21

This answer is a small attempt to address your third question. You may find the article, Jethro van Ekeren, The six-vertex model, $R$-matrices, and quantum groups, to be helpful. One of the main sources in the development of quantum groups was the field of exactly solvable models in statistical mechanics. A number of simple mathematical models were ...


20

If you want to study a mathematical object, whether it is a set, manifold, group, vector space, whatever, it is often fruitful to look at natural collections of functions on that space. Roughly, the purpose of the tensor product, $\otimes$, is to make the following statement true: $$\text{functions}(X \times Y) = \text{functions}(X)\otimes \text{functions}(...


19

I know two-ish answers to this question. Representation-theoretic: The category of representations of a group has both tensor products and duals, but the category of representations of a general algebra generally has neither (or at least there is no obvious way to define them). Since the category of representations of a group $G$ is equivalent to the ...


11

Ok, a couple of months have passed since I posted this question, and I have begun this project on quantum groups. I am now in the condition of answering a couple of the questions I posed. There are many different definitions of quantum groups, all of which are related somehow. A possible definition is: a (Drinfeld-Jimbo) quantum group is a $1$-parameter ...


8

Recall that $$ \small U=\frac{\mathbb{k} \{X,Y,H\}}{([X,Y]-H, [H,X]-2X, [H,Y]+2Y)}\\ \small { U_1'=\frac{\mathbb{k}\{E,F,L,K,K^{-1}\}}{\begin{pmatrix}KK^{-1}-1& K^{-1}K-1& KE-EK& KF-FK\\ [E,F]-L& K^2-1& [L,E]-EK-K^{-1}E& [L,F]+FK+K^{-1}F\end{pmatrix}} } $$ Now we proceed to construction of isomorphisms. First isomorphism Note that $$ ...


8

Here's an algorithm for finding such a reduced word decomposition. Start with $w=e$ (=the empty word) and $\lambda=-\rho=-\sum_i\lambda_i$, where the $\lambda_i$ are the fundamental dominant weights determined by $\langle \lambda_i,\check{\alpha_j}\rangle=\delta_{ij}$. Here $\alpha_1,\ldots,\alpha_n$ are simple positive roots. If there exists an index $i$ ...


8

A bit long for a comment: There's a "more standard" way of making the quantum integers into a ring via so-called quantum addition: $[x]_q\oplus_q[y]_q=[x]_q+q^x[y]_q$. If you work it out, this will give you $[x]_q\oplus_q [y]_q=[x+y]_q$. There's a similar definition for multiplication: $[x]_q\otimes_q [y]_q=[x]_q[y]_{q^x}$, which if you work out gives you ...


7

If I recall - and I seem to have misplaced my copy of Kassel - what you have written is the definition of the (quantum) Yang-Baxter equation. What I know of this stuff is from the quantum group point of view. There's a bunch of physics that this comes from but unfortunately I know little of this. An $R$-matrix in this setting is the following. Let $(H,\mu,\...


7

Before Drinfeld's work in the 1980s there was only marginal interest by mathematicians in general (noncommutative, noncocommutative) Hopf algebras, so it could be difficult to honestly interest students in the Hopf algebra axioms without mentioning quantum groups. On this view the place to start is Drinfeld's ICM lecture and papers, or books on quantum ...


7

They're (co)group-objects, for one. That's always fun!


6

The converse is not true, Sweedler's 4-dimensional Hopf algebra (first example of a non-comm and non-cocomm Hopf algebra) is a counter example. This example is completely taken from section 4.3.6 of "Hopf Algebras" by Dascalescu, Nastasescu and Raianu. Let $k$ be a field of characteristic $\neq 2$. As a vector space, define $H = \langle 1, c, x, cx \rangle$...


6

Brain dump: Kassel is good. Jantzen (Lectures on Quantum Groups) is nicely written and clear, it's my favourite. Majid (Foundations of Quantum Group Theory) is an option, I haven't read it. Lusztig's book has a reputation of being tough to read. Chari and Pressley (A Guide To...) is comprehensive and has some interesting material on knots, 3-manifolds, ...


6

I think Qiaochu's answer is the best, but I thought that two topological reasons should be given. The first is that the Steenrod squares form a Hopf algebra, and their structure is one of the most important aspects of homology theory. The second is the diagrammatic interpretation of the axioms. Letting $\lambda$ denote the multiplication operator, and $Y$ ...


5

Sure. If $A$ is a finite-dimensional algebra over a field $k$ and $V$ a finite-dimensional $A$-module (so we have a morphism $\rho : A \to \text{End}_k(V)$), we can define a character $\chi_V(a) = \text{tr}(\rho(a)) : A \to k$ just as in the case of finite groups (where $A = k[G]$). Characters are additive in short exact sequences, so if $\chi_V$ has a ...


5

A complex representation of a finite group $G$ whose character vanishes on all non-unit elements of $G$ is a multiple of the regular representation. That follows immediately by standard character theory, as set up in, for example, Serre's book on the Representation Theory of Finite Groups. Later: You asked for an proof of this... The hypothesis you have ...


5

In an associative algebra, if an element has a left inverse and a right inverse, these two are equal. Now $\rho\star\mu=\eta\epsilon$ and $\mu\star\nu=\eta\epsilon$ mean precisely that $\rho$ and $\nu$ are left and right inverse to $\mu$, respectively, in the convolution algebra $\hom(H\otimes H,H)$ (recall that $\eta\epsilon$ is the unit element in that ...


5

Often we have an object and we desire to find analogues of it in other settings. Sometimes the definition of the object does not directly translate from the original setting to the new setting. The solution is to rewrite the definition in a way that it can be translated directly into the new setting. Consider the Fourier transform. This I think is prudent ...


5

Quantum groups are not groups, nor are these Hopf algebras (which is what they are) directly associated to any groups (as it the case of Lie algebras).


5

This is making things too hard. This is the same as a $\mathfrak g$ action on $\mathbb C[G]$, which is given by differentiation by right invariant vector fields. Extending to $U(\mathfrak g)$ is given by taking the associated differential operator. The pairing you want is applying this operator and then evaluating at $e$.


5

Yes it is true. But not for the commutative hopf algebras in general. You need some more assumptions: first we need the field to be algebraically closed and of characteristic zero. You also need cocommutativity and finite dimensionality of the hopf algebra to have a full "correspondence". To be more precise: it can be shown that there is an ...


4

Your exponent $|x_2|\cdot|x_1'|+|x_2x_2'|\cdot|x_1''|$ is correct. The exponent for the other grouping is $|x_2'|\cdot|x_1''|+|x_2|\cdot|x_1'x_1''|$. To show that these are equal, you need the two facts that the product is bilinear and that $|xy|=|x|+|y|$ (corresponding to $'\mathbf f_\nu'\mathbf f_{\nu'}\...


4

In ordinary group representation theory, a character is a homomorphism $\chi:G\to \mathbb{C}^\times$. A character associated to a representation $\rho: G\to GL(V)$ is $\chi_\rho: g\mapsto trace(\rho(g))$. Character of Lie algebra arising from differentiating a Lie group is the same thing in the sense that character of Lie algebra representation is ...


4

On the one hand, I'd suggest taking that description of $U_q(\mathfrak{sl}_2)$ as a formal definition. That is, think of $U_q(\mathfrak{sl}_2)$ as being the free associative algebra $k\langle E, F, K, K^{-1}\rangle$ modulo the relations you gave. $E$, $F$, $K$ are just formal symbols. You can just as well call them $x$, $y$, and $z$. You can procede in ...


4

In general, if it is the ordinary vector space tensor product we are talking about, the canonical map $C(G) \otimes C(G) \to C(G \times G)$ that takes $\phi \otimes \psi$ to the function $(g, h) \mapsto \phi(g)\psi(h)$ is not an isomorphism. It does however give an isomorphism onto a dense subspace of $C(G \times G)$ (equipped as usual with the topology of ...


4

$k(G)$ cocommutative means $\tau \circ \Delta = \Delta$ with $\tau: k^{G \times G} \to k^{G\times G},\,\tau(f)(x,y)=f(y,x)$. Let $f: G \to k$. Then $(\tau \circ \Delta)(f)=\Delta f$ means $f(xy)=f(yx)$ for all $x,y \in G$. For fixed x,y choose $f=\delta_{xy}$, i.e. $\delta_{xy}(xy)=1$ and $\delta_{xy}(g)=0$ for $g \ne xy$. Hence $\delta_{xy}(yx)=\delta_{...


4

I can mainly say something about compact quantum groups from the C*-algebraic point of view... This C*-algebraic definition of compact quantum groups seems to be quite established by now. For an introduction from this starting point, including the basic results on existence of a Haar measure and representations, you can have a look at the paper by S.L. ...


4

The simplest thing to do is just to define and algebra with an element like that: $R=F\langle X, Y\rangle/(YX-qXY)$ where, say, $q\in F$. In $M_2(\mathbb R)$, setting $X=Y=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ gives you an example of $XY=qYX$ for any $q\in\mathbb R$ at all.


3

You have a sign error we have $EK^{-1}\otimes KF=K^{-1}E\otimes FK$. This is because of the following two relations: \begin{align*} KEK^{-1}&=q^2E\\ KFK^{-1}&=q^{-2}F \end{align*} Hence we have \begin{align*} EK^{-1}=q^2K^{-1}E\\ KF=q^{-2}FK \end{align*} This gives the above equality (since $q^2q^{-2}=1$ and you can put scalars in front of the tensor ...


3

Let $m_1\otimes m_2\otimes m_3\in M\otimes M\otimes M$, then $$ \begin{align} R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)&= (\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(1_M\otimes\tau)(m_1\otimes m_2\otimes m_3)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_1\otimes m_3\otimes m_2)\\ &=(\...


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