[math-fun] Stupid question about geometrical transformations
What is generated by the union of projective and conformal (Moebius) groups? Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it? How should such transformations be represented for computational purposes? Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...] Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode. [Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.] Fred Lunnon
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n . One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity. WFL On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
It's infinite-dimensional. We can certainly represent these things as maps of the form: (x, y, z) --> (f(x,y,z), g(x,y,z), h(x,y,z)) [for two dimensions; generalise as appropriate] where f, g and h are homogeneous rational functions, where deg(numerator) = deg(denominator) + 1.
Sent: Thursday, August 28, 2014 at 8:34 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
"represent these things" seems further to imply that the correspondence goes both ways, or at least that the degree of such functions is unbounded. Is that the case? Where does one learn about such matters? WFL On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
It's infinite-dimensional.
We can certainly represent these things as maps of the form:
(x, y, z) --> (f(x,y,z), g(x,y,z), h(x,y,z)) [for two dimensions; generalise as appropriate]
where f, g and h are homogeneous rational functions, where deg(numerator) = deg(denominator) + 1.
Sent: Thursday, August 28, 2014 at 8:34 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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Yes, the degree is unbounded. Let's say that an algebraic curve has 'crossing number' k if we can find a straight line which intersects it at K distinct points, where K is finite and K >= k. In particular, by Bezout's theorem, crossing number <= degree. Now, I will describe a composition of transformations capable of turning a curve C of crossing number k into a curve of crossing number 2k. 1. Find a line L1 which intersects C in (at least) k distinct points. 2. Find another such line L2 with that property, which is not parallel to L1 (e.g. by slightly perturbing L1). 3. Slightly perturb this pair of lines to produce a hyperbola H which intersects C in at least 2k distinct points. 4. Apply a projective transformation to turn H into a circle H'. 5. Apply a Möbius transformation to turn H' into a line H''. The image of C under this pair of maps thus has crossing number 2k. Consequently, we can 'boost' the degree of an algebraic curve arbitrarily high using projective transformations and Möbius maps. Sincerely, Adam P. Goucher
Sent: Thursday, August 28, 2014 at 11:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
"represent these things" seems further to imply that the correspondence goes both ways, or at least that the degree of such functions is unbounded.
Is that the case? Where does one learn about such matters? WFL
On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
It's infinite-dimensional.
We can certainly represent these things as maps of the form:
(x, y, z) --> (f(x,y,z), g(x,y,z), h(x,y,z)) [for two dimensions; generalise as appropriate]
where f, g and h are homogeneous rational functions, where deg(numerator) = deg(denominator) + 1.
Sent: Thursday, August 28, 2014 at 8:34 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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That's a very neat construction! WFL On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Yes, the degree is unbounded.
Let's say that an algebraic curve has 'crossing number' k if we can find a straight line which intersects it at K distinct points, where K is finite and K >= k.
In particular, by Bezout's theorem, crossing number <= degree.
Now, I will describe a composition of transformations capable of turning a curve C of crossing number k into a curve of crossing number 2k.
1. Find a line L1 which intersects C in (at least) k distinct points.
2. Find another such line L2 with that property, which is not parallel to L1 (e.g. by slightly perturbing L1).
3. Slightly perturb this pair of lines to produce a hyperbola H which intersects C in at least 2k distinct points.
4. Apply a projective transformation to turn H into a circle H'.
5. Apply a Möbius transformation to turn H' into a line H''.
The image of C under this pair of maps thus has crossing number 2k.
Consequently, we can 'boost' the degree of an algebraic curve arbitrarily high using projective transformations and Möbius maps.
Sincerely,
Adam P. Goucher
Sent: Thursday, August 28, 2014 at 11:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
"represent these things" seems further to imply that the correspondence goes both ways, or at least that the degree of such functions is unbounded.
Is that the case? Where does one learn about such matters? WFL
On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
It's infinite-dimensional.
We can certainly represent these things as maps of the form:
(x, y, z) --> (f(x,y,z), g(x,y,z), h(x,y,z)) [for two dimensions; generalise as appropriate]
where f, g and h are homogeneous rational functions, where deg(numerator) = deg(denominator) + 1.
Sent: Thursday, August 28, 2014 at 8:34 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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Have you any you (or has anybody else) a handle on what Bertram calls "Jordan theory", apparently some kind of algebra for representing such transformations? It looks powerful stuff, but I'd run out of puff by page 3 ! See eg. http://molle.fernuni-hagen.de/~loos/jordan/archive/gpg1/index.html http://molle.fernuni-hagen.de/~loos/jordan/archive/gpg2/index.html WFL On 8/29/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
That's a very neat construction! WFL
On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Yes, the degree is unbounded.
Let's say that an algebraic curve has 'crossing number' k if we can find a straight line which intersects it at K distinct points, where K is finite and K >= k.
In particular, by Bezout's theorem, crossing number <= degree.
Now, I will describe a composition of transformations capable of turning a curve C of crossing number k into a curve of crossing number 2k.
1. Find a line L1 which intersects C in (at least) k distinct points.
2. Find another such line L2 with that property, which is not parallel to L1 (e.g. by slightly perturbing L1).
3. Slightly perturb this pair of lines to produce a hyperbola H which intersects C in at least 2k distinct points.
4. Apply a projective transformation to turn H into a circle H'.
5. Apply a Möbius transformation to turn H' into a line H''.
The image of C under this pair of maps thus has crossing number 2k.
Consequently, we can 'boost' the degree of an algebraic curve arbitrarily high using projective transformations and Möbius maps.
Sincerely,
Adam P. Goucher
Sent: Thursday, August 28, 2014 at 11:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
"represent these things" seems further to imply that the correspondence goes both ways, or at least that the degree of such functions is unbounded.
Is that the case? Where does one learn about such matters? WFL
On 8/28/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
It's infinite-dimensional.
We can certainly represent these things as maps of the form:
(x, y, z) --> (f(x,y,z), g(x,y,z), h(x,y,z)) [for two dimensions; generalise as appropriate]
where f, g and h are homogeneous rational functions, where deg(numerator) = deg(denominator) + 1.
Sent: Thursday, August 28, 2014 at 8:34 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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I don't think incompatible compactifications is a problem. The conformal group on n dimensions is generated by inversions in (n-1)-spheres. It naturally lives on the n-sphere S^n. The projective group on n+1 dimensions naturally lives on projective space P^n, by the action of linear transformations on lines through the origin in R^(n+1). But by just looking at the action on rays instead of lines (or equivalently, taking the double cover of P^n), we get the projective group acting on S^n also. So, both the projective group and the conformal group act naturally on S^n. Is Adam's answer compatible with this point of view? I'm not sure I understand it. --Dan On Aug 28, 2014, at 12:34 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On reflection I agree with this. In fact, the algebra I favour for Euclidean geometry in 3-space is the Clifford Cl(3, 0, 1) , which puzzled me initially by representing a finite point in two ways --- the "doppelganger" arising from the polar opposite on the Riemann sphere. Adam shows how to construct algebraic curves of arbitrary degree via sequences of projectivities and Moebius transformations acting on an initial line. I suppose that strictly speaking it doesn't immediately follow that one can construct sufficiently many to force an unbounded dimension ... More pondering may be required? WFL On 8/29/14, Dan Asimov <dasimov@earthlink.net> wrote:
I don't think incompatible compactifications is a problem.
The conformal group on n dimensions is generated by inversions in (n-1)-spheres. It naturally lives on the n-sphere S^n.
The projective group on n+1 dimensions naturally lives on projective space P^n, by the action of linear transformations on lines through the origin in R^(n+1).
But by just looking at the action on rays instead of lines (or equivalently, taking the double cover of P^n), we get the projective group acting on S^n also.
So, both the projective group and the conformal group act naturally on S^n.
Is Adam's answer compatible with this point of view? I'm not sure I understand it.
--Dan
On Aug 28, 2014, at 12:34 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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Yes, but they differ in how we associate (some compactification of) R^n with (some quotient of) S^n. There is, however, a rather nice fact: Let G be the group of projective transformations of R^(n+1) which preserve S^n. Let H be the group of conformal transformations of R^(n+1) which preserve S^n. Then G and H are isomorphic, and we can choose the isomorphism such that they act on S^n in the same way.
Sent: Friday, August 29, 2014 at 12:31 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
I don't think incompatible compactifications is a problem.
The conformal group on n dimensions is generated by inversions in (n-1)-spheres. It naturally lives on the n-sphere S^n.
The projective group on n+1 dimensions naturally lives on projective space P^n, by the action of linear transformations on lines through the origin in R^(n+1).
But by just looking at the action on rays instead of lines (or equivalently, taking the double cover of P^n), we get the projective group acting on S^n also.
So, both the projective group and the conformal group act naturally on S^n.
Is Adam's answer compatible with this point of view? I'm not sure I understand it.
--Dan
On Aug 28, 2014, at 12:34 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Oops --- should read (n+1)^2-1 + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + 3n .
One immediate difficulty is incompatible compactifications --- a hyperplane versus a single point at infinity.
WFL
On 8/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What is generated by the union of projective and conformal (Moebius) groups?
Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it?
How should such transformations be represented for computational purposes?
Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...]
Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode.
[Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.]
Fred Lunnon
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On Aug 29, 2014, at 10:47 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Yes, but they differ in how we associate (some compactification of) R^n with (some quotient of) S^n.
Not a problem. The projective group and the conformal group both *naturally* reside on (the round) S^n. That's where they begin life. So both groups are naturally subgroups of the diffeomorphism group Diffeo^w(S^n) (here w indicates that the diffeomorphism are real analytic). Hence elements of both groups can be naturally composed with each other So it makes perfect sense to ask what subgroup of Diffeo^w(S^n) is generated by the two groups together -- as Fred asked. Neither group naturally resides on R^n. That is, some -- in fact almost all -- group elements of either group on S^n fail to take R^n to itself. To get the effect of any element g of either group on R^n, we just restrict it to S^n - {oo}. But then we also have to be careful to also remove any point taken to oo by g. (When we do this, we identify S^n - {oo} with R^n by inverse stereographic projection, which is conformal.) Which means neither group acts honestly on R^n in the first place. The best we can do is let them be pseudogroups on R^n. (We can always restrict either group to those transformations on S^n that carry oo to itself, and those elements do form a group on R^n.)
There is, however, a rather nice fact:
Let G be the group of projective transformations of R^(n+1) which preserve S^n. Let H be the group of conformal transformations of R^(n+1) which preserve S^n.
Then G and H are isomorphic, and we can choose the isomorphism such that they act on S^n in the same way.
I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known. --Dan
On 8/29/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
There is, however, a rather nice fact:
Let G be the group of projective transformations of R^(n+1) which preserve S^n. Let H be the group of conformal transformations of R^(n+1) which preserve S^n.
Then G and H are isomorphic, and we can choose the isomorphism such that they act on S^n in the same way.
G ~ H is an interesting result --- do you have a reference? Your H is just the Moebius(n) group, so it would follow that Moebius(n) is a subgroup of Project(n+1) ; as is Project(n) of course, though acting on a plane rather than a sphere, so again confounding my ambitions for a finite-dimensional representation of the joint conformal-projective group. Also I ran a simulation generating random 2-space transformations of a fixed point-set which confirms that the dimension exceeds 24 . I've been skimming through a book found online: "A Taste of Jordan Algebras" Kevin McCrimmon (2004) http://math.nsc.ru/LBRT/a1/files/mccrimmon.pdf which proceeds at a considerably more leisurely pace than Bertram's papers. But all the examples so far seem to be of orthogonal/Hermitian groups ... Dan wrote << I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known.
But (qua topological groups) their dimensions are respectively (n+2)(n+1)/2 versus (n+1)^2 - 1 ?!
Fred Lunnon
Let G be the group of projective transformations of R^(n+1) which preserve S^n. Let H be the group of conformal transformations of R^(n+1) which preserve S^n.
Then G and H are isomorphic, and we can choose the isomorphism such that they act on S^n in the same way.
G ~ H is an interesting result --- do you have a reference?
Yes, they're isometries of different models of the same hyperbolic (n+1)-space: G: consider the Beltrami-Klein model. H: consider the Poincare disc model. Lorentz group SO+(n+1,1): consider the hyperboloid model. (You can probably find references for each of these.)
From this, you get isomorphisms such as:
Lorentz group SO+(3,1) ~ Möbius group PSL(2,C) which is intimately entangled with the fact that relativistic boosts cause your celestial sphere to undergo a conformal deformation (stellar aberration).
Your H is just the Moebius(n) group, so it would follow that Moebius(n) is a subgroup of Project(n+1) ; as is Project(n) of course, though acting on a plane rather than a sphere, so again confounding my ambitions for a finite-dimensional representation of the joint conformal-projective group.
Also I ran a simulation generating random 2-space transformations of a fixed point-set which confirms that the dimension exceeds 24 .
I've been skimming through a book found online: "A Taste of Jordan Algebras" Kevin McCrimmon (2004) http://math.nsc.ru/LBRT/a1/files/mccrimmon.pdf which proceeds at a considerably more leisurely pace than Bertram's papers. But all the examples so far seem to be of orthogonal/Hermitian groups ...
Dan wrote << I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known.
But (qua topological groups) their dimensions are respectively (n+2)(n+1)/2 versus (n+1)^2 - 1 ?!
Fred Lunnon
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Well, yes. The projective group PGL(n+1) acting on P^n or S^n has dimension dim(PGL(n+1)) = n^2 + 2n and the conformal group acting on S^n has dimension dim(Conf(S^n)) = (n+1)(n+2)/2 = (1/2)n^2 + (3/2)n + 1 -- just as Fred said. And, any element g of the projective group on S^n satisfies g(-p) = -g(p) for any point p of S^n, whereas that is not true for most elements of the conformal group on S^n. In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where Conf(S^2) = Aut(S^2) = PSL(2,C), the holomorphic automorphism group of S^2. --Dan On Aug 29, 2014, at 5:53 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Dan wrote << I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known.
But (qua topological groups) their dimensions are respectively (n+2)(n+1)/2 versus (n+1)^2 - 1 ?!
Fred Lunnon
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail. The group SO(n) of rotations is considerably smaller.
--Dan
On Aug 29, 2014, at 5:53 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Dan wrote << I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known.
But (qua topological groups) their dimensions are respectively (n+2)(n+1)/2 versus (n+1)^2 - 1 ?!
Fred Lunnon
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Adam, I thought you referred to elements of PGL(n+1) that *preserve* S^n, which is not the same thing. --Dan On Aug 30, 2014, at 1:52 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail.
The group SO(n) of rotations is considerably smaller.
I meant `fix as a whole' rather than `fix pointwise'.
Sent: Saturday, August 30, 2014 at 6:32 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Adam, I thought you referred to elements of PGL(n+1) that *preserve* S^n, which is not the same thing.
--Dan
On Aug 30, 2014, at 1:52 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail.
The group SO(n) of rotations is considerably smaller.
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Yes -- I think you used the word "preserve", which means exactly what you just said -- so I think what you said was clear. Maybe we are using different terminology. Because with the terms I'm using, the group Conf(S^n) is generated by the inversions in circles (okay, even numbers of inversions in circles). The orientation preserving projective group on S^n would be PGL+(n+1,R), which just permutes the lines through the origin in R^(n+1) and so induces a map of S^n to itself. But (take S^2 for instance, where Conf(S^2) == PSL(2,C)) most elements of Conf(S^n) do not take antipodal pairs to antipodal pairs, whereas elements of PGL(n+1,R) always take antipodal pairs to antipodal pairs. With this view, I don't understand: -----
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n).
--Dan On Aug 31, 2014, at 9:32 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
I meant `fix as a whole' rather than `fix pointwise'.
Sent: Saturday, August 30, 2014 at 6:32 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Adam, I thought you referred to elements of PGL(n+1) that *preserve* S^n, which is not the same thing.
--Dan
On Aug 30, 2014, at 1:52 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail.
The group SO(n) of rotations is considerably smaller.
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Correction: On Aug 31, 2014, at 10:20 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Yes -- I think you used the word "preserve", which means exactly what you just said -- so I think what you said was clear.
Maybe we are using different terminology.
The next sentence should have said "(n-1)-spheres", not "circles":
Because with the terms I'm using, the group Conf(S^n) is generated by the inversions in circles (n-1)-spheres (okay, even numbers of inversions in circles (n-1)-spheres).
The orientation preserving projective group on S^n would be PGL+(n+1,R), which just permutes the lines through the origin in R^(n+1) and so induces a map of S^n to itself.
But (take S^2 for instance, where Conf(S^2) == PSL(2,C)) most elements of Conf(S^n) do not take antipodal pairs to antipodal pairs, whereas elements of PGL(n+1,R) always take antipodal pairs to antipodal pairs.
With this view, I don't understand:
-----
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n).
--Dan
You're describing PGL(n, R), not PGL(n+1, R). I was talking about transformations of real projective (n+1)-space, in which S^n is the unit sphere. You were talking about transformations of real projective n-space, in which S^n is the double cover of the entire space RP^n. Apologies for the misunderstanding.
Sent: Sunday, August 31, 2014 at 6:20 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Yes -- I think you used the word "preserve", which means exactly what you just said -- so I think what you said was clear.
Maybe we are using different terminology.
Because with the terms I'm using, the group Conf(S^n) is generated by the inversions in circles (okay, even numbers of inversions in circles).
The orientation preserving projective group on S^n would be PGL+(n+1,R), which just permutes the lines through the origin in R^(n+1) and so induces a map of S^n to itself.
But (take S^2 for instance, where Conf(S^2) == PSL(2,C)) most elements of Conf(S^n) do not take antipodal pairs to antipodal pairs, whereas elements of PGL(n+1,R) always take antipodal pairs to antipodal pairs.
With this view, I don't understand:
-----
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n).
--Dan
On Aug 31, 2014, at 9:32 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
I meant `fix as a whole' rather than `fix pointwise'.
Sent: Saturday, August 30, 2014 at 6:32 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Adam, I thought you referred to elements of PGL(n+1) that *preserve* S^n, which is not the same thing.
--Dan
On Aug 30, 2014, at 1:52 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail.
The group SO(n) of rotations is considerably smaller.
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Okay, I hope we've cleared that up. But as I understand the notation (as in < http://en.wikipedia.org/wiki/Projective_linear_group >), PGL(n+1, R) denotes the group of linear transformations of R^(n+1) factored out by the scalars {cI | c in R}, which is the very group I meant acting on 1-dimensional subspaces of R^(n+1) and thereby inducing an action on S^n [sic]. (Which action is indeed the double cover of the action of PGL(n+1,R) on P^n.) Its dimension would be dim(GL(n+1, R)) - 1 = (n+1)^2 - 1 = n^2 + 2n, as Fred has mentioned. Or restricting to orientation-preserving linear maps, we have PGL+(n+1,R) with the same dimension, and more reasonable to compare with Conf(S^n). If we're only considering orientation-preserving conformal maps of S^n, then we can ignore spherical inversions and just say every such map can be described as a composition of rotations, dilatations, and translations on R^n (corresponding to S^n - {pt} by a stereographic projection). --Dan On Aug 31, 2014, at 11:03 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
You're describing PGL(n, R), not PGL(n+1, R).
I was talking about transformations of real projective (n+1)-space, in which S^n is the unit sphere.
You were talking about transformations of real projective n-space, in which S^n is the double cover of the entire space RP^n.
Apologies for the misunderstanding.
Sent: Sunday, August 31, 2014 at 6:20 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Yes -- I think you used the word "preserve", which means exactly what you just said -- so I think what you said was clear.
Maybe we are using different terminology.
Because with the terms I'm using, the group Conf(S^n) is generated by the inversions in circles (okay, even numbers of inversions in circles).
The orientation preserving projective group on S^n would be PGL+(n+1,R), which just permutes the lines through the origin in R^(n+1) and so induces a map of S^n to itself.
But (take S^2 for instance, where Conf(S^2) == PSL(2,C)) most elements of Conf(S^n) do not take antipodal pairs to antipodal pairs, whereas elements of PGL(n+1,R) always take antipodal pairs to antipodal pairs.
With this view, I don't understand:
-----
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n).
--Dan
On Aug 31, 2014, at 9:32 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
I meant `fix as a whole' rather than `fix pointwise'.
Sent: Saturday, August 30, 2014 at 6:32 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Stupid question about geometrical transformations
Adam, I thought you referred to elements of PGL(n+1) that *preserve* S^n, which is not the same thing.
--Dan
On Aug 30, 2014, at 1:52 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, I wonder if the elements of the conformal group Conf(S^n) that happen to also be elements of [the projective group PLG(n+1) acting on S^n] are just the rotations. (This is certainly true for n = 2, where
Conf(S^2) = Aut(S^2) = PSL(2,C),
the holomorphic automorphism group of S^2.
The [orientation-preserving] elements of the projective group PSL(n+1) acting on S^n are precisely the elements of the conformal group Conf(S^n), as I mentioned in my previous e-mail.
The group SO(n) of rotations is considerably smaller.
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We can all relax --- the universe is saved. For a while longer, at any rate. I can after all demonstrate Mob(n) as a subgroup of Lag(n+1) , using an unexpected geometrical twist in the construction. Starting with (say) Mob(2) acting on circles on the unit sphere in 3-space in the natural fashion, expand the circles to spheres orthogonal to the unit sphere --- these spheres are unoriented, and can be identified with Möbius inversions. Now apply the reflector (a properly Lie-sphere involution) interchanging Möbius inversion with Laguerre eversion (turning oriented spheres inside-out). The result is the subgroup of Lag(3) having reflector base planes meeting the origin; and adjoining Euc(3) translations yields Lag(3) as required. I wish I could show the proof of this --- it is still very neat (4 lines) --- with the algebra verified by computer (this time around). I might try putting up a crash-course explaining the background; however public reception of such attempts in the past has been less than ecstatic, to put it mildly ... It would be easy to overlook the extra involution in the absence of a common framework (the Lie-sphere group) in which to embed everything formally. I wonder if it is actually known to physicists, and if so what possible physical significance it might have? WFL << Unfortunately, another computer has pointed out that my "trivial proof" of Mob(n) + translations = Lag(n+1) is bunkum --- looks like my GA got a little rusty. Instead, it seems obvious that the natural embedding with Mob(2) acting on the unit sphere could not possibly work; from Adam's assertion that Poincaré(?) restricted to the unit sphere equals Möbius, it would follow that Poincaré does NOT equal Laguerre! So for me at any rate, it's back to square one on the relation between these groups when n = 2 . It might help if I had some clue what these people do with them once they're properly nailed down: eg. what group element corresponds to a change in velocity from rest to uniform along the x-axis? In the meantime, if you happen to notice the stars going out one by one tonight, you'll know what happened ...
participants (3)
-
Adam P. Goucher -
Dan Asimov -
Fred Lunnon