Talk:Orthogonal group/Archive 1

Incorrect
The definition of the orthogonal group given here is FALSE. The orthogonal group is defined only over the real numbers. When defined over the complex numbers it is known as the unitary group. Over quaternions it is known as the spin group.

Let $$E$$ the the $$n \times n$$ identity matrix and
 * $$ \mathbb{H} = \{a + ib + jc + kd : (a,b,c,d) \in \mathbb{R}^4, \ i^2 = j^2 = k^2 = -1, \ ij = k, \ jk = i, \ ki = j \} \ . $$

Then the CORRECT definitions are given as


 * $$ O(n) = \{ X \in \mbox{Mat}(n,\mathbb{R}) : X^{\top}X = E \} \, $$
 * $$ U(n) = \{ X \in \mbox{Mat}(n,\mathbb{C}) : \overline{X}^{\top}X = E \} \, $$
 * $$ Sp(n) = \{ X \in \mbox{Mat}(n,\mathbb{H}) : \overline{X}^{\top}X = E \} \ . $$

The definition of the orthogonal group comes from finding all matrices which preserve the standard euclidean dot product, so the preserve distances and angles. The definition of the unitary group comes from finding matrices which preserve the standard Hermitian product. We see that
 * $$ O(n) \subset U(n) \subset Sp(n) \ . $$

In particicular, the statement that $$\det(X) = \pm 1$$ is also INCORRECT. It is true for the othogonal group since the definition $$X^{\top}X = E$$ means that $$\det(X^{\top}X)  = \det(E)$$ which in turn shows that $$\det(X)^2 = 1.$$

For the unitary group and the spin group we see that $$\det(\overline{X}^{\top}X) = \det(E)$$ this means that
 * $$\overline{\det(X)}\det(X) = 1 \, $$ which means that $$|\det(X)| = 1.$$

So for the unitary group, we see that
 * $$ \det(X) \in \{ z \in \mathbb{C} : |z| = 1 \} \neq \{\pm 1\} \ . $$

Since $$\mathbb{H}$$ is not commutative, the idea of a determinant is not well defined!

—The preceding unsigned comment was added by 128.101.10.93 (talk • contribs).


 * You are confusing the complex orthogonal group (preserving a symmetric form) with the unitary group (preserving a hermitean form). R.e.b. 18:52, 5 May 2006 (UTC)

Over a finite field
I think it would be useful to discuss the orthogonal groups over a finite field here. If someone is capable of writing such a section it would be greatly appreciated. TooMuchMath 17:31, 15 April 2006 (UTC)
 * This is absolutely essential. It's a shame that there is no section here on it. My knowledge of these groups isn't good enough to do a section, however. We also need some words on O+ and O- etc. Triangle e 14:26, 9 December 2006 (UTC)
 * I've added the most minimal amount of finite field stuff. I'd be grateful if someone could please expand. Triangle e 14:40, 22 April 2007 (UTC)
 * Things are progressing with orthogonal groups over finite fields. I removed some incorrect order formulas, and am working on giving the orders correctly and legibly.  Here is a first draft, but only for odd characteristic.  I'll work on the characteristic two case next.  Both may require some reorganizing of the finite field sections (so that spinor and dickson come early enough to define SO and Ω).

For an odd prime power q: The orthogonal groups may refer to several particular families of groups, Ω < SO < O, each with index 2 in the next. The group Ω is the derived subgroup both of SO and of O, and Ω is usually a perfect group. One may define SO as the kernel of the determinant map on O, and Ω as the kernel of the spinor norm map on SO. Additionally, one is often interested in the quotients by the center, PΩ ≤ PSO ≤ PO, especially since PΩ is usually a simple group. For an odd prime power q, the orders of the orthogonal groups are given by:
 * Note the constant, messy parts are split off to make it easy to compare the orders. Perhaps some of this is silly, since the center is so small, people should be able to work out things for themselves based on just one of them, but which one depends heavily on the area in which the person is working.  For me, O is useless, and only PΩ and Ω are interesting. JackSchmidt (talk) 23:05, 24 November 2007 (UTC)


 * Here is the even case.

For q a power of two: The orthogonal groups may refer to either of two families of groups, Ω ≤ O.  The group Ω is the kernel of the Dickson pseudodeterminant; note that both the spinor norm and determinant are trivial for finite fields of characteristic 2. The group Ω is the derived subgroup of O, and Ω is usually a simple group. When the dimension of the underlying vector space is odd, then in fact O(2k+1,q) = Sp(2k,q). Note that in characteristic 2, the orthogonal groups have trivial center, so there is no distinction between O and PO, Ω and PΩ. The orders of the orthogonal groups are given by: Small even dimensions and all odd dimensions have additional names:
 * Oε(2,q) = Dih(2(q−ε))
 * Ωε(2,q) is cyclic of order q−ε
 * O+1(4,q) = PSL(2,q) wreath Sym(2)
 * Ω+1(4,q) = PSL(2,q) × PSL(2,q), for q>2, but Ω+1(4,2) is an exception
 * Ω-1(4,q) = PSL(2,q²)
 * Ω+1(6,q) = PSL(4,q)
 * Ω-1(6,q) = PSU(4,q)
 * Ω(2k+1,q) = O(2k+1,q) = PSp(2k,q)
 * I wasn't sure of the best way to handle the weirdo O+(4,2), since its derived subgroup is smaller than expected. At any rate, I'll copy these into the article in some sane way this week so it will be easier to tweak.  I just wanted to put some drafts up first in case someone had major stylistic concerns.  The formulas are very detailed and easy to get wrong (different sources use different notations), so it is probably best not to adjust too much without checking carefully. JackSchmidt (talk) 16:43, 26 November 2007 (UTC)

Incorrect order formulas
Above I posted a suggestion on presenting the order formulas, but my edit was reverted. Just to make sure everyone is on the same page, here are the wrong order formulas: When n=1,q=3, the set of two by two matrices over the field with three elements whose inverse is their transpose has cardinality 8 as can be seen by writing down the 48 such invertible matrices and checking inverses. The first formula clearly doesn't apply, the second does not either but should give the order as 32, not a divisor of 48. The third formula is the one that should apply, but it gives the order as 64, definitely not a divisor of 48.
 * $$|O(2n+1,q)|=2q^n\prod_{i=0}^{n-1}(q^{2n}-q^{2i})$$
 * If $$-1$$ is a square in $$\mathbf{F}_q$$, $$|O(2n,q)|=2(q^n-1)\prod_{i=0}^{n-1}(q^{2n}-q^{2i})$$
 * If $$-1$$ is a nonsquare in $$\mathbf{F}_q$$, $$|O(2n,q)|=2(q^n+(-1)^{n+1})\prod_{i=0}^{n-1}(q^{2n}-q^{2i})$$

At any rate, correct order formula for all orthogonal groups are given above. In particular, O+(2,3) has order 4 and O-(2,3) has order 8. A reference is Grove's Classical Groups and Geometric Algebra, but I was planning on comparing references before including it in the article. JackSchmidt (talk) 21:55, 29 November 2007 (UTC)

Ups, you are right. The zero in the last two formulas should be a 1. —Preceding unsigned comment added by Franklin.vp (talk • contribs) 22:31, 29 November 2007 (UTC)


 * Thanks for checking the talk page. I checked the paper and it looks like this time the typo was not in the paper, but the published literature is filled with typos like this.  That's why I said above to be careful checking the formulas; make sure to work some examples.


 * At some point, I will probably still want to replace the current article's order formulas with something like the tables above. However, your contribution is very sane and concrete and has a sane and concrete reference.  I like {A:AA^t=I} much better than the Clifford algebras approach in Grove.  It would be good to decide clearly when {A:AA^t=I} is O+ and when it is O-, hopefully with a reference (one can basically tell from the order formula). JackSchmidt (talk) 22:53, 29 November 2007 (UTC)

finite field
Is the definition, as stated in the article, for the orthogonal group over an arbitrary field correct? I think that, for example over finite fields the definition is not ``very good'' as the group we obtain in this manner need not be split. —Preceding unsigned comment added by 129.175.50.21 (talk) 10:57, 23 September 2010 (UTC)

Incorrect: GO is not another notation for O
In the literature GO is not simply another notation for O analogous to the GL/SL notation, as is claimed in the article. GO refers to similitude orthogonal matrices, i.e., those which preserve the quadratic form up to a scalar. —Preceding unsigned comment added by 140.180.130.54 (talk) 04:14, 30 November 2010 (UTC)

Development
I have reworded the introduction to give the more general definition first and then specialise into the Euclidean group. I propose to add some more material about groups for forms other than the Euclidean and this makes a better framework. Deltahedron (talk) 06:44, 2 August 2012 (UTC)

Plans for improvement
If an article is large, it is not necessarily good. This article require a significant effort to become a clean article.

First of all, structuring of the content is currently horrific. Subsections Even and odd dimension, Conformal group and Lie algebra have nothing to do, specifically, with real number field. Contrary, the Topology section deals exclusively with $O(R)$, but wrongfully resides on the ==-== level.

Second, I intend to enforce the Lie algebra notation mentioned in []. Objections? Incnis Mrsi (talk) 17:48, 17 September 2012 (UTC)
 * ✅. Also, I explain why I demolished the “3D isometries that leave the origin fixed” section: we already have Rotation group SO(3) article, which is more specific and topical for this content. Incnis Mrsi (talk) 16:51, 12 February 2013 (UTC)

Interpretation of homotopy groups section
Whoops… I missed that H is not a ring, because it is not an associative algebra. Suggestions? Incnis Mrsi (talk) 17:11, 12 February 2013 (UTC)

Even dimension
I'm not too clear on the difference between $$SO(n)$$ and $$O(n)$$ in even dimensions for real spaces, given that reflection through the origin is part of the connected component about the identity in the even case. If they are the same, should that not be made explicit?--Leon (talk) 20:13, 11 September 2010 (UTC)


 * SO(n) and O(n) are not the same in even dimensions, if that is what you are thinking. E.g. in two dimensions SO(2) is the group of all rotations about the origin, O(2) is all such rotations + all reflections along a line through the origin. The groups are closely related: not only is SO(2) a subgroup of O(2) but any two reflections gives a rotation. More generally an even number of reflections gives a rotation, and in n-dimensions all rotations can be generated this way.
 * To get the "reflection through the origin" you can reflect along each of the the axes. Two axes in two dimensions, so two reflections which is a rotation, and the same is true in any dimension. The 'reflection through the origin' is not a reflection in the usual sense in even dimensions, it's a rotation. It's usually a particularly interesting rotation: in 2D it's the only rotation which when done twice gives the identity, so is it's own inverse - this is true in higher dimensions though other rotations have the same property. In 4D it's isoclinic, and if that classification were generalised it would be isoclinic in higher dimensions too. -- JohnBlackburne wordsdeeds 20:33, 11 September 2010 (UTC)


 * added the above (slightly edited to the page WillemienH (talk) 11:36, 22 November 2014 (UTC)

I believe the order formula shown is incorrect! There should be formulae for the plus and minus types. These formulae are given in R.A. Wilson, The Finite Simple Groups in Section 3.7.2. The formulae work for q even or odd and do not depend on whether -1 has a square root in the field. — Preceding unsigned comment added by 1.127.170.144 (talk) 06:04, 4 April 2013 (UTC)

alternating vs. antisymmetric
I think the term "alternating" should be used with more care, especially in the section about char=2. The definition of 'alternating' is different from the definition of 'antisymmetric', symmetric coincides with antisymmetric but is distinct from alternating when char=2. srostami (talk) 17:30, 2 May 2015 (UTC)
 * Are you referring to that the text states that symmetric forms are alternating forms in characteristic 2? This does seem odd, because in characteristic 2 not every symmetric form is alternating, even though every alternating form is symmetric. At a glance, there appears to be an error in the conclusion. —Quondum 20:30, 2 May 2015 (UTC)

Split and merge templates - is the resolved/not-yet-discussed/ ?
I saw the year-old split and move templates (my bad habit of being a "wandering Wikipedaian"), although there aren't any sections on the Math portal talk page. Have these issues been resolved, or just raised-but-never -discussed? Can the templates be removed? Jimw338 (talk) 14:58, 20 April 2015 (UTC)


 * I suspect that this is a case of raised-but-never-discussed. It seems reasonable to remove the templates; I see no compelling reason to split out the special orthogonal group until more material on it is added. I cannot comment on the classical groups. —Quondum 16:15, 20 April 2015 (UTC)


 * Yes, split this into Orthogonal Group and Special Orthogonal Group. Do NOT worry about repetition within the separate articles.  It is highly annoying that Special Orthogonal Group does not exist and redirects here.  — Preceding unsigned comment added by Berrtus (talk • contribs) 05:19, 3 May 2015 (UTC)


 * Very weak support for Split. The original discussion/disagreement is here: Wikipedia talk:WikiProject Mathematics/Archive/2014/Mar and begins by confusing the rotation group with the orthogonal group (they are not the same; rotations are special-orthogonal). Next comes arguments about finite fields, which is interesting to mathematicians but useless to the physics students interested in rotations and/or the classical Lie groups. As it currently stands, the article is already long, making it hard or impossible to further expand it, without getting mired in the length. On the other hand, I've seen some splits go badly, making the subject harder to understand, because some grand unifying concept was lost during the split. There's a huge tangle of inter-related ideas here, making splitting hard.  So care must be taken...  67.198.37.16 (talk) 18:11, 2 September 2016 (UTC)

Dimensions in Low-dimensional topology
I'm a bit confused about the dimensions in this section. The dimension of SO(3) should be 2 but the dimension of RP^3 is 3 so they can not be isomorphic. — Preceding unsigned comment added by 2A02:908:671:E920:F855:8B65:E233:57E7 (talk) 21:55, 2 September 2015 (UTC)


 * You are confused. SO(3) is famously double-covered by SU(2) which is why spinors are rotations. SU(2) is isomorphic to the 3-sphere. But the very easiest way to see that SO(3) is 3-dimensional is to see that you need three numbers - the Euler angles, to describe a rotation. 67.198.37.16 (talk) 18:27, 2 September 2016 (UTC)

Isoclinic ...
In article read: In 4D it is isoclinic...

Isoclinic links to Isoclinism of groups. In that article read: isoclinism is an equivalence relation on groups which generalizes isomorphism.

So, In 4D it is isoclinic to WHAT? Jumpow (talk) 17:12, 2 December 2015 (UTC)


 * It should read that in 4D, the reflections are isoclinic i.e. the reflected 4D space is isoclinic to itself. That whole paragraph is poorly worded, though, and could use clarification/expansion, as the reflection trick is a pretty cool/important trick. It should serve as an introduction to Clifford algebra among other things ...  67.198.37.16 (talk) 18:31, 2 September 2016 (UTC)

It is a major omission that this article fails to describe the matrix of a rotational reflection in 3-space
This article discusses many wonderful advanced and very advanced mathematical topics with great skill.

But it neglects to mention one of the most basic topics -- which must be covered first: what is the matrix form of an orientation-reversing orthogonal matrix in 3-space. And I don't mean "with respect to a suitable basis". I mean with respect to the standard basis.

Not even the matrix form of the simple reflection in a plane in 3-space appears here!Daqu (talk) 19:48, 24 January 2015 (UTC)


 * This is discussed in the section "In even and odd dimension" although poorly and quite incompletely. For the 3D case, it really really needs to go into the O(3) article which is where you should look. 67.198.37.16 (talk) 18:34, 2 September 2016 (UTC)