From:ebuddenh@artsci.wustl.edu (Esther Buddenhagen)Newsgroups:sci.math,geometry.forumSubject:4 lines skewered by 2 linesDate:2 Feb 1996 07:07:13 GMTOrganization: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 (ebuddenh@artsci.wustl.edu) (posting from my wife's account)

From:rusin@washington.math.niu.edu (Dave Rusin)Newsgroups:sci.math,geometry.forumSubject:Re: 4 lines skewered by 2 linesDate:2 Feb 1996 20:32:05 GMTOrganization:NIU Mathematical Sciences

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

Date:Mon, 01 Apr 1996 10:38:10 PSTFrom:alltop@scfe.chinalake.navy.milTo:eppstein@ics.uci.eduSubject:four lines

April 1, 1996 eppstein@ics.uci.edu 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 619-939-3034 alltop@rattler.chinalake.navy.mil Copies to: ebudden@artsci.wustl.edu rusin@washington.math.niu.edu

To:alltop@scfe.chinalake.navy.milSubject:four linesDate:Mon, 01 Apr 1996 11:07:29 -0800From:David Eppstein <eppstein@ICS.UCI.EDU>

From D. Hilbert and S. Cohn-Vossen, "Geometry and the Imagination", p.164: "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 eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/