q-deformation of the permutation group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:29:01Z http://mathoverflow.net/feeds/question/96840 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group q-deformation of the permutation group? John Mangual 2012-05-13T16:22:25Z 2012-11-23T09:47:31Z <p>The only definition of a quantum group I know of involves q-deforming the relation $EF-FE=H$ or for SL(2): $\left[ \left( \begin{array}{cc} 0 &amp; 1 \\ 0 &amp; 0 \end{array} \right), \left( \begin{array}{cc} 0 &amp; 0 \\ 1 &amp; 0 \end{array} \right) \right] = \left( \begin{array}{cr} 1 &amp; 0 \\ 0 &amp; -1 \end{array} \right)$</p> <p>All the axioms I have seen are <a href="http://math.mit.edu/~ptingley/QuantumGroupsSpring2011/lecture1.pdf" rel="nofollow">very confusing</a> and don't help me with much. I also get the sense, these should be called 'quantum lie algebras' rather than quantum groups. And I never understood the point of co-commutativity. <hr> For now, what does a q-deformation of the permutation group look like? Or the dihedral group?</p> http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/96844#96844 Answer by James Griffin for q-deformation of the permutation group? James Griffin 2012-05-13T17:07:18Z 2012-05-13T17:07:18Z <p>The algebras you are looking for are called <em>Iwahori-Hecke algebras</em>. In the case of the symmetric groups the Iwahori-Hecke algebras are generated by `transpositions' $T_i$ which satisfy the braid relations but don't square to zero; instead there is a relation which looks like $$T_i^2 = qT_i + (1-q)$$</p> <p>I'd recommend you read up on the monoidal category of modules for a Hopf algebra. The various properties of a Hopf algebra determine properties of its module category. For instance if the coproduct is cocommutative then the category of modules is symmetric monoidal. Many of these q-deformations aren't cocommutative but their module categories still have structure, they become braided monoidal.</p> <p>Once I understood this everything became much clearer, good luck!</p> http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/96877#96877 Answer by Leandro Vendramin for q-deformation of the permutation group? Leandro Vendramin 2012-05-14T07:36:35Z 2012-05-14T07:36:35Z <p>In </p> <ul> <li>Wang, Shuzhou. Quantum symmetry groups of finite spaces. Comm. Math. Phys. 195 (1998), no. 1, 195--211. MR1637425 (99h:58014), <a href="http://www.springerlink.com/content/e9mpjtmc25mfm0a2/" rel="nofollow">link</a></li> </ul> <p>a quantum version of the symmetric group $\mathbb{S}_n$ is defined. </p> <p>Let me sketch Wang's construction. </p> <p>Let $u_{ij}$ be the characteristic function of the set of $\sigma\in\mathbb{S}_n$ such that $\sigma(j)=i$. </p> <p>Assume that all entries $u_{ij}$ are projections, and on each row and column of $u=(u_{ij})$ these projections are orthogonal, and sum up to $1$. Then the commutative $C^*$-algebra generated by these $u$ is $C(\mathbb{S}_n)$.</p> <p>Now drop the commutativity condition and let $A_s(n)$ be the $C^*$-algebra generated by all the $u_{ij}$. Then we have a <em>quantum</em> analogue of $\mathbb{S}_n$. </p> <p>It turns out that $A_s(n)$ is a finitely generated Hopf algebra.</p> <p>The group $\mathbb{S}_n$ acts on an set $X=[1,2,...,n]$ with $|X|=n$. The corresponding action map $(i,\sigma)\mapsto \sigma(i)$ gives by transposition a certain morphism $\alpha$ ($\alpha$ is called <em>coaction</em>). This coaction can be expressed as $\alpha(\delta_i)=\sum\delta_j\otimes u_{ji}$. Furthermore, $\alpha$ is a sort of universal coaction. </p> <p>It is possible to prove that the following diagram is commutative $$\begin{array}{ccc} C(X) &amp; \to &amp; C(X)\otimes A_s(n)\\ \downarrow &amp; &amp; \downarrow\\ C(X) &amp; \to &amp; C(X)\otimes C(\mathbb{S}_n) \end{array}$$</p> <p>Furthermore, $C(\mathbb{S}_n)=A_s(n)$ if $n=1,2,3$. For $n\geq4$, $A_s(n)$ is not commutative and infinite dimensional.</p> <p>For a nice survey about quantum permutation groups and some applications see the following paper:</p> <ul> <li>Banica, Teodor; Bichon, Julien; Collins, Benoît. Quantum permutation groups: a survey. Noncommutative harmonic analysis with applications to probability, 13--34, Banach Center Publ., 78, Polish Acad. Sci. Inst. Math., Warsaw, 2007. MR2402345 (2009f:46094), <a href="http://arxiv.org/abs/math/0612724" rel="nofollow">link</a></li> </ul> <p>For a quantum version of the automorphism group of finite graphs (and a quantum version of the dihedral group $\mathbb{D}_4$):</p> <ul> <li>Bichon, Julien. Quantum automorphism groups of finite graphs. Proc. Amer. Math. Soc. 131 (2003), no. 3, 665--673 (electronic). MR1937403 (2003j:16049), <a href="http://www.ams.org/journals/proc/2003-131-03/S0002-9939-02-06798-9/home.html" rel="nofollow">link</a></li> </ul> <p>A complete classification of quantum permutation groups acting on 4 points was given in:</p> <ul> <li>Banica, Teodor; Bichon, Julien. Quantum groups acting on 4 points. J. Reine Angew. Math. 626 (2009), 75--114. MR2492990 (2010c:46153), <a href="http://arxiv.org/abs/math/0703118" rel="nofollow">link</a></li> </ul> http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/114228#114228 Answer by Aakumadula for q-deformation of the permutation group? Aakumadula 2012-11-23T09:47:31Z 2012-11-23T09:47:31Z <p>Consider the braid group $B_n$ on $n$ strands. This is generated by $s_1,\cdots, s_{n-1}$ with relations $s_is_j=s_js_i$ if $\mid i-j\mid \geq 2$ and $s_is_js_i=s_js_is_j$ otherwise. The group $B_n$ has a representation called the Burau representation with parameter $q$ given on the generators $s_i$ as follows. If $e_1\cdots ,e_{n-1}$ is the standard basis of ${\mathbb Z}^{n-1}$ then $s_i(e_i)=(-q)e_i$, $s_i(e_{i+1})= e_{i+1}+e_i$ and $s_i(e_{i-1})=e_{i-1}+qe_i$. When you specialise $q=1$, the image is the symmetric group $S_n$. So you may think of the Burau representation at the parameter $q$ as a defomration of the permutation group $S_n$. </p>