From:  (Esther Buddenhagen)
Newsgroups:     sci.math,
Subject:        4 lines skewered by 2 lines
Date:           2 Feb 1996 07:07:13 GMT
Organization:   Arts and Sciences -- Washington University, St. Louis, Missouri, USA

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 (
(posting from my wife's account)

From:  (Dave Rusin)
Newsgroups:     sci.math,
Subject:        Re: 4 lines skewered by 2 lines
Date:           2 Feb 1996 20:32:05 GMT
Organization:   NIU Mathematical Sciences

In article <4esd71$>,
piggybacking-poster Jim Buddenhagen ( 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.


Date:           Mon, 01 Apr 1996 10:38:10 PST
Subject:        four lines

April 1, 1996
David Eppstein,

Your Geometry Junkyard is very interesting and informative.
Below are remarks pertaining to the four lines, and their
possible generalizations to higher dimensions, attributed 
to Jim Buddenhagen.  

I'm assuming a finite set of lines in  3-space to be
in gp (general position) provided no two of them lie in
a plane.  

I don't think that four lines in gp in 3-space always admit
a common skewer, i.e. a line which intersects all four.

For example, consider the four lines  L(i)  represented
parametrically by the four real parameters  u(i):

       L(1) :  ( u(1),     1,   -1 )         

       L(2) :  (   -1,  u(2),    1 )         

       L(3) :  (    1,    -1, u(3) )         

       L(4) :  ( u(4),  u(4), u(4) ) .

The first three lines are parallel to the coordinate axes,
intercepting the coordinate planes in  ( 0, 1,-1), (-1, 0, 1)
and  ( 1,-1, 0), respectively.  The fourth passes through the 
origin with direction ( 1, 1, 1).  None of the six pairs lies
in a plane.

According to my calculations no line intersects all four.
Grinding through the algebra yields the requirement that

                  u(4)**2 = -1/3 .

I suspect that something like the following takes place.  
If we have three lines  L(i), i=1,2,3,  in gp in 3-space, the
union of all lines  K  which intersect all  L(i)  forms a 
(quadratic?) surface. Furthermore, these lines  K  are
disjoint and form a decomposition of the surface. 

A  fourth line  L(4)  may intersect that 
surface in  zero points, one tangent point, two points or
the line  L(4)  itself.  This would imply that four lines in 
gp in 3-space would admit zero, one, two  or an entire 
continuum of common skewers.

If this is roughly correct (whatever that means), generalizations
may not be easy to formulate.

If this is wrong, I plead 'April Fool'.

Bill Alltop  
Code 474400D
China Lake, CA 93555

Copies to:

Subject:        four lines
Date:           Mon, 01 Apr 1996 11:07:29 -0800
From:           David Eppstein <eppstein@ICS.UCI.EDU>

From D. Hilbert and S. Cohn-Vossen, "Geometry and the Imagination",
	"Three skew straight lines define a hyperboloid H.
	 In general, an arbitrary fourth straight line intersects H
	 at two points, althought it may also be tangent to H or lie on H".

I would consider being tangent to H or lying on H to be
not in general position.  Yet another special position occurs
when the three lines are parallel to a common plane; then you
get a hyperbolic paraboloid instead of a hyperboloid
and there are lines that cross it only once.

But you seem to be right: even in the general position there
are lines that miss the hyperboloid altogether.

The usual excuse for this sort of situation is that you should really be
working in complex space (i.e. a space where the three coordinates are
complex numbers rather than real numbers); then the statement really is
true that (in general position, so no coplanarities, no parallel planes,
and no tangencies to the hyperboloid) there are exactly two skewering lines.

So back in the original real-space, you can say that there still are
two skewering lines, you just can't see them because they're complex.
It's the same sort of thing as saying that every quadratic equation
has two solutions, even though the solutions for e.g. the equation
x^2=-1 are not on the real line.

Would you mind if I added your message to my WWW file?
David Eppstein		UC Irvine Dept. of Information & Computer Science