# Do holomorphic mappings always preserve/induce Kähler forms?

I was thinking in several related questions for which I haven't been able to find a proof nor a formal statement (Basically when do holomorphic mappings induce Kähler forms and how does Kählerness behave under holomorphic mappings).

So, let's suppose $$X$$ and $$Y$$ are complex manifolds. Now, I will denote by $$\pi\colon X\to Y$$ and $$f\colon Y\to X$$ some random holomorphic mappings. My questions are

1. If $$(X, \omega_X)$$ is Kähler, is it true that $$(Y, f^*\omega_X)$$ is always Kähler? (it should be true if $$f$$ is an immersion, but I want to know if that holds as well for other conditions, e.g., properness, submersions, etc.)
2. Does the existence of a holomorphic mapping $$f\colon Y\to (X, \omega_X)$$ impose some restrictions on the Kählerness of $$Y$$ or the fibres $$Y_x:= f^{-1}(x)$$? (in case the latter happen to be complex manifolds as well). For example, can we find another Kähler form on $$Y$$ other than $$f^*\omega_X$$ (whenever $$f^*\omega_X$$ is Kähler of course)? In general, does the existence or the properties of the mapping $$f$$ itself impose restrictions on the the Kählerness of $$Y$$ and/or the fibres $$Y_x$$?.
3. If $$\pi\colon (X, \omega_X)\to Y$$ is holomorphic, when can we say that $$Y$$ and/or the fibres $$X_y:=\pi^{-1}(y)$$ are Kähler as well? (and how do we construct their Kähler forms if it can be done?) [this one is particularly relavant when $$\pi$$ is a quotient map or a covering map] Edit: I found this exercise (aptly labelled hard) in some online notes a few days ago. As for the weird counterxample mentioned below (a deformation of Kähler manifolds that is not a Kähler), one can trace it back to Hironaka (An Example of a Non-Kahlerian Complex-Analytic Deformation of Kahlerian Complex Structures)
4. A slightly related question: is it true that the pullback of positive forms is always a positive form, i.e., if $$\omega$$ is a positive $$(p, p)$$-form on $$X$$, does it hold as well for $$f^*\omega$$ on $$Y$$? Edit: This seems to be true at least for vector spaces, as you can see in page 129 of Demailly's Complex analytic and differential geometry.

I will appreciate if you can give me some hints/proofs/counterexamples or point me out to some solid references with fully elaborated proofs/examples, not just some yes/no kind of answers. Regards!

• When you're thinking about different scenarios, don't forget to include the case that $Y$ is the blow-up of $X$ at a point or along a smooth subvariety. Feb 7, 2021 at 0:02
• Also trivial counterexamples arise from products (and the corresponding projection maps), as well as constant maps. I don't know if you want such counterexamples articulated or if you would like to add further conditions to rule these out. Feb 7, 2021 at 0:43

$$1.$$ This is not true. For example, if $$\dim Y > \dim X$$, then $$f^*\omega_X$$ is degenerate (in directions in the kernel of $$f_*$$). Even if $$\dim Y = \dim X$$, this issue can still occur. For example, let $$Y$$ be the blowup of $$X$$ at a point $$x$$ (as Ted Shifrin alluded to in the comments). If $$\dim X \geq 2$$, then $$(f^*\omega_X)_y$$ is degenerate for all $$y$$ with $$f(y) = x$$. To see this, note that for a tangent vector $$v \in T_yY$$ along the exceptional divisor, we have $$f_*(v) = 0$$, so $$(f^*\omega_X)_y(v, w) = (\omega_X)_x(f_*(v), f_*(w)) = (\omega_X)_x(0, f_*(w)) = 0$$ for all $$w \in T_yY$$.

In general, a two-form $$\alpha$$ is the Kähler form of a Kähler metric if and only if it is a real positive $$(1, 1)$$-form, see this question. As $$f$$ is holomorphic, $$f^*\omega_X$$ is a real $$(1, 1)$$-form, so it is the Kähler form of a Kähler metric if and only if it is positive. In the examples above, $$f^*\omega_X$$ is degenerate and hence not positive. If $$f : Y \to X$$ is a holomorphic immersion, then $$f^*\omega_X$$ is positive and therefore gives rise to a Kähler metric on $$Y$$.

$$2.$$ The existence of a holomorphic map $$f : Y \to X$$ with $$X$$ Kähler has no bearing on the existence of Kähler metrics on $$Y$$ or the fibers of $$f$$. One reason for this is that we always have constant maps $$X \to Y$$. Even if $$f : Y \to X$$ is non-constant, it could be the case that neither $$Y$$ nor the fibers of $$f$$ admit Kähler metrics. For example, let $$Y = Y_0\times X$$ where $$Y_0$$ is a Hopf surface and $$X$$ is a compact Kähler manifold, and consider the holomorphic map $$f : Y \to X$$ given by projection onto the second factor. The complex manifold $$Y$$ does not admit a Kähler metric, and neither do the fibers of the map as they are Hopf surfaces.

$$3.$$ The existence of a holomorphic map $$\pi : X \to Y$$ with $$X$$ Kähler does not imply that $$Y$$ is Kähler. For example, take $$X$$ and $$Y$$ as in point $$2$$ and let $$\pi(x) = (y_0, x)$$ where $$y_0 \in Y_0$$ is some chosen point. On the other hand, if $$\pi^{-1}(y)$$ is a complex submanifold of $$X$$ which is the case if $$y$$ is a regular value of $$\pi$$ for example, then the restriction of the Kähler metric on $$X$$ to $$\pi^{-1}(y)$$ is a Kähler metric. More generally, the restriction of a Kähler metric to a complex submanifold is a Kähler metric, see Proposition $$3.1.10$$ of Huybrechts' Complex Geometry: An Introduction for example.

If $$\pi : X \to Y$$ is a finite covering map and $$X$$ admits a Kähler metric, then so does $$Y$$. This is true because there is a finite covering $$\rho : X' \to X$$ such that the induced finite covering $$\pi' : X' \to Y$$ is a normal covering. Then we can average the Kähler metric $$\rho^*\omega_X$$ on $$X'$$ over the deck transformations of $$\pi'$$ to obtain a Kähler form which descends to $$Y$$.

$$4.$$ This is false as the counterexamples in point $$1$$ demonstrate (for the case of $$\omega = \omega_X$$ a positive $$(1, 1)$$-form).

• Thank you very much for the detailed answer. Can you check out my latest edits? As for the fourth point, can we impose conditions on $f$ in order for it to preserve positive forms of any degree? (I guess holomorphic immersions can do the trick for any positive $(p, p)$-forms, but I wanted to see if we can impose a weaker condition) Feb 19, 2021 at 5:30
• Regarding your first edit, the first part of the exercise is difficult (in particular, you need to know about currents). For the second part, I don't know why you're mentioning Hironaka's example, that doesn't seem relevant as all the fibers of $f : X \to Y$ are Kähler. Instead, you can take $X$ to be an elliptic Hopf surface, then there is a holomorphic map $f : X \to \mathbb{CP}^1$ such that every fibre is an elliptic curve. Feb 19, 2021 at 12:07
• Regarding your second edit, note that Demailly is not using the same definition of positive as I thought you intended. He's using positive in the French sense (what I would call non-negative). In particular, Demailly would consider the zero form to be positive. Feb 19, 2021 at 12:08