Re: [math-fun] Wechsler disk problem
I'd say that since a circle in R^2 is determined by its center and radius, the space of circles is naturally isometric to R^2 x (0,oo). For the general equation below to give a circle, it must hold that B^2 + C^2 > 4AD. So the space of coefficients (mod giving the same circle) isn't quite P^3. --Dan << . . . the general equation of a circle is: A(x^2 + y^2) + Bx + Cy + D = 0 which forms a projective 3-space of possible circles.
________________________________________________________________________________________ It goes without saying that .
I'd say that since a circle in R^2 is determined by its center and radius, the space of circles is naturally isometric to R^2 x (0,oo).
Indeed, we can consider a circle to be described by the vertex (in three-dimensional space) of a right circular cone erected upon it. Sometimes it's useful to extend this to R^3, since circles with negative radius (corresponding to upside-down cones) are very elegant for proving Casey's theorem and things of that flavour.
For the general equation below to give a circle, it must hold that B^2 + C^2 > 4AD. So the space of coefficients (mod giving the same circle) isn't quite P^3.
Oh, yes, I've included circles of imaginary radius and suchlike. And probably the union of the line at infinity with an arbitrary straight line, since that's also a degenerate circle (conic passing through the circular points). Thanks for noticing that. I often overlook these matters of diagram dependency. Sincerely, Adam P. Goucher
It's nice that the space of circles in R^2 is, as a metric space, naturally R^2 x (0,oo). For this is an invariant metric, in that every [permutation of the circles in R^2 that is induced by an isometry of R^2] is also an isometry of the configuration space R^2 x (0,oo). What's both amusing and frustrating is that there is no such metric on, e.g., the configuration space of all affine lines in R^2. (I.e., all loci of ax + by + c = 0 where a and b are not both 0.) It's fun to verify (and well-known) that topologically, this space is the (open) Moebius band. But it's both amusing and frustrating that it has no invariant metric! For, the permutations of the affine lines in R^2 that are induced from isometries of R^2 is identical to the isometry group of R^2, which is 3-dimensional. On the other hand, it can be shown that no matter what metric is put on the Moebius band, its isometry group will be at most 1-dimensional. --Dan Adam wrote: << I wrote: << I'd say that since a circle in R^2 is determined by its center and radius, the space of circles is naturally isometric to R^2 x (0,oo).
Indeed, we can consider a circle to be described by the vertex (in three-dimensional space) of a right circular cone erected upon it. . . .
What's both amusing and frustrating is that there is no such metric on, e.g., the configuration space of all affine lines in R^2. (I.e., all loci of ax + by + c = 0 where a and b are not both 0.)
It's fun to verify (and well-known) that topologically, this space is the (open) Moebius band.
Indeed. The projective lines (ax + by + cz = 0), where (a,b,c) is not (0,0,0) and (la,lb,lc) = (a,b,c) form a projective plane P^2, which is a consequence of projective duality. The set of affine lines is simply the set of projective lines with the line at infinity removed. So, its configuration space is a projective plane with a single point removed. We know what the real projective plane looks like: http://en.wikipedia.org/wiki/File:ProjectivePlaneAsSquare.svg So, we can just have some topological fun puncturing and deforming it. When we puncture the square (which we have deformed into a disc), we obtain an annulus with antipodal points on the outer circumference identified with each other. Now, suppose we start with a Möbius strip, and tear along it until it splits into a loop of twice its original length. (We imagine that the points at either side of the tear are still 'connected' by 'wormholes', so this is still topologically a Möbius strip.) This loop is an anullus with a smooth edge and a torn edge. Without loss of generality, we define the smooth edge to be the 'inside' of the annulus and the rough edge to be the 'outside'. Observe that tearing the Möbius strip would have split every point along the tear into two antipodal points on the outer circumference of the annulus, which are still connected via wormholes. Hence, the Möbius strip is topologically equivalent to the punctured real projective plane. Sincerely, Adam P. Goucher
participants (2)
-
Adam P. Goucher -
Dan Asimov