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§3. Fibrations Especially Surfaces Over Curves 391

In the case of a surface over a curve, the fibres are curves or more generally one

dimensional schemes of the form D = n i E i where the E i are irreducible curves.
i
Hence the fibre is this divisor D. Now we study the intersection properties of D and
E i with the following elementary result from linear algebra.
(3.4) Proposition. Let V be an inner product space over Q generated by vectors e i

for i ∈ I with (e i |e j ) ≥ 0 for all i = j. If there exists a vector z = i a i e i with all
a i > 0 such that (z|e j ) = 0 for all j, then we have (z|x) = 0 and (x|x) ≤ 0 for all
x ∈ V.
Proof. Since any x is a linear combination of the e j and (z|e j ) = 0 for all indices j,
it follows that (z|x) = 0. for the negative definite statement, we write any x ∈ V as

c
x = i i a i e i where c i ∈ Q and calculate
1
2 2 2 2

(x|x) − c a (e i |e i ) = c i c j (a i e i |a j e j ) ≤ (c + c )(a i e i |a j e j )
i i 2 i j
i i = j i = j
1 1
2 2
= c (a i e i |z − a i e i ) + c (a j e j |z − a j e j )
i
j
2 2
i j
2 2
=− c a (e i |e i ).
i i
i
Thus the sequence of inequalities gives (x|x) ≤ 0. This proves the proposition.
(3.5) Remark. In the previous inequality (3.3) we can ask when is it true that
(x|x) = 0 as we already know one example (z|z) = 0. The inequality that we
2
2
used above is uv ≤ (1/2)(u + v ) which comes from the positivity of squares
2
0 ≤ (u − v) . Clearly this inequality is an equality if and only if u = v. This means
that 0 = (x|x) if and only if for all i = j we have either (e i |e j ) = 0or c i = c j .
Let I be the set of indices, and define two indices i and j to be connected provided
there exists a sequence of indices i = i(0), ..., i(q) = j with (e i( −1) |e i( ) )> 0 for
all = 1,..., q. This puts an equivalence relation of connectedness on I, and the
equivalence classes are called connected components of I. For a connected compo-

nent K of I, we form the part z K = i∈K a i e i of z, and observe that z = K z K
where the sum is over the connected components of z. With these notations, we have
the following assertion.
(3.6) Assertion. For x ∈ V ,wehave0 = (x|x) if and only if we can write x =

c K z K where the sum again is over the connected components of I and each
K
c K ∈ Q. This sum formula for x is just the condition that for i = j we have either
(e i |e j ) = 0or c i = c j . In the special case where I is connected, this becomes
0 = (x|x) if and only if x = cz for some c ∈ Q.
(3.7) Application I. Let f : X → B be a morphism of a surface onto a curve,

and consider a fibre D = n i E i where the E i are distinct irreducible curves and
i
n i > 0. Then D · E i = 0 because D is algebraically equivalent to any near by fibre,
and E i · E j ≥ 0 for i = j. Form the Q-vector space generated by the curves E i
with the intersection form and apply (3.3) to obtain E i · E i ≤ 0. This is a necessary
condition for any divisor to be the fibre of a fibration to a curve.