Gaining an appreciation for homology class representatives in $\mathbb CP^n$. Given a compact oriented submanifold $N \subset M$ one says that $N$ represents a homology class in $M$ by taking $i_*(\tau_N)$ where $i_*$ is induced by inclusion and $\tau_N$ is the fundamental class of $N$ chosen according to orientation.
There are some cases where this is completely clear. For example, $S^1$ represents a generator in $\mathbb C \setminus \{0\}$, or $\mathbb CP^1$ represents a homology class in $\mathbb CP^2$ by the $CW$-structure.
However, there are some more mysterious cases for me.
For example, a degree $3$ complex projective curve should be $3 \cdot [\mathbb CP^1] \in H_2(\mathbb CP^2).$ But these are tori (when they are elliptic curves) so up to homology, $[T^2] =3 \cdot [\mathbb CP^1]$. 
Probably a satisfactory answer (assuming that I'm thinking about this correctly) would include something like: 

Can one prove that a torus represents $3 \cdot [\mathbb CP^1] \in
 \mathbb H_2(\mathbb CP^2)$ geometrically?

or a reference pointing to how one can begin to do these types of geometric calculations.
 A: Take $n$ generic lines in $\Bbb{CP}^2$. That means that any two of them intersect in a point, and no three of them intersect at a point. Let's take this and smooth it bit by bit, one sphere at a time. The intersections look like the two factors of $\Bbb C$ in $\Bbb C^2$. To resolve this intersection, delete a neighborhood of $0$, and then connect a tube between the two factors. 
This has the effect of a connected sum operation. When doing this with two lines, the result is topologically a sphere. When doing this with three lines, we have to resolve two intersections with the third sphere and the previous two. Resolving one of those just does a connected sum with a sphere (and so, topologically, nothing), but resolving the next performs the connected sum of a sphere with itself - this adds 1 to the genus. 
To go from an embedded representative of $(n-1)C$ this way to one of $nC$, you add $(n-2)$ to the genus - there are $(n-1)$ intersections with previous lines, and resolving the first self-intersection doesn't add to the genus. Inductively, one finds that the genus of such a representative of $nC$ is $(n-1)(n-2)/2$. 
