Quintic threefold

Recall that a homogeneous polynomial ${\displaystyle f\in \Gamma (\mathbb {P} ^{4},{\mathcal {O}}(d))}$ (where ${\displaystyle {\mathcal {O}}(d)}$ is the Serre-twist of the hyperplane line bundle) defines a projective variety, or projective scheme, ${\displaystyle X}$, from the algebra

$${\displaystyle {\frac {k[x_{0},\ldots ,x_{4}]}{(f)}}}$$

where ${\displaystyle k}$ is a field, such as ${\displaystyle \mathbb {C} }$. Then, using the adjunction formula to compute its canonical bundle, we have

$${\displaystyle {\begin{aligned}\Omega _{X}^{3}&=\omega _{X}\\&=\omega _{\mathbb {P} ^{4}}\otimes {\mathcal {O}}(d)\\&\cong {\mathcal {O}}(-(4+1))\otimes {\mathcal {O}}(d)\\&\cong {\mathcal {O}}(d-5)\end{aligned}}}$$

hence in order for the variety to be Calabi-Yau, meaning it has a trivial canonical bundle, its degree must be ${\displaystyle 5}$. It is then a Calabi-Yau manifold if in addition this variety is smooth. This can be checked by looking at the zeros of the polynomials

$${\displaystyle \partial _{0}f,\ldots ,\partial _{4}f}$$

and making sure the set

$${\displaystyle \{x=[x_{0}:\cdots :x_{4}]|f(x)=\partial _{0}f(x)=\cdots =\partial _{4}f(x)=0\}}$$

is empty.

One of the easiest examples to check of a Calabi-Yau manifold is given by the Fermat quintic threefold, which is defined by the vanishing locus of the polynomial

$${\displaystyle f=x_{0}^{5}+x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}}$$

Computing the partial derivatives of ${\displaystyle f}$ gives the four polynomials

$${\displaystyle {\begin{aligned}\partial _{0}f=5x_{0}^{4}\\\partial _{1}f=5x_{1}^{4}\\\partial _{2}f=5x_{2}^{4}\\\partial _{3}f=5x_{3}^{4}\\\partial _{4}f=5x_{4}^{4}\\\end{aligned}}}$$

Since the only points where they vanish is given by the coordinate axes in ${\displaystyle \mathbb {P} ^{4}}$, the vanishing locus is empty since ${\displaystyle [0:0:0:0:0]}$ is not a point in ${\displaystyle \mathbb {P} ^{4}}$.

Another application of the quintic threefold is in the study of the infinitesimal generalized Hodge conjecture where this difficult problem can be solved in this case.[2] In fact, all of the lines on this hypersurface can be found explicitly.

Another popular class of examples of quintic three-folds, studied in many contexts, is the Dwork family. One popular study of such a family is from Candelas, De La Ossa, Green, and Parkes,[3] when they discovered mirror symmetry. This is given by the family[4] ^{pages 123-125}

$${\displaystyle f_{\psi }=x_{0}^{5}+x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}-5\psi x_{0}x_{1}x_{2}x_{3}x_{4}}$$

where ${\displaystyle \psi }$ is a single parameter not equal to a 5-th root of unity. This can be found by computing the partial derivates of ${\displaystyle f_{\psi }}$ and evaluating their zeros. The partial derivates are given by

$${\displaystyle {\begin{aligned}\partial _{0}f_{\psi }=5x_{0}^{4}-5\psi x_{1}x_{2}x_{3}x_{4}\\\partial _{1}f_{\psi }=5x_{1}^{4}-5\psi x_{0}x_{2}x_{3}x_{4}\\\partial _{2}f_{\psi }=5x_{2}^{4}-5\psi x_{0}x_{1}x_{3}x_{4}\\\partial _{3}f_{\psi }=5x_{3}^{4}-5\psi x_{0}x_{1}x_{2}x_{4}\\\partial _{4}f_{\psi }=5x_{4}^{4}-5\psi x_{0}x_{1}x_{2}x_{3}\\\end{aligned}}}$$

At a point where the partial derivatives are all zero, this gives the relation ${\displaystyle x_{i}^{5}=\psi x_{0}x_{1}x_{2}x_{3}x_{4}}$. For example, in ${\displaystyle \partial _{0}f_{\psi }}$ we get

$${\displaystyle {\begin{aligned}5x_{0}^{4}&=5\psi x_{1}x_{2}x_{3}x_{4}\\x_{0}^{4}&=\psi x_{1}x_{2}x_{3}x_{4}\\x_{0}^{5}&=\psi x_{0}x_{1}x_{2}x_{3}x_{4}\end{aligned}}}$$

by dividing out the ${\displaystyle 5}$ and multiplying each side by ${\displaystyle x_{0}}$. From multiplying these families of equations ${\displaystyle x_{i}^{5}=\psi x_{0}x_{1}x_{2}x_{3}x_{4}}$ together we have the relation

$${\displaystyle \prod x_{i}^{5}=\psi ^{5}\prod x_{i}^{5}}$$

showing a solution is either given by an ${\displaystyle x_{i}=0}$ or ${\displaystyle \psi ^{5}=1}$. But in the first case, these give a smooth sublocus since the varying term in ${\displaystyle f_{\psi }}$ vanishes, so a singular point must lie in ${\displaystyle \psi ^{5}=1}$. Given such a ${\displaystyle \psi }$, the singular points are then of the form

$${\displaystyle [\mu _{5}^{a_{0}}:\cdots$$ :\mu _{5}^{a_{4}}]}

such that

$${\displaystyle \mu _{5}^{\sum a_{i}}=\psi ^{-1}}$$

where ${\displaystyle \mu _{5}=e^{2\pi i/5}}$. For example, the point

$${\displaystyle [\mu _{5}^{4}:\mu _{5}^{-1}:\mu _{5}^{-1}:\mu _{5}^{-1}:\mu _{5}^{-1}]}$$

is a solution of both ${\displaystyle f_{1}}$ and its partial derivatives since ${\displaystyle (\mu _{5}^{i})^{5}=(\mu _{5}^{5})^{i}=1^{i}=1}$, and ${\displaystyle \psi =1}$.

• Barth–Nieto quintic
• Consani–Scholten quintic

Herbert Clemens (1984) conjectured that the number of rational curves of a given degree on a generic quintic threefold is finite. (Some smooth but non-generic quintic threefolds have infinite families of lines on them.) This was verified for degrees up to 7 by Sheldon Katz (1986) who also calculated the number 609250 of degree 2 rational curves. Philip Candelas, Xenia C. de la Ossa, and Paul S. Green et al. (1991) conjectured a general formula for the virtual number of rational curves of any degree, which was proved by Givental (1996) (the fact that the virtual number equals the actual number relies on confirmation of Clemens' conjecture, currently known for degree at most 11 Cotterill (2012)). The number of rational curves of various degrees on a generic quintic threefold is given by

2875, 609250, 317206375, 242467530000, ...(sequence A076912 in the OEIS).

Since the generic quintic threefold is a Calabi–Yau threefold and the moduli space of rational curves of a given degree is a discrete, finite set (hence compact), these have well-defined Donaldson–Thomas invariants (the "virtual number of points"); at least for degree 1 and 2, these agree with the actual number of points.