# 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, $[T^2] =3 \cdot [\mathbb CP^1]$.

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.

• For reference, the way I would prove the statement about elliptic curves in $H_2$ is probably by bezout's theorem and using that cup product is dual to intersection, but there should be ways to calculate that using topological information, right? – Andres Mejia Sep 16 '18 at 15:07
• I'm leaving this as a comment since it's a non-answer. If we take homology with rational coefficients, and assume that numerical and homological equivalence of cycles are the same, then the "Bezout" approach is all there is, i.e. the elliptic curve is 3 times [CP^1] precisely because it intersects generic lines in 3 points. – hunter Sep 16 '18 at 15:14
• @hunter what do you mean by numerical and homological equivalence of cycles are the same? Thank you for your comment by the way. – Andres Mejia Sep 16 '18 at 15:37
• it is Grothendieck's "standard conjecture D." Let $X$ be a smooth projective algebraic variet of dimension $n$, and $Z^i(X)$ the (very large) free abelian group of codimension $i$ subvarieties of $X$. There is a map $Z^i(X) \to H_{2n - 2i}(X)$ (the index switch is because of real vs. complex dimension) and the conjecture states roughly that the kernel of this map is exactly the set of (formal sums of) $n-i$-dimensional subvarieties whose intersection number with every subvariety is zero (these guys are in the kernel from the defn, but the conjecture says they're everything). – hunter Sep 16 '18 at 17:47
• I think the easiest thing to my intuition might be to observe that every complex subvariety (even singular!) has a fundamental class, and if you have a homotopy $f_t$ through polynomials, the images of the two fundamental classes will agree. Then you just need to check it in something sufficiently simple, like a union of $d$ lines. But that claim about fundamental classes requires some heavy machinery. I suspect there's an elementary framing of this idea. – user98602 Sep 23 '18 at 21:48

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 $$(n-1)(n-2)/2$$.

• thank you for the answer by the way! Do you have a basic source that covers arguments of this type? – Andres Mejia Nov 8 '18 at 5:44
• Oh, this is clever! I'm wondering now if one can see something like this with equations. For example, your case with lines seems to be topologically the same thing as $y^2=(x−1)(x-2)$. In this case, if $x \neq 1,2$, then you get two solutions, so two copies of the plane, and cutting along $[0,1]$, and making the necessary identifications, I think you get basically the connect sum of two planes so after compactification (a sphere.) – Andres Mejia Nov 8 '18 at 5:51
• @Andres I think the first place I saw this written down is in Gompf and Stipsicz. I am unsure how to do things algebraically here, but that doesn't mean one can't. – user98602 Nov 8 '18 at 10:25