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?

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    $\begingroup$ How about $n = 1, S = 2^a 3^{-b}, (a, b) \in \mathbb{N}$? $\endgroup$
    – user44191
    Commented Jan 18, 2020 at 10:09
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    $\begingroup$ @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 $\mathbf{R}$). $\endgroup$
    – YCor
    Commented Jan 18, 2020 at 22:24
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    $\begingroup$ @YCor Ah, I thought "closed" here was algebraic, not topological. That does make more sense. $\endgroup$
    – user44191
    Commented Jan 18, 2020 at 23:22
  • $\begingroup$ @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 $({\mathbb C}^*)^n$ $\endgroup$
    – Yemon Choi
    Commented Jan 19, 2020 at 22:41
  • $\begingroup$ 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. $\endgroup$
    – user44191
    Commented Jan 20, 2020 at 9:59

1 Answer 1


My idea in the above comments didn't quite work as stated, but the generality of the result Yemon Choi mentioned bridges the gap.

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 $f^{-1} \in \overline{\{\prod_i g_i^{a_i} f^b | a_i, b \in \mathbb{Z}_{\geq 0}\}} \subseteq S$.

  • $\begingroup$ It seems that this set of products is used 6 times, maybe use a symbol for it. $\endgroup$ Commented Feb 21, 2020 at 10:57
  • $\begingroup$ @MartinBrandenburg Most of the sets are different, because of differing restrictions on the exponents; I thought that defining new sets with different indices would be more complicated (and take longer), so I chose not to. I think only on set is used 3 times. $\endgroup$
    – user44191
    Commented Feb 21, 2020 at 11:46

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