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In reference to Showing a subset of the torus is dense, the responders helped show the poster that the image set $f(\mathbb{R})$ is dense in the torus. But, it's not immediately clear to me why the image set is not an embedded submanifold. If $f(\mathbb{R})$ is an embedded submanifold, we must have that it's

  1. a smooth manifold in the subspace topology and

  2. that the inclusion map from $f(\mathbb{R})$ to $T^2$ is a smooth embedding.

It's visually clear to me that under the subspace topology, $f(\mathbb{R})$ is not locally Euclidean (as it's not locally path connected). But, I can't seem to formalize this or any argument, using that $f(\mathbb{R})$ is dense in $T^2$, which says that $f(\mathbb{R})$ fails 1) or 2). Thanks for any help!

Community
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Winston
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1 Answers1

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For convenience, let's call $f(\mathbb{R}) = A$. It's clear that $A$ is not locally path connected. As you point out, if $A$ were an embedded submanifold, it would be a smooth manifold in the subspace topology. If it were a smooth manifold in the subspace topology, every $a\in A$ would have a neighborhood homeomorphic to $\mathbb{R}$. In particular, $A$ would be locally path-connected. However, it's not, so it can't be a submanifold.

Alternatively, if $A$ were a submanifold, every $a\in A$ would have a neighborhood $U$ in $T^2$ with a coordinate chart taking $a$ to zero, $U$ to $\mathbb{R}^2$, and $U\cap A$ to the $x$-axis. This cannot happen, for every open ball $U_\epsilon$ of $a$ in $T^2$ (for $\epsilon$ sufficiently small) has that $U\cap A$ is disconnected (with countably many components).

Neal
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  • It's not true that every neighborhood is disconnected. The whole space is a neighborhood which is connected, as are a number of other "big" neighborhoods. – Cheerful Parsnip Oct 23 '13 at 17:11
  • Agreed, Grumpy, this is a space which is path connected but not locally path connected. – Winston Oct 23 '13 at 18:28
  • @GrumpyParsnip Very true, editing in the clarification. – Neal Oct 23 '13 at 18:49
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    @Winston: if you can show that, then you are done since manifolds are locally path connected. – Cheerful Parsnip Oct 23 '13 at 19:03
  • Quick relevant question: is it possible to have a non-compact submanifold of a compact manifold, in particular a non-compact one dimensional submanifold of a two torus? Sorry if this is elementary but I forgot a lot of my differential geometry... – Learning Math Jun 20 '23 at 09:49
  • @LearningMath yes -- take an open line segment, for example. (Precisely, realize a torus as $[-1, 1] \times [-1, 1]$ quotiented by the usual identification of boundaries. The segment $(-1/2, 1/2)$ is an open submanifold.) – Neal Jun 21 '23 at 22:12