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So I am studying boundary orientation (e.g. Gulliemin and Pollack page 97, Mukherjee page 182) and it is stated that in the first introduction if

$\dim M \geq 1$ is a smooth manifold with boundary and since $codim \partial M= 1$ at each $x \in \partial M$ there exists two orthogonal vectors in the tangent space $T_x(M)$ to $T_x(\partial M)$ (one inward and one outward). I am 100% sure this is just some trivial linear algebra.

Because if the leftover space has cxdimension $1$, then it is a linear subspace. But how do we know the left over would be orthogonal?

Arthur
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Lemon
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  • What means orthogonal? What is the metric on the tangent space T_x M? – Federico Fallucca Apr 30 '19 at 08:23
  • @FedericoFallucca Actually all of these differential topology books, not once was a metric (inner product) is mentioned, but if it helps the normal space (bundle) is defined as the orthogonal complement. So this must mean a bilinear form is assumed which we can take as the metric. – Lemon Apr 30 '19 at 08:28
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    Guillemin and Pollack work in $\Bbb R^n$ from first page to last, and they use the dot product structure of $\Bbb R^n$ throughout. The "trivial linear algebra" is, of course, the fact that the orthogonal complement of a hyperplane is a line. BTW, in your title, that should be two unit vectors .... – Ted Shifrin Apr 30 '19 at 18:36
  • Yeah it just occurred to me this was what I needed. – Lemon May 01 '19 at 01:24

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Let $X^n$ be a smooth manifold with boundary and let $x\in \partial M$. Then let $\phi:U\to X$ be a local parametrization with $\phi(0)=x$ where $U\subset H^n$ is a neighborhood of $0$ in upper half-space (vectors in $\mathbb{R}^n$ with positive $n-$th component). Then, under G&P's definition of tangent space, $T_x(X)$ is the image of $d\phi_0$. Further, if one knows that $d\phi_0(H^n)\subset T_x(M)$ does not depend on $\phi$, then we can define $H_x(X)=d\phi_0(H^n)$.

We want to show that there are only two unit vectors perpendiuclar to $T_x(\partial X)$ and that one points into $H_x(X)$ and the other away from it. First, observe that $T_x(X)=T_x(\partial X)\oplus T_x(\partial X)^{\perp}$. So it is clear that, since $T_x(X)$ has dimension $n$ and $T_x(\partial X)$ has dimension $n-1$, $T_x(\partial X)^{\perp}$ has dimension 1 and therefore only has 2 unit vectors which both span, say $\hat{v}$ and $-\hat{v}$. But $d\phi_0^{-1}$ takes $T_x(\partial X)$ to $\mathbb{R}^{n-1}\times\lbrace 0\rbrace$. So it follows that $d\phi_0^{-1}$ takes $T_x(\partial X)^\perp$ to some one dimensional subspace $W$ of $\mathbb{R}^n$ so that $W\oplus (\mathbb{R}^{n-1}\times\lbrace 0\rbrace)=\mathbb{R}^n$. It follows that $d\phi_0^{-1}(\hat{v})$ is a vector that spans $W$ and hence has some $n-$th component. Thus $\hat{v}$ is the image of some vector in $H^n$ and hence $\hat{v}\in H_x(X)$, as desired.

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J. Moeller
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    Yeah when I thought about it, I came to a similar conclusion, but not all written out like this. – Lemon May 01 '19 at 01:14