# Does $S^1$ has boundary?

According to Guillemin and Pollack's Differential Topology:

The sum of the orientation numbers at the boundary points of any compact oriented one-manifold $X$ with boundary is zero.

By The Classification of One-manifold, every compact, connected, one-dimensional manifold with boundary is diffeomorphic to $[0,1]$ or $S^1$.

I think oriented manifold means a manifold with an orientation. By definition, an orientation of a manifold with boundary is a smooth choice of orientations for the tangent space. (By the way, does "smooth choice of orientation" mean the orientation varies smoothly? Then since orientation is either $0$ or $1$, it means orientation does not change?)

So I assume this means $X$ is connected? Hence, we can apply the classification, and $X$ must be diffeomorphic to $[0,1]$ or $S^1$. So for the $[0,1]$ case, its two ends just cancel out regardless the orientation of $X$; for $S^1$, the boundary is trivial - non exist, correct?

It is really frustrating that I have to work out pages of details for many of the one sentence claims made in this book. But sincere thanks to MathSEers, who essentially are teaching me this subject - without you I could have gave up at Page 10, at most.

• You are correct $S^1$ does not have any boundary. You are also correct that orientation is a continuous choice (vaguely) of basis of each tangent space. To define orientation we assume that the space is connected. By the way there is a (intrinsic) lopological definition of orientation which does not required smoothness (hence topological.) P.S: I love this book, and the way it was written is excellent (at least to me.) – tessellation Jul 19 '13 at 5:30
• By the way there is a general fact. Boundary of a boundary is empty (try to show this.). I think its also there in that book. – tessellation Jul 19 '13 at 5:33
• Oh I love the book too @tessellation, that's why I wouldn't mind the toil going through it. Thank a lot for your help and confirmation. That clears up the cloud in my mind a lot! – 1LiterTears Jul 19 '13 at 5:36
• I would say that the circle has a boundary, namely an empty boundary. – Baby Dragon Jul 19 '13 at 23:11

Orientation has several possible, equivalent, definitions. One for smooth manifolds is a choice of ordered basis of the tangent space for positive dimension, while a point gets a $\pm 1.$
For $\mathbb S^1$ a novanishing tangent vector field along the circle supplies an orientation. For the closed unit interval, an arrow along the segment orients the open segment, then the enpoint at which those arrows point gets orientation $+1,$ the other endpoint gets $-1.$