The next goal in this series of posts is to understand fully the computation of the complex cobordism ring , which is due to Milnor. To start with, though, let’s consider a simpler example: the spectrum
. The spectrum
is obtained in the same way that
is, but with the Thom spaces of the universal
-dimensional real bundles over
. In other words, if
is the universal
-dimensional bundle, we define
(where means Thom space) and define maps
coming from the map classifying
. The homotopy colimit of the sequence
is a spectrum and, for the same reasons as , is in fact a commutative ring spectrum (even an
ring spectrum).
One reason to care about is the following result:
Theorem 1 (Thom) The homotopy groups
(a graded ring) is the cobordism ring of unoriented manifolds.
In other words, to describe , we can also use geometry: an element of
is a compact
-manifold
modulo the relation of cobordism: two
-manifolds
are cobordant if there is an
-manifold-with-boundary
such that
. From a geometric point of view, it is thus interesting to determine
. It turns out that we can do so using homotopy-theoretic, algebraic methods; this was what Thom showed.
How might we compute the homotopy groups? In general, the homotopy groups of any space are extremely difficult to compute, but in this case we can get a complete answer for two reasons: first, it’s stable homotopy groups we’re after, not just homotopy groups; second, much more importantly, is a fairly simple spectrum.
In fact, we have:
Theorem 2 (Thom)
is a wedge of Eilenberg-MacLane spectra
(and its shifts) and
is a polynomial ring on variables
for all
such that
is not a power of
.
This is the theorem I’d like to discuss today. It is definitely a much easier computation than that of , but it will be a toy example.
1. The homology of : part 1
The strategy for showing that splits as a sum of shifts of
is to look at its (co)homology. It turns out that the cohomology of
is a free module over the Steenrod algebra
. In view of this, and the Hurewicz theorem, we’ll be able to see that
itself must be a sum of
‘s.
But first we need to know what the homology of is, and we’ll need as a ring as well as a graded group. Fortunately, we have a fairly explicit homotopy presentation of
as the homotopy colimit of
, and we can work out the homology of each of these explicitly via the Thom isomorphism.
Let’s start by recalling that
where each in degree
is dual to the
th power of the generator
. This handles
. The homology of the spaces
in general can be computed as follows:
Theorem 3 The map
establishes an equivalence
In other words, a basis for
is given by the products
where
.
In the limit, we find that is a polynomial ring
(the element
becomes the unit).
By the Thom isomorphism, the homology of the spectrum is more or less going to be the homology of
, so it seems we’ve answered the first question. But we need a bit more: we need the action of the dual Steenrod algebra
, and this is in fact the main part. To understand the action of
on
, let’s start by working it out on
. It’ll be easier to first do the dual computation.
Proposition 4 Let
be the generator. Let
be a sequence of nonzero elements. Then
unless
, in which case one gets
.
Proof: Induction on (when
it is clear). Suppose
. Then, we clearly have
, so the inductive hypothesis implies that
.
The claim is thus that the only Steenrod square that doesn’t annihilate is
. This follows from the fact that if
is the total Steenrod square, then
This means that is forced to be
.
With this in mind, we can write down what is as a comodule over
. Let’s recall the structure of the dual Steenrod algebra
, which was determined by Milnor. There are elements
of degree
, such that
The are defined as the dual elements to
(with respect to the Serre-Cartan basis of
).
We have a comodule structure map
Another way of phrasing the previous result is that the coefficient of in
is
if
, and zero otherwise. That is,
Here the dots signify terms that do not include or
.
2. The homology of , II
Let’s now finish working out the homology of as a comodule over
. We know that if
is the Thom space of the universal bundle on
, then
(because the
-bundle over
is contractible), and that the Thom class in
corresponds to the generator
.
In other words, we have that
where each is in degree
and corresponds under the Thom isomorphism to
: equivalently, under the equivalence
,
corresponds to
. In particular,
where each is in degree
; this is because of the desuspension in
. Since
is free on the appropriately ordered products of the
, and we have a commutative diagram (where all homology is with
-coefficients)
we find inductively that the vector space is free on the products
with
. Observe that the desuspensions made in
mean that we don’t have to change degrees anywhere. Moreover,
becomes the unit becomes
acts as the unit in
.
So, putting everything together, we find:
Proposition 5 We have an isomorphism of rings
I’ve just been writing reduced homology for emphasis: that’s the natural thing for spectra, anyway.
The next goal is to figure out the action of on this homology. The map
is just the map
and it is in particular a morphism of rings. So we just need to figure out where each goes.
In fact, comes from
where each
is in degree
: the element
corresponds to
. By naturality, we might as well work out the coaction of
on
. Let’s work it out mod decomposables.
Proposition 6 The coaction
sends
The dots signify terms not involving
or
.
Proof: To see this, we just have to remember that is a desuspension of
, and consequently
is just a shift of
. Consequently, the
‘s are corresponding to the
‘s under the identification
. If we keep this in mind, then the formula is a consequence of the previous analysis for
.
Now, if we recall that is just the unit element of
under the map
, then we find that the coaction
sends
where the signify decomposable elements in
.
The main result is:
Theorem 7 As an
-comodule and an algebra, there is an isomorphism
When we analyze as a comodule, we will get something similar, but it will not be quite so simple.
Proof: The strategy is to produce a map of comodule-algebras . We can do this very easily: it’s the same thing as a map of plain algebras
because the tensor product is a “coinduced” comodule and has a similar universal property (for mapping into it) that induced modules have in ordinary algebra. We define this map via
if
and zero otherwise.
The map of comodule-algebras is then obtained via the composition
and it is this composition we want to prove is an isomorphism. So far, we see that it’s a map between graded algebra of the same dimension in each degree, since .
To show that the composite map is an isomorphism, we’re going to have to use a commonly used technique: we’ll look at indecomposable elements. In dimension
, the indecomposable elements in
is spanned the image of
. We know, moreover, from our earlier computations that modulo decomposables,
This is, in either case, indecomposable in degree in
. In particular, the map
sends indecomposables to indecomposables, so by induction on the degree we find that
is a surjection. Since the domain and the codomain have the same dimension in each degree,
must be an isomorphism.
3. The homotopy of
In the previous sections, we worked out completely the homology of as a comodule over the coalgebra
, as well as an algebra in its own right. We know want to work out the spectrum
, and prove the theorems stated at the start of the blog post.
So, we want to show that is a wedge of Eilenberg-MacLane spectra.
Proposition 8 Suppose
is a connective spectrum, whose homotopy groups
are all 2-torsion. Suppose
is a co-induced module over
, i.e.
for a graded
-vector space
whose graded pieces are finite-dimensional.Choose a basis
of the graded piece
. Then we have a decomposition of spectra
Proof: The strategy will be to produce a map which is an isomorphism on
-homology, and which will then be forced to be an equivalence.
In fact, we find that is free as a module over the Steenrod algebra,
. The dual basis elements
are cohomology classes in dimension
which define a map
This map is an isomorphism on cohomology by construction: the cohomology of an Eilenberg-MacLane spectrum is free on .
Dualizing, we find that it is an isomorphism on -homology. It follows that the cofiber, which is a connective spectrum with 2-torsion homotopy groups, has trivial
-homology. This means that the cofiber is contractible: for, if its first nonzero homotopy group were in degree
, then we would have
.
So, in particular, applying this to , we find that
is a wedge of shifts of
. This means in particular that the Hurewicz map
is an injection, and that its image must consist of the primitive elements: these are given by
This completes the proof of Thom’s theorem.
May 23, 2012 at 6:13 pm
[…] of this series of posts is to understand the computation of the complex cobordism ring . In the computation of the unoriented cobordism ring , we did this by working out as a comodule over the dual Steenrod algebra , and observing that it […]
May 24, 2012 at 4:05 pm
[…] proof of this will be analogous to the computation of . In fact, the idea is essentially that, by the Thom isomorphism […]