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Some classical line geometry

  1. Feb 10, 2005 #1

    mathwonk

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    some classical "line geometry"

    Schubert, over 100 years ago, considered questions like this: given several lines Li in space, how many lines M meet all of them?

    For instance if there is only one line L to begin with, we can pass lots of lines M through every point of L. Indeed if we choose a plane P off somewhere, then each point of L and each point of P determine a line meeting L, so there are at least a three dimensional infinity of them, one dimensions for the choice of a point of L, and 2 more for the choice of a point of P.

    If we choose two lines L1 and L2. then for each point of L1 and each point of L2 there is a unique line through both points, hence a doubly infinite family of lines meeting both.

    If we choose three lines L1, L2, L3, then if we fix a point x on L1, and conside all lines through x and also thropugh some point of L2, we get a one dimensional family of lines sweeping out a plane, which polane should meet L3 somewhere, so we get one line throughx meeting both L2 and L3. Since we can do this for each point x on L1, we get a one dimensional infinity of lines meeting all 3 lines.


    Now what if we have 4 lines L1, L2, L3, L4? There will presumably be only a finite number of lines meeting all 4? But how many?

    This is sort of the first problem in "enumerative geometry".
     
  2. jcsd
  3. Feb 10, 2005 #2

    StatusX

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    This might not be the most elegant way to do it, but it should work. Starting in the single line case, given a line in the form:

    [tex] \vec p_1 + x_1 \vec m_1[/tex]

    Where x1 is free to vary, you need to find all lines of the form:

    [tex] \vec p + t \vec m[/tex]

    That intersect that line at some point. That means there must be a (t,x1) such that the two expressions above are equal. In 3D, there are 6 free variables, the components of m and p. The first two components of m and p can be picked arbitrarily and determine a unique(t,x1). The last component gives a constraining equation. Also, since multiples of m and translates of p along m correspond to the same line, we should constrain p to lie in some plane and m to have norm 1, which adds two more constraining equations. This is 3 equations for 6 unknowns, leaving 3 degrees of freedom, as you already found.

    For another line, there will be another constraining equation from the last component (there will be another vector equation with a t2), leaving 2 degrees of freedom. Three lines leaves 1, and 4 lines should give a unique solution, that is, there's a single line that passes through each of the other 4 at some point. Of course, all of this assumes the lines are general, and there are no degeneracies (ie, non-invertibile matrices).
     
    Last edited: Feb 10, 2005
  4. Feb 10, 2005 #3

    mathwonk

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    good beginning,

    unfortunately the equations are not linear, so altho0ugh there are zero degrees of freedom, there is more than one solution. i.e that argument is so crude it only gives the dimension of the solution set, i.e. a zero dimensional solution set is finite, but thats all you can say.

    if it were that easy, no one would have posed it.
     
  5. Feb 10, 2005 #4

    StatusX

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    How are the equations for intersections of lines non linear? Are you talking about non-euclidean space?
     
  6. Feb 10, 2005 #5

    mathwonk

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    you are mistaking the equations for the intersections of lines, with the equations for the lines that intersect.

    i.e. you need a space of lines. then in that space of lines you need to solve simultaneously 4 equations, each equation describing the lines that inetersect a given one of the L's.
     
  7. Feb 10, 2005 #6

    Hurkyl

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    Bah, the problem with studying high-powered stuff is that you forget to try the easy approach! Here I was thinking about algebraic varieties and Gröbner bases, and it's a simple vector geometry problem!
     
  8. Feb 10, 2005 #7

    mathwonk

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    if it is simple, what is your simple solution? or even your simple approach?

    look, to parametrize lines is not so simple. suppsoe you take apair of coordinate planes intersecting along a coordinate axis. then most lines meet each plane once. so you can surmise that the family of lines has a big open set corresponding to pairs of points, one on each plane. that puts a big copy of euclidean 4 space in the family of all lines, but what is the condition in that space, for a pair of points (x,y;u,v) to represent a line which meets another given line?
     
  9. Feb 10, 2005 #8

    Hurkyl

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    There are either none, one, or infinitely many solutions.

    We have an example of each, now just to prove these are the only cases:

    Suppose we have more than one solution. Denote this solution as the n-tuple of collinear points (one per line).

    So, one solution is:

    P = (P1, P2, ..., Pn)

    and another is

    Q = (Q1, Q2, ..., Qn)

    From these, generate a family of solutions:

    t P + (1 - t) Q

    Each coordinate (t Pi + (1 - t) Qi) clearly lies on the i-th line.

    The only thing to check is that these points are still collinear. Hrm, I had convinced myself that was obvious, but I'm not seeing the proof... in fact I'm seeing a counterexample now, so nevermind. :blushing:
     
    Last edited: Feb 10, 2005
  10. Feb 10, 2005 #9

    mathwonk

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    if you would like a hint, the correct answer is 2, for four lines in general position. Of course properly considered, the lines should be in projective space.

    Also there is in fact a sense in which the equations governing the situation can be considered linear. But then you may ask why the answer is two?
     
    Last edited: Feb 10, 2005
  11. Feb 11, 2005 #10

    mathwonk

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    here is another useful piece of data from projective geometry: three lines in general position in P^3, all lie on some unique quadric surface.
     
  12. Feb 11, 2005 #11

    mathwonk

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    if you want to use algebraic varieties to consider the problem, the right variety to look at, is the grassman variety G parametrizing all lines in P^3, and its plucker embedding in P^5.
     
  13. Feb 11, 2005 #12

    mathwonk

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    heres another nice one, easier and more classical: Given three general circles in the plane, how many other circles can be drawn tangent to all three?
     
  14. Feb 11, 2005 #13

    Hurkyl

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    (I haven't looked at your hints yet)

    This one bugs me, since when I model it, I have 14 variables, 8 linear equations, and 6 quadratic equations... so why don't I have 64 solutions? :frown: I wonder if there will be 64 solutions over the complexes?
     
  15. Feb 11, 2005 #14

    mathwonk

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    no there will not.
     
  16. Feb 12, 2005 #15

    mathwonk

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    here is another little puzzler. in three space a degree one surface, i.e. a plane, contains a 2 parameter family of lines. a quadric surface contains a one parameter family of lines, indeed two of them.

    it is thus plausible that a cubic surface usually contains a zero parameter family of lines, i.e. a finite number of lines; how many?

    (i.e. for a line to lie on a cubic surface is 4 conditions, since 4 points of the line must belong to the surface to force the whole line in there. but as we have seen above, the family of all lines in space has dimension 4, being parametrized generically by pairs of points on two planes, so since it is 4 conditions to lie on a given cubic surface and there is a 4 dimensional family of lines, a finite number should lie on the surface.)
     
    Last edited: Feb 12, 2005
  17. Feb 13, 2005 #16

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    Forget the work I posted before, I have a proof now. It gives two solutions, which is what you said, but please tell me if there's a flaw here. (besides the end, which is a little incomplete)

    Identify the first line with the x-axis. The other three lines have the equations yi(x) = ai x + bi and zi(x) = ci x + di for i =1,2,3. Consider the plane given by z=m y. Each of the lines will intersect this plane at one point pi(m), with coordinates:

    [tex] z_i = m y_i [/tex]

    [tex] c_i x_i + d_i = m a_i x_i + m b_i [/tex]

    [tex] x_i = -\frac{m b_i -d_i}{m a_i -c_i} [/tex]

    And yi and zi given by yi(xi) and zi(xi). Now, any line in this plane will intersect the first line (x-axis), so to find a line that intersects all four lines, we need to find an m such that the three pi(m) are collinear. This will be true if the following identity holds:

    [tex] \frac{y_2-y_1}{x_2-x_1} = \frac{y_3-y_1}{x_3-x_1} [/tex]

    or:

    [tex] (y_2-y_1)(x_3-x_1) = (y_2-y_1)(x_3-x_1) [/tex]

    Where the x's and y's are the components of pi. The corresponding equation for z just differs by a factor of m, so it will follow automatically. First we need the explicit formula for y:

    [tex]y_i (x_i) = a_i (-\frac{m b_i -d_i}{m a_i -c_i}) + b_i [/tex]

    [tex] y_i = \frac{- m a_i b_i + a_i d_i + m a_i b_i - b_i c_i}{m a_i - c_i} = \frac{a_i d_i - b_i c_i}{m a_i - c_i} [/tex]

    Now we need to plug this into the collinear constraint. We can factor out the denominators by multiplying by:

    [tex] (m a_1 - c_1)^2 (m a_2 - c_2) (m a_3 - c_3) [/tex]

    This will give:

    [tex] ((a_2 d_2 - b_2 c_2)(m a_1 - c_1) - (a_1 d_1 - b_1 c_1)(m a_2 - c_2) )((m b_3 - d_3)(m a_1 - c_1) - (m b_1 - d_1)(m a_3 - c_3)) =[/tex]
    [tex] ((a_3 d_3 - b_3 c_3)(m a_1 - c_1) - (a_1 d_1 - b_1 c_1)(m a_3 - c_3) )((m b_2 - d_2)(m a_1 - c_1) - (m b_1 - d_1)(m a_2 - c_2)) [/tex]

    This is a sixth degree polynomial, and should have six roots. However, four of these roots are given by the factor we just included, which will leave two roots to the original equation. I'd like to prove these are real and distinct, but I really don't feel like expanding that mess. (if anyone can think of a simpler way, I'd appreciate it). Roughly, the generality of the constants probably guarantees distinctness and the balance of the equation around the equal sign probably guarantees real solutions, but if I think of a more rigorous way I'll post it. So there are two values of m, and correspondingly, two lines that intersect all four of the original lines.
     
  18. Feb 13, 2005 #17

    mathwonk

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    wow! that looks impressive. i will try to read it.

    actually i think i am doing the same thoing below as you, but without coordinates. I.e. I am choosing a pencil of planes containing one of the lines, which you took as the x axis. ...........



    in the meantime, it is easier for me to write a complete alternative solution, than to read one. (I have a bad headache anyway, and a cold.) so here it is, coordinate free.

    This problem should be done in complex projective 3 space, to avoid all reality problems, and imaginary roots.

    Briefly, given three general lines L,M,N, there is a unique quadric surface S containing all three. Then a 4th line R meets that surface in 2 points, say P,Q. Then for a line B to meet all 4 lines L,M,N,R, B must meet S in three points (where B meets L,M,N) hence B must lie on S. Moreover B must belong to the ruling of S by lines transverse to L,M,N, and must pass through either P or Q. There are two such lines, one through P, one through Q..

    In more detail:
    Consider 2 disjoint lines L,M in P^3, hence not coplanar.

    By means of a pencil of planes through a third line N, each plane meeting L and M in one point, set up a linear pencil of corresponding pairs of points on these 2 lines. then pass a line through each pair of points creating by their union a surface S ruled by a pencil of disjoint lines. (If any two of the lines met, there would be a plane containing both of them which would then contain the 4 points of L, M on these two lines. But then the plane would contain both L and M, a contradiction.)

    We claim S is a quadric surface, i.e. it meets a plane in a conic curve. Consider the intersection of the surface S with a plane P containing the line L. P meets M in one point, hence contains the line A joining this point to the corresponding point of L. We claim the full intersection of P and S is the two lines L and A. I.e. P meets no other point of M, hence contains none of the other lines sweeping out S, hence meets each of those other lines only at the point where the line meets L. Since the intersection curve of P with S is 2 lines, i.e. a conic, S is a quadric surface.

    We also claim S contains N. I.e. each line sweeping out S meets L, M at a pair of points which both lie in a plane containing N. Hence the line itself joining these 2 points lies also in that plane, hence meets N. Thus every line sweeping out S as their union, meets the line N. But since those lines are disjoint they all meet N at a different point, so their union contains N.

    Now since S is a quadric surface containing the three general lines L,M,N, if R is any 4th general line, then R meets S at precisely two points P,Q, and any line B meeting all 4 lines, L,N,M,R, must meet S in three points, hence must also lie in S. Furthermore, B must pass through either P or Q.

    One can also see that the only lines meeting all three of L, N, M are the pencil of disjoint lines sweeping out S. (Any line B meeting N determines a unique plane with N, and if B also meets both L, M, then B must be the unique line joining the two points where that plane meets L, M.)

    Now in the pencil of disjoint lines sweeping out S, exactly one passes through P and one passes through Q. So there are two lines meeting L,M,N,R.
     
    Last edited: Feb 13, 2005
  19. Feb 15, 2005 #18

    mathwonk

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    since interest has subsided, i will give a quicky "solution". the grasmannian variety parametrizing all lines in 3 space embeds via the plucker embedding, as a 4 dimensional quadric in projective 5 space. The set of lines incident to a given line is a tangent hyperplane section of that quadric. hence lines incident to 4 given lines form a section of that quadric by 4 hyperplanes, i.e. a line section of that quadric, hence either 1 or 2 intersections exist.

    To show in general it is two, it suffices to show that it is sometimes 2.

    However, it we specialize 2 of our 4 lines so they meet, and also specialize the other 2 so they meet, then there are exactly two lines incident to all 4, namely the line through both intersection points of the two pairs of lines, and the line of intersection of the two planes spanned by the two pairs of lines.

    this last is Schubert's own calculation of why the number is two, using his method of specialization.
     
  20. Feb 15, 2005 #19

    Hurkyl

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    Sigh, Magma wouldn't compute my Gröbner basis in any reasonable amount of time. I'm actually going to have to think this one out. :smile:


    I think I caught that you said I could take as a given that, given three generic lines, the surface comprised of lines intersecting all three forms the quadric surface... a surface whose defining polynomial is degree two.

    So, if I take a parametrization of the fourth generic line and substitute into the equation for the quadric surface, then I have a degree two equation in the parameter, thus the fourth line intersects the quadric surface in two places.

    (Insert some reason why the solutions aren't complex)

    This means there is at least two solutions to the original problem: there must be at least one solution line for each of the two intersection points. However, each intersection point can only admit a single solution line, otherwise the first three lines would have to lie in a plane.

    QED
     
  21. Feb 17, 2005 #20

    Hurkyl

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    BTW, more questions are always good. :smile:
     
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