# Orientable with respect to complex cobordism?

I have learned that an orientation of a manifold $$M$$ with respect to ordinary cohomology is an ordinary orientation, that an orientation with respect to complex K-theory is a Spin$$^c$$ structure, and that an orientation with respect to real K-theory is a spin structure. I think this is a very beautiful picture and I am wondering if orientations with respect to other theories like elliptic cohomology, G-equivariant cohomology, quaternionic K-theory, or spin cobordism correspond to interesting and well-studied differential-geometric structures.

Complex manifolds should be oriented with respect to any complex-oriented cohomology theory. Indeed, if $$E$$ is a complex-oriented cohomology theory then all complex vector bundles carry $$E$$-orientations. In particular, if $$X$$ is a complex manifold then its tangent bundle $$TX$$ has a complex structure making $$X$$ an $$E$$-oriented manifold.

Given that complex cobordism is the universal complex-oriented cohomology theory, I would guess that an orientation with respect to complex cobordism is a complex structure. I have been unable to find any literature on this and I am unsure how to approach the problem rigorously on my own. Maybe someone knows?

• It isn’t really right to say that an “an orientation with respect to real K theory is a spin structure”. It’s the other way around: if you admit a Spin structure then you have a Thom isomorphism in KO-theory. May 29, 2019 at 15:29
• map.mpim-bonn.mpg.de/Complex_bordism#Stably_complex_structures May 29, 2019 at 15:41
• In fact the relationship between ordinary orientability and ordinary cohomology is sort of a coincidence that doesn’t occur in the other examples nor in the situation of your question. (In particular having a stably almost complex structure is not the same as having an orientation wrt MU). May 29, 2019 at 16:08
• @DylanWilson Is it not true to say that the data of a $KO$-Thom class for the stable normal bundle of a manifold $X$ is equivalent to the data of a spin structure on $X$?
– user137162
May 29, 2019 at 18:16
• @DylanWilson: I'm having trouble seeing why a stably complex structure on a manifold $M$ isn't the same as having an $MU$-orientation. An $E$-orientation on an manifold $M$ is defined as a choice of Thom class for the stable normal $\nu$ bundle of $M$. But then an $MU$-orientation of $M$ is a Thom class $t\in \tilde{MU}^k(T\nu)$, which is represented by a map $T\nu\to MU(k)$. Is the problem that this might not be the Thomification of a classifying map? May 29, 2019 at 18:26

Let me expand a bit on my comments. If $$E$$ is a nice enough ring spectrum (e.g. an $$\mathbb{E}_2$$-ring spectrum; there is also a slightly modified version that works for an $$\mathbb{E}_1$$-ring) then the story of orientations work like this:

If you have a vector bundle, or, more generally, a stable spherical fibration (of rank 0, say) on a space $$X$$, this will be classified by a map $$X \to \mathrm{BGL}_1(S^0)$$ where $$\mathrm{GL}_1(S^0)$$ is the space of self-equivalences of the sphere spectrum. Let $$\mathrm{GL}_1(E)$$ denote the union of those components of $$\Omega^{\infty}E$$ corresponding to units in $$\pi_0\Omega^{\infty}E = \pi_0E$$. An $$E$$-orientation is a nullhomotopy of the composite $$X \to \mathrm{BGL}_1(S^0) \to \mathrm{BGL}_1(E)$$. (Again, there are some slight modifications if $$E$$ is less nice; there are also definitions one can make without anything more than a homotopy ring structure on $$E$$, but that is a slightly less intuitive picture I think.)

To summarize in a more informal way: if you have a vector bundle $$V$$ of rank $$n$$ on $$X$$, you can form the corresponding stable spherical fibration which, intuitively, means you associate to each point of $$x$$ the spectrum $$\Sigma^{-n}S^{V_x}$$; to each path $$x \to y$$ in $$X$$ you get an equivalence $$\Sigma^{-n}S^{V_{x}} \to \Sigma^{-n}S^{V_y}$$; a homotopy of paths gives a homotopy between equivalences, etc. etc. This spells out a map $$X \to \mathrm{BGL}_1(S^0)$$. This is a local system of spectra which all look like $$S^0$$ up to equivalence. You can fiberwise smash with $$E$$ to get a local system of spectra that all look like $$E$$ up to equivalence, and that gives you the map $$X\to \mathrm{BGL}_1(E)$$. An $$E$$-orientation is a trivialization of this local system. It's saying that, through the eyes of $$E$$, the bundle looks like the trivial bundle (whence the Thom isomorphism: the Thom spectrum looks like a suspension (i.e. the Thom spectrum of a trivial bundle) through the eyes of $$E$$).

Now, if you want to $$E$$-orient all spin bundles, or spin-c bundles, etc., then you just have to do the above in the universal case, when $$X$$ is the classifying space for such bundles.

So, for example, to show that every oriented vector bundle is $$\mathrm{H}\mathbb{Z}$$-oriented, we consider the map $$\mathrm{BSO} \to \mathrm{BGL}_1(\mathrm{H}\mathbb{Z})$$. This factors as $$\mathrm{BSO} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{H}\mathbb{Z})$$. But $$\mathrm{GL}_1(\mathrm{H}\mathbb{Z}) = \mathbb{Z}/2=O(1)$$, the discrete group, so its classifying space is $$\mathrm{BO}(1)$$ and the sequence $$\mathrm{BSO} \to \mathrm{BO} \to \mathrm{BO}(1)$$ is the defining sequence for $$\mathrm{BSO}$$. In other words: not only is it the case that every oriented bundle is $$\mathrm{H}\mathbb{Z}$$-oriented, but the converse also holds because a nullhomotopy of the composite $$X \to \mathrm{BO} \to \mathrm{BO}(1)$$ is exactly the data of an orientation.

But this is a happy accident. For example, it is not the case that we have fiber sequences $$\mathrm{BSpin} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{KO})$$, nor do we have fiber sequences $$\mathrm{BU} \to \mathrm{BO} \to \mathrm{BGL}_1(\mathrm{MU})$$. Instead, in each instance the composite has a nullhomotopy (which is, in the first case, a difficult theorem of Atiyah-Bott-Shapiro, and, in the latter case, sort of tautological) but the first term is not the fiber of the second map.

Some added stuff in response to the OP and Mark:

Suppose you've got some classifying space for vector bundles with extra structure, $$\mathrm{BG}$$, and you provide a nullhomotopy for $$\mathrm{BG} \to \mathrm{BO} \to \mathrm{BGL}_1(E)$$. This buys you a map $$\mathrm{BG} \to \mathrm{GL}_1(E)/\mathrm{O}$$ (where you should interpret this symbol carefully- really it's just notation for the fiber of $$\mathrm{BO} \to \mathrm{BGL}_1(E)$$ but you can realize it as a sort of homotopy quotient if you want). The failure of this map to be an equivalence will be the failure of "$$E$$-orientation" to be the same as "$$G$$-structure".

As an explicit example, let's consider the difference between $$U$$-structures and $$\mathrm{MU}$$-orientations. The nontrivial map $$S^9 \to \mathrm{BO}$$ certainly doesn't lift to $$\mathrm{BU}$$ (since $$\pi_9\mathrm{BU} = 0$$), but it does become nullhomotopic in $$\pi_9\mathrm{BGL}_1(\mathrm{MU})$$. Indeed, every bundle on every sphere $$S^n$$ for $$n>1$$ is $$\mathrm{MU}$$-orientable because the map $$\pi_n\mathrm{BGL}_1(S^0) \to \pi_n\mathrm{BGL}_1(\mathrm{MU})$$ is trivial for $$n>1$$ (since the source is torsion and the target is torsion-free when $$n>1$$).