Hopf structure of Uq(sl(2))

Question

Given the universal enveloping algebra, $U(\mathfrak{sl}(2))$ the coalgebra structure is defined such that the generators $X,Y$ and $H$ are primitive elements. From this, is there a "nice" way to motivate the coproduct for $U_{q}(\mathfrak{sl}(2))$? Of course, this question can be generalized to: given $U(\mathfrak{g})$ how does one discover what the coproduct for $U_{q}(\mathfrak{g})$ should be based on what the coproduct for $U(\mathfrak{g})$ is? Most texts simply plunk down the structure without really motivating where it comes from or how it was arrived at (or the various ways it could be arrived at).

Given that the generators for $U_{q}(\mathfrak{sl}(2))$ are $E,F,K,K^{-1}$, I surmised that since $KK^{-1} = K^{-1}K = 1$ and $\Delta$ is to be an algebra morphism, then $$\Delta(KK^{-1}) = \Delta(K)\Delta(K^{-1})=1\otimes 1 = \Delta(K^{-1})\Delta(K) = \Delta(K^{-1}K)$$ seems to "naturally" suggest the assignment $\Delta(K) = K\otimes K$ and $\Delta(K^{-1}) = K^{-1}\otimes K^{-1}$. With this I then looked at $$\Delta(EF-FE) = \Delta\left(\frac{K-K^{-1}}{q-q^{-1}}\right)$$ and was able to use some algebraic manipulation to, sort of, "coax out" that $$\Delta(E) = E\otimes K+1\otimes E \qquad \Delta(F) = F\otimes 1+K^{-1}\otimes F$$ This all seems well and good, but nevertheless pedestrian (and a tad artificial). Is there or are there some more elegant ways (revealing deeper connections) to motivate the coalgebra structure (or the entire Hopf structure) of $U_{q}(\mathfrak{sl}(2))$ or more generally $U_{q}(\mathfrak{g})$? Thanks in advance.

Answer 1

Let me add a few motivations to the nice ones already provided.

The first goes through the dual Hopf algebra $O_q(SL_2)$; it is more roundabout, but each step is more naturally motivated. (this approach is explained in, e.g. Kassel's book on Quantum Groups. First, let me admit the "quantum plane", whose algebra of functions is defined as

$\mathbb{C}_q[x,y]:=\mathbb{C}\langle x,y\rangle / (yx-qxy).$

If we choose to regard this as an algebra of functions on a quantum $\mathbb{C}^2$, then we should expect that whatever our definition of quantum matrices on $\mathbb{C}^2$, its algebra $O_q(Mat_2)$ should have a co-action, i.e. a map $\Delta:C_q[x,y]\to C_q[x,y]\otimes O_q(Mat_2)$. Let us define elements $a,b,c,d$ by the equations

$\Delta(x)=x\otimes a + y\otimes b$, $\Delta(y)=x\otimes c+y\otimes d$.

Note that these are the same equations as for the usual coaction of $O(Mat_2)$ on $\mathbb{C}[x,y]$, and are just the formula for matrix multiplication written in a funny way.

Now ask that $O_q(Mat_2)$ be generated by those elements, subject to the requirement that $\Delta$ is a map of algebras. You find relations on $a,b,c,d$ as follows:

$\Delta(xy)=\Delta(x)\Delta(y)=x^2\otimes ac + xy\otimes ad + yx\otimes cb + y^2\otimes bd$.

Setting this equal to $1/q\Delta(yx)$ gives you relations like $ca=qac$, $bc=cb$, $ad-da = (q-q^{1})bc$,

and so on, precisely the defining relations of $O_q(Mat_2)$. Direct computation tells you that $det_q=ad-q^{-1}bc$ (if i remember correctly) is the unique central element in degree 2, and so quotienting by $det_q-1$ defines $O_q(SL_2)$. Now, define $U_q(sl_2)$ as the dual Hopf algebra w.r.t to $O_q(SL_2)$, and using this, recover the relations for $U_q(sl_2)$ uniquely.

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Now let me give another motivation for the relations in $U_q(sl_2)$ coming not from the quantum geometry point of view, but from braided tensor categories. First, for any number $q$, note that there is a braided tensor category structure on the category of $\mathbb{Z}$-graded vector spaces, where $\sigma(v_k\otimes v_l)=q^{kl} v_l\otimes v_k,$ for $v_k, v_l$ in degrees $k$ and $l$. Ranging over $q$ this essentially exhausts the possible braidings on $Z$-graded vector spaces. Let's call this category $C_q$. Let's denote by $V_k$ the one dimensional vector space concentrated in degree $k$.

Now, consider the tensor algebra of $V_1$, $T(V_1)$ in this category. As any tensor algebra, it admits the free coproduct $\Delta: V_1\to V_1\otimes V_0 + V_0\otimes V_1$, making $T(V_1)$ into a Hopf algebra in $C_q$. Now if we have modules $M,N$ in $C_q$, we can act on their tensor product $M\otimes N$ by the co-product, but we have to be careful!

$v . (m \otimes n) = \Delta(v) m \otimes n$, but now $T(V_1)\otimes T(V_1)$ acts on $m\otimes n$, by first braiding the second factor of $T(V_1)$ past $m$, then acting in the obvious way:

$T(V_1)\otimes T(V_1)\otimes m \otimes n \xrightarrow{\sigma} T(V_1)\otimes m \otimes T(V_1)\otimes n$

In particular if $v\in V_1$, then the braiding adds a factor of $q^{|m|}$ to one of the summands of $\Delta(v)$ when you braid past.

Now in this story, $T(V_1)$ is the free algebra on $E$, which is $U_q(n_+)$ in this case. A $\mathbb{Z}$-graded vector space is the same as an integral $\mathbb{C}[K,K^{-1}]$-module, since the grading is determined by eigenspaces of $K$, and a $U_q(n_+)$-module in the category $C_q$ is the same as an integral $U_q(b)$-module. So in this latter way of phrasing things the extra $q^{|m|}$ we found has to be put in by hand, by putting a $K$ in the second term of the co-product: $\Delta(E)=E\otimes 1 + K\otimes E$.

On the one hand, this second explanation probably seems rather artificial; on the other hand it is saying that there is nothing at all mysterious about the appearance of $K$'s in the formula for the co-product, it's just that you were working in the symmetric category of vector spaces, rather than the category of graded vector spaces, where you should have worked. The penalty you pay is putting $K$'s in places to keep track of what the braiding was keeping track of for you.

Answer 2

Hi Ryan,

Let me elaborate on my answer to your previous question. Somehow, deforming only the algebra structure is easy, in the sense that if you give some generators and some relations you're sure to get... an algebra. So just take the same generators as for $U(\mathfrak{sl}_2)$, some random relations whose quasi-classical limit gives you the defining relations of $U(\mathfrak{sl}_2)$ and you're done. Of course it's not a very interesting approach, but at least you are sure to get something matching your requirements.

On the other hand, deforming the Hopf structure is hard: constructing a well defined algebra map $\Delta$ is easy, but coassociativity is a hard condition which is of course not guaranted at all if you pick something at random.

First of all, it's strictly speaking not completely true that $K=q^H$ since $H \not\in U_q(\mathfrak{sl}_2)$, but it is still the right intuition, justified by the "formal" setting below. Then the same argument as in my previous answer apply:

$$\Delta(K)=q^{\Delta(H)}=q^{H\otimes 1+1\otimes H}=(q^H \otimes 1)(1 \otimes q^H)=K\otimes K$$

All of this is more natural if you take the formal version $U_{h}(\mathfrak{sl}_2)$ which is a topological $\mathbb{C}[[h]]$-Hopf algebra. In that case $H$ is a true generator and there are many reasons for which you want to let the "Cartan part" really undeformed. It is, indeed, a general fact: for a semi simple Lie algebra $\mathfrak{g}$, the sub-Hopf algebra of $U_h(\mathfrak{g})$ generated by the $H_i$'s is isomorphic as an Hopf algebra to $U(\mathfrak{h})[[h]]$, and the isomorphism is just the identity $H_i\mapsto H_i$. In fact, it is known that $U_{h}(\mathfrak{g})$ is isomorphic as an algebra to $U(\mathfrak{g})[[h]]$, but now the isomorphism is far from being trivial. Still, it clearly show that the important things is the whole Hopf algebra structure.

Therefore we know the coproduct for $K$, or at least we expect that there exists an Hopf structure such that $\Delta(K)=K\otimes K$. Now you can do some computation, write down the constraints imposed by co-associativity, still trying to get something "familiar" (ie s.t. $E$ and $F$ are close from being primitive). I may be wrong but I'm not sure that there is another way of finding something explicit, so what you did was right.

Now for general $\mathfrak g$ you can try to "glue together" copies of $U_q(\mathfrak{sl}_2)$. Again there is nothing obvious here, and finding these formulas was a great achievement. It seems to me, for example, that no similar deformation for non semisimple Lie algebras are known except in a few cases.

As I said in the abovementionned answer, the very much existence of a non-trivial deformation of the Hopf structure is a highly non-trivial fact, which "could have been false". The deep reason for which such object exists is really related to the theory of Drinfeld associators.

Answer 3

Since the question is on how to motivate such structure I think one should mention relations with Lie bialgebras. A quantized universal eneveloping algebra $U_q(\mathfrak g)$ quantizes a Lie bialgebra structure on $\mathfrak g$, which is the same as a coPoisson-Hopf algebra structure on $U(\mathfrak g)$ (definitions may be found in a standard reference as Chari-Pressley Quantum Groups).

The basic (additional) object in a Lie bialgebra is a Lie coproduct $\delta:\mathfrak g \to \wedge^2\mathfrak g$.

The basic (additional) object in a coPoisson Hopf algebra is a cobracket $\Delta_1:U(\mathfrak g)\to \otimes^2U(\mathfrak g)$.

One could understand properties of this map, formally, writing an $\hbar$ expansion for the deformed coproduct $\Delta=\Delta_0+\hbar \Delta_1+{\cal O}(\hbar^2)$ and looking for properties implied by coassociativity etc.

Now Lie bialgebra structures may be classified (at least in some cases). On $\mathfrak{sl}(2)$ essentially there is only one which on the usual $h,e,f$ generators is given by

$\delta(h)=0$, $\delta(e)=h\wedge e$, $\delta(f)=f\wedge h$

This tells us that the deformed coproduct should not, in fact, be deformed in the direction of the Cartan and suggest an expression for the deformed coproduct of the other generators. This bialgebra structure can be generalized to all simple Lie algebras: on simple roots it is given exactly by the same formulas as in $\mathfrak{sl}(2)$ and this is extended to non simple roots via the cocycle condition. This is, again, what is usually done in quantized universal enveloping algebras where the coproduct deformation is twisted primitive on simple generators and increasingly more complicated on non simple ones (which, in fact, can be defined as iterated $q$-commutators).