Talk:Groupoid
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Would it be agreeable to write
- In Wikipedia, we reserve the term "groupoid" for the first meaning, and use "magma" for the second.
Otherwise, we can never use "groupoid" anywhere without a half sentence explanation. AxelBoldt 17:47 Oct 29, 2002 (UTC)
That would be nice. But, what if someone carelessly or innocently writes [[groupoid]] when (s)he means magma ? We could keep "groupoid" as a disambiguation page as it is now, and have articles called "groupoid category" and "magma (mathematics)" (or better names). FvdP 18:10 Oct 29, 2002 (UTC)
The thing is that the two meanings are similar enough that it is not always immediately clear from the context whether the link groupoid was meant to refer to groupoid (category) or to magma. So the innocent groupoid link would have to be fixed no matter what. AxelBoldt 22:01 Oct 29, 2002 (UTC)
I agree with Axel; we can have a disambiguation block here. — Toby 02:14 Nov 3, 2002 (UTC)
[edit] Disambigation revisited
I have the impression that "groupoid" in the algebraic sense, a set with binary operation, is still common. In that case it seems to me we ought to have a disambiguation page with links to "Magma" and "Groupoid (category theory)" (this page, renamed). I'm willing to do it as time allows, and I'm willing to be convinced it's not a good idea. Any comments? Zaslav 04:33, 18 September 2006 (UTC)
- A disambiguation page is acceptable. It is true that occasionally people link to Groupoid when they mean Magma, but this is usually fixed pretty quickly. I disagree that the magma sense of "groupoid" is more algebraic than the Brandt groupoid sense. The redirect for Groupoid (algebra) is probably harmless, though, since I suspect nothing will link to it.
- Besides the links for those who want to move on to one or the other meaning, a disambiguation page could briefly mention the origins of the two uses of the term. "Groupoid" in the magma sense is due to Oystein Ore sometime in the late 1930s, I believe. "Groupoids" in the categorical sense used to be called Brandt groupoids, but were not originally defined as categories with inverses, of course. I am not sure when Brandt's name was dropped, perhaps in Bourbaki? Certainly by the time Lie groupoids became all the rage in Poisson geometry, Brandt's name had vanished. "Magma" is a Bourbaki-ism. Michael Kinyon 12:23, 18 September 2006 (UTC)
I removed the sentence that said that the collapse from groupoids to collections of groups doesn't lose information; in fact it does. If you view a groupoid as a category, and you only apply category theoretical concepts, then indeed equivalent categories are pretty much identical for all practical purposes. However, if you view a groupoid as an algebraic structure, a group with partially defined group law, then you want to ask many algebraic questions, and you need to distinguish between non-isomorphic equivalent groupoids. I could for instance ask "how many elements of order 2 does this groupoid have?". AxelBoldt 20:33 Nov 4, 2002 (UTC)
Yes, just as I might ask of a group "how many elements of this group are real numbers?". If the group is the group of real numbers under addition, then the answer is "all 2א0 of them", whereas if the group is the group of purely imaginary numbers under addition, then the answer is "only one of them, the identity". The answer is irrelevant from certain points of view (the groups are isomorphic), but relevant from others (they're different subgroups of the additive group of the field of complex numbers). — Toby 09:35 Nov 17, 2002 (UTC)
From one who doesn't actually know any category theory, this is a nice point to clarify, since I think I almost "get" it :). But it still seems to me like in both your examples, the collapse into groups doesn't so much lose information as add a layer of obfuscating information - can't you always create the same groupoid from algebraic structures which differ only by arbitrary choice of x when making G(x)? (as opposed to trying to figure out which specific algebraic structure was used to make a particular groupoid) Chas zzz brown
I don't understand the question, specifically "can't you always create the same groupoid from algebraic structures which differ only by arbitrary choice of x when making G(x)". You "create" a groupoid by taking a bunch of groups and connecting some of them by a bunch of isomorphisms. The labeling of the groups as G(x) is indeed arbitrary. But which and how many isomorphisms you take to connect those groups matters.
Toby, I just thought of a better example: the fundamental groupoid of the reals is equivalent, but not isomorphic, to the fundamental groupoid of two separate copies of the reals. AxelBoldt 16:17 Nov 18, 2002 (UTC)
These are definitely not equivalent. If they are according to the definitions in Wikipedia, then something's wrong (or ambiguous) with them somewhere.If x and y aren't (path-)connected, then G(x) and G(y) can't be identified. -- Toby 06:24 Feb 2, 2003 (UTC)
Here's my thinking: the category * which has only one object and one morphism (the identity) is equivalent to the category * * which has two objects and two morphisms (both identities). The proof of this equivalence should carry over to the proof of the equivalence of the two fundamental groupoids discussed above. AxelBoldt 23:30 Feb 2, 2003 (UTC)
I suppose that such a proof would carry over, but there is no such proof (or had better not be!), since these categories are not equivalent. If you throw in two more morphisms (to * *), one going from each object to the other (with composition defined the only way possible), then you get a category equivalent to * -- and this is analogous to throwing in a path from a point in one copy of the real line to a point in the other. (IOW, * and *↔* are equivalent, but * * is different.) -- Toby 00:05 Feb 3, 2003 (UTC)
Yes, you're right. AxelBoldt 18:36 Feb 4, 2003 (UTC)
I said "algebroid" instead of "algebraoid", since I've only ever seen the former. So I don't know if it's a mistake on your part, or if it appears both ways. Feel free to change it back if the latter! -- Toby Bartels 02:27 6 Jun 2003 (UTC)
This paragraph is not really accurate:
- Note that the isomorphism described above is not unique, and there is no natural choice. Choosing such an isomorphism for a connected groupoid essentially amounts to picking one object x0, a group isomorphism h from G(x0) to G, and for each x other than x0 a morphism in G from x0 to x.
While such a choice is sufficient to constuct the isomorphism, it's not necessary. In particular, the general case won't pick out any particular object x0. -- Toby Bartels 03:13 6 Jun 2003 (UTC)
I have added comments about the various types of morphisms of groupoids, and also the example of the classification of groupoids with one endomorphism. I am not sure if this has or can be done! Recall that the classification of vector spaces is easy; there is just the dimension. The classification of vector spaces with one endomorphisms involves theorems on normal forms of square matrices, and so is very interesting. I am told the classification of vector spaces with two endomorphisms is hard, and with three has not been done!
I also mention that there are recent applications of groupoids to combinatorics, shown by papers on the arXiv in 2005; and that the objects of a groupoid give a spatial component to group theory. -- Ronnie Brown 22:25 27 April, 2006 GMT
