Otdfbt

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?

Let me expand a bit on my comments. If $E$ is a nice enough ring spectrum (e.g. an $mathbbE_2$-ring spectrum; there is also a slightly modified version that works for an $mathbbE_1$-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 $X to mathrmBGL_1(S^0)$ where $mathrmGL_1(S^0)$ is the space of self-equivalences of the sphere spectrum. Let $mathrmGL_1(E)$ denote the union of those components of $Omega^inftyE$ corresponding to units in $pi_0Omega^inftyE = pi_0E$. An $E$-orientation is a nullhomotopy of the composite $X to mathrmBGL_1(S^0) to mathrmBGL_1(E)$. (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 $Sigma^-nS^V_x$; to each path $x to y$ in $X$ you get an equivalence $Sigma^-nS^V_x to Sigma^-nS^V_y$; a homotopy of paths gives a homotopy between equivalences, etc. etc. This spells out a map $X to mathrmBGL_1(S^0)$. 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 $Xto mathrmBGL_1(E)$. 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 $mathrmHmathbbZ$-oriented, we consider the map $mathrmBSO to mathrmBGL_1(mathrmHmathbbZ)$. This factors as $mathrmBSO to mathrmBO to mathrmBGL_1(mathrmHmathbbZ)$. But $mathrmGL_1(mathrmHmathbbZ) = mathbbZ/2=O(1)$, the discrete group, so its classifying space is $mathrmBO(1)$ and the sequence $mathrmBSO to mathrmBO to mathrmBO(1)$ is the defining sequence for $mathrmBSO$. In other words: not only is it the case that every oriented bundle is $mathrmHmathbbZ$-oriented, but the converse also holds because a nullhomotopy of the composite $X to mathrmBO to mathrmBO(1)$ 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 $mathrmBSpin to mathrmBO to mathrmBGL_1(mathrmKO)$, nor do we have fiber sequences $mathrmBU to mathrmBO to mathrmBGL_1(mathrmMU)$. 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, $mathrmBG$, and you provide a nullhomotopy for $mathrmBG to mathrmBO to mathrmBGL_1(E)$. This buys you a map $mathrmBG to mathrmGL_1(E)/mathrmO$ (where you should interpret this symbol carefully- really it's just notation for the fiber of $mathrmBO to mathrmBGL_1(E)$ 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 $mathrmMU$-orientations. The nontrivial map $S^9 to mathrmBO$ certainly doesn't lift to $mathrmBU$ (since $pi_9mathrmBU = 0$), but it does become nullhomotopic in $pi_9mathrmBGL_1(mathrmMU)$. Indeed, every bundle on every sphere $S^n$ for $n>1$ is $mathrmMU$-orientable because the map $pi_nmathrmBGL_1(S^0) to pi_nmathrmBGL_1(mathrmMU)$ is trivial for $n>1$ (since the source is torsion and the target is torsion-free when $n>1$).