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The parallels between the formulas in Schubert calculus and in the theory of the representations of symmetric groups (par Geissinger-Zelevinsky) are so apparent (e.g. Giambelli formula), that one must wonder how to directly define the co-multiplication on Schubert (co)cells.

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up vote 1 down vote accepted

If you're talking about Grassmannians, then you can use the direct sum maps

$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$

to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.

According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.

Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).

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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. – George Jun 6 '13 at 2:27
Another attempt to get a nice formula: $Grass(\infty,\infty) \times Grass(\infty,\infty) \to Grass(\infty, \infty)$ – George Jun 6 '13 at 2:32

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