9
$\begingroup$

I was reading about Calabi-Yau manifolds and thought of this problem. Suppose we consider $Z$ as the zero locus of a degree $n+2$ homogeneous polynomial on $(\mathbb{P}^{n+1}, \omega_{FS})$, where $\omega_{FS}$ is the Fubini-Study metric. Then we know that $Z$ is a Calabi-Yau manifold. All the books/lecture notes prove that $Z$ is a Calabi-Yau manifold by using the Adjunction Formula (so they show that Z has trivial canonical bundle).

My question is that since $Z$ is Calabi-Yau, so it must have a nowhere vanishing holomorphic volume form. How do we explicitly calculate that ?

I know that $\omega_{FS}|_Z$ is $not$ a Ricci-flat K$\ddot{a}$hler metric. So merely restricting the volume form from $\mathbb{P}^{n+1}$ is not the right choice.

Thanks !!

$\endgroup$
2
  • 1
    $\begingroup$ Do you want a nowhere-vanishing $(n,0)$-form, or a Ricci-flat Kahler metric? You can write down the first, but as far as I know no one's ever written down an explicit Ricci-flat Kahler metric on a compact Calabi-Yau manifold. $\endgroup$ Commented Jun 27, 2017 at 7:06
  • $\begingroup$ @GunnarÞórMagnússon Initially, I was thinking about explicit Ricci-Flat Kahler metric, but then I got to know that there are numerical results on approximation of RFK metrics. $\endgroup$
    – seeker
    Commented Jun 28, 2017 at 23:46

1 Answer 1

12
$\begingroup$

For ease of notation, I'll explain how to construct the nowhere-vanishing holomorphic volume form in the case $n = 2$, where the Calabi-Yau hypersurface is a quartic K3 surface. This method generalises for all $n$.

Let's first focus on the affine patch: $$ U = \{ [1 : z_1 : z_2 : z_3] \ \vert \ (z_1, z_2, z_3) \in \mathbb C^3 \} \subset \mathbb{CP}^3.$$ On this affine patch $U$, the hypersurface is defined by a polynomial equation: $$ f_{\rm quartic}(z_1, z_2, z_3) = 0.$$

For each $i \in \{1,2,3 \}$, we define the open subset $$ V_i = \left\{ (z_1, z_2, z_3) \in \mathbb C^3 \ \big\vert \ \frac{\partial f_{\rm quartic}}{\partial z_i}(z_1, z_2, z_3) \neq 0\right\} \subset U.$$ Since $f_{\rm quartic}(z_1, z_2, z_3)$ is non-singular, we have $$ U = V_1 \cup V_2, \cup V_3.$$

Now, observe that $z_2$ and $z_3$ form a set of holomorphic coordinates on $V_1$, by the holomorphic implicit function theorem (with $z_1$ determined holomorphically in terms of $z_2$ and $z_3$). Similar statements hold for $V_2$ and $V_3$.

We can therefore define non-vanishing holomorphic two-forms $\Omega_{V_1}$, $\Omega_{V_2}$ and $\Omega_{V_3}$ on $V_1, V_2, V_3$ respectively: $$ \Omega_{V_1} = \frac{dz_2 \wedge dz_3}{\partial f_{\rm quartic}/\partial z_1}, \ \ \ \ \ \Omega_{V_2} = \frac{dz_3 \wedge dz_1}{\partial f_{\rm quartic}/\partial z_2}, \ \ \ \ \ \Omega_{V_3} = \frac{dz_1 \wedge dz_2}{\partial f_{\rm quartic}/\partial z_3},$$

It is easy to see that $$\Omega_{V_1} = \Omega_{V_2}$$ on $V_1 \cap V_2$. [One can verify this by multiplying the equation $$ \frac{\partial f_{\rm quartic}}{\partial z_1} dz_1 + \frac{\partial f_{\rm quartic}}{\partial z_2} dz_2 + \frac{\partial f_{\rm quartic}}{\partial z_3} dz_3 = 0$$ by $\wedge dz_3$.]

Similarly, $\Omega_{V_2} = \Omega_{V_3}$ on $V_2 \cap V_3$, and $\Omega_{V_3} = \Omega_{V_1}$ on $V_3 \cap V_1$.

So $\Omega_{V_1}$, $\Omega_{V_2}$ and $\Omega_{V_3}$ glue together to define a non-vanishing holomorphic two-form $\Omega$ on the whole of the affine patch $U$.

There remains a possibility that $\Omega$ may vanish, or diverge, on the hyperplane $$ H = \{ [0 : x_1 : x_2 : x_3] \ \vert \ [x_1 : x_2 : x_3] \in \mathbb{CP}^2 \}$$ which is not covered by the affine patch $U$. We must show that this does not happen.

Expressing this in the language of divisors, we know that ${\rm div}(\Omega) = nH$ for some $n \in \mathbb Z$, and our task is to show that $n = 0$. But this is obvious: since $\Omega$ is a meromorphic section of the canonical bundle, which is trivial for the quartic by adjunction, the divisor class ${\rm div}(\Omega)$ is the trivial divisor, hence $n = 0$. [Or if you want to avoid divisors, then just compute what $\Omega$ is after the change of coordinates and you'll see...]


As Gunnar pointed out, finding a Ricci-flat metric on the quartic K3 is a much more difficult problem. As far as I'm aware, my colleagues in string theory only know how to approximate this numerically.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for your crystal clear answer. $\endgroup$
    – seeker
    Commented Jun 28, 2017 at 23:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .