I am reading Spivak's Calculus on Manifolds, and just need a bit of clarification when it comes to definitions.

He defines a manifold as some space $M$ that satisfies:

$(M)$: $\forall x \in M, \exists U,V \subset \mathbb{R}^n$, (both open), with $x\in U$ and a diffeomorphism $h: U\rightarrow V$ such that $h(U \cap V)=V\cap (\mathbb{R}^k \times \{0\})=\{y\in V | y^{k+1}=\dots=y^n=0 \}$

He then defines a manifold with boundary with the similar definition:

Let $\mathbb{H}^k=\{x\in \mathbb{R}^k|x^k\geq 0\}$

$(M')$: $\forall x \in M, \exists U,V \subset \mathbb{R}^n$, (both open), with $x\in U$ and a diffeomorphism $h: U\rightarrow V$ such that $h(U\cap M)=V\cap (\mathbb{H}^k \times \{0\})=\{y\in V | y^k \geq 0, y^{k+1}=\dots=y^n=0\}$

He then goes on to say that "It is important to note that conditions $(M)$ and $(M')$ cannot both hold for the same $x$"...However, isn't $\mathbb{H}^k\subset\mathbb{R}^k$, so of course if $(M')$ holds, wouldn't $(M)$ hold as well? Also why does this boundary have to have a positive $k^{th}$ component?

Is this $\partial M$ the intersection between $M$ and $\{y\in \mathbb{H}^k | y^k=0\}\subset \mathbb{H}^k$? If we consider a finite line in $\mathbb{R}^2$ as a one dimensional manifold, then the ends would surely be its boundary...however I see no reason why $x_2$ (aka $y$) has to have zero values.

  • $\begingroup$ It's possible you have an older edition that has an error, but looking at my copy it seems like you need to read around this paragraph more carefully. In $M'$ the point $x$ has to have $k$-th coordinate equal to $0$, so transporting to $\mathbb R^k$ it's on the boundary of the half-space. $\endgroup$ – Hoot Jun 28 '16 at 19:06
  • $\begingroup$ @Hoot $x$ has to have $k$-th coordinate $\geq 0$, I'm not sure where you're getting the equality to satisfy $(M')$. I'm not sure why boundary points require $x_k=0$, which is why I added the bit about one dimensional manifolds. $\endgroup$ – malxmusician212 Jun 28 '16 at 19:10
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    $\begingroup$ imgur.com/HkwDDfe The last line. $\endgroup$ – Hoot Jun 28 '16 at 19:13
  • $\begingroup$ ah so for this example of a two dimensional manifold with boundary (which he illustrates in figure 5.3), all he is saying is that there is some diffeomorphism $h$ that sends the open face on the torus to $x_2=y=0\in \mathbb{R}^2$? Couldn't we just use that diffeomorphism and translate it to make any point on our manifold evaluate to 0? See I'm still confused because $\partial M$ is still a manifold, so every point on $\partial M$ must satisfy $(M)$... $\endgroup$ – malxmusician212 Jun 28 '16 at 19:22

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