How to orient a manifold in the Euclidean space?

Question

I learned that the orientation of a smooth manifold is a smooth choice of an orientation for bases of tangent space.

Also I sometimes read that an embedded manifold in $\mathbb{R^3}$ inherites an orientation from the standard orientatio from $\mathbb{R^3}$.

I understand what the standard orientation for $\mathbb{R^3}$. But I am not sure how to induce the orientation for an embedded manifold in $\mathbb{R^3}$.

For example, I would like to know how to orient an embedded solid torus in $\mathbb{R^3}$.

In the interior of the solid torus, the tangent space of the torus and $\mathbb{R^3}$ agree and it seems OK for me. But the orientation for boundary is kind of counterintuitive. (Well, even though tangent spaces are the same on the boundary?)

I think I need some local coordinate from the upper half space $H^3$ to a neighborhood of the boundary of the torus. But I don't see why this "agrees" with the induced orientation from $\mathbb{R^3}$.

I am sorry for a vague question. I hope someone can read my confusion well and give a clear explanation of the inherit orientation.

Answer 1

If we allow the manifold to have a boundary, an embedded manifold in $\mathbb{R}^3$ may be nonorientable, for example the Mobius strip. Closed manifolds like the boundary of the torus also need not inherit an orientation as we can see from the fact that we can embed nonorientable closed manifolds in 4 dimensions. However, an open set in $\mathbb{R}^n$ inherits an orientation because its tangent space can be considered a subspace of the tangent space of the surrounding space, as you have noted.

Answer 2

An embedded submanifold in $\mathbb{R}^3$ of dimension 3 inherits an orientation from the standard orientation of $\mathbb{R}^3$. This is not generally true for 2-dimensional submanifolds, for example the Mobius band.

However, if $M \subset \mathbb{R}^3$ is a compact 3-dimensional manifold with boundary, and if $\partial M$ denotes its 2-dimensional boundary, then $\partial M$ does indeed inherit an orientation. Namely, at each $x \in \partial M$ choose an outward pointing normal vector $\vec n$, and then given a basis $\vec v,\vec w$ for $T_x \partial M$ say that $\vec v,\vec w$ are positively oriented in $T_x \partial M$ if and only if $\vec v,\vec w,\vec n$ are positive oriented in $T_x M$.

Answer 3

As explained in @Lee Mosher answer, a manifold with boundary is orientable if and only if its interior is so, hence we can restrict ourselves to boundaryless manifolds. And as @Matt Samuel says, any manifold can be embedded in some $\mathbb R^n$, in particular all nonorientable are and they cannot inherit any orientation. Thus, we need some condition to produce an induced orientation on a boundaryless manifold $M\subset\mathbb R^n$. A well known one is to be a local complete intersection, namely, that there is a submersion $f:W\to\mathbb R^k$ defined on some open nbhd $W$ of $M$ such that $M=f^{-1}(0)$ (and $\dim(M)=n-k$); in other words, the normal bundle is trivial (generated by the gradients $\nabla f_j$). Now with such an $f$, we have the inverse image orientation: for $p\in M$ write $\mathbb R^n=T_pM\oplus E_p$ where $E_p\subset\mathbb R^n$ is a linear subspace such that $d_pf|:E_p\to\mathbb R^k$ is an isomorphism. Via this restriction we transport the standard orientation in $\mathbb R^k$ to an orientation $\xi_p$ in $E_p$, and choose an orientation $\zeta_p$ in $T_pM$ for $\zeta_p\oplus\xi_p$ to be the standard orientation in $\mathbb R^n=T_pM\oplus E_p$. This is an example of orientation of inverses images by transversal mappings: the $f$ above is transversal to the point $\{0\}$. In any case smoothness of the asignment $p\mapsto\zeta_p$ follows once seen the construction does not depend on the choices involved. This is a very instructive Linear Algebra exercise (although rarely seen in texts...).