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Hello,

I'm looking for an argument that the n-dimensional stereographic projection maps circles (intersections of affine two-dimensional subspaces with S^n) to circles in R^n. I've looked around and the only argument I saw for the n-dimensional case is a generalization of the geometric proof for n = 2 (with the tangent cone) which I don't really feel comfortable with, even when n = 2. Is it possible to reduce it to the n = 2 case somehow or give a "direct", algebraic, proof?

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It is of course possible to do a direct computation. It might be ugly, though. –  Benoît Kloeckner May 19 '10 at 17:38
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2 Answers 2

The circle and the north pole (or wherever the origin of the stereographic projection is) span a 3-dimensional subspace generically, such that the restriction to this subspace is the 2-dimensional stereographic projection. If the circle goes through the north pole, then it is actually sent to a line under stereographic projection, and this is in some sense a reduction to the 1-dimensional case.

Yana Mohanty has a nice proof that stereographic projection sends circles to circles.

A more sophisticated approach is to notice that stereographic projection is the restriction of inversion through a sphere orthogonal to $S^n$ in $R^{n+1}\subset S^{n+1}$. Then one needs to see that inversions send circles to circles, or more generally that Mobius transformations of $S^n$ do. The group of Mobius transformations of $S^n$ is $PO(n+1,1)$ or $Isom(\mathbb{H}^{n+1})$, the isometry group of hyperbolic $n+1$-space. This groups preserves the cone $x_0^2+x_1^2 +\cdots - x_{n+1}^2=0$. The sphere at infinity (in the projectivization) of this cone is $S^n$, and the action is by the Mobius group. A circle is the intersection of the projective closure of a 3-dimensional subspace with the sphere at infinity. Since $PO(n+1,1)$ consists of linear transformations, it permutes 3-dimensional subspaces of $R^{n+1,1}$, and therefore sends circles to circles in the projectivization.

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I don't know what is the "tangent cone" argument you mention, but anyway here is my favorite proof of the fact.

The stereographic projection is the restriction of an inversion $I$ from $\mathbb R^{n+1}\setminus \{p\}$ to itself, where $p$ is the pole from which you project (or maybe $I$ is a map from $\mathbb R^{n+1}\cup\{\infty\}$ to itself if you prefer it this way). So it suffices to prove that $I$ sends circles to circles and lines.

First observe that $I$ sends $n$-spheres and hyperplanes to $n$-spheres and hyperplanes. (This follows from the planar case with circles and lines and rotation symmetry. The planar case is elementary). In $\mathbb R^3$, every circle is an intersection of two spheres. Hence its image is an intersection of two sets each of which is a sphere or a plane. Such set is a circle again.

In higher dimensions, either restrict to a suitable 3-dimensional subspace, as Agol said in his answer, or represent a circle as an intersection of several spheres.

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I'm guessing the tangent cone approach amounts to that in section 36 of the book by David Hilbert and S. Cohn-Vossen, in English "Geometry and the Imagination," the original "Anschauliche Geometrie." (1932) –  Will Jagy May 20 '10 at 2:26
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