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  Why does this quintic CY3 have 16 nodes?

+ 4 like - 0 dislike

Let's consider the following quintic CY 3-fold in $\mathbb{P}^4$: $x_0f(x)+x_1g(x)=0$ where $f(x)$ and $g(x)$ are generic homogeneous polynomial of degree 4. It is known that this CY 3-fold is singular at the points $x_0=x_1=f(x)=g(x)=0$, each of which gives a node. 

I would like to confirm this fact, but Jacobian criteria does not really help me. Could anyone show me how to show this? (I see they are nodes).

asked Oct 17, 2014 in Mathematics by anonymous [ no revision ]

1 Answer

+ 4 like - 0 dislike

The Jacobian criteria is all what you need. A point x of the quintic is singular if and only if all the partial derivatives of the defining equation of the quintic vanish at x. First write the partial derivatives with respect to variables different from $x_0$ and $x_1$. As $f$ and $g$ are generic, the vanishing of these derivatives imply $x_0 = x_1 =0$. Then compute the partial derivatives with respect to $x_0$, $x_1$ and use the fact that we already know that $x_0 = x_1=0$ to conclude that $f(x)=g(x)=0$ for a singular point. This shows that a singular point $x$ necessarely satisfies $x_0 = x_1 = f(x)=g(x)=0$. As conversely any $x$ satisfying these conditions is on $X$, the solutions to these conditions are exactly the singular points of $X$.

As $f$ and $g$ are generic of degree 4, the equations $f=g=0$ define a surface in $\mathbb{P}^4$ of degree 4.4=16 and so its intersection with the plane $x_0=x_1=0$ is made of 16 points.

As $f$ and $g$ are generic, they have some non-trivial linear term around each singular point, in particular the defining equation of the quintic has a leading non-trivial quadratic form around each singular point, i.e. each singular point is a node. 

answered Oct 21, 2014 by 40227 (5,140 points) [ revision history ]

Does this singular CY3 admit a global Kaehler crepant resolution? If yes (no), is it easy to see why (not)?

I think that the answer is yes: simply blow-up in $\mathbb{P}^4$ the plane $x_0=x_1=0$ (which is contained in the 3CY and contains the 16 singular points) and look at the preimage of the 3CY. It is again a 3CY and the resolution is crepant because locally around each singular point, it is exactly the standard minimal resolution of a node, which is crepant.

Is it possible to blow-up along $f(x)=g(x)=0$ as well? I wonder what makes a resolution non-Kahler.

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