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Let $M$ be a smooth $n$-manifold. A pointwise orientation is to specify an orientation of each tangent space $T_pM$. A local frame $(E_i)$ on $U\subset M$ for $TM$ is said to be oriented, if $(E_1|_p,\cdots,E_n|_p)$ is a positively oriented basis for $T_pM$ at any $p\in U$.

By John Lee's Introduction to Smooth Manifolds, a continuous pointwise orientation means that every point $p\in M$ is in the domain of a certain oriented local frame.

My question: Is the oriented local frame above in the definition of continuous pointwise orientation merely continuous?

Here is why I doubt this: In John Lee's proof of Proposition 15.5 (The Orientation Determined by an $n$-Form), one specifies a local frame $(E_i)$ on a connected neighbourhood $U$ of $p$, and $(\mathcal{E}^i)$ be its dual coframe, $f$ a nonvanishing continuous function on $M$. Then he claimed that $\omega:=f\mathcal{E}^1\wedge\cdots\wedge\mathcal{E}^n$ is a nonvanishing $n$-form. But everything here is merely continuous, then why would $\omega$ be a smooth differential $n$-form?

user835469
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1 Answers1

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You are correct in that for the proof of proposition 15.5 one requires a smooth local frame, but any continuous local frame can be made smooth by arbitrarily small (in local coordinates) perturbations without changing the induced pointwise orientation. So it does not matter for the definition of orientation: any continuous orientation is also a smooth orientation.

Edit: More precisely if we choose a coordinate neighbourhood $U$, a local frame $\{E_i\}$ for $U$ is equivalent to a continuous function $f:U \to GL_n(\mathbb{R})$ given by the matrix whose columns are $E_i$ with respect to the basis $\partial /\partial x^j$. If $U$ is connected then $f$ takes its image in one of the two connected components of $GL_n(\mathbb{R})$. By the Whitney approximation theorem (theorem 6.24 in Lee's book) $f$ is homotopic to a smooth map, which must take its image in the same path component of $GL_n(\mathbb{R})$ (else for some $x \in U$ there would be a path between the two path-components given by the homotopy). Therefore a continuous orientation frame yields a smooth one. Alternatively, without loss of generality replace $U$ with a connected pre-compact $V \subset U$ and use theorem 6.21 - then $f$ can be approximated by a smooth map $g:V \to \mathbb{R}^{n^2}$ to arbitrary precision in local coordinates. $GL_n(\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$, and by compactness of $\bar V$ there is an $\epsilon > 0$ such that any approximation $g$ that is closer than $\epsilon$ to $f$ will still have its image within $GL_n(\mathbb{R})$.

It would make life easier if at the outset you assume that the definition of orientation is in terms of smooth local frames.

abhi01nat
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