Can you distinguish topological properties of (sub)spaces by embedding them in an ambient space?

Question

I'm curious, but I can't seem to reason through it myself.

For instance, is it logical to deduce things like "[0,1] is not homeomorphic to (0,1)" by considering both as subspaces of $\mathbb{R}$ with the usual topology, where we have tools like the Heine-Borel theorem? Can this be done in general with other spaces and topological invariants?

Thanks,

Edit: For what it's worth, I'm not concerned about the particular example (but thank you for your answers anyway). I just chose something familiar and simple to illustrate the point. I just wanted to know if it's possible (or useful) to determine topological properties this way.

Depends on what you're trying to do with them. In this particular case, the Heine-Borel theorem would tell you that $[0, 1]$ and $(0, 1)$ are and are not compact, respectively, but that's nothing new; the proof of the theorem boils down to showing that $[0, 1]$ is compact. — anomaly, Jul 21 '17 at 18:55

score 5 · Answer 1 · answered Jul 21 '17 at 19:14

You don't need Heine-Borel to see that $[0,1]$ is compact (it's an ordered space that is complete and has a minimum and a maximum), and $(0,1)$ is not compact (it's ordered and has no maximum). So it pays more to look at the internal structure, instead of how they are embedded into a larger space.

It's hard to deduce non-homeomorphism from embeddings in general. My hypothesis is that if we could prove it with an embedding argument, then there also is a good internal reason for it.

score 3 · Answer 2 · answered Jul 21 '17 at 18:45

It depends on the topology of the original spaces. If the intervals $[0,1]$ and $(0,1)$ are equipped with the subspace topology from $\mathbb{R}$, then embedding them in $\mathbb{R}$ seems perfectly natural and right. If they have some other topology, then not so much. For example, if both spaces are equipped with the discrete topology, then they are homeomorphic.

Assuming they have the subspace topology from $R$, I would just point out that you can remove a point from $[0,1]$ without disconnecting it, which is not true for $(0,1)$. That seems easier than using some big theorem.

Elle Najt · Accepted Answer · 2017-07-22T02:04:57.687

You can certainly use the Heine-Borel theorem to show that $[0,1]$ is compact, which distinguishes it topologically from $(0,1)$.

You can use embeddings to compute various invariants. For example, let $i : S^n \to \mathbb{R}^{n+1}$ be the usual embedding of the unit $n$ sphere. Then we get a decomposition of vector bundles $i^*(T(\mathbb{R}^{n+1}) = T(S^n) \oplus N$, where $N$ is the normal bundle.

There is a map $w : Vect(X) \to H^*(X ; \mathbb{Z}/2\mathbb{Z})$ that sends an isomoprhism class of real vector bundle to a certain cohomology class called the Steifel Whitney class. Among other properties, it satisfies $w(E \oplus E') = w(E) \cup w(E')$ an for $F$ a trivial bundle, $w(F) = 1$.

Thus, from the above formula $i^*(T(\mathbb{R}^{n+1}) = T(S^n) \oplus N$, where we note that $i^*(T(\mathbb{R}^{n+1})$ an $N$ are trivial bundles, we get that $w(T(S^n)) = 1$.

(Note: This does not imply that $T(S^n)$ is trivial. Indeed, it generally is not, see the examples here: https://en.wikipedia.org/wiki/Parallelizable_manifold )

This presumably lets you distinguish the sphere from something. I'm not sure what though.

Using the same argument, we can show that we cannot embed $\mathbb{RP}^n$ into $\mathbb{R}^{n + 1}$, for certain $n$. Let take $n = 2$ for simplicity, to try to get the result that we cannot embed the real projective plane into 3-space.

If we could, then we would get that $w(T(\mathbb{RP}^2)) = 1$, by the same argument as for the sphere.

Major Edit: We need a lemma to ensure that the normal bundle to a closed codimension one manifold $M$ in $\mathbb{R}^n$ is trivial. If $M$ is the zero set of a smooth function $f$, which has non-vanishing gradient along $M$, the normla bundle is trivial, for then the gradient of $f$ provides a nowhere vanishing normal vector field. This is how we know that the sphere's normal bundle is trivial, for instance.

Now, it is proven here: Smooth surfaces that isn't the zero-set of $f(x,y,z)$ that every smooth codimension 1 submanifold $S$ of Euclidean space is cut out by an equation whose gradient is nonzero along that submanifold.

End Major Edit.

However, we can compute $w(T(\mathbb{RP}^2))$ by separate means, and see that it is not $1$. In fact, it suffices to know that for any bundle $E$, the first Steifel Whitney class of $E$ (which is the degree 1 component of $w(E)$) is zero iff $E$ is orientable. Since $T(\mathbb{RP}^2)$ is non-orientable, $w(T(\mathbb{RP}^2) \not = 1$.

The same style of argument would presumably show that any non-orientable closed $n$ manifold cannot be embedded in $\mathbb{R}^{n + 1}$, modulo that lemma.

https://en.wikipedia.org/wiki/Stiefel%E2%80%93Whitney_class#Axiomatic_definition

You can presumably run the above proof without mentioning Stiefel-Whitney classes once; you can find direct proofs on this webpage that closed hypersurfaces of Eucliean space are orientable, so this gives an application of an embedding to proving an intrinsic statement.

A bit of philosophy: Embeddings of a compact manifold $M$ into $\mathbb{R}^n$ have tubular neighborhoods, and these tubular neighborhoods are like vector bundles on $M$. So, in some sense which I do not know enough about to make precise, studying embeddings of $M$ and vector bundles on $M$ are closely related. The study of vector bundles on $M$ contains a lot of information about $M$. Maybe someone more knowlegable can say something about this?

@MathIsArt I think there is an error or gap in my answer, because I don't see why the normal bundle to the hypothesized embedding would be to trivial. Will update later. — Elle Najt, Jul 21 '17 at 22:54
@MathIsArt Okay, updated. I think it's correct now, but I also realized the whole thing about Steifel-Whitney classes is probably overkill just for questions of orientability. Maybe a more succinct answer would be to point out that if you can embed your closed manifold as a hypersurface, then it is necessarily orientable. This might be easy to see from an embedding, but hard to see from an abstract definition of the manifold. — Elle Najt, Jul 22 '17 at 02:22

Can you distinguish topological properties of (sub)spaces by embedding them in an ambient space?

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