# Closed submonoid of $(\mathbb{C}^*)^n$

The answer of this question might be known but I was not able to find any answer. Let $$n\geq 1$$ and $$S$$ be a closed submonoid of $$(\mathbb{C}^*)^n$$, that is, a closed and stable by product subset of $$(\mathbb{C}^*)^n$$ which also contains the unit $$(1,\ldots,1)$$. Let also $$R_1,\ldots,R_d\geq1$$, and denote by $$A:=\{(r_ie^{i\theta_i})_{i=1}^{d}:\text{ for every }i=1,\ldots,d\ , 1\leq r_i\leq R_i, \theta_1,\ldots,\theta_d\in[0;2\pi]\}$$. We assume that (the product of the two sets has to be understood component by component):$$A\times S=(\mathbb{C}^*)^n.$$ Whatever $$R_1,\ldots,R_d$$ and $$n$$, is it true that $$S$$ has to be a subgroup?

• How about $n = 1, S = 2^a 3^{-b}, (a, b) \in \mathbb{N}$? Commented Jan 18, 2020 at 10:09
• @user44191 This is not a closed subset. Indeed its log is the subsemigroup generated by $\log(2)$ and $-\log(3)$. But for any $u<0<v$ with $u/v$ irrational, the additive subsemigroup generated by $u$ and $v$ is not discrete (it's dense in $\mathbf{R}$).
– YCor
Commented Jan 18, 2020 at 22:24
• @YCor Ah, I thought "closed" here was algebraic, not topological. That does make more sense. Commented Jan 18, 2020 at 23:22
• @MarkSapir I agree that the OP should clarify some of his or her terminology, but I think a reasonable guess would be that he or she seeks a closed subset of $({\mathbb C}^*)^n$ with its usual topology, which is also a submonoid for the natural pointwise product on $({\mathbb C}^*)^n$ Commented Jan 19, 2020 at 22:41
• With the new modification, I'm pretty sure the problem reduces to the question of whether every closed submonoid of a torus is a subgroup. Commented Jan 20, 2020 at 9:59

Define $$\log^n |\cdot|: \mathbb{C}^{*n} \rightarrow \mathbb{R}^n, \log^n |(z_1, z_2, \dots, z_n)| = (\log |z_1|, \log |z_2|, \dots, \log |z_n|)$$. This is a map of topological groups, where the "multiplication" on $$\mathbb{R}^n$$ is addition. Then the $$A_R$$ condition can be rewritten as: $$\log^n |S| + [0, \log R]^n = \mathbb{R}^n \left(1\right)$$
Let $$f \in S$$; we want to prove that $$f^{-1} \in S$$. Condition $$\left(1\right)$$ can be used to show that there is some set $$\{g_i\}$$ such that $$\{log^n |g_i|\}$$ forms a basis of $$\mathbb{R}^n$$ and such that the coordinates of $$\log^n |f|$$ are negative with respect to that basis. Then I claim that $$f^{-1} \in \overline{\{\prod_i g_i^{a_i} f^b | a_i, b \in \mathbb{Z}_{\geq 0}\}}$$. Equivalently, $$e \in \overline{\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}}$$.
Assume otherwise. Consider the map $$p: \mathbb{C}^{*n} \rightarrow \mathbb{C}^{*n}/\{\prod g_i^{a_i} | a_i \in \mathbb{Z}\} \simeq \mathbb{T}^{2n}$$. This is a map of topological groups, so $$p(\{f^b | b \in \mathbb{Z}_{\geq 1}\})$$ is a subsemigroup. It is not necessarily closed; however, its closure is - so by Yemon Choi's comment, it must be a group. Specifically, it must include the identity.
Let $$e \in U \subseteq \mathbb{C}^{*n}$$ be an open neighborhood of the identity; we want to show that $$U$$ contains an element of $$\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}$$. We can assume WLOG that U is both symmetric and "small enough" (more on this later). Then $$p(U)$$ is an open neighborhood of the identity in $$\mathbb{T}^{2n}$$, so it must contain some element of $$p(\{f^b | b \in \mathbb{Z}_{\geq 1}\})$$. Equivalently, $$U$$ must contain some element of the form $$\prod_i g_i^{a_i} f^b$$ such that $$a_i \in \mathbb{Z}, b \in \mathbb{Z}_{\geq 1}$$. But because the coordinates of $$\log^n |f|$$ are all negative with respect to $$\log^n |g_i|$$, by choosing $$U$$ small enough, we can guarantee that all of the $$a_i$$ are positive - so we have proven that $$U$$ contains an element of $$\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}$$. We are therefore done: $$e \in \overline{\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}}$$, so $$f^{-1} \in \overline{\{\prod_i g_i^{a_i} f^b | a_i, b \in \mathbb{Z}_{\geq 0}\}} \subseteq S$$.