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Oct 15, 2015 at 3:21 vote accept QSR
Oct 5, 2015 at 13:18 comment added Mark Grant 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.
Sep 5, 2015 at 2:52 history edited QSR CC BY-SA 3.0
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Sep 4, 2015 at 14:11 answer added Igor Rivin timeline score: 0
Sep 4, 2015 at 12:53 history edited QSR CC BY-SA 3.0
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Sep 4, 2015 at 8:49 comment added QSR Sorry, Prof. I type wrongly.
Sep 4, 2015 at 8:46 history edited QSR CC BY-SA 3.0
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Sep 4, 2015 at 8:41 comment added user43326 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?
Sep 4, 2015 at 7:52 history asked QSR CC BY-SA 3.0