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If I want to show that the spere $\mathbb{S}^1$ is a topological manifold. Graphically it's clear that we have a chart $x:\mathbb{R}\rightarrow \mathbb{S}^1$ since we can "cut" $\mathbb{S}^1$ and and form it to a line right? But somehow I can't imagine how to show this mathematically.

To have the context, I want to show the following statement

The n-dimensional torus $\mathbb{T}^n \subset \mathbb{C}^n$ with it's subspace topology is a topological manifold of dimension n.

I wrote $\mathbb{T}^n=\mathbb{S}^1\times ...\times \mathbb{S}^1$ as an n-dimensional product. Then i thought if I could show that $\mathbb{S}^1$ is a topological manifold, also the product is one.

Another question is, why is the subspace topology so important in this case, don't I need to use the product topology since I rewrote the torus as a product?

thank you!

user123234
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  • The reason that the subspace topology is so important is that the statement says the torus is "nicely" sitting inside $\mathbb{C}^n$. If it weren't the subspace topology, then the torus could have been a manifold in itself, but it could not have been made to sit inside $\mathbb{C}^n$. Since you want to write the torus as a product space, you will also have to prove that the resulting topology due to the product is the same as the subspace topology. To get this, you could show that the product topology is coarser than the subspace topology but the inclusion is still contnuous. – Aniruddha Deshmukh Oct 26 '21 at 09:21
  • As for the proof of $\mathbb{S}^1$ being a topological manifold, you can try stereographic projection by removing $\left(0, 1 \right)$ and $\left( 0, -1 \right)$. That will give you the precise charts you want. – Aniruddha Deshmukh Oct 26 '21 at 09:22

1 Answers1

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If you define $T^n$ as the product of circles, then it has the product topology and no reference to the subspace topology is necessary. Note, however, that both topologies agree (which is a little lemma: If you have subspaces $A_i \subset X_i$, then $\prod_i A_i$ with the product topology is a subspace of $\prod_i X_i$). Here you have $S^1 \subset \mathbb C$.

If you know that $S^1$ is a $1$-manifold, then of course you also know that $T^n$ is an $n$-manifold. See your last question.

To show that $S^1$ is a manifold, you can use stereographic projection (see here) or the following four charts (where $J = (-1,1)$): $$a_1 : \{ z \in \mathbb C \mid \Re (z) > 0 \} \to J, a_1(z) = \Im(z)$$ $$a_2 : \{ z \in \mathbb C \mid \Re (z) < 0 \} \to J, a_2(z) = \Im(z)$$ $$a_3 : \{ z \in \mathbb C \mid \Im (z) > 0 \} \to J, a_3(z) = \Re(z)$$ $$a_4 : \{ z \in \mathbb C \mid \Im (z) < 0 \} \to J, a_4(z) = \Re(z)$$

You can translate this to your chart concept by considering $a^{-1}_i$ and composing with a homeomorphism $h : \mathbb R \to J$.

Paul Frost
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  • why do I need to compose $a_i^{-1}$? – user123234 Oct 26 '21 at 11:45
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    Because charts in your sense have domain $\mathbb R^n$. At least I understood this from your last question. If you do not have this requirement, it is unnecessary. – Paul Frost Oct 26 '21 at 12:17
  • But then what are exaclty the inverses of $a_i$ are $a_i^{-1}:J\rightarrow {z\in \mathbb{C}|...}$? Because we take an integer an this should be the imaginary part of z for example in $a_i$ but then where should we send this? – user123234 Oct 26 '21 at 12:32
  • another question, on $\mathbb{S}^1$ do we take the metric topology, since $\mathbb{C}$ is isomorph to $\mathbb{R}^2$ – user123234 Oct 26 '21 at 12:47
  • or do we take the subspacetopology of the metric topology? – user123234 Oct 26 '21 at 13:10
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    Yes, $S^1$ has the metric inherited from $\mathbb C = \mathbb R^2$. This makes $S^1$ a metric and topological subspace. – Paul Frost Oct 26 '21 at 14:15
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    The domains of the maps $a_i$ are the four open half-circles (upper, lower, right, left). We project them to the real resp. imaginary axis. The inverses have the form $a^{-1}_1(t) = t + i\sqrt{1-t^2}$ etc. – Paul Frost Oct 26 '21 at 14:29
  • thanks a lot for your help – user123234 Oct 26 '21 at 18:58