While working on a problem of differential topology I stumbled on a question of algebraic geometry that seems pretty basic, but that I'm unable to answer because I know very little algebraic geometry. I hope somebody could help:

It's known that if $X$ and $Y$ are (algebraic) subvarieties of $d$-dimensional (complex) projective space $P^d$ with complementary dimensions (i.e. $\dim X + \dim Y = d$) then $X \cap Y \neq \emptyset$ (in particular, any two curves in the projective plane $P^2$ always intersect).

My question is: Does this theorem hold if $P^d$ is replaced by the Grassmannian $G=G(m,n)$ (the set of $m$-dimensional subspaces of $\mathbb{C}^n$)?

In other words: If $X$, $Y$ are subvarieties of $G=G(m,n)$ whose dimensions sum to $\dim G(m,n) = m(n-m)$, does it follow that they have non-empty intersection?

In fact, in the situation I'm interested in, $Y$ is just a immersed smaller grassmannian $G(m-k,n-k)$. [More precisely, $Y$ is the set of $m$-dimensional subspaces of $\mathbb{C}^n$ that contain a given fixed $k$-dimensional subspace, where $k < m < n$.]

If the answer to the question is no (even for this particular $Y$), I pose a sharper question: What is the least $s=s(m,n)$ such that for any subvariety $X$ of $G$, if $\dim X + \dim Y \geq s$ (where $Y$ is as above) then $X \cap Y \neq \emptyset$?

Thanks for any help!


3 Answers 3


The answer to your first question is no.

In fact, consider the Grassmannian $G(2,4)$ of $2$-planes in $\mathbb{C}^4$, which has dimension $4$. It is isomorphic to the projective Grassmannian $\mathbb{G}(1,3)$ of lines in $\mathbb{P}^3$, and the Plücker embedding realizes it as a quadric hypersurface $X \subset \mathbb{P}^5$.

Now take a point $p$ and a plane $H$ in $\mathbb{P}^3$ such that $p \notin H$.

Let $\Sigma_p$ be the set of lines in $\mathbb{P}^3$ containing $p$ and let $\Sigma_H$ be the set of lines in $\mathbb{P}^3$ lying in $H$.

Then under the Plücker embedding $\Sigma_p$ and $\Sigma_H$ are carried to two linear subspaces of dimension $2$ in $X$ such that

$\Sigma_p \cap \Sigma_H= \emptyset$.

  • 5
    $\begingroup$ Gah, too slow writing the same counterexample. Anyway the basic point is that the Betti numbers of projective space are all 1, but $b_4$ of G(2,4) is 2, with $\Sigma_p,\Sigma_H$ giving a basis. $\endgroup$ Jan 14, 2011 at 14:32
  • $\begingroup$ Dear Allen, I think this is really a standard counterexample. I was just lucky to sit in front of my computer at the right moment :-) $\endgroup$ Jan 14, 2011 at 14:36
  • $\begingroup$ Ok. I think I can reread this as follows (applying back the isomorphism): Take a splitting $\C^4 = E \oplus F$ where $\dim E = 1$, $\dim F = 3$. Take $Y$ (resp. $X$) as the set of $2$-spaces of $\C^4$ that contain $E$ (resp. are contained in $F$). Then $X$ and $Y$ are disjoint and $\dim X + \dim Y = 2 + 2 = \dim G(2,4)$. Thank you very much. Any idea about the second question? $\endgroup$ Jan 14, 2011 at 14:46

As Franscesco's answer shows, there is a standard counterexample to your problem. However, if you put some mild restrictions on the subvarieties, the answer is yes.

One of Hartshorne's old conjectures asks if the following is true:

If $X$ and $Y$ are subvarieties of $Z$ with ample normal bundles $N_{X|Z}$, $N_{Y|Z}$ and $\dim X+\dim Y=\dim Z$, then $X$ and $Y$ have non-empty intersection.

In $\mathbb{P}^n$ the above conditions on the normal bundles are automatic (since they are a quotient of the tangent bundle of $\mathbb{P}^n$, which is ample), but in the general case, the ampleness of the normal bundles is a natural requirement. Hartshorne's conjecture has been proved by Lubke for any homogeneous variety, so in particular for any Grassmannian. For the proof, see

Martin Lübke, Beweis einer Vermutung von Hartshorne für den Fall homogener Mannigfaltigkeiten., Crelles Journal, 1980.

The general Hartshorne's conjecture is however still open, although there is more evidence supporting it. For a survey, see the recent paper by Peternell: Submanifolds with ample normal bundles and a conjecture of Hartshorne

  • 4
    $\begingroup$ Ah, this is interesting! In fact, by a result of Papantonopoulou, see jstor.org/stable/2042706, the normal bundle of a smooth subvariety $X$ in the Grassmannian $\mathbb{G}(1,3)$ is not ample if and only if $X$ is a Schubert cycle. $\endgroup$ Jan 14, 2011 at 14:43

Well, after a crash course on cohomology of the grassmannians, Schubert calculus, etc, I think I've found the answer to the second question: If the codimension of $X$ in $G_m(\mathbb{C}^n)$ is less than or equal to $m-k$ then $X$ should intersect the (Schubert cell) $Y$.

Luckily this should be sufficient for the application I have in mind.

Thanks for Francesco for pointing in the right direction.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.