Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.