1
$\begingroup$

Hi, everyone: I am reading a small expository paper on properties of CP2, in which the intersection form is defined as an integral of the wedge of two forms $w_1$, $w_2$, and these forms $w_1$, $w_2$ (no problem with compact support, since CP2 is compact) seem to have been obtained from the fundamental class [z] of H2(CP2)-- a copy of CP1 (embedded in CP2), after which we integrate $w:=w1\wedge w2$ to get the intersection number.

I am curious on whether I am reading the above correctly, i.e., that the volume form in CP2 is obtained by using the fund. class [z] in H2. If not, would someone explain; if this is correct, if we are we using some form of deRham's theorem to turn a purely topological object like [z] into an object like $w$, for which we must have a differentiable structure defined)?

Thanks in Advance.

$\endgroup$
2
  • $\begingroup$ I took the liberty of editing the formatting of your question a little - hope that's OK, if not let me know. $\endgroup$
    – Yemon Choi
    Apr 28, 2010 at 23:46
  • $\begingroup$ Short answer - for all compact oriented manifolds. $\endgroup$ Apr 29, 2010 at 1:36

1 Answer 1

1
$\begingroup$

Consider a suitably small tubular neighbourhood $\mathcal{N}$ of $\mathbb{CP}^1$, thought of as sitting inside $\mathbb{CP}^2$. Then $\mathcal{N}$ locally looks like $\mathbb{CP}^1\times D_2$. The volume form $\omega$ of $\mathbb{CP}^1$ is not necessarily a $2$-form in $\mathbb{CP}^2$. However, one can imagine changing it so that we have new $2$-form $\widetilde{\omega}$ in $\mathbb{CP}^2$, supported in $\mathcal{N}$, such that $\widetilde{\omega}$ restrcited to $\{p\}\times D_2$ (for $p\in\mathbb{CP}^1$) looks like a smooth bump function which integrates to $1$. This can be taken to be $w_1$ in your case. Now assume you take your copy of $\mathbb{CP}^1$ inside $\mathbb{CP}^2$ and perturb it a bit (i.e., make it transversal to itself) to get another copy. Apply what we said before and get $w_2$ supported in a suitable tubular neighbourhood of this perturbed copy. Now integrating $w_1\wedge w_2$ over $\mathbb{CP}^2$ gives you an integration over balls around points where self-intersections occur. The normalization were so chosen that it counts the intersection number of $\mathbb{CP}^1$ with itself.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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