![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
-- Beland 16:27, 18 December 2005 (UTC)
Regarding this from the introduction: "A length-preserving transformation which reverses orientation is called an improper rotation. Every improper rotation of three-dimensional Euclidean space is a reflection in a plane through the origin." If I have a transformation T that reflects in the z-axis (negates z) and translates 2 units along in the x-axis, then according to the first sentence T is an improper rotation. The second sentence tells me that T, being an improper rotation, is a reflection in a plane through the origin. Yet I can think of no such reflection that reproduces the effect of T. Where am I going wrong here? Oioisaveloy ( talk) 10:44, 9 August 2009 (UTC)
In the article it is mentioned that SO(3) is homeomorphic to and .
However, the first homology group of is , and for the product, it's just , because S^2 is simply connected.
Can someone clarify that? —Preceding
unsigned comment added by
77.127.154.232 (
talk)
21:38, 1 November 2009 (UTC)
Next question: why a topological space homeomorphic' to the rotation group, but this manifold is diffeomorphic to the rotation group? why we use in one case the word homeomorphic, but in other - diffeomorphic? Explane, please! Vyrivykh ( talk) 14:50, 23 October 2010 (UTC)
For A(t), a one-parameter subgroup of SO(3) parametrised by t, differentiating with respect to t gives
Matekosor ( talk) 05:57, 1 August 2012 (UTC)
Shouldn't this page be renamed something like
because there are lots of rotation groups out there, but this article is about one very specific one. Jheald ( talk) 07:55, 13 September 2008 (UTC)
If one had done any checking/intuiting of Engø's formulas for compositions of arbitrary rotations, he would quickly supplant his ponderous notation for a and b with the much simpler, and intuitive, a=c cotφ/2, b=c cot θ/2. This would immediately shorten several of his formulas, notably for d, now proportional to c, and illuminate α,β,γ. I don't see any compulsion to stick to his original notation. But it is a purely aesthetic point... Cuzkatzimhut ( talk) 14:08, 3 January 2015 (UTC)
I added a physics project template in order for someone to rate the article. Someone did, but the importance of the topic was set to low. This is something I find ridiculous. The importance lies somewhere between high and top. Little Nothing in physics is more important than symmetries, rotational symmetry being the most important symmetry of them all. (Important = useful + much more (physical truth as far as we know)) I reverted the complete rating because of this misjudgement, hoping that someone else drops by to rate the article.
YohanN7 (
talk)
11:21, 8 January 2015 (UTC)
Summary: Extend the final sentence of the fourth para under the subheading Topology, with the part following the first comma here below:
"This circle can be shrunk to the north pole without problems, as an "antipodal circle" forms and simultaneously shrinks to the south pole." [revised again, 30-8-08]
Detail: As formerly shown in the main article, I don’t see how a single great circle can shrink, unless the antipodal point of every point on the shrinking circle described is also included. The article doesn’t seem to hint at this. However, after looking at Greg Egan’s applet:
http://gregegan.customer.netspace.net.au/APPLETS/21/21.html
... I am emboldened to suggest [a suitable] extension of the final sentence. Unless someone can explain comprehensibly (to someone at my level) how I’m in error, then after seven days I’ll post my suggested extension. OK? PaulGEllis ( talk) 19:46, 4 August 2008 (UTC)
Thank you for your reply. What I have a problem with is why you can shrink the great circle, as if it was on a plane. In my understanding (which I admit is not graduate mathematician level - I'm an interested physical chemist, thinking purely geometrically ): I can see how you get to a great circle including the north and south poles. And for that great circle, every point still has its equivalent antipodal point on the great circle. But when you start to shrink the great circle away from the south pole, then no point on [a smoothly; inserted 30-8-08] shrinking circle has an antipodal point represented. Since every point on the surface of the pi-radius ball is equivalent to its antipodal point, I can't see how you can start to shrink the great circle, unless there is a corresponding circle (representing all the points antipodal to the original path). Moreover, Greg Egan's applet appears to represent a "side view" of the same idea.
Still having had no further reply, it seems none can or will explain in simple terms why my proposed short clarificatory insertion is invalid, so I believe I am free to insert it (some parts of my contribution on this page edited out as redundant 30-8-08, having inserted the suggested clarification) PaulGEllis ( talk) 13:30, 30 August 2008 (UTC)
Neat! Thankyou.
As a non-mathematician, I agree with your recommendation of an overhaul, but appeal for any such to be helpful to the non-expert. All too often, IMHO, the Wikipedia maths entries (which along with the science entries generally seem to be of the highest standard, as compared with entries on other areas) are written more for the expert than to help the student and interested enquirer whom I would expect to be a larger part of the readership (and potentially able to benefit far more) than fellow experts. PaulGEllis ( talk) 06:17, 31 August 2008 (UTC)
Late update (2017), just in case anyone else has a similar difficulty seeing how one can start shrinking the great circle (from 4-pi rotation) *without* maintaining the antipodal points: Imagine it in reverse by starting from the central identity point of the ball and expanding a loop outwards until it touches a boundary point. If it can be expanded to the boundary, it can also be contracted from the boundary. PaulGEllis ( talk) 13:54, 19 May 2017 (UTC)
In the article there is a sentence : "Hence, any length-preserving transformation in R3 preserves the dot product" which is not exactly true ! The transformation has to satisfy more constraints ! ( linearity ? ... it has to preserve the length of the sum too!) —Preceding unsigned comment added by 62.121.113.97 ( talk) 10:02, 27 December 2008 (UTC)
You know that this page is almost impenetrable for anyone not already familiar with rotation matrices? — Preceding unsigned comment added by 131.181.33.96 ( talk) 02:26, 13 June 2018 (UTC)
I found the first notation below (coded as \frac{\langle X,Y\rangle}{||X|| ||Y||}) in this article and changed it to the second (coded as \frac{\langle X,Y\rangle}{\|X\|\|Y\|}):
This should make it clear why the latter is standard usage. Michael Hardy ( talk) 19:45, 9 May 2019 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
-- Beland 16:27, 18 December 2005 (UTC)
Regarding this from the introduction: "A length-preserving transformation which reverses orientation is called an improper rotation. Every improper rotation of three-dimensional Euclidean space is a reflection in a plane through the origin." If I have a transformation T that reflects in the z-axis (negates z) and translates 2 units along in the x-axis, then according to the first sentence T is an improper rotation. The second sentence tells me that T, being an improper rotation, is a reflection in a plane through the origin. Yet I can think of no such reflection that reproduces the effect of T. Where am I going wrong here? Oioisaveloy ( talk) 10:44, 9 August 2009 (UTC)
In the article it is mentioned that SO(3) is homeomorphic to and .
However, the first homology group of is , and for the product, it's just , because S^2 is simply connected.
Can someone clarify that? —Preceding
unsigned comment added by
77.127.154.232 (
talk)
21:38, 1 November 2009 (UTC)
Next question: why a topological space homeomorphic' to the rotation group, but this manifold is diffeomorphic to the rotation group? why we use in one case the word homeomorphic, but in other - diffeomorphic? Explane, please! Vyrivykh ( talk) 14:50, 23 October 2010 (UTC)
For A(t), a one-parameter subgroup of SO(3) parametrised by t, differentiating with respect to t gives
Matekosor ( talk) 05:57, 1 August 2012 (UTC)
Shouldn't this page be renamed something like
because there are lots of rotation groups out there, but this article is about one very specific one. Jheald ( talk) 07:55, 13 September 2008 (UTC)
If one had done any checking/intuiting of Engø's formulas for compositions of arbitrary rotations, he would quickly supplant his ponderous notation for a and b with the much simpler, and intuitive, a=c cotφ/2, b=c cot θ/2. This would immediately shorten several of his formulas, notably for d, now proportional to c, and illuminate α,β,γ. I don't see any compulsion to stick to his original notation. But it is a purely aesthetic point... Cuzkatzimhut ( talk) 14:08, 3 January 2015 (UTC)
I added a physics project template in order for someone to rate the article. Someone did, but the importance of the topic was set to low. This is something I find ridiculous. The importance lies somewhere between high and top. Little Nothing in physics is more important than symmetries, rotational symmetry being the most important symmetry of them all. (Important = useful + much more (physical truth as far as we know)) I reverted the complete rating because of this misjudgement, hoping that someone else drops by to rate the article.
YohanN7 (
talk)
11:21, 8 January 2015 (UTC)
Summary: Extend the final sentence of the fourth para under the subheading Topology, with the part following the first comma here below:
"This circle can be shrunk to the north pole without problems, as an "antipodal circle" forms and simultaneously shrinks to the south pole." [revised again, 30-8-08]
Detail: As formerly shown in the main article, I don’t see how a single great circle can shrink, unless the antipodal point of every point on the shrinking circle described is also included. The article doesn’t seem to hint at this. However, after looking at Greg Egan’s applet:
http://gregegan.customer.netspace.net.au/APPLETS/21/21.html
... I am emboldened to suggest [a suitable] extension of the final sentence. Unless someone can explain comprehensibly (to someone at my level) how I’m in error, then after seven days I’ll post my suggested extension. OK? PaulGEllis ( talk) 19:46, 4 August 2008 (UTC)
Thank you for your reply. What I have a problem with is why you can shrink the great circle, as if it was on a plane. In my understanding (which I admit is not graduate mathematician level - I'm an interested physical chemist, thinking purely geometrically ): I can see how you get to a great circle including the north and south poles. And for that great circle, every point still has its equivalent antipodal point on the great circle. But when you start to shrink the great circle away from the south pole, then no point on [a smoothly; inserted 30-8-08] shrinking circle has an antipodal point represented. Since every point on the surface of the pi-radius ball is equivalent to its antipodal point, I can't see how you can start to shrink the great circle, unless there is a corresponding circle (representing all the points antipodal to the original path). Moreover, Greg Egan's applet appears to represent a "side view" of the same idea.
Still having had no further reply, it seems none can or will explain in simple terms why my proposed short clarificatory insertion is invalid, so I believe I am free to insert it (some parts of my contribution on this page edited out as redundant 30-8-08, having inserted the suggested clarification) PaulGEllis ( talk) 13:30, 30 August 2008 (UTC)
Neat! Thankyou.
As a non-mathematician, I agree with your recommendation of an overhaul, but appeal for any such to be helpful to the non-expert. All too often, IMHO, the Wikipedia maths entries (which along with the science entries generally seem to be of the highest standard, as compared with entries on other areas) are written more for the expert than to help the student and interested enquirer whom I would expect to be a larger part of the readership (and potentially able to benefit far more) than fellow experts. PaulGEllis ( talk) 06:17, 31 August 2008 (UTC)
Late update (2017), just in case anyone else has a similar difficulty seeing how one can start shrinking the great circle (from 4-pi rotation) *without* maintaining the antipodal points: Imagine it in reverse by starting from the central identity point of the ball and expanding a loop outwards until it touches a boundary point. If it can be expanded to the boundary, it can also be contracted from the boundary. PaulGEllis ( talk) 13:54, 19 May 2017 (UTC)
In the article there is a sentence : "Hence, any length-preserving transformation in R3 preserves the dot product" which is not exactly true ! The transformation has to satisfy more constraints ! ( linearity ? ... it has to preserve the length of the sum too!) —Preceding unsigned comment added by 62.121.113.97 ( talk) 10:02, 27 December 2008 (UTC)
You know that this page is almost impenetrable for anyone not already familiar with rotation matrices? — Preceding unsigned comment added by 131.181.33.96 ( talk) 02:26, 13 June 2018 (UTC)
I found the first notation below (coded as \frac{\langle X,Y\rangle}{||X|| ||Y||}) in this article and changed it to the second (coded as \frac{\langle X,Y\rangle}{\|X\|\|Y\|}):
This should make it clear why the latter is standard usage. Michael Hardy ( talk) 19:45, 9 May 2019 (UTC)