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.

## 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 1, 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 1.

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 1, 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 $$(a, b)\neq (0,0)$$.

## 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 1. 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. $$$(D_1\cdots D_n)^{n-1} =(\prod_{i\neq a}D_i|_{D_a})^{n-1}\\ \geq \prod\limits_{i\neq a}(D_i|_{D_a}^{n-1})\\ = \prod\limits_{i\neq a}(D_a\cdot D_i^{n-1}) \tag{1}$$$

Then $$$(D_1\cdots D_n)^{n(n-1)}\geq \prod_{a=1}^n\prod\limits_{i\neq a}(D_a\cdot D_i^{n-1}). \tag{2}$$$

Similarly to the proof of inequality (1), we get \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} \tag{3} Therefore, \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} \tag{4} 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 (4) and the above equality, we see that $$$\prod_{a=1}^n\prod\limits_{i\neq a}D_a\cdot D_i^{n-1} \geq \prod\limits_{i\neq a}({D_i}^{n})^{(n-1)} \tag{5}$$$

The inequality $(D_1\cdots D_n)^n\geq \prod_{i=1}^nD_i^n$ follows from the inequalities (2) and (5).

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 3 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.
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.

### Author's bio

Fei Ye (https://yfei.page) is an assistant professor at QCC-CUNY.

### Reuse

Text and figures are licensed under Creative Commons Attribution CC BY-NC-SA 4.0.

### Citation

Fei Ye (2020). Hodge Index Theorem. Fei Ye's Math Blogs. /post/2020/11/20/hodge_index/
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