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A Why are conics indistinguishable in projective geometry?

  1. May 19, 2016 #1

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    It is said that curves of the second order which we usually refer to as ellipse, parabola and hyperbola, i. e. conics, are all represented on projective plane by closed curves (oval curve), which means there is no distinction between them. Why is it?

    Projective space can, in principle, be thought of as Euclidean space supplemented with points at infinity. Say, we have a hyperbola on projective plane. It has two asymptotes, one for each branch. And those asymptotes are not parallel lines (in projective geometry they themselves may cross at any angle, but zero). It means that they can not cross at the same point of infinity, which would give us a nice closed curve. Instead, they both cross the same (and the only) line at infinity at different points. That makes a closed hyperbola consisting of two parts: the curved segment and the line segment at infinity. How can it be considered as a qualitative equivalent of ellipse that is curved everywhere?

    The same is true for parabola, there are no asymptotes here, but tangent lines are not parallel anywhere, which means a parabola can not be closed without some kind of similar linear construction, can it?
     
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  3. May 19, 2016 #2

    micromass

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    There are some neat ways to visualize the projective plane. The intuition is of course that the proective place is Euclidean space plus a line at infinity. But that is hard to visualize. There are beter tools. I recommend looking through the first two chapters of "A guide to plane algebraic curves" by Kendig. http://smile.amazon.com/Algebraic-Curves-Dolciani-Mathematical-Expositions/dp/0883853531

    My favorite way to visualize curves in projective plane is by a sphere with antipodal points identified. The line at infinity there is just the equator. Visualizing curves on the sphere elucidates the entire situation.
     
    Last edited by a moderator: May 7, 2017
  4. May 20, 2016 #3

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    Thank you for pointing (projecting :-)) me in the right direction.

    I have found the book you recommended and I like it much. Exactly kind of book I needed.

    Join to Micromass in sincere recommending this book to everyone who is looking for elucidation for subjects like the one discussed in this thread and nice introduction to algebraic geometry in general.
     
  5. May 20, 2016 #4

    mathwonk

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    just ask yourself if the three curves you describe can have isomorphic compactifications. I.e. obviouly adding one point to a parabola makes it look like an ellipse, and adding two points to a hyperbola does the same. Note that projective space is an enlagement of affine space in which the extra points added correspond to families of parallel lines in affine space. E.g. since the two sides of a parabola go off to infinity both becoming more and more vertical, the added point is the one corresponding to the family of vertical lines. Can you see why there are two points added at infinity to a hyperbola?

    To be technically precise, a "conic" is any curve of degree two, and there are two more degenerate conics which are not equivalent to these three, namely a pair of distinct lines, and a double line. The invariant that distinguishes conics is their "rank", and the fact you refer to is that all conics of maximal rank are equivalent. In algebraic terms, this is because you can diagonalize their symmetric matrices. Then the rank is the number of non zero diagonal entries. The three conics you consider are of rank 3 in the projective plane, and the degenerate ones have ranks 2 and 1.

    To see algebraically the equivalence of ellipse, parabola, and hyperbola in the projective plane, we can use the process of “homogenizing” their equations. To homogenize an equation, you replace each letter by its quotient by a new letter, and then clear denominators. This gives the equation of the closure in projective space of the graph of the original affine equation. So if the homogenizations are the same then the affine curves have the same projective closures.


    The hyperbola XY = 1, becomes (X/Z)(Y/Z) = 1, then XY = Z^2, and by changing variables, with X = U-V and Y = U+V, we get (U-V)(U+V) = Z^2, or U^2 -V^2 = Z^2, or U^2 = Z^2 + V^2.

    Now for the parabola Y = X^2, we get (Y/Z) = (X/Z)^2 or ZY = X^2, and changing variables with U-V = Z and U+V = Y, gives U^2-V^2 = X^2 or U^2 = X^2 + V^2, which is the same equation essentially as for a hyperbola.

    Finally the ellipse X^2 + Y^2 = 1, becomes (X/Z)^2 + (Y/Z^2) = 1, or X^2 +Y^2 = Z^2, which is again the same form as before.
     
    Last edited: May 23, 2016
  6. May 24, 2016 #5
    I'll just add to what mathwonk wrote that each of the three equations in three variables: U^2 = Z^2 + V^2, U^2 = X^2 + V^2, and X^2 + Y^2 = Z^2 should be thought of as describing a curve in 2-dimensional projective space, which is the same as the space of all lines through the origin (0, 0, 0) in 3-space.

    So in a sense it is a curve whose points are lines! These equations each describe a cone in 3-space, which can easily be seen as a closed curve formed of lines though the origin.

    Since all such lines pass through the origin, one of them is completely described by any point in 3-space that is not the origin. And any two such points that are constant multiples of each other, like (x, y, z) and (cx, cy, cz) for any c ≠ 0, describe the same line. (No other points describe the same line.)

    So this is the fundamental definition of the projective plane: 3-space minus the origin, with any two points of this considered to be "the same" if one is a multiple of the other by a non-zero constant.

    The discovery that the apparently different ellipse, parabola, and hyperbola are really the same things in the projective plane was what led mathematicians to understand that the projective plane (and in higher dimensions, projective n-space) is really the right setting in which to think of algebraic equations.
     
    Last edited: May 24, 2016
  7. May 24, 2016 #6

    mathwonk

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    as mentioned above, those are equations in post #4 are of cones in 3 space, each line on the cone being thought of as one point in projective space.

    Thus to recover most of the points, we can intersect the cone with an affine plane that meets each line of the cone at most once. However no plane can meet every line of the cone so we lose some points of the projective conic by passing back to the affine conic.

    To see this intersection algebraically, we set one variable equal to say 1, and starting from X^2 +Y^2 = Z^2, we get the ellipse X^2 + Y^2 = 1 by setting Z=1;
    This seems not to miss any (real) lines, but we have missed the (complex) points (1, i, 0) and (1,-i,0), where Z=0.

    similarly by setting Y=1 we get the equation X^2 + 1 = Z^2, or 1 = Z^2-X^2, which is a hyperbola. Here we have missed the points where Y=0, namely (1,0,1) and (1,0,-1).

    From X^2 + Y^2 = Z^2, we can pass to X^2 = Z^2 - Y^2, then change variables to get X^2 = UV, and finally set V=1, to get a parabola X^2 = U. This misses only the point V=0, U=1, X=0.


    Thinking in projective geometry has huge simplifying advantages over affine geometry since there are much simpler rules for intersections. Namely any two curves in the projective plane always meet, and if the curves has degrees n and m, and no common components, they meet in nm points, if multiplicities (tangential intersections) are counted properly. Of course I am also giving myself the simplification of working over complex numbers,
     
    Last edited: May 26, 2016
  8. May 26, 2016 #7

    mathwonk

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    It is easy this way to see that the projective plane is larger than the affine plane and just how much larger. Each line in the affine X,Y,Z space passing through the origin is a point of the projective plane. If we consider the affine plane Z = 1, it meets exactly once each such line except those lying in the plane Z=0. Thus the projective plane consists of one point for each point in the affine plane Z=1, plus the additional set of points corresponding to lines through the origin in the affine plane Z=0. But this additional set corresponds to the points of a “projective line”, i.e. the set of lines in the X,Y plane and passing through the origin. Of these lines through the origin in the X,Y plane, we can consider those which meet the line Y=1, and here each such line meets that line exactly once except for the line Y=0. So the projective line consists of a copy of the affine line Y=1, plus one additional point corresponding to the line Y=0.

    Thus a projective line is obtained by adding one point to an affine line, and a projective plane consists of an affine plane plus one additional projective line. I.e. the projective plane consists of one copy of an affine plane, plus one copy of an affine line, plus one point.

    Moreover the set of lines through the origin of affine three space is symmetrical, so we can choose in any way we wish, which affine plane through the origin to remove, i.e. to be the additional projective line. In the case of the cone X^2 + Y^2 = Z^2, if we choose to remove the lines in the plane Z=0, these meet that (real) cone only at the origin, so all the real lines of this cone meet the parallel affine plane Z=1, so we get the full real ellipse X^2+Y^2=1 lying entirely in the affine plane. I.e. removing the projective line Z=0 does not remove any points of this real ellipse.

    But if we remove the projective line corresponding to lines in the plane X=0, this removes the two lines Y-Z=0 and Y+Z=0 (in the plane X=0) of the cone X^2+Y^2=Z^2, i.e. 2 points of the projective ellipse, so in the parallel plane X=1, we are left with the ellipse minus two points, i.e. the real hyperbola 1+ Y^2=Z^2, or 1 = (Z+Y)(Z-Y), or 1 = UV.

    Finally the plane Z=Y, or Z-Y=0, meets the cone X^2+Y^2 = Z^2 only in the line spanned by the vector (0,1,1). Hence the parallel plane Z-Y = 1 meets it in the curve X^2 = Z+Y, or X^2 = U, a parabola, which is the projective ellipse minus the one point (0,1,1).

    One can see from this construction why these curves are called “conic sections” since they occur by intersecting the cone X^2+Y^2 = Z^2 with planes. What was not explained to me in high school was to observe that these various planes each fail to meet a different number of lines of the (real) cone, i.e. the plane Z=1 meets them all and we get an ellipse, the plane X=1 misses the two lines in the parallel plane X=0 and we get a hyperbola, while the plane Z-Y=1 misses only the one line in the parallel plane Z=Y and we get a parabola.

    If you think of your ellipse as lying in the projective plane and removing the points where it meets a particular projective line, you can see (in the real case) there are three cases, the projective line we remove meets the projective ellipse either not at all or in one point or in two. Thus the case of one point occurs only when the projective line we remove is tangent to the ellipse, which is a rather special case. In the complex setting we are down to only two cases, one intersection point or two, so there is no such thing as a complex affine ellipse, i.e. complex affine curves are never bounded. (In the non singular case this follows from the maximum modulus principle, since a closed bounded curve would be compact, but a holomorphic function like an affine coordinate, cannot have a local maximum.)
     
    Last edited: May 26, 2016
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