I'm trying to find an example of two divisors $D_{1} \ D_{2}$ on a complex algebraic projective surface $S$ such that:

$D_{1}\equiv D_{2}$ where the equivalence relation is the numerical equivalence relation:

Let $[A],[B] \in Pic(S)$ the cocycle associated to $A,B\in Div(S)$, $A \equiv B$ if and only if $[A][C]=[B][C]$ for every $[C]\in Pic(S)$.

with $D_{1}$ very ample and $D_{2}$ is not very ample.

  • $\begingroup$ yes the correct relation is $[A][C]=[B][C]$ $\endgroup$
    – dario
    Sep 20, 2014 at 16:38

1 Answer 1


You ask for a projective surface $S$ and two divisors $H_1, H_2$ on S, which are numerically equivalent, but not linear equivalent. In addition, the corresponding line bundle $\mathscr L_1 := \mathscr L_{H_1}$ shall be very ample but $\mathscr L_2 :=\mathscr L_{H_2}$ not.

The general problem behind is the question: Is very-ampleness of a line bundle $\mathscr L$ on an algebraic variety $X$ a numeric condition, depending only on numeric properties of $\mathscr L$ like its Chern and intersection numbers?

First, for a smooth curve $X$: Very-ampleness of a line bundle on $X$ cannot be characterized numerically.

Consider $X \subset \mathbb P^2$ a smooth hypersurface of degree $d=4$. Then X has genus $g = 3$ and $\mathscr L_1 := \mathscr O_X(1) = \kappa _X$ is very ample. On the other hand, choose a non-trivial line bundle $\mathscr L_0 \in Pic^0(X)$ on $X$ with degree $deg \ \mathscr L_0 = 0.$ The choice is possible because the Jacobi torus $Pic^0(X)$ has dimension $g = 3$. Define $\mathscr L_2 := \kappa_X \otimes \mathscr {(L^0)^{-1}}$. We have

$$c_1(\mathscr L_2) = deg \ \mathscr L_2 = 4 = deg \ \mathscr L_1 = c_1(\mathscr L_1),$$

$h^1(X, \mathscr L_2) = h^0(X, \mathscr L_0)$ by Serre duality and

$$h^0(X, \mathscr L_0) = 0$$

because $deg \ \mathscr L_0 = 0$ but $\mathscr L_0 \neq \mathscr O_X.$ By Riemann-Roch we get

$$h^0(X, \mathscr L_2) = h^0(X, \mathscr L_2) - h^1(X, \mathscr L_2) = deg \ \mathscr L_2 + 1 - g = 4 +1 -3 = 2.$$

As a consequence,

$$\phi _{\mathscr L_2}: X \longrightarrow \mathbb P^1$$

cannot be an embedding, i.e. $\mathscr L_2$ is not very ample, q.e.d.

Secondly, for a smooth surface $S$: Very-ampleness of a line bundle on $S$ cannot be characterized numerically.

To prove the claim I follow Asal's advice to extend the curve example above to a surface example. Define the surface $S := X \times \mathbb P^1$ with the curve $X$ from above. On $S$ consider the two line bundles

$$\mathscr L_1(1) := \pi _1^* \mathscr L_1 \otimes \pi_2^* \mathscr O_{\mathbb P^1}(1), \mathscr L_2(1) := \pi _1^* \mathscr L_2 \otimes \pi_2^* \mathscr O_{\mathbb P^1}(1)$$

with $\mathscr L_i$ from above and $\pi_i$ the canonical projections from $S, i=1,2$. Then, e.g. using the Künneth formula for a product,

$$h^0(S, \mathscr L_i(1)) = h^0(X, \mathscr L_i) * h^0(\mathbb P^1, \mathscr O_{\mathbb P^1}(1)) = 2 * h^0(X, \mathscr L_i), i = 1,2.$$

As a consequecen $h^0(S, \mathscr L_1(1)) = 2*3 = 6$, while $h^0(S, \mathscr L_2(1)) = 2*2 = 4$. The map

$$\phi_{\mathscr L_1(1)}:S \longrightarrow \mathbb P^5$$

is the composition of the product $\phi_{\mathscr L_1} \times id_{\mathbb P^1}: X \times \mathbb P^1 \longrightarrow \mathbb P^2 \times \mathbb P^1$ with the Segre embedding $\mathbb P^2 \times \mathbb P^1 \longrightarrow \mathbb P^5$. Hence $\mathscr L_1(1)$ is very ample.

On the other hand, the map

$$\phi_{\mathscr L_2(1)}:S \longrightarrow \mathbb P^3$$

cannot be an embedding: Otherwise it would embedd $S \subset \mathbb P^3$ as a hypersurface of degree $d = (\mathscr L_2(1), \mathscr L_2(1)) = 2 * 4 = 8.$ Then

$$\kappa_S = (\kappa_{\mathbb P^3} \otimes \mathscr O_{\mathbb P3}(8))|S = \mathscr O_S(-4+8) = \mathscr O_S(4)$$

is very ample, notably $h^0(S, \kappa_S) \neq 0$. On the other hand - using again the Künneth formula - $h^0(S, \kappa_S) = h^0(X, \kappa_X) * h^0(\mathbb P^1, \kappa_{\mathbb P^1}) = 3 * 0 = 0$. The contradiction shows that $\mathscr L_2(1)$ is not very ample.

Eventually, we prove that both line bundles $\mathscr L_1(1)$ and $\mathscr L_2(1)$ are numerically equivalent, i.e. $(\mathscr L_1(1), \mathscr L_C) = (\mathscr L_2(1), \mathscr L_C)$ for any divisor $C$ on $S$. Because $h^1(\mathbb P^1, \mathscr O_{\mathbb P^1}) = 0$ we have $Pic(S) \cong Pic(X) \times Pic(\mathbb P^1)$ and we may assume $\mathscr L_C$ the inverse image under the canonical projection of a line bundle on $X$ or a line bundle on $\mathbb P^1$. Because a divisor on a smooth algebraic curve is ample iff it has positive degree, we may restrict to effective divisors. For an effective divisor $C$ on X we have

$$(\mathscr L_1(1), \pi_1^*\mathscr L_C) = (\mathscr L_2(1), \pi_1^*\mathscr L_C) = deg \ C > 0.$$

For an effective divisor $C$ on $\mathbb P^1$ we have

$$(\mathscr L_1(1), \pi_2^*\mathscr L_C) = (\mathscr L_2(1), \pi_2^*\mathscr L_C) = 4* deg \ C > 0,$$

which completes the proof, q.e.d.

  • $\begingroup$ +1 for the amended version. I guess you should now be able to soup this up to a surface example, just by taking product with $\mathbf P^1$ say. $\endgroup$
    – user64687
    Sep 25, 2014 at 9:25

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