The $n$-dimensional torus is $\mathbb T^n=\mathbb R^n/\mathbb Z^n$. Let $|x|$ be the Euclidian norm. What is $|x|$ for $x\in \mathbb T^n$?
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The answer depends on how you embed the torus into Euclidean space. – Potato Mar 06 '14 at 15:41
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I usually see $\mathbb T^n$ as $[0, 1]^n$ with opposite sides indetified. – PtF Mar 06 '14 at 15:44
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My comment above was stupid, that is a consequence of $\mathbb T^n:=\mathbb R^n/\mathbb Z^n$.. – PtF Mar 06 '14 at 15:50
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@Potato I have never thought on how to norm it.. – PtF Mar 06 '14 at 15:57
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Since $\mathbb T^n$ is not a vector space, it makes little sense to talk about norms. Let's consider the metric induced on $\mathbb R^n$ by the norm, namely $d(x,y)=|x-y|$. This metric naturally induces the quotient metric on $\mathbb T^n$: it's just the minimal distance between representatives of equivalence classes $\bar x,\bar y\in\mathbb T^n$: $$\tilde d(\bar x, \bar y) = \min_{z\in \mathbb Z^n}d(x+n,z)$$ Here it helps that we take the quotient by a group of isometries.
Since the metric is invariant under group translation, it makes perfect sense to write it as $|\bar x-\bar y|$. Here $$|\bar x| = \min_{z\in\mathbb Z^n} |x+n|$$ which is the closest thing to a "norm" that you can have on the torus.
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