$\pi_* MU$, 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?)

  • $\begingroup$ One could identify the E_2 term of the Adams spectral sequence, but I think you are looking for a simpler/more geometric answer. I haven't spent much time thinking about MU in terms of cobordism, so I am very curious to see an answer to this question. $\endgroup$ Dec 29, 2011 at 6:30
  • $\begingroup$ I don't think there is a mathematical reason (aka easy proof), only some "philosophical" (aka unconvincing) ones. $\endgroup$
    – Grigory M
    Dec 29, 2011 at 8:41

2 Answers 2


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:

  1. 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!
  2. 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.
  3. 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.
  4. 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.

  • $\begingroup$ (I should mention I purposefully avoided mentioning anything about the Steenrod algebra or Adams spectral sequence... because once you do that- you may as well just calculate the whole thing. Everything collapses anyway.) $\endgroup$ Dec 29, 2011 at 12:59
  • $\begingroup$ Thanks! This is very nice (and you're right, I was being silly about the unoriented part of it). The paper by Quillen sounds fun. Incidentally, do you know where the usual proof via the Adams spectral sequence of $\pi_*MU$ (that it is the Lazard ring, not just a polynomial ring) was done? Everyone keeps pointing to Adams's blue book. $\endgroup$ Dec 29, 2011 at 15:26
  • $\begingroup$ Of course any polynomial ring like that "is" the Lazard ring :) But showing the formal group law of $\pi_*MU$ is the universal one... That was originally announced by Quillen in "On the formal group laws of unoriented and complex cobordism theory," so I imagine the first way he figured out how to prove this was by knowing the computation of $\pi_*MU$ already, and then showing some stuff commuted. But his published proof was the elementary one. The references I know of that use the ASS + formal group theory are Adams book and Ravenel's book. $\endgroup$ Dec 29, 2011 at 15:34
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    $\begingroup$ Basically the proof goes like this: (1) Identify the Lazard ring $L$ and its relationship to a ring $M \subset L \otimes \mathbb{Q}$ using formal group law stuff. (2) Identify $\pi_*MU$ and its relationship to $H_*MU$ using homotopy theory. (3) Construct a commutative square that says the relationship between the corresponding rings is (basically) what you think and use it to show the map you want is an isomorphism. $\endgroup$ Dec 29, 2011 at 15:37
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    $\begingroup$ Sounds like you go to some fun parties! (I'm not even being sarcastic) $\endgroup$ Dec 30, 2011 at 19:45

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!

  • 2
    $\begingroup$ This article is really nice, by the way. $\endgroup$ Dec 30, 2011 at 14:01
  • $\begingroup$ I agree! I found it originally in I.M. James' History of Topology $\endgroup$
    – Juan S
    Dec 30, 2011 at 23:21

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