# Simply connected manifolds are orientable

For a simply connected $n$-manifold $M\subseteq\Bbb{R}^k$, I want to show that $M$ is orientable.

Take a point $p\in M$ and take an $n$-disc, $D^n$, around $p$ (we can take it as small as we please). Since $S^{n-1}$ is orientable and $M$ (and consequently $TM$) is simply connected, the orientation map $S^{n-1}\to TM$ can be extended to $D^n\to TM$. So around every point there is such a disc. We can construct an atlas (an orientation) out of these.

Does this suffice to prove the claim?

• Why exactly can the map of the sphere be extended to the whole ball when $n\geq 3$? May 8, 2014 at 14:25
• Oh I misread this: en.wikipedia.org/wiki/… So these are not true then, are they?
– Xena
May 8, 2014 at 14:28
• If you know about covering spaces, I would consider the oriented double cover of a manifold. For connected non-orientable $M$, the oriented double cover $M^*$ is orientable and connected. May 8, 2014 at 14:33

Let $M$ be a connected non-orientable manifold of dimension $n$ and let $M^*$ be its oriented double cover which is connected as $M$ is non-orientable. By general covering space theory, there is a short exact sequence $$0\to\pi_1(M^*)\to \pi_1(M)\stackrel{f}{\to}\mathbb{Z}/2\mathbb{Z}\stackrel{g}{\to} 0$$ and so if $\pi_1(M)$ is trivial, then $\mathbb{Z}/2\mathbb{Z} = \ker g =\operatorname{im}f = 0$ which is a contradiction.

Here is one way to argue. An orientation of a smooth $n$-manifold is a nowhere vanishing section of $\Lambda^n M$ (i.e., a volume form). This is a real line bundle $L\to M$. Put a Riemannian metric on $M$. This defines a metric on $L$ as well. Constructing a nowhere vanishing section of this bundle is the same as constructing a section of the unit sphere bundle $U\to M$ of $L$ (unit sphere in ${\mathbb R}$ is of course the set $\pm 1$). The unit sphere bundle $U\to M$ is a covering map (since the fiber is zero-dimensional). Since $M$ is simply-connected, the bundle $U\to M$ is trivial. Hence, it admits a section.

I realize that the question is old, but I wanted to present a more elementary approach, which only uses the fact that homotopically trivial loops lift to loops on a covering space.

Suppose $$M$$ is not orientable. Consider the oriented double cover $$p:X \to M$$ and pick $$x_0 \in X$$. Now, let $$\lambda$$ be a path connecting $$x_0$$ and the other point in the fiber, call it $$x_1$$ (here we are using the fact that the double cover of a non-orientable $$M$$ is connected - since it is a manifold, it is also path-connected). Now, $$p \circ \lambda$$ is a loop in $$M$$, which lifts to $$\lambda$$. Since $$M$$ is simply connected, $$p \circ \lambda$$ is homotopically trivial, which is an absurd (since $$\lambda$$ is not a loop).

• You took $\lambda$ to be an arbitrary path in the fiber. Then why is $p \circ \lambda$ a loop? Nov 14, 2018 at 17:06
• @Error404 $\lambda$ is a path in $X$ which begins in $x_0$ and ends in the other point in the fiber. Thus, $p(\lambda(0))=p(\lambda(1))$, since both $\lambda(0)$ and $\lambda(1)$ are in the same fiber. Nov 14, 2018 at 17:25
• @AloizioMacedo, why $p\circ \lambda$ being homotopically trivial implies $\lambda$ homotopically trivial? In order to prove that, we should be able to find a homotopy for $\lambda$ from a homotopy for $p\circ\lambda$, right? I don't know how to do that Jan 8, 2019 at 16:39
• @rmdmc89 coverings satisfy the homotopy lifting property Jun 11, 2020 at 7:10

I realise this question has been answered twice and is now quite old, but just to correct Daniel's answer:

You don't quite get the short exact sequence you describe, but you get the following;

the oriented double cover $M^*$ can be viewed as a fibration sequence $\mathbf{Z}/2\mathbf{Z}\to M^*\to M$, whose long exact sequence has a piece which looks like $0\to \pi_1M^*\to \pi_1M\to \mathbf{Z}/2\mathbf{Z}\to \pi_0M^*\to0$. Now if $M$ is non-orientable, then $M^*$ is connected, and so $\pi_0M^*=0$. If $M$ were also simply connected, then we would have an exact sequence $0\to \mathbf{Z}/2\mathbf{Z}\to 0$, which cannot happen.

I realize this question is pretty old, but I do not resist writing an argument I feel is elementary and, at least for me, transmits the idea why it is true.

We may assume $$M$$ is a connected and simply-connected smooth manifold. Fix a point $$p\in M$$ and an orientation $$o_p$$ on $$T_pM$$. We will show how to continuously propagate'' $$o_p$$ along any smooth curve $$\gamma:[0,1]\to M$$ with $$\gamma(0)=p$$.

If the image of $$\gamma$$ is contained in a connected coordinate domain $$U$$ of $$M$$, this is clear: let $$\varphi:U\to\mathbb R^n$$ be a local chart and define $$o_t$$ to be the orientation of $$T_{\gamma(t)}M$$ that makes $$(d\varphi_{\gamma(t)})^{-1}\circ d\varphi_p:T_pM\to T_{\gamma(t)}M$$ orientation-preserving (note that $$o_0=o_p$$). In case $$\gamma$$ is arbitrary, we cover its image by connected coordinate domains and use the Lebesgue number lemma to partition $$[0,1]$$ so that each subinterval $$[t_{i-1},t_i]$$, has image under $$\gamma$$ contained in a connected coordinate domain $$U_i$$, where $$i=1,\ldots,m$$; then we apply the previous case and induction.

We end up with an orientation $$o_1$$ of $$T_qM$$, and we define $$o_q$$ to be $$o_1$$, where $$q=\gamma(1)$$. Next we show that $$o_q$$ is independent of the covering $$\mathcal U=\{U_i\}_{i=1}^m$$. Let $$\mathcal V=\{V_j\}_{j=1}^k$$ be another covering of $$\gamma([0,1])$$ such that $$\gamma([t_{j-1},t_j])\subset V_j$$ and $$V_j$$ is a connected coordinate domain of $$M$$ for $$j=1,\ldots,k$$. Denote by $$o^{\mathcal U}_t$$ and $$o^{\mathcal V}_t$$ the orientations of $$T_{\gamma(t)}M$$ obtained by using $$\mathcal U$$ and $$\mathcal V$$, respectively. Consider $$I=\{\,t\in[0,1]\;|\; o^{\mathcal U}_t=o^{\mathcal V}_t\,\}$$. Of course $$I\neq\varnothing$$ as $$0\in I$$. Further, $$I$$ is open: if $$\bar t\in I$$ and $$\gamma(\bar t)\in U_i\cap V_j$$ then the connected component of $$\gamma^{-1}(U_i\cap V_j)$$ containing $$\bar t$$ is also contained in $$I$$. Similarly, one sees that $$I$$ closed. Hence $$I=[0,1]$$.

Finally, we show that $$o_q=o^\gamma_q$$ is independent of the curve $$\gamma$$ joining $$p$$ to $$q$$. Given a homotopy $$\{\gamma_s\}$$ of $$\gamma=\gamma_0$$ keeping the endpoints fixed and $$s_0$$, we may find $$\epsilon>0$$ so that $$\gamma_s([t_{i-1},t_i])$$ is contained in a connected coordinate domain $$U_i$$ for all $$s\in(s_0-\epsilon,s_0+\epsilon)$$ and $$i=1,\ldots,m$$; fix $$s_1\in(s_0-\epsilon,s_0+\epsilon)$$. We proceed by induction on $$i:1,\ldots,m$$. By construction $$(d\varphi_{\gamma_{s_0}(t_i)})^{-1}\circ d\varphi_{\gamma_{s_0}(t_{i-1})} \quad\mbox{and}\quad (d\varphi_{\gamma_{s_1}(t_i)})^{-1}\circ d\varphi_{\gamma_{s_1}(t_{i-1})}$$ are orientation-preserving. By the induction hypothesis on $$i$$, $$d\varphi_{\gamma_{s_1}(t_{i-1})}^{-1}\circ d\varphi_{\gamma_{s_0}(t_{i-1})}$$ is orientation-preserving. Therefore $$d\varphi_{\gamma_{s_1}(t_{i})}^{-1}\circ d\varphi_{\gamma_{s_0}(t_{i})}$$ is also orientation-preserving. Since $$t_m=1$$ and $$\gamma_{s_0}(1)=q=\gamma_{s_1}(1)$$, we deduce that $$o_q^{\gamma_s}$$ is locally constant on $$s$$; hence it is constant on $$s$$. Since any two curves joining $$p$$ to $$q$$ are homotopic due to the simple-connectedness of $$M$$, we are done.