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In this question $\varphi$ in $\operatorname{Hom}{(S^1, S^1)}$ are of the form $z^n$ all the continuous homomorphisms from $S^1$ to $S^1$ are characterized. However I am looking for non continuous homomorphisms.

My observations are $S^1$ is isomorphic to $\frac{\mathbb{R}}{\mathbb{Z}}$.

Any subgroup of $\frac{\mathbb{R}}{\mathbb{Z}}$ is in the form $\frac{H}{\mathbb{Z}}$ where $H$ is a subgroup of real numbers containing integers.

If $H$ is generated by $1/n$ then we get $\frac{H}{\mathbb{Z}}$ as the kernel of continuous homomorphisms.

So it is possible to show the existence of non continuous homomorphisms if we can find a non cyclic subgroup $H$ of real numbers such that $\mathbb{R}/H$ is subgroup of $\mathbb{R}/\mathbb{Z}$. But I am not able to find any such H explicitly.

Is there a way to improve this or any alternative ways?

Infinity_hunter
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  • For example, it is easy to see that the subgroup $H={a+b\sqrt2\mid a,b\in\mathbb{Z}}$ is not cyclic. – kabenyuk Jul 28 '22 at 15:36
  • @kabenyuk Yes, it's easy to find examples of subgroups of real numbers. But is it easy to see whether their quotient is a subgroup of $\mathbb{R}/\mathbb{Z}$? – Infinity_hunter Jul 28 '22 at 15:46
  • Here's another example. There is an injective homomorphism $\mathbb{R}/\mathbb{Q}\to\mathbb{R}/\mathbb{Z}$. See the theory of abelian divisible groups. – kabenyuk Jul 28 '22 at 16:13
  • @If you can provide some more details it'll be helpful or even you could write an answer. – Infinity_hunter Jul 28 '22 at 16:43
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    Also posted to MO, https://mathoverflow.net/questions/427460/non-continuous-homomorphism-from-unit-circle-to-unit-circle without notice of crosspost to either site – an abuse of both sites. – Gerry Myerson Jul 28 '22 at 23:20

1 Answers1

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$\mathbb{R}$ is an uncountable-dimensional vector space over $\mathbb{Q}$, so (assuming the axiom of choice) it has an uncountable basis and hence can be written as an uncountable direct sum $\bigoplus_I \mathbb{Q}$. One of these copies of $\mathbb{Q}$ can be chosen to contain $\mathbb{Z}$. The result is that we have an abstract isomorphism

$$\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_J \mathbb{Q}$$

where $J$ is $I$ with one element removed. There are no nonzero homomorphisms from the first summand to the second. This gives

$$\text{End}(\mathbb{R}/\mathbb{Z}) \cong \left[ \begin{array}{cc} \text{End}(\mathbb{Q}/\mathbb{Z}) & \text{Hom}(\bigoplus_J \mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \\ 0 & \text{End}(\bigoplus_J \mathbb{Q}) \end{array} \right].$$

All three of these can be calculated as follows.

  • $\text{End}(\mathbb{Q}/\mathbb{Z})$ turns out to be the profinite integers $$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p.$$

  • $\text{End}(\bigoplus_J \mathbb{Q})$ is an uncountable-dimensional (column-finite) matrix algebra $$M_J(\mathbb{Q}).$$

  • And $\text{Hom}(\bigoplus_J \mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \prod_J \text{Hom}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$. $\text{Hom}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$ is a bit tricky to describe but you can see here that one possible description is $\mathbb{Q} \otimes \widehat{\mathbb{Z}}$, and you can find another description working one prime at a time towards the end of these notes.

So this is a more or less complete description of all endomorphisms, nearly all of which are highly discontinuous.

Qiaochu Yuan
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  • Since this was bumped by the edit, a comment: I think the existence of discontinuous endomorphisms essentially relies on the axiom of choice (so we won't have any ZFC-provable "explicit" formula for a discontinuous endomorphism). More precisely, I think the following is true: in ZF, every measurable endomorphism of the circle is continuous, and there are models of ZF in which every function from the circle to itself is measurable. – tomasz Feb 26 '25 at 22:17
  • @tomasz: yes, that's right. – Qiaochu Yuan Feb 26 '25 at 22:24
  • What about the homomorphism $\phi_i: \mathbb R /\mathbb Z \to \mathbb R /\mathbb Z$ that is $(\phi_i){|\mathbb Q /\mathbb Z}=\operatorname {Id}$ and $0$ on $\bigoplus{J_i} \mathbb Q$ (where $\mathbb Q_i$ is the copy of $\mathbb Q$ chosen to contain $\mathbb Z$ and $J_i:= I\setminus {i}$). Shouldn't this be an injection from $I$ to the set of groupendomorphisms on the torus, while the set of all such continuous homomorphisms is countable? – Tina Feb 27 '25 at 09:34
  • @Tina: the OP said they were looking for discontinuous homomorphisms and, as I said in the last sentence, almost every homomorphism constructed here is highly discontinuous. – Qiaochu Yuan Feb 27 '25 at 16:03
  • Yes i see. But i think the decomposition $\mathbb R/ \mathbb Z = \mathbb Q / \mathbb Z \oplus \bigoplus_J \mathbb Q$ immediately points out that there are uncountable many, probably $| \mathcal P(\mathbb R)|$-many, of such discontinuous characters, with a lot less structural theorems required. I still think your result is pretty cool, i just wanted to show a different/ "easier" solution. – Tina Feb 27 '25 at 18:08
  • @Tina: if the only goal is to exhibit a single discontinuous homomorphism you can just consider the map which is the identity on $\mathbb{Q}/\mathbb{Z}$ and $0$ on the free part. – Qiaochu Yuan Feb 27 '25 at 18:17