16
$\begingroup$

A complex-oriented cohomology theory $E^*$ is a multiplicative cohomology theory with a choice of Thom class $x\in\tilde{E}^2(\mathbb{C}P^\infty)$ for the universal complex line bundle (which can be used to define generalised Chern classes for all complex vector bundles).

A real-oriented cohomology theory $F^*$ a multiplicative cohomology theory with a choice of Thom class $x\in \tilde{F}^1(\mathbb{R}P^\infty)$ for the universal real line bundle (which can be used to define generalised Stiefel-Whitney classes for all real vector bundles).

Question 0: Is this correct?

Question 1: Are there any examples of cohomology theories which are both real and complex orientable?

Question 2: (Assuming a yes to Question 1) Are there any results/papers where the interaction between a real and complex orientation is used in an essential way? (I'm thinking perhaps about non-immersion results for real projective spaces.)

Thanks.

Update: Neil's answer and Johannes' comments have answered my original questions: every real oriented cohomology theory is complex oriented, and in fact is a wedge of $H\mathbb{Z}/2$'s. Then let me ask a follow up question. Are there any complex-oriented theories which are not real-orientable but have $E^1(P^\infty)\neq 0$?

Improve this question
$\endgroup$
4
  • 9
    $\begingroup$ I think a real oriented theory is an algebra over the orthogonal Thom spectrum $MO$, for the same reason as a complex oriented one is an algebra over $MU$. Thom has proven that $MO$ is a wegde of Eilenberg-Mac-Lane spectra. Therefore, any real-oriented theory is a wedge of Eilenberg Mac Lanes as well. Consequence: such theories are boring for stable homotopists. They are, though, interesting for the study of real vector bundles, but all the information they detect can also be detected with ordinary $Z/2$-homology. $\endgroup$
    – Johannes Ebert
    Commented Feb 16, 2011 at 10:21
  • 3
    $\begingroup$ Btw: $HZ/2$ is both, real and complex oriented and there is the relation $w_{2i}=c_i mod 2$ between Stiefel-Whitney classes and the mod $2$ reductions of Chern classes. This is sometimes important. $\endgroup$
    – Johannes Ebert
    Commented Feb 16, 2011 at 10:21
  • 2
    $\begingroup$ you might be interested in looking at the work of Romie Banjeree math.jhu.edu/~banerjee $\endgroup$
    – Sean Tilson
    Commented Feb 17, 2011 at 14:59
  • $\begingroup$ @Sean: Thanks, it certainly seems to be close to what I was thinking of. $\endgroup$
    – Mark Grant
    Commented Feb 18, 2011 at 7:54

1 Answer 1

Reset to default
17
$\begingroup$

A real-orientable ring spectrum $F$ admits a ring map from $MO$, and there is a straightforward ring map $MU\to MO$, so $F$ is also complex orientable. Moreover, $MO$ is a wedge of $H/2$'s, so $MO\wedge F$ is also a wedge of $H/2$'s, and $F$ is a retract of $MO\wedge F$ (by using the ring structure etc) so it is again a wedge of $H/2$'s. Thus, you don't expect to learn anything about (non)immersion from $F$ that you could not already learn from $H/2$ (although some kinds of bookkeeping may be simplified).

For the follow-up question:

The mapping spectrum $E=F(S^1_+,MU)$ is complex-orientable but not real-orientable and has $$ \tilde{E}^1(\mathbb{R}P^\infty) = \tilde{MU}^1(\mathbb{R}P^\infty) \oplus \tilde{MU}^0(\mathbb{R}P^\infty) $$ Here $MU^{\ast}(\mathbb{R}P^\infty)=MU^*[[x]]/{[2]}(x)$ with $|x|=2$ and $MU^{-2}=MU_2\simeq\mathbb{Z}$ so the first summand is zero but the second is not.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Couldn't the ring structure on $F$ give you new information? $\endgroup$
    – Oscar Randal-Williams
    Commented Feb 17, 2011 at 9:15
  • 5
    $\begingroup$ As $F$ is a wedge of $H/2$'s, the same is true of $F\wedge F$, so the multiplication map $F\wedge F\to F$ decomposes into maps $H/2\to H/2$ (of various degrees) ie Steenrod operations. Thus, the ring structure is determined by the Steenrod module structure of the mod 2 homology, but it may package the information in a more convenient way. $\endgroup$
    – Neil Strickland
    Commented Feb 17, 2011 at 10:39
  • 2
    $\begingroup$ So this canonical map $MU \to MO$ will surely factor over $MSO$, right? Can one describe what formal group law it classifies? Do you know any references where this map is studied? $\endgroup$
    – Markus Land
    Commented Mar 7, 2013 at 17:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .