# Hodge Index Theorem

Algebraic Geometry

Hodge index theorem and its variations is a fundamental tool in study of smooth projective varieties. We aim at proving Hodge index theorem on higher dimensional varieties over an algebraically closed field of arbitrary characteristic.

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November 20, 2020

In this post, we will explore Hodge index theorem and its variations on a smooth projective variety $$X$$ over an algebraically close field $$\mathbb{k}$$ of arbitrary characteristic.

For simplicity, we assume divisors are integral. However, all results hold true for $$\mathbb{Q}$$-divisors.

## 1 Hodge Index Theorem on Surfaces

In this section, we denote by $$S$$ a smooth projective surface over an algebraically close field $$\mathbb{k}$$ of arbitrary characteristic.

A very good reference for algebraic surfaces in arbitrary characteristic is .

Let $$D$$ and $$H$$ be divisors on $$S$$. Assume that $$D^2>0$$. By Riemann-Roch theorem, either $$nD$$ or $$-nD$$ is effective for some sufficiently large $$n$$. If in addition that $$H\cdot D>0$$ for an ample divisor $$H$$, then $$nD$$ is effective. Using this fact, we can prove easily the following version of Hodge index theorem.

Lemma 1 Let $$H$$ an ample divisor on $$S$$. For a divisor $$D$$ on $$S$$, if $$H\cdot D=0$$, then $$D^2\leq 0$$. Moreover, $$D^2=0$$ if and only if $$D$$ is numerically trivial, that it $$D\cdot C=0$$ for any integral curve $$C$$.

Proof. Assume on the contrary that $$D^2>0$$. Then, either $$nD$$ or $$-nD$$ is effective. Because $$H$$ is ample. Then either $$H\cdot D>0$$ or $$H\cdot D<0$$ which is a contradiction with the assumption $$H\cdot D=0$$.

It suffices to show that $$D$$ is numerically trivial if $$H\cdot D=0$$ and $$D^2=0$$.

Assume on the contrary that $$D$$ is not numerically trivial. There would exist a curve $$C$$ such that $$D\cdot C\neq 0$$. Let $$E=(H^2)C-(H\cdot C)H$$. Then $$H(nD+E)=0$$ for any $$n$$. Hence, $(nD+E)^2=n^2D^2+2nD\cdot E+E^2=2nD\cdot E+E^2\leq 0$ If $$D\cdot C>0$$, then $$DE=H^2D\cdot C>0$$ and $$2nD\cdot E+E^2>0$$ for a sufficiently large $$n$$. That’s a contradiction. If $$D\cdot C<0$$, then $$2nD\cdot E+E^2>0$$ for a sufficiently large $$-n$$. Again, there is a contradiction.

Therefore, if $$H\cdot D=0$$ and $$D^2=0$$, then $$D$$ must be numerically trivial.

Apply the above theorem to $$A=D-\frac{H\cdot D}{H^2}H$$, where $$H$$ is ample, we get the following corollary.

Corollary 1 Let $$H$$ be an ample divisor on $$S$$. For a divisor $$D$$ on $$S$$, $$(H\cdot D)^2\geq H^2D^2$$.

In the above lemma and corollary, the condition that $$H$$ is ample can be generalized to $$H^2>0$$.

Theorem 1 (Hodge index theorem) The signature of the intersection pairing on $$S$$ is $$(1,\rho(S)-1)$$, where $$\rho(S)$$ is the Picard number.

In particular, for any two divisors $$D$$ and $$E$$ on $$S$$, if $$D^2>0$$ and $$D\cdot E=0$$, then $$E^2\leq 0$$ and $$E^2=0$$ if and only if $$E$$ is numerical trivial.

Proof. It suffices to prove the particular case. Because the Neron-Severi group $$NS(S)$$ can be decompose into $$D\oplus D^{\perp}$$. If $$D^2>0$$, then for any $$E\in D^\perp$$, as $$D\cdot E=0$$, $$E^2<0$$. So the intersection pairing on $$D^\perp$$ is negative definite.

Let $$A$$ be an ample divisor. If $$A\cdot E=0$$, by Lemma @ref(lem:HodgeIndex), we know that $$E^2\leq 0$$. Assume that $$A\cdot E\neq 0$$. Note that $$D\cdot A\neq 0$$. Let $$F=(A\cdot E)D-(A\cdot D)E$$. Then $$A\cdot F=0$$ which implies that $F^2=((A\cdot E)D-(A\cdot D)E)^2=(A\cdot E)^2D^2+(A\cdot D)^2E^2\leq 0.$ Consequently, $$E^2\leq 0$$.

If in addition $$E^2=0$$, then the same proof for that $$D$$ is ample works well in this case and shows that $$E$$ is numerical trivial.

Remark. For any divisors $$D_1$$, $$D_2$$, …, $$D_r$$, if $$D_1^2>0$$ and the intersection matrix $$\begin{pmatrix}D_i\cdot D_j\end{pmatrix}$$ is non-degenerate, then by the Hodge index theorem, the determinate $$\begin{vmatrix}D_i\cdot D_j\end{vmatrix}$$ has the sign $$(-1)^{r-1}$$.

The following proposition may be viewed as a justification of the claim in the remark.

Proposition 1 Let $$D_1$$ and $$D_2$$ be divisors on $$S$$. If $$(a D_1+b D_2)^2>0$$ for some numbers $$a$$ and $$b$$, then $\begin{vmatrix} D_1^2 & D_1\cdot D_2\\ D_1\cdot D_2 & D_2^2 \end{vmatrix}\leq 0$

Proof. By the assumption, one of $$a$$ and $$b$$ must be nonzero. If one of them is zero, the proposition is nothing but Corollary @ref(cor:FirstCor).

Now assume that $$b=0$$. Moreover, we may assume that $$a=1$$. Write $$D$$ for $$D_1$$ and $$E$$ for $$D_2$$ Let $$F=D^2E-(D\cdot E)D$$. Then $$D\cdot E=0$$. By Hodge index Theorem @ref(thm:HodgeIndexNefBig), we see that $E^2=(D^2)^2E^2-(D^2)(D\cdot E)^2\leq 0.$ Consequently, $\begin{vmatrix} D^2 & D\cdot E\\ D\cdot E & E^2 \end{vmatrix}\leq 0$

Now assume both $$a$$ and $$b$$ are nonzero. Let $$D=aD_1+bD_2$$ and $$E=D_2$$. Then $\begin{vmatrix} (aD_1+bD_2)^2 & (aD_1+bD_2)\cdot D_2\\ (aD_1+bD_2)\cdot D_2 & D_2^2 \end{vmatrix}= \begin{vmatrix} a^2D_1^2+2abD_1\cdot D_2+b^2D_2^2 & aD_1\cdot D_2+bD_2^2\\ aD_1\cdot D_2+bD_2^2 & D_2^2 \end{vmatrix}\leq 0$ Equivalently, $\begin{vmatrix} D_1^2 & D_1\cdot D_2\\ D_1\cdot D_2 & D_2^2 \end{vmatrix}\leq 0$

Conversely, if $$D_1^2<0$$ and $\begin{vmatrix} D_1^2 & D_1\cdot D_2\\ D_1\cdot D_2 & D_2^2 \end{vmatrix}>0,$ Then the intersection matrix $\begin{pmatrix} D_1^2 & D_1\cdot D_2\\ D_1\cdot D_2 & D_2^2 \end{pmatrix}$ is negative definite. Hence $$(aD_1+bD_2)^2<0$$ if $$(a, b)\neq (0,0)$$.

## 2 Hodge Index Theorem in Higher Dimension

In higher dimension, using inductions, we may obtain some generalizations of Hodge index theorem.

In this section, we denote by $$X$$ a $$n$$-dimensional smooth projective variety over an algebraically close $$\mathbb{k}$$.

Proposition 2 Let $$D$$ and $$D_i$$, $$i=1, 2, ... k$$, be divisors on $$X$$. Assume that $$D_i$$ are nef. Let $$n_i$$, $$i=1, 2, ... k$$, be nonnegative integers. If $$n_1+n_2+\cdots n_k=n-1\geq 1$$ and $$n_1\geq 1$$, then $(D\cdot D_1^{n_1}\cdots D_k^{n_k})^2\geq (D^2\cdot D_1^{n_1-1}\cdots D_k^{n_k})( D_1^{n_1+1}\cdots D_k^{n_k}).$

Proof. We first prove the theorem under the assumption that $$D_i$$ are all very ample. Let $$H_i$$ be general hyperplane sections that are linearly equivalent to $$D_i$$ respectively. By Bertini Theorem, we may assume that the intersection $$D_1^{n_1-1}\cdots D_k^{n_k}$$ is a smooth surface $$S$$. The inequality is then follows from the Hodge index theorem on surfaces. Indeed, $(D\cdot D_1^{n_1}\cdots D_k^{n_k})^2=(D|_S\cdot D_1|_S)^2\geq (D|_S^2)(D_1|_S)^2=(D^2\cdot D_1^{n_1-1}\cdots D_k^{n_k})( D_1^{n_1+1}\cdots D_k^{n_k}).$

The proof of the theorem can be reduced to the above particular case using a limiting trick with possibly necessary scaling.

Let $$H$$ be an ample divisor on $$X$$. Then by Nakai–Moishezon-Kleiman criterion for ampleness, all divisors $$D_i+tH$$ are ample for any $$t>0$$. Scale those divisors with if necessary, we may assume that $$D_i+tH$$ are very ample. Therefore, the divisor $$D$$ and $$D_i+tH$$ satisfy the inequality. Taking limits of both sides as $$t$$ goes to $$0$$, we end up with the desired inequality.

Remark. The case that $$n_i=1$$ was first proved by Fujita in . The version above can be found in Section 2.5 .

Note that in the above result, the divisor $$D$$ was not assumed to be nef.

If all divisors are assumed to be nef in the proposition, we can get a result involve the $$n$$-th power.

Theorem 2 (Hodge Index Theorem in Higher Dimensions) Let $$D_1$$, …, $$D_k$$ be nef divisors on $$X$$. If $$n_1+\cdots+n_k=n\geq 2$$ and $$n_i\geq0$$ for all $$i$$, then $(D_1\cdots D_k)^n\geq (D_1^n)^{n_1}\cdots (D_k^n)^{n_k}.$

Proof. We follow the argument in .

It suffices to show the theorem holds true if $$n_i=1$$ and $$D_i$$ are very ample for $$i=1,..., n$$. Moreover, by Bertini’s theorem, we may assume that $$D_i$$ are smooth hyperplane sections and the intersections $$D_{i_1}\cdots D_{i_r}$$ are smooth varieties of dimension $$n-r$$.

We prove the theorem by induction on $$n$$. The case $$n=2$$ is nothing but Corollary @ref(cor:FirstCor). Assume that the theorem holds true on varieties of dimension at most $$n-1$$.

For any integer $$a$$ such that $$1\leq a\leq n$$, restricting to $$D_a$$, we obtain the following inequality. $\begin{equation} (D_1\cdots D_n)^{n-1} =(\prod_{i\neq a}D_i|_{D_a})^{n-1}\\ \geq \prod\limits_{i\neq a}(D_i|_{D_a}^{n-1})\\ = \prod\limits_{i\neq a}(D_a\cdot D_i^{n-1}) (\#eq:ineq-a) \end{equation}$

Then $\begin{equation} (D_1\cdots D_n)^{n(n-1)}\geq \prod_{a=1}^n\prod\limits_{i\neq a}(D_a\cdot D_i^{n-1}). (\#eq:ineq-b) \end{equation}$

Similarly to the proof of inequality @ref(eq:ineq-a), we get \begin{equation} \begin{aligned} (D_a\cdot D_i^{n-1})^{(n-1)} =&D_a|_{D_i}\cdot D_i|_{D_i}^{n-2}\\ \geq & ((D_a|_{D_i})^{n-1})({D_i|_{D_i}}^{n-1})^{n-2}\\ =&(D_i\cdot {D_a}^{n-1})({D_i}^{n})^{n-2} \end{aligned} (\#eq:ineq-c) \end{equation} Therefore, \begin{equation} \begin{aligned} \prod_{a=1}^n\prod\limits_{i\neq a}(D_a\cdot D_i^{n-1})^{n-1} \geq & \prod_{a=1}^n\prod\limits_{i\neq a}((D_i\cdot {D_a}^{n-1})({D_i}^{n})^{n-2}\\ \geq & (\prod_{a=1}^n\prod\limits_{i\neq a}(D_i\cdot {D_a}^{n-1}))(\prod\limits_{i\neq a}({D_i}^{n})^{(n-1)(n-2)}) \end{aligned} (\#eq:ineq-d) \end{equation} By switching index names, we see that $\prod_{a=1}^n\prod\limits_{i\neq a}(D_i\cdot {D_a}^{n-1})= \prod_{i=1}^n\prod\limits_{i\neq a}(D_a\cdot {D_i}^{n-1}).$

From the inequality @ref(eq:ineq-d) and the above equality, we see that $\begin{equation} \prod_{a=1}^n\prod\limits_{i\neq a}D_a\cdot D_i^{n-1} \geq \prod\limits_{i\neq a}({D_i}^{n})^{(n-1)} (\#eq:ineq-e) \end{equation}$

The inequality $(D_1\cdots D_n)^n\geq \prod_{i=1}^nD_i^n$ follows from the inequalities @ref(eq:ineq-b) and @ref(eq:ineq-e).

Therefore, by mathematical induction, the theorem is proved.

Apply the Hodge index theorem, one can get the following corollary which is used to prove the Hodge index theorem in higher dimensions in .

Corollary 2 Let $$A$$ and $$B$$ be nef divisors on $$X$$. Then $(A^{n-1}\cdot B)(A\cdot B^{n-1})\geq A^n\cdot B^n.$

Proof. By Hodge index theorem, $(A^{n-1}\cdot B)^n\geq (A^n)^{n-1}B^n$ and $(B^{n-1}\cdot A)^n\geq (B^n)^{n-1}A^n$ The desired inequality follows from taking products and then the $$n$$-th root.

Here are some other very useful corollaries.

Corollary 3 Let $$A_1$$, …, $$A_p$$, $$B_1$$, …, $$B_{n-p}$$ be nef divisors on $$X$$. Then $(A_1\cdots A_p\cdot B_1\cdots B_{n-p})^p\geq \prod_{i=1}^p(A_i^p\cdot B_1\cdots B_{n-p})$

In particular, if $$1\leq q\leq p\leq n$$, then $(A^p\cdot B^{n-p})^p\geq (A^p\cdot B^{n-p})^q\cdot (B^n)^{p-q}.$

Proof. Similar to the proof of the Hodge index theorem, we may assume that $$B_1\cdots B_{n-p}$$ is a smooth projective variety $$Y$$ of dimension $$p$$. Apply Hodge index theorem to $$A_i|_Y$$, we get the inequality.

Let $$A_1=\cdots A_q=A$$ and $$A_{q+1}=\cdot A_p=B_1=\cdots B_{n-p}=B$$. Then the second inequality follows from the first one.

Corollary 4 Let $$A$$ and $$B$$ be nef divisors on $$X$$. Then $((A+B)^n)^{1/n}\geq (A^n)^{1/n}+(B^n)^{1/n}.$

Proof. The inequality follows byapplying the above Corollary @ref(cor:variation) with $$p=n$$ to the expansion of the left side and then taking the $$n$$-th root.

Remark. The results in this post are still valid if $$X$$ is simply an irreducible complete variety.

## References

Badescu, Lucian Silvestru. 2001. Algebraic Surfaces. Universitext. New York: Springer-Verlag. https://doi.org/10.1007/978-1-4757-3512-3.
Beltrametti, Mauro C., and Andrew J. Sommese. 1995. The Adjunction Theory of Complex Projective Varieties. Vol. 16. De Gruyter Expositions in Mathematics. Walter de Gruyter & Co., Berlin. https://doi.org/10.1515/9783110871746.
Fujita, Takao. 1982. “Theorems of Bertini Type for Certain Types of Polarized Manifolds.” Journal of the Mathematical Society of Japan 34 (4): 709–18. https://doi.org/b3xnm4.
Lazarsfeld, Robert. 2004. Positivity in Algebraic Geometry. I. Vol. 48. Ergebnisse Der Mathematik Und Ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics]. Springer-Verlag, Berlin. https://doi.org/10.1007/978-3-642-18808-4.