0
$\begingroup$

It is easy to see that in $3$-dimensional Euclidean space, given $4$ lines in general position, there exists precisely one line who intersects with each of the $4$ lines. We call the $4$ lines determined this line.

I want to know the analogue of this statement in higher dimension, i.e. for given positive integers $N=n+m+1$, how many $n$-planes in general position will determine one $m$-plane in $N$-dimensional total space? If we denote the number by $d(n,m)$, for example the first paragraph just says $d(1,1)=4$. Then is there a formula for $d(n,m)$?


Edit

I just realized that we may not get one $m$-plane in general. It should be finite instead.


Edit'

Now I found the finite intersection may also not happen. The question now we should ask is, for which $n,m$ we will have the finite interstion case?

My approach is to compute the dimension of $X(V^n)\subset \mathbb G(m,N)$ where $X(V^n)$ consists of all $m$-plane in $\mathbb P^N$ which intersects with the $n$-plane $V^n$. But I don't know how to do it.

$\endgroup$
  • $\begingroup$ Interesting question! What are your thoughts on the problem? What did you try and where did you get stuck? $\endgroup$ – Servaes Nov 3 '18 at 17:59
  • $\begingroup$ @Servaes I think one way is to consider the subset $X(V^n)\subset \mathbb G(m,N)$ which intersects with the $n$-plane $V^n$ and compute its dimension. And I don't know how to compute it. $\endgroup$ – User X Nov 3 '18 at 18:18
  • 1
    $\begingroup$ "It is easy to see that in 3-dimensional Euclidean space, given 4 lines in general position, there exists precisely one line who intersects with each of the 4 lines. We call the 4 lines determined this line." That is not true --- in fact, 4 lines determine exactly 2 lines. Update: OK, I see you mentioned this in the edit. $\endgroup$ – Art Nov 3 '18 at 19:41
  • $\begingroup$ @Art Yes you are right, I have also realized this. See the first "Edit" below. $\endgroup$ – User X Nov 3 '18 at 19:43
  • $\begingroup$ Anyway, under the assumption that $N=n+m+1$, the dimension of the set you call $X(V^n)$ is always just one less than the dimension of $\mathbf G(m,N)$. So the number of $n$-planes you need is exactly the dimension of $\mathbf G(m,N)$, namely $(m+1)(N-m)=(m+1)(n+1)$. $\endgroup$ – Art Nov 3 '18 at 19:45
1
$\begingroup$

The unresolved question in the comments is why intersection with a linear space of dimension one less than complementary defines a codimension-1 subset of the Grassmannian.

So suppose we have linear spaces of dimensions $m$ and $n$ in a space of dimension $m+n+1$. Passing to the associated vector spaces, these correspond to vector subspaces $U_{m+1}$ and $V_{n+1}$ in a space $W$ of dimension $m+n+2$.

Now the condition that the linear spaces intersect in $\mathbf P^{m+n+1}$ translates to the condition that $U$ and $V$ have nonzero intersection in $W$ --- in other words, that they fail to span $W$.

We can change basis on $W$ so that $U$ is simply the subspace spanned by the first $m+1$ basis vectors. Choosing a basis for $V$ and forming the matrix whose first $m+1$ columns are the basis vectors of $U$ and whose last $n+1$ columns are the basis vectors of $W$ we get something of the form

$$ \begin{pmatrix} I & A \\ 0 & B \end{pmatrix} $$ where the diagonal blocks have sizes $(m+1) \times (m+1)$ and $(n+1) \times (n+1)$.

The fact that $U$ and $V$ together fail to span $W$ is equivalent to the condition that this matrix has determinant 0. On the other hand, due to the block form the determinant is evidently equal to $\operatorname{det}(B)$, so again our condition is equivalent to $$\operatorname{det}(B)=0.$$

But the Plücker coordinates of $\mathbf P(W)$ are precisely the $(n+1)\times(n+1)$ minors of the matrix $$ \begin{pmatrix} A \\ B \end{pmatrix} $$ and $\operatorname{det}(B)$ is one of those minors. So the condition of intersecting $\mathbf P(V)$ nontrivially is precisely the vanishing of one Plücker coordinate, and in particular the corresponding subset of $\mathbf G$ has codimension 1.


Let me finish by saying that this is the tip of the iceberg of the beatiful and important topic of Schubert calculus. You can read about this in Principles of Algebraic Geometry by Griffiths and Harris, or for a friendlier exposition the new book 3264 And All That by Eisenbud and Harris.

$\endgroup$
  • $\begingroup$ Thank you! This is very helpful. $\endgroup$ – User X Nov 10 '18 at 9:40

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.