The intersection form on the nontrivial $S^2$-bundle over $S^2$, denoted $S^2 \tilde \times S^2$, can be written as $\left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ with respect to a suitable basis for $H_2(S^2 \tilde \times S^2)$.$^\mathbf{1}$

Things get confusing when I try to use another basis. Let $[F]$ denote the homology class of the $S^2$ fiber and let $[B]$ come from a section $S^2 \to S^2 \tilde \times S^2$ of the bundle, which are linearly independent$^\mathbf{2}$. Since $[F] \cdot [B]=1$, they should form a basis for $H_2(S^2 \tilde \times S^2)$. It's clear that $[F]\cdot [F]=0$, so the intersection form is given by $\left[ \begin{smallmatrix} 0 & 1 \\ 1 & * \end{smallmatrix} \right]$, where $*=[B]\cdot [B]$. Now, regardless of what $[B]\cdot [B]$ is, there seems to be an issue: The conjugacy class of $\left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ over $\mathbb{Z}$ contains matrices of the form $$\begin{bmatrix} 1 + 2ad & -2ab \\ 2cd & -1 -2ad\end{bmatrix},$$ and we can't have $0=1+2ad$ over $\mathbb{Z}$. So there can't be a change of basis connecting these forms.

What is wrong with the computation of the intersection form using the class of the fiber $[F]$ and base $[B]$?


  1. $S^2 \tilde \times S^2$ is diffeomorphic to $\mathbb{CP}^2 \# - \mathbb{CP}^2$. Each summand of $\mathbb{CP}^2 \# - \mathbb{CP}^2$ has its second homology generated by $\mathbb{CP}^1$, which has self-intersection $\pm 1$ in $\pm \mathbb{CP}^2$.

  2. One "direct" way to see that sections exist is to use the homotopy lifting property, and one "indirect" way would be to use the fiber bundle LES in homotopy groups to see that $\pi_2(S^2 \tilde \times S^2) \to \pi_2(S^2)$ is surjective. For linear independence, we note that $p_*[F]=0$ while $p_*[B]$ generates $\pi_2(S^2)$.

  • 1
    $\begingroup$ There are two very obvious sections for you to use. The usual way you construct $S^2 \tilde \times S^2$ it is explicitly fibered over a line in $\Bbb CP^2$ representing homology class $ H$ and the exceptional divisor representing $E$ in the blowup. The fiber will necessarily have intersection number $1$ with the line and intersection number $1$ with the exceptional divisor. Hence it will represent the class $H-E$. Notice the self-intersection number of the section depends on the section chosen. $\endgroup$ – PVAL-inactive Jul 21 '15 at 0:57

The intersection matrix transforms like a quadratic form under change of basis, not like an endomorphism. That is, if $P$ denotes the change of basis matrix, the matrix $Q$ of the intersection form transforms by $Q \mapsto P^{t}QP$, not by the similarity $Q \mapsto P^{-1}QP$.

| cite | improve this answer | |
  • $\begingroup$ Ah, there we go! $\endgroup$ – Kyle Jul 21 '15 at 4:59

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.