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I am trying to work through the following problem: Let $M$ be the cylinder $S^1 \times [0,1]$ with counterclockwise orientation when viewed from the exterior. I am trying to figure out the boundary orientation on the $C_0 = S^1 \times \{0\}$ and on $C_1 = S^1 \times \{1\}$.

I am not sure how to find an orientation form on the cylinder, as it seems to be different from the other manifolds I have worked with. Also, what would be the outward pointing vectors on the boundary?

This is the image associated with the problem:

enter image description here

Tex
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3 Answers3

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  • Take your cylinder $S^1 \times[0,1]$ and draw a small disk on the barrel part.

  • Next, place a curved arrow on it to represent the counter clockwise direction.

  • Now just simply slide your disk to the boundary curves and you'll see the induced boundary. They will be opposite.

The top should be clockwise with your right index finder and the bottom is counter-clockwise with your left index finger.

Faraad Armwood
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  • The book I am using says the opposite, that the top is clockwise and the bottom is counterclockwise. – Tex Nov 16 '17 at 22:02
  • Also, is there a way to do this in a more "rigorous" fashion by using differential forms, without relying as much on the geometry? – Tex Nov 16 '17 at 22:04
  • I don't think they are right and ok, so what do you want? It's not very instructive to do this kind of thing with differential forms every single time. If you can visualize it, why not use that to help? Also, I know differential forms. – Faraad Armwood Nov 16 '17 at 22:05
  • Thanks, I really appreciate your helping me better understand this subject. – Tex Nov 16 '17 at 22:19
  • No problem. I'll actually have to let this wait until later. I have to leave out. – Faraad Armwood Nov 16 '17 at 22:20
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This same exercise is 22.8 in the second edition of Loring Tu's book An Introduction to Manifolds. Here is my solution:

In order to describe the induced orientation on the boundary, we first need an outward-pointing vector field on the boundary. Consider the component $C_1 = S^1\times \{1\}$ at the top of the cylinder. If $t$ is the coordinate for $[0,1]$, then an outward-pointing vector field along $C_1$ is the vector field $\partial/\partial t$. Now, we take the interior product of $\partial/\partial t$ with the $2$-form $d\theta\wedge dt$: \begin{align} \iota_{\partial/\partial t}(d\theta\wedge dt) &= -\iota_{\partial/\partial t}(dt\wedge d\theta) \\ &= -d\theta. \end{align} The corresponding orientation for this orientation $1$-form on $C_1\cong S^1$ is the reverse of the usual orientation. Hence $C_1$ is oriented clockwise when viewed from the exterior as in the figure. Similarly, to figure out the orientation of the bottom of the cylinder, we take the interior product of the outward-pointing vector field $-\partial/\partial t$ on $C_0=S^1\times\{0\}$ with $d\theta\wedge dt$, to get \begin{align} \iota_{-\partial/\partial t}(d\theta\wedge dt) &= -\iota_{\partial/\partial t}(d\theta\wedge dt) \\ &= \iota_{\partial/\partial t}(dt\wedge d\theta)\\ &= d\theta. \end{align} Thus the orientation on $C_0$ is the usual orientation of $S^1$, so it is counterclockwise when viewed from the exterior as in the figure.

Alex Ortiz
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You can use either vectors or differential forms to work out orientations, but for a surface in $3$-space, using vectors is easier.

First, you only need to figure it out at one point on each connected component of the surface or its boundary.

The man-made rule is the following: If $n$ is the outward normal of a surface, $(t_1,t_2)$ is a positively oriented oriented basis of the surface if $(n,t_1,t_2)$ is positively oriented on $\mathbb{R}^3$, i.e., has the standard orientation. At the point $(R,0,0)$ on the cylinder, the outward unit normal is $e_1$ and therefore $(e_2,e_3)$ has positive orientation at $(R,0,0)$ on the cylinder. In general, $(x,y,0)$ is an outward normal at each $(x,y,z)$ on the cylinder.

Now if $n$ is a vector on the boundary of an oriented surface $S$ that is tangent to $S$ but normal to the boundary (which is a curve) and $t$ is a vector tangent to $S$ such that $(n,t)$ is positively oriented on $S$, then the man-made convention says that $t$ is positively oriented with respect to the induced boundary orientation.

At $(R,0,h) \in C_1$, then you can see that $e_3$ is tangent to the cylinder that is an outward normal of $C_1$. Since $(e_3,-e_2)$ has the same orientation as $(e_2,e_3)$ on the cylinder, $-e_2$ has positive orientation at $(R,0,h) \in C_1$. Looking from above, this is clockwise around $C_0$. On the other hand, at $(R,0,0) \in C_0$, the outward normal is $-e_3$ and therefore $e_2$ is positively oriented. Therefore, $e_2$ is positively oriented at $(R,0,0) \in C_0$. Looking from above, this is counterclockwise around $C_1$.

Deane
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