Is π(L) a Submanifold of the Torus?

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SUMMARY

The discussion centers on the proof that the image of the line L under the canonical projection map π from ℝ² to the torus ℝ²/ℤ² is not a submanifold of the torus when the ratio b/a is irrational. Specifically, the user is attempting to demonstrate that π(L) is not locally Euclidean, concluding that every neighborhood of the point π(0) is disconnected. The user references John Lee's book for a more elegant solution, while also mentioning a brute force method utilizing the Hurwitz Theorem that ultimately provides a valid proof.

PREREQUISITES
  • Understanding of submanifolds in differential geometry
  • Familiarity with canonical projection maps, specifically π: ℝ² → ℝ²/ℤ²
  • Knowledge of the concept of local Euclidean spaces
  • Acquaintance with the Hurwitz Theorem in mathematics
NEXT STEPS
  • Study the properties of submanifolds in differential geometry
  • Learn about canonical projection maps and their applications in topology
  • Explore local Euclidean spaces and their significance in manifold theory
  • Investigate the Hurwitz Theorem and its implications in proving topological properties
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Mathematicians, particularly those specializing in differential geometry and topology, as well as students seeking to understand the relationship between lines in Euclidean space and their projections onto manifolds.

jgens
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I am trying to prove the following result: Fix a,b \in \mathbb{R} with a \neq 0. Let L = \{(x,y) \in \mathbb{R}^2:ax+by = 0\} and let \pi:\mathbb{R}^2 \rightarrow \mathbb{T}^2 be the canonical projection map. If \frac{b}{a} \notin \mathbb{Q}, then \pi(L) (with the subspace topology) is not a submanifold of \mathbb{T}^2.

I am having difficulty however showing that \pi(L) is not locally Euclidean. From drawing a few pictures, I think every neighborhood of \pi(0) is disconnected (which would be enough to complete the proof), but I am having difficulty showing this. Any help?
 
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Look at p.158 of the book of John Lee.
 
quasar987 said:
Look at p.158 of the book of John Lee.

Thanks! I (finally) figured out a brute force method using the Hurwitz Theorem that works, but Lee's solution is much cleaner.
 

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