# $\S21.2$: A sufficient condition for finite-dimensionality (Monday, November 21) :::{.remark} Last time: if $V\in \lmod^\fd$ then $V = L( \lambda)$ for some dominant weight $\lambda\in \Lambda^+$, yielding a necessary condition for finite-dimensionality. Today: a sufficient condition. ::: :::{.lemma title="?"} Write \( \Delta = \tsl \alpha 1 n \) and set $x_i \da x_{ \alpha_i}, y_i \da y_{\alpha_i}$. For $k\geq 0$ and $1\leq i,j\leq \ell$, the following relations hold in $U(L)$: a. $[x_j, y_i^{k+1}] = 0$ for $i\neq j$ b. $[h_j, y_i^{k+1}] = - (k+1) \alpha_i(h_j) y_i^{k+1}$ c. $[x_i, y_i^{k+1}] = (k+1) y_i^k(h_i - k\cdot 1)$ ::: :::{.proof title="of (c)"} Use that $\ad$ acts by derivations: \[ [x_i, y_i^{k+1}] &= x_i y_i^{k+1} - y_i^{k+1} x_i \\ &= x_i y_i y_i^k - y_i x_i y_i^k + y_i x_i y_i^k - y_i y_i^k x_i \\ &= [x_i y_i] y_i^k + y_i[x_i y_i^k] \\ &= h_i y_i^k + y_i [x_i y_i^k] \\ &= \qty{ y_i^k h_i - k \alpha_i (h_i) y_i^k } + y_i \qty{k y_i^{k-1} (h_i - (k-1)\cdot 1) } \qquad\text{by (b) and induction} \\ &= y_i^k h_i - 2k y_i^k + ky_i^k (h_i - (k-1)\cdot 1) \\ &= (k+1)y_i^k h_i - (2k + k(k-1)) y_i^k \\ &= (k+1) y_i^k h_i - k(k+1) y_i^k \\ &= (k+1)y_i^k (h_i - k\cdot 1) .\] ::: :::{.theorem title="?"} Given $V\in \lmod$, let $\Pi(V) \da \ts{ \lambda\in H\dual \st V_ \lambda\neq 0}$ be the set of weights. If \( \lambda\in \Lambda^+ \) is a dominant weight, then $V \da L(\lambda)\in \lmod^\irr$ is finite-dimensional and $\Pi(V)$ is permuted by $W$ with $\dim V_ \mu = \dim V_ { \sigma \mu}$ for all $\sigma\in W$. ::: :::{.proof title="?"} The main work is showing the last part involving equality of dimensions. It STS this for a simple reflection $s_i \da s_{\alpha_i}$ since $\sigma$ is a product of such reflections. Write $\phi: L\to \liegl(V)$ be the representation associated to $V$ -- the strategy is to show that $\phi(x_i)$ and $\phi(y_i)$ are locally nilpotent endomorphisms of $V$. Let $v^+ \in V_ \lambda\smz$ be a fixed maximal vector and set $m_i \da \lambda(h_i)$ so $h_i.v^+ = m_i v^+$. 1. Set $w \da y_i^{m_i+1}. v^+$, then the claim is that $w=0$. Supposing not, we'll show $w$ is a maximal vector of weight not equal to \( \lambda \), and thus not a scalar multiple of $v^+$. We have $\wt( w) = \lambda- (m_i +1) \alpha_i < \lambda$ (a strict inequality). If $j\neq i$ then $x_j.w = x_j y_i^{m_i + 1} v^+ = y_i^{m_i + 1} x_j v$ by part (a) of the lemma above, and this is zero since $v^+$ is highest weight and thus maximal (recalling that these are distinct notions). Otherwise \[ x_i w &= x_i y_i^{m_i + 1} v^+ \\ &= y_i^{m_i+1} x_i v^+ + (m_i + 1) y_i^{m_i} (h_i - m_i\cdot 1)v^+ \\ &= 0 + (m_i+1)y_i^{m_i}(m_i - m_i) v^+ = 0 .\] So $w$ is a maximal vector of weight distinct from $\lambda$, contradicting corollary 20.2 since this would generate a proper submodule. $\contradiction$ 2. Let $S_i = \gens{x_i, y_i, h_i} \cong \liesl_2$, then the claim is that $v^+, y_i v^+, \cdots, y_i^{m_i} v^+$ span a nonzero finite-dimensional $S_i\dash$submodule of $V$. The span is closed under (the action of) $h_i$ since all of these are eigenvectors for $h_i$, and is closed under $y_i$ since $y_i$ raises generators and annihilates $y_i^{m_i} v^+$, and is closed under $x_i$ by part (c) of the lemma (since it lowers generators). 3. The sum of two finite-dimensional $S_i\dash$submodules of $V$ is again a finite-dimensional $S_i\dash$submodule, so let $V_i$ be the sum of all finite-dimensional $S_i\dash$submodules of $V$ (which is not obviously finite-dimensional, since we don't yet know if $V$ is finite-dimensional). The claim is $V'$ is a nonzero $L\dash$submodule of $V$. Let $w\in V'$, then $w$ is a finite sum and there exists a finite-dimensional $S_i\dash$submodule $W$ of $V$ with $w\in W$. Construct $U \da \sum_{ \alpha \in \Phi} x_ \alpha . W$ where $x_ \alpha \da y_{- \alpha}$ if $\alpha\in \Phi^-$, which is a finite-dimensional vector subspace of $V$. Check that \[ h_i (x_ \beta . W) &= x_ \beta(h_i.W) + [h_i x_ \beta].W \subseteq x_ \beta.W \subseteq U \\ x_i (x_ \beta.W) &= x_ \beta(x_i.W) + [x_i x_ \beta] .W \subseteq x_ \beta .W + x _{ \beta + \alpha_i}.W \subseteq U \\ y_i(x_ \beta. W) &= x_{ \beta}(y_i.W) + [y_i x_ \beta].W \subseteq x_ \beta W + x_{ \beta - \alpha_i}.W \subseteq U ,\] and so $U$ is a finite-dimensional $S_i\dash$submodule of $V$ and thus $U \subseteq V'$. So if $w\in V'$ then $x_ \alpha .w \in V'$ for all \( \alpha\in \Phi \) and $V'$ is stable under a set of generators for $L$, making $V' \leq V$ an $L\dash$submodule. Since $V'\neq 0$ (since it contains at least the highest weight space), it must be all of $V$ since $V$ is irreducible. 4. Given an arbitrary $v\in V$, apply the argument for $w$ in step 3 to show that there exists a finite-dimensional $S_i\dash$submodule $W \subseteq V$ with $v\in W$. The elements $x_i, y_i$ act nilpotently on any finite-dimensional $S_i\dash$module, and so in particular they act nilpotently on $v$ and we get local nilpotence. 5. Now $\tau_i \da e^{\phi(x_i)} \circ e^{\phi(-y_i)} \circ e^{\phi(x_i)}$ is well-defined. As seen before, $\tau_i(V_ \mu) = V_{s_i \mu}$, i.e. $\tau_i$ is an automorphism that behaves like $s_i$, and so $\dim V_ \mu = \dim V_{ \sigma \mu}$ for all $\mu\in \Pi(V)$ and $\sigma\in W$. Now any $\mu\in \Pi(V)$ is conjugate under $W$ to a unique dominant weight $\mu^+$, and by (4) $\mu^+\in \Pi(V)$ and since the weights in $V = L(\lambda)$ has only weights smaller than $\lambda$, we have $\mu^+ \leq \lambda$. Note $\lambda \in \Lambda^+$ is dominant, and so by 13.2B there are only finitely many such weights $\mu^+$. Now there are only finitely many conjugates of the finitely many possibilities for $\mu^+$, so $\size \Pi(V) < \infty$. By the general theory of highest weight modules, all weight spaces $V_{\mu} \leq L( \lambda)$ are finite-dimensional. Since $\dim V_\mu < \infty$ for all $\mu\in\Pi(V)$, we have $\dim V < \infty$. ::: :::{.remark} Skipping the next two sections $\S 21.3$ on weight strings and weight diagrams, and $\S 21.4$ on generators and relations for $L( \lambda)$ for \( \lambda\in \Lambda^+ \) dominant. :::