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Given a (real or almost complex) manifold $M$, Let the 2-nd unordered configuration space be the quotient space $$ B(M,2)=(M\times M\setminus\ \Delta)/\ \mathbb{Z}_2 $$ where $$ \Delta=\{(m,m)\mid m\in M \} $$ and $\mathbb{Z}_2$ acts by reversing the coordinates order. Then $B(M,2)$ is a (real or almost complex) manifold. Suppose the characteristic classes of $TM$ is given. How to obtain the characteristic classed of $TB(M,2)$? For example,

(1). How to compute the Stiefel-Whitney class $$ w(TB(\mathbb{R}P^n,2))? $$

(2). How to compute the Stiefel-Whitney class $$ w(TB(\mathbb{C}P^n,2))? $$

(2). How to compute the Chern class $$ c(TB(\mathbb{C}P^n,2))? $$

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  • $\begingroup$ Don't we have the characteristic classes of $M\times M$ by the product formula? In which case, can't we simply pull-it back to $M\times M\setminus \Delta$? Besides, why do you call this "symmetric product"? Usually symmetric product would mean $M\times M/(\Sigma _2)$, or am I missing something here? $\endgroup$ – user43326 Sep 4 '15 at 8:41
  • $\begingroup$ Sorry, Prof. I type wrongly. $\endgroup$ – QSR Sep 4 '15 at 8:49
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    $\begingroup$ To answer this question you first need to describe the cohomology of the unordered configuration space. This can be done using the map $B(M,2)\to S^\infty\times_{\mathbb{Z}/2} M\times M$, as in my answer to your previous question mathoverflow.net/questions/193982/… , but it's a bit messy. Then the next step would be to try to understand the Stiefel-Whitney classes of the quadratic construction on a vector bundle. This is an interesting question which should be approachable; I don't know if it's been done before. $\endgroup$ – Mark Grant Oct 5 '15 at 13:18
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An excellent discussion of symmetric powers in general is given by Pavle Blagojevich - one of the first things he mentions is that the $k$-th symmetric power of $\mathbb{R}P^n$ is $\mathbb{R}P^{kn},$ a result due to Arnold. This is NOT true of $\mathbb{C}P^n,$ but for $\mathbb{C}P^\infty$ this is studied by Gusein-Zade et al. This might be enough for what you want.(that paper seems quite nice also).

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    $\begingroup$ The question is not actually about symmetric powers (see user43326's comment)... $\endgroup$ – Dan Petersen Sep 4 '15 at 14:33

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