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It's not possible to put an atlas (of manifold with boundary) on $[0,1]$ because any chart whose domain contains $0$ has to have a positive derivative while any chart whose domain contains $1$ has to have a negative derivative (taking charts of $n$-manifold with boundary to have values in $\mathbb{R}_+\times\mathbb{R}^{n-1}$). Yet it has to be orientable: what's the catch?

There's an answer here: Is $[0,1]$ an *oriented* manifold with boundary? (and Stokes theorem), but I find it very distasteful. Is there a problem with defining an orientation of a manifold with boundary to be an orientation of the manifold of its interior points? Wouldn't that be a more kosher approach?

  • Actually my proposition isn't so good because it doesn't give a way to integrate $\mathrm{d}t$ on $[0,1]$ without resorting to minus signs –  Jun 17 '20 at 18:02

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This is, afaik, the usual approach to defining an orientation on a manifold with boundary: take an orientation on the interior, which then induces an orientation on the boundary. The problem that you ran into when trying to find an oriented atlas for a manifold with boundary on $[0, 1]$ is reflective of how the induced orientation is determined. The points $0$ and $1$ end up having opposite orientations.

An orientation on an orientable manifold can be described by an ordered basis of the tangent space to a point. Say, if $(v_1, ..., v_n)$ is a basis for $T_pM,$ where $M$ is orientable, one orientation of $M$ is given by $[v_1, ..., v_n];$ the other is given by $[-v_1, ..., v_n] = [v_2, v_1, v_3, ..., v_n].$ You may have a definition of an orientation on an $n$-dimensional manifold $M$ as a nowhere-vanishing $n$-form $\omega.$ These definitions are equivalent - the orientation determined by $\omega$ is the orientation $[v_1, ..., v_n]$ iff $\omega(v_1, ..., v_n)>0.$

Then, to define the induced orientation on the boundary of $M,$ let $p\in\partial M$ and let $\varphi: U\to \mathbb{H}^n$ be a chart from a neighborhood of $p$ to the $n$-dimensional upper half-space $\mathbb{H}^n,$ i.e. $\mathbb{R}^{n-1} \times \mathbb{R}_{\ge 0}.$

Then, the orientation on $\partial M$ is $[u_1, ..., u_{n-1}],$ where $(u_1, ..., u_{n-1})$ is a basis of $T_p\partial M$ such that $[\varphi^*(0, 0, ..., 0, -1), u_1, ..., u_{n-1}]$ is the orientation on $M.$ ($\varphi^*(0, 0, ..., 0, -1)$ is the pullback of the vector $(0, ..., 0, -1),$ which can be visualized as a vector in the tangent space $T_pM$ pointing away from $T_p\partial M.$ The "outward normal vector" can also be used in place of this vector in this construction.) It takes a little bit of work to show that this is well-defined, but it is.

In your specific example, the vector $\varphi^*(0, 0, ..., 0, -1)$ points in different directions when $\varphi$ is a chart in a neighborhood of $0$ or $1,$ so the induced orientations at $0$ and $1$ are different.

anon
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  • O.K. that would be good but how do you integrate with that definition? For instance to integrate $\mathrm{d}t$ on a segment you need to put a minus sign somewhere. –  Jun 17 '20 at 19:43
  • I don't know exactly what you mean by "need to put a minus sign somewhere." Are you looking to apply Stokes' Theorem? – anon Jun 17 '20 at 20:09
  • No just to integrate. I mean to say that you can't find an oriented atlas to integrate $\mathrm{d}t$. What you can do is take a chart about $0$ ($x$) and one about $1$ ($1-x$), the second one with the wrong orientation, take a subordinate partition of unity $\varphi_1$, $\varphi_2$ and put $\int \mathrm{d}t = \int \varphi_1\mathrm{d}t - \int \varphi_2\mathrm{d}t$. –  Jun 17 '20 at 20:19
  • Actually, I don't think that your definition works for $n=1$. For the orientation on the interior to induce an orientation on the boundary you need to do the following: given an oriented atlas of the interior (which doesn't give an atlas of the boundary, for instance if all the domains of the charts are all in the interior), chose for $x$ in the boundary a boundary map about $x$, and fix it if its jacobian doesn't agree with the oriented atlas on the interior. You can do that for $n>1$ because you can do a reflection relative to one of the $n-1$ extra dimensions, but you can't for $n=1$, as... –  Jun 17 '20 at 20:36
  • every move would make the chart not have value in $\mathbb{R}_+$ anymore, as in my orignal question. –  Jun 17 '20 at 20:37
  • I don't think this is a huge problem, although it is an "edge case" (accidentally good pun) that should be addressed - if we can't find $\varphi$ compatible with the orientation on the interior of the manifold, we can use an orientation-reversing $\varphi$ instead and just reverse the orientation we obtain at the end. To your point about integration, yes, a minus sign will end up coming in, but this occurs naturally - integration is well-defined only up to orientation of a manifold, and if the boundary of a manifold has an induced orientation different from the canonical one, and you want to – anon Jun 17 '20 at 21:04
  • integrate with the canonical orientation on that manifold, your answer would be off by -1, so a minus sign would end up coming in. I'll edit my response to address the point you brought up about $n=1.$ – anon Jun 17 '20 at 21:06
  • So what you're suggesting is that, for $n=1$, we keep count of the points of the boundary about which we fail to find an orientation-preserving map and, when integrating, that we put minus signs appropriately? Also I'm sorry but I don't really get what you're saying about canonical orientations. –  Jun 17 '20 at 22:54
  • Let us continue this discussion in chat. – anon Jun 18 '20 at 00:01