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Could someone tell me why and how Hilbert schemes and relative Hilbert schemes are important and useful in algebraic geometry?

Could anyone give me some applications of this notion in concrete terms?

Thanks a lot for your answers.

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    $\begingroup$ It depends what your bend is? They are extremely useful in showing many natural stacks are algebraic. But, what type of answer are you looking for? $\endgroup$ – Alex Youcis Feb 29 '16 at 2:25
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As Alex Youcis said, there are many ways Hilbert schemes can be useful. What follows is my personal favorite: Kollár's counterexample to the Integral Hodge Conjecture.

I will try to be elementary for the motivation, but for the actual example I will have to be more technical.

The Hodge Conjecture

You might've heard of the Hodge Conjecture (one of the Clay Millenium problems). Fix $X$ a non-singular complex projective variety. We can ask: what cohomology classes of $X$ are linear combinations of Poincaré duals of subvarieties of $X$? Only some can be expected to be such linear combinations: these are called Hodge classes and are the elements of $H^{2k}(X,A) \cap H^{k,k}(X)$, where our coefficients $A$ will be either $\mathbf{Q}$ or $\mathbf{Z}$. If a Hodge class is indeed a linear combination of Poincaré duals of homology classes of subvarieties in $X$, we say it is algebraic.

We can now state the conjecture:

Hodge Conjecture. If $X$ is a non-singular complex projective variety, then every rational Hodge class is algebraic.

Now this is not the original form of the Hodge Conjecture: the original conjecture concerns integral Hodge classes, i.e., elements in $H^{2k}(X,\mathbf{Z}) \cap H^{k,k}(X)$:

Integral Hodge Conjecture. If $X$ is a non-singular complex projective variety, then every integral Hodge class is algebraic.

This version of the Hodge Conjecture is false: Atiyah and Hirzebruch in 1962 found examples of torsion integral Hodge classes that are not algebraic. Maybe we can at least ask for all non-torsion Hodge classes to be algebraic? But the answer to this question is also "no": Kollár in 1990 constructed non-torsion integral Hodge classes that are not algebraic. These are the first known non-torsion counterexamples to the Integral Hodge Conjecture; see Colliot-Thélène–Voisin, Totaro, Diaz, and Ottem–Suzuki for other examples.

Kollár's idea was the following: for "very general" hypersurfaces $X$ of certain degree, you can degenerate the pair $(C,X)$ where $C$ is any curve in $X$, to a special pair $(C_0,X_0)$ using the relative Hilbert scheme, and then compute the degree of $C$ there to prove certain one-dimensional Hodge classes in $X$ cannot be classes associated to any curve $C$.

Kollár's Counterexample

I'll now present Kollár's counterexample. For simplicity, I will do a very special case of the construction. You can modify the numerics in many ways and still get counterexamples to the integral Hodge conjecture. See Voisin and Soulé for details.

Let $X \subseteq \mathbf{P}^4$ be a hypersurface of degree $125$. The Lefschetz hyperplane theorem together with Hodge duality implies $H^2(X,\mathbf{Z}) = \mathbf{Z} \cdot h$, where $h$ is the restriction of the hyperplane section $H$ on $\mathbf{P}^4$, and $H^4(X,\mathbf{Z}) = \mathbf{Z} \cdot \alpha$ where $$ \int_X \alpha \cap h = 1. $$ Note that $125\alpha$ satisfies $$ \int_X 125\alpha \cap h = 125 = \int_X h^3, $$ so $125h = \alpha$ in cohomology. We then claim

Theorem (Kollár). Let $X$ be a very general hypersurface of degree $125$ in $\mathbf{P}^4$. Then, any curve $C \subseteq X$ has degree divisible by $5$, and so the Hodge class $\alpha$ from above is not algebraic.

Note that since $125h = \alpha$ in cohomology, this does not give a counterexample to the (rational) Hodge Conjecture.

Here "very general" means that $X$ is in the complement of a union of countably many Zariski closed subsets of the space $\mathbf{P}^N$ parametrizing all degree $125$ polynomials on $\mathbf{P}^4$. The exact description of this set will be in the proof.

Step 1. For such hypersurfaces $X_0$ of a certain type, any 1-dimensional subscheme $C_0 \subseteq X_0$ has degree divisible by $5$.

Proof of Step 1. We first describe these $X_0$. Consider $Y$ a hyperplane in $\mathbf{P}^4$, and its image $X_0$ via the composition of a $5$-uple embedding, followed by a generic projection back down to $\mathbf{P}^4$:

$\hskip2.5in$Generic Projection

By the theory of generic projections (the reference I found is Roberts), the map $\phi\colon Y \to X_0$ then satisfies the following properties:

  1. $X_0$ is a degree $125$ hypersurface in $\mathbf{P}^4$;
  2. $\phi$ is generically one-to-one;
  3. $\phi$ is two-to-one generically over a surface in $X_0$;
  4. $\phi$ is three-to-one generically over a curve in $X_0$.

Now let $C_0 \subseteq X_0$ be a curve with associated cycle $z_0$. By the properties above, there is a cycle $\tilde{z}_0$ on $Y$ such that $\phi_*(\tilde{z}_0) = 6z_0$, and so $6 \deg z_0 = \deg \phi_*(\tilde{z}_0)$. On the other hand, we can compute $$ \deg \phi_*(\tilde{z}_0) = \int_{X_0} c_1(\mathcal{O}_{X_0}(1)) \cap \phi_*(\tilde{z}_0) = \int_Y c_1(\phi^*\mathcal{O}_{X_0}(1)) \cap \tilde{z}_0 = \int_Y c_1(\mathcal{O}_{Y}(5)) \cap \tilde{z}_0, $$ by the projection formula [Fulton, Thm. 2.5] and so $5 \mid \deg \phi_*(\tilde{z}_0) = 6 \deg z_0$. But $5$ and $6$ are coprime, so $5 \mid \deg z_0$. $\blacksquare$

Step 2. Any very general hypersurface $X$ of degree $125$ with a chosen curve $C \subseteq X$ can be degenerated into a hypersurface $X_0$ and a curve $C_0 \subseteq X_0$ such that $\deg C = \deg C_0$.

Proof of Step 2. Let $\mathbf{P}^N$ be the space parametrizing all degree $125$ polynomials on $\mathbf{P}^4$. Let $\mathcal{X} \to \mathbf{P}^N$ be the universal hypersurface. Then, consider the relative Hilbert schemes $$ \mathcal{H}_v \to \mathbf{P}^N $$ parametrizing pairs $\{(Z,X) \mid Z \subseteq X\}$, where $Z$ is a one-dimensional subscheme with Hilbert polynomial $v$. The Hilbert polynomials $v$ encode all possible values for the degree and genus of $Z$, and so there are only countably many choices of $v$. Now, we use the following facts about the relative Hilbert scheme [Kollár, Thm. 1.4]:

  • The morphism $\rho_v\colon \mathcal{H}_v \to \mathbf{P}^N$ is projective;
  • There exists a universal subscheme $$\mathcal{Z}_v \subseteq \mathcal{H}_v \times_{\mathbf{P}^N} \mathcal{X}$$ which is flat over $\mathcal{H}_v$.

Now let $U$ be the set $$\mathbf{P}^N \setminus \bigcup_{v \in I} \rho_v(\mathcal{H}_v)$$ where $I$ is the set of Hilbert polynomials for which the map $\rho_v$ is not dominant. This set $U$ will parametrize the "very general" hypersurfaces $X$ of degree $125$.

Suppose $X \subseteq \mathbf{P}^4$ is a very general non-singular hypersurface $X$ of degree $125$, parametrized by $x \in U$. Let $C \subseteq X$ be a curve; then, giving $C$ the reduced subscheme structure, $(C,X)$ parametrizes a point $c_x \in \mathcal{H}_v$ over $x$ for some $v$. By definition of $U$, we have $\rho_v(c_x) = x$, hence the map $\rho_v$ is surjective since it is dominant and projective. We then have some point $c_0 \in \mathcal{H}_v$ such that $\rho_v(c_0) = x_0$, where $x_0$ is the point parametrizing the hypersurface $X_0$ constructed above. The fibre $Z_0$ of the universal subscheme $\mathcal{Z}_v$ over $c_0$ gives a subscheme $Z_0 \subseteq X_0$, which by flatness has the same degree as $C$. $\blacksquare$

Finally, we have shown the integral Hodge class $\alpha$ is not algebraic, since it has degree $1$, but any curve $C$ would have an associated cycle whose degree is divisible by $5$. $\blacksquare$

An Open Question

I want to conclude with some remarks about what is open (in addition to the Hodge Conjecture). Notice that Kollár's example constructed above is a hypersurface very high degree, hence is of general type. We can ask:

Open Question. Let $X$ be a Fano (or more generally, rationally connected) variety of dimension $n$. Does the integral Hodge conjecture hold for cohomology classes of degree $4$ or $2n-2$?

The reason we cannot hope for better is because we can construct rationally connected varieties that fail the integral Hodge conjecture for cohomology classes of degree $2n-2k$ for any $n-2 > k > 1$ by blowing up Kollár's example embedded in a larger projective space; see Voisin and Soulé, p. 113. On the other hand, the question above has a positive answer for rational varieties [Voisin and Soulé, p. 113], uniruled threefolds [Voisin 2006], and cubic fourfolds [Voisin 2007].

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  • $\begingroup$ Excellent answer! Thanks for making it! $\endgroup$ – Alex Youcis Feb 29 '16 at 7:11
  • $\begingroup$ Thank you @AlexYoucis! Now I just have to hope I didn't say anything wrong… $\endgroup$ – Takumi Murayama Feb 29 '16 at 7:15
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    $\begingroup$ Thank you very much for your clean answer. This is exactly the main purpose of my two questions above. My real purpose was to understand the Kollar counterexample refuting Integral Hodge conjecture. :) $\endgroup$ – Lina45 Feb 29 '16 at 9:55
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    $\begingroup$ @Lina, next time make expicit what you want to know in your questions! $\endgroup$ – Mariano Suárez-Álvarez Feb 29 '16 at 17:12
  • $\begingroup$ Ok ! Sorry Sir. ;-) $\endgroup$ – Lina45 Feb 29 '16 at 18:58

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