52

Just as in the case of smooth manifolds, rank can degenerate down but not up. That is, there's always a neighborhood of $p$ on which $\text{rk} f\geq \text{rk}_p f$. This is just because the rank is witnessed by a certain minor of the Jacobian having nonzero determinant, and this determinant is a continuous function of $p$. For an example when rank degenerates down, look no further than $z^2:\mathbf{C}\to\mathbf{C}$ at zero.

Rank of a Holomorphism

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