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For an orientable $n$-manifold $M$ and its (orthonormal) frame bundle classifying map $\tau_M : M \to BO(n)$, we have a lift diagram of the following sort:

Lift diagram here: the map from M to BO(n) has an up-to-homotopy lift through the projection BSO(n) to BO(n)

There are two orientations on $M$. Is it ever true that they correspond to two different witnesses $H$ in the above diagram?

This question is motivated by the observation that an orientation on $M$ is sometimes considered an example of an "$SO(n)$-structure" on $M$, a special case of a $G$-structure on $M$, part of the data of which one takes the 2-morphism witnessing the lift of $\tau_M$ to $BG$.

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  • $\begingroup$ An aside example: if I take the circle $S^1$, whose (oriented orthonormal) frame bundle $Fr^+(S^1) \cong S^1 \times \text{SO($1$)}$ is in the same isomorphism class as its orientation-reversed "$Fr^-(S^1)$" (which follows from $[M, B\text{SO($1$)}] \cong \ast$), is there a way to identify these two orientations on the frame bundle via e.g. different choices of $H$ in the question? $\endgroup$
    – Arnav Das
    Commented Nov 18 at 9:03
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    $\begingroup$ The space of orientations on the circle is too trivial. $\endgroup$
    – David Roberts ♦
    Commented Nov 18 at 10:46
  • $\begingroup$ @DavidRoberts Can you make your sentiment precise? (Are you implying there are sufficiently nontrivial spaces of orientations on other manifolds that encode different lifts/witnesses?) $\endgroup$
    – Arnav Das
    Commented Nov 18 at 13:28
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    $\begingroup$ I don't think I understand the terminology here. What is a "witness"? A 2-morphism? If so, then couldn't the answer to the question in the first paragraph be yes, simply because the source 1-morphism is different? $\endgroup$
    – Mark Grant
    Commented Nov 18 at 21:21
  • $\begingroup$ Is the question whether for any chosen $\tau_{M,\pm}$ there are different (i.e. non-homotopic) homotopies $H$ that could fill the triangle? You should note that there is a model for $Bi$ that is a 2-sheeted covering space. So one can consider what happens in that case. $\endgroup$
    – David Roberts ♦
    Commented Nov 18 at 21:36

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Up to homotopy, there is a fibration $$ BSO_n \to BO_n \to B\mathbb Z_2. $$ The space of orientations of $M$ is the (homotopy) fiber of the induced map of mapping spaces $$ \text{map}(M,BSO_n) \to \text{map}(M,BO_n) $$ at the basepoint defined by the tangent bundle of $M$. This homotopy fiber is empty when $M$ is unorientable, and has the homotopy type of the space of maps $$ \Omega \text{map}(M,B\mathbb Z_2) \simeq \text{map}(M,\pm 1) \cong \text{functions}(\pi_0(M),\mathbb Z_2) $$ when $M$ is orientable. A point in the homotopy fiber is the same thing as a lift $M\to BSO_n$ together with a choice of commuting homotopy, as in the diagram appearing in the OP.

The space of orientations is therefore homotopically discrete with contractible components. This resolves your question: the answer is that for a given lift, there is just one commuting homotopy up to homotopy. In fact, we've established that even more is true: for a given lift, the space of commuting homotopies for that lift is contractible.

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