From: ebuddenh@artsci.wustl.edu (Esther Buddenhagen)
Newsgroups: sci.math,geometry.forum
Subject: 4 lines skewered by 2 lines
Date: 2 Feb 1996 07:07:13 GMT
Given four "generic" lines in R^3, there are exactly two lines which pass
through all four. How does this generalise to higher dimensions?
--Jim Buddenhagen (ebuddenh@artsci.wustl.edu)
(posting from my wife's account)
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From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,geometry.forum
Subject: Re: 4 lines skewered by 2 lines
Date: 2 Feb 1996 20:32:05 GMT
In article <4esd71$q2u@newsreader.wustl.edu>,
piggybacking-poster Jim Buddenhagen (ebuddenh@artsci.wustl.edu) wrote:
>Given four "generic" lines in R^3, there are exactly two lines which pass
>through all four. How does this generalise to higher dimensions?
Interesting. I can think of several layers of generalization, all of
which I will answer in the charming language of "genericness" which
lets one avoid ever actually proving anything (and which, therefore,
may be wrong!). There are some meaty computational questions raised at
the end to which I hope to return in a later post.
Let's see, a generic line in R^3 is characterized by the
point p where it crosses the xy-plane and the unit vector v in it pointing
up (towards positive z), giving 4 degrees of freedom. Each line determines
a unique linear function f: R -> R^3 which parameterizes the line:
f(t) = p + t v. Two such lines intersect if there exist values t and t'
with f(t)=f'(t'), which sets 3 linear equations on the 2 unknowns and
so in general has a unique solution iff some single equation in
a, p, a', and p' is met. If a' and p' are given, this requires that
(a, p) lie in an algebraic set of codimension 1 in the family of all possible
(a, p) 's. Since this family has dimension 4, as noted above, we should
expect to be able to do this four times before we're down to a discrete set:
generically, any set of four lines has common intersection with only
finitely many different other lines. The original question claims this
finite number is 2.
In R^n we likewise see that the collection of lines has dimension 2n-2;
the condition that a line meets some other fixed line imposes n equations
on the set of variables made of all these parameters and two more (the
t and t') and so imposes n-2 equations on the set of (2n-2) variables.
For large n we see this is generically impossible to arrange as soon as
3 such intersections are required, although we would expect finitely many
lines to intersect any three lines in R^4 in general position. It is
not clear from this analysis what that finite number is.
In dimensions higher than four, you can only expect to start with 2
lines, but clearly given any two lines, the possible other lines which
meet both of these are parameterized by the positions on the two fixed
lines where the intersections are to occur, so that the collection of
all of them is precisely the 2-dimensional affine space, which is
consistent with the above dimension counts.
We could generalize a little differently by allowing the subspaces to
have a dimension k different from 1. The collection of k-dimensional
subspaces of R^n has dimension N=(k+1)(n-k). Fix two of them by
specifying the N parameters for each. A point on each subspace is
determined by specifying k coordinates, so the condition that the
subspaces meet is expressed with n equations in 2k unknowns, which
will not all be satisfied in general; indeed, we expect this to force
n-2k equations on the 2N parameters. So if we view just one of the
two subspaces as really fixed and the other as variable, then the
condition that the two meet forces n-2k equations to be satisfied
by the N varying parameters. We expect to be able to do this m
times as long as m(n-2k) <= N; if equality holds, we expect only a
finite number of possibilities.
So it looks to me that given any generic set of m subspaces of dimension k
in R^n, you can find another such subspace which intersects all m of
them iff m <= (k+1)(n-k)/(n-2k). The earlier paragraphs considered the
case k=1. (The case k=0 is trivial: two points don't intersect unless
equal.) For any k, it's clear that you can usually only find an
intersecting subspace if the number of starting subspaces is at most k+1
(any two lines can be joined, any three planes, etc.); but for each k
you can possibly join more subspaces if the ambient dimension n is small
(you can join an extra subspace as long as n <= k(k+3). The number of
extra subspaces you can join goes up as n goes down, until you reach
n=2k; for dimensions that low or lower you can join any number of subspaces
of dimension k; for example, you can find a line passing through any
countable collection of planes in R^3).
Finally, one could generalize by allowing the subspaces to have different
dimensions. This looks like a combinatorial nightmare, but one case is
perhaps worthwhile: to let all the fixed subspaces have the same dimension
k1 and to let the intersecting ones have a different dimension k2.
For example, this includes the question of the number of subspaces of
dimension k2 pass through any k2 distinct points (k1=0); of course
the answer there is one. It's easy enough to incorporate this variation in
the preceding analysis.
The more challenging question is to determine the number of possible
intersecting subspaces when that number is finite (but nonzero). That is,
for any k and any factorization k(k+1) = d1 d2, we look for the generic
number of subspaces of dimension k which meet each of k+1+d1 other
subspaces of dimension k inside R^(2k+d2).
k=1: How many lines pass through a generic set of 4 lines in R^3? (Two?)
How many lines pass through a generic set of 3 lines in R^4?
k=2: How many planes pass through a generic set of 9 planes in R^5?
How many planes pass through a generic set of 6 planes in R^6?
How many planes pass through a generic set of 5 planes in R^7?
How many planes pass through a generic set of 4 planes in R^10?
and so on. It is probably possible to answer any one of these with sufficient
calculations in algebraic geometry (I'll give it a go) but I don't
see right off the bat how to tackle all of them at once.
dave
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