From: baez@galaxy.ucr.edu (John Baez)
Subject: Algebraic topology for physicists
Date: 10 Oct 2000 17:52:56 GMT
Newsgroups: sci.physics.research
Summary: [missing]
In article <8rn3hc$2ar$1@nnrp1.deja.com>, wrote:
>Oops... Yeah, now I talked to a person I know, and realized I was
>thinking of the Whitehead theorem which requires an actual map between
>the spaces establishing an isomorphism between the homotopy groups.
Right, it's sorta sneaky: a map between spaces that induces an
isomorphism on homotopy groups must be a homotopy equivalence, but it
ain't true that spaces with the same homotopy groups are homotopy
equivalent.
(Here by "spaces" I mean reasonably nice spaces, like CW complexes.)
>> In addition to the homotopy groups you need "Postnikov data", which
>> says how the homotopy groups talk to each other.
>
>What's the Postnikov data? Sounds like something cool. :)
It's VERY cool. In the game called "Postnikov towers", we describe
any connected space X as follows. First we work out the homotopy groups
pi_n(X). Then for each n we cook up a space K(pi_n(X),n), called an
Eilenberg-MacLane space, which has vanishing homotopy groups except
for the nth one, which equals pi_n(X). Then the trick is to put
these spaces together to form a space homotopy equivalent to X.
We could just take the product of all of them, and get a space with
the same homotopy groups as X. But that would not be right, because
it would be forgetting how the homotopy groups of X interact.
So what we do instead is start with X_1 = K(pi_1(X),1).
Then we form a space X_2 which is the total space of a bundle whose
base is X_1 and whose fiber is K(pi_2(X),2).
Then we form a space X_3 which is the total space of a bundle whose
base is X_2 and whose fiber is K(pi_3(X),3).
Then we form a space X_4 which is the total space of a bundle whose
base is X_3 and whose fiber is K(pi_4(X),4).
... and we go on forever, and when we're done we have a space that's
homotopy equivalent to X!
If all these bundles were trivial, X would just be a big product of
Eilenberg-MacLane spaces --- a very dull sort of space where the
homotopy groups don't talk to each other at all.
But usually these bundles are not trivial. So to describe these
bundles we need extra information: POSTNIKOV DATA!
It's like a layer cake: the Eilenberg-MacLane spaces are the layers,
but we also need to say how each layer is stuck on top of the preceding
layers. Bundle upon bundle upon bundle... an infinitely tall layer cake.
So: together with the homotopy groups, the Postnikov data says what
the space X is really like, up to homotopy equivalence.
>Almost forgot: ain't there a purely group theoretical definition too? I
>think I read something about it once.
Of group cohomology, you mean? Sure! You said what it was, too: you
take your group G, you form a free resolution, you hom that resolution
into A, then you take the cohomology of the resulting cochain complex
to get the cohomology groups H^n(G,A). This yoga can be found in any
book on homological algebra - just go to your local bookstore. This
kind of stuff leads to some good ways to actually *compute* the cohomology
of groups. In fact, these days you can get computer programs that do
this for you - just ask at your local computer software store. (Yeah, right.)
The only problem with this purely algebraic approach to group cohomology
is that, taken by itself, it doesn't offer enough intuition about the
deep inner meaning of it all. Throwing some topology in the mix helps.
For example, it lets you actually visualize what's going on.
==============================================================================
From: baez@galaxy.ucr.edu (John Baez)
Subject: Re: Algebraic topology for physicists
Date: 12 Oct 2000 18:46:33 GMT
Newsgroups: sci.physics.research
Summary: [missing]
In article <8s2dpn$onu$1@nnrp1.deja.com>, wrote:
>John Baez took a nice space X and then said:
>> ... we start with X_1 = K(pi_1(X),1).
>>
>> Then we form a space X_2 which is the total space of a bundle whose
>> base is X_1 and whose fiber is K(pi_2(X),2).
>>
>> Then we form a space X_3 which is the total space of a bundle whose
>> base is X_2 and whose fiber is K(pi_3(X),3).
>>
>> Then we form a space X_4 which is the total space of a bundle whose
>> base is X_3 and whose fiber is K(pi_4(X),4).
>>
>> ... and we go on forever, and when we're done we have a space that's
>> homotopy equivalent to X!
>> But usually these bundles are not trivial. So to describe these
>> bundles we need extra information: POSTNIKOV DATA!
>Cool. Is there a good way to classify such bundles, then?
Yes! That's the cool part. You may recall that G-bundles over X
are secretly the same thing as maps from X to a space BG, called
the "classifying space" of G. If you don't, well, take a peek at
this:
http://math.ucr.edu/home/baez/week151.html
We can use this trick to classify the possible bundles at each
stage of this Postnikov tower game. It's not very hard... can you
guess how it goes???
>> ...You said what it was, too: you take your group G, you form a free
>> resolution, you hom that resolution into A, then you take the
>> cohomology of the resulting cochain complex to get the cohomology
>> groups H^n(G,A).
>Sounds almost like the definition of derived functor.
You bet! In fact this is probably the very first example of a derived
functor that anyone ever thought of. Group cohomology was invented by
Eilenberg and MacLane in the 1940s. In the process, they invented the
basics of category theory. Then Cartan and Eilenberg generalized the
heck out of group cohomology in their massive tome "Homological
Algebra". They showed you could define cohomology not only for groups
but also for modules, rings, algebras, Lie algebras - you name it! At
some point, someone realized that you could systematize all this stuff
using the concept of "derived functor". For all I know, it was Cartan
and Eilenberg who did it.
>But we have an injective rather than free resolution there. Eh, what's a free
>resolution anyway? :)
For specificity, let me talk about the case where we're playing with
modules of a given ring. It generalizes...
An injective resolution of the module M is an exact sequence
0 -> M -> E1 -> E2 -> ...
where each of the E's is injective. Dually, a projective resolution
of M is an exact sequence
0 <- M <- E1 <- E2 <- ...
where each of the E's is projective. Both injective and projective
resolutions are equally important when you're trying to cook up
derived functors. Basically, a projective resolution of M is like a
puffed-up version of M that's good to map *out* of, while an injective
resolution of M is like a puffed-up version of M that's good to map
*into*. So we use projective resolutions when cooking up "left
derived functors" and injective resolutions when cooking up "right
derived functors". It's one of those "turn around all the arrows"
games - no big deal.
A special case of a projective resolution is a free resolution, where
each of the E's is a free module. I mentioned "free resolutions"
instead of "projective resolutions" because more people know about
free modules than projective ones. How was I to guess that you
actually preferred *injective* resolutions? I'm sort of shocked.
==============================================================================
From: baez@galaxy.ucr.edu (John Baez)
Subject: Re: Algebraic topology for physicists
Date: 18 Oct 2000 08:39:55 GMT
Newsgroups: sci.physics.research
In article <8s5do8$9ag$1@nnrp1.deja.com>, wrote:
>In article <8s39dv$4to$1@Urvile.MSUS.EDU>,
> baez@galaxy.ucr.edu (John Baez) wrote:
>> In article <8s2dpn$onu$1@nnrp1.deja.com>, wrote:
>>> John Baez took a nice space X and then said:
>>>> ... we start with X_1 = K(pi_1(X),1).
>>>>
>>>> Then we form a space X_2 which is the total space of a bundle whose
>>>> base is X_1 and whose fiber is K(pi_2(X),2).
>>>>
>>>> Then we form a space X_3 which is the total space of a bundle whose
>>>> base is X_2 and whose fiber is K(pi_3(X),3).
>>>>
>>>> ... and we go on forever, and when we're done we have a space
>>>> that's homotopy equivalent to X!
>>> Cool. Is there a good way to classify such bundles, then?
>> Yes! That's the cool part. You may recall that G-bundles over X
>> are secretly the same thing as maps from X to a space BG, called
>> the "classifying space" of G.
>Okay, so at first we need to specify an element of
>
>[K(pi_1(X), 1), BK(pi_2(X), 2)] = [K(pi_1(X), 1), K(pi_2(X), 3)] =
>= H^3(K(pi_1(X), 1), pi_2(X))
>
>and in general of
>
>H^(n+1)(Y, pi_n(X))
>
>where Y is the space we cooked up already.
[You wrote "Hom^n", but I changed it to "H^n" - the nth cohomology group.]
RIGHT! Good!
These are the famous "Postnikov data" or "k-invariants"!
There are actually some subtleties which you overlooked. In fact, I
sort of tricked you into overlooking these subtleties. Things can get
a little more complicated than you describe here: we may need to use
cohomology with twisted coefficients. Postnikov data of the sort you
describe will get us *some*, but not *all* spaces. But I'll leave it
to the homotopy theory experts - like Dan Christensen! - to explain
what we're missing.
>Looks really funny, all these homology and homotopy stuff mixed.
Yes - but no nobler use of these ideas could be imagined, than the
grand project of CLASSIFYING ALL SPACES up to homotopy.
>Okay, I listened to a lecture about group cohomology in a seminar, and
>ordered my head a bit. We, however, couldn't come to a consensus about
>what functor is being derived here. It looks as if it's the Hom of the
>category of G-modules, but then, why call it H^n(G, A) rather than
>Ext^n(G, A), where Ext is taken, again, in the G-module category?
Well, it depends what you mean by "G-module".
If by "G-module" you mean a *set* with an action of the group G, then
G is naturally a G-module. Unfortunately, G-modules of this sort
don't form an abelian category, so you can't play the usual derived
functor game, so "Ext^n(G, A)" doesn't make sense.
If by "G-module" you mean an *abelian group* with an action of the
group G, then G-modules do form an abelian category, and you can play
the derived functor game. Great! Unfortunately, G is not a G-module
in this sense - so "Ext^n(G, A)" still doesn't make sense.
So hopefully you see why we don't talk about "Ext^n(G, A)".
So what *should* we do? We should define a "G-module" be an abelian
group with an action of G, and let Z be the integers with the trivial
action of G, and define
H^n(G, A) = Ext^n(Z, A)
This definition of H^n(G, A) is equivalent to the more topological
one I already told you about: the nth cohomology of the Eilenberg-
MacLane space K(G, 1) with coefficients in A.
Alternatively, suppose you're one of those people who like modules
of *rings*. Then you can form the group ring ZG and work with modules
of that! Any G-module (i.e. abelian group with action of G on it)
naturally becomes a module of this ring ZG, and vice versa. In other
words, the category of G-modules is equivalent to the category of
ZG-modules. So we can equivalently define
H^n(G, A) = Ext^n(Z, A)
where A is a ZG-module and Z is the trivial ZG-module. This is just
yet another way of packaging the same idea; I mention it only because
lots of people spend a lot of time talking about Ext and Tor in the
context of modules of *rings*.
Personally, I think the most comprehensible definition of H^n(G, A) is
the first one I gave you: create a space whose fundamental group is G,
and for which all the other homotopy groups vanish; then take its nth
cohomology with coefficients in A.