Kähler manifold structure on fibres

Question

I have read in my course notes the following statement :

Let $p : X \longrightarrow Y$ be a holomorphic map between two complex manifolds. If $Y$ is a Kähler manifold then $\forall y\in Y$, $p^{-1}(y)$ is a Kähler manifold.

It is very vague, and assumptions are probably missing. I am trying to see when this might be true and what assumptions on $p$, $Y$ and $X$ should be added. For instance, it is clear that we should also require $p$ non constant.

First problem : $p^{-1}(y)$ isn't a submanifold of $X$ a priori. One case where it works, is when $X$ is a holomorphic fibre bundle on $Y$, but can we take something weaker ?

Second problem : Suppose now we have the first problem settled and every fibre is a complex submanifold of $X$. Since the Kähler metric we would like to put on $p^{-1}(y)$ for all $y \in Y$ is the pullback of the one on $Y$, shouldn't we also require for $p$ to be an immersion?

Thanks in advance for your answers!

Answer 1

What you wrote will not work.

1. You should assume that $X$, not $Y$, is Kähler.

2. You should assume that $p$ is surjective and consider preimages of generic points in $Y$ (generic fibers), otherwise some of the preimages will not be submanifolds.

With this modification, the claim follows from Sard’s theorem (generic fibers are submanifolds) and the fact that complex submanifolds of a Kähler manifold are all Kähler.

See here.

Edit: 1. The Kähler metric (as well as the complex structure) on the fibers is not obtained by pull-back from the one on $Y$, it is obtained via pull-back from $X$ via inclusions $p^{-1}(y)\subset X$.

2. Yes, one way to avoid dealing with singular fibers is to assume that $p$ is a submersion: If you assume that $p$ is an immersion then all preimages $p^{-1}(y)$ are discrete and the entire discussion becomes trivial.