Closed curve in a torus diffeomorphic to a circle?

Asked Dec 05 '23 at 06:18

Active Dec 05 '23 at 08:18

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In Example 15.9 of Tu's "An Introduction to Manifolds: Second Edition", it is written

Let G be the torus $R^2/Z^2$ and L a line through the origin in R2. The torus can also be represented by the unit square with the opposite edges identified. The image H of L under the projection π : $R^2$ → $R^2/Z^2$ is a closed curve if and only if the line L goes through another lattice point, say (m,n) ∈ $Z^2$. This is the case if and only if the slope of L is n/m, a rational number or ∞; 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 $R^2/Z^2$

Intuitively, it is understandable the closed curve mentioned above is diffeomorphic to a circle. I also read several related questions like this and this and this, and understood this closed curve is homeomorphic to a circle, however, I can't prove the smoothness rigorously.

Any help would be appreciated!

edited Dec 05 '23 at 08:18

asked Dec 05 '23 at 06:18

Foo

The map from $L$ to its image is a local diffeomorphism. – Ted Shifrin Dec 05 '23 at 06:29
Can you describe the smooth structure on $\mathbb R^2/\mathbb Z^2$? Do you know that the quotient map is smooth and a local diffeomorphism? – Paul Frost Dec 05 '23 at 17:58

Closed curve in a torus diffeomorphic to a circle?

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