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The only definition of a quantum group I know of involves q-deforming the relation $EF-FE=H$ or for SL(2): \[ \left[ \left( \begin{array}{cc} 0 & 1 \\\\ 0 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\\\ 1 & 0 \end{array} \right) \right] = \left( \begin{array}{cr} 1 & 0 \\\\ 0 & -1 \end{array} \right) \]

All the axioms I have seen are very confusing 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.

For now, what does a q-deformation of the permutation group look like? Or the dihedral group?

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  • $\begingroup$ With regards to your comment before your question. A very nice introduction to quantum groups can be found here: math.ucdavis.edu/~rthomas/quantumgroups.pdf . This does a nice job of comparing the universal enveloping algebra of sl2 with its quantized enveloping algebra. This allows one to see exactly how the relations change. $\endgroup$
    – B. Bischof
    May 13, 2012 at 18:01
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    $\begingroup$ As James answered Hecke algebra is standard answer. However it is an answer in some informal way. If we will ask is there some "generalized quantization" procedure which can be applied to Sn and get Hecke, then i think the answer may be does not exist... $\endgroup$
    – Alexander Chervov
    May 13, 2012 at 18:10
  • $\begingroup$ mathoverflow.net/questions/55647/… discuss different quantizations $\endgroup$
    – Alexander Chervov
    May 13, 2012 at 18:13

3 Answers 3

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The algebras you are looking for are called Iwahori-Hecke algebras. 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 $$T_i^2 = qT_i + (1-q)$$

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.

Once I understood this everything became much clearer, good luck!

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    $\begingroup$ As a remark. As algebras over C Hecke is isomorphic to group algebra Sn. This is just deformation theory. Group algebra is sum of matrix algebras which are simple so no cohomolgy no deformations. The same is true for quantum Lie algebras they differ from classical only via coproducct. $\endgroup$
    – Alexander Chervov
    May 13, 2012 at 17:21
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    $\begingroup$ mathoverflow.net/questions/4547/definitions-of-hecke-algebras nice discussions about Hecke algebras. What is amasing that Hecke algebra appeared before quantum groups but appeared to be related with them e.g. in q-SchurWeyl duality. $\endgroup$
    – Alexander Chervov
    May 13, 2012 at 18:04
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    $\begingroup$ Alexander, in reply to your first comment. Don't you have to be careful with this? The Hecke algebra isn't a formal deformation in the sense that it isn't defined over the power series ring, but rather over polynomials in q and perhaps q^-1. Enlarging the base ring using something like q=e^t you do have this isomorphism, but that's a necessary step. Over C you also have to be careful, if q were a root of unity then your statement wouldn't be true. $\endgroup$
    – James Griffin
    May 17, 2012 at 9:49
  • $\begingroup$ Having said that, it is still a good comment. But the deformation theoretic view sort of misses the point. The Hopf algebra is there to represent a monoidal category in some sense, in a related sense it's a decategorification of something. Both viewpoints are lost when you view it in terms of formal deformations. I suppose the `moral' is that q-deformations are somewhat different from formal deformations. $\endgroup$
    – James Griffin
    May 17, 2012 at 9:53
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In

  • Wang, Shuzhou. Quantum symmetry groups of finite spaces. Comm. Math. Phys. 195 (1998), no. 1, 195--211. MR1637425 (99h:58014), link

a quantum version of the symmetric group $\mathbb{S}_n$ is defined.

Let me sketch Wang's construction.

Let $u_{ij}$ be the characteristic function of the set of $\sigma\in\mathbb{S}_n$ such that $\sigma(j)=i$.

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 $C(\mathbb{S}_n)$.

Now drop the commutativity condition and let $A_s(n)$ be the $C^*$-algebra generated by all the $u_{ij}$. Then we have a quantum analogue of $\mathbb{S}_n$.

It turns out that $A_s(n)$ is a finitely generated Hopf algebra.

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 coaction). This coaction can be expressed as $\alpha(\delta_i)=\sum\delta_j\otimes u_{ji}$. Furthermore, $\alpha$ is a sort of universal coaction.

It is possible to prove that the following diagram is commutative $$ \begin{array}{ccc} C(X) & \to & C(X)\otimes A_s(n)\\\\ \downarrow & & \downarrow\\\\ C(X) & \to & C(X)\otimes C(\mathbb{S}_n) \end{array} $$

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.

For a nice survey about quantum permutation groups and some applications see the following paper:

  • 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), link

For a quantum version of the automorphism group of finite graphs (and a quantum version of the dihedral group $\mathbb{D}_4$):

  • Bichon, Julien. Quantum automorphism groups of finite graphs. Proc. Amer. Math. Soc. 131 (2003), no. 3, 665--673 (electronic). MR1937403 (2003j:16049), link

A complete classification of quantum permutation groups acting on 4 points was given in:

  • Banica, Teodor; Bichon, Julien. Quantum groups acting on 4 points. J. Reine Angew. Math. 626 (2009), 75--114. MR2492990 (2010c:46153), link
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  • $\begingroup$ it is nice for this construction to be mentioned. However let me mention that it is NOT similar to quantization or q-analog. This is very different, it is something like "making free", it is NOT deformation in the sense that size of the algebra becomes very huge, while in usual deformations the size of algebra is free. Also the idea of this construction is due to Yuri Manin. It is also related to Manin's black product and more particular this is "non-commutative endomorphisms of rings". This construction can be applied to any ring instead of C(X)... continued $\endgroup$
    – Alexander Chervov
    May 14, 2012 at 7:43
  • $\begingroup$ ...continued... for example one can take polynomial ring C[x_1... x_n] instead of C(X). What you will get is NOT Mat_n ! but the algebra of much bigger size, which we propose to call "Manin matrices" arxiv.org/abs/0901.0235 . The motivation for Manin's original work that if one take the algebra of q-commuting polynoms x_i : x_i x_j = q x_j x_i i<j , then similar construction will give half of the relations of quantum group GL_q(n). One can also get the second half of the relations imposing similar condition for anticommuting variables. $\endgroup$
    – Alexander Chervov
    May 14, 2012 at 7:47
  • $\begingroup$ The link to springerlink.com is broken, but the article can be found at arXiv:math/9807091 [math.OA] and doi:10.1007/s002200050385 (Zbl 1013.17008). $\endgroup$
    – The Amplitwist
    Apr 24, 2023 at 18:21
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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 $S_n$.

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