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Let $V_N$ denote the $N$-dimensional representation of the quantum group $U_q(\mathfrak s\mathfrak l_2)$. I am told that in the limit $N\to\infty$ with $q=e^{2\pi i/n}$ and $N/n\to\alpha\in(0,1)$, the representation $V_N$ of $U_q(\mathfrak s\mathfrak l_2)$ "converges" in some precise sense to an infinite-dimensional representation of the complex Lie group $\operatorname{SL}_2(\mathbb C)$.

What is the precise formulation of this result, and where is it discussed in the literature? [My searches so far have not found anything substantial.]

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    $\begingroup$ I think arxiv.org/abs/hep-th/0312282v2 contains something like what you want, but for $N=n.$ Unfortunately I wasn't able to find anything more general/precise by looking through papers that reference it. $\endgroup$
    – dhy
    Oct 1, 2018 at 3:03

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This is a very interesting question. I have also made some search but i have not found this result explicitly mentioned somewhere in the literature. However, i remember i have heard such a claim in the past. So i will try to record my thoughts on the problem. Here is the way i understand it:

Since you are considering the limits $N\to \infty$ and $N/n\to\alpha\in(0,1)$, this implies that $n\to\infty$ thus $q\to 1$. In this limit: $U_q(\mathfrak{sl}_2)\stackrel{q\to 1}{\longrightarrow} U(\mathfrak{sl}_2)$ and so you get a representation of the UEA and thus of the lie algebra $\mathfrak{sl}_2$. Since the corresponding lie group $SL(2,\mathbb{C})$, is simply connected, its category of representations is equivalent to the category of representations of the lie algebra $\mathfrak{sl}_2$ (in the sense that there is a bijection -up to isomorphism- between the lie group and the lie algebra representations). Thus, we finally arrive at a representation (up to isomorphism) of the lie group $SL(2,\mathbb{C})$. And since $N\to\infty$, the limit -if it exists- will be an infinite dim representation.

Let us try to examine the situation a little closer and attempt to establish whether -and when- the limit is well defined and exists :

The "convergence" of the quantum group to the lie algebra: A non-trivial point in the preceding argument is how to obtain a rigorous formulation of the limit:

$U_q(\mathfrak{sl}_2)\stackrel{q\to 1}{\longrightarrow} U(\mathfrak{sl}_2)$

This is a subtle point, which -imo- does not have a unique answer; it depends on the description one uses for the quantum group $U_q(\mathfrak{sl}_2)$. Generally, we can either view $U_q(\mathfrak{sl}_2)$ as a $\mathbb{C}[[h]]$-algebra and write $q=e^{h/2}$ and $K=q^H$ (in which case $q\to 1\Leftrightarrow h\to 0$ and we use Del' Hospital for the $E$-action limit and power series expansion for the $K$-action limit), or we can view $U_q(\mathfrak{sl}_2)$ as a suitable quotient of a larger $\mathbb{C}$-algebra, which is however well defined at $q=1$. Both methods are well-known. See for example:
- How $U_{q}(\mathfrak{sl}_{2})$ becomes the universal enveloping algebra $U(\mathfrak{sl}_{2})$ of $\mathfrak{sl}_{2}$,
- Quantized Enveloping Algebras at $q=1$,
- Exponentiations over the quantum algebra $U_{q}(\mathfrak{sl}_{2})$ (see the discussion and the computations of section 10, p. 65)

The "convergence" of the representation: Let us study the limit on the irreducible, f.d. representations of $U_{q}(\mathfrak{sl}_{2})$ with $q=e^{2\pi i/n}$:
(First recall that if $q$ is not a root of unity, then the f.d., irred, are highest weight representations. They are parameterized by $\varepsilon=\pm 1$ and the positive integers, i.e. we will denote such $N$-dim, irred, modules as $V_{\varepsilon, N-1}$. The notion of the $N\to\infty$ limit here is well defined in the following sense: The $V_{1,N-1}$ modules here have matrix elements which are not generally continuous functions of $q$ but they can be handled with methods similar to those mentioned above to show that

$V_{1,N-1}\stackrel{q\to 1}{\longrightarrow}V_{N-1}$

where $V_{N-1}$ are the usual $N$-dim, irred, highest weight reps of $\mathfrak{sl}_2$. The $V_{-1,N-1}$ modules "vanish" when $q\to 1$. One can be more precise at that point but it would be irrelevant to the rest. We can equally well consider that we are taking the $N\to\infty$ limit considering the $V_{1,N-1}$ family only).
Now, let us come back to the root of unity case: $q=e^{2\pi i/n}$. It is well known that, the irreducible modules now have an upper bound in their dimension: ${\small e=\begin{cases} n, & \text{$n$: odd} \\ n/2, & \text{$n$: even.} \end{cases}}$. It is also well known that the $N$-dim irred, representations, now fall into two classes: First, representations for which $N<e$. These are of the form $V_{\varepsilon,N-1}$, with $\varepsilon=\pm 1$, i.e. they are exactly the same modules as in the non-root of unity case; i.e. the highest weight modules. However, the second class includes irred. reps with dimension $N=e$. These look quite different than the highest weight modules; they generally fall into distinct subclasses including cyclic, semi-cyclic, etc modules. So it would be wise to exclude these $\dim V_{N-1}=N=e$ modules from the limitting procedure (if we want some reasonable sense of convergence).
Thus, while $N\to\infty$, it is important to keep in mind that $N<e$. This is -imo- the meaning of the $\alpha$ parameter: $N/n\to\alpha\in(0,1)$, which in some sense expresses the "relative rate of divergence" between the dimension $N$ of the rep and the order $n$ of the root of unity $q$.

Let us now attempt to compute these limits; i will not include details on the handling of the order of the limits $N/n\to\alpha$, $n\to\infty$, since i admit i have not -yet- been able to obtain a rigorous formulation of the following but here is what i have got:

  • First consider the limit $N/n\to\alpha$. We get the result:

    $V_{1,N-1}\stackrel{N/n\to \alpha}{\longrightarrow}V\big(q^{n\alpha}\big)=V\big(e^{2\pi i\alpha}\big)$

where $V_{1,N-1}$ is the $N$-dimensional, irreducible, $U_q\big(\mathfrak{sl}(2)\big)$-module of highest weight $\lambda=q^{N-1}$ and the $V\big(q^{n\alpha}\big)$ is the $U_q\big(\mathfrak{sl}(2)\big)$-Verma module of highest weight $\lambda=q^{n\alpha}=e^{2\pi i\alpha}$.
The $U_q\big(\mathfrak{sl}(2)\big)$-action for the $V_{1,N-1}$ module is given by: $$ K\cdot v_p=q^{N-2p-1}v_p, \ \ E\cdot v_p=[N-p]_qv_{p-1}, \ \ F\cdot v_p=[p+1]_qv_{p+1} $$ whereas its $N/n\to\alpha$ limit, gives the $U_q\big(\mathfrak{sl}(2)\big)$-action for the $V\big(q^{n\alpha}\big)$-Verma module: $$ K\cdot v_p=q^{n\alpha-2p-1}v_p, \ \ E\cdot v_p=[n\alpha-p]_qv_{p-1}, \ \ F\cdot v_p=[p+1]_qv_{p+1} $$ where $[x]_q=\frac{q^x-q^{-x}}{q-q^{-1}}$, as usual.

  • The next step will be to compute the Verma module $V\big(q^{n\alpha}\big)$ limit at $q\to 1\Leftrightarrow n\to\infty$;
    Conjecture: This produces a $U\big(\mathfrak{sl}(2)\big)$-Verma module $V({\small 4\pi i \alpha-1})$:

    $V\big(q^{n\alpha}\big)\stackrel{n\to\infty}{\longrightarrow}V({\small 4\pi i \alpha-1})$

with $U\big(\mathfrak{sl}(2)\big)$-action given by: $$ H\cdot v_p=-(4\pi i\alpha-2p-1)v_p, \ \ E\cdot v_p=-(4\pi i\alpha-p)v_{p-1}, \ \ F\cdot v_p=(p+1)v_{p+1} $$ This is an infinite dimensional, Verma module, of highest weight $4\pi i \alpha-1$, for the (undeformed) universal enveloping algebra $U\big(\mathfrak{sl}(2)\big)$.

In any case, if there is not some silly mistake in my conjecture (the result comes from a mixture of computations and intuition), we finally arrive at an infinite dimensional $U\big(\mathfrak{sl}(2)\big)$-Verma module and from there we can get the corresponding infinite dimensional, $SL(2,\mathbb{C})$ representation. I will try to come back if -and when- i will be able to obtain a somewhat more rigorous formulation of the last limit.

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    $\begingroup$ A couple of comments: (1) in order for the limit to "converge", it is important to fix the parity of $N$ as it tends to $\infty$. (2) In order for the limit to be a representation as opposed to a representation-up-to-isomorphism one can fix, throughout the limiting process, a trivialization of one of the (one-dimensional) weight spaces -- e.g. the zero-weight space (assuming $N$ has been chosen to always be odd). $\endgroup$ Apr 22, 2019 at 11:13
  • $\begingroup$ 37 edits in as many days seems excessive $\endgroup$
    – Yemon Choi
    May 24, 2019 at 4:38

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