I've recently been to a seminar on quantum matrices. In particular the speaker introduced these objects as the coordinate ring of $2$ by $2$ matrices modulo some odd looking relations (see start of Section 2 here). As a theoretical physicist, I'm struggling to understand in what sense these objects are quantum!

Has anyone got any references or knowledge which might help answer

  1. Where do these odd looking relations come from?

  2. In what sense are the matrices quantum (can the non-commutativity in the coordinate ring be understood as emerging from some quantisation procedure...)?

  3. Should I think of quantum groups as a controlled mechanism for introducing non-commutativity into coordinate rings in general?

Many thanks in advance for your expertise!

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    $\begingroup$ have you looked at the Wikipedia article? en.wikipedia.org/wiki/Quantum_group $\endgroup$ Jan 7, 2016 at 16:39
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    $\begingroup$ @CarloBeenakker - yes I have taken a look, but the intuitive explanation is rather vague, and the examples rather technical. I'm really looking for something in between! $\endgroup$ Jan 7, 2016 at 16:43
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    $\begingroup$ One survey article that might have enough of physics intuition to be useful is scitation.aip.org/content/aip/journal/jmp/45/10/10.1063/… - are you aware of it? $\endgroup$ Jan 7, 2016 at 17:11
  • $\begingroup$ @VladimirDotsenko - I hadn't come across that, many thanks! $\endgroup$ Jan 7, 2016 at 18:07
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    $\begingroup$ Just had an enlightening discussion on this topic with Tom Lenahan - seems like the particular commutation relations in (1) can be seen as a consequence of the natural definition of the $2$-dimensional quantum plane $ab = qba$. Apparently there's also a derivation of these relations from a universal enveloping algebra; if anyone has a reference, I'd be more than happy to hear! $\endgroup$ Jan 7, 2016 at 21:36

4 Answers 4


Typically in math "quantum X" means a deformation of "X" which is in some sense "less commutative." So quantum groups should be deformations of groups which are "less commutative." Interpreting this is slightly tricky since groups are already non-commutative, but nonetheless they do have some "commutativity" built in which you can see either by noting:

  1. The ring of functions on the group is a commutative ring.
  2. The tensor product of representations of the group has a symmetric isomorphism $V \otimes W \rightarrow W \otimes V$.

You can use either of these to motivate versions of quantum groups. A quantum group is:

  1. A Hopf algebra which deforms the Hopf algebra of functions on a group, but where the ring structure is non-commutative.
  2. Something whose category of representations deforms the category of representations of a group, but where the tensor product structure is not symmetric.
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    $\begingroup$ According to your definition, a Hopf algebra which is not a deformation of the Hopf algebra of functions on a group, should not be called a quantum group, right? $\endgroup$ Jan 8, 2016 at 3:58
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    $\begingroup$ Many thanks for your answer. I think it accurately addresses my points (2) and (3). Have you got a slightly more technical answer to my question (1)? In particular do you know how to derive the definition of quantum matrices from some natural deformation of a commutative ring? To me it still seems a bit arcane from that perspective! I'd be more than happy to accept your answer if you can clear this up. $\endgroup$ Jan 8, 2016 at 9:35
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    $\begingroup$ @SébastienPalcoux, I'd agree, but some people (say, Majid) call any quasitriangular Hopf algebra a quantum group. $\endgroup$ Jan 8, 2016 at 12:39
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    $\begingroup$ I don't think there's any benefit to being excessively rigid about what is or is not a quantum group. The question was about the motivation and meaning of the term, not about all situations people use it in. $\endgroup$ Jan 8, 2016 at 13:59
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    $\begingroup$ Interesting how the meaning of "quantum" has now dramatically evolved twice, from "discretised into quanta" to "relating to quantum mechanics" and finally to "being less commutative than before". $\endgroup$
    – Terry Tao
    Feb 3, 2016 at 21:32

For quantum matrices and related objects specifically, I heartily recommend the opening chapters of "Lectures on Algebraic Quantum Groups" by Brown and Goodearl. I hesitate to write any more details, since they do such a good job of it, IMHO.


Yes it can be understood as a quantization procedure. You are probably familiar with the first step to giving a quantization is to give a Poisson algebra. In this case you want the algebra in question to be better than the coordinate ring of the group, you want it to be the coordinate ring of a Poisson-Lie group.

Let's focus on the identity. Now I'm giving you a Lie bialgebra. Many examples come from looking at QR or LU factorizations. Here you see Etingof-Kazhdan.

I recommend Semenov-Tian-Shansky's lecture notes http://arxiv.org/pdf/q-alg/9703023.pdf

Edit: the deformation parameter has a different interpretation not necessarily $\hbar$, so be careful when you have multiple deformations.

  • $\begingroup$ Thanks @AHusain. So I'm right to think that a quantum group should be thought of as a deformation of a type of algebra. Is the procedure unique in general? There seems to be a consensus about the definition of a quantum matrix in my question (1). But I still can't see what motivates the slightly odd form for the relations. Is that obvious from the quantisation somehow? $\endgroup$ Jan 7, 2016 at 18:18
  • $\begingroup$ Sorry. I didn't look at the link to see you were interested in giving it in cluster coordinate presentation. You start off writing down the Poisson brackets for the obvious functions on the group from matrix coefficients, but those aren't quite the best to work with. Think canonical commutation relations vs some other function on phase space. For reference, see "Cluster Algebras and Poisson Geometry" $\endgroup$
    – AHusain
    Jan 7, 2016 at 19:01

Let me add a few words referring in particular to the paper you cited: "From totally nonnegative matrices to quantum matrices and back, via Poisson geometry" S. Launois and T.H. Lenagan

So we have a matrix $$( a \quad b )$$

$$ ( c \quad d ) $$

and the relations

1) Four relations of the type XY = q YX $$ ab = q ba, ac = q ca, bd = q bd , cd = q cd $$ they cover four out of 6 relations between basic elements, relations between diagonal terms a,d and anti-diagonal b,c are not covered.

2) relation for b,c is just commutativity: $$ bc = cb $$

3) and the last one is really "odd looking" here I will agree with you:

$$ ad - da = (q-q^{-1} ) bc $$

One way to motivate such relations is Manin's point of view - "coaction on a quantum place" - as explained by Theo Johnson-Freyd here: Intuition behind the definition of quantum groups

Alternative point of view appeared before by Drinfeld: Look at the Section 3.1 page 11 from the paper you cited: here you can see "classical limit" i.e. Poisson brackets for these relations :

1) Four relations of the type {X, Y} = XY $$ \{a, b \} = ab, \{a, c \} = ca, \{b, d \}= bd , \{c, d \} = cd $$ they cover four out of 6 relations between basic elements, relations between diagonal terms a,d and anti-diagonal b,c are not covered.

2) $$ \{ b, c \} = 0$$

3) and the last one: $$ \{a,d \} = 2bc $$

The claim is that quantum relations above are quantization of the Poisson bracket relations here. I.e. non-commutative algebra of observables is deformation quantization of the Poisson algebra here. Deformation quantization is now quite understood by efforts of Kontsevich et.al. However without refering to big theories, as you a physicists you probably can easily see that relations {X,Y} = XY should be quantized to XY = q YX - that are relations of the type 1. Why this is true: look at canonical quantization {p,q} = 1, quantized to [p,q]= h. Consider exponentials: {exp(p), exp(q) } = exp(p) exp(q) - we see exactly the same Poisson bracket as (1) - so it is just the exponents of canonical variables, so quantization is easy just exp(p) and exp(q) which in quantum world will give $$ exp(p) exp(q) = e^h exp(q) exp(p)$$, so exactly the quantum relation (1) which we saw above.

{b,c} = 0, is quantized to bc = cb - is natutal

About the third relation - well I do not know easy way to quantize it from basic principles. If you know - tell me.

So your question "Where do these odd looking relations come from? " can be translated to the question: where do these Poisson relations come from ?

That is Drinfeld's insight. He asked the following question - consider a Lie group and invariant Poisson bracket on it, we want Poisson structure would be COMPATIBLE with the group structure (in certain easy to guess sense). And from that compatibility you will get that kind of Poisson relations. Such groups are called Poisson-Lie groups as mentioned in answer by AHusain above. The group is GL(2) for examples above.

Concerning question 2: In what sense are the matrices quantum (can the non-commutativity in the coordinate ring be understood as emerging from some quantisation procedure...)?

Yes as it was discussed above we have Poisson brackets which are quantized to that kind of noncommuative relations.

Question 3: Should I think of quantum groups as a controlled mechanism for introducing non-commutativity into coordinate rings in general?

NO, that is not the case. Quantization needs only the Poisson bracket as an input. Only in special case when Poisson bracket is somehow naturally related to some semisimple group you will get something related to quantum group. In general you can take arbitrary Poisson bracket , quantize it and it will have nothing to do with the quantum groups.


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