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I'm trying understand one point about Example 15.9 in the Introduction to Manfolds by Tu (p. 167). I'll reproduce the example here. See this post, which addressed a separate question I had.

Example 15.9 (Lines with irrational slope in a torus). Let $G$ be the torus $\mathbb{R}^2/\mathbb{Z}^2$ and $L$ a line through the origin in $\mathbb{R}^2$. The torus can also be represented by the unit square with the opposite edges identified. The image $H$ of $L$ under the projection $\pi:\mathbb{R}^2\longrightarrow\mathbb{R}^2/\mathbb{Z}^2$ is a closed curve if and only if the line $L$ goes through another lattice point, say $(m,n)\in\mathbb{Z}^2$. This is the case if and only if the slope of $L$ is $n/m$, a rational number or $\infty$; then $H$ is the image of finitely many line segments on the unit square. It is a closed curve diffeomorphic to a circle and is a regular submanifold of $\mathbb{R}^2/\mathbb{Z}^2$ (Figure 15.1).

If the slope of L is irrational, then its image H on the torus will never close up. In this case the restriction to L of the projection map, $f := \pi\big|_L : L \rightarrow \mathbb{R}^2/\mathbb{Z}^2$, is a one-to-one immersion. We give H the topology and manifold structure induced from f.

  1. I don't understand why $f$ is an immersion. How would I prove this?
  2. Why is the subspace topology of H in $\mathbb{R}^2/\mathbb{Z}^2$ a strict subset of the topology on $H$ induced from $ f : L \xrightarrow{\sim} H$? For the induced topology, I understand that a basic open set of H is the image under $f$ of an open set in $L$. I also understand that basic open sets in the subspace topology will be the intersection of H with the image of an open ball in $\mathbb{R^2}$.
IsaacR24
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    What is your definition of immersion? If it's a differentiable function whose is everywhere injection, it's just a matter of showing that the tangent map $T_x \pi\vert_L$ is not zero for any $x$. – Travis Willse Jul 15 '22 at 18:06
  • Yes that's the definition I intended, that the pushforward is injective. How do I show this? – IsaacR24 Jul 15 '22 at 18:07
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    Here's one possible outline: (1) The inclusion $\iota: L \hookrightarrow \Bbb R^2$ is an immersion. (2) The quotient map $\pi: \Bbb R^2 \to \Bbb R^2 / \Bbb Z^2$ has discrete kernel so its differential is an isomorphism, and in particular $\pi$ is an immersion. (3) The composition of two immersions is an immersion. – Travis Willse Jul 15 '22 at 18:21
  • Your first point is very clear thanks! But why does a quotient map with a discreet kernel imply the differential is an isomorphism? Can you address my question 2? – IsaacR24 Jul 16 '22 at 12:26
  • Since the kernel is discrete, around every point $x \in \Bbb R^2$ is a neighborhood $U$ such that $\pi\vert_U$ is a diffeomorphism, so $\pi$ is a local diffeomorphism. As for (2), your answer is essentially contained in your question: You just have to show that every open set in the subspace topology is also open in the induced topology, and it suffices to show the claim for an arbitrary intersection $f(L) \cap f(B)$. – Travis Willse Jul 16 '22 at 13:52
  • (1) I see that if the kernel is discrete, for each $x \in \mathbb{R^2}$, there's a neighborhood U where either $\pi\vert_U$ is injective (if $x$ is not in the kernel) or otherwise has only one nonzero element in the kernel. Why does this make the Jacobian invertible? (2) Re induced topology. For L, I believe we use the subspace topology under $\mathbb{R^2}$. Then if $U \in \tau_L$, $U = L \cap V$ some $V \in \mathbb{R^2}$. But then $f(U) = f(L) \cap f(V) = H \cap f(V)$, which equals the union of basic open sets in the subspace topology of H -- I must be confused somewhere here? – IsaacR24 Jul 16 '22 at 17:43
  • For (2), I meant to ask why $\tau_{subspace} \subsetneq \tau_{induced}$. – IsaacR24 Jul 17 '22 at 10:57
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    @TravisWillse: Regarding question 1, your comments would make a good answer. – Lee Mosher Jul 17 '22 at 13:31
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    Regarding question 2, you can find an answer here. – Lee Mosher Jul 17 '22 at 13:31
  • @Lee Mosher, can you help address my question about (1)? I still don't quite follow (2), per the point I made yesterday. – IsaacR24 Jul 17 '22 at 14:27

1 Answers1

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  1. I don't understand why f is an immersion. How would I prove this?

Hint Here's a brief outline:

  1. The inclusion $\iota :L \hookrightarrow \Bbb R^2$ is an immersion.
  2. The map $\pi: \Bbb R^2 \to \Bbb R^2 / \Bbb Z^2$ is a covering map. In particular, it is a local diffeomorphism and hence an immersion.
  3. The composition of immersions is again an immersion.
  1. Why is the subspace topology of H in $\mathbb{R}^2/\mathbb{Z}^2$ a strict subset of the topology on $H$ induced from $ f : L \xrightarrow{\sim} H$? For the induced topology, I understand that a basic open set of H is the image under $f$ of an open set in $L$. I also understand that basic open sets in the subspace topology will be the intersection of H with the image of an open ball in $\mathbb{R^2}$.

See the answer to this question mentioned by Lee Mosher. In particular, if we pick an open ball $B \subset \Bbb R^2$ that intersects $L$, then $\pi(B \cap L)$ is by definition open in the subset topology, but any neighborhood (in the subspace topology) of any point in $\pi(B \cap L)$ contains the image of points in $L$ that are not in $\pi(B \cap L)$, hence $\pi(B \cap L)$ is not open in the subspace topology.

Travis Willse
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  • Your explanation of part 2 finally clicked, thanks! I still don't understand why the map in 2. is a local diffeomorphism, however. I see that if the kernel is discrete, for each $x \in R^2$, there's a neighborhood $U$ where either $\pi\vert_U$ is injective (if $x$ is not in the kernel) or otherwise has only one nonzero element in the kernel. Why does this make the Jacobian invertible? – IsaacR24 Jul 17 '22 at 18:11
  • The map $\pi \vert_U$ is injective regardless of whether $x$ is in the kernel of the quotient map (i.e., has integer coordinates). Injectivity alone is not enough to make the $\pi \vert_U$ smooth, let alone a diffeomorphism (consider $\Bbb R \to \Bbb R$, $x \mapsto x^{1/3}$), though it's adequate for quotients of Lie groups. Are you familiar with the language of covering maps? If so, it's probably most straightforward to think of $\pi$ as one. – Travis Willse Jul 17 '22 at 19:40
  • As best I can tell, the Tu Manifolds book doesn't address covering maps, so I'm not familiar with this. I think I have 2 points of confusion. (1) $\pi\vert_U$ is injective. If U is a neighborhood around a point in $\mathbb{Z^2}$, then the kernel is non-trivial so the map can't be injective. (2) I couldn't find a proof to this claim: If a quotient of a lie group (mod a closed normal subgroup) has an injective projection, its projection is locally diffeomorphic. – IsaacR24 Jul 18 '22 at 12:17
  • So I don't understand (1), but fairly sure I understand (2) now. Represent $\pi$ by a right group action by the additive group $\mathbb{Z^2}$ on $\mathbb{R^2}$. I know this makes $\pi$ an open map. Thus $\pi$ is bicontinuous and surjective, so you only need injectivity to have a diffeomorphism. – IsaacR24 Jul 18 '22 at 14:30
  • Ah of course, $\pi\vert_{U}$ has at most one element in the kernel, so it is injective. I understand everything now, so accepting the answer. @Travis Willse thanks so much for the patient explanations! – IsaacR24 Jul 18 '22 at 16:07
  • @IsaacR24 You're welcome, I'm glad you found it helpful. To be precise, you need that $\pi\vert_U$ is bicontinuous (and also smooth), but those are both nearly automatic. – Travis Willse Jul 18 '22 at 17:17