# Boundary of a one-dimensional manifold - choosing an oriented atlas

I am trying to understand how to assign an orientation to the boundary of one-dimensional manifolds using Loring Tu's book on manifolds.

This is what I got so far: Let $M$ be an oriented manifold. If $M$ is one-dimensional, we see that it is always possible to give $\partial M$ an orientation, since $\Omega^0(M)=\mathbb{C}^\infty(M)$ clearly has non-vanishing elements. Exactly like for the $n\ne 0$ case, we can divide $\Omega^0(M)$ in two classes $\{+,-\}$. The standard choice is to give orientation $-$ to the point $0\in\partial\mathbb{H}^1$. Now, let $p\in\partial M$. In the book (section 21.6) he says that: if there is an orientation-preserving(reversing) chart $\phi:U\rightarrow\mathbb{H}^1$ around $p$, then we assign the orientation $-(+)$ to $p$.

But my guess would be that such $\phi$ cannot be just any chart in the (maximal) atlas of $M$ - if we allowed for this, then we would get both orientation-preserving and orientation-reversing charts around $p$, leaving the orientation at $p$ badly defined. So I would say that $\phi$ must belong to the oriented atlas corresponding to the orientation of $M$, instead of the atlas of $M$.

Question 1: Is this correct?

In the case of the closed interval, for example, we cannot use the chart $\phi(x)=-x$ (defined on $[0,1)$) to orient $\{0\}$ nor we can use $\psi(x)=1-x$ (defined on $(0,1]$) to orient $\{1\}$, and so I would say we can use instead $\phi(x)=x$ and $\psi(x)=x-1$. This gives orientation $-$ for $0$ and $+$ for $1$. This agrees with the result in the book.

Question 2: If I use $\phi(x)=-x$ and $\psi(x)=-x+1$ (which belong to an oriented atlas that gives the opposite orientation $[-dx]$), I still get orientation $-$ for $0$ and $+$ for $1$. So, does it even matter which oriented atlas I take my charts from?

• It's true that every one-dimensional smooth manifold is orientable, but the argument is more complicated than the one you gave in your second paragraph. Nonvanishing sections of $\Omega^0(M)$ have nothing to do with orientability -- for a $1$-manifold, what you need is a nonvanishing section of $\Omega^1(M)$. – Jack Lee Sep 14 '18 at 17:50
• @JackLee I wanted to orient the boundary of $M$, not $M$ itself - I rewrote the question. – Soap Sep 15 '18 at 12:49
• @JackLee Can you answer your competitor's example without using the classification theorem in your book? – Selene Auckland Mar 11 at 11:39