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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMay 21st 2012

felt like creating double cover

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeMay 21st 2012

I do not understand. The entry orthogonal structure talks about Riemannian metric on a vector bundle which is not necessarily the tangent bundle. Hence one can consider it even on the bundles whose total space is nonorientable, e.g. Moebius band. OK, fine. But then the diagram at double cover depicts lifting the cocycle corresponding to the tangent bundle (at least the notation suggests this). So I do not see how this is possible in nonorientable case (the entry says that it makes sense even then),

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 21st 2012

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeMay 21st 2012

I still do not understand. The orthogonal structure is by the corresponding entry a smooth choice depending on $x$, hence a smooth section, hence after quotienting to $\mathbf{Z}_2$ one gets a section of the orientation sheaf. Surely orientation sheaf exists always, but its section, nor the section of the sheaf of all metrics, don’t unless $X$ is orientable. Maybe I do not understand the terminology…

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeMay 21st 2012

A choice of orientation is a lift all the way through $B SO \to BO \to B GL$.

Notice that you can always lift through $\mathbf{B} O \to \mathbf{B} GL$, since the underlying spaces are homotopy equivalent. So if you forget smooth structure for a second, then $B O \to B GL$ is even an equivalence and a lift through it means nothing.