Let $\mathbb{S}^2\to E\to B$ be an $\mathbb{S}^2$-bundle. We may assume that it is the fibre-wise one-point compactification of a complex line bundle, so there is a zero section $s_0:B\to E$ and an $\infty$-section $s_\infty:B\to E$. Identify $B_\infty:=s_\infty(B)$. Now look at the $h$-fold symmetric fibre product $\mathbb{C}P^h\to \mathrm{SP}^hE\stackrel{p}{\to} B$. The following is clear:
- The bundle is homologically trivial, i.\,e. there is a class $\chi_h\in H^2(\mathrm{SP}^hE)$ which gives us an isomorphism $H^*(\mathbb{C} P^h)\otimes H^*(B)\to H^*(\mathrm{SP}^hE), \mu^i\otimes\alpha\mapsto \chi^i_h\smile p^*\alpha$ where $\mu\in H^2(\mathbb{C}P^h)$ is a generator.
- We have a bundle morphism $\jmath:E=\mathrm{SP}^1E\to \mathrm{SP}^hE$ and $\jmath^*\chi_h=\chi_1$.
- Let $\tau\in H^2(E,B_\infty)$ be the Thom class and $\imath:E\to(E,B_\infty)$ the inclusion. Then $\imath^*\tau=\chi_1$. In particular, $s_0^*\jmath^*\chi_h = s_0^*\chi_1=s_0^*\imath^*\tau\in H^2(B)$ is the Euler class of $E\to B$.
Now consider a section $s:B\to \mathrm{SP}^hE$ which unfortunately does not satisfy $s=\jmath\circ s_0$, but has the property that for each $x\in B$, the zero element $s_0(x)$ is part of the configuration $s(x)$ (if you think of $s(x)\in \mathrm{SP}^h(\mathbb{S}^2)$ as a polynomial, this means that $0$ is a root of zero of $s(x)$.).
Question: Can we say anything about $s^*\chi_h$? Maybe it is a multiple of the Euler class?