Schubert class in the Grassmannian G(3,6)












1












$begingroup$


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$.






algebraic-geometry intersection-theory grassmannian schubert-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Jan-Magnus Økland
    Sep 15 '18 at 11:54










  • $begingroup$
    Thanks, but I want to work by hand, if it is not too complicated.
    $endgroup$
    – Strongart
    Sep 15 '18 at 14:05










  • $begingroup$
    These lecture notes perform this calculation (see figure 5).
    $endgroup$
    – Jan-Magnus Økland
    Sep 17 '18 at 6:06
















1












$begingroup$


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$.






algebraic-geometry intersection-theory grassmannian schubert-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Jan-Magnus Økland
    Sep 15 '18 at 11:54










  • $begingroup$
    Thanks, but I want to work by hand, if it is not too complicated.
    $endgroup$
    – Strongart
    Sep 15 '18 at 14:05










  • $begingroup$
    These lecture notes perform this calculation (see figure 5).
    $endgroup$
    – Jan-Magnus Økland
    Sep 17 '18 at 6:06














1












1








1


0



$begingroup$


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$.






algebraic-geometry intersection-theory grassmannian schubert-calculus







share|cite|improve this question











$endgroup$




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$.






algebraic-geometry intersection-theory grassmannian schubert-calculus




algebraic-geometry intersection-theory grassmannian schubert-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 13:17









Matt Samuel

38.5k63768




38.5k63768










asked Sep 15 '18 at 4:59









StrongartStrongart

1,8991430




1,8991430












  • $begingroup$
    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.
    $endgroup$
    – Jan-Magnus Økland
    Sep 15 '18 at 11:54










  • $begingroup$
    Thanks, but I want to work by hand, if it is not too complicated.
    $endgroup$
    – Strongart
    Sep 15 '18 at 14:05










  • $begingroup$
    These lecture notes perform this calculation (see figure 5).
    $endgroup$
    – Jan-Magnus Økland
    Sep 17 '18 at 6:06


















  • $begingroup$
    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.
    $endgroup$
    – Jan-Magnus Økland
    Sep 15 '18 at 11:54










  • $begingroup$
    Thanks, but I want to work by hand, if it is not too complicated.
    $endgroup$
    – Strongart
    Sep 15 '18 at 14:05










  • $begingroup$
    These lecture notes perform this calculation (see figure 5).
    $endgroup$
    – Jan-Magnus Økland
    Sep 17 '18 at 6:06
















$begingroup$
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.
$endgroup$
– Jan-Magnus Økland
Sep 15 '18 at 11:54




$begingroup$
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.
$endgroup$
– Jan-Magnus Økland
Sep 15 '18 at 11:54












$begingroup$
Thanks, but I want to work by hand, if it is not too complicated.
$endgroup$
– Strongart
Sep 15 '18 at 14:05




$begingroup$
Thanks, but I want to work by hand, if it is not too complicated.
$endgroup$
– Strongart
Sep 15 '18 at 14:05












$begingroup$
These lecture notes perform this calculation (see figure 5).
$endgroup$
– Jan-Magnus Økland
Sep 17 '18 at 6:06




$begingroup$
These lecture notes perform this calculation (see figure 5).
$endgroup$
– Jan-Magnus Økland
Sep 17 '18 at 6:06










1 Answer
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$begingroup$

Note that $sigma_{2,1} = sigma_2cdotsigma_1 - sigma_3$, hence
$$
sigma_{2,1}^2 = sigma_{2,1}cdotsigma_2cdotsigma_1 - sigma_{2,1}cdotsigma_3.
$$
By Pieri rule
$$
sigma_{2,1}cdotsigma_2cdotsigma_1 = (sigma_{3,2} + sigma_{3,1,1} + sigma_{2,2,1})cdotsigma_1 = sigma_{3,3} + 3sigma_{3,2,1}+sigma_{2,2,2},
$$
while
$$
sigma_{2,1}cdotsigma_3 = sigma_{3,2,1}.
$$
Subtracting, you get the result.



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






share|cite|improve this answer









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    $begingroup$

    Note that $sigma_{2,1} = sigma_2cdotsigma_1 - sigma_3$, hence
    $$
    sigma_{2,1}^2 = sigma_{2,1}cdotsigma_2cdotsigma_1 - sigma_{2,1}cdotsigma_3.
    $$
    By Pieri rule
    $$
    sigma_{2,1}cdotsigma_2cdotsigma_1 = (sigma_{3,2} + sigma_{3,1,1} + sigma_{2,2,1})cdotsigma_1 = sigma_{3,3} + 3sigma_{3,2,1}+sigma_{2,2,2},
    $$
    while
    $$
    sigma_{2,1}cdotsigma_3 = sigma_{3,2,1}.
    $$
    Subtracting, you get the result.



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






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Note that $sigma_{2,1} = sigma_2cdotsigma_1 - sigma_3$, hence
      $$
      sigma_{2,1}^2 = sigma_{2,1}cdotsigma_2cdotsigma_1 - sigma_{2,1}cdotsigma_3.
      $$
      By Pieri rule
      $$
      sigma_{2,1}cdotsigma_2cdotsigma_1 = (sigma_{3,2} + sigma_{3,1,1} + sigma_{2,2,1})cdotsigma_1 = sigma_{3,3} + 3sigma_{3,2,1}+sigma_{2,2,2},
      $$
      while
      $$
      sigma_{2,1}cdotsigma_3 = sigma_{3,2,1}.
      $$
      Subtracting, you get the result.



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






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Note that $sigma_{2,1} = sigma_2cdotsigma_1 - sigma_3$, hence
        $$
        sigma_{2,1}^2 = sigma_{2,1}cdotsigma_2cdotsigma_1 - sigma_{2,1}cdotsigma_3.
        $$
        By Pieri rule
        $$
        sigma_{2,1}cdotsigma_2cdotsigma_1 = (sigma_{3,2} + sigma_{3,1,1} + sigma_{2,2,1})cdotsigma_1 = sigma_{3,3} + 3sigma_{3,2,1}+sigma_{2,2,2},
        $$
        while
        $$
        sigma_{2,1}cdotsigma_3 = sigma_{3,2,1}.
        $$
        Subtracting, you get the result.



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






        share|cite|improve this answer









        $endgroup$



        Note that $sigma_{2,1} = sigma_2cdotsigma_1 - sigma_3$, hence
        $$
        sigma_{2,1}^2 = sigma_{2,1}cdotsigma_2cdotsigma_1 - sigma_{2,1}cdotsigma_3.
        $$
        By Pieri rule
        $$
        sigma_{2,1}cdotsigma_2cdotsigma_1 = (sigma_{3,2} + sigma_{3,1,1} + sigma_{2,2,1})cdotsigma_1 = sigma_{3,3} + 3sigma_{3,2,1}+sigma_{2,2,2},
        $$
        while
        $$
        sigma_{2,1}cdotsigma_3 = sigma_{3,2,1}.
        $$
        Subtracting, you get the result.



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 17 '18 at 6:54









        SashaSasha

        5,053139




        5,053139






























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