Intuition behind the definition of quantum groups

Question

Being far from the field of quantum groups, I have nevertheless made in the past several (unsuccessful) attempts to understand their definition and basic properties. The goal of this post is to try to improve the situation.

The definition of the quantum group I saw is that it is a Hopf algebra given by some explicit generators and relations. Though I have heard that at least the case of the quantum $sl_2$ was motivated by physics, this was not explicit enough. If someone defined the classical (i.e. non-quantum) Lie algebra $gl_n$ or the symmetric group $S_n$ using generators and relations rather than operators acting on a vector space or on a finite set respectively, such a definition would be equally unclear to me. The abstract approach to quantum groups as deformations of a universal enveloping algebras in the class of Hopf algebras is very useful, but still not very intuitive and explicit.

While any clarifying remarks would be appreciated, I can ask the following more specific questions. (1) In simple examples of quantum groups, such as quantum $sl_2, sl_n$, are there "natural" examples of their representations, like the standard representation of the classical $sl_n$, its dual representation and their tensor powers? (2) Are there mathematical problems which are not about quantum groups, but whose solution does require this notion?

Answer 1

In algebraic combinatorics, there is an important concept of a "$q$-analogue". Surprisingly often when you have a counting problem with a good integer answer, you realize that it can be refined to a (finite) generating function with an equally good polynomial answer. A simple example of this is the $q$-analogue of the number of permutations, which is of course $n!$. If you define the weight of a permutation to be $q^k$, where $k$ the inversion number of the permutation, then the total weight is then the important and beautiful formula

$$1(q+1)(q^2+q+1)\cdots(q^{n-1}+\ldots+1).$$

This expression is called a $q$-factorial the factors are called $q$-integers. Interesting q-analogues usually involve $q$-integers. Note that $q$-integers are closely related to cyclotomic polynomials: Every $q$-integer is a (unique) product of cyclotomic polynomials, and every cyclotomic polynomial is a (unique) ratio of products of $q$-integers.

Gaussian binomial coefficients are among the most important $q$-analogues.

The best way to think of a quantum group $U_q(\mathfrak{g})$ is that it is an algebraic $q$-analogue of a simple Lie group $G$ or its universal enveloping algebra $U(\mathfrak{g})$. For generic values of $q$, it has exactly the same (names of) representations as $G$ that tensor in the same way, plus (depending on conventions) possibly some other representations that are less important. But, what has changed is the positions of the sub-representations and the specific representation matrices. In general, when you see expressions such as integers, binomial coefficients, and factorials in the formulas for representation matrices, you see $q$-integers, Gaussian binomial coefficients, and $q$-factorials in the quantum group version. The only asterisk to this is the preferred convention of using centered Laurent polynomials (which may have half-integer exponents) rather than standard polynomials with non-negative integer exponents.

The only non-generic values of $q$ for quantum groups are roots of unity. In this case a new and also fundamental effect appears: The representation theory acquires features shared with representations of algebraic groups in positive characteristic. They have been used to strengthen or at least clarify the representation theory of algebraic groups.

The main application of quantum groups: Topological invariants. Eventually when studying the representation theory of a Lie group, you consider tensor networks, i.e., invariant tensors combined with tensor products and contractions. Because of the extra non-commutativity of a quantum group (or any non-commutative, non-cocommutative Hopf algebra), an invariant tensor network of a quantum group needs to be embedded in $\mathbb{R}^3$ in order to be interpreted or evaluated as an algebraic expression. And then the remarkable outcome is that you obtain the Jones polynomial, when the quantum group is $U_q(\text{sl}(2))$, and its well-known generalizations for other quantum groups.

Quantum groups are the main algebraic way to understand the quantum polynomial invariants of knots and links; and quantum groups at roots of unity are the main algebraic way to understand the corresponding 3-manifold invariants. In fact, this is closely connected to why they were first defined.

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In response to Semyon's question in the comments: The concept of a $q$-analogue in combinatorics has never been entirely rigorous, and if anything the construction of quantum groups has been clarifying. The rough idea is that a counting problem in combinatorics is interesting when it has a "nice" answer, which often (but not by any means always) means an efficient product formula. So then a $q$-analogue is a weighted enumeration or finite generating function in which every weight is a power of $q$, and the enumeration still has all favorable numerical properties, and $q$-integers or cyclotomic factors arise.

In the case of quantum groups, first of all they are Hopf algebras. A Hopf algebra is an algebra together with all necessary extra apparatus to define the tensor product of two representations as a representation (i.e., comultiplication) and the dual of a representation as a representation (i.e., the antipode map). A universal enveloping algebra $U(\mathfrak{g})$ is of course a Hopf algebra. In this case, any deformation of $U(\mathfrak{g})$ as a Hopf algebra is potentially interesting. There is a cohomology result that if $\mathfrak{g}$ is complex and simplex, then there is only one non-trivial deformation, and you might as well call its parameter $q$ with $q=1$ at the undeformed point. (Sometimes the logarithm of $q$ is used and is called $h$, in reference to Planck's constant.) Since this is the only deformation, it is an analogue of some kind, and it is interesting. Moreover, there is a parametrization of the deformation so that $q$-integers (or quantum integers, in centered form) and cyclotomic polynomials arise in the structure of the Hopf algebra and its representations. The analogue thus deserves to be called a $q$-analogue.

Theo's explanation illustrates this more explicitly. The quantum plane is a non-commutative algebraic space (in the sense that you can interpret it as a purely formal "Spec" of the quantum plane ring) that is associated to the non-cocommutative algebraic group version of $U_q(\text{sl}(2))$. So, then, in the ring of the quantum plane, if you just expand $(x+y)^n$, you get a $q$-analogue of the binomial coefficient theorem using Gaussian binomial coefficients. (Where the $q$-exponent of a word in $x$ and $y$ is its inversion number, just as with the $q$-enumeration of permutations.) This is one of many examples where $q$-analogues that were considered long before quantum groups appear in the theory of quantum groups.

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As for knots and links: In order for $U(\mathfrak{g})$ to have a non-trivial deformation as a Hopf algebra, you have to allow comultiplication to be non-commutative, even though $U(\mathfrak{g})$ itself is cocommutative. So then if $V$ and $W$ are two representations, $V \otimes W$ and $W \otimes V$ are not isomorphic via the usual switching map $v \otimes w \mapsto w \otimes v$, because that switching map is not in the category. (Due to non-cocommutativity, it is not equivariant, i.e., not an intertwiner.) However, $V \otimes W$ and $W \otimes V$ are still isomorphic, just by an adjusted version of the switching map. (You see the same theme in the category of super vector spaces, where there is a sign correction when $v$ and $w$ both have odd degree.) However, there are two natural, in-category deformations of the switching map, not just one. It so happens that they should be interpreted as left- and right- half twists in a braid group, so that you get braid representations and ultimately knot and link invariants.

The point is that ordinary tensors (and tensor networks) live in spaces that have natural actions of symmetric groups, because you can permute indices of tensors. The whole theme of quantum group definitions is deformations, and it so happens that the symmetric group action deforms into a braid group action.

This explains topological invariants such as the Jones polynomial in a one- and two-dimensional sort of way, from braids to diagrams of knots to knots themselves. It is more satisfying to have a more intrinsically 3-dimensional definition. (Actually, what counts as intrinsically 3-dimensional is somewhat debatable, but never mind that.) This is why Witten provided a "definition" of the Jones polynomial and related invariants using Chern-Simons quantum field theory. It is not really a rigorous definition, but it is very credible as a "physics definition" or even an incomplete, but maybe one-day rigorous, mathematical definition. This leads to the basic association between Chern-Simons quantum field theory and quantum groups, that they are two ways to describe the same topological invariants.

Answer 2

Here is an answer to question (1). I recommend that you split of question (2) as a separate question.

Define the quantum plane to be the "spectrum" of the noncommutative ring $\mathbb K\langle x,y\rangle / (xy = qyx)$, where $\mathbb K$ is some ground commutative ring in which $q$ is invertible (e.g. $\mathbb K = \mathbb C(q)$). So a "point" in this "plane" is determined by the "values" of the two "coordinates" $x$ and $y$. Note that these "values" don't commute: they are "valued" in some space of "noncommutative numbers".

Now, recall that $2\times 2$ matrices act on the usual plane. Are there "$2 \times 2$ matrices" that act on the quantum plane?

Well, in the usual case, a matrix $\bigl( \begin{smallmatrix} a & b \\ c & d\end{smallmatrix}\bigr)$ takes the vector $\bigl( \begin{smallmatrix} x \\ y \end{smallmatrix}\bigr)$ to: $$ \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.$$ Let's try to keep the same rule. For this to work, the coordinates of the new point need to satisfy the same "$q$-mmutation" (as opposed to "commutation") law as $\{x,y\}$, which is to say:

$$ (ax + by)(cx+ dy) = q(cx+dy)(ax+by). $$

If we suppose that variables $a,b,c,d$ commute with variables $x,y$, then we get:

$$ ac = qca \quad bd = qdb \quad ad + q^{-1}bc = qcb + da. \quad\quad (\star)$$

What about, say, how $a$ and $b$ commute? Well, I should have mentioned that the quantum plane has a "dual quantum plane". Recall that in classical linear algebra, the dual to the space of column vectors is the space of row vectors. Let's, then, take the dual quantum plane to consist of row vectors $(v,w)$ whose coordinates satisfy $vw = qwv$. Then by letting quantum matrices act from the right on quantum row vectors, you can compute:

$$ ab = qba \quad cd = qdc \quad ad + q^{-1}cb = qbc + da. \quad\quad (\star\star)$$

Note that the last lines in each are slightly different. Together, the rules $(\star)$ and $(\star\star)$ define the space of "quantum $2\times 2$ matrices". Let me simplify the last two rules:

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

Now here's an exercise: suppose that $\bigl( \begin{smallmatrix} a_1 & b_1 \\ c_1 & d_1\end{smallmatrix}\bigr)$ and $\bigl( \begin{smallmatrix} a_2 & b_2 \\ c_2 & d_2\end{smallmatrix}\bigr)$ are each quantum $2\times 2$ matrices, by which I mean that their coordinates each independently satisfy the rules $(\star,\star\star)$, and suppose that the $r_1$s commute with the $r_2$s. Then their product $$ \begin{pmatrix} a_1a_2 + b_1 c_2 & a_1 b_2 + b_1 d_2 \\ c_1 a_2 + d_1 c_2 & c_1 b_2 + d_1 d_2 \end{pmatrix}$$ is again a quantum $2\times 2$ matrix.

Thus our "quantum space" of quantum $2\times 2$ matrix (defined by $(\star,\star\star)$, which just determines if at some space in which each point is determined by the values of 4 noncommuting coordinates) is in face a "ring", or at least a "monoid". Oh, also you should check that (exercise) $\bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$ is a quantum matrix, and is the unit. So now always do the coordinates have to evaluate to noncommuting numbers.

Now, an important property of matrices is the "determinant". Exercise: the number $\Delta = ad - qbc$ commutes with all coordinates.

Exericse: If $X = \bigl( \begin{smallmatrix} a & b \\ c & d\end{smallmatrix}\bigr)$ is a quantum $2\times 2$ matrix in which $\Delta^{-1}$ exists, then $X$ is invertible, with inverse $\Delta^{-1}\bigl( \begin{smallmatrix} d & -b \\ -qc & a\end{smallmatrix}\bigr)$.

Anyway, by definition, the group quantum $SL(2)$ is the subgroup of the quantum $2\times 2$ matrices consisting of those matrices for which $\Delta = 1$. By definition, the group quantum $GL(2)$ is the subgroup for which $\Delta^{-1}$ exists (thus you could say that a point in quantum $GL(2)$ has five coordinates, $a,b,c,d,\Delta^{-1}$, satisfying the equations $(\star,\star\star)$ and $\Delta^{-1}(ad - qbc) = 1$).

Similar constructions give other quantum groups as well, by starting with higher-dimensional quantum spaces and/or quantum versions of symmetric or skew-symmetric pairings.

Finally, you might have seen "groups" like $U_q \mathfrak{sl}(2)$, which really are the universal enveloping algebras of "lie algebras" of these groups.

Answer 3

To follow up on Greg's answer, here is a simple explanation as to how quantum groups can be thought of as $q$-analogs of semisimple Lie algebras. Let's just take $ sl_2 $ for simplicity.

Recall that $sl_2 $ has a basis $ E, H, F $. Suppose that we have a finite-dimensional representation $ V $ of $ sl_2 $. Then $ V = \oplus_{n \in \mathbb Z} V_n $ where $ V_n $ is the $n$-eigenspace for the action of $ H$. Then for any $ v \in V_n $, we have $ E(v) \in V_{n+2}, F(v)\in V_{n-2}$ and $ EF(v) - FE(v) = n v $.

Similarly, a representation of $ U_q sl_2 $ is a vector space $ V $ with a direct sum decomposition as above and linear operators $ E, F : V \rightarrow V $ such that for any $ v \in V_n $, we have $ E(v) \in V_{n+2}, F(v)\in V_{n-2}$ and $ EF(v) - FE(v) = [n] v, $ where $ [n] = q^{n-1} + q^{n-3} + \cdots + q^{1-n} $ is the $q$-analog of $ n$.

Answer 4

Here is a possible affirmative answer to question (2). (Which is to say, I am not aware of any non-quantum way to answer this question.) I assume from the original post that it is sufficient motivation to say that the complex simple Lie algebras $\mathfrak{g}$ (like, say, $\mathfrak{sl}_n$) act on interesting spaces, and they have a well-understood finite-dimensional representation theory.

Let's assume we have fixed some preferred triangular decompostion $\mathfrak{g}=\mathfrak{n}^-\oplus \mathfrak h\oplus \mathfrak{n}$. Then we have the well-known fact that the simple finite-dimensional modules are the highest weight modules $V(\lambda)$ for some dominant weight $\lambda$. While studying these simple representations, you might notice that the obvious bases (i.e. descending from a highest weight vector via multiplication by elements of $U(\mathfrak{n}^-)$) tend to be similar, in the sense that the corresponding elements of $U(\mathfrak{n}^-)$ are the same (or rather, can be chosen to be the same). In other words, given a finite number of modules, we can more or less determine by hand some elements of $U(\mathfrak{n}^-)$ which induce a basis on all our modules (via acting on the highest weight vector and throwing away zeros).

A trivial example of this is how the simple $n+1$-dimensional $\mathfrak{sl}_2$-module has the basis $F^iv_n$ for $0\leq i\leq n$, where $v_n$ is a vector satisfying $Ev_n=0$ and $Hv_n=nv_n$. This is plainly generated via acting on $v_n$ with the obvious basis $F^i$ of $U(\mathfrak{n}^-)$ and throwing away those which annihilate the module. (We can obviously be more clever about this by using divided powers so that everything is nice and integral, but let me ignore this.) In this sense, the basis of $U(\mathfrak{n}^-)$ is canonical and induces a uniform, canonical family of bases on all simple finite-dimensional modules. (Here, I mean canonical in the sense that, well, how else would you pick the basis in a non-contrived way?) It would be nice if this were always the case, so a natural question is:

Q: Can we cook up a basis of $U(\mathfrak n^-)$ that gives us a basis for every simple finite-dimensional highest weight module $V(\lambda)$? (Bonus points if this basis has nice properties, like living in the integral form or behaving well under tensor products.)

But in higher rank and other types, things are of course less clear; for instance, the Serre relations making various monomials dependent makes monomials a bad guess in general, plus we would still need a way to uniformly pick which monomial is "best". A uniform family of bases can be (and has been) constructed in some special cases using various strategies, but as far as I know there is no construction for all types using the vanilla theory of simple complex Lie algebras.

However, it turns out that if we take the quantum version of the enveloping algebra, the increased asymmetry of the quantized enveloping algebra (which is manifest in the non-cocommutativity of the coproduct, but also influential in the structure of the algebra and the representations) introduces some inherent preferences independent of how we choose to set things up. As a result, we can always construct (in a canonical way) a nice basis of $U_q(\mathfrak{n}^-)$ which induces a basis on any of the simple highest weight module, and which moreover can be specialized to the classical case by "setting $q=1$''. (In fact, it is so powerful that it also works for the simple integrable highest weight modules of symmetrizable Kac-Moody Lie algebras!)

What's remarkable about this story is that it is somewhat underselling the reality. These canonical bases (originating in the work of Lusztig and Kashiwara around 1990) both motivate and contribute to many other cool stories in combinatorics, geometry, categorification, and (modular) representation theory. In other words, there are many other reasons to be interested in this construction, but it does seem to be an example of a quantum solution to a non-quantum problem.