Holomorphic function is zero on an analytic set then $df=0$.

Question

Assume we have an homomorphic function $f:U\rightarrow \mathbb{C} $ which is holomorphic on the open set $U$ of $\mathbb{C}^n$. Assume there is $V\subset U$ analytic and that $f$ restricted to $V$ equals zero ($V$ is obviously not open in $U$). Is it true that this implies that $df=0$? Here $d= \partial +\bar{\partial}$. If so, I need a hint on how to prove it. Or a reference. Thanks in advance.

Answer 1

If $V = \{(0,z):z \in \mathbb{C}\} \subset \mathbb{C}^2$, then $V$ is a complex analytic submanifold of $\mathbb{C}^2$.

The function $f(z_1,z_2) = z_1$ is holomorphic on $\mathbb{C}^2$, and vanishes on $V$. Nevertheless, $df = dz_1$, which is nonzero everwhere, even on $V$. So the "theorem" in the question is false.

It is true that $df$ vanishes when restricted to the tangent space of $V$. Maybe this is what you were thinking?

A related computation is that $d(fg) = fdg+gdf$, so if $f$ and $g$ both vanish on $V$, we would have $d(fg)$ vanishing on $V$.