3

Letting $C_1$ and $C_2$ be two algebraic curves sitting in $\mathbb CP^2$, they both realize homology classes in $H_2(\mathbb CP^2)$ and are in fact represented by $n \cdot [\mathbb CP^1]$ and $m \cdot [\mathbb CP^1]$.

If the curves intersect transversely, then we have that $n \cdot m=[C_1]^* \smile [C_2]^*=[C_1 \cap C_2]^* \in H^4(\mathbb CP^2)$.

By dualizing, one realizes that $n \cdot m=[C_1 \cap C_2] \in H_0(\mathbb CP^2)$ which proves the case for Bézout's theorem for transverse intersections.

Can one generalize this argument for curves intersecting non-transversely?

My guess is that since this number depends only on homology class, one can perturb such cases in order to specialize back to this proof.

Bernard
  • 175,478
Andres Mejia
  • 20,977
  • A comment on the correctness of this proof (and maybe a reference?) would be nice as well. I got the idea from an IUM problem from the topology II course offered there. There is a reference here – Andres Mejia Jun 11 '18 at 01:44

1 Answers1

4

This is a good question, and you're right that you can perturb without changing intersection numbers while making them transverse. The crucial question is why doing this changes a multiplicity $n$ intersection - an ($n-1$)-fold tangency - to $n$ positive intersections. The point is that near intersections of curves in 2-space, we may model the intersection as the zero set of a holomorphic map $\Bbb C \to \Bbb C$ (near zero). Then we know that holomorphic maps are, up to a coordinate change, given by $z \mapsto z^k$; a multiplicity $n$ intersection corresponds to the map $z \mapsto z^n$. Then consider the homotopy $f_t(z) = z^n - t$, for $t \in [0, \varepsilon]$. For $t > 0$, this has $n$ distinct zeroes $t^{1/n} e^{2\pi i/n}$, all of which are positive (holomorphic zeroes always are).

Now observing that this homotopy never crosses zero outside of $|z|\leq t^{1/n}$, we may 'dampen out its effects' near $\infty$ to get a compactly supported homotopy from $z^n$ to a function $f(z)$ which is $z^n - t$ for small $|z|$ and $z^n$ for large $|z|$ (and the homotopy is through functions identical to $z^n$ for large $|z|$); I don't want to write out a formula, but it is not hard to see this is possible. In particular, we may change the non-transverse multiplicity $n$ zeroes to $n$ positive transverse zeroes without changing the curve far away from the zero in question.

Now your argument shows that there are $nm$ of these points.

  • 3
    Note that you can make any two smooth submanifolds of an ambient manifold transverse; the magic here is being able to count how many transverse intersection points come from a given non-transverse one. –  Jun 11 '18 at 11:57
  • This is a really nice argument; do you know of any reference for it? – Steve D Jun 11 '18 at 15:25
  • @MikeMiller this is very nice indeed. Could you suggest a case (smooth manifolds, analytic manifolds etc.) where such an argument would fail? It seems at least to me that the holomorphic structure is very essential here if you would like to keep the same total intersection number (using the local structure of holomorphic maps.) – Andres Mejia Jun 11 '18 at 17:27
  • @SteveD I don't, but that's a sign of my ignorance more than anything else; I think this argument must appear many times in the literature, and I am told it is implicit in the work of the Italian algebraic geometers. But I do not personally know a reference. –  Jun 11 '18 at 17:30
  • 1
    @AndresMejia For a partially illuminating example, I believe this works mod 2 but not integrally for real analytic manifolds. The essential point is to consider the variations $f_\epsilon(x) = x^k - \epsilon$; $f_0$ has a single zero, while if $k$ is odd there is a unique real solution when $\epsilon \neq 0$, but if $k$ is even there are two real zeroes for $\epsilon > 0$ and no real zeroes for $\epsilon < 0$. So you can read off the parity of the number of transverse zeroes from the zeroes and their multiplicity, but not the actual count. I have not actually checked but I imagine this... –  Jun 11 '18 at 17:35
  • @MikeMiller thank you! How interesting... – Andres Mejia Jun 11 '18 at 17:36
  • ...observation continues to the realm of zeroes of analytic functions $\Bbb R^k \to \Bbb R^k$. –  Jun 11 '18 at 17:36
  • For smooth manifolds, unfortunately, we may have intersection points of infinite order, and so have no mechanism to count by. What I did not say explicitly in my answer is that whenever you have intersection points of infinite order of connected analytic submanifolds, one must actually be contained inside the other. This is not true in the land of smooth functions (think graphs of non-analytic functions on $\Bbb R$ and horizontal lines). –  Jun 11 '18 at 17:38
  • Hi @MikeMiller I actually realized recently that my argument is cyclical in that I don't know how to prove the cohomology class that a curve represents without bezout's theorem. I've asked a new question here. I hope that you might have an answer! – Andres Mejia Sep 23 '18 at 21:38