Intereting Posts

If $a_{1}=1$ and $a_{n+1}=1+\frac{n}{a_{n}}$, then $a_{n}=\sqrt{n}+\frac{1}{2}-\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)$
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Finitely generated modules over a Noetherian ring are Noetherian
An interesting geometry problem about incenter and ellipses.

Let $S^1$ be the subspace of $R^2$ given the usual topology. How many group structures make $S^1$ a topological group?

- The group $E(\mathbb{F}_p)$ has exactly $p+1$ elements
- Normal subgroup where G has order prime
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- For abelian groups: does knowing $\text{Hom}(X,Z)$ for all $Z$ suffice to determine $X$?
- Set of All Groups
- What's the smallest exponent to give the identity in $S_n$?
- Let $G$ a group of order $6$. Prove that $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$.

Up to conjugating by homeomorphisms, the only topological group structure on $S^1$ is the standard one. Here is the lowest-tech proof of this fact I can see.

Suppose we have a continuous group structure on $S^1$; we may assume its identity is $1\in S^1\subset\mathbb{C}$. This group structure then lifts to a group structure on the universal cover $\mathbb{R}\to S^1$ ($x\mapsto \exp(2\pi i x)$), such that translation by preimages of the identity correspond to deck transformations. Thus we may assume we have a continuous group operation $*$ on $\mathbb{R}$ which has the property that if $n\in\mathbb{Z}$ and $x\in\mathbb{R}$, $n*x=n+x$. We will show that $*$ is conjugate to addition by a homeomorphism $\mathbb{R}\to\mathbb{R}$ that commutes with the map $x\mapsto x+1$; it then follows that our group operation on $S^1$ is conjugate to the standard group structure by the corresponding homeomorphism $S^1\to S^1$.

First, fix any $x\in\mathbb{R}$ and consider the map $R_x(y)=y*x$. This is a homeomorphism $\mathbb{R}\to\mathbb{R}$, so it is either increasing or decreasing. Since $R_x(n)=n+x$ for $n\in\mathbb{Z}$, we find that $R_x$ must be increasing. That is, $y<z$ implies $y*x<z*x$.

Now consider the map $f(x)=x*x$ from $\mathbb{R}\to\mathbb{R}$. We know that $f(0)=0$ and $f(1)=2$, so by the intermediate value theorem there is some $x_1\in (0,1)$ such that $f(x_1)=1$. There is then some $x_2\in (0,x_1)$ such that $f(x_2)=x_1$, and so on. In this way we can find elements $x_n$ which generate a subgroup $Q$ which is (algebraically) isomorphic to the dyadic rationals $D$ via a map $\varphi:D\to Q$ such that $\varphi(1/2^n)=x_n$. Furthermore, repeatedly applying the implication $y<z\implies y*x<z*x$ gives that $\varphi$ is order-preserving.

Furthermore, the elements $x_n$ are positive and decreasing, so they converge to some value $y$. This $y$ will be a fixed point of $f$; i.e., $y=y*y$, so $y=0$.

I now claim that $Q$ is dense in $\mathbb{R}$. Since $1\in Q$ and $1*x=1+x$ for all $x$, it suffices to show $Q$ is dense in $[0,1]$. Now note that the group operation $*$ is uniformly continuous on the compact set $[0,1]\times[0,1]$. In particular, this implies that for any $\epsilon>0$, there exists $\delta>0$ such that if $0\leq x\leq\delta$, then $y\leq y*x\leq y+\epsilon$ for all $y\in [0,1]$. Since the numbers $x_n$ converge to $0$, this means that if $n$ is sufficiently large then consecutive powers of $x_n$ in $[0,1]$ differ by at most $\epsilon$. Since these powers are elements of $Q$ and range from $0$ to $1$, this means every number between $0$ and $1$ is within $\epsilon/2$ of an element of $Q$. Since $\epsilon$ is arbitrary, this means $Q$ is dense in $[0,1]$ and hence in $\mathbb{R}$.

Thus we have an order-preserving group-isomorphism $\varphi:D\to Q$ from $D$ to a dense subset $Q$ of $\mathbb{R}$. Since $\mathbb{R}$ is the Dedekind-completion of both $D$ and $Q$, $\varphi$ extends uniquely to an order-isomorphism $\bar{\varphi}:\mathbb{R}\to\mathbb{R}$. By continuity of the group operations on both sides, $\bar{\varphi}$ is also a group-isomorphism. It follows that $\bar{\varphi}$ is an isomorphism of topological groups from $(\mathbb{R},+)$ to $(\mathbb{R},*)$. Furthermore, $\bar{\varphi}$ is the identity on $\mathbb{Z}$ and $x*1=x+1$, so $\bar{\varphi}$ commutes with the map $x\mapsto x+1$.

It seems the following. Let $G$ be an abelian topogical group whose space is $S^1$. Then $G$ is a compact locally connected second countable topological group, so by [Pon, Th. 49] it is topologically isomorphic to a direct product $(S^1)^r Z$, where $r\le\omega$ and the group $Z$ is finite. Since $G$ is connected and has dimension $1$, we have $r=1$ and the group $Z$ is trivial, so the group $G$ is topologically isomorphic to the group $S^1$.

*References*

[Pon] Lev S. Pontrjagin, *Continuous groups*, 2nd ed., M., (1954) (in Russian).

If $S^1$ is a topological group, it is, in a unique way, a Lie group in a compatible way, and with that structure, isomorphic to the usual $S^1$ by the structure theory of compact abelian Lie groups.

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