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Is the following statement true: a differentiable manifold is orientable if and only if the identity map on the manifold has a topological degree of 1?

No proofs needed; just confirmation.

If the statement isn't true, are there any modified version(s) of it that are, and what are they. Again, no proofs needed.

Thanks in advance!

MCS
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    This is nonsense. You need an orientation for degree to make sense as an integer instead of mod 2 in the first place, and the degree of the identity map is always 1 (either integrally, when it makes sense, or mod 2 when it doesn't). –  Dec 27 '16 at 21:42

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I pointed out in the comment that this is not true. Maybe the following is what you're thinking of.

Let $M$ be a closed oriented manifold and $G$ a finite group acting freely on $M$. Then the manifold $M/G$ is orientable if and only if $g_*: M \to M$ has degree $1$ for all $g$. (That is, the action is orientation preserving.)

  • @MCS Then this answer is relevant. It's the quotient of the sphere by the action of $\Bbb Z/2$, the nontrivial element acting by the antipodal map. The degree of the antipodal map is $(-1)^{n+1}$. –  Dec 27 '16 at 22:59
  • The context of the issues is in proving the equivalence of the orientability of a real projective space and the odd parity of its dimension.

    My argument is this:

    The antipode map A on the n-sphere is known to have degree (-1)^(n+1). Since RP^n is the quotient of S^n modulo the action of Z2 embodied BY the map A, A projects down to the identity map on RP^n. Thus, the identity map has degree 1 when n is odd and degree -1 when n is even. Hence, RP^n is orientable if and only if n is odd.

     – MCS Dec 27 '16 at 23:00
  • Your "thus" makes no sense. Everything before then is fine. –  Dec 27 '16 at 23:01
  • If A projects down to the identity, isn't deg(proj o A) = deg(Identity on RP^n)? If not, then what's the conclusion here? (I feel like I should be epsilon close to a valid result. xD) – MCS Dec 27 '16 at 23:06
  • @MCS That's correct for maps $M \to M$ that respect the action of $G$, so give well-defined maps $M/G \to M/G$, provided that $M/G$ is oriented and $M$ is oriented so the covering map is orientation-preserving. Degree does not make sense as an integer on non-orientable manifolds, and the degree of the identity is always one. All of these facts should be straightforward to prove using whatever your definition of degree is. –  Dec 27 '16 at 23:43