Simply connected manifolds are orientable

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).