Let be a ring spectrum, and a vector bundle. Then is -oriented if there is a “Thom class” such that the induced map of -modules
is an isomorphism.
This induces isomorphisms
induced by cupping and capping with (using the Thom diagonal ).
Let be a map of ring spectra. Then an -orientation of gives rise to an -orientation of by composition.
[Proof sketch] The key input here is that an
functorially induces an
via the composite
Crucially, this construction does not make use of any coherence; if we had some sort of coherence, we could simply apply . One checks that the Thom isomorphism for is induced from that for via this procedure. Functoriality then ensures the resulting map is also an equivalence.
A cohomology theory is complex oriented if there is a choice of a Thom class for every complex vector bundle that
is functorial under pullbacks, i.e. ; and
sends direct sums to products, i.e. .
is complex oriented. Indeed, if is a complex vector bundle, it is classified by a map , and is the pullback of the universal bundle. Applying gives a map
We claim this is a Thom class. We have to show that the composite
is an isomorphism. This map is obtained by applying to
Here is the tautological bundle, and when we write , we really mean .
We can describe the bottom map as sending to . This has an inverse given by , so is an isomorphism. Hence the induced map on Thom spaces is also an isomorphism.
The requirement that comes from the fact that the multiplication map is induced by .
Let be a ring spectrum. Then there is a bijection between
ring maps ; and
complex orientations of .
is complex oriented, a ring map
gives a complex orientation of
. Conversely, if
is complex oriented, then since
is the Thom spectrum of a complex vector bundle, we get a Thom class
. This is a ring map since the product
is induced by the direct sum, and the requirement that the Thom class sends direct sums to products is exactly the statement that
is a ring map.
Let be complex oriented. Then there is a canonical isomorphism and . Note that the choice of depends on the complex orientation of .
We first construct the class
, we know that
is the Thom space of the tautological bundle over
. So there is a preferred class
, the Thom isomorphism theorem tells us this generates
. This proves the lemma for
. Note that inductively applying the theorem proves that
as groups, but we want the multiplicative structure as well.
For , consider the Atiyah–Hirzebruch spectral sequence for . We claim that is represented by a generator of . Indeed, if it weren't, then its image when pulled back along would also not be a generator, which contradicts our previous observation.
Now the page of the Atiyah–Hirzebruch spectral sequence for is generated as a ring by and , both of which are permanent. So the Atiyah–Hirzebruch spectral sequence degenerates and gives the desired isomorphism.
The case follows from taking the limit.
There is a bijection between
complex orientations of ; and
classes that restrict to an -module generator of .
[Proof sketch] Our previous computation showed that a complex orientation of
induces such a class. The other direction is more roundabout. As in our previous argument, the class
forces the Atiyah–Hirzebruch spectral sequence for
to degenerate at
. By the Kronecker pairing, the AHSS for
also has to degenerate at
Now since is generated as a ring by the image of , we know the AHSS for also has to degenerate at , and so does that for . This gives us a preferred element in which is equivalently a map as desired. Playing the same game with shows that this map is a homotopy ring map.
A even homotopy ring spectrum is always complex orientable.
The Atiyah–Hirzebruch spectral sequence for
for degree reasons.
Complex -theory is complex orientable.