Connected components of this topological group

Question

I was confused: today some undergraduate asked me the following exercise

let $m,n\in \mathbb{Z}_{\geq 1}$ and $$G:=\{(a,b)\in \mathbb{C}^\times \times\mathbb{C}^\times: a^m b^n=1\}$$ as a topological group, with the "usual" topology (as a subgroup of $\mathbb{C}^\times \times\mathbb{C}^\times$). Question: what's the number of the connected components of $G$?

If $m=1$, then the answer is $1$: $G\rightarrow \mathbb{C}^\times, (a,b)\mapsto b$ is a homeomorphism.

If $m=n=2$, then $G=\{(a,a^{-1})\in \mathbb{C}^\times \times\mathbb{C}^\times\}\cup \{(a,-a^{-1})\in \mathbb{C}^\times \times\mathbb{C}^\times\}$, and the answer is $2$.

Is the answer for the general case $\min\{m,n\}$? Thanks a lot in advance!

Answer 1

We will show that $\rho:\mathbb C^\times\times\mathbb C^\times\rightarrow \mathbb C^\times$ where $\rho(a,b)=a^mb^n, \gcd(m,n)=1$ has connected fibres, i.e. for any $z\in\mathbb C$, $\rho^{-1}(z)$ is connected.

If this is true, then for general $m,n$ with $d=\gcd(m,n)$, we have $\rho(a,b)=(a^{m/d}b^{n/d})^d=1$ indicates $\zeta_d=a^{m/d}b^{n/d}$ is a $d$-th root of unity, and for fixed $\zeta_d$, we have $\{(a,b) | a^{m/d}b^{n/d}=\zeta_d\}$ is connected (and non-empty). So there are exactly $d$ connected components.

---

Now we prove the claim, which is purely topological. Assume $z=\rho(z_1,z_2)=\rho(z_3, z_4)$, we shall find a path to connect $(z_1, z_2)$ and $(z_3, z_4)$ within $\rho^{-1}(z)$. We may assume $z_1=e^{a}, z_2=e^b, z_3=e^c, z_4=e^d$, then $\rho(z_1, z_2)=\rho(z_3, z_4)$ implies $$am+bn\equiv cm + dn \mod 2\pi i$$

or equivalently there is an integer $l\in\mathbb Z$ such that $$(a-c)m+(b-d)d=2\pi l i $$

As $\gcd(m,n)=1$, we have $mx+ny=l$ for some $x,y\in\mathbb Z$. So we get $$(a-c)m+(b-d)n=2\pi i (mx+ny)$$

$$am + bn = (c+2\pi i x)m + (d+2\pi i y)n$$

If $c$ is replaced by $c+2\pi i x$ and $d$ by $d+2\pi i y$, obviously $e^c, e^d$ won't be changed. Hence WLOG, we may asume $am+bn=cm+dn$. Now it's pretty easy to construct the path from $(e^a, e^b)$ to $(e^c, e^d)$: $$p(t)=(e^{(1-t)a+tc}, e^{(1-t)b+td})$$ It's easy to check that $\rho(p(t))$ is a constant.

Answer 2

This is a very interesting problem, in my opinion. As I see it, the answer here is.

Let $m,n\in\mathbb{Z}$, $m\geq1$, $n\geq1$. The number of connected components of $$ G=\{(u,v)\in\mathbb{C}^*\times\mathbb{C}^*\mid u^mv^n=1\}\tag1 $$ is $d=\gcd(m,n)$.

Let us denote the group $\mathbb{C}^*\times\mathbb{C}^*$ by $F$, for brevity. First of all, note that $G$ is a closed subgroup of the group $F$. Next, $$ G_0=\{(z^n,z^{-m})\mid z\in\mathbb{C}^*\}\tag2 $$ is a connected subgroup of $G$ and $\dim G_0=2$. It follows that $\dim G\geq2$. On the other hand, if $H=\{(1,v)\mid v\in\mathbb{C}^*\}$, then the intersection $G\cap H$ is a finite set. Hence $\dim F/G\geq2$. Thus, $\dim G=2$ and $G_0$ is the component of $1$ in $G$.

Consequently, the number of connected components of $G$ is equal to $|G:G_0|$ (the index of $G_0$ in $G$).

Now let's look at a few special cases.

1. Let $m=n$. This is a case from @Chad K's comment. In this case $G_0=\{(z,z^{-1})\}$ and $|G:G_0|=n$.

The last equality follows from the fact that if $$ D=\{(1,\varepsilon_n^k)\mid k=0,1\ldots,n-1\}, $$ where $\varepsilon_n$ is a primitive $n$th root of unity, then $D\cap G_0=\{e\}$, $|D|=n$ ($e$ is an identity element of a group, here $e=(1,1)$), and $D<G$. Hence $DG_0\leq G$. In fact, $DG_0=G$, since if $(u,v)\in G$, then $(uv)^n=1$ i.e. $uv=\varepsilon_n^k$ for some $k\in\{0,1,\ldots,n-1\}$ and $$ (u,v)=(1,\varepsilon_n^k)(z,z^{-1}), $$ where $z=u$.

It follows that $|G:G_0|=n$.

2. If $\gcd(m,n)=1$, then $G$ is connected.

We will prove that in this case $G=G_0$. Let $(u,v)\in G$ and let $u_0$ (resp. $v_0$) be an $n$th (resp. $m$th) root of $u$ (resp. $v$). We have $$ (u_0v_0)^{mn}=u^mv^n=1\ \Rightarrow\ u_0v_0=\varepsilon_{mn}^k\tag3 $$ for some $k\in\{0,1,\ldots,mn-1\}$. Since $m$ and $n$ are relatively prime there exist such integers $r$ and $s$ that $$ \varepsilon_{mn}=\varepsilon_{m}^r\varepsilon_n^s. $$ Hence and from $(3)$ it follows that $$ u_0v_0=\varepsilon_{m}^{kr}\varepsilon_n^{ks}\ \Rightarrow\ u_0\varepsilon_{n}^{-ks}v_0\varepsilon_{m}^{-kr}=1. $$ If $z=u_0\varepsilon_{n}^{-ks}$, then $z^{-1}=v_0\varepsilon_{m}^{-kr}$ and $(z^n,z^{-m})=(u,v)\in G_0$ (see $(2)$).

3. If $\gcd(m,n)=1$ and $d>1$ is a integer and $d$ is relatively prime to $m$, and $$ G=\{(u,v)\in\mathbb{C}^*\times\mathbb{C}^*\mid u^{dm}v^{dn}=1\}\tag4 $$ and $G_0$ is defined by $(2)$, then $|G:G_0|=d$.

As proved in item 2 $G_0=\{(u,v)\in F\mid u^mv^n=1\}$.

If $$ D=\{(\varepsilon_d^k,1)\mid k=0,1\ldots,d-1\}, $$ then $|D|=d$, $D<G$, $D\cap G_0=\{e\}$, and $DG_0=G$. The latter is proved in the same way as in 1. It follows that $|G:G_0|=d$.

4. General case. Let $\gcd(m,n)=1$ and $d>1$ be integer. If $G$ is defined by $(4)$ and $G_0$ is defined by $(2)$, then $|G:G_0|=d$.

Let us represent $d$ in the form $$ d=p_1^{\alpha_1}\ldots p_s^{\alpha_s}q_1^{\beta_1}\ldots q_t^{\beta_t}d'. $$ Here $p_i$, $q_j$ are prime, $p_i$ is a divisor of $m$, $q_j$ is a divisor of $n$ for every $i,j$ and $d'$ is relatively prime to $m$ and $n$. Denote by $$ p=p_1^{\alpha_1}\ldots p_s^{\alpha_s},\ q=q_1^{\beta_1}\ldots q_t^{\beta_t}d'. $$ It is clear $p$ is relatively prime to $n$ and $q$ is relatively prime to $m$.

Let $$ D_1=\{(1,\varepsilon_p^k)\mid k=0,1,\ldots,p-1\},\ D_2=\{(\varepsilon_q^k,1)\mid k=0,1,\ldots,q-1\}, $$ and $D=D_1D_2$. We have $D<G$, $|D|=d$, $D\cap G_0=\{e\}$, and $G=DG_0$. All the checks are similar to those discussed in $1$-$3$. It follows that $|G:G_0|=d$.

Conclusion.
If $M$ and $N$ are integers not less than $1$ and $d=\gcd(M,N)$, then assuming $m=M/d$, $n=N/d$ we get relatively prime $m$ and $n$ and further apply one of the considered items.

PS This ended up being pretty lengthy, so hopefully I didn't miss anything. I am happy to answer any questions you may have.