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This is Theorem VI.5.5 in Kassel's Quantum Groups:

Let $q$ be a root of unity. Prove: Any simple $U_q$-module of dimension $e$ is isomorphic to a module of the following list:

  1. $V(\lambda, a, b)$ with $b\neq 0$,
  2. $V(\lambda,a,0)$ where $\lambda\neq \pm q^{j-1}$, $1\leq j\leq e-1$
  3. $\widetilde{V}(\pm q^{1-j},c)$ with $c\neq 0$ and $1\leq j\leq e-1$

Here $e$ is the smallest integer s.t. $q^e=\pm 1$. The $U_q$-module structures are defined on an $e$-dimensional vector space $V$ with basis $\{v_0,\ldots,v_{e-1}\}$ like so.

  1. $V(\lambda\neq0,a,b)$:

    $Ev_0=av_{e-1}$, $\,Fv_{e-1}=bv_0$, $Kv_p = \lambda q^{-2p}v_p$, $Fv_p=v_{p+1}$, and $$ Ev_{p+1}=\left([p+1]\frac{q^{-p}\lambda - q^p\lambda^{-1}}{q-q^{-1}}+ab\right)v_p$$

  2. $\widetilde{V}(\mu\neq0,c)$:

    $Fv_0=0$, $Ev_{e-1}=cv_0$, $Kv_p = \mu q^{2p}v_p$, $Ev_p=v_{p+1}$, and $$Fv_{p+1}=[p+1]\frac{q^{-p}\mu^{-1} - q^p\mu}{q-q^{-1}} v_p $$


My approach to the proof is the following: Take an arbitrary simple $e$-dimensional $U_q$-module $W$. This module has a nonzero element $w$ s.t. $Kw=\alpha w$ for some $\alpha\in\mathbb{C}^x$. Then see if you can find the relations from above in there under some additional assumptions.

I found that if $Fw\neq 0$ and $F^e\neq 0$, then $W\cong V(\lambda, a, b)$ with $b\neq0$.

Next I wanted to look at a module with $Fw\neq0$ but $F^e = 0$, but I cannot see why $Ew_0=aw_{e-1}$ should be true --- here, $w_p \equiv Fw_{p-1}$ with $w_0\equiv w$. In the first case, I proved it using $$Ew_0 = \frac{1}{b}EFw_{e-1}$$ and then plugging in the Casimir element, but now that first step doesn't work anymore. Is the following possible? \begin{align*} Ew_0 \propto E^2 w_1 \propto \ldots\propto E^e w_{e-1}, \end{align*} but $E^e$ is in the center so acts by multiplication by a scalar. Thus, $Ew_0 = aw_{e-1}$. [Problem: what if $Ew_i$ vanishes somewhere?] Furthermore, assuming that $\alpha=\pm q^{j-1}$ leads to a contradiction: If $j=p+1$, then the form of $\alpha$ implies $$0=Ew_{p+1} = (FE + [E,F])w_p,$$ which is only possible if $\alpha =\pm q^{\frac{3}{2}}$.

For the last one I simply assume $Fw=0$, which implies $Ew\neq 0$, otherwise span of $w$ would be a 1-d submodule. Here we define $w_p = Ew_{p-1}$. It is easy to verify $Fw_{p+1}$, but again I am stuck with $Ew_{e-1}=cw_0$, and in particular I don't see why $c$ is nonzero.

Now, I would greatly appreciate any hint towards solving this. Thanks!


EDIT: To make it clear, my questions are basically:

  1. Where does $Ew_0 = a w_{e-1}$ come from in the second part?

  2. Where does $Ew_{e-1} = c w_0$ come from in the third part?

  3. Why does $c$ have to be nonzero?

  4. Why is the eigenvalue $\alpha$ of the form $\pm q^{j-1}$ in the last part?

These are the things that need to be accounted for, then the proof is finished.


EDIT #2:

I think the answer to question 1 is as follows. This derivation: \begin{align*} Ew_0 \propto E^2 w_1 \propto \ldots\propto E^e w_{e-1}, \end{align*} of $Ew_0=aw_{e-1}$ is valid iff not $\exists p$ s.t. $Ew_p=0$. And $Ew_p=0$ is equivalent to \begin{align*} q^{-(p-1)}\lambda - q^{p-1}\lambda^{-1} = 0. \end{align*} The existence of such a $p$ would also violate simplicity, btw.


EDIT #3:

Answers to the other questions: $KEw_{e-1}=\alpha Ew_{e-1}$ implies $Ew_{e-1}=cEw_0$, like before. If $\alpha=\pm q^{1-j}$, then $Fw_{p+1}=0$ for some $p$. The vectors $\{w_{p+1},..., w_{e-1}\}$ then span a $U_q$-submodule unless $c\neq 0$.

The only open question is now: Why is $\alpha$ of that form? Or rather: Is it true that, if $\alpha$ is not of that form, then the submodule is isomorphic to one of the first two?

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    $\begingroup$ If you don't receive an answer on Math Stack Exchange after a while, you might consider cross-posting this to Math Overflow (with a link to this original post). $\endgroup$
    – Tom Church
    Feb 4, 2018 at 20:01

1 Answer 1

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I will summarize my results in this answer.


If we assume $Fw\neq0$ and $F^e=0$, then the derivation of our module being isomorphic to $V(\lambda, a,0)$ is almost the same as the first one, but we need to derive $Ew_0=aw_{e-1}$ differently, since now $Fw_{e-1}=0$. We know \begin{align*} Ew_0 \propto E^2w_1\propto\ldots\propto E^ew_{e-1}, \end{align*} and since $E^e$ is in the center, this means $Ew_0 = aw_{e-1}$ for some (possibly zero) scalar $a$. Here it is important to note that we cannot have $Ew_p=0$ for some $w_p$, since then the subspace spanned by $\{w_p,...,w_{e-1}\}$ would be closed under $U_q$-action, contradicting simplicity.

But $Ew_p=0$ is equivalent to $\lambda = \pm q^{p-1}$ -- you should convince yourself of this if you don't see it. Thus we're done for this part.

For the last part, we of course assume the only possible case left: $Fw=0$. Necessarily, $Ew\neq 0$, otherwise $W$ would not be simple. And now I leave it as an easy exercise to verify that \begin{align*} \widetilde{V}(\pm q^{1-j},c) \cong V(\pm q^{j-1},0,c) \end{align*} via the Cartan automorphism, that is, the map \begin{align*}uv\mapsto \omega(u)v \end{align*} where $u\in U_q$, $v\in W$, and \begin{align*} \omega(E) = F, \quad \omega(F) = E \quad \omega(K) = K^{-1}, \end{align*} is an isomorphism of $U_q$-modules. Simplicity then implies $c\neq 0$

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