# Monday March 16th Proposition (Chains of Containments of Vermas for Dominant Integral) : Suppose $\lambda + \rho$ is dominant integral, then - $M(w\cdot \lambda) \subset M(\lambda)$ for all $w\in W$ - $[M(\lambda): L(w\cdot \lambda)] > 0$ for all $w\in W$ More precisely, if $w = s_1 \cdots s_\ell$ is reduced with $s_i = s_{\alpha_i}$ with $\alpha_i \in \Delta$ and $\lambda_k = s_k \cdots s_1 \cdot \lambda$, then $$ M(w\cdot \lambda) = M(\lambda_n) \subset M(\lambda_{n-1}) \subset \cdots \subset M(\lambda_0) = M(\lambda) $$ Moreover $(\lambda_k + \rho, \alpha_{k+1}\dual) \in \ZZ^+$ for $0\leq k \leq n-1$ and so $$ \lambda_n \leq \lambda_{n-1} \leq \cdots \leq \lambda_0 .$$ Proof : By induction on $n = \ell(w)$. The $n=0$ case is obvious. For $\ell(w) = k+1$, write $w'= s_k \cdots s_1$. From section 0.3, $(w')\inv \alpha_{k+1} > 0$. We can compute \begin{align*} (\lambda_k + \rho, \alpha_{k+1}\dual) &= (w' \cdots \lambda + \rho, \alpha_{k+1}\dual) \\ &= (w'(\lambda + \rho), \alpha_{k+1}\dual) \\ &= (\lambda + \rho, (w')\inv \alpha_{k+1}\dual) \\ &= (\lambda + \rho, ((w')\inv \alpha_{k+1})\dual) \\ &\in \ZZ^{+} \end{align*} since $\lambda + \rho \in \Lambda^+$ and $(w')\inv \alpha_{k+1} \in \Phi^+$. This means that $\lambda_{k+1} = s_{k+1} \lambda_k \leq \lambda_k$. By proposition 1.4, reformulated in terms of the dot action, we have a map $M(\lambda_{k+1}) \injects M(\lambda_k)$, and nonzero morphisms are injective by 4.2a. Exercise (4.3) : If $\lambda + \rho \in \Lambda^+$, $\soc M(\lambda) = M(w_o \cdot \lambda)$, and moreover if $\lambda \in \Lambda_0^+$ then the inclusions in the proposition are all proper. Remark : For general $\mu \in \Lambda$, it is not so easy to decide when $M(w\cdot \mu) \subset M(\mu)$. The basic problem is that Proposition 1.4 only works for *simple* roots, whereas we can have $s_\gamma \cdot \mu < \mu$ for $\gamma \in \Phi^+\setminus \Delta$ with no obvious way to constrct an embedding $M(s_\gamma \cdot \mu) \subset M(\mu)$. See the following example. Example : Let $\lieg = \liesl(3, \CC)$. \begin{center} \begin{tikzpicture} \pgfplotsset{every x tick label/.append style={font=\tiny, yshift=0.5ex}} \pgfplotsset{every y tick label/.append style={font=\tiny, xshift=0.5ex}} \begin{axis}[ xmin=-10, xmax=10, ymin=-2, ymax=2, xtick = {0}, ytick = {0}, axis equal, axis lines=middle, disabledatascaling] \node[font=\tiny] at (axis cs:-4,5) [anchor=north west] {$\beta$}; \node[font=\tiny] at (axis cs:5,0) [anchor=north west] {$\alpha$}; \node[font=\tiny] at (axis cs:5,5) [anchor=north west] {$\alpha + \beta$}; \node[font=\tiny] at (axis cs:6,8) [anchor=north west] {$\liesl(3, \CC)$}; \begin{scope}[thick, draw=blue] \draw[-][opacity=0.9, postaction={decorate}] (axis cs:-5.0, -5.0) -- (axis cs:5,5); \draw[-][opacity=0.9, postaction={decorate}] (axis cs:-5.0, 5.0) -- (axis cs:5,-5); \draw[-][opacity=0.9, postaction={decorate}] (axis cs:-5.0, 0.0) -- (axis cs:5,0); \end{scope} \end{axis} \end{tikzpicture} \end{center} We don't know if there's a diagonal map indicated by the question mark in the following diagram: \begin{center} \begin{tikzcd} & & \lambda \in \Lambda^+ & & \\ & & & & \\ \mu = s_\alpha \cdot \lambda \arrow[rruu] & & & & s_\beta \cdot \lambda \arrow[lluu] \\ & & & & \\ & & s_\alpha s_\beta \cdot \lambda = s_{\alpha + \beta} \lambda = s_\gamma \lambda \arrow[lluu, "?", dashed] \arrow[rruu] & & \end{tikzcd} \end{center} Next few sections: any root reflection that moves downward through the ordering induces a containment of Verma modules. ## (4.4) Simplicity Criterion: The Integral Case Theorem (Vermas Equal Quotients iff Antidominant Weight) : Let $\lambda \in \lieh\dual$ be any weight. Then $M(\lambda) = L(\lambda) \iff \lambda$ is antidominant. The proof for $\lambda$ integral is fairly easy, because antidominance reduces to a condition involving simple roots, where we can use our Verma module embedding criterion from Proposition 1.4. Proof (Integral Case) : Assume $\lambda \in \Lambda$. $\implies$: Assume $M(\lambda)$ is simple but $\lambda$ is not antidominant. Then since $\lambda \in \Lambda$, $(\lambda + \rho, \alpha\dual)$ is a positive integer for some $\alpha \in \Delta$. But then $s_\alpha \lambda < \lambda$ so $M(s_\alpha \cdot \lambda) \subset M(\lambda)$ by 1.4 and 4.2. But then $N(\lambda) \neq 0$, which contradicts irreducibility. $\impliedby$: Assume $\lambda$ is antidominant. By proposition 3.5, $\lambda < w\cdot \lambda$ for all $w\in W$. Since all composition factors of $M(\lambda)$ and $L(w\cdot \lambda)$ where $w\cdot \lambda \leq \lambda$. This can only happen if $w\cdot \lambda = \lambda$, and so the only possible composition factor is $L(\lambda)$. Since $[M(\lambda) : L(\lambda)]$ is always equal to one, $M(\lambda)$ is simple. Remark : The reverse implication still works in general if $W$ is replaced by $W_{[\lambda]}$. To extend the forward implication, we need to understand embeddings $M(s_\beta \cdots \lambda) \injects M(\lambda)$ when $\beta$ is not simple. ## Existence of Embeddings (Preliminaries) Lemma (Commuting Nilpotents) : Let $\mfa$ be a nilpotent Lie algebra (e.g. $\lien^-$) and $x\in \mfa, u\in U(\mfa)$, then for every $n\in \ZZ^+$ there exists a $t\in \ZZ^+$ such that $x^t u \in U(\mfa) x^n$. > See Engel's theorem Proof : Use the fact that $\ad x$ acts nilpotently on $U(\mfa)$, so there exists a $q\geq 0$ such that $\qty{\ad x}^{q+1}u = 0$. Let $\ell_x, r_x$ be left and right multiplication by $x$ on $U(\mfa)$. Then $\ad x = \ell_x - r_x$, and $\ell_x, r_x \ad x$ all commute. Choosing $t \geq q + n$, we have \begin{align*} x^t u &= \ell_x^t u \\ &= (r_x + \ad x)^t u \\ &= \sum_{i=0}^t {t \choose i} r_x^{t-i} \qty{\ad x}^i u \\ &= \sum_{i=0}^q {t \choose i} \qty{\qty{\ad x}^iu }x^{t-i} \\ &\in U(\mfa) x^{t-q} \\ &\subset U(\mfa) x^n \end{align*} > This will be useful when moving things around by positive roots that are not simple.