14
$\begingroup$

Consider the two types of Grassmannians Gr(2,7) and Gr(3,6) having their plucker embeddings in $\mathbb P^{20}$ and $\mathbb P^{19}$ respectivley. The first one is 10-dimensional and latter is 9-dimensional, so each having codimension 10. We can easily compute their defining equations and both of them are defined by 35 equations.

By using computer algebra system(such as Macaulay 2 or Magma) we can also compute their Hilbert series and so the Hilbert numerator. If we do the computation then both varieties have basically same numerator, which suggests that both idels have the same(kind of) free resolution.

This raises the question whether Gr(3,6) is a linear section of Gr(2,7) or not.?

Improve this question
$\endgroup$
8
  • $\begingroup$ Can you give more details on these Hilbert series? $\endgroup$
    – Sasha
    Commented Nov 3, 2011 at 16:55
  • $\begingroup$ @Sasha.. Sorry, I Didn't get what exactly may I let you know about them? $\endgroup$
    – M I
    Commented Nov 3, 2011 at 17:43
  • 2
    $\begingroup$ I know what I want the answer to be. Let $\Delta$ be the Schubert divisor in $G(2,7)$ -- this is the set of $2$-planes in $7$-space which meet a fixed $5$-plane nontransversely. This is a hyperplane section of $G(2,7)$. I want there to be a flat degeneration from $G(3,6)$ to $\Delta$. No idea about how to find it though... $\endgroup$
    – David E Speyer
    Commented Nov 3, 2011 at 18:31
  • $\begingroup$ @M I - I mean, what is the precise relation between the numerators and denominators of Hilbert series? $\endgroup$
    – Sasha
    Commented Nov 4, 2011 at 9:21
  • $\begingroup$ @Sasha..In fact the Hilbert numerator is $1-35t^2+140t^3-189t^4-112t^5+735t^6-1080t^7+735t^8-112t^9-189t^{10}+140t^{11}-35t^{12}+t^{14},$ for both varieites. The denomenator is obviously $(1-t)^{21}$ and $(1-t)^{20}$. $\endgroup$
    – M I
    Commented Nov 4, 2011 at 23:12

3 Answers 3

Reset to default
11
$\begingroup$

I don't know whether the ideals have the same kind of free resolutions, but $Gr(3,6)$ is definitely not a hyperplane section of $Gr(2,7)$. Otherwise, their $H^{\leq 8}$ would be the same by the Lefschetz hyperplane theorem. However, $H^6(Gr(3,6))$ is 3-dimensional and is spanned by $c_1^3, c_1c_2$ and $c_3$, and $H^6(Gr(2,7))$ is 2-dimensional: it is spanned by $c_1^3$ and $c_1c_2$. Here $c_i$'s are the Chern classes of the respective tautological bundles.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Can you please give me a reference for this computation? $\endgroup$
    – M I
    Commented Nov 3, 2011 at 17:44
  • 1
    $\begingroup$ M I -- the cohomology of the Grassmannians is described in many places. in particular, the rank of $H^i(Gr(j,k))$ is the number of partitions of $i$ into $\leq j$ positive integers $\leq k-j$ (see e.g. Milnor-Stasheff, Characteristic classes, corollary 6.7 for the real case; the complex case is similar. Another method: use the fact that the cohomology is generated as an algebra by the Chern classes of the universal bundle, with one relation: the total Chern class of the universal bundle times the total Chern class of the universal quotient bundle is 1 (c.f. Milnor-Stasheff, problem 7B). $\endgroup$
    – algori
    Commented Nov 3, 2011 at 20:00
9
$\begingroup$

There is a theorem of Fujita saying that Grassmannian is never a hyperplane section of a smooth variety unless it is $P^n$ or $Gr(2,4)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think the theorem is about the Grassmannian corresponding to the special or general linear group. This is not true if you take the isotropic Lagrangiann Grassmannians LGr(2,6), which is a linear section of Gr(2,6). $\endgroup$
    – M I
    Commented Nov 3, 2011 at 17:28
  • 3
    $\begingroup$ This theorem is about usual Grassmannians --- $Gr(k,n)$ is a hyperplane section of a smooth variety if and only if $k = 1$, or $k = n-1$, or $(k,n) = (2,4)$. $\endgroup$
    – Sasha
    Commented Nov 3, 2011 at 19:07
1
$\begingroup$

EDIT: The following idea doesn't work, for kind of obvious reasons (see the comments).

As other answers have pointed out, the answer to the original question is no. However, taking up the idea from David Speyer's comment, both $Gr(3,6)$ and the Schubert divisor $\Delta$ in $Gr(2,7)$ are linear sections of $Gr(3,8)$, by linear subspaces of the same dimension. Specifically, as Schubert varieties inside $Gr(3,8)$, $$Gr(3,6) = \Omega_{(2,2,2)}$$ and $$\Delta = \Omega_{(5,1)}.$$ (I'm using notation where $\Omega_\lambda$ has codimension $|\lambda|$, for $\lambda$ a partition inside the $3 \times (8-3)$ rectangle.) One checks that these two Schubert varieties are both defined by the vanishing of 36 Plücker coordinates (on $Gr(3,8)$).

Taking any curve in the Grassmannian of codimension-36 subspaces inside ${\Bbb P}^N$, where $N = \binom{8}{3}-1$, connecting the two linear spaces cutting out $Gr(3,6)$ and $\Delta$, you should get a flat family having these two as fibers, explaining why they have the same Hilbert polynomial.

Improve this answer
$\endgroup$
7
  • 3
    $\begingroup$ This is a nice idea, but it doesn't work. First I'll explain what's wrong with the argument, then how to know it can't work. $G(3,8)$ only has dimension $15$, so a generic linear space of codimension $36$ misses $G(3,6)$ entirely. It is not clear that you can move from one of the codim $36$ spaces to the other without passing through some intermediate spaces that miss $G(3,8)$, in which case your family isn't flat. And, in fact, you can't. $G(3,6)$ and $G(2,7)$ have different cohomology classes in $G(3,8)$ so they can't be linked by a flat family within $G(3,8)$. $\endgroup$
    – David E Speyer
    Commented Nov 3, 2011 at 20:37
  • $\begingroup$ Whoops! Thanks for the correction... after posting, I got a feeling something was off... $\endgroup$
    – Dave Anderson
    Commented Nov 3, 2011 at 20:42
  • $\begingroup$ @Dave & David..Is there any other possible reason for them to have same Hilbert Series? $\endgroup$
    – M I
    Commented Nov 4, 2011 at 8:40
  • $\begingroup$ @ M I, in some sense, no, this is the only possible reason. If two subvarieties of ${\Bbb P}^n$ have the same Hilbert polynomial, then they are related by a flat family (by connectivity of the Hilbert scheme). $\endgroup$
    – Dave Anderson
    Commented Nov 4, 2011 at 17:30
  • 2
    $\begingroup$ @Dave & M I, the connectivity of the Hilbert scheme means that there is a flat family over a connected, put possibly reducible base connecting the two varieties. To connect them by a single curve, and for that curve to be rational is a stronger statement and is not true for arbitrary pairs of varieties with the same Hilbert series. $\endgroup$
    – Dustin Cartwright
    Commented Nov 4, 2011 at 19:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .