Are orientations on an orientable manifold locally equivalent?

Question

Reading up on orientable manifolds in Milnor, and he defines a manifold to be orientable in the following way:

-Two bases $b=(b_{1},...,b_{m})$ and $c=(c_{1},...,c_{m})$ have the same orientation as long as $B=C\cdot A$, where $B$ and $C$ are the matrices whose columns comprise of the basis vectors of $b$ and $c$, respectively, and $det(A)>0$.

-An orientation for an $m$-manifold $M \subseteq \mathbb{R}^{k}$ assigns for each $T_{x}M$ an equivalence class $C_{x}$ of bases.

-Then, a manifold $M$ is orientable as long as for each $x \in M$, there exists an open set $U$ containing $x$ and a diffeomorphism $g:U \rightarrow \mathbb{R}^{m}$ such that for each $y \in U$, and each $V \in C_{y}$, $Dg(y)\cdot V = D$, where $det(D)>0$.

I have proven that this definition is equivalent to a few others in the literature.

QUESTION: Must it be the case that for all $x \in M$, there exists an open set $\mathcal{O}$ containing $x$ such that for all $y \in \mathcal{O}$, $C_{x}$ = $C_{y}$ ?

It is not proven in the book, but I think I've proven this. Here is a proof sketch:

Step 1: If $Z$ is a vector space, and $I : Z \rightarrow \mathbb{R}^{m}$ is any isomorphism, then two basis matrixes $V$ and $V'$ for $Z$ are of the same orientation as long as $I(V)$ and $I(V')$ are of the same orientation in $\mathbb{R}^{m}$.

Step 2: Let $x \in M$ be given, and let the set $U$ containing $x$ and the diffeomorphism $g:U\rightarrow \mathbb{R}^{m}$ be as described above according to the definition of $M$ being orientable. We first show that there exists a set $\mathcal{O}$ containing $x$ such that for each $V \in C_{x}$, and $y \in \mathcal{O} \cap U$, $V$ is a basis for $T_{y}M$. To this end, since $Dg(x)\cdot V = M$ for some matrix $M$ such that $det(M)>0$ by definition, it follows from the continuity of determinant that for some open set $\mathcal{O}$ containing $x$, $Dg(y)\cdot V = M_{y}$ for some matrix $M_{y}$ such that $det(M_{y})>0$. Let $z \in T_{y}M = Dg^{-1}(g(y))(\mathbb{R}^{m})$ be given. Then for some $z' \in \mathbb{R}^{m}$, $z=Dg^{-1}(g(y)) \cdot z'$. But since $M_{y}$ is a basis for $\mathbb{R}^{m}$, then $z'=M_{y} \cdot z''=Dg(y)\cdot V \cdot z''$ for some $z'' \in \mathbb{R}^{m}$. Putting this all together, we have that

$$ z=Dg^{-1}(g(y)) \cdot z' = Dg^{-1}(g(y))Dg(y)\cdot V \cdot z''=V \cdot z''$$

so that $V$ spans $T_{y}M$, and hence is a basis for $T_{y}M$.

Step 3. Since each $Dg(y) :T_{y}M \rightarrow \mathbb{R}^{m}$ is an isomorphism, we have by Step 1 and 2 above that for each $y \in \mathcal{O} \cap U$, since $Dg(y) \cdot V$ has positive determinant for the $V \in C_{x}$ described above, and by definition $Dg(y) \cdot V'$ has positive determinant for each $V' \in C_{y}$, it follows that $V \in C_{x}$ is equivalent to each $V' \in C_{y}$. It then follows that $C_{x}=C_{y}$.

Does this look like it holds? Thanks in advance!

Answer 1

I'm sorry, but your claim doesn't even make sense. $C_x$ is an equivalence class of bases for the vector space $T_xM$. $C_y$ is an equivalence class of bases for the vector space $T_yM$. These are two distinct vector spaces. They share nothing in common. Even their $0$s are distinct elements. So $C_x$ and $C_y$ cannot be equal, because they are built from different building blocks.

You go wrong here:

We first show that there exists a set $\mathcal O$ containing $x$ such that for each $V \in C_x$, and $y \in \mathcal O \cap U$, $V$ is a basis for $T_yM$.

$V$ cannot be a basis for $T_yM$ because the vectors in $V$ are elements of $T_xM$, not $T_yM$. In your argument, you simply assume that $V$ consists of vectors of $T_yM$ for every $y$ in $\mathcal O$, but that is false.

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(Edit - added a more general definition of manifolds and tangent spaces that is not restricted to subsets of $\Bbb R^n$).

A common modern definition of a finite-dimensional manifold and its tangent spaces is this. Note that there are many variations, but the variations are either equivalent, or allow somewhat looser (or sometimes tighter) topologies and/or degrees of smoothness.

A differential manifold is a triple $(M, \tau, \mathcal A)$ where

• $M$ is a set,
• $\tau$ is a second-countable locally compact Hausdorff topology on $M$, and
• $\mathcal A$ is an atlas on $(M, \tau)$. That is, for some fixed integer $m > 0, \mathcal A$ is a maximal collection of continuous open injective maps $\phi : U \to \Bbb R^m$ for non-empty open sets $U \in \tau$ whose domains $U$ cover $M$, and such that for all $\phi, \psi \in \mathcal A, \phi\circ \psi^{-1}$ is smooth on $\psi(\operatorname{dom}(\phi) \cap \operatorname{dom}(\psi))$

Note that $\phi\circ \psi^{-1}$ is a map between two open sets of $\Bbb R^m$. "Smooth" usually is taken to mean $C^\infty$ i.e., being infinitely differentiable. The maps in $\mathcal A$ are called charts.

Functions $f: U \to \Bbb R, U\in \tau$ are then defined as smooth if for each $x \in U$, there is a chart $\phi$ with $x \in \operatorname{dom}(\phi)$ such that $f\circ \phi^{-1}$ is smooth. The collection of all smooth maps on $U$ is denoted by $\mathfrak X(U)$ (most commonly, $U = M$). Note that if $f, g \in \mathfrak X(U)$ and $a, b \in \Bbb R$, then $af + bg \in \mathfrak X(U)$. So $\mathfrak X(U)$ is a real vector space.

Next, for a point $x \in M$, consider the set $S_x = \{ f \in \mathfrak X(U) \mid x \in U \in \tau\}$ and for $f,g \in S_x$, define $f \sim g$ if $f = g$ on $\operatorname{dom}(f) \cap \operatorname{dom}(g)$. $\sim$ is an equivalence relation on $S_x$ which defines a set of equivalence classes $\tilde S_x = S_x / \sim$. If $f \sim f', g \sim g'$, then $af + bg \sim af' + bg'$ for all $a, b \in \Bbb R$, and $fg \sim f'g'$. This makes $\tilde S_x$ an algebra (a vector space with a multiplication that distributes over the addition). Note that if $f \sim g$, then we can be sure that $f(x) = g(x)$, but for any other point $y \in M$, even if $y$ is in the domains of both $f$ and $g$, there is no guarantee that $f(y)$ will be the same as $g(y)$. Similarly, any limiting behavior of $f$ and $g$ at $x$ must be the same, since they agree on some neighborhood of $x$. This includes any differential operators applied to $f\circ \phi^{-1}$ and $g\circ \phi^{-1}$ for some chart $\phi$ about $x$.

A linear operator $D : \tilde S_x \to \tilde S_x$ can be pulled back to $D : S_x \to \tilde S_x$ by defining $Df = D[f]$, where $[f]$ is equivalence class of $f$. Note that as an element of $\tilde S_x, Df$ has a well-defined real numeric value at $x$. That value is denoted by $Df|_x$. Such a linear operator is called Leibnitzian if it also satisfies $$\forall f, g \in S_x, Dfg = g(x)Df + f(x)Dg$$.

The tangent space to $M$ at $x$ is defined to be the collection $T_xM$ of all Leibnitzian operators on $\tilde S_x$. If $D_1, D_2$ are two such Leibnitzian operators, then it is easy to check that for all $a,b \in \Bbb R$, so is $aD_1 + bD_2$, making $T_xM$ a vector space. One can use charts to show that $T_xM$ has the same dimension as $M$.

In fact, you can show that if $\phi$ is a chart about $x$ and $D \in T_xM$, then there is some vector $v \in \Bbb R^m$ such that for all $f \in S_x$, $$D_v(f\circ \phi^{-1})|_{\phi^{-1}(x)} = Df|_x$$ where $D_v$ is the directional derivative with respect to $v$. By this, we can just consider $D$ as representing a tangent vector (a direction & magnitude) in $M$ at $x$, and the action of $D$ on a smooth function $f$ to be the directional derivative of $f$ with respect to the tangent vector.

One other comment. If you are not already aware of it, you should picture $T_xM$ as being a vector space tangent to $M$ at $x$. I normally picture a manifold as an undulating surface. At each point $x$ on that surface, $T_xM$ will be the plane tangent to the surface at $x$, treated as a vector space whose origin is at the point $x$. Tangent planes at two different points will be different. They likely will intersect at some line, but since the points in each represent vectors in different vector spaces, these planes should not be considered to actually intersect. Rather, the intersection is merely an artifact of our forcing the depiction into $\Bbb R^3$. Like in depictions of the Klein bottle, the actual planes "slide past" each other without intersecting.

Answer 2

Milnor's "Topology from the Differentiable Viewpoint" is still a valuable reading. His approach is limited to submanifolds $M \subset \mathbb R^k$ and does not cover the abstract concept of a manifold. Nevertheless you can learn a lot from the book.

For submanifolds $M \subset \mathbb R^k$ Milnor defines the tangent space at $x \in M$ as a suitable linear subspace $T_x M \subset \mathbb R^k$. Although it may by chance be the case that $T_x M = T_y M$ for $x \ne y$, we have in general $T_x M \ne T_y M$ for $x \ne y$. As an example take $M = S^2 \subset \mathbb R^3$. The tangent space $T_x S^2$ at $x \in S^2$ is the two-dimensional orthogonal complement of the vector $x$, and the affine subspace $x + T_x S^2$ is the tangent plane of $S^2$ at $x$. We have $T_x S^2 = T_y S^2$ iff $x = \pm y$, but in a small neigborhood $U$ of $x$ we definitely have $T_x S^2 \ne T_y S^2$ for $y \in U \setminus \{x\}$.

This shows that your question is not well-posed. In general you have to compare distinct vector spaces $T_x M$ and $T_y M$ so that $C_x$ and $C_y$ can never be the same. You need a specific linear isomorphism $T_x M \to T_y M$ to relate $C_x$ and $C_y$ and this is what Milnor does via the diffeomorphism $g$.

The only special case where your question makes sense is when $x$ has a neighborhood $W \subset M$ which is contained in an $m$-dimensional affine subspace $P \subset \mathbb R^k$. In that case all tangent spaces of points in $W$ agree. In fact, we have $P = x + E$ for some $m$-dimensional linear subspace of $\mathbb R^k$, thus $T_x M = E$ for all $x \in W$.

In this special case the answer to your question is "yes".