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?

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    $\begingroup$ 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. $\endgroup$ May 29, 2019 at 15:29
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    $\begingroup$ map.mpim-bonn.mpg.de/Complex_bordism#Stably_complex_structures $\endgroup$
    – Mark Grant
    May 29, 2019 at 15:41
  • $\begingroup$ 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). $\endgroup$ May 29, 2019 at 16:08
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    $\begingroup$ @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 $X$? $\endgroup$
    – user137162
    May 29, 2019 at 18:16
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    $\begingroup$ @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? $\endgroup$
    – Mark Grant
    May 29, 2019 at 18:26

1 Answer 1


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 $n>1$).


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