Intersecting Subspaces in N dimensions

planck2

Hey guys,

I have a little problem here:

given two subspaces U and W both of dimension two of an N dimensional space show in general that if N = 3 the intersection of U and W forms a curve; if N = 4 a finite number of points; and N > 4 they do not in general intersect at all.

I can kind of visualize the answer for the first two cases based on analagous cases lines viewed in R2 and R3, but I am not really satisfied with the answers.

I would like a really rigorous maths proof.

please help

charlieroot

I may be wrong on this but, suppose the N=3 and U = V. Then the intersection would be 2 dimensional and hence a plane?

If N = 4, then if you have one point in the intersection surely you have infinitely many points - you couldn't have a finite number of points?

If N > 4, if U = V then of course they'd intersect?

We are talking about vector spaces with external multiplication by Q right?
I'm guessing we're not? Are we talking about finitely generated modules?

planck2

Just before I answer, what do you mean by external multiplication Q ? Do you mean external multiplication by the set of rational numbers?

I don't know what finitely generated modules means since I haven't done maths since the end of my second year in university? Any way I don't think they have any relevance since this question appears in a book on tensor calculus by J.L. Synge and A. Schild published in the 1950's, old, but definitely not out of date.

Now back to the question. As you know a space of N dimensions requires N co-ordinates to determine the position of a point in that space. A subspace of this N dimensional, lets call it B ( the N dimensional space being called A),i is formed when the N co-ordinates of the space A can be written as a function of parameters, the number of these parameters can be anything from 1,....,N-1. If the number of the parameters is N-1 then a hypersurface is formed in A, i.e. B forms a hypersurface in A.

Now if I have a space A of N dimensions, and two subspaces B and C both of dimension 2 we know that the co-ordinates of A can be written in the following way:

Ai = f i(B1,B2) and Ai = gi ( C1, C2).

Now if B and C intersect then we can only have f i(B1,B2) = gi ( C1, C2), with Bj = Ck.
Note that j doesn’t not necessarily equal k, and at the same time we cannot have Bk= Cj since if we did B would equal C. If N = 3, i.e. the dimension of A is 3, then the B ∩ C simply gives the parameterized equation of a curve. However for me the problem lies in the extension of the proof to higher dimensions for the same line of reasoning gives the same result in spaces of dimension N ≥ 3.

mrhappy42

curve = straight line
"finite number of points" = single point

would be interested in your proof.