Small doubling in ordered semigroups

Abstract

Let $\mathbb {A} = (A, \cdot )$ be a semigroup. We generalize some recent results by Freiman, Herzog and coauthors on the structure theory of set addition from the context of linearly orderable groups to linearly orderable semigroups, where we say that $\mathbb {A}$ is linearly orderable if there exists a total order $\le$ on $A$ such that $xz < yz$ and $zx < zy$ for all $x,y,z \in A$ with $x < y$. In particular, we find that if $S$ is a finite subset of $A$ generating a non-abelian subsemigroup of $\mathbb {A}$, then $|S^2| \ge 3|S|-2$. On the road to this goal, we also prove a number of subsidiary results, and most notably that for $S$ a finite subset of $A$ the commutator and the normalizer of $S$ are equal to each other.

Appendix 1: Examples

We conclude the paper with a few examples. As mentioned in the introduction, the basic goal is to show that [linearly] orderable semigroups and related structures are far from being “exotic”.

We start with an orderable semigroup which is not linearly orderable. Next, we mention some notable classes of linearly orderable groups and a linearly orderable monoid which is not a linearly orderable group (we do not know if it embeds into a linearly ordered semigroup).

Example 4.1

Every set $A$ can be turned into a semigroup by the operation $\cdot {}: A \times A \rightarrow A: (a,b) \rightarrow a$; see, for instance, [8, p. 3]. Trivially, if $\le$ is a total order on $A$ then $(A, \cdot {}, \le )$ is a totally ordered semigroup. However, $(A, \cdot {})$ is not linearly orderable for $|A| \ge 2$ (e.g., because it is not cancellative).

Example 4.2

An interesting variety of linearly orderable groups is provided by abelian torsion-free groups, as first proved by Levi in [11], and the result can be, in fact, extended to abelian cancellative torsion-free semigroups with no substantial modification; see the comments following Remark 1 in Sect. 3 and Corollary 3.4 in Gilmer’s book on commutative semigroup rings [5].

In a similar vein, Iwasawa [9], Malcev [13] and Neumann [15] established independently that torsion-free nilpotent groups are linearly orderable.

Save for the semigroup analogue of Levi’s result, all of the above is already mentioned in [3], where the interested reader can find further references to existing literature on the subject. Two more examples (of linearly orderable groups) which are not included in [3] are pure braid groups [16] and free groups [9].

Example 4.3

As for linearly orderable monoids which are not linearly orderable groups, consider, for instance, the free monoid [8, Sect. 1.6] on a well-ordered alphabet $(X, \le )$ together with the “shortlex ordering”: Words are primarily sorted by length, with the shortest ones first, and words of the same length are then sorted into lexicographical order.

The next example seems interesting per se. Not only it gives a family of linearly ordered semigroups which are neither abelian nor groups (at least in general); it also shows that, for each $n$, certain subsemigroups of $\mathrm{GL}_n(\mathbb R)$ consisting of triangular matrices are linearly orderable.

Example 4.4

We let a semiring be a triple $(A,+,\cdot )$ consisting of a set $A$ and associative operations $+$ and $\cdot$ from $A \times A$ to $A$ (referred to, respectively, as the semiring addition and multiplication) such that

• 1. $(A,+)$ is an abelian monoid, whose identity we denote by $0$;

• 2. $0$ annihilates $A$, that is $0 \cdot a = a \cdot 0 = 0$ for every $a \in A$;

• 3. multiplication distributes over addition, that is $a(b+c) = ab + ac$ and $(a + b) c = ac + b c$ for all $a,b,c \in A$.

(In other words, a semiring is just a ring where elements do not need have an additive inverse.) We call $(A, +)$ and $(A, \cdot )$, respectively, the additive monoid and the multiplicative semigroup of $(A,+,\cdot )$, which in turn is termed a unital semiring if $(A, \cdot )$ is a monoid too; see [7, Ch. II] and [6, Ch. 1, p. 1].

A semiring $(A, +, \cdot )$ is said to be orderable if there exists a (total) order $\le$ on $A$ such that $(A, +, \le )$ and $(A, \cdot , \le )$ are ordered semigroups, in which case $(A, +, \cdot , \le )$ is referred to as an ordered semiring. If, on the other hand, the following hold:

• 4. $(A, +, \le )$ is a linearly ordered monoid;

• 5. $a c < b c$ and $c a < c b$ for all $a,b,c \in A$ with $a < b$ and $0 < c$,

then $(A, +, \cdot )$ is said to be linearly orderable and $(A, +, \cdot , \le )$ is called a linearly ordered semiring; cf. [6, Ch. 20]. Common examples of linearly ordered semirings are the [non-negative] integers, the [non-negative] rational numbers, and the [non-negative] reals with their usual addition, multiplication, and order.

With that said, let $\mathbb A = (A, +, \cdot )$ be a fixed semiring. We write $\mathcal M_n(A)$ for the set of $n$-by-$n$ matrices with entries in $A$. Endowed with the usual operations of entry-wise addition and row-by-column multiplication induced by the structure of $\mathbb A$, here respectively denoted by the same symbols as the addition and multiplication of the latter, $\mathcal M_n(A)$ becomes itself a semiring, which we call the semiring of $n$-by-$n$ matrices over $\mathbb A$ and write as $\mathcal M_n(\mathbb A)$; see [6, Ch. 3].

Suppose now that $\mathbb A$ is linearly ordered by a certain order $\le$, in such a way that $\mathbb A_\sharp = (A, +, \cdot , \le )$ is a linearly ordered semiring, and denote by $\mathrm{U}_n(\mathbb A_\sharp ^+)$ the subsemigroup of the multiplicative semigroup of $\mathcal M_n(\mathbb A)$ consisting of all upper triangular matrices whose entries on or above the main diagonal belong to

$$\begin{aligned} \mathbb A_\sharp ^+ = \{a \in A: a > 0\}. \end{aligned}$$

We observe that $\mathrm{U}_n(\mathbb A_\sharp ^+)$ is not a group (and not even a monoid) for $n \ge 2$. But what is perhaps more interesting is the following:

Theorem 3

$\mathrm{U}_n(\mathbb A_\sharp ^+)$ is a linearly orderable semigroup.

Proof

Set $I_n = \{1, 2, \ldots , n\}, \Xi _n = \{(i,j) \in I_n \times I_n: i \le j\}$ and define a binary relation $\le _n$ on $\Xi _n$ by $(i_1, j_1) \le _n (i_2, j_2)$ if and only if (i) $j_1 - i_1 < j_2 - i_2$ or (ii) $j_1 - i_1 = j_2 - i_2$ and $j_1 < j_2$. It is seen that $\le _n$ is a well-order, so we can define a binary relation $\le _{n,\mathrm U}$ on $\mathrm{U}_n( \mathbb A_\sharp ^+)$ by taking, for $\alpha = (a_{i,j})_{i,j=1}^n$ and $\beta = (b_{i,j})_{i,j=1}^n$ in $\mathrm{U}_n(\mathbb A_\sharp ^+), \alpha \le _{n,\mathrm U} \beta$ if and only if (i) $\alpha = \beta$ or (ii) there exists $(i_0,j_0) \in \Xi _n$ such that $a_{i_0,j_0} < b_{i_0,j_0}$ and $a_{i,j} = b_{i,j}$ for all $(i,j) \in \Xi _n$ with $(i,j) <_n (i_0, j_0)$.

It is straightforward that $\le _{n,\mathrm U}$ is an order. To see, in particular, that it is total: Pick $\alpha = (a_{i,j})_{i,j=1}^n$ and $\beta = (b_{i,j})_{i,j=1}^n$ in $\mathrm{U}_n(\mathbb A_\sharp ^+)$ with $\alpha \ne \beta$. There then exists $(i_0, j_0) \in \Xi _n$ such that $a_{i_0,j_0} \ne b_{i_0,j_0}$, where $(i_0, j_0)$ is chosen in such a way that $a_{i,j} = b_{i,j}$ for every $(i,j) \le _n (i_0, j_0)$. Since $\le$ is total, we have that either $\alpha <_{n,\mathrm U} \beta$ if $a_{i_0,j_0} < b_{i_0,j_0}$ or $\beta <_{n,\mathrm U} \alpha$ otherwise, and we are done.

It remains to prove that $\mathrm{U}_n(\mathbb A_\sharp ^+)$ is linearly ordered by $\le _{n,\mathrm U}$. For, let $\alpha$ and $\beta$ be as above and suppose $\alpha <_{n,\mathrm U} \beta$, viz there exists $(i_0, j_0) \in \Xi _n$ with $a_{i_0,j_0} < b_{i_0,j_0}$ and $a_{i,j} = b_{i,j}$ for all $(i,j) \in \Xi _n$ with $(i,j) <_n (i_0, j_0)$. Given $\gamma = (c_{i,j})_{i,j=1}^n$ in $\mathrm{U}_n(\mathbb A_\sharp ^+)$ we then have $a_{i,k} c_{k,j} \le b_{i,k} c_{k,j}$ and $c_{i,k} a_{k,j} \le c_{i,k} b_{k,j}$ for all $(i,j) \in \Xi _n$ and $k \in I_n$ such that $(i,k) \le _n (i_0, j_0)$ and $(k,j) \le _n (k,j_0)$, and in fact $a_{i_0,j_0} c_{j_0,j_0} < b_{i_0,j_0} c_{j_0,j_0}$ and $c_{i_0,i_0} a_{i_0,j_0} < c_{i_0,i_0} b_{i_0,j_0}$ since $(A, +, \cdot , \le )$ is a linearly ordered semiring. It follows that, for all $(i,j) \in \Xi _n$ with $(i,j) \le _n (i_0, j_0)$,

$$\begin{aligned} \sum _{k=1}^n a_{i,k} c_{k,j} = \sum _{k=i}^j a_{i,k} c_{k,j} \le \sum _{k=i}^j b_{i,k} c_{k,j} = \sum _{k=1}^n b_{i,k} c_{k,j} \end{aligned}$$

and, similarly, $\sum _{k=1}^n c_{i,k} a_{k,j} \le \sum _{k=1}^n c_{i,k} b_{k,j}$. In particular, these majorations are equalities for $(i,j) <_n (i_0, j_0)$ and strict inequalities if $(i,j) = (i_0, j_0)$. So $\alpha \gamma <_{n,\mathrm U} \beta \gamma$ and $\gamma \alpha <_{n,\mathrm U} \gamma \beta$, and the proof is complete.

We refer to the order $\le _{n,\mathrm U}$ defined in the proof of Theorem 3 as the zig–zag order on $\mathrm{U}_n(\mathbb A_\sharp ^+)$. If $\mathrm{L}_n(\mathbb A_\sharp ^+)$ is the subsemigroup of the multiplicative semigroup of $\mathcal M_n(\mathbb A)$ consisting of all lower triangular matrices whose entries on or below the main diagonal are in $\mathbb A_\sharp ^+$, it is then easy to see that $\mathrm{L}_n(\mathbb A_\sharp ^+)$ is itself linearly orderable: It is, in fact, linearly ordered by the binary relation $\le _{n,\mathrm L}$ defined by taking $\alpha \le _{n,\mathrm L} \beta$ if and only if $\alpha ^\top \le _{n,\mathrm U} \beta ^\top$, where the superscript ‘$\top$’ stands for ‘transpose’. If $\mathrm{T}_n(\mathbb A_\sharp ^+)$ is the subsemigroup of $(\mathcal M_n(A), \cdot )$ generated by $\mathrm{U}_n(\mathbb A_\sharp ^+)$ and $\mathrm{L}_n(\mathbb A_\sharp ^+)$, it is hence natural to ask the following:

Question 2

Is $\mathrm{T}_n(\mathbb A_\sharp ^+)$ a linearly orderable semigroup?

While at present we do not have an answer to this, it was remarked by Carlo Pagano (Università di Roma Tor Vergata, Italy) in a private communication that $\mathcal M_n(\mathbb A_\sharp ^+)$, namely the subsemigroup of $(\mathcal M_n(A), \cdot )$ consisting of all matrices with entries in $\mathbb A_\sharp ^+$, is not in general linearly orderable: For a specific counterexample, let $\mathbb A_\sharp$ be the real field together with its usual order, and take as $\alpha$ the $n$-by-$n$ real matrix whose entries are all equal to $1$ and as $\beta$ any $n$-by-$n$ matrix with positive real entries each of whose columns has sum equal to $n$. Then $\alpha ^2 = \alpha \beta$.

Apparently, the question has not been addressed before by other authors, although the ordering of $\mathcal M_n(\mathbb A)$, in the case where $\mathbb A$ is a partially orderable semiring, is considered in [6, Example 20.60].

Example 4.5

In what follows, we let $\mathbb K = (K, +, \cdot )$ be a semiring (see Example 4.4 for the terminology) and $\mathbb A = (A, \diamond )$ a semigroup, and use $K[A]$ for the set of all functions $f: A \rightarrow K$ such that $f$ is finitely supported in $\mathbb K$, namely $f^{-1}(0_K)$ is a finite subset of $A$, where $0_K$ is the additive identity of $\mathbb K$.

In fact, $K[A]$ can be turned into a semiring, here written as $\mathbb K[\mathbb A]$, by endowing it with the operations of pointwise addition and Cauchy product induced by the structure of $\mathbb A$ and $\mathbb K$ (these operations are denoted below with the same symbols as the addition and the multiplication of $\mathbb K$, respectively). We have:

Theorem 4

Suppose $\mathbb K$ is a linearly orderable semiring and $\mathbb A$ is a linearly orderable semigroup. Then $\mathbb K[\mathbb A]$ is a linearly orderable semiring too.

Proof

The claim is obvious if $A = \emptyset$, so assume that $A$ is non-empty, and let $\le _K$ and $\le _A$ be, respectively, orders on $A$ and $K$ for which $(K, +, \cdot , \le _K)$ is a linearly ordered semiring and $(A, \diamond , \le _A)$ a linearly ordered semigroup.

Then, given $\alpha \in A$ and $f \in K[A]$, we let $f_{\downarrow \alpha }$ (respectively, $f_{\uparrow \alpha }$) be the function $A \rightarrow K$ taking $a$ to $f(a)$ if $a <_A \alpha$ (respectively, $\alpha \le _A a$), and to $0_K$ otherwise, in such a way that $f = f_{\downarrow \alpha } + f_{\uparrow \alpha }$. Also, we denote by $\mu$ the map $K[A] \times K[A] \rightarrow A \cup \{A\}$ sending a pair $(f,g)$ to $\min \{a \in A: f(a) \ne g(a)\}$ if $f \ne g$ (the minimum is taken with respect to $\le _A$, and it exists by consequence of the definition itself of $K[A]$), and to $A$ otherwise.

We define a binary relation $\le$ on $K[A]$ by letting $f \le g$ if and only if either $f = g$ or $f \ne g$ and $f(\mu (f,g)) <_K f(\mu (f,g))$. It is clear that $\le$ is a total order on $K[A]$, and we want to prove that it is also compatible with the algebraic structure of $\mathbb K[\mathbb A]$, in the sense that $\mathbb K[\mathbb A]$ is linearly ordered by $\le$.

For, pick $f,g,h \in K[A]$ with $f < g$. Since the additive monoid of $\mathbb K$ is linearly ordered by $\le _K$, we have $\mu (f, g) = \mu (f + h, g + h)$, and thus $f + h < g + h$. That is, $(K[A], +, \le )$ is a linearly ordered monoid in its own right. On another hand, assume $\Theta < h$, where $\Theta$ is the function $A \rightarrow K: a \mapsto 0_K$, and set $\alpha = \mu (f,g)$ and $\beta = \mu (\Theta , h)$. We have $f_{\downarrow \alpha } = g_{\downarrow \alpha }$ and $h = h_{\uparrow \beta }$, with the result that $f h < g h$ if and only if $f_{\uparrow \alpha } h_{\uparrow \beta } < g_{\uparrow \alpha } h_{\uparrow \beta }$, and the latter inequality is certainly true, since on the one side $f_{\uparrow \alpha } h_{\uparrow \beta }(a) = g_{\uparrow \alpha } h_{\uparrow \beta }(a) = 0_K$ for $a <_A \alpha \diamond \beta$, and on the other side

$$\begin{aligned} f_{\uparrow \alpha } h_{\uparrow \beta }(\alpha \diamond \beta ) = f_{\uparrow \alpha }(\alpha ) h_{\uparrow \beta }(\beta ) <_K g_{\uparrow \alpha }(\alpha ) h_{\uparrow \beta }(\beta ) = g_{\uparrow \alpha } h_{\uparrow \beta }(\alpha \diamond \beta ). \end{aligned}$$

In a similar way, it is seen that $h f < h g$. So, by the arbitrariness of $f, g$, and $h$, we get that $(K[A], +, \cdot , \le )$ is a linearly ordered semiring.

So taking $\mathbb A$ to be the free commutative monoid (respectively, the free monoid) on a certain set and recalling that free groups (and hence free monoids) are linearly orderable (Example 4.2), we have:

Corollary 2

The semiring $\mathbb K$ is linearly orderable if and only if the same is true for the semiring of polynomials over $\mathbb K$ depending on a given set of pairwise commuting (respectively, non-commuting) variables.