What are oriented embeddings of zero-dimensional manifolds

differential-topologyorientationsmooth-manifolds

I'm trying to understand what it should mean for a submanifold of an oriented manifold to have a compatible orientation.

For $1$-dimensional manifolds embedding into a $2$-dimensional manifold, I think I can visualise how this works. i.e. an oriented circle embedding into a $2$-sphere just has to follow the orientation of the sphere:

embedding of a circle into a 2-sphere

I'd like to have a similar picture for what an oriented embedding of a $0$-dimensional manifold into a $1$-dimensional manifold would look like. However, for an embedding of points into a circle for example I can't see any obvious way that the polarity of a point should relate to an orientation of a circle:

embedding of 2 points into a circle

This is complicated by the fact that there are various ways to define the orientation of a manifold, and some of them define the orientation of $0$-dimensional manifolds as a special case.

In this MathOverflow post I can see that one way to define orientation in a way that consistently works for $0$-dimensional manifolds is using the reduction of the structure group of the top exterior-power of the tangent bundle. I would guess from this that for an embedding $f : M \to N$ of connected manifolds, there is an induced map
$$\hat f : \mathrm{GL}_1(\mathbb R)/\mathrm{GL}_1^+(\mathbb R) \to \mathrm{GL}_1(\mathbb R)/\mathrm{GL}_1^+(\mathbb R)$$
and we can say $f$ preserves orientation if the induced map
is the identity on the chosen orientation of $M$. In the case that $M$ is not connected, I guess I can just ask that this holds for the induced map on each connected component. So I can try to work out what that means for an embedding of a point:

A map $f : \{*\} \to M$ just picks out a point on the circle, and since $T\{*\} = {\mathbb R}^0 = \{0\}$, the induced map on tangent bundles $Tf : \mathbb R^0 \to T M$ must be constant on $0 \in TM$.
Now the exterior algebra $\Lambda \mathbb R^0$ is trivial past degree $0$, so if $M$ has dimension $m > 0$ then the top exterior power $\Lambda^m(TM)$ must lie in the kernel of $\Lambda(Tf)$. From this I deduce that, if it makes sense to have an induced map between top exterior-powers, then this must be constantly $0$ for $f$. Hence $\hat f$ must constantly choose one orientation.

If the above is correct then I think this tells me that an embedding of a zero-dimensional manifold is orientation preserving only if all points in the domain have the same orientation as the manifold in the codomain. This seems surprising since, in one dimension up, we can embed a disjoint union of circles with opposite orientations into a 2-sphere by "turning one of them upside down".

My question therefore is, whether this is a correct characterisation of oriented embeddings, or if there is a better way to think about how to embed zero-dimensional manifolds into higher dimensions. And if this is the case, is there some geometric intuition for why points with opposite orientations should not embed into the same manifold?

Edit: On reflection I think my characterisation of oriented embeddings must not be correct, because I could draw the same conclusion about any embedding of a manifold into a manifold with strictly greater dimension. At the least, I think the notion I guessed for the induced morphism of reduced structure groups cannot be right. So I'm back to the point where I simply don't know how to define an oriented embedding of $0$-manifolds..

Best Answer

There is no natural — or even obvious — way to orient a submanifold $X$ of a given oriented manifold $M$. You need extra structure. For example, you might have an orientation on in its normal bundle. Or $X$ might arise as the preimage of a regular value of a smooth mapping $f\colon M\to N$, where $N$ is oriented (and, ostensibly more generally, the preimage of an oriented submanifold $Z\subset N$ where $f$ is transverse to $Z$. Or you might have the boundary $\partial M$, in which case there is a canonical "boundary orientation." [This generalizes our familiar experience, as Deane alluded to above, of saying that the boundary of $[a,b]$ consists of $+b$ and $-a$.]

See, for example, Section 2 of Chapter 3 of Guillemin and Pollack or pp. 103-108 and Chapter 5 of Hirsch's Differential Topology.

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