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Suppose $M$ is an $m$ dimensional real smooth manifold (analytic if necessary) and $\exp_a\colon \mathbb{R}^m\to M$ is the exponential map at the point $a\in M$. Further suppose that $\exp_a$ is actually well defined on $\mathbb{R}^m$ (i.e. $M$ is complete) and is surjective. Then there is an induced topological space $\mathbb{R}^m/\sim$ where $x\sim y$ if and only if $\exp_a(x) = \exp_a(y)$.

Under what conditions does $\mathbb{R}^m/\sim$ have an induced smooth (analytic) manifold structure?

As an example, if $\exp\colon \mathbb{R}\to S^1\subset \mathbb{R}^2$ is given by $\exp(\theta) = \begin{bmatrix}\cos(\theta) \\ \sin(\theta) \end{bmatrix}$; I believe this is the exponential map at the point $(1,0)\in S^1$, however I'm not even entirely sure if it is indeed the exponential map. The quotient space $\mathbb{R}/\sim$ is homeomorphic to $S^1$ as a topological space. Moreover, if $\psi \sim \phi$ then we may additionally identify $v\in T_\psi \mathbb{R}$ and $w\in T_\phi \mathbb{R}$ if $\frac{d\exp}{d\theta}(\psi)(v) = \frac{d\exp}{d\theta}(\phi)(w)$. By this means we may induce a manifold structure on $\mathbb{R}/\sim$, or at least of some notion of a tangent space.

I'm still very new to Riemann Manifolds and exponential maps so I might have said something false and am likely using poor notation. However this seems like a natural thing to extend to other manifolds. Is there a reference that discusses such a quotient manifold as I have constructed, as apposed the Quotient Manifold Theorem which I'm sure is related.

Ted Shifrin
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RyanK
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    Maybe you should think about $S^2$. In the $S^1$ case you have a bona-fide covering map. – Ted Shifrin Apr 09 '23 at 17:59
  • I have thought about $S^2$ and it seems to work, although this is where I may be conflating a couple of ideas. I have obtained the map $\mathbb{R}^2\to S^2$ given by $$(\theta_1,\theta_2)\mapsto \begin{bmatrix} \theta_1\frac{\sin(\sqrt{\theta_1^2 + \theta_2^2})}{\sqrt{\theta_1^2 + \theta_2^2}} \ \theta_2\frac{\sin(\sqrt{\theta_1^2 + \theta_2^2})}{\sqrt{\theta_1^2 + \theta_2^2}} \ \cos(\sqrt{\theta_1^2 + \theta_2^2})\end{bmatrix}.$$ I think a similar idea works even when $\sqrt{\theta_1^2 + \theta_2^2} = \pi$, just the identification of tangent vectors is a bit weird. @TedShifrin – RyanK Apr 09 '23 at 18:07
  • Where do points on circles of radius $k\pi$ in $\Bbb R^2$ map? Maybe you should use polar coordinates on $\Bbb R^2$ instead of two $\theta_i$? – Ted Shifrin Apr 09 '23 at 18:11
  • Yes, think I fixed the typo. If $k$ is odd they map to the south pole, and if $k$ is even they map to the north pole. The polar form is exactly the standard parameterization given at Sphere. However this form is the one I obtained by attempting to construct an exponential map at the north pole. – RyanK Apr 09 '23 at 18:12
  • Right. So you do not have a covering map. Does the quotient space have a decent structure? – Ted Shifrin Apr 09 '23 at 18:19
  • I don't immediately see why not (but again I am very new to this area of study). It feels very similar to thinking of the Torus as a quotient of $\mathbb{R}^2$. Admittedly it is not as simple in this case, i.e when identifying tangent vectors, anything tangent to a circle of radius $k\pi$ is identified with the zero vector. It felt like it would work out though haha... – RyanK Apr 09 '23 at 18:26
  • It would appear that the answer to the question Proving a topological manifold homeomorphic to a smooth manifold is smooth. does in fact provide a method to give a smooth manifold structure to the quotient space in my question. – RyanK Apr 09 '23 at 20:25
  • That is not relevant here. You should look in some standard sources (like Lee's Differentiable Manifolds or Kobayashi-Nomizu's Differential Geometry). When you have a group action of $G$ on $M$, there is a smooth structure on $M/G$ so that the projection is a submersion if and only if ${(p,q): q=g\cdot p \text{ for some }g\in G}\subset M\times M$ is a closed submanifold. In this case, you have the union of the diagonal with various hyperspheres, which is most definitely not. – Ted Shifrin Apr 09 '23 at 21:10
  • Pardon my naivete, but I am not making the connection. It feels like what I am doing is slightly different than constructing a group action and hence I am not working with the quotient $M/G$. In particular, I am not sure which group you are suggesting I am using (either $SO(3)$, $O(3)$ or possibly $\mathbb{R}^2$), and in any case I have fixed a particular point, then looked at a particular surjective map. As far as I am aware, a continuous surjective function of topological spaces induces a quotient space and a quotient topology, which is homeomorphic to the co domain of the function ... – RyanK Apr 10 '23 at 00:08
  • ... Then by the linked post, my quotient space is homeomorphic to $S^1$ so has some smooth structure induced. Again I am sorry to inconvenience you, If what I am doing is blatantly obvious and false, then I can delete the OG question and move on with my studies. @TedShifrin – RyanK Apr 10 '23 at 00:11
  • No need to apologize. I haven’t thought about this sort of thing in a long time. However, saying that being homemorphic to the sphere means there is a diffeomorphism compatible with the smooth structure on $\Bbb R^n$ seems very different. We need someone more expert than I to weigh in. – Ted Shifrin Apr 10 '23 at 00:17
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    At least if $M$ is connected and compact, your quotient space will be homeomorphic to $M$, hence, has a smooth manifold structure. – Moishe Kohan Apr 10 '23 at 01:03
  • The definition of a covering map includes the condition that the map be a local diffeomorphism. So the exponential map for a sphere is not a covering map. In general, if a manifold has too much positive curvature, there are conjugate points for geodesics, which implies that the exponential cannot be a covering map. – Deane Apr 10 '23 at 01:06
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    You might want to clarify what you mean by "induced smooth structure". General equivalence relations do not induce a smooth structure on the quotient, though there are such notions for some "sufficiently nice" relations. – Kajelad Apr 10 '23 at 03:43

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