Exercise 1.15.2 of Burns and Gidea's differential geometry/topology states that:

Exercise 1.15.2: Consider a bijection between the real line $\Bbb R$ and the sphere $\Bbb S^2$ (such a bijection exists since these are sets with same cardinality). Show that the composition of the local parametrizations of $\Bbb S^2$ from above with this bijection defines a smooth structure on $\Bbb R$. Show that $\Bbb R$ endowed with this smooth structure is diffeomorphic to the sphere $\Bbb S^2$. With this smooth structure, the real line is a sphere! The point of this exercise is to stress that a manifold is not just a set that can be endowed with some structure, but the set together with that structure.

If so what is the role of invariance of dimensions? It seems that this exercise is a serious mistake by authors!!

In the page 67, exercise 1.15.5 claims that

Exercise 1.15.5: Provide the unit cube $Q\subset \Bbb R^{n+1}$ with a smooth structure. The point of this exercise it to illustrate that a smooth manifold may not look smooth! Of course this smooth structure is not compatible with the smooth structure of $\Bbb R^{n+1}$.

Is the claimed statement correct? I have no idea about $n>2$ but in $n=1,2$ I think it is wrong by unique differential structure in dim$<4$!

  • $\begingroup$ They are not even homeorphic. What am I missing here? Dont we need at least a homeo to construct a diffeo? $\endgroup$ – Garmekain Oct 23 '20 at 7:45
  • $\begingroup$ I think like you. $\endgroup$ – C.F.G Oct 23 '20 at 7:49
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    $\begingroup$ $\mathbb{S}^2$ and $\mathbb{R}$ aren't homeomorphic if they both have their usual topologies, but surely this construction induces a non-standard topology on $\mathbb{R}$? $\endgroup$ – Hans Lundmark Oct 23 '20 at 8:10
  • $\begingroup$ Invariance of domain says no such bijection can be a homeomorphism. The construction given here equips $\mathbb{R}$ with a completely different topology. The exercise is correct, the point is to force you to think about this sort of thing. $\endgroup$ – Qiaochu Yuan Oct 23 '20 at 8:33
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    $\begingroup$ The point is that dimension is not a property of the set, but of the topology. The underlying set is $\mathbb{R}$, but this set is endowed with a 2-dimensional topology. $\endgroup$ – Kajelad Oct 23 '20 at 9:13

The point is that you endow the set $\mathbb R$ with a topology (and a smooth structure) which has nothing to do with its usual topology (and smooth structure).

In fact, the general construction is this:

Let $M$ be a smooth manifold, $X$ be set and $h : X \to M$ be a bijection. Then there exists a unique topology $\tau$ on $X$ such that $h$ becomes a homeomorphism. Moreover, there exists a unique smooth structure on $(X,\tau)$ such that $h$ becomes a diffeomorphism.

If you reflect upon this, there is no surprise in it. What is confusing you is this: We start with a smooth manifold $N$ and take a bijection $h : N \to M$. This map is not subject to any restrictions, it may not even be continuous. But it induces a smooth structure on the set $N$ such that $h$ becomes a diffeomorphism. But, as we have seen above, this smooth structure is not related to the original smooth structure of $N$.

  • $\begingroup$ So $\Bbb S^1\simeq \Bbb S^2\simeq \Bbb S^3\simeq \dots$ with appropriate topologies? $\endgroup$ – C.F.G Oct 23 '20 at 10:25
  • $\begingroup$ @C.F.G Yes. Any two sets with the same cardinality can be endowed with topologies making the resulting toplogical spaces homeomorphic. $\endgroup$ – Paul Frost Oct 23 '20 at 10:28
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    $\begingroup$ The confusing thing, is to use the "wrong" notation. If you endow the underlying set of $\mathbb{S}^2$ (or any set with the same cardinality of the real numbers for that matter) with a topology (or a smooth structure) making it isomorphic to $\mathbb{S}^1$, you should name it $\mathbb{S}^1$, not $\mathbb{S}^2$. After all, from the pure set theoritical point of view, $\mathbb{R}$, $\mathbb{S}^n$ etc... are all the "same" set. When you write $\mathbb{S}^n$, a lot of implicit assumptions are made about the topology, smooth structure... $\endgroup$ – Ahr Oct 23 '20 at 10:39
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    $\begingroup$ Yes, it is the same. The authors only want to show that something which "obviously" seems to have corners may nevertheless be endowed with a smooth structure. $\endgroup$ – Paul Frost Oct 23 '20 at 11:44
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    $\begingroup$ @C.F.G You are not stupid. Quote from Shakespeare's "Hamlet": There are more things in Heaven and Earth, Horatio, than are dreamt of in your philosophy. $\endgroup$ – Paul Frost Oct 23 '20 at 12:26

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