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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 6, 2013 at 2:32 comment added George Another attempt to get a nice formula: $Grass(\infty,\infty) \times Grass(\infty,\infty) \to Grass(\infty, \infty)$
Jun 6, 2013 at 2:29 vote accept George
Jun 6, 2013 at 2:27 comment added George Steve - thanks! Actually, I wanted something more elementary, in terms of intersections. And from your answer it is clear how to obtain it: in Schubert calculus multiplication is intersection, co-multiplication is the inverse image of $Grass(\infinity, \infinity) \times Grass(\infinity, \infinity) \to (\infinity, \infinity)$. Apparently, inverse image commutes with the intersection.
Jun 6, 2013 at 2:16 history answered Steven Sam CC BY-SA 3.0