It is well known that a smooth cubic surface $X\subset \mathbb{P}^3$ has exactly 27 lines in it. Furthermore, it is easy to check that Picard group $$Pic(X)\cong \mathbb{Z}^7$$ Here the generators are lines which are $\mathbb{P}^1$ topologically. Furthermore, it is easy to check that $$\chi_{top}(X)=2+H_2(X)=9$$Topologically speaking, notice that as smooth manifold $X$ has no 1-skeleton. This makes the 2-dimensional cells glue to points along their bounday, getting spheres $\mathbb{P}^1$ as the result of this gluing process.

*My first question is why the other 20 lines do not contribute to the Euler characteristic of $X$.*

Going further, if $X\subset \mathbb{P}^3$ has degree 4, it is known that $X$ sometimes has lines, sometimes it does not. However, $$\chi_{top}(X)=2+H_2(X)=24$$ meaning that, despite the fact that $X$ can perfectly have no lines, we still have homology $H_2$ which are spheres topologically! meaning, there are in fact spheres (due to the argument above which says that $X$ has no 1-skeleton). Besides $\chi_{top}$ is constant even though $X$ may have $64$, $32$ or even $0$ number of lines in it. There are spheres whose existence is not being noticed by $\chi_{top}$ at all. Here let me be vague please. What is going on!?

Any type of editing to make this clearer will be welcome. References highly appreciated.