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From my lecture notes: "The notation $\mathbb T$ will be used for the additive circle and $S^1$ for the multiplicative circle."

What I understand: As a topological group, $S^1$ has the subspace topology of $\mathbb R^2$ and multiplication is defined as $(e^{ia}, e^{ib}) \mapsto e^{i(a + b)}$.

My guess is that $\mathbb T$ as an additive group should then be something like $(a,b) \mapsto (a + b) \mod 1$. The problem with that is that the space would look like $[0,1)$ but that's not compact.

But I'm confused: "mod 1" seems to be the same as $\mathbb R / \mathbb Z$ which is $S^1$. But I can't add complex numbers on the unit circle and stay on the unit circle.

So: What's $\mathbb T$? How are elements in it added?

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The map $\mathbb R \to S^1$, $t \mapsto e^{2\pi it}$ descends to a map $\mathbb T \to S^1$ which is an algebraic and topological isomorphism. [Remember that $\mathbb T$ is an algebraic and topological quotient. What is the inverse of this map?] So you can indeed pass freely between the two, and in particular $\mathbb T$ is compact. If you want to see this without reference to $S^1$, note that the restriction of the quotient map $\mathbb R \to \mathbb R/\mathbb Z$ to the compact subspace $[0, 1]$ is still surjective.

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  • $\begingroup$ "topological isomorphism" = "homeomorphism"? $\endgroup$ – Rudy the Reindeer Jul 20 '12 at 19:40
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    $\begingroup$ @MattN Yep yep. $\endgroup$ – Dylan Moreland Jul 20 '12 at 19:41
  • $\begingroup$ To answer your question: if $\pi: \mathbb R \to \mathbb R / \mathbb Z$ is the projection then $\pi^{-1} x = x + \mathbb Z$ for $x \in [0,1)$? $\endgroup$ – Rudy the Reindeer Jul 20 '12 at 19:41
  • $\begingroup$ @MattN. A topological isomorphism is an algebraic homomorphism which is also a homeomorphism. $\endgroup$ – Alex Becker Jul 20 '12 at 19:41
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    $\begingroup$ @MattN. You know about quotient groups, right? That's where the group law on $\mathbb T = \mathbb R/\mathbb Z$ comes from. You can always choose representatives from $[0, 1)$, but you don't have to. $\endgroup$ – Dylan Moreland Jul 21 '12 at 16:39
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In my experience the two are used interchangeably, as the two groups are isomorphic, even as topological groups, and this isomorphism (as topological groups) is unique! You are right that "$\bmod 1$" is the same as $\mathbb R/\mathbb Z$, which is best viewed as $[0,1]$ with the endpoints identified. The only differences I can see are one of notation (elements of $\mathbb T$ are added while those in $S^1$ are multiplied) and that $S^1$ is more naturally considered a subset of $\mathbb C$.

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  • $\begingroup$ But if they were the same addition would go outside $S^1$. $\endgroup$ – Rudy the Reindeer Jul 20 '12 at 19:18
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    $\begingroup$ @MattN. Like I said, the operations are different but the groups are effectively the same. $\endgroup$ – Alex Becker Jul 20 '12 at 19:20

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