5
$\begingroup$

In their paper "Calabi-Yau 3-folds and Moduli of Abelian Surfaces I", Gross and Popescu calculate the hodge numbers of smooth CY 3-folds obtained in the following way (see Remark 4.11 of their paper): suppose $X$ is a complete intersection CY 3-fold (i.e. dualizing sheaf is trivial) in $\mathbb P^{n+3}$ with only ODP's (ordinary double points) as singularities such that the $n$ equations $f_1,...,f_n$ which define $X$ all have the same degree $d$. Assume further that there exists a projective small resolution $\tilde{X}\rightarrow X$ of the singularities of $X$. Notice that $\tilde{X}$ is then a smooth CY 3-fold. Let $S=\mathbb C[x_0,...,x_{n+3}]/(f_1,...,f_n)$ be the homogeneous coordinate ring of $X$.

Now they claim that there is a natural isomorphism $T^1\cong \left(\frac{S^n}{\{(\partial f_1/\partial x_i,...,\partial f_n/\partial x_i)\}}\right)_d$, where the subscript $d$ refers to the degree $d$ homogeneous part of this module and $T^1\cong \operatorname{Ext}^1(\Omega_X^1,\mathcal O_X)$ is the space of infinitesimal deformations of $X$, or equivalent the tangent space to the deformation space of $X$. Why is there such a natural isomorphism?

They further state that the space of infinitesimal deformations of the small resolution $\tilde{X}$ is obtained as the kernel of the natural map $T^1 \rightarrow T^1_{\text{loc}}$, where $T^1_{\text{loc}}$ is the tangent space of the deformation space of the germ of the singular locus of $X$. Why is this the case?

Finally, they remark that this kernel is given by elements $(g_1,...,g_n)\in S^n$ representing an element of $T^1$ such that adding this as a new column to the Jacobian matrix gives a matrix which has rank $<n$ at the singular points of $X$. Why does the kernel have this description?

Thanks

$\endgroup$

1 Answer 1

1
$\begingroup$

Since nobody has answered you, let me give you some sort of answer. It will be somewhat lacking in rigour...

First, let me frame the discussion in terms of $H^1(X, TX) \cong \mathrm{Ext}^1(\Omega^1_X, \mathcal{O}_X)$, and also write $P \equiv \mathbb{P}^{n+3}$. Start with the short exact sequence \begin{equation*} 0 \longrightarrow TX \longrightarrow TP\vert_X \stackrel{df}{\longrightarrow} \mathcal{N}_{X\vert P} \longrightarrow 0 ~. \end{equation*} Taking cohomology, this gives in part \begin{equation*} 0 \longrightarrow H^0(X, TP\vert_X) \stackrel{df}{\longrightarrow} H^0(X, \mathcal{N}_{X\vert P}) \longrightarrow H^1(X, TX) \longrightarrow 0 ~. \end{equation*} The global sections of $\mathcal{N}_{X\vert P}$ are just $n$-tuples of degree $d$ polynomials in the homogeneous coordinate ring $S$, and you can check that $H^0(X, TP\vert_X)$ is just the space of homogeneous linear polynomials. Since $df$ is represented by the matrix of first derivatives of $(f_1,\ldots, f_n)$, the specified isomorphism follows. The intuition is that polynomial deformations which are proportional to the first derivatives are realised by coordinate changes, and are therefore trivial.

This is actually a special case of an old theorem of Green and Hübsch (Commun.Math.Phys. 113 (1987) 505), which gives conditions under which counting such 'polynomial deformations' gives the value of $h^{1,2}(X)$, for $X$ a complete intersection Calabi-Yau in a product of projective spaces.

For the other parts, I don't know the details (e.g. the definition of $T^1_\mathrm{loc}$), but note that the moduli space of $\tilde X$ (the small resolution) coincides locally with the subspace of the moduli space of $X$ over which $X$ has the specified configuration of nodes. The kernel of the map $T^1 \to T^1_\mathrm{loc}$ must consist of those deformations which preserve the nodes, which can therefore be interpreted as deformations of $\tilde X$.

$\endgroup$
2
  • $\begingroup$ Actually we don't have $H^1(X,TX)\cong Ext^1(\Omega_X^1,\mathcal O_X)$. The deformation theory exact sequence (aka local-to-global spectral sequence) gives $0\rightarrow H^1(X,TX)\rightarrow Ext^1(\Omega_X^1,\mathcal O_X)\rightarrow H^0(X,\mathcal T^1)$. Since $\mathcal T^1$ is a skyscraper sheaf support at the ODPs and of dimension 1 at each of them, we find that $H^0(X,\mathcal T^1)=\mathbb C^k$, where $k$ is the number of ODPs. Moreover, this last map is nonzero, so those two v. spaces are not isomorphic. Moreover this last space is $T^1_loc$ as it is the deformation space of the ODPs. $\endgroup$
    – HNuer
    Commented Jul 26, 2012 at 2:43
  • 2
    $\begingroup$ I have given a full proof of this with all of the deformation theoretic details in Section 7 of my paper "On (2,4) Complete Intersection threefolds containing an Enriques Surface" arxiv.org/abs/1210.1903. $\endgroup$
    – HNuer
    Commented Oct 11, 2012 at 13:26

You must log in to answer this question.

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