When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:02:23Z http://mathoverflow.net/feeds/question/108643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108643/when-a-semigroup-extension-s-times-a-of-a-semigroup-s-with-a-commutative-gro When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? Marius Buliga 2012-10-02T15:13:26Z 2012-10-02T18:10:04Z <p>Maybe is a trivial question, but I don't know how to handle it. </p> <p><em><strong>Setting:</em></strong> Let $S$ be a semigroup (i.e. has an associative operation with neutral element $e$) and let $(A,+)$ be a commutative group (with neutral element $0$). </p> <p>A semigroup extension of $S$ with $A$ is any operation on $S \times A$, of the form </p> <p>$(s,a)(s',a') = (s s' , a+a'+ \lambda(s,s')$ </p> <p>where $\lambda: S \times S \rightarrow A$ is a function such that $S \times A$ with the mentioned operation is a semigroup with neutral element $(e,0)$. Obviously, the operation is encoded by the function $\lambda$, which satisfies certain equations. </p> <p><em><strong>Problem</em></strong>: Describe the class of functions $\lambda$ with the property: there is an injective morphism of semigroups from $S$ to $S \times A$ (with the semigroup extension operation induced by $\lambda$). Is there an elegant way to describe this class? </p> <p><em><strong>Motivation:</em></strong> I am puzzled by the following two examples, I cannot put the finger on the essential difference between those. </p> <p>Let $X$ and $Y$ be topological, locally convex, real vector spaces of dual variables $x \in X$ and $y \in Y$, with the duality product $\langle \cdot , \cdot \rangle : X \times Y \rightarrow \mathbb{R}$. </p> <p>The spaces $X, Y$ have topologies compatible with the duality product, in the sense that for any continuous linear functional on $X$ there is an $y \in Y$ which puts the functional into the form $x \mapsto \langle x,y\rangle$ (respectively any continuous linear functional on $Y$ has the form $y \mapsto \langle x,y\rangle$, for a $x \in X$). </p> <p><em><strong>Example 1:</em></strong> (Heisenberg group) Let $S = X \times Y$ with the operation of addition of pairs of vectors and let $A = \mathbb{R}$ with addition. We may define a Heisenberg group over the pair $(X,Y)$ as $H(X,Y) = S \times A$ with the operation </p> <p>$(x,y,a)(x',y',b) = (x+x', y+y', a+a'+ \langle x, y'\rangle - \langle x', y \rangle)$</p> <p>There is no injective morphism from $(X\times Y, +)$ to $H(X,Y)$. </p> <p><em><strong>Example 2:</em></strong> Let this time $S = (X \times Y)^{*}$, the free semigroup generated by $X \times Y$, i.e. the collection of all finite words with letters from $X \times Y$, together with the empty word $e$, with the operation of concatenation of words. </p> <p>Let $A$ be the set of bi-affine real functions on $X \times Y$, i.e. the collection of all functions $a: X \times Y \rightarrow \mathbb{R}$ which are affine and continuous in each argument. $A$ is a commutative group with the addition of real valued functions operation. </p> <p>We define the function $\lambda: S \times S \rightarrow A$ by: </p> <ul> <li>$\lambda(e, c)(x,y) = \lambda(c,e)(x,y)=0$ for any $c \in S$ and any $(x,y) \in X \times Y$. </li> <li>if $c, h \in S$ are words $c = (x_{1},y_{1})...(x_{n}, y_{n})$ and $h = (u_{1},v_{1})...(u_{m}, v_{m})$, with $m,n \geq 1$, then </li> </ul> <p>$\lambda(c,h)(x,y) = \langle u_{1} - x , y_{n} - y \rangle$ for any $(x,y) \in X \times Y$. </p> <p>This $\lambda$ induces a semigroup extension operation on $S \times A$ and there is an injective morphism $F: S \rightarrow S \times A$, with the expression $F(c) = (c, E(c))$. Here, for any $c = (x_{1},y_{1})...(x_{n}, y_{n})$ the expression $E(c)(x,y)$ is the well known circular sum associated to the "discrete cycle" $(x_{1},y_{1})...(x_{n}, y_{n})(x,y)$, namely: </p> <p>$E(c)(x,y) = \langle x_{1},y\rangle + \langle x,y_{n}\rangle - \langle x,y \rangle + \sum_{k=1}^{n-1}\langle x_{k+1}, y_{k}\rangle - \sum_{k=1}^{n} \langle x_{k},y_{k} \rangle$</p> <p>dear to convex analysis. </p> http://mathoverflow.net/questions/108643/when-a-semigroup-extension-s-times-a-of-a-semigroup-s-with-a-commutative-gro/108644#108644 Answer by Benjamin Steinberg for When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? Benjamin Steinberg 2012-10-02T15:25:03Z 2012-10-02T15:25:03Z <p>First of all I would not call what you write a semidirect product. It will be such in your setting precisely when the natural maps to $S$ splits, in which case it will actually be isomorphic to a direct product.</p> <p>More precisely, you are viewing $A$ here as a trivial module over $\mathbb ZS$. Then $\lambda$ is a 2-cocycle and you want it to be equivalent in $H^2(S,A)$ with the trivial cocycle.</p>