# Wednesday February 5th Recall the Harish-Chandra morphism $\xi$: \begin{center} \begin{tikzcd} \mcz(\lieg) \arrow[rr, hook] \arrow[rrdd, "\xi", dashed] & & U(\lieg) = U(\lieh) \oplus (\lien^- U(\lieg) + U(\lieg)\lien) \arrow[dd, "\mathrm{pr}"] \\ & & \\ & & U(\lieh) \end{tikzcd} \end{center} If $M$ is a highest weight module of highest weight $\lambda$ then $z\in \mcz(\lieg)$ acts on $M$ by scalar multiplication. Note that if we have $\chi_\lambda(z)$ where $z\cdot v = \chi_\lambda(z) v$ for all $v\in M$, we can identify $\lambda(\mathrm{pr}(z)) = \lambda(\xi(z))$. ## Central Characters and Linkage The $\chi_\lambda$ are not all distinct -- for example, if $M(\mu) \subset M(\lambda)$, then $\chi_\mu = \chi_\lambda$. More generally, if $L(\mu)$ is a subquotient of $M(\lambda)$ then $\chi_\mu = \chi_\lambda$. So when do we have equality $\chi_\mu = \chi_\lambda$? Given $\lieg \supset \lieh$ with $\Phi \supset \Phi^+ \supset \Delta$, then define $$\rho = \frac 1 2 \sum_{\beta \in \Phi^+} \beta \in \lieh\dual.$$ Note that $\alpha \in \Delta \implies s_\alpha \rho = \rho - \alpha$. Definition (Dot Action) : The *dot action* of $W$ on $\lieh\dual$ is given by $$w\cdot \lambda = w(\lambda + \rho) - \rho,$$ which implies $(\rho, \alpha\dual) = 1$ for all $\alpha \in \Delta$. Then $\rho = \sum_{i=1}^\ell w$. Exercise : Check that this gives a well-defined group action. Definition (Linkage Class) : $\mu$ is *linked* to $\lambda$ iff $\mu = w\cdot \lambda$ for some $w\in W$. Note that this is an equivalence relation, with equivalence classes/orbits where the orbit of $\lambda$ is $\theset{w\cdot \lambda \suchthat w\in W}$ is called the *linkage class* of $\lambda$. Note that this is a finite subset, since $W$ is finite. Orbit-stabilizer applies here, so bigger stabilizers yield smaller orbits and vice-versa. Example : $w\cdot (-\rho) = w(-\rho + \rho) - \rho = -\rho$, so $-\rho$ is in its own linkage class. Definition (Dot-Regular) : $\lambda \in \lieh\dual$ is *dot-regular* iff $\abs{W\cdot \lambda } = \abs{W}$, or equivalently if $(\lambda + \rho, \beta\dual) \neq 0$ for all $\beta \in \Phi$. To think about: does this hold if $\Phi$ is replaced by $\Delta$? We also say $\lambda$ is *dot-singular* if $\lambda$ is not dot-regular, or equivalently $\stab_{W\cdot}\lambda \neq \theset{1}$. > I.e. lying on root hyperplanes. Exercise : If $0\in \lieh\dual$ is regular, then $-\rho$ is singular. ![Image](figures/2020-02-05-09:26.png)\ Proposition (Weights in Weyl Orbit Yield Equal Characters) : If $\lambda \in \Lambda$ and $\mu \in W\cdot \lambda$, then $\chi_\mu = \chi_\lambda$. Proof : Start with $\alpha \in \Delta$ and consider $\mu = s_\alpha \cdot \lambda$. Since $\lambda \in \Lambda$, we have $n\definedas (\lambda ,\alpha\dual) \in \ZZ$ by definition. There are three cases: 1. $n\in \ZZ^+$, then $M(s_\alpha \cdot \lambda) \subset M(\lambda)$. By Proposition 1.4, we have $\chi_\mu =\chi_\lambda$. 2. For $n=-1$, $\mu = s_\alpha \cdot \lambda = \lambda + \rho -(\lambda + \rho, \alpha\dual)\alpha - \rho = \lambda + n+1 = \lambda + 0$. So $\mu = \lambda$ and thus $M_\mu = M_\lambda$. 3. For $n\leq -2$, \begin{align*} (\mu, \alpha\dual) &= (s_\alpha \cdot \lambda , \alpha\dual) \\ &= (\lambda i (n+1)\alpha, \alpha\dual) \\ &= n - 2(n+1) \\ &= -n-2 \\ &\geq 0 ,\end{align*} so $\chi_\mu = \chi_{s_\alpha \cdot \mu} = \chi_{s\alpha \cdot (s_\alpha \cdot \lambda)} = \chi_\lambda$. Since $W$ is generated by simple reflections and the linkage property is transitive, the result follows by induction on $\ell(w)$. Exercise (1.8) : See book, show that certain properties of the dot action hold (namely nonlinearity). ## 1.9: Extending the Harish-Chandra Morphism We want to extend the previous proposition from $\lambda \in \Lambda$ to $\lambda \in \lieh\dual$. We'll use a density argument from affine algebraic geometry, and switch to the Zariski topology on $\lieh\dual \subset \CC^n$. Fix a basis $\Delta = \theset{a_1, \cdots, a_\ell}$ and use the Killing form to identify these with a basis for $\lieh = \theset{h_1, \cdots, h_\ell}$. Similarly, take $\theset{w_1, \cdots, w_\ell}$ as a basis for $\lieh\dual$, and we'll use the identification \begin{align*} \lieh\dual &\iff \AA^\ell \\ \lambda &\iff (\lambda(h_1), \cdots, \lambda(h_\ell)) .\end{align*} We identify $U(\lieh) = S(\lieh) = \CC[h_1, \cdots, h_\ell]$ with $P(\lieh\dual)$ which are polynomial functions on $\lieh\dual$. Fix $\lambda \in \lieh\dual$, extended $\lambda$ to be a multiplicative function on polynomials. For $f\in \CC[h_1, \cdots, h_\ell]$, we defined $\lambda(f)$. Under the identification, we send this to $\tilde f$ where $\tilde f(\lambda) = \lambda(f)$. > Note: we'll identify $f$ and $\tilde f$ notationally going forward and drop the tilde everywhere. Then $W$ acts on $P(\lieh\dual)$ by the dot action: $(w\cdot \tilde f)(\lambda) = \tilde f(w\inv \cdot \lambda)$. Exercise : Check that this is a well-defined action. Under this identification, we have \begin{align*} \lieh\dual &\iff \AA^\ell \\ \Lambda &\iff \ZZ^\ell .\end{align*} Note that $\Lambda$ is discrete in the analytic topology, but is *dense* in the Zariski topology. Proposition (Polynomials Vanishing on a Lattice Are Zero) : A polynomial $f$ on $\AA^\ell$ vanishing on $\ZZ^\ell$ must be identically zero. Proof : For $\ell = 1$: A nonzero polynomial in one variable has only finitely many zeros, but if $f$ vanishes on $\ZZ$ it has infinitely many zeros. For $\ell > 1$: View $f\in \CC[h_1, \cdots, h_{\ell-1}][h_\ell]$. Substituting any fixed integers for the $h_i$ for $i\leq \ell - 1$ yields a polynomial in one variable which vanishes on $\ZZ$. By the first case, $f \equiv 0$, so the coefficients must all be zero and the coefficient polynomials in $\CC[h_1, \cdots ,h_{\ell-1}]$ vanish on $\ZZ^{\ell-1}$. By induction, these coefficient polynomials are identically zero. Corollary (Lattices Are Zariski-Dense in Affine Space) : The only Zariski-closed subset of $\AA^\ell$ containing $\ZZ^\ell$ is $\AA^\ell$ itself, so the Zariski closure $\bar{\ZZ^\ell} = \AA^\ell$ and $\ZZ^\ell$ is dense in $\AA^\ell$.