Timeline for Reference request: affine transforms + circle inversion?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 19, 2013 at 12:40 | comment | added | Misha | Thank you, Robert, this looks quite plausible. At least, this seems to show that for every $k$, every orbit is dense in $(S^n)^k$ minus diagonals. | |
| May 19, 2013 at 11:47 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added some information and clarified some statements
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| May 19, 2013 at 11:40 | comment | added | Robert Bryant | (cont) Let $\frak{l}$ be the union of the ${\frak{l}}_i$. Then any $m$-jet of a vector field on $S^n$ is the $m$-jet of some vector field in $\frak{l}$. Since, under the conformal group, any $k$ points are equivalent to $k$ points that lie within some $\epsilon$-ball of a given point, it may be possible to use this 'jet density' to prove that the orbit under the non-Lie group of any $k$-tuple of distinct points in $S^n$ is open in $(S^n)^k$. Since the compliment of the 'diagonals' in $(S^n)^k$ is connected (assuming that $n>1$), it would follow that there is only one such orbit. | |
| May 19, 2013 at 11:30 | comment | added | Robert Bryant | @Misha: I don't know the answer to your question above about $k$-transitivity for this latter (non-Lie) group, but, like you, I suspect that it is $k$-transitive for all $k$. Aside from the obvious observation that this holds for $k\le3$, I don't have a proof in hand, but, perhaps a proof could be devised along the following lines: Let ${\frak{l}}_0={\frak{so}}(n{+}1,1)+{\frak{sl}}(n{+}1)$ be the (finite-dimensional) space of vector fields on $S^n$ belonging to the above two Lie transformation groups and, for $i\ge0$, define ${\frak{l}}_{i+1}={\frak{l}}_i+[{\frak{l}}_i,{\frak{l}}_i]$. (cont) | |
| May 18, 2013 at 13:30 | comment | added | Robert Bryant | @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 $\mathrm{SL}(n{+}1)$? | |
| May 18, 2013 at 8:24 | comment | added | Ryan Budney | Ah, right. I got really off-track there. | |
| May 17, 2013 at 22:56 | comment | added | Misha | 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. | |
| May 17, 2013 at 22:14 | comment | added | Benoît Kloeckner | @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. | |
| May 17, 2013 at 22:00 | comment | added | Ryan Budney | 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 $\mathbb R^n$. | |
| May 17, 2013 at 21:50 | history | answered | Robert Bryant | CC BY-SA 3.0 |