Question

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 $E^1(P^\infty)\neq 0$?

Answer 1

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.