Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p \in M$ be a point of a non-orientable smooth manifold, $M$. Does there exist a diffeomorphism $f: M \rightarrow M$ with $p \mapsto p$ and such that $df : T_pM \rightarrow T_pM$ is orientation reversing? My feeling is yes. I was thinking about trying to take an embedding $\gamma : S^1 \rightarrow M, \star \mapsto p$ such that parallel translation around $\gamma$ reverses orientation, then pushing forward the vector field $d/d\theta$ and extending it. Then taking the flow at time $2\pi$. However I wasn't sure about the existence of such a $\gamma$ and the whole approach seems a bit contrived. Is it true and if so is there an easier way? Thank you for your time.

P.S.this is motivated by the question of the well-definedness of connect-sum for non-orientable manifolds.

share|improve this question
Concerning the existence of $\gamma$: consider $\widetilde{M}$, the oriented double cover of $M$. The point $p$ has two preimages under the covering map. Take a simple curve that connects these preimages, and project it to $M$. If the image has self-intersections, you can let $p$ be such a self-intersection, and restrict to a simple loop based at $p$. The smoothness at $p$ is not guaranteed, but can be achieved by modifying the curve near $p$. –  user53153 Jan 13 '13 at 6:38
@PavelM :Thanks! Maybe I'm picturing this incorrectly but why can't the restriction have still more self intersections? Maybe you can choose $p=\gamma(t_i)= \gamma (t_j)$ such that $t_i-t_j$ is minimal among such points of self-intersection or something? –  Tim kinsella Jan 13 '13 at 6:53
@PavelM It seems to me that the question chooses $p$ first and you want that $p$ to be fixed by $f$. Moreover, if $M$ is not compact then an orientation-reversing path may have infinite self-intersections, or am I wrong? –  Zango Lotino Jan 13 '13 at 10:46
@ZangoLotino : I worried about both of those issues but: I think if you can do it for $q$ then you can do it for $p$ by pre and post-composing with any diffeomorphism taking $p$ to $q$. As for the self-intersections, I think what I wrote in the comment above yours works. The set of all such $t_i - t_j$ must be closed (by continuity and sequential compactness of the interval) and bounded away from 0 since the path is locally injective because it is a covering projection of an injective loop. Does this make sense? I'm not totally certain. –  Tim kinsella Jan 13 '13 at 11:27
@Timkinsella True, in general there will be more self-intersections under self-intersection. One has to choose $p$. For example, fix $t$ so that $\gamma(t)$ is within a loop and consider the restrictions of $\gamma$ to $(t-\epsilon,t+\epsilon)$. For small $\epsilon$ this is an embedding, but for some $\epsilon>0$ is ceases to be. That means either $\gamma(t+\epsilon)$ or $\gamma(t-\epsilon)$ hits another point in $\gamma([t-\epsilon,t+\epsilon])$. Now restrict to a smaller interval, and you have a simple closed curve. –  user53153 Jan 13 '13 at 14:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.