Why is a Kähler form not always holomorphic?

• nonequilibrium
In summary: But if "holomorphic top-form" means "there exists a coordinate system (z') such that \overline{\partial}h'_{ij}=0 at p', then that's a little more tricky.
nonequilibrium
Any Kähler form (?) can be written in local coordinates as $\omega = \frac{i}{2} \sum h_{ij} dz^i \wedge d z^j$ with $h_{ij}(z) = \delta_{ij} + \mathcal O(z^2)$.

But that would mean $\bar \partial \omega = \frac{i}{2} \sum \frac{\partial h_{ij}}{\partial \bar z_k} d \bar z^k \wedge dz^i \wedge d z^j = 0$ since $\frac{\partial h_{ij}}{\partial \bar z_k} = \mathcal O(z)$ is zero at our point $z = 0$.

I must be making a stupid mistake. What is it?

Your mistake is that the Kahler form is a (1,1) form:

$$\omega = \frac{i}{2} \, h_{i \bar \jmath} \, dz^i \wedge d \bar z^j$$

nonequilibrium said:
But that would mean $\bar \partial \omega = \frac{i}{2} \sum \frac{\partial h_{ij}}{\partial \bar z_k} d \bar z^k \wedge dz^i \wedge d z^j = 0$ since $\frac{\partial h_{ij}}{\partial \bar z_k} = \mathcal O(z)$ is zero at our point $z = 0$.

Also, this reasoning only applies at a point. The property of holomorphicity requires an open set around a point.

Thanks for pointing out my typo with the bar, but that doesn't change the argument though.

As for your second post: the reasoning is as follows: I check that $\bar \partial \omega$ is zero at every point (but to calculate it at a point, I indeed need a neighbourhood around that point). So fix a point z = 0. Then around that point we can write $\omega$ as I did with the expansion of $h$. Then I act with $\bar \partial$ and afterwards I put $z = 0$ to get that at that point $\bar \partial \omega = 0$. Since our original point was arbitrary, this is true for every point!

The Kähler form is (1,1) and $(dz^i\wedge d\overline{z}^j)$ does not define a holomorphic structure on the bundle of (1,1) forms. Only on the bundle of (p,0) forms do the complex coordinates on M define a holomorphic structure in this way.

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So is my mistake in interpreting $\bar \partial \omega = 0$ to mean that my form is holomorphic? I thought that was the definition of holomorphic. What is the correct meaning?

To speak of holomorphicity of a section of a complex vector bundle you need first to put a holomorphic structure on your bundle. That is, you need to specify trivializing local frames (e_i) for it such that the transition maps between these frames are holomorphic. Then, and only then, will it be well-defined to say a section $s=\sum_i s_ie_i$ is holomorphic if $\overline{\partial}s_i=0$. Otherwise, we may have $\overline{\partial}s_i=0$ but $\overline{\partial}s_i'\neq 0$ wrt another local frame (e_i').

Also, you seem to think your argument shows the functions h_ij to be holomorphic but that is not so:

Fix a point p in your manifold and coordinates (z) in a nbhd of p. As Ben said, holomorphicity of h_ij at p means $\overline{\partial}h_{ij}=0$ in a nbhd of p. What your argument shows is that for any point p', you can always find a coordinate system (z') wrt which $\overline{\partial}h'_{ij}=0$ at p'. But this does not imply that $\overline{\partial}h_{ij}=0$ at p' wrt (z) because the transition function btw $(d\overline{z})$ and $(d\overline{z'})$ (which appear when you write h_ij in terms of h'_ij) are not holomorphic. (Same reason why (dzi∧dz¯j) does not define a holomorphic structure on the bundle of (1,1) forms.)

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I'm confused, I'm not using coordinate expressions in my definition, I'm just using the fact that the coordinate independent notion $\bar \partial \eta$ (defined in terms of the exterior derivative $d$ etc) with $\eta = \eta_{IJ} dz^I \wedge d \bar z^J$ can be written in coordinates as $\bar \partial \eta = \frac{\partial \eta_{IJ}}{\partial \bar z^k} d \bar z^k \wedge dz^I \wedge d\bar z^J$

I should have said explicitely that your definition of a form being holomorphic iff $\overline{\partial}\eta=0$ is wrong. That is not how holomorphicity is defined. See paragraph 1 of post #7.

Ah and also it is the bundle of (p,0) forms that admit a natural holomorphic structure, not the (0,p) forms as I said in post #5, so I edited that post as well as #7 accordingly.

Ah okay, thanks. So I see what I did wrong, but still trying to understand what the correct notion of holomorphicity of (p,q)-forms is. I think I get what you are saying, but that would seem to suggest that we need to make an extra choice before we have that notion. Is there no canonical notion of holomorphicity? E.g. when people say "a Calabi-Yau mfd is a compact Kähler mfd with holomorphic top-form", how am I to interpret that?

EDIT: I just realized another way to see my original thought was wrong, since if "holomorphic top-form" meant "$\bar \partial \omega = 0$", then that's trivial because $\bar \partial$ of a top-form is always zero! (since $\Omega^{2n+1} = 0$))

There is no natural/canonical notion of holomorphic (p,q)-form if q>0. On the other hand, on $\Omega^{(p,0)}$, there is a natural holomorphic structure given by the local frames $dz^{i_1}\wedge\ldots dz^{i_p}$ for every local coordinate system (z) on M (check that the transition functions between these local frames are indeed holomorphic). A holomorphic p-form is then one of type (p,0) such that $\overline{\partial}\eta=0$.

A Calabi-Yau manifold is (depending on your definition...) a compact Kähler manifold with trivial canonical bundle. The canonical bundle of a complex n-manifold is the bundle $\Omega^{(n,0)}$ with its natural holomorphic structure. Triviality of this bundle then means that there exists a global nowhere vanishing holomorphic n-form. Note that $\Omega^{(n,0)}$ may be trivial as a complex line bundle but not as a holomorphic line bundle in which case we don't have Calabi-Yau.

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1 person
nonequilibrium said:
EDIT: I just realized another way to see my original thought was wrong, since if "holomorphic top-form" meant "$\bar \partial \omega = 0$", then that's trivial because $\bar \partial$ of a top-form is always zero! (since $\Omega^{2n+1} = 0$))

Note that a holomorphic top form is an element of $\Omega^{(n,0)}\subset \Omega^n$, and not of $\Omega^{2n}$. That is because $\Omega^{(p,0)} = \wedge^p\Omega^{(1,0)}$ and $\Omega^{(1,0)}$ has complex dimension 1, so we can wedge it with itself only n times before it vanishes. So while $\partial \eta=0$ for a (n,0)-form, $\bar\partial \eta\in\Omega^{(n,1)}$ does not vanish in general.

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A last remark: While it is true that for (p,0)-forms, holomorphicity in the sense of post #7 is the same as the condition $\bar \partial \eta=0$, for a general (p,q)-form, the condition $\bar \partial \eta=0$ has nothing to do with holomorphicity of anything.

As an illustration, consider a (0,1)-form on a cx 2-manifold: $\eta=\eta_1d\bar z_1 + \eta_2 d\bar z_2$. Then
$$\bar\partial \eta = \left(\frac{\partial \eta_1}{\partial \bar z_1} - \frac{\partial \eta_1}{\partial \bar z_2}\right)d\bar z_1\wedge d\bar z_2$$
What does the vanishing of this has to do with holomorphicity ?

Thank you very much!

1. Why is a Kähler form not always holomorphic?

A Kähler form is a differential form that satisfies certain properties, including being closed and being compatible with a Hermitian metric. However, not all Kähler forms are holomorphic because they may not satisfy the additional condition of being purely imaginary. This condition is necessary for a form to be considered holomorphic.

2. What is the difference between a Kähler form and a holomorphic form?

A Kähler form is a differential form that satisfies certain properties, including being closed and being compatible with a Hermitian metric. Holomorphic forms, on the other hand, are differential forms that satisfy the Cauchy-Riemann equations and are purely imaginary. While all holomorphic forms are also Kähler forms, not all Kähler forms are holomorphic.

3. Can a non-holomorphic Kähler form still be useful in mathematics?

Yes, a non-holomorphic Kähler form can still be useful in mathematics. Kähler forms are important in complex geometry and symplectic geometry, and they have applications in physics as well. Even though they may not be holomorphic, they still possess useful properties that make them valuable in various areas of mathematics.

4. What is the relationship between Kähler forms and complex structures?

Kähler forms and complex structures are closely related. A complex structure is a linear transformation that preserves the complex structure of a manifold, while a Kähler form is a differential form that satisfies certain properties, including being compatible with a Hermitian metric. In fact, a Kähler form uniquely determines a complex structure on a manifold.

5. Can a Kähler form be non-holomorphic in some regions of a manifold and holomorphic in others?

Yes, it is possible for a Kähler form to be non-holomorphic in some regions of a manifold and holomorphic in others. This is because the condition of being holomorphic is dependent on the region being considered. A Kähler form may satisfy the Cauchy-Riemann equations in some regions, making it holomorphic, but not in others.

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