# Schubert class in the Grassmannian G(3,6)

How to compute the Schubert class $\sigma$$^2$$_2$$_1 in the Grassmannian G(3,6)? I remember the result is \sigma$$_3$$_3 + 2\sigma$$_3$$_2$$_1$ + $\sigma$$_2$$_2$$_2. • In maple with(schubert) grass(3, 6, c) Gc[relations_] yields the ideal I use in M2 below. sigma33 := schur([3, 3], Qc) sigma321 := schur([3, 2, 1], Qc) sigma222 := schur([2, 2, 2], Qc) sigma33+2*sigma321+sigma222 yields c2^3. sigma21 := schur([2, 1], Qc) and sigma21^2 yields (c1*c2-c3)^2. Now in M2 R=ZZ[c_1,c_2,c_3,Degrees=>{1,2,3}] I=ideal(2*c_1*c_3+c_2^2-3*c_2*c_1^2+c_1^4, 2*c_3*c_2-3*c_3*c_1^2-3*c_1*c_2^2+4*c_2*c_1^3-c_1^5, c_3^2-6*c_3*c_1*c_2+4*c_3*c_1^3-c_2^3+6*c_2^2*c_1^2-5*c_2*c_1^4+c_1^6) (c_2^3-(c_1*c_2-c_3)^2)%I yields 0: at least your memory checks out. – Jan-Magnus Økland Sep 15 '18 at 11:54 • Thanks, but I want to work by hand, if it is not too complicated. – Strongart Sep 15 '18 at 14:05 • These lecture notes perform this calculation (see figure 5). – Jan-Magnus Økland Sep 17 '18 at 6:06 ## 1 Answer Note that \sigma_{2,1} = \sigma_2\cdot\sigma_1 - \sigma_3, hence$$ \sigma_{2,1}^2 = \sigma_{2,1}\cdot\sigma_2\cdot\sigma_1 - \sigma_{2,1}\cdot\sigma_3. $$By Pieri rule$$ \sigma_{2,1}\cdot\sigma_2\cdot\sigma_1 = (\sigma_{3,2} + \sigma_{3,1,1} + \sigma_{2,2,1})\cdot\sigma_1 = \sigma_{3,3} + 3\sigma_{3,2,1}+\sigma_{2,2,2}, $$while$$ \sigma_{2,1}\cdot\sigma_3 = \sigma_{3,2,1}.$\$ Subtracting, you get the result.

Alternatively, one can directly use the Littlewood-Richardson rule.