How to define CW-complex structure on cubic surface in $CP^3$? I have read roughly this blog and I have following question. I changed my original question to following.
How to define CW-complex structure on cubic hypersurface $M$ in $\mathbb CP^3$ defined by equation $x^3+y^3+z^3+t^3=0$. The second Betti number of this 4-manifold is 7. I expect that CW-complex structure is: couple of 2-cells and one 4-cell.
Intersection $T$ of $M$ with $\mathbb CP^2=\{t=0\}$ is surface genus $1$ which is torus, I conlude from the blog. But this doesn't help much.
The next misterious thing is 27 lines on the surface which I read in wikipedia article on cubic surface. I don't exactly understand what it means.
 A: I don't know what the CW-complex structure on the cubic hypersurface looks like, but I can talk about the $27$ lines : if you see the cubic hypersurface as embedded in $\mathbb CP^3$, then one can consider lines in $\mathbb CP^3$, i.e. subspaces of $\mathbb CP^3$ which are linearly isomorphic to $\mathbb CP^1$ (the embedding of $\mathbb CP^1$ to $\mathbb CP^3$ must be induced by a $\mathbb C$-linear map). In the case of the cubic hypersurface $M$, there are $27$ such lines which lie completely within $M$. 
An easy way to see the lines is to re-write the equation as 
$$
x^3 + y^3 = (-z)^3 + (-t)^3
$$
and note that $x^3 + y^3 = (x+ y)(x+\rho y)(x + \rho^2 y)$ where $\rho$ is a third root of unity, i.e. $\rho^2 + \rho + 1 = 0$ (or in other words, $\rho^3 = 1$ and $\rho \neq 1$). You can do the same with $-z$ and $-t$, so that a product of three linear factors equals a product of three linear factors, the variables being distinct on each side. Setting a pair of factors to zero gives a line ; an example would be 
$$
x=-y, \quad z=-t
$$
(this is the intersection of two complex hyperplanes in $\mathbb CP^3$, thus a complex line, i.e. a linear embedding of $\mathbb CP^1$ in $\mathbb CP^3$). 
Counting the number of pairs, you get $9$ lines. Repeating with each possible separation of variables (i.e. $( (x,y) , (z,t) )$ is the one we did above, but we could do $( (x,z),(y,t))$ and $( (x,t),(y,z))$), you get $9 \times 3 = 27$ lines. 
The big deal is that there are no other lines, and this is a non-trivial and very interesting algebro-geometric result.
Hope that helps,
