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My question is related to the following question by Mark Grant here on math overflow:

Formal group law of unoriented cobordism

There it is stated that $MO_*$ has a formal group law $F_0$, universal for formal group laws in characteristic $2$ which satisfy $F(x,x) = 0$.

Of course this formal group law is classified by a map $MU_* \to MO_*$ and in view of the model of $MU$ as unitary ring spectrum and $MO$ as orthogonal ring spectrum this map is indeed induced by a canonical map of (unitary) ring spectra $$c: MU \to MO,$$

see also the answer by Neil Strickland here: Which cohomology theories are real- and complex-orientable? .

Now I have two questions:

Q1) Does the map $c$ factor over $MSO$ ?

Q2) If so, can one say anything about the corresponding formal group law over $MSO$?

In view of how I think one can construct the map $MU \to MO$ I think it should be obvious that the map factors through $MSO$, since the map $U(n) \to O(2n)$ factors over $SO(2n)$. But I have not been able to find a reference that gives the example of $MSO$ being complex oriented.

So I guess Q2) is more interesting, already for the simple reason that the ring $MSO_\ast$ is much more delicate than the ring $MO_\ast$. Moreover (as Johannes Ebert also pointed out in a comment in the link above) real oriented theories come equipped with a map from $MO$ and hence split in Eilenberg-MacLane spectra, whereas $MSO$ behaves quite differently.

In addition, I have one last question:

Supposing Q1) and Q2) are true, then as I have already mentioned that $MSO$ is complex oriented. In particular we have a notion of abstract chern classes for $MSO$-cohomology. Moreover we have a canonical map $MSO \to H\mathbb{Z}$. The composite $MU \to H\mathbb{Z}$ is just the usual orientation of integral cohomolgy, in particular the abstract chern classes of this complex oriented theory are just the usual chern classes, but since oriented real bundles have pontryagin classes I ask myself:

Q3) Is there any way to view pontryagin classes as special case of abstract chern classes?

(It seems not to be the case for the cohomology theory $MSO$ which I would have thought of first).

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Answer to Q1 is certainly yes (this is just a forgetful map, geometrically.) I believe the answer to Q3 is "yes" and I would proceed as follows: $MSO^*(-)$ has a pushforward map (either directly from the geometric definition, or using the Thom isomorphism for orianted bundles, etc.) and so we can define Euler classes for oriented bundles (and in particular complex bundles) by pushing forward the unit along the zero section and then pulling back. In the standard Grothendieckian way this gives rise to a theory of "Chern classes" once you know that $MSO^*(\mathbb{P}E)$ is free over $MSO^*(X)$... –  Dylan Wilson Mar 8 '13 at 14:09
    
... where $X$ is the base of the bundle $E$ and $\mathbb{P}E$ is its projectivization. I believe this is true, via the usual proof (say, by considering some AHSSs). The trouble is that this procedure only works for defining characteristic classes for complex bundles! If I tried to do the same thing for oriented real bundles I would have to assign an invariant to oriented line bundles- but those are trivial. So to get Pontryagin classes I would have to at least develop the theory in the unoriented case, at least. But the usual way of doing this gives me the Stiefel-Whitney classes. That doesn't –  Dylan Wilson Mar 8 '13 at 14:17
    
mean that some slightly modified technique won't work... but I dunno if I'd call them abstract chern classes any more. –  Dylan Wilson Mar 8 '13 at 14:17
    
Regarding a similarity between Pontryagin classes and Chern classes, see this very interesting mathoverflow answer by Johannes Ebert: mathoverflow.net/questions/16632/… . However, that answer does not address any possibility of describing Pontryagin classes as Chern classes for some complex oriented theory. –  Ricardo Andrade Mar 9 '13 at 3:47
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1 Answer

up vote 6 down vote accepted
  • MSO is certainly complex-oriented
  • The resulting formal group law satisfies $[-1]_F(x)=-x$
  • Let $R_*$ be the universal example of a graded ring with a formal group with the above property, so there is a natural map $R_*\to MSO_*$. I think this is injective, and it becomes an isomorphism after inverting $2$.
  • Let $U$ be the set of all positive integers not of the form $2^j-1$, let $U_0$ be the subset of all even integers, and put $U_1=U\setminus U_0$. One can show that $R_*$ has a polynomial generator $a_i$ of degree $2i$ for each $i\in U$, and the only relations are $2a_i=0$ for $i\in U_1$.
  • The additive structure of $MSO_*$ is known by old work of Wall and Atiyah. I don't know a really convincing interpretation in terms of formal groups. All the torsion is killed by $2$, and there is a lot of stuff in odd degrees.
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Hi @Neil, I'm aware of a preprint by Andy Baker and Jack Morava showing that there is some kind of universal odd fgl over $MSp[1/2]$. Does the fact that $MSO[1/2]$ also supports a universal fgl come from the Novikov's determination that $\Omega^{SO}_\ast\cong\Omega^{Sp}_\ast$ when 2 is inverted? Is there a reference for your statement about this universal odd formal group law? –  Jon Beardsley Oct 15 '13 at 23:05
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