Talk:Rotation group SO(3)
|WikiProject Mathematics||(Rated Start-class, Mid-importance)|
Rotations and inner product spaces
In "properties" it is said that the standard inner product can be written purely in terms of lengths; this is (the so-called polar identity (?)) valid for any inner product space (in fact, for any symmetric bilinear form, I think).
Btw, I didn't find much on this on any of the pages I searched, just some quite well hidden remarks on quadratic form. If s.o. has an idea, it would be nice to have this added somewhere (eg in bilinear form). — MFH: Talk 21:32, 21 Jun 2005 (UTC)
- I don't understand the question. Can you rephrase it? What is it that you want more information on? In a standard real vector space, the "purely in terms of lengths" is just some simple vector space math, trigonometry, nothing deeper than that? linas 22:55, 21 Jun 2005 (UTC)
The length of a vector is given by its norm, which is given by the scalar product (or dot product), which in turn is defined by a symmetric positive definite bilinear form. But reciprocally, a norm defines a quadratic form which has an associated bilinear form (which I knew as its "polar form", but this term does not seem common here).
So one could define the rotation group w.r.t. any other (positive?) symmetric bilinear form. It seems to me if we take the bilinear form of index (3,1) i.e. with matrix diag(1,1,1,-1) we get the Lorentz group SO(3,1) pertaining to special relativity. — MFH: Talk 17:44, 22 Jun 2005 (UTC)
- Yes, all of those statements appear to be correct; I detect no question. Traditional, narrow formal usage has the rotation group being SO(n) only, although you will occasionally find the broader usage you refer to, e.g. SO(3,1) might be called the "group of hyperbolic rotations". I've seen other non-SO(N) things called roatations as well, although the usage is informal and speaker-listener-context-dependent. linas 00:29, 23 Jun 2005 (UTC)
Request for technical explanation
- The first meaning of "rotation group" should probably just be merged with symmetry group, or at least direct readers there for a fuller explanation. Otherwise, it's a bit confusing.
- It's at least a little sad that this article about geometrical relationships doesn't have any diagrams. It would be a lot more accessible if each distinct concept were illustrated with a diagram. For example, the basic concept of what a rotation group is could be illustrated by showing some rotations which are and some rotations which are not in the rotation group. These pictures could demonstrate which properties of the same are and are not preserved. An example of an improper rotation could also be diagrammed; the "properties", "axis of rotation" ,and "topology" sections are definitely ripe for obvious illustrations.
- It's not clear in the introduction whether "the rotation group" and "SO(3)" are the same thing.
- Even having studied matrices in college well enough to be able to find determinants and eigenvectors and whatnot, I find the "orthogonal matrices" quite dense, and in some places very difficult to understand. I was not familiar with the term standard basis, and trying to read that article did not help. But come to basis, and it turns out that this is essentially the notion of an XYZ coordinate system, which most readers are actually already familiar with. This connection should be explained. This will also help give them an example to understand what it's saying about orthonormality, at a glance. The discussion of matrices would definitely benefit from a concrete example showing a specific matrix representing a specific rotation (preferably with a picture showing which properties the numbers in the matrix correspond to).
-- Beland 16:27, 18 December 2005 (UTC)
Suggested clarification re topological path shrinkage
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:
... 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)
- that particular sentence in the article is fine. it's not clear what it is exactly that you have a problem with. the precise meaning of "shrink" here is homotopy, with the starting point of a loop fixed throughout the homotopy. the great circle is shrinkable to a point for the same reason any circle in the plane is shrinkable. Mct mht (talk) 21:44, 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)
- I don't know that it is invalid, but I do find the entire description could use an overhaul. If you have a path going from the north pole to the south pole and back, the *obvious* thing to do turns out to be the right one: just move the bottom point of the loop up through the interior of the ball along the polar axis until it hits the north pole. There is your homotopy. The present description is needlessly complicated, irrespective of your contribution to it (which, I should say, adds to rather than reduces the complexity of the description). siℓℓy rabbit (talk) 21:13, 30 August 2008 (UTC)
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)
Shouldn't this page be renamed something like
- Rotation Group SO(3), or
- [[Rotation Group of R3]], or
- Rotation Group (3D)
- I'm neutral about the proposed change. Other rotation groups are covered in the article orthogonal group. I have added a note to the top of the article to this effect. siℓℓy rabbit (talk) 11:41, 13 September 2008 (UTC)
- There are many rotation groups, SO3 is just one of them. It may be considered the most important, as it is the one acting on the R3 in which most Wikipedia users happen to be embedded – but mathematically, it is just one among many. So I believe
Preservng a dot product
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 188.8.131.52 (talk) 10:02, 27 December 2008 (UTC)
- No. This is true as stated. Linearity follows from the fact that the transformation preserves lengths. (Additivity follows essentially by the parallelogram law, and homogeneity follows because any continuous additive transformation is also homogeneous.) siℓℓy rabbit (talk) 15:54, 27 December 2008 (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 184.108.40.206 (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)
- Here are some answers:
- The topology section states that SO(3) is homeomorphic to , which is correct. Where is it stated that it is homeomorphic to ?
- The word homeomorphic makes sense for any pair of topological spaces, while the stronger notion of being diffeomorphic is only defined for differentiable manifolds. Because the first construction given is inherently topological (involving gluing along a boundary), it makes sense to say that the result is "homeomorphic" to . This gluing also induces a diffeomorphism between manifolds, but this requires additional proof. Jim (talk) 19:16, 23 October 2010 (UTC)
- Thank you very much! Vyrivykh (talk) 20:04, 23 October 2010 (UTC)
Lie algebra section
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)