Calabi-Yau $3$veces da como elípticamente fibrado colector de más de $\mathbb{C}P^1 \times \mathbb{C}P^1$

Question

Considere la posibilidad de un Calabi-Yau tres veces se da como una elíptica fibrado colector de más de $\mathbb{C}P^1 \times \mathbb{C}P^1$$$y^2 = x^3 + f(z_1, z_2)x + g(z_1, z_2),$$where $z_1$, $z_2$ represent the two $\mathbb{C}P^1$s and $f$, $g$ are polynomials in $f$ in $(z_1, z_2)$.

1. ¿Cuál es el grado de los polinomios $f$$g$?
2. ¿Cuál es el número de independientes de la compleja estructura de las deformaciones de esta Calabi-Yau, y ¿cuál es la Hodge número $h^{2, 1}$?
3. Cuántos Kähler deformaciones están ahí, y ¿qué implica esto para $h^{1, 1}$?

Answer 1

1. Supongamos que tenemos un complejo colector de $M$ descrito por la ecuación$$F(z_1, z_2, \ldots, z_n) = 0.$$Now, without imposing this equation, the holomorphic form locally is just$$dz_1 \wedge dz_2 \wedge \ldots \wedge dz_n.$$The simplest way to impose the constraint is to just multiply this by a delta function and integrate over one dimension:$$\Omega = \int \delta(F(z_1, z_2, \ldots, z_n))\,dz_1 \wedge dz_2 \wedge \ldots \wedge dz_n.$$As a result, we will effectively get a holomorphic form on our manifold $M$. We can choose any direction, it is just a matter of convenience. For example, if we integrate over $z_1$,$$\Omega = {{dz_2 \wedge \ldots \wedge dz_n}\over{\partial F/\partial z_1}}.$$Now, if we go back to our Calabi-Yau example$$F = y^2 - x^3 - f(z_1, z_2)x - g(z_1, z_2),$$we see that it is convenient to choose the $y$-direction, since in this case $\partial F/\partial y = 2y$. Thus, we get a holomorphic form$$\Omega = {{dx \wedge dz_1 \wedge dz_2}\over y}.$$Suppose that we have the following scaling at infinity:$$z_{1, 2} \to \lambda z_{1, 2}, \quad f \to \lambda^af, \quad g \to \lambda^b g.$$Consider the $x = 0$ locus. We have$$y \to \lambda^{b/2}y, \quad dx \to \lambda^{b - a}dx.$$In order for the holomorphic form not to have any poles or singularities at infinity, we require that it is invariant under this scaling, which implies $b = 2a - 4$. We should also require that the function $F$ that defines the manifold should be homogeneous, which implies that each term scales the same. That is, if we consider $z_1 = z_2 = 0$,$$F = \lambda^b y^2 - \lambda^{3(b - a)}x^3,$$which implies $b = 3(b - a)$, hence $2b = 3a$. These two equations give $ = 8$, $b = 12$.
2. Que nos permiten contar el número de coeficientes de los polinomios$$f(z_1, z_2) = \sum_{i, j = 0}^8 f_{ij}z_1^i z_2^j, \quad g(z_1, z_2) = \sum_{i, j = 0}^{12} g_{ij}z_1^i z_2^j.$$We have $9 \cdot 9 + 13 \cdot 13 = 250$. One parameter can be killed by the rescaling $f \a \lambda^8f$, $g \a \lambda^{12}g$. Furthermore, we can use the $\text{SL}(2, \mathbb{C}) \times \text{SL}(2, \mathbb{C})$ symmetry of $\mathbb{C}P^1 \times \mathbb{C}P^1$ to kill $3 + 3 = 6$ more parameters. This leaves us$$h^{2, 1} = 250 - 1 - 6 = 243$$parámetros complejos.
3. Tenemos una Kähler parámetro para cada una de las $\mathbb{C}P^1$'s, y uno para la elíptica de la fibra. Esto nos da $h^{1, 1} = 3$.