suppose X is an elliptic surface with a section an...

Let $\pi:X\to B$ be the elliptic fibration, $S$ the chosen zero-section, and $F$ the class of a fibre. Assume $X$ is smooth, projective, and relatively minimal. Since $\rho(X)=2$, the two independent classes $S,F$ form a basis of $N^1(X)_{\mathbb R}$.

Put

$$\chi=\chi(\mathcal O_X).$$

Then

$$F^2=0,\qquad S\cdot F=1,\qquad S^2=-\chi.$$

So for

$$D=aS+bF$$

we have

$$D\cdot F=a,\qquad D\cdot S=b-a\chi,\qquad D^2=2ab-\chi a^2.$$

The cones are:

$$\boxed{\overline{\mathrm{Eff}}(X)=\mathbb R_{\ge 0}[S]+\mathbb R_{\ge 0}[F].}$$

Equivalently, an effective divisor class is in the closed cone spanned by the section and the fibre.

The nef cone is the dual cone:

$$\boxed{\mathrm{Nef}(X)=\mathbb R_{\ge 0}[F]+\mathbb R_{\ge 0}[S+\chi F].}$$

Thus

$$\boxed{\mathrm{Amp}(X)=\{\,aS+bF: a>0,\ b>a\chi\,\}.}$$

Equivalently,

$$\mathrm{Amp}(X)=\mathbb R_{>0}[F]+\mathbb R_{>0}[S+\chi F].$$

Why: by Shioda–Tate, $\rho=2$ forces no reducible fibre components contributing to $NS(X)$; the “trivial lattice” is just generated by the zero-section and fibre, up to finite Mordell–Weil torsion. The standard Shioda–Tate formula is

$$\operatorname{rank} MW = \rho(X)-2-\sum_v(m_v-1),$$

where $m_v$ is the number of components of the fibre over $v$. (PlanetMath)
For any irreducible curve $C$, if $C$ is vertical then $C\equiv rF$; if $C$ is horizontal and $C\ne S$, then $C\cdot F>0$ and $C\cdot S\ge 0$, which puts its class in the cone generated by $S,F$. Since $S$ and $F$ themselves are effective, this gives the effective cone. The ample cone then follows from Nakai–Moishezon/Kleiman: on a surface, ampleness is positivity against curves plus positive self-intersection; here that becomes $a>0$ and $b-a\chi>0$. For a modern reference on Nakai–Moishezon for real line bundles, see Fujino–Miyamoto. (Springer Link)

References for the elliptic-surface background: Miranda’s The Basic Theory of Elliptic Surfaces is a standard source; its bibliographic record explicitly lists the zero section, $NS(X)$, Mordell–Weil group, and self-intersection among the topics. (Google Books) Also see Schütt–Shioda, Elliptic Surfaces, Adv. Stud. Pure Math. 60 (2010), 51–160, a survey of elliptic surfaces with section and Mordell–Weil lattices. (research.uni-hannover.de)

Let $\pi:X\to B$ be the elliptic fibration, $S$ the zero-section, $F$ a fibre, and

$$L=(R^1\pi_*\mathcal O_X)^{-1},\qquad \deg L=\chi(\mathcal O_X)=:\chi .$$

Let $M$ be the line bundle on $B$ corresponding to the vertical part $bF$, so $\deg M=b$. Then

$$\mathcal O_X(aS+bF)=\mathcal O_X(aS)\otimes \pi^*M.$$

For $a\ge 0$,

$$h^0(X,\mathcal O_X(aS+bF)) = h^0\bigl(B,\pi_*\mathcal O_X(aS)\otimes M\bigr).$$

The key standard decomposition is

$$\pi_*\mathcal O_X(aS) \simeq \begin{cases} \mathcal O_B, & a=0,1,\\[4pt] \mathcal O_B\oplus L^{-2}\oplus L^{-3}\oplus\cdots\oplus L^{-a}, & a\ge 2. \end{cases}$$

Therefore:

$$\boxed{ h^0(X,\mathcal O_X(aS+bF)) = h^0(B,M) + \sum_{i=2}^a h^0(B,M\otimes L^{-i}) }$$

with the convention that the sum is empty for $a=0,1$.

So explicitly:

$$\boxed{ h^0(aS+bF) = \begin{cases} h^0(B,M), & a=0,1,\\[4pt] h^0(B,M)+\displaystyle\sum_{i=2}^a h^0(B,M\otimes L^{-i}), & a\ge 2. \end{cases} }$$

Since

$$\deg(M\otimes L^{-i})=b-i\chi,$$

this becomes especially explicit when $B=\mathbb P^1$. Then $L\simeq \mathcal O_{\mathbb P^1}(\chi)$, $M\simeq\mathcal O_{\mathbb P^1}(b)$, and

$$h^0(\mathbb P^1,\mathcal O(d))=\max\{d+1,0\}.$$

Hence

$$\boxed{ h^0(X,\mathcal O_X(aS+bF)) = \begin{cases} b+1, & a=0,1,\\[4pt] (b+1)+\displaystyle\sum_{i=2}^a \max\{b-i\chi+1,0\}, & a\ge 2. \end{cases} }$$

Equivalently, for $a\ge 2$, if

$$m=\min\left(a,\left\lfloor \frac{b}{\chi}\right\rfloor\right)$$

when $\chi>0$, then

$$h^0(aS+bF) = b+1+ \sum_{i=2}^m (b-i\chi+1).$$

So if $m\ge 2$,

$$\boxed{ h^0(aS+bF) = m(b+1)-\chi\left(\frac{m(m+1)}2-1\right). }$$

For a general base curve $B$ of genus $g$, the formula in terms of $a,b,\chi$ alone is not determined unless the line bundles involved are nonspecial. If

$$b-a\chi\ge 2g-1,$$

then every $M\otimes L^{-i}$ appearing is nonspecial, and Riemann–Roch gives

$$\boxed{ h^0(aS+bF) = a(b+1-g)-\chi\left(\frac{a(a+1)}2-1\right) }$$

for $a\ge 1$. For $a=0$, it is simply $h^0(B,M)$, equal to $b+1-g$ if $b\ge 2g-1$.

A standard reference for the decomposition of $\pi_*\mathcal O_X(aS)$ is Miranda, The Basic Theory of Elliptic Surfaces, especially the discussion of Weierstrass models and the fundamental line bundle. Another useful reference is Schütt–Shioda, Elliptic Surfaces, in Algebraic Geometry in East Asia—Seoul 2008, Adv. Stud. Pure Math. 60.

Let $X\to B$ have zero-section $S$, fibre $F$, and

$$S^2=-\chi,\qquad S\cdot F=1,\qquad F^2=0,$$

where $\chi=\chi(\mathcal O_X)$. Suppose

$$C=\sum_{i=1}^m n_i C_i\in |D|,\qquad D=d(aS+bF),$$

and

$$C_i\equiv a_iS+b_iF$$

is integral. Then the class equality gives

$$\sum_i n_i a_i=da,\qquad \sum_i n_i b_i=db.$$

Since $n_i\ge 1$,

$$\sum_i a_i\le da,\qquad \sum_i b_i\le db.$$

The product dimension is

$$\boxed{ \dim \bigl(|C_1|\times\cdots\times |C_m|\bigr) = \sum_{i=1}^m \dim |C_i| = \sum_{i=1}^m \bigl(h^0(X,\mathcal O_X(C_i))-1\bigr). }$$

For each $C_i=a_iS+b_iF$, let $M_i\in \operatorname{Pic}^{b_i}(B)$ be the line bundle on $B$ giving the vertical part, and let $L$ be the fundamental line bundle with $\deg L=\chi$. Then

$$h^0(X,\mathcal O_X(C_i)) = h^0(B,M_i) + \sum_{j=2}^{a_i} h^0(B,M_i\otimes L^{-j}).$$

So the exact expression is

$$\boxed{ \dim \prod_i |C_i| = \sum_{i=1}^m \left( h^0(B,M_i) + \sum_{j=2}^{a_i} h^0(B,M_i\otimes L^{-j}) -1 \right). }$$

This is the cleanest general formula. If $B$ has positive genus, the numerical data $a_i,b_i$ alone do not determine these $h^0$’s.

In the common case $B=\mathbb P^1$, the formula becomes explicit. Since

$$h^0(\mathbb P^1,\mathcal O(k))=\max\{k+1,0\},$$

we get

$$\boxed{ \dim |C_i| = b_i+ \sum_{j=2}^{a_i}\max\{b_i-j\chi+1,0\}. }$$

Because $C_i$ is integral, there are two main cases:

1. $C_i=F$, a fibre. Then

$$\dim |C_i|=\dim |F|=1.$$

2. $C_i$ is horizontal. If $C_i\ne S$, then $C_i\cdot S\ge 0$, hence

$$b_i-a_i\chi\ge 0.$$

Thus no truncation occurs in the max terms, and

$$\boxed{ \dim |C_i| = a_i(b_i+1) - \chi\left(\frac{a_i(a_i+1)}2-1\right) -1. }$$

This formula also gives $\dim |S|=0$ when $a_i=1,b_i=0$.

Therefore, for $B=\mathbb P^1$,

$$\boxed{ \dim \prod_i |C_i| = \#\{i:C_i\equiv F\} + \sum_{C_i\not\equiv F} \left[ a_i(b_i+1) - \chi\left(\frac{a_i(a_i+1)}2-1\right) -1 \right]. }$$

For a rough asymptotic estimate in terms of $d$, use

$$C_i^2=2a_i b_i-\chi a_i^2.$$

Then

$$\dim |C_i| \le \frac{C_i^2}{2}+O(a_i+1).$$

Summing and using $\sum n_iC_i=D$, one obtains

$$\boxed{ \dim \prod_i |C_i| \le \frac{D^2}{2}+O(d). }$$

Since

$$D^2=d^2(aS+bF)^2=d^2(2ab-\chi a^2),$$

this is

$$\boxed{ \dim \prod_i |C_i| \le d^2\left(ab-\frac{\chi a^2}{2}\right)+O(d). }$$

More explicitly, one may take a crude bound of the form

$$\boxed{ \dim \prod_i |C_i| \le d^2\left(ab-\frac{\chi a^2}{2}\right) + O_{a,b,\chi}(d). }$$

The multiplicities $n_i$ only enter through the constraints

$$\sum_i n_i a_i=da,\qquad \sum_i n_i b_i=db;$$

they do not affect $\dim |C_1|\times\cdots\times |C_m|$ directly.

Let $g=p_a(B)>0$, and let

$$L=(R^1\pi_*\mathcal O_X)^{-1},\qquad \deg L=\chi.$$

For

$$C_i\equiv a_iS+b_iF,$$

write the vertical line bundle as $M_i\in \operatorname{Pic}^{b_i}(B)$. Then

$$\mathcal O_X(C_i)=\mathcal O_X(a_iS)\otimes \pi^*M_i,$$

and

$$h^0(X,\mathcal O_X(C_i)) = h^0(B,M_i)+\sum_{j=2}^{a_i}h^0(B,M_i\otimes L^{-j}).$$

So the problem is to bound the terms

$$N_{ij}:=M_i\otimes L^{-j},\qquad \deg N_{ij}=b_i-j\chi,$$

with $j=0,2,\dots,a_i$, where $N_{i0}=M_i$.

Strictly speaking, $N_{ij}$ lives on $B$, not on $C_i$. Applying Clifford to the base curve $B$, assuming $B$ is smooth or more generally Gorenstein in a setting where Clifford holds, gives the following bound. For a line bundle $N$ of degree $e$ on $B$,

$$h^0(B,N)\le \Phi_g(e),$$

where

$$\Phi_g(e)= \begin{cases} 0, & e<0,\\[4pt] \left\lfloor \dfrac e2\right\rfloor+1, & 0\le e\le 2g-2,\\[8pt] e+1-g, & e\ge 2g-1. \end{cases}$$

The middle range is exactly Clifford’s theorem; the last range is Riemann–Roch plus nonspeciality.

Therefore

$$\boxed{ \dim |C_i| \le -1+\Phi_g(b_i)+\sum_{j=2}^{a_i}\Phi_g(b_i-j\chi). }$$

Hence

$$\boxed{ \dim\bigl(|C_1|\times\cdots\times |C_m|\bigr) \le \sum_{i=1}^m \left[ -1+\Phi_g(b_i)+\sum_{j=2}^{a_i}\Phi_g(b_i-j\chi) \right]. }$$

This is the clean Clifford-type upper bound in terms of the $a_i,b_i$.

If every term satisfies

$$0\le b_i-j\chi\le 2g-2,$$

then Clifford gives the simpler estimate

$$h^0(B,M_i\otimes L^{-j}) \le \frac{b_i-j\chi}{2}+1.$$

Thus

$$\dim |C_i| \le \frac12\left[ a_ib_i-\chi\left(\frac{a_i(a_i+1)}2-1\right) \right] +a_i-1.$$

So in the fully special range,

$$\boxed{ \dim\prod_i |C_i| \le \sum_i \left\{ \frac12\left[ a_ib_i-\chi\left(\frac{a_i(a_i+1)}2-1\right) \right] +a_i-1 \right\}. }$$

On the other hand, if

$$b_i-a_i\chi\ge 2g-1,$$

then all the terms $M_i\otimes L^{-j}$ are nonspecial, and

$$h^0(B,M_i\otimes L^{-j})=b_i-j\chi+1-g.$$

Then

$$\boxed{ \dim |C_i| \le a_i(b_i+1-g) - \chi\left(\frac{a_i(a_i+1)}2-1\right) -1. }$$

There is also a useful coarser estimate directly on the Gorenstein curve $C_i$. Since $C_i$ is an integral Cartier divisor on the smooth surface $X$, it is Gorenstein. From

$$0\to \mathcal O_X\to \mathcal O_X(C_i)\to \mathcal O_{C_i}(C_i)\to 0$$

we get

$$\dim |C_i| = h^0(X,\mathcal O_X(C_i))-1 \le h^0(C_i,\mathcal O_{C_i}(C_i)).$$

Now

$$\deg \mathcal O_{C_i}(C_i)=C_i^2=2a_ib_i-\chi a_i^2.$$

By adjunction,

$$\omega_{C_i}\otimes \mathcal O_{C_i}(-C_i) \simeq \mathcal O_{C_i}(K_X).$$

Since

$$K_X\equiv (2g-2+\chi)F,$$

we have

$$K_X\cdot C_i=a_i(2g-2+\chi)\ge 0.$$

Thus $\mathcal O_{C_i}(C_i)$ lies in the Clifford range. Applying Clifford on the Gorenstein curve $C_i$,

$$h^0(C_i,\mathcal O_{C_i}(C_i)) \le \frac{C_i^2}{2}+1.$$

Therefore

$$\boxed{ \dim |C_i|\le \frac{C_i^2}{2}+1 = a_ib_i-\frac{\chi a_i^2}{2}+1. }$$

Summing gives

$$\boxed{ \dim\prod_i |C_i| \le \sum_i\left(a_ib_i-\frac{\chi a_i^2}{2}+1\right). }$$

Using

$$\sum_i n_iC_i=D=d(aS+bF)$$

and $n_i\ge 1$, this yields the asymptotic bound

$$\boxed{ \dim\prod_i |C_i| \le \frac{D^2}{2}+O(d) = d^2\left(ab-\frac{\chi a^2}{2}\right)+O(d). }$$

More explicitly,

$$D^2=d^2(2ab-\chi a^2),$$

so the leading term is

$$\boxed{ d^2\left(ab-\frac{\chi a^2}{2}\right). }$$

The $O(d)$ term comes from the number of components $m$, since horizontal components contribute at least one unit of $S$-coefficient and vertical components are fibres, so

$$m=O(d).$$

For references on Clifford for singular/Gorenstein curves, see Franciosi–Tenni, On Clifford’s theorem for singular curves, which treats 2-connected Gorenstein curves contained in a smooth surface and gives a Clifford-type inequality for Cartier clusters; their Theorem 3.9 gives the relevant form for Cartier data. (pagine.dm.unipi.it) More broadly, Franciosi’s later work extends Clifford-index language to reduced curves with planar singularities via rank-one torsion-free sheaves. (Springer)

Put

$$A=da,\qquad B=db,\qquad K=2B+4-cA=d(2b-ca)+4.$$

Then

$$a_1+a_2=A,\qquad b_1+b_2=B.$$

Define

$$u_i=2b_i+2-ca_i.$$

Then

$$u_1+u_2=2B+4-cA=K.$$

Also

$$(a_1+1)+(a_2+1)=A+2.$$

Hence

$$f=(a_1+1)(a_2+1)u_1u_2.$$

By the elementary inequality $xy\le \frac{(x+y)^2}{4}$,

$$(a_1+1)(a_2+1)\le \frac{(A+2)^2}{4},$$

and

$$u_1u_2\le \frac{K^2}{4}.$$

Therefore

$$\boxed{ f\le \frac{(A+2)^2K^2}{16}. }$$

Substituting $A=da$, $B=db$, this gives

$$\boxed{ f\le \frac{(da+2)^2\bigl(d(2b-ca)+4\bigr)^2}{16}. }$$

With integrality, the sharper integer upper bound is

$$\boxed{ f\le \left\lfloor \frac{(da+2)^2}{4}\right\rfloor \left\lfloor \frac{\bigl(d(2b-ca)+4\bigr)^2}{4}\right\rfloor . }$$

This is the clean maximal estimate.

If one wants the exact integer maximum, there are parity effects. For each $a_1=x$, with $0\le x\le A$, choose $b_1=y$. Then

$$a_2=A-x,\qquad b_2=B-y.$$

For fixed $x$, the two linear factors have constant sum $K$, and their difference is

$$u_1-u_2 = 4y-2B+c(A-2x).$$

So for fixed $x$, the best choice of $y$ is the integer $0\le y\le B$ minimizing

$$\left|4y-2B+c(A-2x)\right|.$$

Thus the exact maximum is

$$\boxed{ \max_{0\le x\le A} (x+1)(A-x+1) \cdot \frac{ K^2-\epsilon_x^2 }{4}, }$$

where

$$\epsilon_x = \min_{0\le y\le B} \left|4y-2B+c(A-2x)\right|.$$

Here $x,y$ range over integers.

In particular, if $da$ and $db$ are both even, then the upper bound is attained by

$$a_1=a_2=\frac{da}{2},\qquad b_1=b_2=\frac{db}{2},$$

and the exact maximum is

$$\boxed{ \frac{(da+2)^2\bigl(d(2b-ca)+4\bigr)^2}{16}. }$$

For general parity, the exact value is the finite maximum above, while the floor bound is usually the most useful closed-form estimate.

Let

$$E:=B-cA.$$

The constraints $b_i>ca_i$ are equivalent to

$$b_i=ca_i+e_i,\qquad e_i\ge 1,\qquad e_i\in \mathbb Z.$$

Since $b_1+b_2=B$ and $a_1+a_2=A$, we get

$$e_1+e_2=B-cA=E.$$

So a feasible solution exists only if

$$\boxed{E=B-cA\ge 2.}$$

Now

$$f=a_1b_2+a_2b_1-ca_1a_2.$$

Substituting $b_i=ca_i+e_i$,

$$f = ca_1a_2+a_1e_2+a_2e_1.$$

For fixed $a_1,a_2$, since $e_1+e_2=E$, the maximum is obtained by putting as much excess as possible on the larger $a_i$-side. Thus, after possibly swapping indices, write

$$a_1=r,\qquad a_2=A-r,\qquad 0\le r\le \left\lfloor \frac A2\right\rfloor,$$

and choose

$$e_1=E-1,\qquad e_2=1.$$

Then

$$f(r)=c r(A-r)+r+(A-r)(E-1).$$

Equivalently,

$$f(r)=A(E-1)+(2cA-B+2)r-cr^2.$$

Put

$$H:=2cA-B+2.$$

So the exact maximum is

$$\boxed{ f_{\max} = A(E-1)+\max_{0\le r\le \lfloor A/2\rfloor} \left(Hr-cr^2\right). }$$

Since the quadratic is concave, the maximizing integer is the integer nearest to $H/(2c)$, clipped to the interval $[0,\lfloor A/2\rfloor]$. Explicitly,

$$\boxed{ r_*= \min\left\{ \left\lfloor \frac A2\right\rfloor, \max\left\{0,\left\lfloor \frac{H+c}{2c}\right\rfloor\right\} \right\}. }$$

Therefore

$$\boxed{ f_{\max} = A(B-cA-1) + (2cA-B+2)r_* - cr_*^2. }$$

That is the exact integer maximum.

A maximizing choice is, up to swapping indices,

$$a_1=r_*,\qquad a_2=A-r_*,$$ $$b_1=cr_*+(B-cA-1),\qquad b_2=c(A-r_*)+1.$$

Indeed,

$$b_1+b_2=B,$$

and

$$b_i>ca_i$$

because the excesses are $B-cA-1$ and $1$.

In particular, if

$$B\ge 2cA+2,$$

then $H\le 0$, so $r_*=0$, and

$$\boxed{ f_{\max}=A(B-cA-1). }$$

If

$$cA+2\le B<2cA+2,$$

then $H>0$, and the formula above with

$$r_*= \min\left\{ \left\lfloor \frac A2\right\rfloor, \left\lfloor \frac{2cA-B+2+c}{2c}\right\rfloor \right\}$$

gives the maximum.