In the book Discriminants, Resultants, and Multidimensional Determinants of Andrei Zelevinsky,M.M. Kapranov and Izrail' Moiseevič Gel'fand, the authors give the following definition of degree of a hypersurface in a Grassmannian.

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As they say, in generale a hypersurfaces in a projective variety is not given by the vanishing of a polynomial in its coordinate ring, but for Grassmannians this is true, since its coordinate ring is a UFD, therefore every height-one prime is principal by Krull Theorem.

However I'm stuck on the definition of degree of a hypersurface in a Grassmannian. To be more precise...I wuold prove that this definition is well posed, as in the case of projective hypersurfaces:

The set of pencils $P_{NM}$ is parametrized by the flag variety $\mathcal{Fl}(n;k-1,k+1):=\lbrace N\subset M \: : \: \dim N=k-1, \dim M=k+1 \rbrace$. I would check that there exists a nonempty open subset $U$ of $\mathcal{Fl}(n;k-1,k+1)$ such that the intersection number of $Z$ with any pencil $P_{NM}$ in $U$ is equal to the maximal number of intersection points of $Z$ with a pencil $P\not\subset Z$.

Any help?

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    $\begingroup$ This book has three authors. I think it's unfair to M.M. Kapranov to omit his name. $\endgroup$ Apr 18, 2018 at 12:55
  • $\begingroup$ Oh, You are completely right. $\endgroup$ Apr 18, 2018 at 20:09

2 Answers 2


Consider the locus in $Z \times \mathcal{Fl}(n;k-1,k+1) $ consisting of pairs of a point of $Z$ and a flag whose associated line intersects this point non-transversely. It suffices to show that the dimension of this is less than $\dim \mathcal{Fl}(n;k-1,k+1)$ as then we can take any flag not in the image of the projection.

To do this, it suffices to show that the fiber over each smooth point of $Z$ has dimension at most $ \dim \mathcal{Fl}(n;k-1,k+1) - \dim Z -1$ and the fiber over each singular point of $Z$ has dimension at most $ \dim \mathcal{Fl}(n;k-1,k+1) - \dim Z$.

The second is a simple dimension count - the fiber has dimension $(k-1) + (n-k-1)$ because it parameterizes pairs of a $k-1$-dimensional subspace and a $k+1$-dimensional superspace of a fixed $k$-dimensional vector sapce, $\dim Z$ is $k(n-k)-1$, and the flag variety has dimension $k (n-k) +(k-1) + (n-k-1) -1$.

For the first, we need only observe that the subspace of the fiber whose tangent vector is contained in the tangent space of the hypersurface has codimmension $1$, because the tangent space consists of $k \times (n-k)$ matrices, the possible tangent vectors are all the matrices of rank one, and these span the space.

In characteristic zero, a more direct argument is possible - the locus in $Z \times \mathcal{Fl}(n;k-1,k+1)$ consisting of pairs of a point $Z$ and a flag whose associated line intersects this point is smooth over $Z$, hence generically smooth, and thus its projection map to $ \mathcal{Fl}(n;k-1,k+1)$ is generically etale.


I am a novice, but just to start you off, I would ask myself what happens to the number of intersections when you deform slightly such a pencil? It seems if there are any tangential intersections or intersections through a singularity, that the number should go up. Thus if we start from a pencil with a maximal number of intersections, it seems that these are all transversal, and hence the number should remain constant under small deformation. Since all such pencils form a single "family", the generic one should have the maximal number of intersections.

I would also try to gain insight from an example, such as the grassmann of lines in P^3. In general there is a standard (Plucker) embedding of the grassmann into projective space, and a hypersurface of degree one in the grassmannian should be one that is cut by a hyperplane in that embedding. Maybe you can use your ufd arguments to argue that every hypersurface is cut by some projective hypersurface for that embedding. It would seem to follow at least that every such pencil does meet every hypersurface.

But I would start thinking about lines in P^3 and note that there it seems the only obvious hypersurface defined by a Schubert condition, is the lines that meet a given line. Then it seems this set has indeed degree one by your definition, since fixing a general point and general plane through that point, exactly one line in the given plane and through the given point, also meets the given line.

Generalizing from this example should give a Schubert cycle - hypersurface in a general grassmannian which does meet your pencil in exactly one point. This suggests that your pencil is the right curve to measure degree. After seeing that every hypersurfce must meet your pencils, I would see if I can show that some such pencil is not contained in a given hypersurface. In the example case, take any line not representing a point in the given hypersurface and just choose a point on it and a plane containing it, to define a pencil not lying wholly in the hypersurface. Thus if every pencil meets a given hypersurface, and some pencil meets it a finite number of times, and the family of all pencils is irreducible, then it seems a generic pencil meets it the maximal number of times.

  • $\begingroup$ Nice explanation! :) The surprizing fact is that the authors give this definition without any explanation or such. I opened this topic just to check that this definition of degree is consistent. $\endgroup$ Apr 20, 2018 at 2:45

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