User talk:YohanN7/Classical groups

Initial queries

 * I'm mildly puzzled. Presumably this is intended as generalization to say division rings. Why restrict it to real division rings? All fields should be covered, finite and otherwise, before division rings are considered. One must take care to state it correctly for char 2 fields (probably alternating, not skew-symmetric).
 * This is supposed to be a treatment of bilinear forms on $R, C, H$ only since the classical groups use these fields. YohanN7 (talk) 13:40, 9 July 2014 (UTC)

—Quondum 02:23, 26 March 2014 (UTC)
 * The choice of bilinear form seems to unmotivated (sourced or not) for division rings. As defined, it is inherently zero for all division rings that are not fields. Rather restrict the definition to fields, or define a bilinear form as
 * \phi(\alpha u, v\beta) = \alpha\phi(u, v)\beta, \quad \forall u,v \in V, \forall \alpha,\beta \in F
 * The sesquilinear form is defined wrong on H (it is inherently zero):
 * $$\phi(u\alpha, v\beta) = \alpha\phi(u, v)\overline{\beta}, \quad \forall u,v \in V, \forall \alpha,\beta \in F.$$
 * It should be one of :
 * \phi(\alpha u, \beta v) = \alpha\phi(u, v)\overline{\beta}, \quad \forall u,v \in V, \forall \alpha,\beta \in F
 * $$\phi(u\alpha, v\beta) = \overline{\alpha}\phi(u, v)\beta, \quad \forall u,v \in V, \forall \alpha,\beta \in F.$$
 * Also, just as a matter of style, why are right vector spaces chosen?
 * The latter is used. (Right vector spaces are used primarily to enable matrix multiplication from the left as usual.)YohanN7 (talk) 13:40, 9 July 2014 (UTC)


 * I see EOM has a nice article Sesquilinear form. It completely defines it, including what an antiautomorphism is and how a sesquilinear form is defined in terms of it for any right vector space. I've taken the liberty of making some tweaks.
 * EOM's approach is to define a general form using $σ$; if it produces an antiautomorphism, it is a sesquilinear form, and if it is the identity map, it produces a bilinear form. —Quondum 06:49, 26 March 2014 (UTC)
 * The problem with this generalization is that it is not made in the reference I'm working on. I can't be entirely sure all group theory based on the invariance of our forms holds true when generalized this way. It probably does, but the generalization (if it works) is probably only mildly more interesting (from the perspective of classical groups) than a change of basis. Else it would be found in the literature. YohanN7 (talk) 13:40, 9 July 2014 (UTC)

You must be aware of the article Classical group? —Quondum 03:00, 10 July 2014 (UTC)
 * Certainly. I intend more or less to overwrite it. It is truly horrible and vastly incorrect. YohanN7 (talk) 03:26, 10 July 2014 (UTC)

Rossmann
The more I get acquainted with Rossmann's book, the more impressed I am by how good it is. However, I have known that there are more than two, three errors per page. Now, when I do this..., Holy Moses. His formulas aren't trustworthy, though his line of thought is. YohanN7 (talk) 05:03, 10 July 2014 (UTC)

Tall vertical bar | in LaTex
How to produce one? There is a formula for the Lie algebra $Sp(m, R)$. In that the "such that" is now a semi-colon. I'd like it to be a tall vertical bar sizing to the matrix on the left. YohanN7 (talk) 05:07, 10 July 2014 (UTC)
 * Done.  produces $$\left. \sum \cdots \right|$$ —Quondum 05:43, 10 July 2014 (UTC)
 * Thanks. Ha, I though I tried just that. Well well... YohanN7 (talk) 06:37, 10 July 2014 (UTC)

Why only the special groups?
The Lorentz group appears to be $O(1,3)$, not $SO(1,3)$. One needs to stick to the modern definitions of terms (on which I'm no authority), rather than only one author's presentation thereof. The restriction to the special subgroups would probably only be of historical interest. Also, a statement like "The rotation group SO(3) is the symmetry of Euclidean space, the Lorentz group SO(3,1) is the symmetry group of spacetime of special relativity" is incorrect, but replace the groups by dropping the 'S', and it becomes correct for the tangent vector space, not for the space itself). It is the Euclidean group that is isometry group of Euclidean space, and the Poincaré group that is the isometry group of spacetime in special relativity. —Quondum 20:13, 10 July 2014 (UTC)


 * Right. Didn't spend too many seconds thinking about that section. I have worse problems with Rossmann's typos. I have seen so many now that I probably see typos even where there aren't any. YohanN7 (talk) 21:38, 10 July 2014 (UTC)

Done
Article now at Classical group. YohanN7 (talk) 18:43, 11 July 2014 (UTC)


 * I see you're still editing here. I presume you'll overwrite article with what is here again. —Quondum 20:35, 13 July 2014 (UTC)
 * Yes, that is the plan. It turned out that there were a number of small errors and formatting issues. Instead of doing 50 edits there, I'll simply replace it when done. Should be okay since only I have touched it.


 * Thanks for the fixes. Appreciated. B t w, Hermitian is always with a capital H. (Should logically be Hermitean. Dudes name wasn't Hermiti)YohanN7 (talk) 22:04, 13 July 2014 (UTC)


 * Cool, just watch in case someone has made edits there, in which case you'll need to merge. I tend to take liberties with lower-case, so feel free to fix. I'm a fan of words derived from names being in lower-case (and they go there eventually anyway). I was being lazy, not checking that this one was still not at least sometimes written as lower-case. —Quondum 22:38, 13 July 2014 (UTC)


 * Move made. I don't know of any direct errors now, and I hope to edit "normally" in the future. Thanks for all the help. What do you think about the article? YohanN7 (talk) 23:39, 13 July 2014 (UTC)
 * Nice, the changes show in the history as they should. I expected something to break... YohanN7 (talk) 23:42, 13 July 2014 (UTC)


 * It's looking good, though a lot of the detail is a bit beyond me. But it strikes me as readable, and something I could use to try to learn about the concepts.  The bold in the "Compact form" column of the table has no purpose that I can see, though I was unsure of this so I left it. The word "hamiltonian" might be like "Hermitian" – always with a capital 'H', and "Hermitian"/"Hermitean" should use a consistent spelling.  The lead is good, though I'm not entirely comfortable with the definition in the first sentence.  You specify special linear groups, but I'm not convinced that this is universal and you hint that it is not later.  I "nicer" definition would be as GL(R, n) with det = ±1 and the subgroups thereof – I guess that this actually encompasses all the C and H cases, but may be wrong.  The most mathematically useful category of groups would make sense to me here, rather than what was historically defined, but I suppose that is a bad argument on WP. It comes across slightly like a list-class article, which suggests that a category theory-oriented approach to classification may add to the article. But generally, I'd say it is of a pretty decent quality already. —Quondum 02:10, 14 July 2014 (UTC)


 * The bold compact groups are the compact classical groups. They are the compact real forms of the complex classical groups. This is mentioned in the article. The problem with defining classical groups is that it is not really a mathematical concept. The definition in terms of groups that preserve bilinear and sesquilinear forms is probably the closest to a rigorous definition you'll get. The presence of $R, C$ and $H$ can be motivated by that the groups are the only ones that can be embedded (not uniquely, but "invariantly") as non-discrete closed subgroups of some $GL(n, R)$, making them Lie groups. This doesn't work for any other (non-commutative) field as far as I know. The special linear groups are exceptions to this pattern, they don't preserve forms on $R^{n}$, but there is nothing to do about that. (They do preserve $R$-linear forms on their Lie algebras.) A definition like $SL(n, R$ with subgroups wouldn't work because it would bring along all linear groups with determinant one. Not all of these are Lie groups, even some fairly nice ones aren't.


 * One thing is for sure. The approach using preservation of forms is the only approach that doesn't just produce a bewildering maze that is hard to understand for those taking a first course (beyond abstract group theory). It also makes it comparatively very simple to compute the Lie algebras. This too is usually presented in an ad hoc fashion in most text books. YohanN7 (talk) 05:29, 14 July 2014 (UTC)


 * Okay, I have in mind the group of transformations that preserve a bilinear form on a real vector space. When the bilinear form is symmetric, these transformation groups should be all the orthogonal groups of all dimensions and signatures.  When the bilinear form is alternating, perhaps it would be all the symplectic groups?  And one can presumably also have bilinear forms that are not reflexive, and those that are degenerate may introduce further funnies. I'm obviously missing something, because there is a wealth of classification that does not seem to be covered in this. I'm rather interested in this construction, because it is a way of constructing Klein geometries. —Quondum 06:30, 14 July 2014 (UTC)


 * You should not see the classical groups in a serious classification scheme. As you said, the article comes across as a list-class article. This is because it is a list-class article (of sorts). In addition, not everyone agrees precisely which groups are supposed to be on the list. But this doesn't really matter, not in the article and not in the literature. The determinant = 1 condition doesn't present any problems, either it is there (together with a tracelessness condition of the Lie algebra) or it is not.
 * Yes, the automorphism groups of non-degenerate bilinear forms on finite-dimensional real vector spaces are all (pseudo-)orthogonal and all symplectic transformation groups (over $R$). This should really be clear from the article: Classical group#Automorphism groups#Bilinear case#Real case.
 * Degenerate forms do not produce groups, at least not this way because the adjoint with respect to the form isn't defined. Then I don't know what you mean by non-reflexive forms. YohanN7 (talk) 12:19, 14 July 2014 (UTC)
 * You should not see the classical groups in a serious classification scheme – Cool. But I would like to understand groups of this general construction better, and it is useful to be able to characterize these (i.e find categories that at least contain them).
 * They are categorized as precisely as the literature allows - general linear + automorphism groups of bilinear and sesquilinear forms, with definite reflexivity properties, with or without the determinant 1 condition. YohanN7 (talk) 19:22, 14 July 2014 (UTC)
 * Special, not general, I presume you mean. An SL group is the automorphism group of a vector space with determinant, which allows us to unify language a bit. The automorphism groups of R-, C- and H-vector spaces with the additional structure of a determinant, a nondegenerate bilinear form or a nondegenerate semi-bilinear form. It probably makes sense to separate out the H case, because nondegenerate bilinear H-forms do not exist, and determinants of H-vector space linear transformations are ill-defined (though you seem to think otherwise just below), leaving only the semi-bilinear form case for H.  Then the uncertainty of what to include then amounts to what subgroups of these to include, e.g. intersections of these groups.  So we may have our overall category, just not all groups in the category are regarded as classical groups. —Quondum 04:18, 15 July 2014 (UTC)


 * I mean general, possibly (usually) with the determinant 1 condition making it special because that's how it is formulated in neutral references. Why go through all this trouble when it isn't done in the literature? It will serve only to confuse. Determinant 1 is much more straightforward than "automorphism group of determinant function". Never heard of it before, and it can't preserve determinants, its elements has determinants with value 1 (and they preserve volume in a certain sense, but that is another story). Bilinear forms on $H$-spaces are excluded in the article because they are zero. I do not only think the determinant is well-defined for quaternionic groups, I know it. Why this is so can be found in the literature and in the article. YohanN7 (talk) 05:58, 15 July 2014 (UTC)
 * Ah, and that volume is a determinant,
 * $$\mathrm{det}(Av_1, \ldots, Av_n) = \mathrm{det}(v_1, \ldots, v_n), \quad A \in \mathrm{SL}(n, \mathbb{R}).$$
 * I can buy that, but it is still going over the bridge for water. YohanN7 (talk) 06:14, 15 July 2014 (UTC)


 * First point (determinant 1) conceded. But as to determinants of transforms of H-linear transforms, I think this is a subject to be checked. Non-commutativity presents severe difficulties to defining a determinant, at least of matrices, but I suppose that is not what we are referring to here. Come to think of it, constructing the automorphism group of an H-vector space is quite a challenge, even before we try to make sense of what a determinant might mean. —Quondum 06:16, 15 July 2014 (UTC)


 * Once again you are arguing about a subject you don't know anything about. You automatically assume that the things you do not know about nobody else knows about either. And, you are taking on a lecturing tone hinting that I am naive in believing that matrices over non-commutative fields can have well-defined determinants. Also, the automorphism groups for the $H$-vector spaces are probably just a dream I've had and they should go out of the article?


 * It is pointless to try to convince you about quaternion groups and their determinants. Have you actually read the article? Have you actually read the relevant literature? Do you know how the determinant is defined? Obviously, you don't since you are talking about problems constructing automorphism groups that must be solved before we attack the determinant problem. I have news for you. These simple problems are solved since 100 years ago. The solutions are presented in the article (often with citations), i.e. how to construct automorphism groups and how the determinant is to be defined. Thus we don't stand before any severe difficulties in defining a determinant.


 * Like I said, these problems have simple solutions. I'll not tell you what they are and I'll not again urge you to read the article because neither of these approaches on my part work. You need to find out for yourself. YohanN7 (talk) 14:30, 15 July 2014 (UTC)


 * The determinant = 1 condition – it seems to me that the preservation of a nondegenerate bilinear form (i.e. $b(Tu,Tv) = b(u,v))$ guarantees that $det T = ±1$; am I right? So the restriction $det T = ±1$ (Strikeout by YohanN7) $det T = +1$ would merely be to drop half the group?
 * Symmetric bilinear ⇔ $det T = ±1$, skew-symmetric bilinear ⇔ $det T = +1$ from the outset. But you also have Hermitian forms. In this case, $1=|det T| = 1$. The quaternionic groups come out with $det T = +1$ ($Sp(p, q)$) and $det T = ±1$ ($O*(2n)$). This is, of course, before any enforcement of the determinant = 1 condition. YohanN7 (talk) 19:22, 14 July 2014 (UTC)
 * the automorphism groups of non-degenerate bilinear forms on finite-dimensional real vector spaces are all (pseudo-)orthogonal and all symplectic transformation groups (over $R$) – This is only true if interpreted as "are all" → "includes all". Whenever the bilinear form is not reflexive, other groups that do not fit into these categories will be generated. I was using "non-reflexive bilinear forms" as shorthand for "neither symmetric nor antisymmetric", because (over a base field of characteristic not 2), they are the same.
 * As the article states, a bilinear form is uniquely a sum of a symmetric and a skew-symmetric form. Likewise for Hermitian and skew-Hermitian. For such a mixed form, the automorphism group is simply the intersection of the automorphism groups of the individual terms. For examples, see the last two sections in the article. YohanN7 (talk) 19:22, 14 July 2014 (UTC)
 * Interesting. So the automorphism group of symmetric and antisymmetric bilinear forms is much larger than for a mix. —Quondum 04:18, 15 July 2014 (UTC)
 * Degenerate forms do not produce groups: The preservation of a degenerate bilinear form can define a group. Example: a degenerate symmetric bilinear form.  Stretches of the vector space in the null directions become part of the group.
 * Can you formalize that? I think I might know what you mean. If so, then the resulting group would be $Aut_{n, m}(φ)$, a group with the first $n − m$ rows and columns arbitrary (assuming a suitable ordering of the basis) (edit: subject to invertibility of course) and then a $m × m$ block corresponding to the maximal non-degenerate subspace of the form. Existing, but not interesting and excluded by the definition of the classical groups. Classical groups are supposed to be useful. YohanN7 (talk) 19:22, 14 July 2014 (UTC)
 * Yes, essentially. The resulting group is presumably the direct product of a GL and a classical group, and as you say, it makes sense to exclude them as valueless. —Quondum 04:18, 15 July 2014 (UTC)
 * No, not a direct product. Just GL-matrices except for an Aut-block. YohanN7 (talk) 05:58, 15 July 2014 (UTC)
 * On third thought, block-diagonal matrices with a GL-block and an Aut-block. Degenerate and non-degenerate subspaces may not mix. YohanN7 (talk) 05:58, 15 July 2014 (UTC)
 * Yup, essentially. Except that it is more complicated than this matrix description. The subspaces may not mix, but the bases can, meaning that under a general basis transformation such a pattern may not be evident from the transformation matrix. —Quondum 13:11, 15 July 2014 (UTC)
 * A coordinate transformation does not change the dimension of the degenerate and non-degenerate spaces. The new matrices, upon reoredring the new basis, will have the same appearance. YohanN7 (talk) 14:30, 15 July 2014 (UTC)
 * An aside: it seems to me that by requiring two symmetric bilinear forms to be preserved by the transformations, we can construct groups such as the Euclidean group and the Poincaré group.
 * I really don't see how. Perhaps we can, but they aren't classical groups. (Translation groups preserve only the zero forms.) YohanN7 (talk) 19:22, 14 July 2014 (UTC)
 * Oops. No, I take that back. Doing maths in my head sometimes results in mistakes. —Quondum 04:18, 15 July 2014 (UTC)
 * BTW, I'm still very weak on Lie algebras and compactness, so have not even thought about these concepts properly. —Quondum 16:27, 14 July 2014 (UTC)
 * Think of the Lie algebra as representing elements of the group near the identity where the groups behavior can be modeled directly by that of the Lie algebra. This viewpoint is very very non-Bourbaki (non-rigorous), but also very very useful conceptually. YohanN7 (talk) 19:22, 14 July 2014 (UTC)