Why is $\pi_* MU$ concentrated in even degrees? $\pi_* MU$, which is the cobordism ring of manifolds with a complex structure on the stable normal bundle, is a polynomial ring $\mathbb{Z}[x_2, x_4, \dots]$. I'm probably being silly here, but is there some obvious reason why everything is in even degrees? Is it obvious that an odd-dimensional manifold whose normal bundle in some imbedding gets a complex structure is cobordant to zero? (For that matter, is it even obvious that such a manifold is unorientedly cobordant to zero?)
 A: I don't think either result is "obvious" geometrically. As Grigory says, it is somehow morally true. Here is what I understand of this part of the story:


*

*You can prove directly from Thom's isomorphism between complex cobordism and $\pi_*MU$ together with some Serre mod C stuff that $MU_* \otimes \mathbb{Q}$ is concentrated in even degrees (you can even get the multiplicative structure pretty easily). As a consequence one gets that every odd-dimensional complex manifold becomes null-bordant (in the complex sense) after taking disjoint unions enough times. Is there a way to do this geometrically? I'd love to see it!

*If you want an almost purely geometric proof that the odd parts vanish, see Quillen's paper "Elementary proofs of some results of cobordism theory using Steenrod operations" (it's awesome) Corollary 5.2. Again- it doesn't seem too obvious.

*If you already know everything about unoriented cobordism, then it is easy (I hate to call anything obvious) to show that any odd-dimensional stably complex manifold is null-bordant. Proof: By the basic result, we need only check that the Stiefel-Whitney numbers are all zero. But when we compute a Stiefel-Whitney number, we'll have to use  a product of Stiefel-Whitney classes where at least one of the Stiefel-Whitney classes lives in odd degree. But these are all zero; the proof of this I leave as an exercise.

*Another reason why this is non-obvious: Every framed manifold is canonically stably complex and so we have a "forgetful" map $\pi_*S \rightarrow MU_*$ (it's the same as the Hurewicz map). Now, we happen to know that the image of this map is zero in positive dimensions, since $\pi_kS$ is torsion for $k>0$, and if we also knew that $MU_*$ was torsion-free we'd conclude that every positive-dimensional framed manifold bounds a stably complex one. Now, this is probably just my ignorance, but I have no idea how to prove this geometrically. So, for example, take some odd-dimensional class in $\pi_*S$ lying in the kernel of $\pi_*S \rightarrow \pi_*MO$ (this is basically to rule out stuff generated by Hopf invariant elements).


Anyway, maybe this doesn't answer your question... and maybe the answer is yes and I'm just being silly... but I should really go to bed.
A: According to Peter May's Stable Algebraic Topology 1945-1966 
...there is no known geometric reason why the 
complex cobordism ring should be concentrated in even degree

(This is on page 23)
So really, as others have alluded to, I guess this is just an algebraic result. It would be interesting to find some geometric meaning behind it!
