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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.

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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? –  Z. L. 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

1 Answer 1

Let $M$ be a non-orientable smooth manifold. Non-orientability means that there are two open sets $U,V\subset M$ diffeomorphic to $\mathbb R^n$, orientations $\zeta$ on $U$, $\xi$ on $V$ and two points $p,q\in U\cap V$ such that $\zeta_p\ne\xi_p$ and $\zeta_q=\xi_q$. Now pick a compact connected nghd $K\subset U$ of the pair $p,q$ and a diffeotopy $f_t:M\to M$ that is the identity off $K$ and moves $p$ to $q$: $f_1(p)=q$. Since $f_t|U$ is a diffeotopy of $U$ and diffeotopies preserve orientations, we have $f_{1,*}(\zeta_p)=\zeta_q$. Similarly, pick a compact connected nghd $L\subset V$ of $p,q$ and a difeotopy $g_t:M\to M$, the identity off $L$ with $g_1(q)=p$, to have $g_{1,*}(\xi_q)=\xi_p$. Then the diffeo $h=g_1\circ f_1$ does the job: $$ h_*(\zeta_p)=g_{1,*}(f_{1,*}(\zeta_p))=g_{1,*}(\zeta_q)=g_{1,*}(\xi_q)=\xi_p\ne\xi_p. $$ For a flip at any other point $x\in M$, take a diffeo $\varphi:M\to M$ with $\varphi(x)=p$ and $\ \varphi^{-1}\!\circ h\circ\varphi$ reverses orientation at $x$.

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