7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Andres Mejia Sep 16 '18 at 15:07
  • 1
    $\begingroup$ 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. $\endgroup$ – hunter Sep 16 '18 at 15:14
  • $\begingroup$ @hunter what do you mean by numerical and homological equivalence of cycles are the same? Thank you for your comment by the way. $\endgroup$ – Andres Mejia Sep 16 '18 at 15:37
  • $\begingroup$ 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). $\endgroup$ – hunter Sep 16 '18 at 17:47
  • 1
    $\begingroup$ 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. $\endgroup$ – user98602 Sep 23 '18 at 21:48
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ thank you for the answer by the way! Do you have a basic source that covers arguments of this type? $\endgroup$ – Andres Mejia Nov 8 '18 at 5:44
  • $\begingroup$ 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.) $\endgroup$ – Andres Mejia Nov 8 '18 at 5:51
  • $\begingroup$ @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. $\endgroup$ – user98602 Nov 8 '18 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.