[math-fun] Something I figured out
For 2 x 2 matrix M define F(M) = f : R->R : x => (M11 x + M12) / (M21 x + M22). Then F(AB) = F(A) o F(B) So the composition of unreduced order 1 rational functions is isomorphic to the product of 2x2 matrices. I assume this is well known, but I thought it was pretty cool. This means that finding, say, all order 1 rational functions f with f(f(x)) = x would reduce to finding all 2x2 matrices M with M^2 = I.
I agree, it's very cool. It features prominently in the study of the modular group (and therefore all things continued fraction). On Thu, Feb 11, 2016 at 5:42 AM, David Wilson <davidwwilson@comcast.net> wrote:
For 2 x 2 matrix M define
F(M) = f : R->R : x => (M11 x + M12) / (M21 x + M22).
Then
F(AB) = F(A) o F(B)
So the composition of unreduced order 1 rational functions is isomorphic to the product of 2x2 matrices.
I assume this is well known, but I thought it was pretty cool.
This means that finding, say, all order 1 rational functions f with
f(f(x)) = x
would reduce to finding all 2x2 matrices M with
M^2 = I.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Yeah, amazing fact, isn't it? This is the action of 2x2 matrices on the set of lines in the plane through the origin, where you identify a line with its intersection with the line y=1. That is, let your matrix act on the point (x,1), taking it to (ax+b, cx+d), and then project back onto the line y=1 to get (ax+b/cx+d, 1). As Mike says, this way lies modular functions on the upper half plane. --Michael On Thu, Feb 11, 2016 at 10:31 AM, Mike Stay <metaweta@gmail.com> wrote:
I agree, it's very cool. It features prominently in the study of the modular group (and therefore all things continued fraction).
On Thu, Feb 11, 2016 at 5:42 AM, David Wilson <davidwwilson@comcast.net> wrote:
For 2 x 2 matrix M define
F(M) = f : R->R : x => (M11 x + M12) / (M21 x + M22).
Then
F(AB) = F(A) o F(B)
So the composition of unreduced order 1 rational functions is isomorphic to the product of 2x2 matrices.
I assume this is well known, but I thought it was pretty cool.
This means that finding, say, all order 1 rational functions f with
f(f(x)) = x
would reduce to finding all 2x2 matrices M with
M^2 = I.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
Yes, this is very interesting. If the function (az + b) / (cz + d) have real coefficients a, b, c, d, and z is allowed to be complex, these linear fractional transformations (LFTs) are (as alluded to below) precisely the conformal bijections of the upper half plane (as long as ad - bd is nonzero). Call this group of LFTs by Aut(H). They're almost exactly the group of SL(2,R) of 2x2 real matrices with determinant = 1, except that negating all the matrix elements gives the same LFT. And so: Aut(H) = SL(2,R) / {I, I} If instead the a, b, c, d can be complex and again ad - bc is nonzero, these LFTs are precisely the conformal automorphisms of the Riemann sphere S^2 = C u {oo}, and we have likewise Aut(S^2) = SL(2,C) / {I, -I} * * * As for the compositional square-root function of an LFT, one can also ask whether a nonconstant LFT L: S^2 —> S^2 lies in a one-parameter subgroup of Aut(S^2), i.e., a family of LFTs of the form L_t: S^2 —> S^2 such that L_0 = the identity, L_1 = L, and L_s o L_t = L_(s+t) for all real s, t. Then of course L_(1/2) o L_(1/2) = L gives a compositional square-root function for L. In fact all invertible LFTs do lie in such subgroups, and it's fun to work out the explicit formulas for them. Each such one-parameter subgroup is generated by a vector field defined by (d/dt L_t)(0), i.e., the one-parameter subgroup is the flow of the vector field. It turns out that the vector fields obtained this way from SL(2,C) / {I, -I} are the nonzero quadratic polynomial vector fields on C u {oo}, i.e., of the form V(z) = a z^2 + b z + c —Dan
On Feb 11, 2016, at 7:48 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Yeah, amazing fact, isn't it? This is the action of 2x2 matrices on the set of lines in the plane through the origin, where you identify a line with its intersection with the line y=1. That is, let your matrix act on the point (x,1), taking it to (ax+b, cx+d), and then project back onto the line y=1 to get (ax+b/cx+d, 1).
As Mike says, this way lies modular functions on the upper half plane.
--Michael
On Thu, Feb 11, 2016 at 10:31 AM, Mike Stay <metaweta@gmail.com> wrote:
I agree, it's very cool. It features prominently in the study of the modular group (and therefore all things continued fraction).
On Thu, Feb 11, 2016 at 5:42 AM, David Wilson <davidwwilson@comcast.net> wrote:
For 2 x 2 matrix M define
F(M) = f : R->R : x => (M11 x + M12) / (M21 x + M22).
Then
F(AB) = F(A) o F(B)
So the composition of unreduced order 1 rational functions is isomorphic to the product of 2x2 matrices.
I assume this is well known, but I thought it was pretty cool.
This means that finding, say, all order 1 rational functions f with
f(f(x)) = x
would reduce to finding all 2x2 matrices M with
M^2 = I.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If you find this topic interesting, you should have a look at Lester R. Ford's "Automorphic Functions", published by AMS Chelsea. I might say that this is one of my all-time favorite books. -- Gene From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, February 11, 2016 10:46 AM Subject: Re: [math-fun] Something I figured out Yes, this is very interesting. If the function (az + b) / (cz + d) have real coefficients a, b, c, d, and z is allowed to be complex, these linear fractional transformations (LFTs) are (as alluded to below) precisely the conformal bijections of the upper half plane (as long as ad - bd is nonzero). Call this group of LFTs by Aut(H). They're almost exactly the group of SL(2,R) of 2x2 real matrices with determinant = 1, except that negating all the matrix elements gives the same LFT. And so: Aut(H) = SL(2,R) / {I, I} If instead the a, b, c, d can be complex and again ad - bc is nonzero, these LFTs are precisely the conformal automorphisms of the Riemann sphere S^2 = C u {oo}, and we have likewise Aut(S^2) = SL(2,C) / {I, -I} * * * As for the compositional square-root function of an LFT, one can also ask whether a nonconstant LFT L: S^2 —> S^2 lies in a one-parameter subgroup of Aut(S^2), i.e., a family of LFTs of the form L_t: S^2 —> S^2 such that L_0 = the identity, L_1 = L, and L_s o L_t = L_(s+t) for all real s, t. Then of course L_(1/2) o L_(1/2) = L gives a compositional square-root function for L. In fact all invertible LFTs do lie in such subgroups, and it's fun to work out the explicit formulas for them. Each such one-parameter subgroup is generated by a vector field defined by (d/dt L_t)(0), i.e., the one-parameter subgroup is the flow of the vector field. It turns out that the vector fields obtained this way from SL(2,C) / {I, -I} are the nonzero quadratic polynomial vector fields on C u {oo}, i.e., of the form V(z) = a z^2 + b z + c —Dan
On Feb 11, 2016, at 7:48 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Yeah, amazing fact, isn't it? This is the action of 2x2 matrices on the set of lines in the plane through the origin, where you identify a line with its intersection with the line y=1. That is, let your matrix act on the point (x,1), taking it to (ax+b, cx+d), and then project back onto the line y=1 to get (ax+b/cx+d, 1).
As Mike says, this way lies modular functions on the upper half plane.
--Michael
On Thu, Feb 11, 2016 at 10:31 AM, Mike Stay <metaweta@gmail.com> wrote:
I agree, it's very cool. It features prominently in the study of the modular group (and therefore all things continued fraction).
On Thu, Feb 11, 2016 at 5:42 AM, David Wilson <davidwwilson@comcast.net> wrote:
For 2 x 2 matrix M define
F(M) = f : R->R : x => (M11 x + M12) / (M21 x + M22).
Then
F(AB) = F(A) o F(B)
So the composition of unreduced order 1 rational functions is isomorphic to the product of 2x2 matrices.
I assume this is well known, but I thought it was pretty cool.
This means that finding, say, all order 1 rational functions f with
f(f(x)) = x
would reduce to finding all 2x2 matrices M with
M^2 = I.
participants (5)
-
Dan Asimov -
David Wilson -
Eugene Salamin -
Michael Kleber -
Mike Stay