View Full Version : Spinors, 4-pi rotation & orientation in topology?
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Having obtained no reply from *.particle, advancedphysics.org, nor\neven a google search, can I try here please?\n\n\nAfter reading some elementary topology, can anyone tell me whether the\n4-pi rotation for the identity of spinors is (closely?) related to the\nnon-orientability of the projective plane? I can see that it takes a\n4-pi\ntransit of a Moebius band to regain the original orientation of a\nchiral object.\n\nI\'ve satisfied (& puzzled) myself that the orientation-entanglement\nrelation in real 3space (eg in MTW, p. 1148-9) shows that a 4-pi\nrotation can be required for identity in E3, but I\'m less clear on why\nit should apply to the complex projective plane (for electron spin\nstates etc.), which (in my ignorance of advanced math) I would expect\nto be orientable. Any (simply explained) advice would be much\nappreciated.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Having obtained no reply from *.particle, advancedphysics.org, nor
even a google search, can I try here please?
After reading some elementary topology, can anyone tell me whether the
4-\pi rotation for the identity of spinors is (closely?) related to the
non-orientability of the projective plane? I can see that it takes a
4-\pi
transit of a Moebius band to regain the original orientation of a
chiral object.
I've satisfied (& puzzled) myself that the orientation-entanglement
relation in real 3space (eg in MTW, p. 1148-9) shows that a 4-\pi
rotation can be required for identity in E3, but I'm less clear on why
it should apply to the complex projective plane (for electron spin
states etc.), which (in my ignorance of advanced math) I would expect
to be orientable. Any (simply explained) advice would be much
appreciated.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>pellis@london.edu (Paul E) wrote in message news:<659defa1.0406220505.3540c9d2@posting.google. com>...\n> Having obtained no reply from *.particle, advancedphysics.org, nor\n> even a google search, can I try here please?\n>\n>\n> After reading some elementary topology, can anyone tell me whether the\n> 4-pi rotation for the identity of spinors is (closely?) related to the\n> non-orientability of the projective plane? I can see that it takes a\n> 4-pi\n> transit of a Moebius band to regain the original orientation of a\n> chiral object.\n>\n> I\'ve satisfied (& puzzled) myself that the orientation-entanglement\n> relation in real 3space (eg in MTW, p. 1148-9) shows that a 4-pi\n> rotation can be required for identity in E3, but I\'m less clear on why\n> it should apply to the complex projective plane (for electron spin\n> states etc.), which (in my ignorance of advanced math) I would expect\n> to be orientable. Any (simply explained) advice would be much\n> appreciated.\n\nHaving now received a couple of answers from sci.math, I should\nclarify that the "4-pi transit of the Moebius band" was with respect\nto an embedding space. And I\'d still be interested in any further\nanswers from the physics, as distinct from the math, perspective.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>pellis@london.edu (Paul E) wrote in message news:<659defa1.0406220505.3540c9d2@posting.google.com>...
> Having obtained no reply from *.particle, advancedphysics.org, nor
> even a google search, can I try here please?
>
>
> After reading some elementary topology, can anyone tell me whether the
> 4-\pi rotation for the identity of spinors is (closely?) related to the
> non-orientability of the projective plane? I can see that it takes a
> 4-\pi
> transit of a Moebius band to regain the original orientation of a
> chiral object.
>
> I've satisfied (& puzzled) myself that the orientation-entanglement
> relation in real 3space (eg in MTW, p. 1148-9) shows that a 4-\pi
> rotation can be required for identity in E3, but I'm less clear on why
> it should apply to the complex projective plane (for electron spin
> states etc.), which (in my ignorance of advanced math) I would expect
> to be orientable. Any (simply explained) advice would be much
> appreciated.
Having now received a couple of answers from sci.math, I should
clarify that the "4-\pi transit of the Moebius band" was with respect
to an embedding space. And I'd still be interested in any further
answers from the physics, as distinct from the math, perspective.
tessel@tum.bot
Jul2-04, 04:32 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Sun, 27 Jun 2004, Paul E wrote:\n\n> can anyone tell me whether the 4-pi rotation for the identity of spinors\n> is (closely?) related to the non-orientability of the projective plane?\n> I can see that it takes a 4-pi transit of a Moebius band to regain the\n> original orientation of a chiral object.\n\nI wish I could draw a picture here, but my short answer---not often can I\nsay this, alas--- is YES!\n\nCut out a strip of paper ruled down the center like so:\n__________________\n| |\n|__________________|\n| |\n|__________________|\n\nMark the central ruling 0 (left hand endpoint), pi (midpoint), 2 pi (right\nhand endpoint). Wrap up the strip a la Moebius. Note that the ruling is\nnow a circle; call it "the central circle". Now try to draw a a small\nlinesegment parallel to the central circle. Extend it carefully; you\nshould get another circle, but with periodicity 4 pi! This is indeed\n-exactly- what spinors are all about. Notice the "natural projection" is\ndefined in the strip; the embedding in E^3 is actually irrelevant for most\npurposes.\n\nThe picture I wish I could draw is the E(2) version of the SO(3) picture\nof Olinde Rodrigues. Instead I can only give some directions which leave\na lot up to your own initiative/insight:\n\n1. Read a book like Coxeter, Geometry, to learn\nthe basic ideas of transformation geometry, in particular\n\n(a) iterating the reflections in two parallel lines in the plane gives a\ntranslation in that plane,\n\n(b) iterating reflections in two intersecting lines gives a rotation in\nthat plane, about the point of intersection.\n\n2. Find and study the discussion of quaternions in MTW, Gravitation,\nFreeman, 1973, in particular the double cover SU(2) -->> SO(3).\n\n3. Find and study old posts here on the E(2) of this. Here, SO(3) is the\nthree-dimensional Lie group of isometries of S^2; E(3) is the\nthree-dimensional Lie group of isometries of E^2. And of course there is\nSO(1,2), the three-dimensional Lie group of isometries of H^2.\n\n4. Find and study\n\nauthor = {Roger Carter and Graeme Segal and Ian MacDonald},\ntitle = {Lectures on {L}ie Groups and {L}ie Algebras},\nseries = {London Mathematical Society Student Texts},\nvolume = 32,\npublisher = {Cambridge University Press},\nyear = 1995}\n\nto learn a bit more about spinors.\n\n5. Find and study many highly relevant posts here by John Baez (This Week\nseries) over the past ten years.\n\n> I\'ve satisfied (& puzzled) myself that the orientation-entanglement\n> relation in real 3space (eg in MTW, p. 1148-9)\n\nOh! You\'ve already done (2)! Hurrah!!!! Like any stimulating reading,\nthis obviously left you hungry for more. If you liked (2) I am sure\nyou\'ll enjoy following my other suggestions.\n\n(Hmm... OK, the embedding in E^3 is probably not irrelevant for "Dirac\'s\nspanner", but never mind that for the moment...)\n\n> shows that a 4-pi\n> rotation can be required for identity in E3, but I\'m less clear on why\n> it should apply to the complex projective plane (for electron spin\n> states etc.), which (in my ignorance of advanced math) I would expect\n> to be orientable.\n\nOh--- so you want to learn -physics-?! :-/\n\nWell, I\'ll let others address your real question, then, but to finish the\nthought: I guess you know that the Moebius strip is nonorientable. So is\nRP^2. And by "sewing a disk around the boundary" of the M. strip you\nget--- RP^2! But the point is that a picture of the double cover circle\n---> central circle describe above fits inside a picture of the double\ncover SU(2) --> SO(3). Here, SO(3) is both nonorientable and\nnon-simply-connected, while SU(2) is both orientable and simply-connected,\nin fact, quite remarkably, SU(2) is homeomorphic to S^3! That\'s the\nconnection between the spinorial representation of rotations (about the\norigin) in E^3 and nonorientable manifolds.\n\n"T. Essel" (hiding somewhere in cyberspace)\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sun, 27 Jun 2004, Paul E wrote:
> can anyone tell me whether the 4-\pi rotation for the identity of spinors
> is (closely?) related to the non-orientability of the projective plane?
> I can see that it takes a 4-\pi transit of a Moebius band to regain the
> original orientation of a chiral object.
I wish I could draw a picture here, but my short answer---not often can I
say this, alas--- is YES!
Cut out a strip of paper ruled down the center like so:
__{________________}
| |
|__{________________}|
| |
|__{________________}|
Mark the central ruling (left hand endpoint), \pi (midpoint), 2 \pi (right
hand endpoint). Wrap up the strip a la Moebius. Note that the ruling is
now a circle; call it "the central circle". Now try to draw a a small
linesegment parallel to the central circle. Extend it carefully; you
should get another circle, but with periodicity 4 \pi! This is indeed
-exactly- what spinors are all about. Notice the "natural projection" is
defined in the strip; the embedding in E^3 is actually irrelevant for most
purposes.
The picture I wish I could draw is the E(2) version of the SO(3) picture
of Olinde Rodrigues. Instead I can only give some directions which leave
a lot up to your own initiative/insight:
1. Read a book like Coxeter, Geometry, to learn
the basic ideas of transformation geometry, in particular
(a) iterating the reflections in two parallel lines in the plane gives a
translation in that plane,
(b) iterating reflections in two intersecting lines gives a rotation in
that plane, about the point of intersection.
2. Find and study the discussion of quaternions in MTW, Gravitation,
Freeman, 1973, in particular the double cover SU(2) -->> SO(3).
3. Find and study old posts here on the E(2) of this. Here, SO(3) is the
three-dimensional Lie group of isometries of S^2; E(3) is the
three-dimensional Lie group of isometries of E^2. And of course there is
SO(1,2), the three-dimensional Lie group of isometries of H^2.
4. Find and study
author = {Roger Carter and Graeme Segal and Ian MacDonald},
title = {Lectures on {L}ie Groups and {L}ie Algebras},
series = {London Mathematical Society Student Texts},
volume = 32,
publisher = {Cambridge University Press},
year = 1995}
to learn a bit more about spinors.
5. Find and study many highly relevant posts here by John Baez (This Week
series) over the past ten years.
> I've satisfied (& puzzled) myself that the orientation-entanglement
> relation in real 3space (eg in MTW, p. 1148-9)
Oh! You've already done (2)! Hurrah!!!! Like any stimulating reading,
this obviously left you hungry for more. If you liked (2) I am sure
you'll enjoy following my other suggestions.
(Hmm... OK, the embedding in E^3 is probably not irrelevant for "Dirac's
spanner", but never mind that for the moment...)
> shows that a 4-\pi
> rotation can be required for identity in E3, but I'm less clear on why
> it should apply to the complex projective plane (for electron spin
> states etc.), which (in my ignorance of advanced math) I would expect
> to be orientable.
Oh--- so you want to learn -physics-?! :-/
Well, I'll let others address your real question, then, but to finish the
thought: I guess you know that the Moebius strip is nonorientable. So is
RP^2. And by "sewing a disk around the boundary" of the M. strip you
get--- RP^2! But the point is that a picture of the double cover circle
---> central circle describe above fits inside a picture of the double
cover SU(2) --> SO(3). Here, SO(3) is both nonorientable and
non-simply-connected, while SU(2) is both orientable and simply-connected,
in fact, quite remarkably, SU(2) is homeomorphic to S^3! That's the
connection between the spinorial representation of rotations (about the
origin) in E^3 and nonorientable manifolds.
"T. Essel" (hiding somewhere in cyberspace)
Thomas Mautsch
Jul2-04, 11:21 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn news:<40e52b9c@news.sentex.net> schrieb tessel@tum.bot <tessel@tum.bot>:\n> On Sun, 27 Jun 2004, Paul E wrote:\n>\n>> can anyone tell me whether the 4-pi rotation for the identity of spinors\n>> is (closely?) related to the non-orientability of the projective plane?\n>> I can see that it takes a 4-pi transit of a Moebius band to regain the\n>> original orientation of a chiral object.\n>\n> I wish I could draw a picture here, but my short answer---not often can I\n> say this, alas--- is YES!\n>\n> Cut out a strip of paper ruled down the center like so:\n> __________________\n> | |\n> |__________________|\n> | |\n> |__________________|\n>\n> Mark the central ruling 0 (left hand endpoint), pi (midpoint), 2 pi (right\n> hand endpoint). Wrap up the strip a la Moebius. Note that the ruling is\n> now a circle; call it "the central circle". Now try to draw a a small\n> linesegment parallel to the central circle. Extend it carefully; you\n> should get another circle, but with periodicity 4 pi! This is indeed\n^^^^^^^^^^^^^^\n> -exactly- what spinors are all about. Notice the "natural projection" is\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n> defined in the strip; the embedding in E^3 is actually irrelevant for most\n> purposes.\n\nI beg your pardon. --\nThis is what ORIENTABILITY is about, not SPINORS, and neither SPIN.\n\nWhat you demonstratete here is that\nto a non-orientable manifold, like the Moebius band,\nthere exists a two-fold cover which is orientable.\n\nThis has nothing directly to do with spin\n(except that the obstructions to orientability and spinnability\nare cohomology classes with values in Z_2).\n\nSPINnability is a HIGHER kind of orientability,\nand SPINORS are in fact modules ("representations") of CLIFFORD ALGEBRAS\nand do not really require knowledge about orientations or spins,\nonly about metrics.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In news:<40e52b9c@news.sentex.net> schrieb tessel@tum.bot <tessel@tum.bot>:
> On Sun, 27 Jun 2004, Paul E wrote:
>
>> can anyone tell me whether the 4-\pi rotation for the identity of spinors
>> is (closely?) related to the non-orientability of the projective plane?
>> I can see that it takes a 4-\pi transit of a Moebius band to regain the
>> original orientation of a chiral object.
>
> I wish I could draw a picture here, but my short answer---not often can I
> say this, alas--- is YES!
>
> Cut out a strip of paper ruled down the center like so:
> __{________________}
> | |
> |__{________________}|
> | |
> |__{________________}|
>
> Mark the central ruling (left hand endpoint), \pi (midpoint), 2 \pi (right
> hand endpoint). Wrap up the strip a la Moebius. Note that the ruling is
> now a circle; call it "the central circle". Now try to draw a a small
> linesegment parallel to the central circle. Extend it carefully; you
> should get another circle, but with periodicity 4 \pi! This is indeed
^^^^^^^^^^^^^^
> -exactly- what spinors are all about. Notice the "natural projection" is
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> defined in the strip; the embedding in E^3 is actually irrelevant for most
> purposes.
I beg your pardon. --
This is what ORIENTABILITY is about, not SPINORS, and neither SPIN.
What you demonstratete here is that
to a non-orientable manifold, like the Moebius band,
there exists a two-fold cover which is orientable.
This has nothing directly to do with spin
(except that the obstructions to orientability and spinnability
are cohomology classes with values in Z_2).
SPINnability is a HIGHER kind of orientability,
and SPINORS are in fact modules ("representations") of CLIFFORD ALGEBRAS
and do not really require knowledge about orientations or spins,
only about metrics.
Thomas Mautsch
Jul2-04, 12:08 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn news:<659defa1.0406220505.3540c9d2@posting.google. com> schrieb Paul E:\n> Having obtained no reply from *.particle, advancedphysics.org, nor\n> even a google search, can I try here please?\n>\n> After reading some elementary topology, can anyone tell me whether the\n> 4-pi rotation for the identity of spinors is (closely?) related to the\n> non-orientability of the projective plane? I can see that it takes a\n> 4-pi\n> transit of a Moebius band to regain the original orientation of a\n> chiral object.\n>\n> I\'ve satisfied (& puzzled) myself that the orientation-entanglement\n> relation in real 3space (eg in MTW, p. 1148-9) shows that a 4-pi\n> rotation can be required for identity in E3, but I\'m less clear on why\n> it should apply to the complex projective plane (for electron spin\n> states etc.), which (in my ignorance of advanced math) I would expect\n> to be orientable. Any (simply explained) advice would be much\n> appreciated.\n\n\nThe existence of non-orientable manifolds has to do with the fact\nthat the group GL(n) of linear transformations\nof an n-dimensional vector space (with n >= 1)\nhas TWO components, of which one contains the identity element\nand all the other orientation-preserving transformations,\nand the other contains all orientation-reversing transformations.\n\nThat the DOUBLE-cover of the Moebius band is orientable\nhas to do with the occurrence of the number "TWO" above.\n\n\nSpin can *only* be discussed for oriented vector spaces.\nOne looks at paths in the CONNECTED group GL^+(n)\nof orientation-preserving linear transformations.\nSince every two elements in GL^+(n) can be connected by a continuous path,\ni.e. every "rotation" can be realized by a continuous motion,\none tries to distinguish between different paths in GL^+(n):\n\nAs it turns out, for n>=2\nthere do exist paths of orientation-preserving transformations\nwhich one can not continuously deformed into each other.\nFor n>=3 there are exactly TWO sorts of these paths:\nSome that can be contracted to a constant path, and some that can not.\nThis is, as you have already seen in the literature,\nthe reason for the existence of spin groups and algebraic spinors.\n\nBUT the fact that the number "TWO" turns up in both\nthe discussion of orientation and the discussion of spin\nlooks from this viewpoint more like A COINCIDENCE...\n\n\nThe Moebius band and projective planes\nonly appear much later after this ALGEBRAIC discussion\nwhen you try to put TOPOLOGICAL structures\nlike orientations or spin structures on manifolds...\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In news:<659defa1.0406220505.3540c9d2@posting.google.com> schrieb Paul E:
> Having obtained no reply from *.particle, advancedphysics.org, nor
> even a google search, can I try here please?
>
> After reading some elementary topology, can anyone tell me whether the
> 4-\pi rotation for the identity of spinors is (closely?) related to the
> non-orientability of the projective plane? I can see that it takes a
> 4-\pi
> transit of a Moebius band to regain the original orientation of a
> chiral object.
>
> I've satisfied (& puzzled) myself that the orientation-entanglement
> relation in real 3space (eg in MTW, p. 1148-9) shows that a 4-\pi
> rotation can be required for identity in E3, but I'm less clear on why
> it should apply to the complex projective plane (for electron spin
> states etc.), which (in my ignorance of advanced math) I would expect
> to be orientable. Any (simply explained) advice would be much
> appreciated.
The existence of non-orientable manifolds has to do with the fact
that the group GL(n) of linear transformations
of an n-dimensional vector space (with n >= 1)
has TWO components, of which one contains the identity element
and all the other orientation-preserving transformations,
and the other contains all orientation-reversing transformations.
That the DOUBLE-cover of the Moebius band is orientable
has to do with the occurrence of the number "TWO" above.
Spin can *only* be discussed for oriented vector spaces.
One looks at paths in the CONNECTED group GL^+(n)
of orientation-preserving linear transformations.
Since every two elements in GL^+(n) can be connected by a continuous path,
i.e. every "rotation" can be realized by a continuous motion,
one tries to distinguish between different paths in GL^+(n):
As it turns out, for n>=2
there do exist paths of orientation-preserving transformations
which one can not continuously deformed into each other.
For n>=3 there are exactly TWO sorts of these paths:
Some that can be contracted to a constant path, and some that can not.
This is, as you have already seen in the literature,
the reason for the existence of spin groups and algebraic spinors.
BUT the fact that the number "TWO" turns up in both
the discussion of orientation and the discussion of spin
looks from this viewpoint more like A COINCIDENCE...
The Moebius band and projective planes
only appear much later after this ALGEBRAIC discussion
when you try to put TOPOLOGICAL structures
like orientations or spin structures on manifolds...
Serenus Zeitblom
Jul4-04, 07:39 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\ntessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...\nBut the point is that a picture of the double cover circle\n> ---> central circle describe above fits inside a picture of the double\n> cover SU(2) --> SO(3). Here, SO(3) is both nonorientable and\n> non-simply-connected,\n\nI beg your pardon. SO(3) is most certainly orientable! RP^{even} is\nnot orientable, but RP^{odd} is. RP^5,9 etc are not spin, but RP^3,7 etc\nare spin. So RP^3 is about as nice as you could wish.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>tessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...
But the point is that a picture of the double cover circle
> ---> central circle describe above fits inside a picture of the double
> cover SU(2) --> SO(3). Here, SO(3) is both nonorientable and
> non-simply-connected,
I beg your pardon. SO(3) is most certainly orientable! RP^{even} is
not orientable, but RP^{odd} is. RP^5,9 etc are not spin, but RP^3,7 etc
are spin. So RP^3 is about as nice as you could wish.
Stephen Blake
Jul4-04, 07:39 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\ntessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...\n> On Sun, 27 Jun 2004, Paul E wrote:\n>\n> > can anyone tell me whether the 4-pi rotation for the identity of spinors\n> > is (closely?) related to the non-orientability of the projective plane?\n> > I can see that it takes a 4-pi transit of a Moebius band to regain the\n> > original orientation of a chiral object.\n>\n> I wish I could draw a picture here, but my short answer---not often can I\n> say this, alas--- is YES!\n>\n> Cut out a strip of paper ruled down the center like so:\n> __________________\n> | |\n> |__________________|\n> | |\n> |__________________|\n>\n> Mark the central ruling 0 (left hand endpoint), pi (midpoint), 2 pi (right\n> hand endpoint). Wrap up the strip a la Moebius. Note that the ruling is\n> now a circle; call it "the central circle". Now try to draw a a small\n> linesegment parallel to the central circle. Extend it carefully; you\n> should get another circle, but with periodicity 4 pi! This is indeed\n> -exactly- what spinors are all about. Notice the "natural projection" is\n> defined in the strip; the embedding in E^3 is actually irrelevant for most\n> purposes.\n>\nThe effect that you describe is nothing to do with the spinor reps and is\nsimply the effect of an ordinary rotation acting in 2-d elliptic space. The\nMobius strip with angles and/or distances defined is really 2-d elliptic space\nwith a hole cut out of it because angles and distance do not exist in 2-d\nprojective space. Consider an equilateral triangle with vertices a1, a2, a3\nin 2-d elliptic space.\n\na1\n/\\\n/ \\\n/ \\\n/ \\\n_________/________\\_________\n-a3 a2 a3 -a2\nFig 1\n\n2-d elliptic space looks like the surface of a sphere with antipodal points\nidentified. Consequently, the straight lines in the diagram are great circles\non the sphere and the interior angles of the equilateral triangle are each\npi/2 radians. If we start out from point a2 along the line a2a3 we eventually\nencounter the antipodal point from a2 which I have written as -a2. The length\nof a straight line in elliptic space turns out to be pi. Consequently, the\ndistance from a2 to -a2 is pi and this is a trip right around elliptic space\nbecause a2 and -a2 are the same point.Consider a rotation U(theta) about\npoint a1. The action of the rotation on the vertices turns out as,\nU(theta)a1=a1\nU(theta)a2=a2*cos(theta)+a3*si n(theta)\nU(theta)a3=a3*cos(theta)-a2*sin(theta) .\nIn particular consider a rotation by pi. The action is,\nU(pi)a1=a1, U(pi)a2=-a2, U(pi)a3=-a3.\nNow consider how to draw a line parallel to a2a3. Start from point p=a1+a2.\n\n-a2\n/\n/q\n/\na1\n/\\\n/ \\\n/____\\_______\n/p \\ q\n_________/________\\_________\n-a3 a2 a3 -a2\n/\n______ /-q\n/\n/\n-a1\nFig 2\n\nWe can move point p parallel to line a2a3 by acting on it with rotation U.\nIf we act with U(pi) then by linearity,\nU(pi)p=U(pi)(a1+a2)=a1-a2=q. So, in figure 2 p goes to q by the action\nof rotation/parallel-transport by angle/distance pi. Notice that if we look\nat the antipodal -q in figure 2, it is just the action of drawing your\nparallel line once around the Mobius strip. Also notice that from the\ncentre of the rotation at point a1, p -> q looks like a rotation through pi.\nWe have to parallel-transport by another pi to get q -> p.\nU(pi)q=U(pi)(a1-a2)=a1+a2=p. Hence U(pi)U(pi)=U(2*pi)=1 and so we are dealing\nwith ordinary reps (NOT spinor reps) of the congruence group of 2-d elliptic\nspace. The above explanation can also be gleaned from section 6.3 of Coxeter\'s\nbook "Non-Euclidean Geometry", 6th edition. Also, my homepage contains some\nsoftware for playing around with 2-d elliptic geometry which can be downloaded.\n\nStephen Blake\n--\nhttp://homepage.ntlworld.com/stebla\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>tessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...
> On Sun, 27 Jun 2004, Paul E wrote:
>
> > can anyone tell me whether the 4-\pi rotation for the identity of spinors
> > is (closely?) related to the non-orientability of the projective plane?
> > I can see that it takes a 4-\pi transit of a Moebius band to regain the
> > original orientation of a chiral object.
>
> I wish I could draw a picture here, but my short answer---not often can I
> say this, alas--- is YES!
>
> Cut out a strip of paper ruled down the center like so:
> __{________________}
> | |
> |__{________________}|
> | |
> |__{________________}|
>
> Mark the central ruling (left hand endpoint), \pi (midpoint), 2 \pi (right
> hand endpoint). Wrap up the strip a la Moebius. Note that the ruling is
> now a circle; call it "the central circle". Now try to draw a a small
> linesegment parallel to the central circle. Extend it carefully; you
> should get another circle, but with periodicity 4 \pi! This is indeed
> -exactly- what spinors are all about. Notice the "natural projection" is
> defined in the strip; the embedding in E^3 is actually irrelevant for most
> purposes.
>
The effect that you describe is nothing to do with the spinor reps and is
simply the effect of an ordinary rotation acting in 2-d elliptic space. The
Mobius strip with angles and/or distances defined is really 2-d elliptic space
with a hole cut out of it because angles and distance do not exist in 2-d
projective space. Consider an equilateral triangle with vertices a1, a2, a3
in 2-d elliptic space.
a1
/\
/ \/ \/ \__{_______}/__{______}\__{_______}
-a3 a2 a3 -a2
Fig 1
2-d elliptic space looks like the surface of a sphere with antipodal points
identified. Consequently, the straight lines in the diagram are great circles
on the sphere and the interior angles of the equilateral triangle are each
\pi/2 radians. If we start out from point a2 along the line a2a3 we eventually
encounter the antipodal point from a2 which I have written as -a2. The length
of a straight line in elliptic space turns out to be \pi. Consequently, the
distance from a2 to -a2 is \pi and this is a trip right around elliptic space
because a2 and -a2 are the same point.Consider a rotation U(\theta) about
point a1. The action of the rotation on the vertices turns out as,
U(\theta)a1=a1U(\theta)a2=a2*cos(\theta)+a3*sin(\t heta)U(\theta)a3=a3*cos(\theta)-a2*sin(\theta) .
In particular consider a rotation by \pi. The action is,
U(\pi)a1=a1, U(\pi)a2=-a2, U(\pi)a3=-a3.
Now consider how to draw a line parallel to a2a3. Start from point p=a1+a2.
-a2
/
/q
/
a1
/\
/ \/__{__}\__{_____}/p \ q__{_______}/__{______}\__{_______}
-a3 a2 a3 -a2
/
__{____} /-q
/
/
-a1
Fig 2
We can move point p parallel to line a2a3 by acting on it with rotation U.
If we act with U(\pi) then by linearity,
U(\pi)p=U(\pi)(a1+a2)=a1-a2=q. So, in figure 2 p goes to q by the action
of rotation/parallel-transport by angle/distance \pi. Notice that if we look
at the antipodal -q in figure 2, it is just the action of drawing your
parallel line once around the Mobius strip. Also notice that from the
centre of the rotation at point a1, p -> q looks like a rotation through \pi.
We have to parallel-transport by another \pi to get q -> p.U(\pi)q=U(\pi)(a1-a2)=a1+a2=p. Hence U(\pi)U(\pi)=U(2*\pi)=1 and so we are dealing
with ordinary reps (NOT spinor reps) of the congruence group of 2-d elliptic
space. The above explanation can also be gleaned from section 6.3 of Coxeter's
book "Non-Euclidean Geometry", 6th edition. Also, my homepage contains some
software for playing around with 2-d elliptic geometry which can be downloaded.
Stephen Blake
--
http://homepage.ntlworld.com/stebla
tessel@tum.bot
Jul4-04, 07:40 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Fri, 2 Jul 2004, Thomas Mautsch wrote:\n\n> SPINnability is a HIGHER kind of orientability,\n\nOh dear. Well, I did say that I have a -picture- in my mind, which I was\ntrying to "translate" on the fly into a verbal description. If my words\nwere confusing or even misleading, I apologize, but I confident that the\n-picture- I have in mind is not--- at least in low dimensions--- and I\nhope the OP will follow up on my suggestions regardless of my own\nexpository inadequacies!\n\nCome to think of it, surely I can\'t have misunderstood his question -too-\nbadly, because I did immediately think of the very picture which (later in\nhis post) he implies had prompted his question, namely the figure in MTW\nillustrating Rodriques\'s way of understanding what mathematicians now call\nthe quaternionic representation of rotations about the origin in E^3\n(thought of as isometries of S^2).\n\nIn my reply, I offered directions for working out a minor generalization\nof this picture to "spinorial" representations of isometries of H^2 and\nE^2. (Maybe your objection is really to my characterization of the second\nof these as a kind of "spinorial" covering?) This doesn\'t raise\ndimension, but the other books I mentioned sketch further generalizations,\nwhich are not, I think, inconsistent with what you are saying.\n\nIn your second reply you say:\n\n> BUT the fact that the number "TWO" turns up in both the discussion of\n> orientation and the discussion of spin looks from this viewpoint more\n> like A COINCIDENCE...\n\nMaybe we can agree that I was talking about a way of thinking about spin\nin the low dimensions, but which according to your point of view is\nmisleading about what happens in higher dimensions?\n\nThe ideas which I attempted to outline in my own reply make a connection\nwith reflection groups. Many old timers here have, like me, followed with\ngreat interest a recurring theme in John Baez\'s "This Week" series over\nthe year, relating Coxeter/Dynkin stuff (reflection groups, regular\npolytopes, simple Lie algebras, and all that), invariant theory,\nrepresentation theory, and much more. Spin groups are certainly part of\nthe Coxeter picture, which certainly involves reflection groups. I hope\nI\'m not entirely misrepresenting this particular "theme" in the Weeks---\nif John happens to see this and thinks I am, I trust he will speak up.\n\nCome to think of it, I probably should have mentioned another wonderful\n"picture book", which offers much information on the topology of compact\nLie groups:\n\nauthor = {Hatcher, Allen},\ntitle = {Algebraic Topology},\npublisher = {Cambridge University Press},\nyear = 1992}\n\n"T. Essel" (hiding somewhere in cyberspace)\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Fri, 2 Jul 2004, Thomas Mautsch wrote:
> SPINnability is a HIGHER kind of orientability,
Oh dear. Well, I did say that I have a -picture- in my mind, which I was
trying to "translate" on the fly into a verbal description. If my words
were confusing or even misleading, I apologize, but I confident that the
-picture- I have in mind is not--- at least in low dimensions--- and I
hope the OP will follow up on my suggestions regardless of my own
expository inadequacies!
Come to think of it, surely I can't have misunderstood his question -too-
badly, because I did immediately think of the very picture which (later in
his post) he implies had prompted his question, namely the figure in MTW
illustrating Rodriques's way of understanding what mathematicians now call
the quaternionic representation of rotations about the origin in E^3
(thought of as isometries of S^2).
In my reply, I offered directions for working out a minor generalization
of this picture to "spinorial" representations of isometries of H^2 and
E^2. (Maybe your objection is really to my characterization of the second
of these as a kind of "spinorial" covering?) This doesn't raise
dimension, but the other books I mentioned sketch further generalizations,
which are not, I think, inconsistent with what you are saying.
In your second reply you say:
> BUT the fact that the number "TWO" turns up in both the discussion of
> orientation and the discussion of spin looks from this viewpoint more
> like A COINCIDENCE...
Maybe we can agree that I was talking about a way of thinking about spin
in the low dimensions, but which according to your point of view is
misleading about what happens in higher dimensions?
The ideas which I attempted to outline in my own reply make a connection
with reflection groups. Many old timers here have, like me, followed with
great interest a recurring theme in John Baez's "This Week" series over
the year, relating Coxeter/Dynkin stuff (reflection groups, regular
polytopes, simple Lie algebras, and all that), invariant theory,
representation theory, and much more. Spin groups are certainly part of
the Coxeter picture, which certainly involves reflection groups. I hope
I'm not entirely misrepresenting this particular "theme" in the Weeks---
if John happens to see this and thinks I am, I trust he will speak up.
Come to think of it, I probably should have mentioned another wonderful
"picture book", which offers much information on the topology of compact
Lie groups:
author = {Hatcher, Allen},
title = {Algebraic Topology},
publisher = {Cambridge University Press},
year = 1992}
"T. Essel" (hiding somewhere in cyberspace)
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nThomas Mautsch <mautsch@math.ethz.ch> wrote in message news:<slrnceb54c.lha.mautsch@pisano.math.ethz.ch>. ..\n\nTHANKS to both T. Essel and Thomas Mautsch for numerous helpful\nsuggestions and clues. It\'ll take me a while to catch up.\n\nMeantime, one rather elementary point I\'m hung up on (apologies for\nthe pedestrian presentation) if you find it of any interest:\n\nMost mentions of SO(3) suggest picturing it as a solid ball and seem\nto stick to the case where the unit vectors are given length pi. Fine.\nI can see that antipodal points are equivalent (and, after your\nadvice, think I now understand the "double cover" by SU(2), which\npresumably means (in my naive terminology) that what were equivalent\nantipodal points in SO(3) become distinct in SU(2)).\n\nBut suppose one considers rotations of other than pi (actually, other\nthan n*pi, n>=1), and thus for vector lengths other than n*pi?\nPresumably the centre point of the ball represents a null rotation,\nwhich sounds like the unity element of the group; while rotations of,\nsay, 3*pi/2 would be represented by points on the surface of the ball\nof correspondingly larger radius.\n\nBut doesn\'t the antipodeal equivalence simply disappear for rotations\nother than n*pi?\n\nThanks - Paul\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thomas Mautsch <mautsch@math.ethz.ch> wrote in message news:<slrnceb54c.lha.mautsch@pisano.math.ethz.ch>...
THANKS to both T. Essel and Thomas Mautsch for numerous helpful
suggestions and clues. It'll take me a while to catch up.
Meantime, one rather elementary point I'm hung up on (apologies for
the pedestrian presentation) if you find it of any interest:
Most mentions of SO(3) suggest picturing it as a solid ball and seem
to stick to the case where the unit vectors are given length \pi. Fine.
I can see that antipodal points are equivalent (and, after your
advice, think I now understand the "double cover" by SU(2), which
presumably means (in my naive terminology) that what were equivalent
antipodal points in SO(3) become distinct in SU(2)).
But suppose one considers rotations of other than \pi (actually, other
than n*\pi, n>=1), and thus for vector lengths other than n*\pi?
Presumably the centre point of the ball represents a null rotation,
which sounds like the unity element of the group; while rotations of,
say, 3*\pi/2 would be represented by points on the surface of the ball
of correspondingly larger radius.
But doesn't the antipodeal equivalence simply disappear for rotations
other than n*\pi?
Thanks - Paul
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\non 06/07/2004 7:47 pm, Paul E at pellis@london.edu wrote:\n\n>\n> Thomas Mautsch <mautsch@math.ethz.ch> wrote in message\n> news:<slrnceb54c.lha.mautsch@pisano.math.ethz.ch>. ..\n>\n> THANKS to both T. Essel and Thomas Mautsch for numerous helpful\n> suggestions and clues. It\'ll take me a while to catch up.\n>\n> Meantime, one rather elementary point I\'m hung up on (apologies for\n> the pedestrian presentation) if you find it of any interest:\n>\n> Most mentions of SO(3) suggest picturing it as a solid ball and seem\n> to stick to the case where the unit vectors are given length pi. Fine.\n> I can see that antipodal points are equivalent (and, after your\n> advice, think I now understand the "double cover" by SU(2), which\n> presumably means (in my naive terminology) that what were equivalent\n> antipodal points in SO(3) become distinct in SU(2)).\n>\n> But suppose one considers rotations of other than pi (actually, other\n> than n*pi, n>=1), and thus for vector lengths other than n*pi?\n> Presumably the centre point of the ball represents a null rotation,\n> which sounds like the unity element of the group; while rotations of,\n> say, 3*pi/2 would be represented by points on the surface of the ball\n> of correspondingly larger radius.\n>\n> But doesn\'t the antipodeal equivalence simply disappear for rotations\n> other than n*pi?\n\nThe center of the ball is the identity. Points at a distance x from the\ncentre represent rotations by an angle of x, and diametrically opposite\npoints represent rotations about the same axis but in the opposite\ndirection. For x < pi, these are not at all the same thing -- I hope that is\nobvious. At x = pi, we are at the surface of the ball. Since a rotation of\npi about an axis is the same thing as a rotation of -pi about the same axis,\ndiametrically opposite points on the surface of the ball are the same.\nObviously a rotation of more than pi in one direction are the same as a\nrotation of less than pi in the opposite direction, so having come up to the\nsurface, we \'jump\' to the diametrically opposed point (which is secretly the\nsame point) and then head down towards the centre again on the other side.\nObviously a rotation of 2 pi is the same as no rotation at all, and brings\nus back to the centre.\n\nDoes this help?\n\nTim\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>on 06/07/2004 7:47 pm, Paul E at pellis@london.edu wrote:
>
> Thomas Mautsch <mautsch@math.ethz.ch> wrote in message
> news:<slrnceb54c.lha.mautsch@pisano.math.ethz.ch>...
>
> THANKS to both T. Essel and Thomas Mautsch for numerous helpful
> suggestions and clues. It'll take me a while to catch up.
>
> Meantime, one rather elementary point I'm hung up on (apologies for
> the pedestrian presentation) if you find it of any interest:
>
> Most mentions of SO(3) suggest picturing it as a solid ball and seem
> to stick to the case where the unit vectors are given length \pi. Fine.
> I can see that antipodal points are equivalent (and, after your
> advice, think I now understand the "double cover" by SU(2), which
> presumably means (in my naive terminology) that what were equivalent
> antipodal points in SO(3) become distinct in SU(2)).
>
> But suppose one considers rotations of other than \pi (actually, other
> than n*\pi, n>=1), and thus for vector lengths other than n*\pi?
> Presumably the centre point of the ball represents a null rotation,
> which sounds like the unity element of the group; while rotations of,
> say, 3*\pi/2 would be represented by points on the surface of the ball
> of correspondingly larger radius.
>
> But doesn't the antipodeal equivalence simply disappear for rotations
> other than n*\pi?
The center of the ball is the identity. Points at a distance x from the
centre represent rotations by an angle of x, and diametrically opposite
points represent rotations about the same axis but in the opposite
direction. For x < \pi, these are not at all the same thing -- I hope that is
obvious. At x = \pi, we are at the surface of the ball. Since a rotation of
\pi about an axis is the same thing as a rotation of -\pi about the same axis,
diametrically opposite points on the surface of the ball are the same.
Obviously a rotation of more than \pi in one direction are the same as a
rotation of less than \pi in the opposite direction, so having come up to the
surface, we 'jump' to the diametrically opposed point (which is secretly the
same point) and then head down towards the centre again on the other side.
Obviously a rotation of 2 \pi is the same as no rotation at all, and brings
us back to the centre.
Does this help?
Tim
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nTim S <Tim@timsilverman.demon.co.uk> wrote in message news:<BD11F793.DD60%Tim@timsilverman.demon.co.uk>. ..\n> on 06/07/2004 7:47 pm, Paul E at pellis@london.edu wrote:\n>\n> >\n> > Meantime, one rather elementary point: Most mentions of SO(3) suggest picturing it as a solid ball and seem\nSNIPS\n> >\n> > But doesn\'t the antipodeal equivalence simply disappear for rotations\n> > other than n*pi?\n\n> Obviously a rotation of more than pi in one direction are the same as a\n> rotation of less than pi in the opposite direction, so having come up to the\n> surface, we \'jump\' to the diametrically opposed point (which is secretly the\n> same point) and then head down towards the centre again on the other side.\n> Obviously a rotation of 2 pi is the same as no rotation at all, and brings\n> us back to the centre.\n>\n> Does this help?\n>\n> Tim\n\nYes - Gottit (but it\'s hard to visualise - seems trickier than RP^2?)\nMany thanks - Paul\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Tim S <Tim@timsilverman.demon.co.uk> wrote in message news:<BD11F793.DD60%Tim@timsilverman.demon.co.uk>...
> on 06/07/2004 7:47 pm, Paul E at pellis@london.edu wrote:
>
> >
> > Meantime, one rather elementary point: Most mentions of SO(3) suggest picturing it as a solid ball and seem
SNIPS
> >
> > But doesn't the antipodeal equivalence simply disappear for rotations
> > other than n*\pi?
> Obviously a rotation of more than \pi in one direction are the same as a
> rotation of less than \pi in the opposite direction, so having come up to the
> surface, we 'jump' to the diametrically opposed point (which is secretly the
> same point) and then head down towards the centre again on the other side.
> Obviously a rotation of 2 \pi is the same as no rotation at all, and brings
> us back to the centre.
>
> Does this help?
>
> Tim
Yes - Gottit (but it's hard to visualise - seems trickier than RP^2?)
Many thanks - Paul
tessel@tum.bot
Jul11-04, 02:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Sun, 4 Jul 2004, Serenus Zeitblom wrote:\n\n> tessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...\n\n> > SO(3) is both nonorientable and non-simply-connected,\n>\n> I beg your pardon. SO(3) is most certainly orientable!\n\nOoops!-- you are right; this was a brain blip on my part. Apologies to\nall, and thanks for the correction.\n\n> RP^{even} is not orientable, but RP^{odd} is. RP^5,9 etc are not spin,\n> but RP^3,7 etc are spin. So RP^3 is about as nice as you could wish.\n\nI don\'t think my goof affects what I -meant- to say, but to save bandwidth\nI\'ll join the chorus advising the OP (Paul E) to forget whatever I\nactually -did- say--- except for the citations!\n\nMany years ago, someone else asked in this group about the same passage in\nMTW which Paul E is asking about. At that time, the writings I mentioned\nhelped me better understand what MTW had to say, so I hope Paul is not\ndiscouraged from looking them up.\n\nBTW, there is a connection here with the soliton thread (c.f. harmonic\nmaps)!\n\nOK, Paul, with this I leave you in the hands of more capable magicians :-/\n\n(need a link here to Kauffman\'s animated gif of the Indian plate trick)\n\n"T. Essel" (hiding somewhere in cyberspace)\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sun, 4 Jul 2004, Serenus Zeitblom wrote:
> tessel@tum.bot wrote in message news:<40e52b9c@news.sentex.net>...
> > SO(3) is both nonorientable and non-simply-connected,
>
> I beg your pardon. SO(3) is most certainly orientable!
Ooops!-- you are right; this was a brain blip on my part. Apologies to
all, and thanks for the correction.
> RP^{even} is not orientable, but RP^{odd} is. RP^5,9 etc are not spin,
> but RP^3,7 etc are spin. So RP^3 is about as nice as you could wish.
I don't think my goof affects what I -meant- to say, but to save bandwidth
I'll join the chorus advising the OP (Paul E) to forget whatever I
actually -did- say--- except for the citations!
Many years ago, someone else asked in this group about the same passage in
MTW which Paul E is asking about. At that time, the writings I mentioned
helped me better understand what MTW had to say, so I hope Paul is not
discouraged from looking them up.
BTW, there is a connection here with the soliton thread (c.f. harmonic
maps)!
OK, Paul, with this I leave you in the hands of more capable magicians :-/
(need a link here to Kauffman's animated gif of the Indian plate trick)
"T. Essel" (hiding somewhere in cyberspace)
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