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The empty set can be regarded as an object in the category of smooth manifolds, at least for technical considerations.

Is the empty set an orientable manifold?

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Just a note: The authors of this book… stipulate in the footnote on page 327 that the empty set is orientable. –  Fredrik Meyer Oct 27 '12 at 14:26

2 Answers 2

up vote -2 down vote accepted

EDIT: This is wrong. See Henning Makholm's answer below.

An $n$-manifold is orientable if and only if it has a nonzero differential form taking $n$ arguments. The empty manifold has no nonzero forms whatsoever, so it is not orientable in any dimension.

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I'm not sure if I like this answer. What about using the empty map? The form only has to be non-zero on every point of the manifold which is trivally satisfied for the emptyset. –  M.B. Oct 27 '12 at 13:39
The relevant condition is not at the differential form must be different from the everywhere zero form, but that at each point in the manifold, the differential form's restriction to that point's tangent space must be nonzero. This is true of the empty map, because there are no points to check it for. –  Henning Makholm Oct 27 '12 at 13:41

Contradicting Espen's answer: The empty map does (vacuously) provide a nonzero differential form at every point on the manifold, and does so continuously. Therefore the empty manifold is orientable in every dimension.

(But really this will depend on the exact definition you're working with, and among the various usually-assumed-equivalent definitions for "orientable" there are some that are only really equivalent when the manifold is assumed to be nonempty).

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Ok, this appeared at the very same time of my reply to the other answer. And I agree with this. –  M.B. Oct 27 '12 at 13:39
But what if we view the emptyset as a closed manifold. It most certainly does not have $\mathbb{Z}$ as its top homology group. This gives that the emptyset is not orientable for any $n$. –  M.B. Oct 27 '12 at 13:43
@M.B.: But the orientable double cover of the empty set is the empty set itself, which is connected. Therefore the empty set is orientable in every dimension. –  Henning Makholm Oct 27 '12 at 13:52
But... Is the empty set really connected? Also, a manifold is orientable if and only if the orientable double cover is disconnected. So I'm not sure if your last comment is an argument for or against orientability of the empty set :-) –  commenter Oct 27 '12 at 13:55
@commenter: Hm, oops. I'll have to claim that the empty set is disconnected then. But that it certainly is, because the set of its connected components has cardinality $\ne 1$. :-) –  Henning Makholm Oct 27 '12 at 13:57

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