Thompson's group F and monoidal categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:15:58Z http://mathoverflow.net/feeds/question/100802 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100802/thompsons-group-f-and-monoidal-categories Thompson's group F and monoidal categories AlexPof 2012-06-27T19:56:43Z 2012-06-29T09:16:37Z <p>(This is a cross-post from MathSE, as someone remarked that the question would be more appropriate on MO)</p> <p><a href="http://arxiv.org/pdf/math/0508617v2.pdf" rel="nofollow">Fiore and Leinster</a> have proved that if $\mathcal{A}$ is a monoidal category freely generated by one object $A$ and an isomorphism $\alpha: A \otimes A \to A$, then for every object $X \in \mathcal{A}, Aut(X)$ is isomorphic to the Thompson group $F$.</p> <p>My question is the following: if we assume instead that $\alpha: A \otimes A \to A$ is not necessarily an isomorphism, and that there exist a morphism $\beta: A \to A \otimes A$ such that $\alpha \circ \beta = id$, is the result of Fiore and Leinster still true ? </p> <p>I have a feeling we at least have $F \subset Aut(X)$. Loosely speaking, my approach is that since every element of $F$ can be represented as a pair $(R,S)$ of forests, we can always represent $R$ by a suitable composition of $\beta$ maps, then $S$ by a composition of $\alpha$ maps, the identity $\alpha \circ \beta = id$ ensuring that every facing caret gets cancelled to form a reduced forest diagram, i.e a unique element of $F$.</p> http://mathoverflow.net/questions/100802/thompsons-group-f-and-monoidal-categories/100810#100810 Answer by Buschi Sergio for Thompson's group F and monoidal categories Buschi Sergio 2012-06-27T21:23:46Z 2012-06-29T09:16:37Z <p><strong>Edit</strong> I noticed that in Fiore-Leinster preliminate the condition (free monoidal category of an isomorphism $\alpha: A \otimes A \to A$) is different from what is written in the preliminary question, so I reworked my answer substantially.</p> <p>In a Monoidal category $\mathcal{C}$ consider (a non empty) class of sections of the type $\beta: A\to A\otimes A$ and let $\Sigma$ its tensor product closure (finite tensor products of some morphisms of type $\beta$ of the choose class and some identities).</p> <p>From the article "Note on monoidal localizations " by B. Day (<a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;aid=4865952" rel="nofollow">link text</a>) the category of fraction $\mathcal{C}_\Sigma$ is (naturally) a monoidal category.</p> <p>let $P: \mathcal{C}\to \mathcal{C}_\Sigma$ the natural functor.</p> <p>The elements of $\Sigma$ are all monomorphisms (are sections) , and if $\Sigma$ admits a calculus of left fractions the canonical functors $P$ is faithful (see "Categories" H Shubert, 12.9.6(a), p.261). THen $\mathcal{C}$-$Aut(X)$ is a subgroup of $\mathcal{C}_\Sigma$-$Aut(X)$ (because $P$ is faithful).</p> <p>Now consider the Monoidal category $[A, \alpha, \beta]$ free on (the condition):</p> <p>"one object $A$ and on two morphisms $\alpha: A\otimes A\to A$, $\beta: A\to A\otimes A$, with $\alpha\circ \beta=1_A$". </p> <p>This category has the following universal property: for any monoidal categories $\mathcal{C}$ with choose morphisms $a: X\otimes X\to X,\ b: X\to X\otimes X$ with $a\circ b=1_X$ there exists a unique strict monoidal functor $F_{a,b}: [A, \alpha, \beta]\to \mathcal{C}$ with $F(\alpha)=a,\ F(\beta)=b$.</p> <p>Now in $[A, \alpha, \beta]$ consider the tensor closure $\Sigma$ of the section $\beta$,</p> <p>and let $P:[A, \alpha, \beta]\to [A, \alpha, \beta]_\Sigma$ the category of fractions.</p> <p>the category $[A, \alpha, \beta]_\Sigma$ has the universal property of the monoidal category on one isomorphisms</p> <p>$\beta: A\to A\otimes A$ as in the FIore-Leinster article, then $F\cong [A, \alpha, \beta]_\Sigma$-$Aut(A)$. </p> <p>Now, <strong>IF</strong> $\Sigma$ admit a calculus of left fraction then $P$ is faithful and $[A, \alpha, \beta]$-$Aut(A)$ is isomorphic to a subgroup of $F$.</p> <p>P.S. I seems that $\Sigma$ admit a calculus of left fraction, but I have not checked it in detail </p>