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Why quintics are Calabi-Yau? Is there a explicit formula of the holomorphic volume form?

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The adjunction formula tells you that the canonical bundle of a quintic threefold vanishes. You can write down an explicit holomorphic volume form using residues (see, e.g., Chapter 5 of Griffiths and Harris). –  Alberto García-Raboso Apr 6 at 2:00
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I find the question "Why quintics are Calabi-Yau" rather cryprtic. Would you mind to give some more information? –  Dirk Apr 10 at 0:02

1 Answer 1

In general if $X\subset\mathbb{P}^N$ is a smooth complete intersection of hypersurfaces of degree $d_1,...,d_c$. Then $\omega_{X}\cong\mathcal{O}_{X}(d_1+...+d_c-N-1)$. In you case $d = 5, N=4$. So $\omega_{X}\cong\mathcal{O}_{X}$ and $X$ is Calabi-Yau.

To prove this formula you have to use the adjunction formula:

http://en.wikipedia.org/wiki/Adjunction_formula_(algebraic_geometry)

If $Y\subset X$ is a smooth subvariety of codimension one then $K_Y = (K_X+Y)_{|Y}$. If $Y$ is an hypersurface of degree $d$ in $X = \mathbb{P}^{N}$, then $K_Y = (d-N-1)H_{|Y}$ where $H$ is the hyperplane section. Here I am using $\omega_{\mathbb{P}^N}\cong\mathcal{O}_{\mathbb{P}^N}(-N-1)$. To prove the formula for complete intersections just proceed by induction.

In general any smooth hypersurface $X\subset\mathbb{P}^{N}$ of degree $N+1$ is Calabi-Yau.

Just for sake of completeness. If $X\subset\mathbb{P}^N$ is a smooth complete intersection of hypersurfaces of degree $d_1,...,d_c$ one can consider $Y = H_{d_1}\cap H_{d_2}\cap...\cap H_{d_{c-1}}$. Then $Y$ is a complete intersection of codimension $c-1$ and $X\subset Y$. One may consider the exact sequence $$0\mapsto N_{X/Y}\rightarrow N_{X/\mathbb{P}^N}\rightarrow N_{Y/\mathbb{P}^N|X}\mapsto 0.$$ Now, $X$ is a smooth divisor in $Y$ cut out by an equation of degree $d_c$. Therefore $\mathcal{I}_{X/Y}=\mathcal{O}_{Y}(-d_c)$ and $N_{X/Y} = \mathcal{O}_{X}(d_c)$. By induction we assume that $N_{Y/\mathbb{P}^N} = (\mathcal{O}_{Y}(d_1)\oplus...\oplus \mathcal{O}_{Y}(d_{c-1}))_{|X} = \mathcal{O}_{X}(d_1)\oplus...\oplus \mathcal{O}_{X}(d_{c-1})$. Now, the above exact sequence clearly splits, and we get $$N_{X/\mathbb{P}^N} = \mathcal{O}_{X}(d_1)\oplus...\oplus \mathcal{O}_{X}(d_c).$$ In particular if $X\subset\mathbb{P}^4$ is a quintic $3$-fold we have $N_{X/\mathbb{P}^4} = \mathcal{O}_{X}(5)$. Now, we may use the adjunction formula (Proposition $8.10$, pag. $182$, in Hartshorne). We have $$\omega_{X}\cong \omega_{\mathbb{P}^N}\otimes \bigwedge^{c}N_{X/\mathbb{P}^{n}}\cong \mathcal{O}_{\mathbb{P}^N}(-N-1)\otimes \mathcal{O}_X(d_1+...+d_c)\cong \mathcal{O}_X(d_1+...+d_c-N-1).$$

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