enumerative geometry - Interaction of topology and the Picard group of Algebraic surfaces

This answer addresses the cubic surface case. It may not tell you anything you don't already know, but just in case: one way to obtain $X$, which is useful for analyzing the lines,
and also the topology, is as the blow-up of ${mathbb P}^2$ at six (generically positioned)
points.

Each point blows up to a ${mathbb P}^1$, or topologically, an $S^2$, giving the six extra
spheres in $X$ that make the $H^2$ have dimension 7 (1 from the class of a line in
${mathbb P}^2$, and the rest from the six blow-ups).

Lable the points $P_i$, and the $S^2$s on $X$ obtained by blowing up $E_i$ (for ``exceptional divisor'').

Now the other 21 of the 27 lines are constructed as follows:

(a) draw lines between distinct pairs of six points (giving 15 lines), and take the proper
transform of each of these. Here "proper transform'' means the following: if $H_{ij}$ is the line joining $P_i$ and $P_j$ (here $H$ is for "hyperplane'', which in this context
is just a line), we pull-back $H_{ij}$ to $X$, which gives the union of
a line $l_{ij}$ intersecting each of $E_i$ and $E_j$ together with $E_i$ and $E_j$. Now
subtract $E_i$ and $E_j$, to obtain just the line $l_{ij}$.

Now in the cohomology ring of $X$, we see that the pull-back of $H_{ij}$ lies in
the ``first'' copy of ${mathbb Z}$ in ${mathbb Z}^7$, namely the one coming from
the class of a line in ${mathbb P}^2$, and so $l_{ij}$ maps to the class which
is $1$ in the first copy of ${mathbb Z}$, $-1$ in the copies spanned by $E_i$ and
$E_j$, and $0$ in the other copies.

More succinctly, the cohomology class of $l_{ij}$ is a linear combination of some of
the seven classes we already know.

(b) draw a conic through five of the six points, say all but $P_i$; denote
this $C_i$. Now take the proper transform of $C_i$; i.e. pull-back $C_i$
to obtain a line $l_i$ together with the five $E_j$ ($jneq i$), and then subtract
off these five $E_j$.

I have now accounted for 6 + 15 + 6 = 27 lines (the $E_i$, the $l_{ij}$, and the $l_i$)
using just the 7 independent cohomology classes.

Since in the cohomology of ${mathbb P}^2$ the class of a conic is just twice that
of a line, we see that again the class of $l_i$ is in the span of the known classes,
and we don't get anything beyond the ${mathbb Z}^7$ we already had.

In the case of a quartic surface, the situation will be similar; if one has a line
$l$ on the surface, the class of this line in cohomology will coincide with some
(linear combinination of) class(es) that we already know.