Last time, I described the construction which assigns to every compact -space
(for
a compact Lie group) the equivariant K-group
. We saw that this was a functor from the (equivariant) homotopy category to commutative rings, using more or less the same arguments as in ordinary homotopy theory, only with small alterations.
The purpose of this post is to describe more of Segal’s paper. Actually, I won’t be covering any legitimate K-theory in this post; that’ll have to wait for a third. I’ll mostly be describing various classical constructions for vector bundles in the equivariant setting.
In the classical theory of (ordinary) vector bundles on compact spaces, a basic result is the Serre-Swan theorem, which identifies the category of (complex) vector bundles with the category of projective modules over the ring
of complex-valued continuous functions on
. This is essentially a reflection of the fact that any vector bundle on
, say
, can be obtained as a retract of some trivial bundle
. Taking retracts corresponds to choosing idempotents in the ring of
-by-
matrices in
, and this description via idempotents applies as well to projective modules over
(or, in fact, any commutative ring).
The crucial statement here, that any vector bundle is a retract of a trivial one, fails in the equivariant case, simply because a vector bundle on which acts nontrivially can’t be a retract of a vector bundle with trivial action. But we have something reasonably close to it.
Definition 1 Given a
-representation
and a
-space
, we can form a vector bundle
, which his naturally
-equivariant.
This bundle is, equivalently, formed by taking the equivariant map .
-vector bundles on
are identified with
-representations, so we just have to pull back.
Anyway, the claim is:
Theorem 2 (Equivariant Serre-Swan (Segal)) Any
-vector bundle
is a direct summand of a bundle
for some
-representation
.
The proof of this result turns out to be significantly harder than the non-equivariant case. There, one has a plain vector bundle over a plain space
, and one just needs to find a finite number of global sections of
that generate
. Then, that gives a surjection from a trivial bundle to
, which necessarily splits.
In order to handle the equivariant case, however, one needs more work. Let be a
-vector bundle over the
-space
. We can always find a finite-dimensional space of sections
such that
generates all the fibers; that is, the image of
in each fiber
is all of
. This has nothing to do with equivariance. Now if
were a
-invariant subspace of the
-vector space
, then we could just think of
as a
-representation and consider a map
which would be the desired surjection.
Unfortunately, this is not the case. It is not even true necessarily that is contained in a finite-dimensional
-invariant vector space of sections (which would also be enough). We’ll actually need something topological.
1. The Peter-Weyl theorem
For a compact group , the classical Peter-Weyl theorem gives a decomposition of
as a
-module: namely, if the
are the finite-dimensional irreducible representations of
, then we have:
Theorem 3 (Peter-Weyl) The Hilbert space
(where
is given the Haar measure) decomposes as an orthogonal sum
.
This is similar to (and a generalization of) the decomposition for a finite group
, as the
range over the irreducible representations of
. The difference, of course, is that the Peter-Weyl theorem gives a topological decomposition of a Hilbert space, rather than a vector space decomposition.
Anyway, we won’t actually need just the Peter-Weyl theorem, but a generalization thereof.
Theorem 4 (Mostow) If
is a
-Banach space, then the elements
such that
is finite-dimensional are dense.
The Peter-Weyl theorem gives a (Hilbert space) decomposition of into a collection of finite-dimensional
-invariant subspaces, so this is a generalization. To prove Mostow’s theorem, one argues as follows. One consequence of the proof of the Peter-Weyl theorem is that the statement of Mostow’s theorem is true not only for
, but for
with the sup norm. That is, any continuous function on
can be uniformly approximated by a continuous function
whose translates by elements of
span a finite-dimensional space.
Now, because is a Banach space, one may integrate an
-valued continuous function over any nice probability space. (This is probably not the most general case under which one may do this.) So, this means we can integrate an
-valued function over
.
We thus define a pairing as follows. Given
,
, we define
When , this is just averaging, for instance.
Let us fix , and try to approximate
by elements of
whose
-translates span a finite-dimensional space. To do so, we take
Now, note that if is such that
, then
In particular, if is chosen such that
is supported in a small neighborhood of the identity
, then
is very close to
. So we can think of these “generalized averages”
as good approximations to
. Note moreover that
is well-behaved with respect to translation; that is,
if the action of on
is appropriately defined.
With these facts, we can easily prove the theorem. Fix , and fix
. If we choose the continuous function
whose support is very near
and whose total integral is one, then
Next, let be a function, approximating
by in
-norm, say such that
, and we can also assume that the
-translates of
span a finite-dimensional subspace of
. In this case, we have:
where . In particular, we find
though by construction of , we have also seen that the translates of
span a finite-dimensional subspace of
. So we have gotten the desired approximation
to
.
2. Proof of the equivariant Serre-Swan theorem
With Mostow’s theorem proved, we can now give the proof of the equivariant Serre-Swan theorem. Let be a
-vector bundle. As we’ve seen, to prove the result (i.e. to express
as a retract of a bundle associated to a
-representation), it suffices to find a finite-dimensional
-invariant subspace
such that the image of in
at each
is all of
.
In order to do this, we start by choosing any subspace (not necessarily
-invariant) with the property that the image of
in
fills
for all
. By the Peter-Weyl theorem, we can find a subspace
which is “close” to
(i.e., by approximating each element of a basis of
very closely) such that
is contained in a finite-dimensional invariant
-representation
. But if
is “very close” to
, then
generates
at each
, since
does, and since
is compact. Thus, we’re done, and the result we wanted is proved.
3. The equivariant Grassmannian
One of the classical facts about vector bundles in topology is the following. If is a compact space (or, more generally, a suitably nice space, e.g. a CW complex), then isomorphism classes of
-dimensional vector bundles on
are in bijection with homotopy classes of maps
from
into the Grassmannian of
-planes in the vector space
. Since isomorphism classes of
-dimensional vector bundles are the same as isomorphism classes of principal
-bundles, it follows that the infinite Grassmannian is a model for the classifying space
.
One consequence of all this discussion is that we can get an analogous result for -spaces. Let
be a compact
-space. Let
be the
-space defined as follows. For each (finite-dimensional)
-representation
, we let
be the
-space of
-dimensional subspaces of
. For an imbedding
, there is an obvious map
. Now since the category of finite-dimensional
-representations is filtered, we define
This is a -space (not compact!) and it is the analog of the infinite Grassmannian in the equivariant case.
The claim is that there is a “universal” -dimensional
-vector bundle on
. To see this, note that there is an
-dimensional
-vector bundle on
for each
(contained in
); this is the “tautological” one, defined in the usual way. These are compatible with the morphisms. Let
be the
-vector bundle on
given by the colimit.
Theorem 5 There is a natural isomorphism between isomorphism classes of
-dimensional
-vector bundles on
and
(that is,
-homotopy classes of
-maps).
Note that . One can prove this as in the real case. Namely, if
is a
-vector bundle, then we can write
for some finite-dimensional
-representation
. Then we get a map
sending
to the image of
in
. This defines a map from
to the infinite Grassmannian, and it’s easy to see that
becomes the pull-back of the universal bundle. The uniqueness is similar as in the non-equivariant case.
Leave a Reply