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This problem cropped up in the context of scale-insensitive methods for generating random variables.

Let $X=R^n \cup \{\infty\}$. Suppose we consider a set of transforms $\cal{T}$ from $X\rightarrow X$. We construct them by concatenating functions chosen from the following set:

  • Invertible linear transform ($x \mapsto Ax$, for $A\in GL_n(R)$)
  • Translation ($x \mapsto x+b$)
  • Circle inversion ($x \mapsto x/|x|^2$; 0 and $\infty$ swap)

The set $\mathcal{T}$ is very similar to the Mobius transformations, which are built from:

  • Rotation and scaling ($x \mapsto sAx$, for $A\in SO_n(R)$ and $s$ a positive scalar)
  • Translation ($x \mapsto x+b$)
  • Circle inversion and reflection ($x \mapsto Mx/|x|^2$, where $M$ reflects through the first coordinate; 0 and $\infty$ swap)

I would like to know if $\cal{T}$ has a standard name, and if any of the properties of the Mobius transformations generalize to $\cal{T}$. For instance, Mobius transformations in $R^2$ preserve generalized circles; are generalized ellipsoids in $R^n$ preserved by $\mathcal{T}$? Is there a property analogous to the cross-ratio? Any references would be greatly appreciated.

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  • $\begingroup$ Do you know if $\cal{T}$ is finite-imensional? $\endgroup$
    – Anton Petrunin
    Commented May 17, 2013 at 17:45
  • $\begingroup$ I think so-- I believe you can write any such transformation as either $b+A(x-c)/|A(x-c)|$ or $b+A(x-c)$, for $A\in GL_n(R)$. $\endgroup$
    – Bill Bradley
    Commented May 17, 2013 at 21:03
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    $\begingroup$ Bill: You cannot, since you have to take compositions as well. $\endgroup$
    – Misha
    Commented May 17, 2013 at 22:23
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    $\begingroup$ Circle inversion doesn't preserve generalized ellipses, or conics. Curves of degree n are typically sent to curves of degree 2n. If you invert an hyperbolas about its center you get a figure-8s, a lemniscates. Ellipses can be sent to dimpled limaçons or hippopedes. Parabolas can be sent to cissoids or cardiods. If the center of inversion is on the conic, though you get a cubic curve like a crunode $y^2 = x^2(x+1)$. en.wikipedia.org/wiki/Inverse_curve xahlee.info/SpecialPlaneCurves_dir/Inversion_dir/inversion.html $\endgroup$
    – Douglas Zare
    Commented May 18, 2013 at 5:53
  • $\begingroup$ Thank you, that's very helpful! Thinking about that zoo of curves makes it even more remarkable that circles are preserved by Mobius transformations. $\endgroup$
    – Bill Bradley
    Commented May 19, 2013 at 17:02

1 Answer 1

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$\mathcal{T}$ is not a Lie group when $n>1$.

Actually, the OP did not say whether he wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if he did, then it is clear that $\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)

In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind the OP mentions because this would define a PDE that $G$ satisfies.

(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)

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  • $\begingroup$ But the projective general linear group is always a Lie group acting on the sphere, and it always contains the conformal group as a proper subgroup. How can the conformal group be maximal? By projective general linear group I mean $GL_n \mathbb R$ acting on the rays out of the origin in $\mathbb R^n$. $\endgroup$
    – Ryan Budney
    Commented May 17, 2013 at 22:00
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    $\begingroup$ @Ryan Budney: the action of $\mathrm{PGL}$ does not contain the conformal group, as the former preserves the antipody relation while the latter doesn't. $\endgroup$
    – Benoît Kloeckner
    Commented May 17, 2013 at 22:14
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    $\begingroup$ Robert: It seems likely that the "conformal+projective" group is transitive on $k$-point sets for all $k$. Do you know if this is indeed the case? Another side remark is that "conformal+projective" transformations frequently appear in proofs of rigidity results, like Mostow rigidity. $\endgroup$
    – Misha
    Commented May 17, 2013 at 22:56
  • $\begingroup$ Ah, right. I got really off-track there. $\endgroup$
    – Ryan Budney
    Commented May 18, 2013 at 8:24
  • $\begingroup$ @Misha: It's not clear to me which group you mean by "conformal+projective". The group $\mathcal{T}$, as defined by the OP, is not smooth on $S^n=\mathbb{R}^n\cup\lbrace\infty\rbrace$ when $n>1$ because the non-conformal affine transformations don't extend smoothly to $\infty$. Are you asking instead about the (non-Lie) group generated by the two (maximal) proper Lie group extensions of $\mathrm{SO}(n{+}1)$ acting on $$S^n=\bigl(\mathbb{R}^{n+1}\setminus\lbrace0\rbrace\bigr) /\mathbb{R}^+,$$ i.e., $\mathrm{SO}(n{+}1,1)$ and $\mathrm{SL}(n{+}1)$? $\endgroup$
    – Robert Bryant
    Commented May 18, 2013 at 13:30

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