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Let $S^m$ be the $m$-sphere and $$F(S^m,2)/\mathbb{Z}_2=\{(a,b)\mid a,b\in S^m, a\neq b\}/(a,b)\sim (b,a)$$ be the $2$-nd unordered configuration space on $S^m$. How to compute the total Stiefel-Whitney class of the tangent bundle of $F(S^m,2)/\mathbb{Z}_2$ $$ w(TF(S^m,2)/\mathbb{Z}_2)? $$

Is $F(S^m,2)/\mathbb{Z}_2$ homeomorphic to $\mathbb{R}P^m\times \mathbb{R}^m$ or not?

Moreover, if we change $S^m$ to other manifolds, for example, projective spaces, Grassmannians, then how to compute the corresponding Stiefel-Whitney class? The question is on characteristic classes of tangent bundle of 2-nd unordered configuration space

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The configuration space $F(S^m, 2)/\mathbb Z_2$ is diffeomorphic to the total space of the $m$-dimensional vector bundle $E=\gamma^\perp$ over $\mathbb RP^m$, where $\gamma\subset \mathbb R^{m+1}$ is the tautological 1-dimensional line bundle and $\gamma^\perp$ is its orthogonal complement.

To define a diffeomorphism $\phi\colon Tot(E)\to F(S^m, 2)/\mathbb Z_2$ let us pick a point $([v], w)\in Tot(E)$ (where $[v]\in\mathbb R P^m$ and $w\in v^\perp$), such that a vector $v\in \mathbb R^{m+1}$ has the unit norm $||v||_2=1$.

Let $\phi \colon ([v],w)\mapsto\bigl(\frac{w+v}{||w+v||_2}, \frac{w-v}{||w-v||_2}\bigr)\in F(S^m, 2)/\mathbb Z_2$. You can easily verify, that $\phi$ is diffeomorphism.

Computation of the total Stiefel-Whitney class is now straighforward. Let $\pi\colon Tot(E)\to \mathbb R P^m$ be the natural projection, then one has the long exact sequence: $$ 0\to \pi^* E\to TTot(E)\to\pi^*T\mathbb RP^m\to 0, $$ hence $$ w(TF(S^m, 2)/\mathbb Z_2)=w(\pi^*E)w(\pi^*T\mathbb RP^m)=\pi^*(1+a)^{-1}\cdot\pi^*(1+a)^{m+1}=(1+\pi^*a)^m,$$ where $a\in H^1(\mathbb RP^m,\mathbb Z_2)$ is the generator.

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  • $\begingroup$ Thanks! How to obtain the (short?) exact sequence? I am not clear.... $\endgroup$
    – QSR
    Commented Sep 8, 2015 at 2:26
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    $\begingroup$ The second map is just the map induced by $\pi_*\colon TTot(E)\to T\mathbb RP^m$. The kernel of this map is the vertical component of $TTot(E)$, which is precisely $\pi^*E$. $\endgroup$
    – Yury Ustinovskiy
    Commented Sep 8, 2015 at 2:33
  • $\begingroup$ Thanks! Why the exact sequence of vector bundles can imply that the Stiefel-Whitney class satisfy the product formula? I am only clear with the special case when the short exact sequence split, then we have the Whitney sum. $\endgroup$
    – QSR
    Commented Sep 8, 2015 at 5:09
  • $\begingroup$ Why the Stiefel-Whitney class's Cartan formula for Whitney sum is also true for short exact sequence? $\endgroup$
    – QSR
    Commented Sep 8, 2015 at 7:00
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    $\begingroup$ Given a short exact sequence of vector bundles $0\to V\to W\to U\to 0$, you can always introduce a metric on $W$ and take the orthogonal complement to $V\subset W$. This gives you a splitting $W=V\oplus U$. $\endgroup$
    – Yury Ustinovskiy
    Commented Sep 8, 2015 at 11:49

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