I have the following question: If $E$ is a ring spectrum, then a complex orientation of $E$ is an element of $E^2(\mathbb{C}P^{\infty})$ that is mapped to $1$ in $E^2(\mathbb{C}P^{1})$. I have read that there is a natural orientation on $MU$ the spectrum of complex bordism. Can you tell me, how this orientation is defined and why it is an orientation?
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4$\begingroup$ This is Example 4.1.3 in Ravenel's ``Complex cobordism and stable homotopy groups of spheres'', where it is quite simply explained. $\endgroup$– Peter MayCommented Jul 17, 2012 at 1:06
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1$\begingroup$ To add to Peter's comment: at first read, I remember finding the unexplained equivalence $\mathbb{C} P^\infty$ somewhat confounding. It is given by the zero section of the tautological Thom spectrum $MU(1) = \mathbb{C} {P^\infty}^\gamma$, and the fact that it is an equivalence is somewhat amazing (and depends upon the fact that the Euler class of $\gamma$ is the generator of $H^2(\mathbb{C} P^\infty)$). $\endgroup$– Craig WesterlandCommented Jul 17, 2012 at 1:47
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1$\begingroup$ Sorry, that was meant to be ``the unexplained equivalence $\mathbb{C} P^\infty \to MU(1)$. $\endgroup$– Craig WesterlandCommented Jul 17, 2012 at 1:48
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2 Answers
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A complex orientation on a homotopy commutative ring spectrum $E$ is sometimes defined to be an element in $\widetilde{E}^2(\mathbb{CP}^\infty)$ which restricts to the canonical element of $\widetilde{E}^2(S^2) = \pi_0 E $.
In fact, there is another definition of an orientation, which generalizes better to other settings: a complex orientation on $E$ is an assignment of a Thom class for each complex vector bundle $V \to X$ (that is, a class in $\widetilde{E}( B(V), S(V))$ which restricts to the canonical generator on each copy of $(B(V_x), S(V_x)) \subset (B(V), S(V))$ for each $x \in X$. (In ordinary homology, this uniquely determines the Thom class, but in generalized homology, this is additional data rather than a merely condition to be satisfied.)
For a complex orientation, this assignment is required to be functorial, and moreover multiplicative: Thom classes for vector bundles $V, W \to X$ determine a Thom class for $V \oplus W$ (because the Thom space of $V \oplus W$ is the smash product of the Thom spaces of $V, W$). If you think about it this way, it is easier to see that a complex orientation on $E$ is the same as a map of ring spectra $MU \to E$. In fact, a complex orientation on $E$ means that the universal vector bundle $\xi_n \to BU(n)$ has a complex orientation, which if you unwind is a map $\mathrm{Th}(\xi_n) \to E$. These give the spaces in the $MU$-spectrum and multiplicativity makes this a map $MU \to E$ (strictly speaking, one has to make some kind of a $\lim^1$-argument to make this rigorous).
One reason that this is a more natural definition is that it generalizes to other types of bundles than complex bundles: one can talk about a real-oriented cohomology theory $E$ (which comes from a map $MO \to E$) or a spin-oriented cohomology theory (coming from a map $M \mathrm{Spin} \to E$), among other things. It takes a little work, though, to show that complex orientability is equivalent to the degeneration of the AHSS for $\mathbb{CP}^\infty$, but one direction is easy: if $E$ has a complex orientation in the sense of this answer, then we get a class in $\widetilde{E}^2(\mathbb{CP}^\infty)$ by regardeding $\mathbb{CP}^\infty$ as the Thom space of the tautological bundle over itself.
You can read more about this in the course notes for "Complex oriented cohomology theories and the language of stacks," among other places. I also found Jacob Lurie's notes on "chromatic homotopy theory" very helpful.
answered Jul 17, 2012 at 2:42
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I don't know how to comment rather than answer, but Craig, the equivalence is really not mysterious: the zero section of the universal complex line bundle gives an equivalence from $\mathbb{C}P^{\infty}$ to the total space of the unit disk bundle $D$. The total space $S$ of the unit sphere bundle is really just $S^{\infty}$ (think of the universal principal circle bundle), which is contractible, and the quotient map $D\to D/S = MU(1)$ is an equivalence. The equivalence $\mathbb{C}P^{\infty}\to MU(1)$ is the composite of these two equivalences.
Let me add a conceptual aside about orientations of fibrations. Let $p\colon X\to B$ be a spherical fibration with fiber $S^n$, such as the fiberwise $1$-point compactification of an $n$-plane bundle and let $E$ be a ring spectrum. In ``Parametrized homotopy theory'', Johann Sigurdsson and I show how to make sense of the parametrized spectrum $X\wedge E$ over $B$ with fibers $E$. An orientation (in the classical cohomological sense) is the same thing as a trivialization of this parametrized fibration, that is, an equivalence from it to the parametrized spectrum $(B\times S^n)\wedge E$ over $B$.
I should admit that my answer is digressive. Section 5 of my paper ``What are $E_{\infty}$-ring spectra'' relates universal $E$-orientations of $G$-bundles to maps of ring spectra $MG \to E$.
answered Jul 17, 2012 at 2:45
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2$\begingroup$ I agree: mysterious is not the right adjective; indeed, the argument that you give is quite clear. However, I find it remarkable that a space can have the same homotopy type as a Thom spectrum over it. In fact, the only way that this is possible is if the cohomology of the base is the ring of power series in the Euler class of the bundle. $\endgroup$ Commented Jul 17, 2012 at 3:14