Talk:Group action/Archive 1

Meaning of invariant vs. stable
If G acts on A, I think that "invariant" means G.A=A while "stable" means that G.A is a subset (possibly proper) of A. — Preceding unsigned comment added by 130.104.2.103 (talk) 07:14, 15 November 2005 (UTC)


 * If so, you can't actually have G.A &ne; A, with G a group. Charles Matthews 16:21, 15 November 2005 (UTC)


 * due to the neutral ? You're true. It you think to the whole matricial group (M) acting on othogonal matrices (O), the group O is not stable because MO is not an orthogonal matrice in general. Then my initial bracket "possibly proper" is actually not necessary. — Preceding unsigned comment added by Moky~enwiki (talk • contribs) 02:27, 18 November 2005 (UTC)

Notation
In the first sentence under definition, $$g$$ is used both as the group action, and as an element of the group $$G$$. Suggest use (say) $$\alpha$$ for the group action instead. Tveldhui 01:38, 17 January 2006 (UTC)tveldhui


 * the dot denotes the action, g is always a group element. -MarSch 16:25, 1 February 2006 (UTC)

Reference
I've added a reference from the group article which I believe is sufficiently familiar and encylopedic to satisfy many people. I propose we use this reference to obtain a uniform set of notations and terms, perhaps even in multiple articles. Perhaps I'll get around to it in a week or so, with consensus, if no one else does. Orthografer 18:53, 29 June 2006 (UTC)

Added reference
I have added a reference to results of Higgins and me on the fundamental groupoid of an orbit space since this is a powerful result, and I hope people will find this useful. RonnieBrown (talk) 04:52, 29 September 2006 (UTC)

similar structure
Does anybody know the following structure, similar to the stabilizer, defined, for A,B\subset 2^G (the power set), by G(A,B) = { x\in G | \forall a\in A \exists b\in B : b x \subset a } (Note: a,b are subsets of G !)

Might this exist in the context of topological groups (where A,B would be neighborhoods of the identity element) ? It ressembles to the Kolmogorov definition of bounded sets

I ran across this in the context of modules over a ring (G would be a module over R, and B\subset 2^R). This seems so basic (and has nice properties) that I'm sure it already exists in literature, but I could not yet find it. Thanks in advance ! &mdash; MFH:Talk 21:50, 12 October 2006 (UTC)

sharply transitive = regular (= simply transitive) ?
As far as I understand the term sharply transitive is the same as regular or simply transitive, isn't it? If so, it would be good to mention this in the article. Florianhe 21:19, 11 January 2007 (UTC)

This is true. It seems to disrupt the flow of the article to mention it. One could delete the part about "sharply transitive" since that term is not widely used (rather, one says "regular"). The part about sharply k-transitive is fine, of course. Similarly, simply transitive is not familiar to my ears; rather simply primitive means primitive and not 2-transitive. JackSchmidt 02:21, 3 July 2007 (UTC)

Convert the formulae to HTML
I think the formulae could be replaced with HTML code, as follows:
 * G &times; X &rarr; X
 * ( g, x ) &#8614; g&middot;x
 * l : G &times; M &rarr; M : ( g, m ) &#8614; r ( m , g &minus;1 )
 * l ( g &middot; h, m ) = r ( m , ( g &middot; h ) &minus;1 ) = m &middot; h &minus;1 &middot; g &minus;1 = l ( h , m )&middot; g &minus;1 = l ( g , l ( h , m ))
 * l ( e, m ) = r ( m , e &minus;1 ) = m &middot; e = m

Does anybody oppose?
 * 157.25.5.68 12:27, 2 July 2007 (UTC)

I think the recent HTML conversion makes this article much harder to edit. While some of the complicated formulas may be just as complicated in wikipedia's &lt;math&gt; tags as in the new spans, certainly one should prefer &#39;&#39;G&#39;&#39; (that is G), to &lt;span class="texhtml"&gt;&lt;var&gt;G&lt;/var&gt;&lt;/span&gt; (that is G ).

I suggest reverting the simplest formulas back to wiki syntax, but am ambivalent about the other changes. Certainly for those with bad eyesight, avoiding the awful .pngs produced by some &lt;math&gt; tags is an advantage.

There are some bad wikilinks in some of the formulas as well. In particular, linking the letter e makes the article harder to read and there is a tradition to link once, and to use a readable description (as done in the second link). I worry that perhaps this article is being rewritten for computers, rather than for human consumption. I'll fix those links in a few days in case you want to adjust any of the HTML in the meantime.

Note that &amp;real; (&real;) is a poor substitute for the more common &amp;#x211d; (&#x211d;) or the tiny &lt;math&gt;\Bbb{R}&lt;/math&gt; ($$\mathbb{R}$$).

JackSchmidt 21:29, 19 July 2007 (UTC)

Reference needed
I need a reference/citation for "Every transitive G action is isomorphic to left multiplication by G on the set of left cosets of some subgroup H of G." can anyone help? (My job doesn't give me good library access) mike40033 (talk) 03:17, 10 December 2007 (UTC)
 * The references in the article by Dummit&amp;Foote or Rotman both contain this. Most abstract algebra texts have a good chance of proving this.  It is often called the orbit-stabilizer theorem.
 * If you are just curious how to prove it, then you might just try it. Say G acts transitively on X, and H is the subgroup of G fixing a point x in X.  For every other point y of X, there is some g in G taking x to y, since G is transitive. For every h in H, applying h to x first, and then applying g, takes you from x to y too.  Assuming you act on the right, that just says that every element of Hg takes x to y.  If some other element k of G takes x to y, then k*g^-1=h takes x to x and so k=h*g is in Hg.  In other words, for every y, there is a unique coset Hg consisting of precisely those elements which take x to y.  This defines a bijection from X to the cosets of H in G.  The actions are the same basically by definition.
 * Note this page is for discussing how to improve the article. There is another page where you can ask general reference questions about mathematics. JackSchmidt (talk) 04:18, 10 December 2007 (UTC)
 * I must have missed it in Rotman. It's easy enough to prove, but I need the citation for an article. Thanks for pointing out the reference desk. mike40033 (talk) 05:49, 10 December 2007 (UTC)
 * Rotman's Theorem 3.19 (4th ed) is orbit-stabilizer, and exercises 3.42 and 3.43i are the part of the proof I outlined above. However, you may prefer Aschbacher, p13-16, especially Prop 5.9 which uses the same language as your quote. I'll add the reference to the article. JackSchmidt (talk) 22:12, 10 December 2007 (UTC)
 * Yes, Aschbacher Prop 5.9 is perfect. Thanks a zillion! mike40033 (talk) 04:52, 12 December 2007 (UTC)

Overlap with Orbit (mathematics)
There's now an overlap with orbit (mathematics).

Charles Matthews 18:52 24 Jun 2003 (UTC)


 * I've now merged in that stuff. Also orbit (group theory) now redirects here. -- Fropuff 17:51, 2004 Aug 23 (UTC)

Looking for definition of "Orbit of a mathematical group"... you jump right into this section..... How is G related to X? is X a subset of G? How can you define operations of g on X.  Confusion!..... you are not aware of the main subject.... Link needs to point to the top of the article...giving the information "Orbit is a type of Group action."

Orbits and stabilizers: Clarity Needs some examples, How about a donut? To aid in understanding, definitions of "invariant" and "fixed" and why one implies the other and not the other way around. "invariant" and "fixed".... At time of reading, There is a mathematical article for invariant, but not for fixed.....the "invariant" one should be linked from here.

Formulae (in graphic), would it be better, if they included the fact  that "x ϵ X" "x belongs to" in mathematical language? LieBugs (talk) 14:27, 17 February 2008 (UTC)

Continuous group actions
"The above statements about isomorphisms for regular, free and transitive actions are no longer valid for continuous group actions." - could you, please, be a bit more specific? Why are these statements no longer valid? Commentor (talk) 05:56, 1 March 2008 (UTC)

Topology
Dear All,

The orbit of an action is a quotient group (as mentioned in the article), but can also be viewed as a quotient space in the topological sense. In fact, if G is a compact topological group, then certain properties of the topological space that G is acting upon imply certain properties of the orbit space. If X happens to equal to G, the orbit space is trivial. I know that this is a group theory article but would it be beneficial to create a new article regarding the topological aspects of the orbit space?

Topology Expert (talk) 10:26, 26 August 2008 (UTC)


 * I agree. Definition of orbit space topology, continuous functions etc.  But we should also cover manifolds, and the partition (stratification) of the orbit space into orbit types? Simplifix (talk) 12:57, 29 August 2008 (UTC)

Introduction
Bijection = Symmetry? The supposedly general introduction and summary have little relation to the rest of the article. Needs to be rewritten by someone who knows the subject.Ht686rg90 (talk) 17:11, 29 June 2008 (UTC)
 * The first paragraph is just meant as a general introduction to the topic and is not supposed to be completely precise. Having said that I don't see a problem in that paragraph although it could probably be worded better. Have removed the disputed tag. MSGJ (talk) 10:42, 21 August 2008 (UTC)
 * I've completely rewritten it. It wrongly identified group actions with permutation groups for example. It still does not connect as well as it should with the rest of the article, but I don't think that can be fixed until someone writes some sort of "motivation" section, which the lead can then summarize more concisely.
 * I was hoping to update the maths rating above, but my comments from over a year ago still apply. Geometry guy 19:41, 14 September 2008 (UTC)

class equation
there should be written something about class equations in general; so this article would seem to be a perfect place for that, class equation redirects to conjugacy class at the moment where it describes only conjugacy class equation… konradek (talk) 15:29, 17 October 2008 (UTC)

Group actions and groupoids section is poorly written
I was reading this article when I noticed that the Group actions and groupoids section seemed like it was hurriedly written by one person as a plug for a commercial website selling the book referenced in that section. What it wikipedia's policy about this? —Preceding unsigned comment added by 71.237.245.96 (talk) 23:13, 14 May 2009 (UTC)

Article is a list of clobbered results and definitions
This article contains way too many definitions and redirects users from Orbits/G-sets/Stabilizers/Orbit-Stabilizer formula. Also, somehow Burnside's lemma has sneaked in. The article should be split into different articles treating each subject separately. Stefan.petrea (talk) 14:19, 30 November 2009 (UTC)

Too hard
Why do we have mathematics that is this abstract within a general encyclopedia? It's not even a description of an overall subject (like a general description of group theory as an entity so an ousider knows what it is as a field (and not the math definition of field), but rather a pretty technical definition.

This would not be in Brittanica or an equivalent.

If someone wants to make a technical description of all of mathematics, they should make it off-wiki. —Preceding unsigned comment added by 72.82.43.222 (talk) 17:04, 6 November 2010 (UTC)

left vs. right
does left or right matter? or just two different notation??? Jackzhp (talk) 05:03, 8 April 2011 (UTC)

It matters. For example we have a group of functions and its group operation is defined as the composition of functions by $$ f \circ g = f(g(x)) $$ we can attempt to define an action of this group on another set of fuctions by saying a function $$f$$ in the group acts on a function $$H(x)$$ by mapping $$H(x)$$ to the function $$H(f(x))$$. As a left action we have $$ f.(g.H(x)) = f.H(g(x)) = H(g(f(x)) $$ versus $$(f \circ g).H(x) = f(g(x)).H(x) = H(f(g(x)) $$ and, in general, $$ H((g(f(x)) $$ need not be the same function as $$ H(f(g(x)) $$  You don't run into that problem as a right action since $$ (H.f).g = H(f(x)).g = H(f(g(x)) $$ and  $$ H.(f \circ g) = H.f(g(x)) = H(f(g(x)) $$

Tashiro (talk) 08:42, 27 June 2013 (UTC)

As the article says, a right action of any group can be converted into a left action of the opposite group, and vice versa. So there is indeed no material difference - any group action that can be achieved on one side can be achieved on the other. As Tashiro illustrates, there is a contravariance (order switching) which happens with right actions, which sometimes makes the notation more elegant on the right than on the left. But there is no new mathematics from right actions. &mdash; Carl (CBM · talk) 12:34, 27 June 2013 (UTC)

Why are actions bijective?
I have a question about actions that is not answered in the article. I am wondering, why is the map $$g : x \mapsto gx$$ bijective?

If $$a, b\in G$$ and $$ax = bx$$ for some $$x$$ (or, equivalently, $$a(x)=b(x)$$), then we know that $$x = (a^{-1}b)x$$. But why is it true that $$gx = x \Longrightarrow g = e$$ (where $$e$$ is the group identity)?

Thanks! —Preceding unsigned comment added by 193.77.126.73 (talk) 17:57, 20 March 2009 (UTC)


 * You asked two different questions, which is probably why you are confused. g:x→gx is bijective, because its inverse is g−1:x→g−1.
 * However, there is no need for g → ( x → gx ) to be bijective. It is certainly possible for a(x) = b(x) without a = b.  For a particular x, the set of all g in G with g(x) = x is called the stabilizer of x and is a subgroup of G.  The intersection of all the stabilizers is a normal subgroup of G called the kernel of the action.  It is precisely those g such that g(x) = x for all x.  Then for a in G and b=ag, one always has a(x) = b(x) for all x. JackSchmidt (talk) 18:22, 20 March 2009 (UTC)


 * Thanks! You're right, I confused the concepts. It's all really simple now :) You were of great help! —Preceding unsigned comment added by 193.77.126.73 (talk) 19:22, 20 March 2009 (UTC)


 * Wait a minute, the action of a particular group element g needs not be a bijection of the set X to itself. It only needs to be an injection because, if gx = gy, then we have x = ex = g-1gx = g-1gy = ey = y. But an injective map from a set to itself needs not be bijective if the set is infinite. For instance, the map $$n \mapsto 2n$$ from the the set of positive integers to itself is injective, but not bijective. This would be the group action of the element 2 of the multiplicative group of positive reals on the set of positive integers.


 * If every such map were bijective, then every group action would be transitive.Cacadril (talk) 17:26, 30 September 2012 (UTC)


 * You can't act with the multiplicative group of positive reals on the set of positive integers. Where would pi send a positive integer, for instance? — Preceding unsigned comment added by 83.104.131.53 (talk) 17:35, 2 September 2013 (UTC)

Terminology question
I'm not sure about the standard terminology here:
 * Is it called "faithful" or "free"?
 * Is "invariant" and "stable" really the same?
 * Are the sets G.x really called "traces"? – — Preceding unsigned comment added by AxelBoldt (talk • contribs) 08:13, 28 October 2001 (UTC)


 * It's called "faithful", or "effective". ("Free" means that only the identity element has a fixed point.)
 * I'm not sure about this one. I would use "invariant" for the sense you were talking about.
 * The sets G.x are usually called "orbits" (as in the Orbit-Stabilizer Theorem).
 * --Zundark, 2001 Oct 28


 * Locally free, in my experience, means that all the stabilizers are discrete, not that their union is separated from the identity. Is there a reference for the given definition of locally free? -- Scot Adams — Preceding unsigned comment added by 128.101.152.39 (talk) 20:51, 23 April 2014 (UTC)

Not sure about this page splitting?
I am not so sure about this page, it is (to my own not very advanced opinion) more about transformation groups than about group actions,

The page doesn't really limit itself to what are group actions? I guess only 10% (maybe even less) of this page is really about that, and examples are missing, the rest is more on group theory and other subjects, while interesting not they are not the subject of this page, and on this page they only confuse the subject. WillemienH (talk) 11:59, 22 November 2014 (UTC)

Good Collective Work!
This is a very good and informative page. In a few lines it informs people with a basic college course in algebra and topology useful and correct facts. It is not correct to claim this article is too hard or would not have found its way as a section of a longer article into the old Encyclopedia Brittanica. — Preceding unsigned comment added by Lkadison (talk • contribs) 09:16, 24 June 2011 (UTC)

Maybe not / techincal tag

 * Actually this article is rather poorly written/organized, particularly in the Group_action section, which is an [if not the] essential part. Invariants get talked of before they are defined etc. No simple example of the invariants of an element etc. So the page shares its ills with many other wiki math pages: if you already know the material, it serves as a somewhat confusingly written recapitulation, and perhaps throws in a few pointers to things you didn't know about. For the neophite I suspect it's rather useless past the Examples section. Some1Redirects4You (talk) 09:05, 25 April 2015 (UTC)


 * While it's fine to have a long list of (initial examples) most of the more refined concepts (orbits, stabilizers, invariants, etc. aren't actually illustrated with any examples. This on top of the confused (and often too divagating) writing that insists on notation peculiarities and connections with cohomology etc. at every turn of the phrase. Some1Redirects4You (talk) 09:14, 25 April 2015 (UTC)

Odd
The second sentence seems to be ungrammatical. — Preceding unsigned comment added by 86.156.239.161 (talk) 14:33, 18 October 2015 (UTC)


 * You are absolutely correct. It is now fixed. — Anita5192 (talk) 16:56, 18 October 2015 (UTC)

Suggested Merge: Properly Discontinuous Action
The definition of a properly discontinuous action on this page contradicts that at Properly discontinuous action. There seem to be two main definitions:
 * 1) $$ g \cdot U \cap U = \emptyset $$
 * 2) $$ g \cdot U \cap U $$ is of finite size.

It seems (from my admittedly limited experience) that the more common of the two is the latter, which is the one employed by Properly discontinuous action, but not by this article. More information about the equivalence (or lack thereof) of these two definitions may be gleaned from this MathOverflow Q&A.

Furthermore, I think that perhaps the article Properly discontinuous action should be merged with this one.

Myridium (talk) 02:45, 8 February 2016 (UTC)

Removed reference to permutation groups
I removed the reference to permutation groups in the first paragraph since a permutation group on M is a subgroups of Sym(M), while a transformation group G on M is given by a (not necessarily injective) homomorphism G &rarr; Sym(M). So they are not the same. AxelBoldt (talk) 17:45, 31 October 2002 (UTC)

Clarify wording
The statement

In every group G with subgroup H, left multiplication is an action of G on the set of cosets G/H: g.aH = gaH for all g,a in G. In particular if H contains no nontrivial normal subgroups of G this induces an isomorphism from G to the a subgroup of the permutation group of size [G : H].

is a little unclear. Are we talking about a subgroup of size [G:H] or a permutation group of that size? I would guess the former, because permutation groups must have factorial order... — Preceding unsigned comment added by 180.165.104.233 (talk) 11:15, 11 December 2016 (UTC)

n-transitivity
Hello,

you define n-transitivity to be "transitivity on $$X^n$$", by which I suppose you mean the action on $$X^n$$ with
 * $$g(x_1,\ldots,x_n) = (gx_1,\ldots,gx_n).$$

But then n-transitivity can not be defined as transitivity on $$X^n$$, because you can never find a $$g$$ satisfying
 * $$g(a, a) = (b, c)$$

for $$b \not = c$$ from $$X$$.

A correct definition would read: For pairwise distinct $$x_1,\ldots,x_n$$ and $$y_1,\ldots,y_n$$ there is a $$g$$ such that
 * $$gx_k = y_k$$

holds for $$1\leq k \leq n$$.

OK, seeing that both the $$x_k$$ and the $$y_k$$ have to be pairwise distinct (otherwise $$g$$ would not be a bijection) we might restrict the action of $$X^n$$ to the subset of $$X^n$$ containing no multiple entries. If that is what you mean by "transitivity on $$X^n$$" I guess it deserves being pointed out explicitly.

— Preceding unsigned comment added by FarSide (talk • contribs) 06:23, 7 July 2006 (UTC)