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By a pointed monoid I mean a monoid $M$ together with an absorbing element $0 \in M$ (i.e. $0x=x0=0$). Equivalently, this is a monoid in the monoidal category $(\mathsf{Set}_*,\wedge)$. By a $\pm$-monoid I mean a pointed monoid $M$ together with a decomposition $M = M^+ \cup M^-$ such that $M^+ \cap M^- = \{0\}$, $1 \in M^+$ and $M^{\pm} \cdot M^{\pm} \subseteq M^+$, $M^{\pm} \cdot M^{\mp} \subseteq M^-$. You can imagine $M^+$ as the set of elements $\geq 0$ and $M^-$ as the set of elements $\leq 0$, even though $M$ doesn't have to be ordered in the usual sense.

As a class of examples, take the multiplicative structure of an ordered (semi)ring, for example $\mathbb{N}$ and $\mathbb{Z}$. These have the $\pm$-submonoids $\{0,1\},\{0,-1\},\{0,1,-1\}$. The "generic" example in terms of representation theory is the following: Take any monoid $G$ and consider $E = E^+ \vee_{0} E^-$, where $EM^+$ is the set of endomorphisms of $G$ and $E^-$ is the set of anti-endomorphisms of $G$. Let $0$ be the trivial endomorphism, and $1$ be the identity. The multiplication in $M$ is the composition of functions. Then $E$ is a $\pm$-monoid, which may be called $\mathrm{End}^{\pm}(G)$. A module over a $\pm$-monoid $M$ is a monoid $G$ together with a morphism $M \to \mathrm{End}^{\pm}(G)$ of $\pm$-monoids.

Question. Have these $\pm$-monoids and their modules already been studied in the literature? Or do they have another common name? What is known about them (classification, properties, etc)?

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According to Wikipedia a signed monoid is a monoid $M$ equipped with a map $M\to\mathbb{Z}/2\mathbb{Z}$, so a $\pm$-monoid is just a signed monoid that is pointed. Unfortunately the term "signed monoid" may have just been invented by a Wikipedian. –  Colin McQuillan Nov 15 '12 at 12:40
    
Well, there are several papers that pop up for "signed monoid," and a couple for "signed monoid with zero," if one Googles. –  Kevin Carlson Nov 15 '12 at 13:34
    
I can only find one paper which mentiones a "signed monoid with zero", but there is no definition. A "signed monoid" in these sense of the Wikipedian means a monoid $M$ together with a disjoint decomposition $M = M^+ \cup M^-$ satisfying $M^{\pm} M^{\pm} \subseteq M^+$, $M^{\pm} M^{\mp} \subseteq M^-$ and $1 \in M^+$. If we adjoin a zero, we get a $\pm$-monoid, but only those satisfying $ab=0 \Rightarrow a=0 \vee b=0$. So let's take a pointed monoid $M$ together with a pointed homomorphism $M \to \{0,\pm 1\}$. But this only gives a $\pm$-monoid when $0$ is the only element mapping to $0$. –  Martin Brandenburg Nov 16 '12 at 11:09
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