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Let h* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle.

  1. Is it true that, if $E$ is orientable with respect to h*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point?

  2. Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$?

Added later: I thought to a possible simple answer using weak orientability, actually proving what stated in the last comment. I wrote this as a comment to answer 3.

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up vote 10 down vote accepted

Assuming $X$ is connected and the dimension $k$ of $\xi$ is positive, here is a proof of the yes answer to (1).

Consider the relative Atiyah-Hirzebruch spectral sequence

$$H^p(D\xi,S\xi;h^q(\mathrm{pt}))\Rightarrow h^\ast(D\xi,S\xi)$$

where $D\xi$ and $S\xi$ are the disk and sphere bundles in some Euclidean metric. The pair $(D\xi,S\xi)$ is $(k-1)$-connected, so $E_2^{p,q}$ is zero for $0<p<k$. Note that $$E_2^{0,k}=H^0(D\xi,S\xi;h^k(\mathrm{pt}))\cong \tilde{H}^0(T\xi;h^k(\mathrm{pt})),$$ the reduced cohomology of the Thom space in dimension zero, which should be trivial by our dimension hypothesis (the Thom space is connected). Hence the edge homomorphism

$$H^k(D\xi,S\xi;h^0(\mathrm{pt}))\to h^k(D\xi,S\xi)$$

is an isomorphism, and you can check that a Thom class corresponds to a Thom class.

Edit Now that I've carefully read Charles' answer, I see that I was over-simplifying things by assuming the spectrum was connective (so that $h^q(\mathrm{pt})=0$ for $q$ negative). If not, either we take the connective cover and appeal to the claim in Charles' answer, or we've got stuff in the lower quadrant of our spectral sequence. Even so, there is still an "edge homomorphism"

$$h^k(D\xi,S\xi)\twoheadrightarrow E^{k,0}_\infty \hookrightarrow E^{k,0}_2 = H^k(D\xi,S\xi;h^0(\mathrm{pt}))$$

which should take a Thom class to a Thom class.

Added later: Here is a proof of a yes answer to (2) (again assuming connected, positive dimension). The homology group $H_k(T\xi;\mathbb{Z})$ is $\mathbb{Z}$ if $\xi$ is orientable (in the usual sense, with $\mathbb{Z}$ coefficients) and $\mathbb{Z}_2$ otherwise. This follows from the twisted Thom isomorphism $H_0(X;\mathbb{Z}^\xi)\cong H_k(T\xi;\mathbb{Z})$, where $\mathbb{Z}^\xi$ denotes the local coefficient system on $X$ determined by $\xi$, which holds regardless of orientability. (I'm getting lazy with tildes here.)

Then the universal coefficient theorem gives $H^k(T\xi;\mathbb{Z}_p)\cong\mathrm{Hom}(H_k(T\xi),\mathbb{Z}_p)\cong 0$ if $\xi$ is non-orientable (and assuming $p$ is odd). So non-orientable in the usual sense implies there can't be a Thom class with $\mathbb{Z}_p$ coefficients.

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(2) No. Let $L\to RP^\infty$ be the tautological line bundle. Then for any odd prime $p$, both $RP^\infty$ and and $Thom(L)$ have trivial cohomology in $p$-adic coefficients (or mod $p$ coefficients). So $L$ is trivially orientable with respect to $H^*(-;\mathbb{Z}_p)$, but is not an orientable bundle.

(1) This is not so easy; I think the answer is yes. I'll write $A^*$ for your multiplicative theory $h^*$, and $B$ for singular cohomology with coefficients in $A^0(pt)$.

(Stuff deleted.)

I believe you should be able to prove it like this: consider the following diagram of generalized cohomology theories (or, I should really say spectra now): $$ D\leftarrow C\to A$$ where $C$ is the connective cover of $A$ (so $\pi_qC=0$ for $q<0$, but $\pi_qC\to \pi_q A$ is an isomorphism for $q\geq0$), and $D$ is the $0$-truncation of $C$ (i.e., $D=H(\pi_0A)$, the Eilenberg-Mac Lane spectrum with coefficient ring $\pi_0A$).

The claim is that if $E\to X$ is $A$-orientable, then it is $C$-orientable (because those negative homotopy groups don't matter ... it's important here that $E\to X$ is an honest bundle, rather than a virtual one, and also I'm assuming that a "Thom class" is always an element in dimension $d=$ dimension of $E$, rather than some other dimension).

If $E\to X$ is $C$-orientable, then it must be $D$-orientable by the map $C\to D$.

Later edit. I tried to answer a more general question originally, but the proof I gave doesn't work, as pointed out in the comments. I've edited out the offending portions; now the proof basically amounts to the argument Mark Grant gives in his answer.

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Sorry, I don't understand one point. You say that, for characteristic bigger than 2, "a bundle E is orientable with respect to B if and only if it is orientable in the conventional sense". But this implies that the answer to the question 2 is yes, while you said that it is no. – Fabio Feb 2 '11 at 17:49
@Charles: why is it true that $A^0(\text{pt}) = B^0(\text{pt})$? Also, I would have thought that orientability over $E$ would imply it for $h\Bbb Z$ with respect to an additional assumption: Wouldn't one also require that $h\Bbb Z$ is an $E$-algebra? – John Klein Feb 2 '11 at 18:07
@ Fabio, I think you have a point there. – Charles Rezk Feb 2 '11 at 18:44
@John. I don't understand. Given a map of multiplicative cohomology theories $A\to B$, $A$-orientability implies $B$-orientability, because a Thom class for $A$ will then necessarily map to a Thom class for $B$. – Charles Rezk Feb 2 '11 at 18:53
@Charles: Thank you for you answer. I have a question more: you say that $Thom(L)$ has trivial $\mathbb{Z}_{p}$-cohomology. But then, how can the Thom class exist? It must be a non-trivial class, because, restricted to each fiber, is non-trivial. – Fabio Feb 2 '11 at 19:37

At the risk of being obvious, let me make some general comments:

For a vector bundle $E\to X$ of rank $n$ let $E_x$ be the fiber over $x\in X$. For a multiplicative cohomology theory $h$, the cohomology $h^\*(E_x,E_x-0)$ is a free module of rank $1$ over the graded ring $h^*$, with generator in $h^n$. Such a generator is called an orientation at the point $x$. A Thom class (also called an orientation) is an element of $h^n(E,E-0)$ that restricts to such a generator for every $x$.

There is also the other sense of orientation: a continuous choice of pointwise orientation for all points. In the special case of ordinary cohomology, in other words when the graded ring $h^*(point)$ is all in degree zero, an orientation in this weak sense implies an orientation in the Thom-class sense; but not in general.

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just a note on notation, when you write $E_x-0$ you mean the fiber minus the zero vector, but $E-0$ means the total space minus the zero section. – Sean Tilson Feb 3 '11 at 6:35
ps, thanks for the above, it is very helpful. – Sean Tilson Feb 3 '11 at 6:36
1. Do you know an example in K-theory of a bundle which is weakly orientable but not orientable? More generally, strong orientability in K-theory is equivalent for the bundle to be spin^c. Do you know the condition for being weakly orientable? 2. When a bundle is weakly orientable, is there a "weak Thom isomorphim"? – Fabio Feb 3 '11 at 7:45
A bundle is orientable in real $K$-theory iff it has Spin structure iff $w_1=w_2=0$. It is orientable in the usual ($H\mathbb{Z}$) sense iff $w_1=0$. – Mark Grant Feb 3 '11 at 8:23
I believe that orientability in the weak sense is always equivalent to orientability with respect to $h^0(point)$, and so for complex K-theory it is equivalent to just $w_1 = 0$. – Tyler Lawson Feb 3 '11 at 13:26

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