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Is any continuous homomorphism $f:\Bbb R\to\mathrm U(1)$ necessarily of the form $f(x)=\mathrm e^{\mathrm ikx}$, for $k\in\Bbb R$? It certainly seems plausible, but I can't nail down a proof. I don't know how to guarantee that an exponential appears without using some analytic argument, which I don't want to do. (So maybe it is false?)

Adam Hughes
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Ryan Unger
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2 Answers2

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The result is true. More generally, any continuous homomorphism $f \colon \mathbb{R} \rightarrow G$ between $\mathbb{R}$ and a Lie group $G$ is of the form $f(t) = e^{itX}$ for $X \in \mathfrak{g}$ (and in particular, automatically smooth).

To see this in your case, note that any continuous map $f \colon \mathbb{R} \rightarrow U(1)$ can be written as $f(x) = e^{i\varphi(x)}$ for a continuous map $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ with $\varphi(0) = 0$ (this follows for example from the theory of covering spaces). Now, the identity $f(x + y) = f(x)f(y)$ implies that $\varphi(x + y) = \varphi(x) + \varphi(y) + 2\pi c(x,y)$ for some $c \colon \mathbb{R}^2 \rightarrow \mathbb{Z}$. Since $c$ is evidently continuous and $c(0,0) = 0$, we must have $c(x,y) \equiv 0$ and so $\varphi(x + y) = \varphi(x) + \varphi(y)$. Now use the well-known result that a continuous additive map $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ must be a linear map to deduce that $\varphi(x) = kx$ for some $k \in \mathbb{R}$.

levap
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  • Interesting, what is a discontinuous additive map that's not linear? – Ryan Unger Dec 09 '16 at 05:24
  • @0celo7: Such examples can be proven to exist assuming the axiom of choice but cannot be constructed explicitly. For details, see http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation/423494#423494. – levap Dec 09 '16 at 05:26
  • This is an old question but can you tell me where I can find something regarding the first statement you made? – user21417 Feb 14 '22 at 05:59
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Yes. The continuity shows we get a connected subset of $U(1)$, and since any neighborhood of the identity generates the whole group, if the homomorphism is trivial, it's $e^{i0x}$ and if it's not, then the map is onto, and we know the kernel is isomorphic to $\Bbb Z$ since all subgroups of $\Bbb R$ are either discrete or dense, and the latter case plus continuity would give a trivial homomorphism. But then let $T>0$ be the smallest positive number for which the map is zero, then we can see that it is periodic with period $T$, and so $e^{2\pi ix/T}\cdot f(x)$ is constant since homogeneity of the fibers over any point we know that it traverses the circle at constant speed.

Adam Hughes
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  • I follow until the very last part, what does "constant since homogeneity of the fibers over any point we know that it traverses the circle at constant speed" mean? – Ryan Unger Dec 09 '16 at 05:18
  • @0celo7 there are many periodic functions, we want to make sure that we're not dealing with one that goes almost all the way around really fast, and then hangs out for a while, and then finishes the period. – Adam Hughes Dec 09 '16 at 05:21
  • I don't see how "homogeneity of the fibers" implies constant speed (or what that even means since we're not talking about derivatives). – Ryan Unger Dec 09 '16 at 05:23
  • @0celo7 You automatically get smoothness from the Lie Group structure, but you can also verify it explicitly. I could outline the argument, but it seems you've already gotten another answer you can understand, so I'll leave it at that. Cheers! – Adam Hughes Dec 09 '16 at 14:30