From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Characteristic classes [LONG]
Date: 21 Dec 1994 09:05:57 GMT
In article ,
Dave Wagner
had the unmitigated gall to change the Subject line from the perfectly
serviceable "Putnam spoilers" I started three weeks ago.
[OK, he didn't, someone else did; but perhaps those continuing that other
discussion might notice that threads, like milk, grow stale and distasteful
after awhile.]
> But I have no geometric picture of what the
>i-th Chern class "measures", much less of what the Todd classes "measure".
>None of the texts I have found address this issue of intuitive understanding,
>which (I feel) is a neccessary complement to technical understanding.
>
>To anyone out there in 'net-land who can offer an explanation of these
>concepts, of the geometrical meaning of Stiefel-Whitney classes, Adams
>operations, Steenrod operations, et cetera, I would be grateful to read
>any of it.
Intuition is hard to come by in this area, but the following perspectives have
helped me. First let me clarify that these are all constructions of
elements ("classes") in a cohomology group or ring, so we're back to
the "what is homology" question that's been popular this year.
One possible answer to this is that a cohomology class represents a
(homotopy class of) map from the space X into, well, something: a
certain space called K(Z,n), if you're looking at an element of H^n(X,Z).
K(Z,1) is the circle, so elements of H^1(X) can be thought of maps into
the circle, and indeed you "understand" H^1 if you just know that
H^1(S^1) is the free abelian group on one generator, call it s_1.
For then you get the cohomology class corresponding to any map
f: X--> S^1 as f_*(s_1), the pullback of s_1. Don't try to figure out
what s_1 is, just remember that all the cohomology classes of H^1(X)
are obtained by taking homotopically distinct maps f, and for every one
of them letting f_* be a pointer carrying "s_1" to different
elements of your set H^1(X).
It's not quite so easy to describe K(Z,n) for n>2. It kinda looks like
the sphere S^n homotopically, but geometrically it's got a lot
of extra stuff hanging on which makes it infinite dimensional (in an
appropriate sense). (The root of the problem is that even perfectly
reasonable spaces like spheres have many non-zero homotopy groups; there
is for example a nice, non-trivial map from S^3 to S^2. You add on
cells to S^n so that all these extra homotopy classes become trivial
in the new space. The details are not important). The key is that you
can tick off each of the elements of H^2(X) by considering each of
the homotopically distinct maps f: X--> K(Z,2); indeed, H^2(K(Z,2))
is the free abelian group on a single generator s_2, and you
get all the elements of H^2(X) of the form f_*(s_2) for some such f.
[Consistency check: since s_2 is an element of H^2(K(Z,2)), to what
map f: K(Z,2) --> K(Z,2) do you suppose _it_ corresponds?]
You can do all this with H^*(X, G) where G is any other coefficient group.
For various technical reasons I'll assume G=Z/2Z below, and I'll
just write K(n) for the space to map into to enumerate H^n(X),
whatever coefficients you want to use.
This is one way to view cohomology classes "geometrically". Using it,
you can make the other constructions "geometric".
For example, what does it mean to add two cohomology classes? Given
x, y in H^n(X), let's find maps f, g: X--> K(n) which represent them.
Now, x+y ought to be back in H^n(X), right? How do we get the
corresponding map h: X --> K(n)? All you have to know is that there is
a map K(n)xK(n) --> K(n) which generalizes the multiplication
S^1 x S^1 --> S^1. Then you can take h to be the composite of the maps
(f,g) : X --> K(n)xK(n) and the preceding "multiplication".
(You may have seen addition in homotopy groups defined in a similar way;
instead of "multiplication" you use "pinching" S^n --> S^n \wedge S^n.
This is dual.)
That was your warmup. What is "multiplication"? Easy: given a map
f: X --> K(n) and another map g: X --> K(m), you can combine them
to get a map h: X --> K(n+m) as soon as I tell you there is a
natural map K(n) x K(m) --> K(n+m). Again the details are not
as significant as the transition I've made: note that I immediately thought
of a cohomology class as a map to a K(n).
OK, now you're ready for the Steenrod algebra. Any time I can point out
a map s: K(n) --> K(m) for some n and m, I've given you a
way to change elements of H^n(X) to elements of H^m(X): just compose
with s. For example, you could have m=n and take s to be the composite
of the diagonal map K(n) --> K(n)xK(n) and the above map
K(n)xK(n) --> K(2n). This is just the squaring map in cohomology: x |--> x^2.
Notice that this _is_ a homomorphism if you take cohomology with
Z/2Z coefficients, which is what Steenrod did, and why you hear all the other
generalizations of this are called the "Steenrod squares". (You can work with
Z/pZ coefficients and begin with the p-th power map, too).
Clearly you can compose some maps between K(n)'s to get more, so the
set of these operations on homology has a "multiplication" (composition),
as well as an "addition" (it's the pointwise addition of functions, but
you ought to think about how to say it with maps into K(n)'s.) It's
also true that you can suspend maps K(n) --> K(m) and get a consistent
family of maps which we usually write as K(n) --> K(n+k) where the n
is no longer important. For example, Sq^1 : K(n) --> K(n+1) gives a
cohomology operation Sq^1: H^n(X) --. H^(n+1)(X) for any X.
One important byproduct of the functional perspective is that it makes it
clear that these constructions are natural, i.e., given a cohomology
class f: X --> K(n) and an element of the Steenrod algebra g: K(n) --> K(n+k)
you get an element gf of H^(n+k)(X). Well, if you have another space Y
and a map h: Y --> X, then you can pull back f to an element
fh of H^n(Y). You can either pull back gf or apply the Steenrod
operation to fh, either way you get (gf)h=g(fh).
In turn an application of naturality is that you can understand the
transformation of any cohomology class if you can somehow compare your space
to a better-known space: if X is well-understood above, and Y is a space
for which you happen to know that an interesting cohomology class can
be written in the form fh, then you can compute the action of the
Steenrod algebra by observing g(fh)=(gf)h, i.e., pretend Y is just X,
do your computation (of g on f) and then interpret the result as if
it were in Y.
This is how Chern first defined the Chern classes, I think, and how Wu
computed the action of the Steenrod algebra on them. Begin with the
classifying spaces BU(n) of n-dimensional vector bundles. If you've
got a bundle over a space X, then this means you've got a map from
f:X --> BU(n). Well, then any cohomology class g:BU(n) --> K(m) of BU
pulls back to a cohomology class gf of X. That's what the Chern classes are.
More precisely, the cohomology of BU(n) turns out to be a polynomial
algebra in certain elements c_1, c_2, ... (deg(c_i)=2i).
I can even tell you what the c_i look like. In the special case n=1,
BU(1) is the same space K(Z,2) discussed earlier. The class c_1 is
the same class I called s_2 in that setting, and the whole cohomology of
BU(1) is the polynomial ring in this class (i.e., H^2n(BU(1)) is the set
of integer multiples of c_1^n.) That makes X=BU(1)^n a space whose
cohomology is a polynomial ring in n variables. Well, OK, there's
a natural n-bundle on X: just takethe product of the n line bundles,
i.e. of the projections x_i: X --> BU(1). This n-bundles must come
from a map X --> BU(n); really it's just the map induced by the
inclusion U(1)^n --> U(n) of diagonal matrices into all of U(n).
OK, so what are the Chern classes of this bundle? It turns out c_i
is just the i-th symmertic polynomial in the x_i.
So here's a partial answer to "what's a Chern class": if you've
got a _split_ line bundle over Y, that means your classifying map
Y --> BU(n) happens to factor thru BU(1)^n ; in that special case,
the chern classes can be found as the sum,..., product of the Chern
classes c_1 in H^2(Y) of the individual line bundles.
This whole discussion is surely tainted by my mod-2 upbringing and
the lateness of the hour, so don't take this as 100% accurate, but the
jist of the idea is still fine: if you don't enjoy lots of algebraic
formulas (there are many) then think of all these cohomology calculations
as homotopy classes of maps between certain fixed, important spaces.
You'll get a real feel for their value and their naturality that way.
Now can we get back to "1=2?" an other critical questions on sci.math?
dave
==============================================================================
Date: Fri, 6 Jan 1995 19:41:13 -0600
From: [Permission pending]
To: rusin@washington.math.niu.edu
Subject: Re: Characteristic classes [LONG]
Your explanation of homology was very interesting. I usually do homology
with local coefficients. Do you know if there is any way to think of
homology with local coefficients as a set of homotopy classes of maps?
Thanks for the (too rare in sci.math) educational post,
[sig deleted -- djr]
==============================================================================
Date: Wed, 25 Jan 95 14:47:06 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Characteristic classes [LONG]
You wrote (long ago):
>Your explanation of homology was very interesting. I usually do homology
>with local coefficients. Do you know if there is any way to think of
>homology with local coefficients as a set of homotopy classes of maps?
Sorry for the long delay.
The short answer is "no". However, I think there ought to be an answer
which melds the two extremes. On the one hand, when you have a complicated
space and trivial module, you can view homology classes as maps in the
way I posted. On the other end, observe that the coefficients are only
local because of the action of pi_1; thus the other extreme case ought
to be when X is a K(pi,1), so that H^n(X,A) is the algebraic cohomology
of groups H^n(pi,A). Of course, this still is best described as the
homology of a chain complex, but if I remember right, there is a way
to describe this as Hom_G (U,V) for a couple of G-modules U,V, one of
which must be the Heller operator \Omega^n(A) or something like that.
It's sort of cheating, really: you take a projective resolution and look
at the kernel of the differenetial at some point; call that \Omega^n.
Then you say "See, I do homology without chain complexes by simply referring
to a Hom-set with this module \Omega", but of course the complex is
hiding in there. Nonetheless, it gives algebraic cohomology with a
coefficient module A the flavor of topological cohomology with flat
coefficients. To combine the two extremes, one presumably invents a
category of pairs (X,A) where A is a module of pi_1(X), then defines
the morphisms in some way to include maps of spaces and homomorphisms of
modules, then defines the universal objects as some sort of common
generalization of the K(pi,n) and the Heller operators. With all this
machinery, cohomology with local coefficients may well be a representable
functor, and the above-defined universal objects would be my guess as to
the objects that represent the functor.
I would be interested to learn of any progress you could make in this
direction. Seems to me I pondered this ages ago but got lost in the
abstraction.
Dave Benson is fond of using the Heller operators so I'd guess his books
and articles would be a place to start reading. It's all algebra, though.
dave