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Context

In its simplest form, the Grothendieck group construction associates an abelian group to a commutative semigroup in a "universal way".

Now I'm interested in the following nilpotent commutative semigroup $N$ consisting of two elements $a$ and $b$ such that $a^2=b^2=ab=ba=a$. The corresponding Grothendieck group is the trivial group with just one element, which is a bit boring. So I asked myself whether it would be possible to construct a "universal enveloping inverse semigroup"(*) in a similar way as the Grothendieck group, and whether it would be more interesting.

I tried to compute the "universal enveloping inverse semigroup" $N_I$ for the semigroup $N=\{a,b\}$. I got $N_I = \{ a, b, b^{-1}, bb^{-1}, b^{-1}b \}$ with $(b^{-1})^2=ab^{-1}=b^{-1}a=a$. What I find surprising is that $N_I$ is not commutative, even so $N$ is commutative. So I tried to compute the "universal enveloping commutative inverse semigroup" $N_C$ instead and got $N_C = \{ a \}$.

(*)Note: In a semigroup $S$, we say that $y\in S$ is an inverse element of $x\in S$, if $xyx=x$ and $yxy=y$. A semigroup $S$ is called an inverse semigroup, if each $x\in S$ has a unique inverse element $x^{-1}\in S$. It's easy to see that $xx^{-1}$ and $x^{-1}x$ are idempotent, that all idempotent elements in an inverse semigroup commute, and that $(xy)^{-1}=y^{-1}x^{-1}$. So at least superficially, inverse semigroups seem to be nice generalization of groups and can have a zero element without being trivial.

Question

Does the "universal enveloping inverse semigroup" (and the "universal enveloping commutative inverse semigroup") of a semigroup $S$ always exist? I guess the answer is yes and this probably follows from some theorem of universal-algebra. Similarly, I guess that the "universal enveloping regular semigroup" doesn't always exist and wonder whether this also follows from some theorem of universal-algebra.

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2 Answers

up vote 4 down vote accepted

I haven't checked the details of this myself, so I can't tell you the correct answers to your questions, but I would suggest that you try to apply the adjoint functor theorem of category theory:

http://en.wikipedia.org/wiki/Adjoint_functors#General_existence_theorem

To translate the problem into categorical language, let $\mathbf{Smgrp}$ denote the category of semigroups (with semigroup homomorphisms as morphisms) and let $\mathbf{Inv}$ denote the category of inverse semigroups (the morphisms are, again, semigroup homomorphisms; it is easy to see that a morphism of semigroups preserve inverses as you've defined them above). Clearly, $\mathbf{Inv}$ is a subcategory of $\mathbf{Smgrp}$, so there is a forgetful functor $F: \mathbf{Inv} \to \mathbf{Smgrp}$. Since the construction here is called the universal enveloping inverse semigroup of a semigroup, we should expect that the construction constitutes a left adjoint functor to $F$.

To spell out the universal property, this means that if $S_{I}$ is the universal inverse semigroup of the semigroup $S$, then there is a semigroup homomorphism $f: S \to S_{I}$ (formally, a morphism $f: S \to F(S_{I})$) such that whenever $g: S \to H$ ($g: S \to F(H)$ is a semigroup homomorphism with $H$ an inverse semigroup, then there exists a unique semigroup homomorphism $h: S_{I} \to H$ such that $g = h \circ f$ (formally, $g = F(h) \circ f$).

It is now a simple matter of checking that the conditions of the theorem are satisfied; the category $\mathbf{Inv}$ is complete (it has products and equalizers, and hence all (small) limits), so the theorem applies. If you want the semigroups involved to be commutative, simply change the categories appropriately.


More details

Proof sketch that the left adjoint exists: First of all, we should look at limits in $\mathbf{Smgrp}$ and $\mathbf{Inv}$; they turn out to be "the same": if $\{ S_{i} \}_{i \in I}$ is some family of semigroups, the product is simply the cartesian product of the underlying sets, with the obvious "pointwise" operation. If all semigroups are inverse semigroups, the result is an inverse semigroup. You can easily check that it has the requisite universal property for a product, see

http://en.wikipedia.org/wiki/Product_(category_theory)

If $f,g: S_1 \to S_2$ are semigroup homomorphisms, their equalizer is $$E = \{ s \in S_1 \,|\, f(s) = g(s) \}$$ together with the inclusion homomorphism $e: E \hookrightarrow S_1$. $E$ is a semigroup in the obvious way, and is an inverse semigroup if $S_1$ and $S_2$ are (if $x \in E$ have inverse $y \in S_1$, we have $f(x) = g(x)$, and it follows that both $g(y)$ and $f(y)$ is the inverse of $f(x) = g(x)$; by uniqueness, $f(y) = g(y)$ so $y \in E$ also). You can again check that $(E,e)$ has the requisite universal property, see

http://en.wikipedia.org/wiki/Equalizer_(mathematics)

Hence both $\mathbf{Smgrp}$ and $\mathbf{Inv}$ are (small-) complete categories, see http://en.wikipedia.org/wiki/Limit_(category_theory) (under Existence of limits).

We get that $F$ preserves limits for free, since limits are constructed in exactly the same manner in both categories, and the morphisms are the same. What remains is the solution set condition. Fix a semigroup $S$, and consider a semigroup homomorphism $f: S \to F(H)$, for some inverse semigroup $H$. The idea is to take the solution set to be the isomorphism classes of inverse semigroups generated by $f(S)$ for some $f$. This will be a set if the cardinality of the inverse semigroup $\langle f(s) \rangle \subseteq H$, generated by $f(S) \subseteq H$ is bounded, for a given $S$. This appears to be the case here, since you can start with $G = f(S) \cup f(S)^{-1}$ (where $f(S)^{-1}$ denotes the set of inverses in $H$ for elements in $f(S)$), and consider all finite products of elements in $G$. This will be an inverse semigroup (because, as you mentioned, the formula $(xy)^{-1} = y^{-1}x^{-1}$ holds). You still need to check that this actually is a solution set, but that should be reasonably easy.

Note that this only answers the question of existence, it doesn't provide an explicit construction. My guess is that coming up with one isn't going to be too difficult (and it appears that you already have some ideas in that direction, at least for specific examples), perhaps the construction of the Grothendieck group can be emulated.    

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Thanks. I will now have to read a bit in order to understand what a "complete category" (or an equalizer) is. Then I can check for each of my three categories (regular semigroups, inverse semigroups and commutative inverse semigroups) whether it is complete. I also wonder what to do with the "regular semigroups" in case it turns out they are not a "complete category". (I hoped for a theorem related to the way I computed the "universal enveloping inverse semigroup" for my example by repeatedly using the "algebraic identities" at my disposal. This method doesn't work for regular semigroups.) –  Thomas Klimpel Feb 2 '12 at 12:58
    
@thomasklimpel: I've added a sketch proof of the existence of a universal enveloping inverse semigroup, which will hopefully be helpful to you. As for regular semigroups, I think you are correct in that they might pose a problem. To get equalizers in the category Inv, I had to use uniqueness of inverses (which, I gather, is not to be expected for regular semigroups). This doesn't necessarily mean they don't exist for regular semigroups, just that you probably can't construct them in the same, straightforward way as above. –  Martin Wanvik Feb 2 '12 at 13:45
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I now found out how to prove that no "universal enveloping regular semigroup" exists for the example given in the question. (The existence of the other two cases has already been proved in the answer by Martin Wanvik.)

Let $a=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, $b=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $b'=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, $c=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$, $c'=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$, $d=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$, $d'=\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$, $e=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, and $f=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. Let $N := \{ a,b \}$ and $N_I := \{ a,b,b',e,f \}$ (as in the question with $b'=b^{-1}$, $e=bb^{-1}$ $f=b^{-1}b$). It is easy to see that $S := \{ a,b,b',c,c',d,d',e,f \}$ is a regular group, and that $N_I$ and $R := \{ a,b,c,d,e \}$ are regular sub-semigroups of $S$. Now $b^{-1}=b'$ in $N_I$ and $c$ is the unique inverse element of $b$ in $R$. So if there would exist a regular semigroup $N_R$ containing $N$ as sub-semigroup, for which it is possible to uniquely extend any homomorphism from $N$ to $N_I$ or $R$, then the extension to $N_R$ of a homomorphism $h$ from $N$ to $S$ (with $h(a)=a$ and $h(b)=b$) won't be unique.

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