Consider the Grassmannian $\operatorname{Gr}(2,n)$. I want to know Hilbert series of $H_1 \cap H_2 \dots \cap H_m \cap \operatorname{Gr}(2,n)$ in the Plücker embedding of $\operatorname{Gr}(2,n)$, here $H_1, H_2, \dots ,H_m$ are very special hyperplanes, that is, each defined by vanishing of one Plücker coordinate.

  • $\begingroup$ The answer depends on he choice of $H_i$. For instance, for $n = 4$, $m = 3$, if you consider the Plucker equations $x_{12} = x_{13} = x_{14} = 0$, the corresponding linear section is $\mathbb{P}^2$, and if instead you consider $x_{12} = x_{23} = x_{34} = 0$, the corresponding linear section is a reducible conic. They have different Hilbert polynomials. $\endgroup$ – Sasha May 15 at 18:33
  • $\begingroup$ Thanks Sasha. I want to know if there are some results (with some conditions on coordinates, if needed) for Gr(2,n). Or any progress in this direction. $\endgroup$ – SAG May 15 at 19:14
  • $\begingroup$ if the linear section is dimensionally transverse (i.e., its dimension is equal to $2(n-2) - m$), then the Hilbert polynomial is equal to $\sum (-1)^i\binom{m}{i} h(t - i)$, where $h(t)$ is the Hilbert polynomial of $Gr(2,n)$. $\endgroup$ – Sasha May 15 at 19:54
  • $\begingroup$ Thanks Sasha. But I am interested in knowing the case when the hyperplanes are very special. $\endgroup$ – SAG May 15 at 20:12
  • $\begingroup$ Then the answer depends on how special they are. Please, try to be more precise. $\endgroup$ – Sasha May 16 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.