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I'm looking at "A Geometric Paradox" by B. H. Brown, in the May--June 1923 issue of The American Mathematical Monthly, pages 193--195. I think people studied advanced Euclidean geometry a lot more then than they do now. The author writes as if for an audience familiar with such material, and I don't think you'd do that in the Monthly now.

Brown reminds us that "a plane bitangent to a torus cuts it in two circles" (see Villarceau circles). That much I knew. He writes:

begin quote

  1. A cone of revolution can be inverted into a torus;
  2. The lines of curvature on a cone of revolution are the rulings and parallel circles;
  3. The lines of curvature on a torus are the meridian and parallel circles;
  4. Inversion carries lines of curvature into lines of curvature;
  5. Inversion carries circles into circles (straight lines being considered circles);
  6. Inversion preserves angles (except for sign).

If we then invert a cone into a torus, where can we find any circles on the cone to invert into the Villarceau circles?

end quote

So here we certainly have a seeming paradox!

Items 1 and 4 were unknown to me, and I find 1 a bit implausible. I tried unsuccessfully, although not very hard, to visualize it. I wondered where the center of inversion should be.

Then Brown gives some details:

begin quote

The inversion with center $(0,0,i)$ and power $-2$, whose equations are $$ \begin{array}{l} \bar x = \dfrac{-2x}{x^2+y^2+(z-i)^2}, \qquad \bar y=\dfrac{-2y}{x^2+y^2+(z-i)^2} \\[6pt] \bar z = i+\dfrac{-2(z-i)}{x^2+y^2+(z-i)^2} = \dfrac{i(x^2+y^2+z^2+1)}{x^2+y^2+(z-i)^2} \end{array}\tag{1} $$ will carry the (imaginary) cone $$ \bar x^2 + \bar y^2 + \frac12\bar z^2 =0\tag{2} $$ into the real torus $$ (x^2+y^2+z^2+1)^1 = 8(x^2+y^2).\tag{3} $$

end quote

I hadn't expected complex numbers to enter in this way, and without some context I'd have guessed that the manifold that he calls an imaginary cone would be a $2$-dimensional manifold over $\mathbb C$ if $\bar x,\bar y,\bar z$ are allowed to be in $\mathbb C$. But the context suggests otherwise. As nearly as I can GUESS, we should have $\bar z$ being real and $\bar x,\bar y$ being pure imaginaries, so that the circle of negative radius $-\bar z/\sqrt{2}$ would be a $1$-dimensional manifold over $\mathbb R$ not over $\mathbb C$, and thus topologically what we normally call a circle.

1. Does my GUESS make sense?

2. Is this way of doing inversive geometry in $\mathbb R^3$ by using polynomials in $\mathbb C[\bar x,\bar y,\bar z]$ in any sense a standard thing? (Brown seems to treat it as standard, so I'd guess people familiar with what went on in geometry in those days might think of it as somewhat standard.)

Note added on the evening of October 8th: The sort of metric geometry I've seen in $\mathbb C^n$ involves an inner product that's linear in one variable and conjugate-linear in the other, so that $\mathbb C^n$ becomes isometric to $\mathbb R^{2n}$. Later in this paper than the part I quoted, Brown writes:

[ . . . ] the inverses of one system of Villarceau circles are given as the intersections of the cone and the planes$$ x+iy=c;\qquad c\ne0,\tag{7} $$ and of the other system by the cone and the planes $$ x-iy=c;\qquad c\ne0,\tag{8} $$

This makes it look as if these are supposed to be circle on the "cone" $x^2+y^2+\frac12z^2=0$ in $\mathbb C^3$. So I'm wondering if Brown intended these "circles" to be $2$-manifolds? And how does one think about metric geometry, including things like curvature and angles, in such a setting? And is it only in that sort of setting that a cone of revolution can be inverted into a torus?

Michael Hardy
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  • Amazing how much silence there is here after six hours. I'm not sure I've seen that before. Maybe I'll have to put a bounty on this when it's eligible for that. – Michael Hardy Oct 07 '12 at 00:30
  • draw the picture if you can :-) – cactus314 Oct 07 '12 at 15:44
  • @johnmangual : The picture in the Wikipedia article I linked to says a lot. Here's the link again: http://en.wikipedia.org/wiki/Villarceau_circles – Michael Hardy Oct 07 '12 at 16:19
  • I've started a bounty. I might add a little bit to the question, at the bottom, leaving largely intact the part that's already there. – Michael Hardy Oct 09 '12 at 01:53
  • A "note added on the evening of October 8th" now appears at the bottom. – Michael Hardy Oct 09 '12 at 03:40
  • My first instinct is to interpret $x$ and $y$ and real, with $z$ purely imaginary. That is, this "imaginary cone" is one with an "imaginary axis". (This seems consistent with the not-unfamiliar use of imaginary axes to allow for an obtuse triangle to have one vertex on each axis.) I don't see how to read (7) and (8) in this context, however. – Blue Oct 09 '12 at 06:30
  • @Blue : I don't know about that particular usage involving obtuse triangles. Can you explain that and point out some sources? – Michael Hardy Oct 10 '12 at 04:13
  • @MichaelHardy: Invoking imaginary axes to create obtuse triangles occurs in an argument that connects the Pythagorean Theorem for Right-Corner Tetrahedra to Heron's Formula for the area of a triangle. (The key observation: a triangle propped in the corner of an orthant --say, with vertices $(a,0,0)$, $(0,b,0)$, $(0,0,c)$-- is necessarily non-obtuse. Taking one of $a$, $b$, $c$ to be purely imaginary brings obtuse triangles into the picture.) See, for instance, Kevin Brown's Math Page on the topic: http://mathpages.com/home/kmath226/kmath226.htm – Blue Oct 10 '12 at 06:04
  • BTW, here's the Brown article in JSTOR: http://www.jstor.org/stable/2298454 . Looking through it, I doubt my observation is helpful. For a while, simply taking $z$ as purely imaginary (and purely real after cone inversion) seemed to satisfy my intuition; but when Brown re-inverts the Villarceau circles, he appears to require fully-complex $x$ and/or $y$ for his equation (5) (or else what can cancel the imaginary terms?). Similarly, if $x$ and $y$ are real, then $x+iy=c$ fixes them and determines a "plane" only if free variable $z$ accounts for two dimensions. My intuition fails me here. – Blue Oct 10 '12 at 06:20
  • I've come to think you're almost certainly right: $z$ is supposed to be a pure imaginary and $x$ and $y$ are supposed to be real. Brown apparently expected his readers to understand that. I surmise that that was conventional when doing this sort of geometry in those days. – Michael Hardy Oct 11 '12 at 17:12

4 Answers4

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The notation seems a bit arcane. Brown's discussion would be done in Hyperbolic 3-space $\mathbb{H}^3$ these days. The images of the torus Villarceau circles are parabolas made by cutting the cone along a vertical plane. The pictures of Villarceau circles are beautiful and more beautiful images along the way :-)

Resources

Old-School Notation

I had to clarify some things before I could understand what Brown was saying or you were asking. This answer may still have a mistake or two.

  • Hyperbolic 3-space is thought of as $\{ (x + iy, z):x,y,z \in \mathbb{R}, z > 0 \} \in \mathbb{C} \times \mathbb{R}^+$ Villarceau's theorem says you can invert the cone $x^2 + y^2 - z^2 = 0$ along a point $(0,0,1)$ and get a torus. In modern notation, the inversion map should be $$ (x,y,z) \mapsto (0,0,1) + \frac{(x,y,z)-(0,0,1)}{||(x,y,z)-(0,0,1)||^2}$$ Brown's paper seems to extrapolate to purely imaginary vertical coordinate $\{ (x + iy, z):x,y \in \mathbb{R}, z \in i\mathbb{R},z > 0 \} \in \mathbb{C} \times \mathbb{R}^+$ and invert at $(0,0,i)$. The "cone" would be $x^2+y^2+z^2=0$ and the inversion map $$ (x,y,z) \mapsto (0,0,i) + \frac{(x,y,z)-(0,0,i)}{||(x,y,z)-(0,0,i)||^2}$$ This amounts to changing the signature of the hyperbolic space to get anti-de Sitter space.
    $$ (x,y,z) \mapsto (0,0,1) + \frac{(x,y,z)-(0,0,1)}{||(x,y,z)-(0,0,1)||^2_{\text{AdS}}}$$ The null-cone is $||(x,y,z)||_{\text{AdS}} = x^2+y^2-z^2=0$ and $z \in \mathbb{R}$.

  • This question has to do with specific Riemannian structures on the cone and torus. "Lines of curvature" refers to the geometry of surfaces - they are the eigenvectors of the Gaussian curvature.

  • The lines of curvature are exactly the ones you'd think of:

    • For the torus they are the equatorial and meridian slices
    • For the cone, you can slide along a vertical plane through the axis of symmetry and get the "rulings" or along a horizontal plane and get circles. enter image description here
  • These days results 5 and 6 are standard properties in conformal geometry

Villarceau circles & Hopf Fibration

The Hopf fibration is a map $(w_1,w_2)\mapsto w_1/w_2 $ from the 3-sphere $S^3 = \{ |w_1|^2+|w_2|^2=1\} $ to $S^2 = \hat{\mathbb{C}}$. We can get a torus by $$ \{ |w_1|^2+|w_2|^2=1\} \cap \{ |w_1|=|w_2|=R\}$$ where the torus is paramterized by $\theta_1 = \arg w_1 , \theta_2 = \arg w_2$. The Villarceau circles are parameterized by $( e^{it} w_1, e^{it} w_2)$ and $( e^{it} w_1, e^{-it} w_2), \hspace{0.1in} t \in [0, 2\pi]$.

Inversion about imaginary point

So in 3D Euclidean space, you can invert the cone thru a small sphere centered at $(0,0,1)$. The rulings of the cone will be mapped to meridian circles the torus and the horizontal circles will also be maps to equatorial circles.

If you invert the cone $x^2-z^2=0$ about the point $(0,1)$ you get 2 intersecting circles and we revolve around to get an immersed torus... Brown inverts around $(0,i)$ to get 2 disjoint circles. Revolve around the $z$ axis & we get a torus.

Parabolic Circles

In terms of "homology", Villiarceau = Meridean + Equatorial so the image on the cone should be a line going from the tip of the cone out to infinity and winding around the vertical axis once.

Towards the end of Brown's article, he explains the image of the Villarceau circles are (Euclidean) parabolas defined intersecting the cone and a vertical plane $$ \{ x^2 + y^2 - z^2 \} \cap \{ x \pm i y = c \}$$ If we slide the cone horizontally, it's clear we get circles. Why should we also consider these vertical slides as circles? It appears, that while inversion interchanges planes and spheres, item 5 can be contradicted for curves.

If I have time, I will try to add some images and the explicit inversion calculation.

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cactus314
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  • If the image on the cone is the curve you describe, then the image of the Villarceau circle is not a circle, contradicting item 5 in the list. – Michael Hardy Oct 11 '12 at 19:13
  • Brown's article says they are parabolas. – cactus314 Oct 11 '12 at 20:47
  • You say item 5 can be contradicted for curves, but item 5 actually says straight lines are considered circles [of infinite radius]. – Michael Hardy Oct 11 '12 at 21:25
  • I've up-voted your answer, but I'm not yet satisfied with it---partly because I haven't yet digested it all, but also: It appears that the torus into which the cone is inverted by the inversion map that you give is in fact immersed rather than embedded in $\mathbb R^3$, i.e. it has self-intersections. And when two circles cross each other, there's no line that's tangent to one of them on one side and to the other on the other side. Hence no plane is bitangent to the torus in the relevant way, so this immersed torus has no Villarceau circles. At this point I'm wondering whether....... – Michael Hardy Oct 11 '12 at 21:29
  • .....that may be the reason why Brown had $z$ taking imaginary values while $x$ and $y$ were real (if in fact that's what he had in mind). – Michael Hardy Oct 11 '12 at 21:29
  • I don't know how using an "imaginary" cone makes a difference. The torus he gets could not come from inversion of a real cone along the vertical axis. It is centered at the origin. So I have to give alternative construction with real cone. – cactus314 Oct 11 '12 at 23:02
  • "Item 5 can be contradicted for curves". So the way out of the paradox is then simply that item 5 is not generally true? – Michael Hardy Oct 13 '12 at 23:10
  • I think it has to do with the signature. Hyperbolic space can be thought of as $\mathbb{C} \times \mathbb{R}^+$ under a certain metric. But in this case he uses imaginary vertical coordinate. $(x+iy,z) \in \mathbb{C} \times i\mathbb{R}^+$. Equivalently, you are changing the signature of the metric on the space to $ds^2 = dx^2 + dy^2 - dz^2$. – cactus314 Oct 13 '12 at 23:29
  • Something similar should work in 2D where the cone is just the pair of lines $x^2 - z^2 = (x-z)(x+z)= 0$. They should map under this "inversion" to a pair of disjoint circles (now in Lorentzian signature). I've been following Brown's algebra with Mathematica... can't quite get the cone to map to the pair of circles. – cactus314 Oct 13 '12 at 23:32
  • Love to see a torus-cone inversion in some space if possible.. with circle of inversion using any " real parameter visualization transformation".. a sort of new space mathematical 3D glasses. cactus314 images they are disjunct. – Narasimham Jan 22 '20 at 04:59
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Here's a Computer Graphics paper looking for conics on Dupin cyclides, which says Fact #1 is well-known.

The Geometry of Dupin Cyclides has something to say about inverting cones to tori and vice versa. They were discovered by Dupin, who was a student of Gaspard Monge. Apparently, James Clerk Maxwell was interested in them, related to separation of variables of Laplace equation in 3-dimensions.

Wikipedia says Dupin cyclides (including the torus itself) are always inverses of cones of revolution. They are also inverses of tori of revolution (as distinct from other surfaces which are topologically tori).

The cone is a degenerate "horn" or "spindle" cyclide. This paper has many pictures of cyclides and talks about Villarceau circles a bit.

Here's a paper on computer aided design, which claims to show all cyclides come from tori.

Consider the analagous problem in 2D, the "cone" $x^2 - z^2 = (x-z)(x+z)=0$ has to be mapped to two disjoint circles $$((x-\sqrt{2})^2+ z^2 - 1)((x+\sqrt{2})^2+ z^2 - 1)= (x^2+z^2+1)^2 - 8x^2= 0$$ Since inversions are conformal maps, they are going to preserve intersections. Perhaps you can invert around a circle of "negative" radius.

cactus314
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  • pondering....... I'll digest these maybe tomorrow (Sunday)..... – Michael Hardy Oct 14 '12 at 03:46
  • I've thrown what I could at it. Learned some interesting geometry in the process. My conclusion is he complexifies the hyperbolic plane an a way that's not well-defined. – cactus314 Oct 14 '12 at 23:08
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Here's how I see it:

If $x,y,z \in \mathbb{R}$ we can set

$$ \begin{array}{l} \bar x = \dfrac{-2x}{x^2+y^2+(z-i)^2}, \qquad \bar y=\dfrac{-2y}{x^2+y^2+(z-i)^2} \\[6pt] \bar z = i+\dfrac{-2(z-i)}{x^2+y^2+(z-i)^2} = \dfrac{i(x^2+y^2+z^2+1)}{x^2+y^2+(z-i)^2} \end{array} $$

this is an inversion of the real torus $$ (x^2+y^2+z^2+1)^2 = 8(x^2+y^2)$$ about the imaginary point $(0,0,i)$ about a circle of imaginary radius, $r = 2i$. For this particular choice of radius you can complete the square so that $\bar x, \bar y, -i \bar z \in e^{i\theta} \mathbb{R} $ for some phase $e^{i\theta} = \arg (x^2+y^2+(z-i)^2)$.

Since they all have the same denominator, it's easy to check:

$$ \bar x^2 + \bar y^2 + \bar{z}^2/2 = 0 \longleftrightarrow (x^2+y^2+z^2+1)^2 = 8(x^2+y^2) $$

I wonder what happens if we project to $y = 0$:

If $x,y,z \in \mathbb{R}$ we can set

$$ \begin{array}{l} \bar x = \dfrac{-2x}{x^2+(z-i)^2}, \hspace{0.25in} \bar z = i+\dfrac{-2(z-i)}{x^2+(z-i)^2} = \dfrac{i(x^2+z^2+1)}{x^2+(z-i)^2} \end{array} $$

this is an inversion of the two disjoint circles to a pair of lines:

$$ \bar x^2 +\bar{z}^2/2 = 0 \leftrightarrow (x^2+z^2+1)^2 = 8x^2 \leftrightarrow ((x - \sqrt{2})^2 + z^2 - 1 )((x + \sqrt{2})^2 + z^2 - 1 ) $$

cactus314
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It's easier to work with a cone, it's a scaled up circle. The circle is $[\sin(x),\cos(y),1]$, scale that up by $\sin(y)$ and you have a cone. $[\sin(x)\sin(y), \cos(x)\sin(y), \sin(y)]$. If you use a function to determine color and relate it to area, you get a very pretty cone, anyway I'll get back to that. Now it seems to me the cones use 2d angles. If you were to take those side lengths and put them inside a semi circle, the diameter would always be 1. A sphere is in 3d. If you took the three side lengths and put them in a circle, it would not fit in a semicircle of diameter $1$. You need to go up on a 90 degree angle into 3d space and you need to use the distance of the other two sides in order to do it. $$\sqrt{(\sin(x)\sin(y))^2+(\cos(x)\sin(y))}=\cos(y)$$

Do the same thing with this function $$[\sin(x)(1+(1/2)\sin(y)), \cos(x)(1+(1/2)\sin(y)), (1/2)\sin(y)]$$ It's a cone with the top chopped off. If you used color as a function of area, you would quickly realize everything is exactly where it should be. So you are using 2d angles, you need to go into 3d space, change the third coordiniate of $(1/2)\sin(y)$ into $(1/2)\cos(y))$.

Anyway, that's an opinion of an uneducated high school level math guy. I know you don't like opinions, but you banned me for asking "what it means" which is a very important question in graphing. So I don't care if you don't want my opinion.

Michael Hardy
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