Where does the univeral $R$-matrix of $U_q(\mathfrak g)$ live?

Question

Let $\mathfrak g$ be a complex simple Lie algebra and let $U_q(\mathfrak g)$ denote the Drinfeld-Jimbo quantum group associated to $\mathfrak g$. I will assume that $U_q(\mathfrak g)$ is a $\mathbb C(q)$-algebra, where $\mathbb C(q)$ is the field of rational functions in the indeterminate $q$.

In the literature, it is usually stated that $U_q(\mathfrak g)$ is almost quasitriangular, but NOT exactly so. Then it is stated that the universal $R$-matrix of $U_q(\mathfrak g)$ lives in some completion of $U_q(\mathfrak g){\otimes} U_q(\mathfrak g)$, and it is a formal sum $R=\sum r\otimes r'$.

However, Chari and Pressley describe the $R$-matrix of $U_q(\mathfrak g)$ in their book (Section 10.1.D) slightly differently, which seems to contradict the above facts. First, they give the formula for the $R$-matrix of the $h$-adic algebra $U_h(\mathfrak g)$: $$ R=\left(e^{h\sum B_{ij}}(H_i\otimes H_j)\right) \left( \prod_{\beta\in\Delta^+} \exp((q_\beta-q_\beta^{-1})E_\beta\otimes F_\beta) \right). $$ Then they define a certain completion of $\widehat{U}_q(\mathfrak g)=\varprojlim U_q/U_qU_q^{+,r}$. They state that the second factor in the above formula is in $\widehat{U}_q(\mathfrak g)\widehat{\otimes} \widehat{U}_q(\mathfrak g)$, but the first factor is NOT there. (Then they explain a workaround due to Tanisaki to circumvent this issue.)

Here are my questions:

1. Is the universal $R$-matrix of $U_q(\mathfrak g)$ actually realizable in some completion of $U_q(\mathfrak g)$? Or an algebra that contains $U_q(\mathfrak g)$? One could say it is realizable in $U_h(\mathfrak g)$. However, it looks like there are two issues with that: (a) strictly speaking, $U_q(\mathfrak g)$ is not a subalgebra of $U_h(\mathfrak g)$ and (b) strictly speaking, modules of $U_q(\mathfrak g)$ are not modules of $U_h(\mathfrak g)$.

2. Is it actually possible to express $R$ as a formal sum $R=\sum r\otimes r'$, such that at least on some well-behaved class of modules one can evaluate the summation term by term, i.e., all but finitely many summands vanish? (The class of modules could be finite dimensional modules of type 1.)

Answer 1

There is a natural choice for such a completion: let $$\bar U=\prod_\lambda End(V_\lambda)$$

where the product runs over simple finite dimensional modules and $End$ means linear (not module) endomorphisms, and likewise let $$\bar{U} \hat \otimes \bar U = \prod_{\lambda,\mu} End(V_{\lambda}) \otimes End(V_\mu).$$

Then the collection of maps $U_q \rightarrow End(V_\lambda)$ given by the action of $U_q$ extends to an injective algebra map $U_q\rightarrow \bar U$, $\bar U$ is a topological Hopf algebra whose coproduct lands in $\bar U \hat\otimes \bar U$, and the $R$-matrix lives in $\bar U \hat\otimes \bar U$ precisely because it acts in a well-defined way on tensor products of f.d. modules (see https://arxiv.org/abs/0707.2248 for a careful construction of this completion). Note this also answers your second question, you can write $R$ as an infinite formal sum of $R_{\lambda,\mu}\in End(V_{\lambda}) \otimes End(V_\mu)$ and all but finitley many of those acts by 0 on $V \otimes W$ for finite dimensional modules $V,W$.

This is a particular case of a general construction: for any (Hopf) algebra $H$ you can take the categorical end $\bar H=\int_V End(V)$ over all finite dimensional modules. Somewhat more abstractly, this is the algebra of endomorphism of the forgetful functor from the category of finite dimensional $H$-modules to $Vect$. So an element of $\bar H$ is a collection of linear maps $l_V$ for all finite dimensional $H$-modules $V$ such that $l_Wf=fl_V$ for all $H$-module map $f:V\rightarrow W$. The kernel of the map $H\rightarrow \bar H$ is the ideal of elements acting by 0 on all finite dimensional modules.

Because the category of f.d. $U_q$ modules is semi-simple this reduces to the formula above in that case, and this is also why the canonical map $U_q \rightarrow \bar U$ is injective (and likewise for its tensor square), which is why you can really think of the $R$-matrix as an actual element of $\bar U \hat \otimes \bar U$.

That being said, I think it's often clearer to think of $F_q$ comodules instead, where $F_q$ is the Hopf dual of $U_q$, and of the co-R-matrix which is thus a pairing on $F_q$. This way there is no need to talk about completion and such, the category of all (non-necessarily finite dimensional) $F_q$-comodules is braided, etc.

From that point of view, note that $\bar U$ is nothing but the profinite algebra obtained by taking the full dual of the coalgebra $F_q$.

Answer 2

I can't answer your question, but I think I can at least highlight the difficulty.

There are two parts to the $R$-matrix, as you noted: let's call them $HH = \exp(h \sum B_{ij} H_i \otimes H_j)$ and $\Theta$. $\Theta$ is relatively well-behaved: it's not in $U_q$ but it does live in an appropriate completion. In addition (2) is possible: because $E$ and $F$ act nilpotently on any finite-dimensional highest weight module (or on their direct sums) $\Theta$ will always converge. However, if you allow any highest weight then you might need arbitrarily many terms in $\Theta$: you won't necessarily be able to truncate it once and for all.

The tricky part is $HH$. For $\mathfrak{g} = \mathfrak{sl}_2$ in the normalization I'm used to it's $\exp(H \otimes H/2)$, so it acts on a tensor product of weight vectors as $$ HH \cdot v_\mu \otimes v_\nu = q^{\mu \nu /2}. $$ It is hard to come up with an element of $U_q \otimes U_q$ (or a completion thereof) that gives a product of weights like this, which I think is the difficulty. Depending on parity $q^{\mu \nu/2}$ might not even lie in $\mathbb{Z}[q,q^{-1}]$.

In a quantum topology context one can fix this by using ribbon elements to change the normalization, but that may not be useful to you.

PS: You are correct that the $R$-matrix does not really live in a completion of $U_q$: it does exist in the $h$-adic quantum group, but that's not a completion of $U_q$. Strictly speaking $U_q$ is not quasitriangular but is instead braided [R]. Because this difference doesn't usually matter for $q$ generic it doesn't come up a lot, but it can matter in some contexts.

Reshetikhin, N., Quasitriangularity of quantum groups at roots of 1, Commun. Math. Phys. 170, No. 1, 79-99 (1995). ZBL0838.17009.