$. Let $\tilde\nn_+$ be the subalgebra of $\tilde\gg$ generated by $\{e_i\}_{i=1}^n$ and let $\tilde\nn_-$ be the subalgebra of $\tilde\gg$ generated by $\{f_i\}_{i=1}^n$; the algebras $\tilde\nn_{\pm}$ are called the ``upper-'' and ``lower-triangular'' subalgebras. \end{lemmadef} \begin{prop} \label{triangulardecompprop} Let $\Delta,\tilde\gg,\tilde\hh,\tilde\nn_\pm$ be as in \cref{tildegdef}. Then $\tilde\gg = \tilde\nn_- \oplus \tilde\hh \oplus \tilde\nn_+$ as vector spaces; this is the ``\indexing{triangular decomposition}'' of $\tilde\gg$. \end{prop} \begin{proof} That $\tilde\nn_-,\tilde\hh,\tilde\nn_+$ intersect trivially follows form the grading, so it suffices to show that $\tilde\gg = \tilde\nn_- + \tilde\hh + \tilde\nn_+$. By inspecting the relations, we see that $(\ad f_i)\tilde\nn_- \subseteq \tilde\nn_-$, $(\ad f_i) \tilde\hh \subseteq \\subseteq \tilde\nn_-$, and $(\ad f_i)\tilde\nn_+ \subseteq \tilde\hh + \tilde\nn_+$. Therefore $\ad f_i$ preserves $\tilde\nn_- + \tilde\hh + \tilde\nn_+$, $\tilde\hh$ does so obviously, and $\ad e_i$ does so by the obvious symmetry $f_i \tofrom e_i$. Therefore $\tilde\nn_- + \tilde\hh + \tilde\nn_+$ is an ideal of $\tilde\gg$ and therefore a subalgebra, but it contains all the generators of $\tilde\gg$. \end{proof} \begin{prop} \label{VermaProp} Let $\Delta,\tilde\gg$ be as in \cref{tildegdef}, and let $\lambda\in \hh^*$. Write $\CC\< f_1,\dots,f_n\>$ for the free algebra generated by noncommuting symbols $f_1,\dots,f_n$ and $M_\lambda \defined \CC\< f_1,\dots,f_n\> v_\lambda$ for its free module generated by the symbol $v_\lambda$. Then there exists an action of $\tilde\gg$ on $M_\lambda$ such that: \begin{align} f_i \left( \prod f_{j_k} v_\lambda \right) & = \left( f_i\prod f_{j_k} \right) v_\lambda \\ h_i \left( \prod f_{j_k} v_\lambda \right) & = \left( \lambda(h_i) - \sum_k \aaa_{i,j_k} \right) \left( \prod f_{j_k} v_\lambda \right) \\ e_i \left( \prod f_{j_k} v_\lambda \right) & = \sum_{k \st j_k = i} f_{j_1} \cdots f_{j_{k-1}} h_i f_{j_{k+1}} \cdots f_{j_l} \,v_\lambda \end{align} \end{prop} \begin{proof} We have only to check that the action satisfies the relations \crefrange{gtilderel1}{gtilderel4}. The $Q$-grading verifies \cref{gtilderel1,gtilderel2,gtilderel4}; we need only to check \cref{gtilderel3}. When $i\neq j$, the action by $e_i$ ignores any action by $f_j$, and so we need only check that $[e_i,f_i]$ acts by $h_i$. Write $\underline{f}$ for some monomial $f_{j_1} \cdots f_{j_n}$. Then $e_if_i(\underline f v_\lambda) = e_i(f_i\underline f v_\lambda) = h_i\underline f v_\lambda + f_ie_i(\underline f v_\lambda)$, clear by the construction. \end{proof} \begin{defn} The $\tilde\gg$-module $M_\lambda$ defined in \cref{VermaProp} is the \emindexing{Verma module} of $\tilde\gg$ with weight\index{weight!of a Verma module} $\lambda$. \end{defn} \begin{cor} The map $\hh \to \tilde\hh$ is an isomorphism, so $\hh \mono \tilde\hh$. The upper- and lower-triangular algebras $\tilde\nn_-$ and $\tilde\nn_+$ are free on $\{f_i\}$ and $\{e_i\}$ respectively. \end{cor} \begin{prop} \label{properidealsofg} Assume that $\Delta$ is an indecomposable system of simple roots, in the sense that the Dynkin diagram of the Cartan matrix of $\Delta$ is connected. Construct $\tilde\gg$ as in \cref{tildegdef}. Then any proper ideal of $\tilde\gg$ is graded, contained in $\tilde\nn_- + \tilde\nn_+$, and does not contain any $e_i$ or $f_i$. \end{prop} \begin{proof} The grading on $\tilde\gg$ is determined by the adjoint action of $\hh = \tilde\hh$. Let $\aa$ be an ideal of $\tilde\gg$ and $a\in \aa$. Let $a = \sum a_qg_q$ where $g_q$ are homogeneous of degree $q \in Q$. Then $[h_i,a] = \sum \ a_qg_q$, and so $[\hh,a]$ has the same dimension as the number of non-zero coefficients $a_q$; in particular, $g_q\in [\hh,a]$. Thus $\aa$ is graded. Suppose that $\aa$ has a degree-$0$ part, i.e.\ suppose that there is some $h\in \hh \cap \aa$. Since the Cartan matrix $\aaa$ is nonsingular, there exists $\alpha_i\in\Delta$ with $\alpha_i(h) \neq 0$. Then $[f_i,h] = \alpha_i(h)f_i \neq 0$, and so $f_i \in \aa$. Now let $\aa$ be any ideal with $f_i\in \aa$ for some $i$. Then $h_i = [e_i,f_i] \in \aa$ and $e_i = -\frac12 [e_i,h_i] \in \aa$. But let $\alpha_j$ be any neighbor of $\alpha_i$ in the Dynkin diagram. Then $\aaa_{ij} \neq 0$, and so $[f_j,h_i] = \aaa_{ij}f_j \neq 0$; then $f_j \in \aa$. Therefore, if the Dynkin diagram is connected, then any ideal of $\tilde\gg$ that contains some $f_i$ (or some $e_i$ by symmetry) contains every generator of $\tilde\gg$. \end{proof} \begin{cor} Under the conditions of \cref{properidealsofg}, $\tilde\gg$ has a unique maximal proper ideal. \end{cor} \begin{proof} Let $\aa$ and $\bb$ be any two proper ideals of $\tilde\gg$. Then the ideal $\aa + \bb$ does not contain $\hh$ or any $e_i$ or $f_i$, and so is a proper ideal. \end{proof} \begin{defn} \label{gdef} Let $\Delta$ be a system of simple roots with connected Dynkin diagram, and let $\tilde\gg = \tilde\gg_\Delta$ be defined as in \cref{tildegdef}. We define $\gg = \gg_\Delta$ as the quotient of $\tilde\gg$ by its unique maximal proper ideal. Then $\\mono \gg$, where by $\ $ we mean the linear span of the generators of $\tilde\gg$. Since we quotiented by a maximal ideal, $\gg$ is simple. \end{defn} \begin{thm}[Serre Relations] \label{SerreThm} Let $\gg$ be as in \cref{gdef}, and $e_i,f_i$ the images of the corresponding generators of $\tilde\gg$. Then: \begin{align} \label{SerreE} (\ad e_j)^{1 - \aaa_{ji}} e_i & = 0 \\ \label{SerreF} (\ad f_j)^{1 - \aaa_{ji}} f_i & = 0 \end{align} \end{thm} \begin{proof} We will check \cref{SerreF}; \cref{SerreE} is exactly analogous. Let $s$ be the left-hand-side of \cref{SerreF}, interpreted as an element of $\tilde\gg$. We will show that the ideal generated by $s$ is proper. When $i=j$, $s = 0$, and when $i\neq j$, $\aaa_{ji} \leq 0$, and so the degree of $s$ is $-\alpha_i - (\geq 1)\alpha_j$. In particular, bracketing with $f_k$ and $h_k$ only moves the degree further from $0$. Therefore, the claim follows from the following equation: \eqn[SerreClaim]{ [e_k,s]_{\tilde\gg} = 0 \, \textrm{ for any }k} When $k \neq i,j$, $[e_k,f_i] = [e_k,f_j] = 0$. So it suffices to check \cref{SerreClaim} when $k = i,j$. Let $m = -\aaa_{ji}$. When $k=j$, we compute: \begin{align} (\ad e_j)(\ad f_j)^{1 +m} f_i & = \bigl[ \ad e_j , (\ad f_j)^{1 +m} \bigr] f_i + (\ad f_j)^{1 +m }(\ad e_j)f_i \\ \label{SerreClaim1} & = \bigl[ \ad e_j , (\ad f_j)^{1 +m } \bigr] f_i + 0 \\ \label{SerreClaim2} & = \sum_{l=0}^{m} (\ad f_j)^{m-l} \bigl(\ad[e_j,f_j]\bigr) (\ad f_j)^{l} f_i \\ & = \sum_{l=0}^{m} (\ad f_j)^{m-l} (\ad h_j) (\ad f_j)^{l} f_i \\ & = \sum_{l=0}^m \bigl( l(-\<\alpha_j,\alpha_j^\vee\>) - \<\alpha_i,\alpha_j^\vee\>\bigr) (\ad f_j)^m f_i \\ & = \left( \sum_{l=0}^m \bigl( -2l + m\bigr) \right) (\ad f_j)^m f_i \\ & = \left( - 2\frac{m(m+1)}2 + (m+1)m\right)(\ad f_j)^m f_i = 0 \end{align} where \cref{SerreClaim1} follows by $[e_i,f_j] = 0$, \cref{SerreClaim2} by the fact that $\ad$ is a Lie algebra homomorphism, and the rest is \cref{gtilderel2,gtilderel3}, that $m = -\aaa_{ji}$, and arithmetic. When $k = i$, $e_i$ and $f_j$ commute, and we have: \begin{align} (\ad e_i)(\ad f_j)^{1 +m} f_i & = \bigl[ \ad e_j , (\ad f_j)^{1 +m} \bigr] f_i + (\ad f_j)^{1 +m }(\ad e_i)f_i \\ & = 0 + (\ad f_j)^{1 +m }(\ad e_i)f_i \\ & = (\ad f_j)^{1 +m }h_i = 0 \end{align} provided that $m\geq 1$. When $m = 0$, we use the symmetrizability of the Cartan matrix: if $\aaa_{ji} = 0$ then $\aaa_{ij} = 0$. Therefore \eqn{ (\ad e_i)(\ad f_j)^{1 - \aaa_{ji}} f_i = (\ad e_i)[f_j,f_i] = -(\ad e_i) (\ad f_i)^{1 - \aaa_{ij}} f_j } which vanishes by the first computation. \end{proof} We have defined for each indecomposable Dynkin diagram $\Delta$ a simple Lie algebra $\gg_\Delta$. If $\Delta = \Delta_1 \times \Delta_2$ is a disjoint union of Dynkin diagrams, we define $\gg_\Delta \defined \gg_{\Delta_1}\times \gg_{\Delta_2}$. \begin{defn} \label{IntegrableDefn} Let $V$ be a (possibly-infinite-dimensional) $\gg$-module. An element $v\in V$ is \emindexing{integrable} if for each $i$, the $\sl(2)_i$-submodule of $V$ generated by $v$ is finite-dimensional. We write $I(V)$ for the set of integrable elements of $V$. \end{defn} \begin{lemma} Let $V$ be a $\gg$-module. Then $I(V)$ is a $\gg$-submodule. \end{lemma} \begin{proof} Let $N \subseteq V$ be an $(n+1)$-dimensional irreducible representation of $\sl(2)_i$; then it is isomorphic to $V_n$ defined in \cref{VnDefn}. It suffices to show that $e_jN$ is contained within some finite-dimensional $\sl(2)_i$ submodule of $V$ for $i\neq j$; the rest follows by switching $e\tofrom f$ and permuting the indices, using the fact that $\{e_j,f_j\}$ generate $\gg$. Then $N$ is spanned by $\{{f_i}^k v_0\}_{k=0}^n$ where $v_0 \in N$ is the vector annihilated by $e_i$; in particular, ${f_i}^{n+1}v_0 = 0$. Since $e_j$ and $f_i$ commute, $e_jN$ is spanned by $\{{f_i}^k e_jv_0\}_{k=0}^n$. It suffices to compute the $\sl(2)_i$ module generated by $e_jv_0$, or at least to show that it is finite-dimensional. The action of $h_i$ on $e_jv_0$ is $h_ie_jv_0 = ([h_i,e_j] + e_jh_i)v_0 = (\aaa_{ij} + n)e_jv_0$. For $k\neq n+1$, $f_i^k e_j v_0 = e_jf_i^k v_0 = 0$. Moreover, by \cref{SerreThm}, $e_i^k e_j v_0 = [e_i^k,e_j]v_0 + e_je_i^kv_0 = (\ad e_i)^k(e_j) v_0 + 0$, which vanishes for large enough $k$. Then the result follows by \cref{PBWthm} and the fact that $[e_i,f_i] = h_i$. \end{proof} \begin{cor} Let $\Delta$ be a Dynkin diagram and define $\gg$ as above. Then $\gg$ is $\ad$-integrable. \end{cor} \begin{proof} Since $\{e_k,f_k\}$ generate $\gg$, it suffices to show that $e_k$ and $f_k$ are $\ad$-integrable for each $k$. But the $\sl(2)_i$-module generated by $f_k$ has $f_k$ as its highest-weight vector, since $[e_i,f_k] = 0$, and is finite-dimensional, since $(\ad f_i)^nf_k = 0$ for large enough $n$ by \cref{SerreThm}. \end{proof} \begin{cor} The non-zero weights $R$ of $\ad: \gg \acts \gg$ form a root system. \end{cor} \begin{proof} Axioms {\bf RS1}, {\bf RS2}, {\bf RS3}, {\bf RS4}, and {\bf Nondeg} of \cref{RootSystemDefn} follow from the $\ad$-integrability. Axiom {\bf Reduced} and that $R$ is finite follow from \cref{HeightInductionLemma}. \end{proof} \begin{thm}[Classification of finite-dimensional simple Lie algebras] The list given in \cref{DynkinClassificationThm} classifies the finite-dimensional simple Lie algebras over $\CC$. \end{thm} \begin{proof} A Lie algebra with an indecomposable root system is simple, because any such system has a highest root, and linear combination of roots generates the highest root, and the highest root generates the entire algebra. So it suffices to show that two simple Lie algebras with isomorphic root systems are isomorphic. Let $\Delta$ be an indecomposable root system, and define $\tilde\gg$ and $\gg$ as above. Let $\gg_1$ be a Lie algebra with root system $\Delta$. Then the relations defining $\tilde\gg$ hold in $\gg_1$, and so there is a surjection $\tilde\gg \onto \gg_1$; if $\gg_1$ is simple, then the kernel of this surjection is a maximal ideal of $\tilde\gg$. But $\tilde\gg$ has a unique maximal ideal, and $\gg$ is the quotient by this ideal; thus $\gg_1 \cong \gg$. \end{proof} \begin{eg} The families $ABCD$ correspond to the classical Lie algebras: $A_n \tofrom \sl(n+1)$\index{sln@$\sl(n)$}, $B_n \tofrom \so(2n+1)$\index{son@$\so(n)$}, $C_n \tofrom \sp(n)$\index{spn@$\sp(n)$}, and $D_n \tofrom \so(2n)$. We recall that we have defined $\sp(n)$ as the Lie algebra that fixes the nondegenerate antisymmetric $2n\times 2n$ bilinear form: $\sp(n) \subseteq \gl(2n)$. The EFG Lie algebras are new. The coincidences in \cref{DynkinCoincidences} correspond to coincidences of classical Lie algebras: $\so(6) \cong \sl(4)$, $\so(5) \cong \sp(2)$, and $\so(4) \cong \sl(2) \times \sl(2)$. The identity $\so(3) \cong \sl(2)$ suggests that we define $B_1 = A_1 = \bullet$, but $\sl(2)$ is not congruent to $\sp(1)$ or to $\so(2)$, so we do not assign meaning to $C_1$ or $D_1$, and justifying the name $B_1$ for $\bullet$ but not $C_1$ is ad hoc. \end{eg} \Section{Exercises} \begin{enumerate} \item \begin{enumerate} \item Show that $\SL(2, \RR)$\index{SL2R@$\SL(2,\RR)$} is topologically the product of a circle and two copies of $\RR$, hence it is not simply connected. \item Let $S$ be the simply connected cover of $\SL(2, \RR)$. Show that its finite-dimensional complex representations, i.e., real Lie group homomorphisms $S \to \GL(n, \CC)$, are determined by corresponding complex representations of the Lie algebra $\Lie(S )^\CC = \sl(2, \CC)$, and hence factor through $\SL(2, \RR)$. Thus $S$ is a simply connected real Lie group with no faithful finite-dimensional representation. \end{enumerate} \item \begin{enumerate} \item Let $U$ be the group of $3\times 3$ upper-unitriangular complex matrices. Let $\Gamma \subseteq U$ be the cyclic subgroup of matrices $$ \left[ \begin{array}{ccc} 1 & 0 & m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right], $$ where $m\in \ZZ$. Show that $G = U/\Gamma$ is a (non-simply-connected) complex Lie group that has no faithful finite-dimensional representation. \item Adapt the solution to Set 4, Problem 2(b) to construct a faithful, irreducible infinite-dimensional linear representation $V$ of $G$. \end{enumerate} \item Following the outline below, prove that if $\hh \subseteq \gl(n,\CC)$ is a real Lie subalgebra with the property that every $X \in \hh$ is diagonalizable and has purely imaginary eigenvalues, then the corresponding connected Lie subgroup $H \subseteq GL(n, \CC)$ has compact closure (this completes the solution to Set 1, Problem 7). \begin{enumerate} \item Show that $\ad X$ is diagonalizable with imaginary eigenvalues for every $X \in \hh$. \item Show that the Killing form of $\hh$ is negative semidefinite and its radical is the center of $\hh$. Deduce that $\hh$ is reductive and the Killing form of its semi-simple part is negative definite. Hence the Lie subgroup corresponding to the semi-simple part is compact. \item Show that the Lie subgroup corresponding to the center of $\hh$ is a dense subgroup of a compact torus. Deduce that the closure of $H$ is compact. \item Show that $H$ is compact --- that is, closed --- if and only if it further holds that the center of $\hh$ is spanned by matrices whose eigenvalues are rational multiples of $i$. \end{enumerate} \item Let $V_n = \Ss^n (\CC^2 )$ be the $(n + 1)$-dimensional irreducible representation of $\sl(2, \CC)$. \begin{enumerate} \item Show that for $m \leq n$, $V_m \otimes V_n \cong V_{n-m} \oplus V_{n-m + 2} \oplus \dots \oplus V_{n+m}$, and deduce that the decomposition into irreducibles is unique. \item Show that in any decomposition of $V^{\otimes n}_1$ into irreducibles, the multiplicity of $V_n$ is equal to $1$, the multiplicity of $V_{n-2k}$ is equal to ${n \choose k} - {n \choose k-1}$ for $k = 1, \dots , \lfloor n/2 \rfloor$, and all other irreducibles $V_m$ have multiplicity zero. \end{enumerate} \item Let $a$ be a symmetric Cartan matrix, i.e.\ $a$ is symmetric with diagonal entries $2$ and off-diagonal entries $0$ or $-1$. Let $\Gamma$ be a subgroup of the automorphism group of the Dynkin diagram $D$ of $a$, such that every edge of $D$ has its endpoints in distinct $\Gamma$ orbits. Define the \emindexing{folding} $D'$ of $D$ to be the diagram with a node for every $\Gamma$ orbit $I$ of nodes in $D$, with edge weight $k$ from $I$ to $J$ if each node of $I$ is adjacent in $D$ to $k$ nodes of $J$. Denote by $a'$ the Cartan matrix with diagram $D'$. \begin{enumerate} \item Show that $a'$ is symmetrizable and that every symmetrizable generalized Cartan matrix (not assumed to be of finite type) can be obtained by folding from a symmetric one. \item Show that every folding of a finite type symmetric Cartan matrix is of finite type. \item Verify that every non-symmetric finite type Cartan matrix is obtained by folding from a unique symmetric finite type Cartan matrix. \end{enumerate} \item An indecomposable symmetrizable generalized Cartan matrix $a$ is said to be of \emindexing{affine type} if $\det(a) = 0$ and all the proper principal minors of $a$ are positive. \begin{enumerate} \item Classify the affine Cartan matrices. \item Show that every non-symmetric affine Cartan matrix is a folding, as in the previous problem, of a symmetric one. \item Let $\hh$ be a vector space, $\alpha_i \in \hh^*$ and $\alpha_i^\vee \in \hh$ vectors such that $a$ is the matrix $\<\alpha_j,\alpha_i^\vee\>$. Assume that this realization is non-degenerate in the sense that the vectors $\alpha_i$ are linearly independent. Define the \emindexing{affine Weyl group}\index{Weyl group!affine} $W$ to be generated by the reflections $s_{\alpha_i}$, as usual. Show that $W$ is isomorphic to the semidirect product $W_0 \ltimes Q$ where $Q$ and $W_0$ are the root lattice and Weyl group of a unique finite root system, and that every such $W_0 \ltimes Q$ occurs as an affine Weyl group. \item Show that the affine and finite root systems related as in (c) have the property that the affine Dynkin diagram is obtained by adding a node to the finite one, in a unique way if the finite Cartan matrix is symmetric. \end{enumerate} \item Work out the root systems of the orthogonal Lie algebras $\so(m, \CC)$ explicitly, thereby verifying that they correspond to the Dynkin diagrams $B_n$ if $m = 2n + 1$, or $D_n$ if $m = 2n$. Deduce the isomorphisms $\so(4, \CC) \cong \sl(2, \CC) \times \sl(2, \CC)$, $\so(5, \CC) \cong \sp(4, \CC)$, and $\so(6, \CC) \cong \sl(4, \CC)$. \item Show that the Weyl group of type $B_n$ or $C_n$ (they are the same because these two root systems are dual to each other) is the group $S_n \ltimes (\ZZ/2\ZZ)^n$ of signed permutations, and that the Weyl group of type $D_n$ is its subgroup of index two consisting of signed permutations with an even number of sign changes, i.e., the semidirect factor $(\ZZ/2\ZZ)^n$ is replaced by the kernel of $S_n$-invariant summation homomorphism $(\ZZ/2\ZZ)^n \to \ZZ/2\ZZ$ \item Let $(\hh, R, R^\vee)$ be a finite root system, $\Delta = \{ \alpha_1,\dots,\alpha_n\}$ the set of simple roots with respect to a choice of positive roots $R_+$, $s_i = s_{\alpha_i}$ the corresponding generators of the Weyl group $W$. Given $w \in W$, let $l(w)$ denote the minimum length of an expression for $w$ as a product of the generators $s_i$. \begin{enumerate} \item If $w = s_{i_1} \dots s_{i_r}$ and $w(\alpha_j) \in R_-$, show that for some $k$ we have $\alpha_{i_k} = s_{i_{k+1}} \dots s_{i_r}(\alpha_j)$, and hence $s_{i_k}s_{i_{k+1}} \dots s_{i_r} = s_{i_{k+1}} \dots s_{i_r} s_j$. Deduce that $l(ws_j ) = l(w) -1$ if $w(\alpha_j ) \in R_-$. \item Using the fact that the conclusion of (a) also holds for $v = ws_j$, deduce that $l(ws_j ) = l(w) + 1$ if $w(\alpha_j ) \not\in R_-$. \item Conclude that $l(w) = |w(R_+) \cup R_-|$ for all $w \in W$. Characterize $l(w)$ in more explicit terms in the case of the Weyl groups of type $A$ and $B$/$C$. \item Assuming that $\hh$ is over $\RR$, show that the dominant cone $X = \{ \lambda \in \hh : \<\lambda,\alpha_i^\vee\> \geq 0 \textrm{ for all } i \}$ is a fundamental domain for $W$, i.e., every vector in $\hh$ has a unique element of $X$ in its $W$ orbit. \item Deduce that $|W |$ is equal to the number of connected regions into which $\hh$ is separated by the removal of all the root hyperplanes $\<\lambda,\alpha^\vee\>$, $\alpha^\vee \in R^\vee$. \end{enumerate} \item Let $h_1, \dots, h_r$ be linear forms in variables $x_1, \dots, x_n$ with integer coefficients. Let $\FF_q$ denote the finite field with $q = p^e$ elements. Prove that except in a finite number of ``bad'' characteristics $p$, the number of vectors $v \in \FF_q^n$ such that $h_i (v) = 0$ for all $i$ is given for all $q$ by a polynomial $\chi(q)$ in $q$ with integer coefficients, and that $(-1)^n \chi(-1)$ is equal to the number of connected regions into which $\RR^n$ is separated by the removal of all the hyperplanes $h_i = 0$. Pick your favorite finite root system and verify that in the case where the $h_i$ are the root hyperplanes, the polynomial $\chi(q)$ factors as $(q - e_1) \dots (q - e_n)$ for some positive integers $e_i$ called the \emindexing{exponents} of the root system. In particular, verify that the sum of the exponents is the number of positive roots, and that (by Problem 9(e)) the order of the Weyl group is $\prod_i (1 + e_i)$ \item The \emindexing{height} of a positive root $\alpha$ is the sum of the coefficients $c_i$ in its expansion $\alpha = \sum_i c_i\alpha_i$ on the basis of simple roots. Pick your favorite root system and verify that for each $k \geq 1$, the number of roots of height $k$ is equal to the number of the exponents $e_i$ in Problem 10 for which $e_i \geq k$. \item Pick your favorite root system and verify that if $h$ denotes the height of the highest root plus one, then the number of roots is equal to $h$ times the rank. This number $h$ is called the \emindexing{Coxeter number}. Verify that, moreover, the multiset of exponents (see Problem 10) is invariant with respect to the symmetry $e_i \mapsto h - e_i$. \item A \emindexing{Coxeter element} in the Weyl group $W$ is the product of all the simple reflections, once each, in any order. Prove that a Coxeter element is unique up to conjugacy. Pick your favorite root system and verify that the order of a Coxeter element is equal to the Coxeter number (see Problem 12). \item The {\em fundamental weights}\index{fundamental weight}\index{weight!fundamental} $\lambda_i$ are defined to be the basis of the weight lattice $P$ dual to the basis of simple coroots in $Q^\vee$, i.e., $\<\lambda_i, \alpha^\vee_j\> = \delta_{ij}$. \begin{enumerate} \item Prove that the stabilizer in $W$ of $\lambda_i$ is the Weyl group of the root system whose Dynkin diagram is obtained by deleting node $i$ of the original Dynkin diagram. \item Show that each of the root systems $E_6$, $E_7$, and $E_8$ has the property that its highest root is a fundamental weight. Deduce that the order of the Weyl group $W(E_k )$ in each case is equal to the number of roots times the order of the Weyl group $W(E_{k-1})$, or $W (D_5 )$ for $k = 6$. Use this to calculate the orders of these Weyl groups. \end{enumerate} \item Let $e_1, \dots, e_8$ be the usual orthonormal basis of coordinate vectors in Euclidean space $\RR^8$. The root system of type $E_8$ can be realized in $\RR^8$ with simple roots $\alpha_i = e_i - e_{i+1}$ for $i = 1,\dots, 7$ and $$ \alpha_8 = \left( -\frac12, -\frac12, -\frac12, \frac12, \frac12, \frac12, \frac12, \frac12 \right). $$ Show that the root lattice $Q$ is equal to the weight lattice $P$, and that in this realization, $Q$ consists of all vectors $\beta \in \ZZ^8$ such that $\sum_i \beta_i$ is even and all vectors $\beta \in \left( \frac12, \frac12, \frac12, \frac12, \frac12, \frac12, \frac12, \frac12 \right) + \ZZ^8$ such that $\sum_i \beta_i$ is odd. Show that the root system consists of all vectors of squared length $2$ in $Q$, namely, the vectors $\pm e_i \pm e_j$ for $i \subseteq \ZZ\}$. A \emindexing{dominant integral weight}\index{weight!dominant integral} is an element of $P_+ \defined \{\lambda \in P \st \<\lambda,\alpha_i^\vee\> \geq 0\,\forall i\}$. \end{defn} We recall \cref{IntegrableDefn}. \begin{prop} \label{IntegrableProp} If $\lambda \in P_+$, then $L_\lambda$ consists of integrable elements. \end{prop} \begin{proof} Since $L_\lambda$ is irreducible, its submodule of integrable elements is either $0$ or the whole module. So it suffices to show that if $\lambda \in P_+$, then $v_\lambda$ is integrable. Pick a simple root $\alpha_i$. By construction, $e_iv_\lambda = 0$ and $h_iv_\lambda = \<\lambda,\alpha_i^\vee\>v_\lambda$. Since $\lambda \in P_+$, $\<\lambda,\alpha_i^\vee\> = m\geq 0$ is an integer. Consider the $\sl(2)_i$-submodule of $M_\lambda$ generated by $v_\lambda$; if $m$ is a nonnegative integer, from the representation theory of $\sl(2)$ we know that $e_if_i^{m+1}v_\lambda = 0$. But if $j\neq i$, then $e_jf_i^{m+1}v_\lambda = f_i^{m+1}e_jv_\lambda = 0$. Recalling the grading, we see then that $f_i^{m+1}v_\lambda$ generates a submodule of $M_\lambda$, and so $f_i^{m+1}v_\lambda \mapsto 0$ in $L_\lambda$. Hence the $\sl(2)_i$-submodule of $L_\lambda$ generated by $v_\lambda$ is finite, and so $v_\lambda$ is integrable. \end{proof} \begin{defn} Let $\gg$ be a semisimple Lie algebra. We define the category $\hat\Oo$\index{O@$\hat\Oo$} to be a full subcategory of the category $\cat{$\gg$-mod}$ of (possibly-infinite-dimensional) $\gg$ modules. The objects $X\in \hat\Oo$ are required to satisfy the following conditions: \begin{itemize} \item The action $\hh \acts X$ is diagonalizable. \item For each $\lambda \in \hh^*$, the weight space $X_\lambda$ is finite-dimensional. \item There exists a finite set $S \subseteq \hh^*$ such that the weights of $X$ lie in $S + (-Q_+)$. \end{itemize} \end{defn} \begin{lemma} The category $\hat\Oo$ is closed under submodules, quotients, extensions, and tensor products. \end{lemma} \begin{proof} The $\hh$-action grades subquotients of any graded module, and acts diagonally. An extension of graded modules is graded, with graded components extensions of the corresponding graded components. Since $\gg$ is semisimple, any extension of finite-dimensional modules is a direct sum, and so the $\hh$-action is diagonal on any extension of objects in $\Oo$. Finally, tensor products are handled by \cref{TensorWeightLemma}. \end{proof} \begin{defn} Write the additive group $\hh^*$ multiplicatively: $\lambda \mapsto x^\lambda$. The \emindexing{group algebra} $\ZZ [\hh^*]$ is the algebra of ``polynomials'' $\sum c_ix^{\lambda_i}$, with the obvious addition and multiplication. I.e.\ $\ZZ[\hh^*]$ is the free abelian group $\bigoplus_{\lambda \in \hh^*} \ZZ x^\lambda$, with multiplication given on a basis by $x^\lambda x^\mu = x^{\lambda + \mu}$. Let $\ZZ[-Q_+]$ be the subalgebra of $\ZZ[\hh^*]$ generated by $\{x^\lambda \st -\lambda \in Q_+\}$. This has a natural topology given by setting $\| x^{-\alpha_i}\| = c^{-\alpha_i}$ for $\alpha_i$ a simple root and $c$ some real constant with $c > 1$. We let $\ZZ[[-Q_+]]$ be the completion of $\ZZ[-Q_+]$ with respect to this topology. Equivalently, $\ZZ[[-Q_+]]$ is the algebra of \emindexing{formal power series} in the variables $x^{-\alpha_1},\dots,x^{-\alpha_n}$ with integer coefficients. Then $\ZZ[-Q_+]$ is a subalgebra of both $\ZZ[\hh^*]$ and $\ZZ[[-Q_+]]$. We will write $\ZZ[h^*,-Q_+]]$ for the algebra $\ZZ[\hh^*] \otimes_{\ZZ[-Q_+]} \ZZ[[-Q_+]]$. \end{defn} The algebra $\ZZ[h^*,-Q_+]]$ is a formal gadget, consisting of formal fractional Laurant series. We use it as a space of generating functions. \begin{defn} \label{chDefn} Let $X \in \hat\Oo$. We define $\ch(X) \in \ZZ[h^*,-Q_+]]$ by: \eqn{ \ch(X) \defined \sum_{\lambda \textrm{ a weight of }X} \dim(X_\lambda)x^\lambda } \end{defn} We remark that every coefficient of $\ch(X)$ is a nonnegative integer, and if $Y$ is a subquotient \begin{eg} \label{MChar} Let $M_\lambda$ be the Verma module with weight $\lambda$, and let $R_+$ be the set of positive roots of $\gg$. Then \eqn{ \ch(M_\lambda) = \frac{x^\lambda}{\prod_{\alpha \in R_+} (1 - x^{-\alpha})} \defined x^\lambda \prod_{\alpha\in R_+} \sum_{l=0}^\infty x^{-l\alpha} } This follows from \cref{PBWthm}, the explicit description of $M_\lambda \cong \Uu\nn_-\otimes \CC v_\lambda$, and some elementary combinatorics. \end{eg} \begin{prop} \label{WeylCharProof2} Let $\gg$ be simple Lie algebra, $P_+$ the set of dominant integral weights, and $W$ the Weyl group. Let $\lambda \in P_+$, and $L_\lambda$ the irreducible quotient of $M_\lambda$ given in \cref{LlambdaDefn}. Then: \begin{enumerate} \item $\ch(L_\lambda)$ is $W$-invariant. \item If $\mu$ is a weight of $L_\lambda$, then $\mu \in W(\nu)$ for some $\nu \in P_+ \cap (\lambda - Q_+)$. \item $L_\lambda$ is finite-dimensional. \end{enumerate} Conversely, every finite-dimensional irreducible $\gg$-module is $L_\lambda$ for a unique $\lambda \in P_+$. \end{prop} \begin{proof} \begin{enumerate} \item We use \cref{IntegrableProp}: $L_\lambda$ consists of integrable elements. Let $\alpha_i$ be a root of $\gg$; then $L_\lambda$ splits as an $\sl(2)_i$ module: $L_\lambda = \bigoplus V_a$, where each $V_a$ is an irreducible $\sl(2)_i$ submodule. In particular, $V_a = \CC v_{a,m} \oplus \CC v_{a,m-2} \dots \oplus \CC v_{a,-m}$ for some $m$ depending on $a$, where $h_i$ acts on $\CC v_{a,l}$ by $l$. But $\ch(L_\lambda) = \sum_a \ch(V_a) = \sum_a \sum_{j= -m,-m+2,\dots,m} \ch(\CC v_{a,m})$. Let $\ch\bigl(\CC (v_a)_l\bigr) = x^{\mu_{a,l}}$; then $\<\mu_{a,l},\alpha_i^\vee\> = l$ by definition, and $v_{a,l - 2} \in f_i \CC v_{a,l}$, and so $s_i \mu_{a,l} = \mu_{a,-l}$. This shows that $\ch(V_a)$ is fixed under the action of $s_i$, and so $\ch(L_\lambda)$ is also $s_i$-invariant. But the reflections $s_i$ generate $W$, and so $\ch(L_\lambda)$ is $W$-invariant. \item We partially order $P$: $\nu \leq \mu$ if $\mu - \nu \in Q_+$. In particular, the weights of $L_\lambda$ are all less than or equal to $\lambda$. Let $\lambda \in P$. Then $s_i(\lambda) = \lambda - \<\lambda,\alpha_i^\vee\>\alpha_i$, and so $W(\lambda) \subseteq \lambda + Q$. If $\lambda \in P_+$ then $\<\lambda,\alpha_i^\vee\>\geq 0$ for every $i$ and so $s_i \leq \lambda$; if $\lambda \in P \sminus P_+$ then there is some $i$ with $\<\lambda,\alpha_i^\vee\>< 0$, i.e.\ some $i$ with $s_i(\lambda) > \lambda$. But $W$ is finite, so for any $\lambda \in P$, $W(\lambda)$ has a maximal element, which must be in $P_+$. This proves that $P = W(P_+)$. Thus, if $\mu$ is a weight of $L_\lambda$, then $\mu \in W(\nu)$ for some $\nu \in P_+$. But by 1., $\nu$ is a weight of $L_\lambda$, and so $\nu \leq \lambda$. This proves statements 2. Moreover, the $W$-invariance of $\ch(L_\lambda)$ shows that if $\lambda \in P_+$, then $W(\lambda) \subseteq \lambda - Q_+$, and moreover that $P_+$ is a fundamental domain of $W$. \item The Weyl group $W$ is finite. Consider the two cones $\RR_{\geq 0} P_+$ and $-\RR_{\geq 0}Q_+$. Since the inner product (the symmetrization of the Cartan matrix) is positive definite and by construction the inner product of anything in $\RR_{\geq 0}P_+$ with anything in $-\RR_{\geq 0}Q_+$ is negative, the two cones intersect only at $0$. Thus there is a hyperplane separating the cones: i.e.\ there exists a linear functional $\eta: \hh^*_\RR \to \RR$ such that its value is positive on $P_+$ but negative on $-Q_+$. Then $\lambda - Q_+$ is below the $\eta = \eta(\lambda)$ hyperplane. But $-Q_+$ is generated by $-\alpha_i$, each of which has a negative value under $\eta$, and so $\lambda - Q_+$ contains only finitely many points $\mu$ with $\eta(\mu) \geq 0$. Thus $P_+ \cap (\lambda - Q_+)$ is finite, and hence so is its image under $W$. \end{enumerate} For the converse statement, let $L$ be a finite-dimensional irreducible $\gg$-module, and let $v \in L$ be any vector. Then consider $\nn_+v$, the image of $v$ under repeated application of various $e_i$s. By finite-dimensionality, $\nn_+v$ must contain a vector $l \in \nn_+v$ so that $e_il = 0$ for every $i$. By the $\sl(2)$ representation theory, $l$ must be homogeneous, and indeed a top-weight vector of $L$, and by the irreducibility $l$ generates $L$. Let the weight of $l$ be $\lambda$; then the map $v_\lambda \to l$ generates a map $M_\lambda \onto L$. But $M_\lambda$ has a unique maximal submodule, and since $L$ is irreducible, this maximal submodule must be the kernel of the map $M_\lambda \onto L$. Thus $L \cong L_\lambda$. \end{proof} \begin{lemmadef} Let $\gg$ be a semisimple Lie algebra, $\hh$ its Cartan subalgebra, and $\Delta = \{\alpha_1,\dots,\alpha_n\}$ its simple root system. For each $i = 1,\dots,n$, we define a \emindexing{fundamental weight}\index{weight!fundamental} $\Lambda_i \in \hh^*$ by $\<\Lambda_i,\alpha_j^\vee\> = \delta_{ij}$. Then $P_+ = \ZZ_{\geq 0}\{\Lambda_1,\dots,\Lambda_n\}$. The following are equivalent, and define the \emindexing{Weyl vector} $\rho$: \begin{enumerate} \item $\rho = \sum_{i=1}^n \Lambda_i$. I.e.\ $\<\rho,\alpha_j^\vee\> = 1$ for every $j$. \item $\rho = \frac12 \sum_{\alpha \in R_+} \alpha$. \end{enumerate} \end{lemmadef} \begin{proof} Let $\rho_2 = \frac12 \sum_{\alpha \in R_+} \alpha$. Since $s_i(R_+\sminus \{\alpha_i\}) = R_+$ but $s_i(\alpha_i) = -\alpha_i$, we see that $s_i(\tilde\rho) = \tilde\rho - \alpha_i$, and so $\<\tilde\rho,\alpha_i^\vee\> = 1$ for every $i$. The rest is elementary linear algebra. \end{proof} \begin{defn} Let $\lambda \in P_+$. We define the \emindexing{character} $\chi^\lambda$ of $\lambda$ to be $\ch(L_\lambda)$. \end{defn} \begin{thm}[Weyl Character Formula] \label{WeylCharThm} Let $\epsilon: W \to \{\pm 1\}$ be given by $\epsilon(w) = \det_\hh w$; i.e.\ $\epsilon$ is the group homomorphism generated by $s_i \mapsto -1$ for each $i$. Let $\lambda \in P_+$. Then $\chi^\lambda$ can be computed as follows: \eqn[WeylCharEqn]{ \chi^\lambda = \frac{\displaystyle \sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) - \rho}}{\displaystyle \prod_{\alpha \in R_+} (1 - x^{-\alpha})} = \frac{\displaystyle \sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) }}{\displaystyle \prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} } \end{thm} The equality of fractions follows simply from the description $\rho = \frac12 \sum_{\alpha \in R_+} \alpha$. \begin{remark} The sum in \cref{WeylCharEqn} is finite. Indeed, the numerator and denominator on the right-hand-side fraction are obiously antisymmetric in $W$, and so the whole expression is $W$-invariant. The numerator on the left-hand-side fraction is a polynomial, and each $(1 - x^{-\alpha})$ is invertible as a power series: $(1-x^{-\alpha})^{-1} = \sum_{n=0}^\infty x^{-n\alpha}$. So the fraction is a $W$-invariant power series, and hence a polynomial. \end{remark} To prove \cref{WeylCharThm} we will need a number of lemmas. In \cref{MChar} we computed the character of the Verma module $M_\lambda$. Then \cref{WeylCharThm} asserts that: \eqn{ \ch(L_\lambda) = \sum_{w\in W} \epsilon(w) \,\ch\bigl(M_{w(\lambda + \rho) - \rho}\bigr) } As such, we will begin by understanding $M_\lambda$ better. We recall \cref{CasimirDefn}: Let $(,)$ be the Killing form on $\gg$, and $\{x_i\}$ any basis of $\gg$ with dual basis $\{y_j\}$, i.e.\ $(x_i,y_j) = \delta_{ij}$ for every $i,j$; then $c = \sum x_iy_i \in \Uu\gg$ is central, and does not depend on the choice of basis. \begin{lemma} Let $\lambda \in \hh^*$ and $M_\lambda$ the Verma module with weight $\lambda$. Let $c \in \Uu\gg$ be the Casimir, corresponding to the Killing form on $\gg$. Then $c$ acts on $M_\lambda$ by multiplication by $(\lambda,\lambda + 2\rho)$. \end{lemma} \begin{proof} Let $\gg$ have rank $n$. Write $R$ for the set of roots of $\gg$, $R_+$ for the positive roots, and $\Delta$ for the simple roots, as we have previously. Recall \cref{KillingNondegCartan}. We construct a basis of $\gg$ as follows: we pick an orthonormal basis $\{u_i\}_{i=1}^n$ of $\hh$. For each $\alpha$ a non-zero root of $\gg$, the space $\gg_\alpha$ is one-dimensional; let $x_\alpha$ be a basis vector in $\gg_\alpha$. Then the dual basis to $\{u_i\}_{i = 1}^n \cup \{x_\alpha\}_{\alpha \in R \sminus \{0\}}$ is $\{u_i\}_{i = 1}^n \cup \{ y_\alpha\}_{\alpha \in R \sminus \{0\}}$, where $y_\alpha = \frac{x_{-\alpha} }{ (x_\alpha,x_{-\alpha})}$. Then: \eqn{ c = \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R\sminus \{0\}} x_\alpha y_\alpha = \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R\sminus \{0\}} \frac{x_\alpha x_{-\alpha}}{(x_\alpha,x_{-\alpha})} = \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R_+} \frac{x_\alpha x_{-\alpha} + x_{-\alpha}x_\alpha}{(x_\alpha,x_{-\alpha})} } Since $M_\lambda$ is generated by its highest weight vector $v_\lambda$, and $c$ is central, to understand the action of $c$ on $M_\lambda$ it suffices to compute $cv_\lambda$. We use the fact that for $\alpha \in R_+$, $x_\alpha v_\lambda = 0$; then \eqn[WeylCharProof2]{ x_\alpha x_{-\alpha} v_\lambda = h_\alpha v_\lambda + x_{-\alpha} x_\alpha v_\lambda = h_\alpha v_\lambda = \lambda(h_\alpha) \, v_\lambda } where for each $\alpha \in R_+$ we have defines $h_\alpha \in \hh$ by $h_\alpha = [x_\alpha,x_{-\alpha}]$. Moreover, $(,)$ is $\gg$-invariant, and $[h_\alpha,x_\alpha] = \alpha(h_\alpha)x_\alpha$, where $\alpha(h_\alpha) \neq 0$. So: \eqn{ (x_\alpha,x_{-\alpha}) = \frac1{\alpha(h_\alpha)} \bigl([h_\alpha,x_\alpha], x_{-\alpha}\bigr) = \frac1{\alpha(h_\alpha)} \bigl(h_\alpha,[x_\alpha, x_{-\alpha}]\bigr) = \frac{(h_\alpha,h_\alpha)}{\alpha(h_\alpha)} } We also have that $u_iv_\lambda = \lambda(u_i)v_\lambda$, and since $\{u_i\}$ is an orthonormal basis, $(\lambda,\lambda) = \sum_{i=1}^n \bigl(\lambda(u_i)\bigr)^2$. Thus: \eqn{ cv_\lambda = \sum_{i=1}^n \bigl(\lambda(u_i)\bigr)^2 v_\lambda + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)}{\frac{(h_\alpha,h_\alpha)}{\alpha(h_\alpha)} } v_\lambda = \left( (\lambda,\lambda) + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} \right) v_\lambda} We recall that $h_\alpha$ is proportional to $\alpha^\vee$, that $(\alpha,\alpha) = 4/(\alpha^\vee,\alpha^\vee)$, and that $\lambda(\alpha^\vee) = (\lambda,\alpha)/(\alpha,\alpha)$. Then $\frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} = (\lambda,\alpha)$, and so: \eqn{ (\lambda,\lambda) + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} = (\lambda,\lambda) + \sum_{\alpha \in R_+} (\lambda,\alpha) = (\lambda,\lambda + 2\rho) } Thus $c$ acts on $M_\lambda$ by multiplication by $(\lambda,\lambda + 2\rho)$. \end{proof} \begin{lemmadef} Let $X$ be a $\gg$-module. We weight vector $v\in X$ is {\em singular}\index{singular vector} if $\nn_+ v = 0$. In particular, any highest-weight vector is singular, and conversely any singular vector is the highest weight vector in the submodule it generates. \end{lemmadef} \begin{cor} Let $\lambda \in P$, and $M_\lambda$ the Verma module with weight $\lambda$. Then $M_\lambda$ contains finitely many singular vectors, in the sense that their span is finite-dimensional. \end{cor} \begin{proof} Let $C^\lambda$ be the set $C^\lambda \defined \{\mu \in P \st (\mu + \rho,\mu + \rho) = (\lambda + \rho,\lambda + \rho)\}$. Then $C^\lambda$ is a sphere in $P$ centered at $-\rho$, and in particular it is a finite set. On the other hand, since $(\mu + \rho,\mu + \rho) = (\mu ,\mu + 2\rho) + (\rho,\rho)$, we see that: \eqn{ C^\lambda = \{\mu \in P \st c\textrm{ acts on }M_\mu \textrm{ by } (\lambda,\lambda + 2\rho) \} } Recall that any module with highest weight $\mu$ is a quotient of $M_\mu$. Let $v \in M_\lambda$ be a non-zero singular vector with weight $\mu$. Then on the one hand $c v = (\lambda,\lambda + 2\rho)v$, since $v\in M_\lambda$, and on the other hand $c v = (\mu,\mu + 2\rho)$, since $v$ is in a quotient of $M_\mu$. In particular, $\mu \in C^\lambda$. But the weight spaces $(M_\lambda)_\mu$ of $M_\lambda$ are finite-dimensional, and so the dimension of the space of singular vectors is at most $\sum_{\mu \in C^\lambda} \dim\bigl((M_\lambda)_\mu\bigr) < \infty$. \end{proof} \begin{cor} \label{WeylCharProof1} Let $\lambda \in P$. Then there are nonnegative integers $b_{\lambda,\mu}$ such that \eqn[KLeqn]{ \ch M_\lambda = \sum b_{\lambda,\mu} \ch L_\mu } and $b_{\lambda,\mu} = 0$ unless $\mu \leq \lambda$ and $\mu \in C^\lambda$. Moreover, $b_{\lambda,\lambda} = 0$. \end{cor} \begin{proof} We construct a filtration on $M_\lambda$. Since $M_\lambda$ has only finitely many non-zero singular vectors, we choose $w_1$ a singular vector of minimal weight $\mu_1$, and let $F_1M_\lambda$ be the submodule of $M_\lambda$ generated by $w_1$. Then $F_1M_\lambda$ is irreducible with highest weight $\mu_1$. We proceed by induction, letting $w_i$ be a singular vector of minimal weight in $M_\lambda / F_{i-1}M_\lambda$, and $F_iM_\lambda$ the primage of the subrepresentation generated by $w_i$. This filters $M_\lambda$: \eqn[MlambdaFiltered]{ 0 = F_0M_\lambda \subseteq F_1M_\lambda \subseteq \dots } Moreover, since $M_\lambda$ has only finitely many weight vectors all together, the filtration must terminate: \eqn{ 0 = F_0M_\lambda \subseteq F_1M_\lambda \subseteq \dots \subseteq F_kM_\lambda = M_\lambda } By construction, the quotients are all irreducible: $F_iM_\lambda / F_{i-1}M_\lambda = L_{\mu_i}$ for some $\mu_i \in C^\lambda$, $\mu_i \leq \lambda$. We recall that $\ch$ is additive for extensions. Therefore \eqn{ \ch M_\lambda = \sum_{i=1}^k \ch(F_iM_\lambda / F_{i-1}M_\lambda) = \sum_{i=1}^k \ch L_{\mu_i}} Then $b_{\lambda,\mu}$ is the multiplicity of $\mu$ appearing as the weight of a singular vector of $M_\lambda$, and we have \cref{KLeqn}. The conditions stated about $b_{\lambda,\mu}$ are immediate: we saw that $\mu$ can only appear as a weight of $M_\lambda$ if $\mu \in C^\lambda$ and $\mu \leq \lambda$; moreover, $L_\lambda$ appears as a subquotient of $M_\lambda$ exactly once, so $b_{\lambda,\lambda} = 1$. \end{proof} \begin{defn} The coefficients $b_{\lambda,\mu}$ in \cref{KLeqn} are the {\em Kazhdan-Luztig multiplicities}\index{Kazhdan-Luztig multiplicity}. \end{defn} \begin{lemma} \label{WeylCharProof3} If $\lambda \in P_+$, $\mu \leq \lambda$, $\mu \in C^\lambda$, and $\mu + \rho \geq 0$, then $\mu = \lambda$. \end{lemma} \begin{proof} We have that $(\mu + \rho,\mu + \rho) = (\lambda + \rho,\lambda + \rho)$ and that $\lambda - \mu = \sum_{i=1}^n k_i\alpha_i$, where all $k_i$ are nonnegative. Then \begin{align*} 0 & = (\lambda + \rho,\lambda + \rho) - (\mu + \rho,\mu + \rho) \\ & = \bigl( (\lambda + \rho) - (\mu + \rho), (\lambda + \rho) + (\mu + \rho) \bigr) \\ & = ( \lambda - \mu, \lambda + \mu + 2\rho) \\ & = \sum_{i=1}^n k_i(\alpha,\lambda + \mu + 2\rho) \end{align*} But $\lambda, \mu+\rho \geq 0$, and $(\alpha_i,\rho) > 0$, so $(\alpha,\lambda + \mu + 2\rho) > 0$, and so all $k_i = 0$ since they are nonnegative. \end{proof} \begin{proof}[of \cref{WeylCharThm}] We have shown (\cref{WeylCharProof1}) that $\ch M_\lambda = \sum b_{\lambda,\mu} \ch L_\mu$, were $b_{\lambda,\mu}$ is a lower-triangular matrix on $C^\lambda = C^\mu$ with ones on the diagonal. Thus it has a lower-triangular inverse with ones on the diagonal: \eqn{ \ch L_\lambda = \sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} \ch M_\mu } But by \cref{WeylCharProof2} statement 1., $\ch L_\lambda$ is $W$-invariant, provided that $\lambda \in P_+$, thus so is $\sum c_{\lambda,\mu} \ch M_\mu$. We recall \cref{MChar}: \eqn{ \ch M_\mu = \frac{ x^\mu}{\prod_{\alpha \in R_+} (1 - x^{-\alpha})} = \frac{ x^{\mu+\rho}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} } Therefore \eqn{ \ch L_\lambda = \frac{\sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} x^{\mu + \rho}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} } But the denominator if $W$-antisymmetric, and so the numerator must be as well: \eqn{ \sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} x^{w(\mu + \rho)} = \sum_{\mu\leq \lambda, \mu \in C^\lambda} \epsilon(w) c_{\lambda,\mu} x^{\mu + \rho} \textrm{ for every $w\in W$}} This is equivalent to the condition that $c_{\lambda,\mu} = \epsilon(w)c_{\lambda,w(\mu +\rho) - \rho}$. By the proof of \cref{WeylCharProof2} statement 2., we know that $P_+$ is a fundamental domain of $W$; since $c_{\lambda,\lambda} = 1$, if $\mu + \rho \in W(\lambda +\rho)$, then $c_{\lambda,\mu} = \epsilon(w)$, and so: \eqn{ \sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} x^{\mu + \rho} = \sum_{w\in W} \left( x^{w(\lambda + \rho)} + \mathop{\sum_{\mu < \lambda, \mu \in C^\lambda}}_{\mu + \rho \in P^+} c_{\lambda,\mu} x^{w(\mu+\rho)} \right)} But the rightmost sum is empty by \cref{WeylCharProof3}. \end{proof} \begin{remark} Specializing to the trivial representation $L_0$, \cref{WeylCharThm} says that \eqn[dimform1]{ 1 = \frac{ \sum_{w\in W} \epsilon(w) x^{w(\rho)}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} } So we can rewrite \cref{WeylCharEqn} as \eqn{ \chi^\lambda = \frac{ \sum_{w\in W} \epsilon(w) x^{w(\lambda + \rho)} }{\sum_{w\in W} \epsilon(w) x^{w(\rho)} } } \end{remark} The following is an important corollary: \begin{thm}[Weyl Dimension Formula] Let $\lambda \in P_+$. Then $\displaystyle \dim L_\lambda = \prod_{\alpha \in R_+} \frac{(\alpha,\lambda + \rho)}{(\alpha,\rho)}$. \end{thm} \begin{proof} The formula $\ch(L_\lambda) = \frac{\sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) }}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} $ is a polynomial in $x$. In particular, it defines a real-valued function on $\RR_{>0}\times \hh$ given by $x^\alpha \mapsto a^{\alpha(h)}$ --- when $a = 1$ or $h = 0$, the formula as written is the indeterminate form $\frac00$, but the function clearly returns $\sum_{\mu} \dim\bigl((L_\lambda)_\mu\bigr) = \dim L_\lambda$. We will calculate this value of the function by taking a limit, using l'H\^opital's rule. In particular, letting $x^\alpha \mapsto e^{t(\alpha,\lambda + \rho)}$ in \cref{dimform1} gives \eqn{ \prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) = \sum_{w\in W} \epsilon(w) e^{t(w(\rho),\lambda + \rho)} = \sum_{w\in W} \epsilon(w) e^{t(\rho,w(\lambda + \rho)) }} where the second equality comes from $w\mapsto w^{-1}$ and $ (w^{-1}x,y) = (x,wy) $. On the other hand, we let $x^\alpha \mapsto e^{t(\alpha,\rho)}$ in \cref{WeylCharEqn}. Then \begin{align} \ch L_\lambda |_{x = e^{t\rho}} & = \frac{ \sum_{w\in W}\epsilon(w) e^{t (w(\lambda + \rho),\rho)} }{\prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right) } \\ & = \frac{\prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{\prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \\ & = \prod_{\alpha \in R_+} \frac{ \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \end{align} Therefore \begin{align} \dim L_\lambda & = \lim_{t\to 0} \prod_{\alpha \in R_+} \frac{ \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \\ & = \prod_{\alpha \in R_+}\lim_{t\to 0} \frac{ \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \\ & \overset{\!\! \textrm{l'H}\!\!}= \prod_{\alpha \in R_+} \lim_{t\to 0} \frac{ \left( (\alpha/2,\lambda + \rho)e^{t(\alpha/2,\lambda + \rho)} + (\alpha/2,\lambda + \rho)e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( (\alpha/2,\rho)e^{t(\alpha/2,\rho)} +(\alpha/2,\rho) e^{-t(\alpha/2,\rho)} \right)} \\ & = \prod_{\alpha \in R_+} \frac{(\alpha,\lambda + \rho)}{(\alpha,\rho)} \end{align} \end{proof} \begin{eg} Let us compute the dimensions of the irreducible representations of $\gg = \sl(n+1)$. We work with the standard the simple roots be $\Delta = \{\alpha_1,\dots,\alpha_{n}\}$, whence $R_+ = \{\alpha_i + \alpha_{i+1} + \cdots + \alpha_j \}_{1 \leq i < j \leq n}$. Let us write $\lambda$ and $\rho$ in terms of the fundamental weights $\Lambda_i$, defined by $(\Lambda_i,\alpha_j) = \delta_{ij}$: $\rho = \sum_{i=1}^n \Lambda_i$ and $\lambda + \rho = \sum_{i=1}^n a_i\Lambda_i$. Then: \begin{align} \dim L_\lambda & = \prod_{\alpha \in R_+} \frac{(\lambda + \rho,\alpha)}{(\rho,\alpha)} \\ & = \prod_{1 \leq i \leq j \leq n} \frac{a_i + a_{i+1} + \dots + a_{j-1} + a_j}{j-i+1} \\ & = \frac1{n!!} \prod_{1 \leq i \leq j \leq n} \sum_{k=i}^j a_k \end{align} where we have defined $n!! \defined n! \,(n-1)! \cdots 3!\,2!\,1!$. For example, the irrep of $\sl(3)$ with weight $\lambda + \rho = 3\Lambda_1 + 2\Lambda_2$ has dimension $\frac1{2!!} 2 \cdot 3 \cdot (2+3) = 15$. \end{eg} \section{Algebraic Lie Groups} We have classified the representations of any semisimple Lie algebra, and therefore the representations of its simply connected Lie group. But a Lie algebra corresponds to many Lie groups, quotients of the simply connected group by (necessarily central) discrete subgroups, and a representation of the Lie algebra is a representation of one of these groups only if the corresponding discrete normal subgroup acts trivially in the representation. We will see that the simply connected Lie group of any semisimple Lie algebra is algebraic, and that its algebraic quotients are determined by the finite-dimensional representation theory of the Lie algebra. \subsection{Guiding example: $\SL(n)$ and $\PSL(n)$} \cite[Lecture 40]{notes} Our primary example, as always, is the Lie algebra $\sl(2,\CC)$\index{sl2C@$\sl(2,\CC)$}, consisting of traceless $2\times 2$ complex matrices. It is the Lie algebra of $\SL(2,\CC)$\index{SL2C@$\SL(2,\CC)$}, the group of $2\times 2$ complex matrices with determinant $1$. \begin{lemmadef} The group $\SL(2,\CC)$ has a non-trivial center: $Z(\SL(2,\CC)) = \{\pm 1\}$. We define the \emindexing{projective special linear group}\index{special linear group!projective} to be $\PSL(2,\CC) \defined \SL(2,\CC)/\{\pm 1\}$\index{PSL2C@$\PSL(2,\CC)$}. Equivalently, $\PSL(2,\CC) = \PGL(2,\CC) \defined \GL(2,\CC) / \{\textrm{scalars}\}$, the \emindexing{projective general linear group}\index{general linear group!projective}\index{PGL2C@$\PGL(2,\CC)$}. \end{lemmadef} \begin{prop} The group $\SL(2,\CC)$ is connected and simply connected. The kernel of the map $\ad: \SL(2,\CC) \to \GL(\sl(2,\CC))$ is precisely the center, and so $\PSL(2,\CC)$ is the connected component of the group of automorophisms of $\sl(2,\CC)$. The groups $\SL(2,\CC)$ and $\PSL(2,\CC)$ are the only connected Lie groups with Lie algebra $\sl(2,\CC)$. \end{prop} \begin{proof} The only nontrivial statement is that $\SL(2,\CC)$ is simply connected. Consider the subgroup $U \defined \left\{ \left[ \begin{matrix} 1 & * \\ 0 & 1\end{matrix}\right] \in \SL(2,\CC) \right\}$. Then $U$ is the stabilizer of the vector $\left[ \begin{matrix} 1 \\ 0 \end{matrix}\right] \in \CC^2 \sminus \{0\}$, and $\SL(2,\CC)$ acts transitively on $ \CC^2 \sminus \{0\}$. Thus the space of left cosets $\SL(2,\CC) / U$ is isomorphic to the space $ \CC^2 \sminus \{0\} \cong \RR^4 \sminus \{0\}$ as a real manifold. But $U \cong \CC$, so $\SL(2,\CC)$ is connected and simply connected. \end{proof} \begin{lemma} The groups $\SL(2,\CC)$ and $\PSL(2,\CC)$ are algebraic. \end{lemma} \begin{proof} The determinant of a matrix is a polynomial in the coefficients, so $\{x\in {\rm M}(2,\CC) \st \det x = 1\}$ is an algebraic group. Any automorphism of $\sl(2,\CC)$ preserves the Killing form, a nondegenerate symmetric pairing on the three-dimensional vector space $\sl(2,\CC)$. Thus $\PSL(2,\CC)$ is a subgroup of ${\rm O}(3,\CC)$. It is connected, and so a subgroup of $\SO(3,\CC)$, and three-dimensional, and so is all of $\SO(3,\CC)$. Moreover, $\SO(3,\CC)$ is algebraic: it consists of matrices $x\in {\rm M}(3,\CC)$ that preserve the nondegenerate form (a system of quadratic equations in the coefficients) and have unit determinant (a cubic equation in the coefficients). \end{proof} Recall that any irreducible representation of $\sl(2,\CC)$ looks like a chain: $e$ moves up the chain, $f$ down, and $h$ acts diagonally with eigenvalues changing by $2$ from $m$ at the top to $-m$ at the bottom: \eqn{ \begin{tikzpicture}[baseline=(v2.base)] \path (0,0) node[label=left:{$v_0$}] (v0) {$\bullet$} ++(0,-1) node[label=left:{$v_1$}] (v1) {$\bullet$} ++(0,-1) node[label=left:{$v_2$}] (v2) {$\bullet$} ++(0,-1) node (dots) {$\vdots$} ++(0,-1) node[label=left:{$ v_{m-1}$}] (vn1) {$\bullet$} ++(0,-1) node[label=left:{$ v_m$}] (vn) {$\bullet$} ; \draw[->] (v0) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m$} (v0); \draw[->] (v1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m-2$} (v1); \draw[->] (v2) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m-4$} (v2); \draw[->] (dots) .. controls +(330:1) and +(30:1) .. (dots); \draw[->] (vn1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = 2-m$} (vn1); \draw[->] (vn) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = -m$} (vn); \draw[->] (v0) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 1$} (v1); \draw[->] (v1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 2$} (v2); \draw[->] (v2) .. controls +(300:.5) and +(60:.5) .. (dots); \draw[->] (dots) .. controls +(300:.5) and +(60:.5) .. (vn1); \draw[->] (vn1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = m$} (vn); \draw[->] (v1) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle m = e$} (v0); \draw[->] (v2) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle m-1 = e$} (v1); \draw[->] (dots) .. controls +(120:.5) and +(240:.5) .. (v2); \draw[->] (vn1) .. controls +(120:.5) and +(240:.5) .. (dots); \draw[->] (vn) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle 1 = e$} (vn1); \end{tikzpicture}} The exponential map $\exp: \sl(2,\CC) \to \SL(2,\CC)$ acts on the Cartan by $th = \left[ \begin{matrix} t & \\ & -t \end{matrix}\right] \mapsto \left[ \begin{matrix} e^t & \\ & e^{-t} \end{matrix}\right] $. Let $T = \exp(\hh)$; then the kernel of $\exp: \hh \to T$ is $2\pi i \ZZ h$. On the other hand, when $t = \pi i$, $\exp(th) = -1$, which maps to $1$ under $\SL(2,\CC) \onto \PSL(2,\CC)$; therefore the kernel of the exponential map $\hh \to \PSL(2,\CC)$ is just $\pi i \ZZ h$. In particular, the $(m+1)$-dimensional representation $V_m$ of $\sl(2,\CC)$ is a representation of $\PSL(2,\CC)$ if and only if $m$ is even, because $-1 \in \SL(2,\CC)$ acts on $V_m$ as $(-1)^m$. We remark that $\ker\{\exp: \hh \to \SL(2,\CC)\}$ is precisely $2\pi i Q^\vee$, where $Q^\vee$ is the coroot lattice of $\sl(2)$, and $\ker\{\exp: \hh \to \PSL(2,\CC)\}$ is precisely the coweight lattice $2\pi i P^\vee$ \begin{remark} This will be the model for any semisimple Lie algebra $\gg$ with Cartan subalgebra $\hh$. We will understand the exponential map from $\hh$ to the simply connected Lie group $G$ corresponding to $\gg$, and we will also understand the map to $G / Z(G)$, the simplest quotient. Every group with Lie algebra $\gg$ is a quotient of $G$, and hence lies between $G$ and $G/Z(G)$. The kernels of the maps $\hh \to G$ and $\hh \to G/Z(G)$ will be precisely $2\pi i Q^\vee$ and $2\pi i P^\vee$, respectively, and every other group will correspond to a lattice between these two. \end{remark} Let us consider one further example: $\SL(n,\CC)$\index{SLnC@$\SL(n,\CC)$}\index{slnC@$\sl(n,\CC)$}. It is simply-connected, and its center is $Z(\SL(n,\CC)) = \{n\textrm{th roots of unity}\}$. We define the \emindexing{projective special linear group}\index{special linear group!projective} to be $\PSL(n,\CC) \defined \SL(n,\CC) / Z(\SL(n,\CC))$\index{PSLnC@$\PSL(n,\CC)$}; the groups with Lie algebra $\sl(n,\CC)$ live between these two, and so correspond to subgroups of $Z(\SL(n,\CC)) \cong \ZZ/n$, the cyclic group with $n$ elements. We now consider the Cartan $\hh \subseteq \sl(n,\CC)$, thought of as the space of traceless diagonal matrices: $\hh = \{ \ \in \CC^n \st \sum z_i = 0\}$. In particular, $\sl(n,\CC)$ is of $A$-type, and so we can identify roots and coroots: $\alpha_i = \alpha_i^\vee = \<0,\dots,0,1,-1,0,\dots,0\>$, where the non-zero terms are in the $(i,i+1)$th spots. Then the coroot lattice $Q^\vee$ is the span of $\alpha_i^\vee$: if $\sum z_i = 0$, then we can write $\ \in \ZZ^n$ as $z_1 \alpha_1 + (z_1 + z_2)\alpha_2 + \dots + (z_1 + \dots + z_{n-1})\alpha_{n-1}$, since $z_n = - (z_1 + \dots + z_{n-1})$. The coweight lattice $P^\vee$, on the other hand, is the lattice of vectors $\ $ with $\sum z_i = 0$ and with $z_i - z_{i+1}$ an integer for each $i \in \{1,\dots,n-1\}$. In particular, $\sum z_i = z_1+ \bigl( z_1 + (z_2 - z_1)\bigr) + \dots + \bigl( z_1 + (z_2 - z_1) + \dots + (z_n - z_{n-1}) \bigr) = nz_1 + \textrm{integer}$. Therefore $z_1 \in \ZZ \frac1n$, and $z_i \in z_1 + \ZZ$. So $P^\vee = Q^\vee \sqcup (\<\frac1n,\dots,\frac1n\> + Q^\vee) \sqcup \dots \sqcup (\< \frac{n-1}n ,\dots,\frac{n-1}n\> + Q^\vee)$. In this way, $P^\vee / Q^\vee$ is precisely $\ZZ/n$, in agreement with the center of $\SL(n,\CC)$. \subsection{Definition and General Properties of Algebraic Groups} \cite[Lecture 41]{notes} We have mentioned already (\cref{AGroupDefn}) the notion of an ``\indexing{algebraic group}\index{group!algebraic}'', and we have occasionally used some algebraic geometry (notably in the proof of \cref{CartanExistsThm}), but we have not developed that story. We do so now. \begin{defn} A subset $X \subseteq \CC^n$ is an \emindexing{affine variety}\index{variety} if it is the vanishing set of a set $P\subseteq \CC[x_1,\dots,x_n]$ of polynomials: \eqn{ X = V(P) \defined \{ x \in \CC^n \st p(x) = 0 \, \forall p \in P\} } Equivalently, $X$ is \emindexing{Zariski closed} (see \cref{ZariskiDefn}). To any affine variety $X$ with associate an ideal $I(X) \defined \{p\in \CC[x_1,\dots,x_n] \st p|_X = 0\}$. The \emindexing{coordinate ring} of, or the {\em ring of polynomial functions}\index{polynomial function} on, $X$ is the ring $\Ooo(X) \defined \CC[x]/I(X)$. \end{defn} \begin{lemma} If $X$ is an affine variety, then $I(X)$ is a radical ideal. If $I \subseteq J$, then $V(I) \supseteq V(J)$, and conversely if $X \subseteq Y$ then $I(X) \supseteq I(Y)$. It is clear from the definition that if $X$ is an affine varienty, then $V(I(X)) = X$; more generally, we can define $I(X)$ for any subset $X\supseteq \CC^n$, whence $V(I(X))$ is the Zariski closure of $X$. \end{lemma} \begin{defn} A {\em morphism of affine varieties}\index{morphism!of affine varieties}\index{affine variety!morphism} is a function $f:X \to Y$ such that the coordinates on $Y$ are polynomials in the coordinates of $X$. Equivalently, any function $f: X\to Y$ gives a homomorphism of algebras $f^\# : \Fun(Y) \to \Fun(X)$, where $\Fun(X)$ is the space of all $\CC$-valued functions on $X$. A function $f: X\to Y$ is a morphism of affine varieties if $f^\#$ restricts to a map $f^\#: \Ooo(Y) \to \Ooo(X)$. \end{defn} \begin{lemmadef} Any point $a\in \CC^n$ gives an \emindexing{evaluation map} $\ev_a: p \mapsto p(a): \CC[x_1,\dots,x_n] \to \CC$. If $X$ is an affine variety, then $a\in X$ if and only if $I(X) \subseteq \ker \ev_a$ if and only if $\ev_a: \Ooo(X) \to \CC$ is a morphism of affine varieties. \end{lemmadef} \begin{cor} The algebra $\Ooo(X)$ determines the set of evaluation maps $\Ooo(X) \to \CC$, and if $\Ooo(X)$ is presented as a quotient of $\CC[x_1,\dots,x_n]$, then it determines $X \subseteq \CC^n$. A morphism $f$ of affine varieties is determined by the algebra homomorphism $f^\#$ of coordinate rings, and conversely any such algebra homomorphism determines a morphism of affine varieties. Thus the category of affine varieties is precisely the opposite category to the category of finitely generated commutative algebras over $\CC$. \end{cor} \begin{lemmadef} The category of affine varieties contains all finite products. The {\em product}\index{product!of affine varieties} of affine varieties $X \subseteq \CC^m$ and $Y \subseteq \CC^l$ is $X\times Y \subseteq \CC^{m+l}$ with $\Ooo(X \times Y) \cong \Ooo(X) \otimes_\CC \Ooo(Y)$. \end{lemmadef} \begin{proof} The maps $\Ooo(X),\Ooo(Y) \to \Ooo(X\times Y)$ are given by the projections $X\times Y \to X,Y$. The map $\Ooo(X) \otimes \Ooo(Y) \to \Ooo(X\times Y)$ is an isomorphism because all three algebras are finitely generated and the evaluation maps separate functions. \end{proof} We recall \cref{AGroupDefn}: \begin{defn} An {\em affine algebraic group}\index{algebraic group} is a group object in the category of affine varieties. We will henceforth drop the adjective ``affine'' from the term ``algebraic group'', as we will never consider non-affine algebraic groups. Equivalently, an algebraic group is a finitely generated commutative algebra $\Ooo(G)$\index{OG@$\Ooo(G)$} along with algebra maps \begin{description} \item[comultiplication] $\Delta : \Ooo(G) \to \Ooo(G) \otimes_\CC \Ooo(G)$ dual to the multiplication $G\times G \to G$ \item[antipode] $\anti: \Ooo(G) \to \Ooo(G)$ dual to the inverse map $G\to G$ \item[counit] $\epsilon = \ev_e : \Ooo(G) \to \CC$ \end{description} The group axioms \cref{GroupAssoc,GroupId,GroupInv} are equivalent to the axioms of a commutative Hopf algebra\index{Hopf algebra!commutative} (\cref{HopfDefn}). \end{defn} \begin{lemmadef} Let $A$ be a Hopf algebra. An algebra ideal $B \subseteq A$ is a \emindexing{Hopf ideal}\index{ideal!Hopf} if $\Delta(B) \subseteq B \otimes A + A\otimes B \subseteq A\otimes A$. An ideal $B \subseteq A$ is Hopf if and only if the Hopf algebra structure on $A$ makes the quotient $B/A$ into a Hopf algebra. \end{lemmadef} \begin{defn} A commutative but not necessarily reduced Hopf algebra is a \emindexing{group scheme}. \end{defn} \begin{defn} An affine variety $X$ over $\CC$ is \emindexing{smooth} if $X$ is a manifold. \end{defn} \begin{prop} An algebraic group is smooth. \end{prop} \begin{proof} Let $E = \ker \epsilon$. Since $e \cdot e = e$, we see that the following diagram commutes: \eqn{ \begin{tikzpicture}[baseline=(X.base),scale=1.5] \path coordinate (X) +(0,.5) node (G) {$\Ooo(G)$} +(2,.5) node (GG) {$\Ooo(G) \otimes \Ooo(G)$} +(0,-.5) node (C) {$\CC$} +(2,-.5) node (CC) {$\CC \otimes \CC$}; \draw[->] (G) -- node[auto] {$\scriptstyle \Delta$} (GG); \draw[->] (G) -- node[auto] {$\scriptstyle \epsilon$} (C); \draw[->] (GG) -- node[auto] {$\scriptstyle \epsilon \otimes \epsilon$} (CC); \draw[<->] (CC) -- node[auto,swap] {$\sim$} (C); \end{tikzpicture} } In particular, \eqn[HopfIdeal]{\Delta E \subseteq E \otimes \Ooo(G) + \Ooo(G) \otimes E,} and so $E$ is a Hopf ideal, and $\Ooo(G) / E$ is a Hopf algebra. Moreover, \cref{HopfIdeal} implies that $\Delta(E^n) \subseteq \sum_{k+l = n} E^k \otimes E^l$, and so $\Delta$ and $\anti$ induce maps $\tilde\Delta$ and $\tilde\anti$ on $R = \gr_E\Ooo(G) \defined \bigoplus_{k\in \NN}E^k / E^{k+1}$. In particular, $R$ is a \emindexing{graded Hopf algebra}\index{Hopf algebra!graded}, and is generated as an algebra by $R_1 = E / E^2$. Moreover, if $x\in R_1$, then $x$ is primitive: $\Delta x = x\otimes 1 + 1\otimes x$. Since $R_1 = E/E^2$ is finitely dimensional, $R$ is finitely generated; let $R = \CC[y_1,\dots,y_n] / J$ where $n = \dim G$ and $J$ is a Hopf ideal of the Hopf algebra $\CC[y_1,\dots,y_n]$ with the generators $y_i$ all primitive. We can take the $y_i$s to be a basis of $R_1$, and so $J_1 = 0$. We use the fact that $\CC[y_1,\dots,y_n] \otimes \CC[y_1,\dots,y_n] = \CC[y_1,\dots,y_n,z_1,\dots,z_n]$, and that the antipode $\Delta$ is given by $\Delta: f(y) \mapsto f(y + z)$. Then a minimal-degree homogeneous element of $J$ must be primitive, so $f(y + z) = f(y) + f(z)$, which in characteristic zero forces $f$ to be homogeneous of degree $1$. A similar calculation with the antipode forces the minimal-degree homogeneous elements $f\in J$ to satisfy $\anti f = -f$. In particular, $\gr_E\Ooo(G)$ is a polynomial ring. We leave out the fact from algebraic geometry that this is equivalent to $G$ being smooth at $e$. But we have shown that the Hopf algebra maps are smooth, whence $G$ is smooth at every point. \end{proof} \begin{cor} An algebraic group over $\CC$ is a Lie group. \end{cor} Recall that if $G$ is a Lie group with $\Ccc(G)$ the algebra of smooth functions on $G$, and if $\gg = \Lie(G)$, then $\Uu\gg$ acts on $\Ccc(G)$ by left-invariant differential operators, and indeed is isomorphic to the algebra of left-invariant differential operators. \begin{defn} Let $G$ be a group. A subalgebra $S \subseteq \Fun(G)$ is {\em left-invariant}\index{left-invariant algebra}\index{algebra!left-invariant} if for any $s\in S$ and any $g\in G$, the function $h\mapsto s(g^{-1}h)$ is an element of $S$. Equivalently, we define the action $G \acts \Fun(G)$ by $gs = s\circ g^{-1}$; then a subalgebra is left-invariant if it is fixed by this action. \end{defn} \begin{lemma} \label{LieAlgAlgLemma} Let $S \subseteq \Fun(G)$ be a left-invariant subalgebra, and let $s\in S$ be a function such that $\Delta s = \{ (x,y) \mapsto s(xy) \} \subseteq \Fun(G\times G)$ is in fact an element of $S \otimes S \subseteq \Fun(G) \otimes \Fun(G) \mono \Fun(G\times G)$. Then let $\Delta s = \sum s_1 \otimes s_2$, where we suppress the indices of the sum. The action $G\acts S$ is given by \eqn{ g: s \mapsto \sum s_1(g^{-1}) s_2 } \end{lemma} \begin{cor} Let $u$ be a left-invariant differential operator and $s\in S$ as in \cref{LieAlgAlgLemma}, where $S \subseteq \Ccc(G)$ is a left-invariant algebra of smooth functions. Then $us \in S$. \end{cor} \begin{proof} The left-invariance of $u$ implies that $u(gs) = g\,u(s)$. Since $s(g^{-1}) \in \CC$, we have: \eqn{ u(gs)(h) = u\left( \sum s_1(g^{-1})s_2\right)(h) = \sum s_1(g^{-1}) \,u(s_2)(h) } Let $h = e$. Then $\sum s_1(g^{-1})\, u(s_2)(e) = u(gs)(e) = g(us)(e) = (us)(g^{-1})$. In particular: \eqn[LieAlgAldEqn1]{ (us)(g) = \sum s_1(g)\,u(s_2)(e)} But $(us_2)(e)$ are numbers. Thus $us \in S$. \end{proof} \begin{cor} Let $G$ be an algebraic group, with Lie algebra $\gg= \Lie(G)$. Then $\Uu\gg$ acts on $\Ooo(G)$ by left-invariant differential operators. Since a differential operator is determined by its action on polynomials, we have a natural embedding $\Uu\gg \mono \Ooo(G)^*$ of vector spaces. \end{cor} \begin{lemma} Let $u,v \in \Uu\gg$, where $G$ is an algebraic group. Then $uv\,s = \sum u(s_1)\,v(s_2)(e)$. \end{lemma} \begin{proof} This follows from \cref{LieAlgAldEqn1}. \end{proof} \begin{cor} \label{LieAlgAldLemma2} For each differential operator $u\in \Uu\gg$, let $\lambda_u \in \Ooo(G)^*$ be the map $\lambda_u:s \mapsto u(s)(e)$. Then $\lambda_{uv}(s) = \sum \lambda_u(s_1) \,\lambda_v(s_2)$. \end{cor} \begin{lemma} Let $A$ be any (counital) coalgebra, for example a Hopf algebra. Then $A^*$ is naturally an algebra: the map $A^* \otimes A^* \to A^*$ is given by $\<\mu \nu,a \> \defined \<\mu \otimes \nu,\Delta a\>$, and $\epsilon: A \to \CC$ is the unit $\epsilon \in A^*$. \end{lemma} \begin{remark} The dual to an algebra is not necessarily a coalgebra; if $A$ is an algebra, then it defines a map $\Delta: A^* \to (A\otimes A)^*$, but if $A$ is infinite-dimensional, then $(A\otimes A)^*$ properly contains $A^* \otimes A^*$. \end{remark} \begin{remark} Following the historical precedent, we take the pairing $(A^* \otimes A^*) \otimes (A\otimes A)$ to be $\<\mu \otimes \nu,a\otimes b\> = \<\mu,a\>\<\nu,b\>$. This is in some sense the wrong pairing --- it corresponds to writing $(A\otimes B)^* = A^* \otimes B^*$ for finite-dimensional vector spaces $A,B$, whereas $B^* \otimes A^*$ would be more natural --- and is ``wrong'' in exactly the same way that the ``$-1$'' in the definition of the left action of $G$ on $\Fun(G)$ is wrong. \end{remark} \begin{prop} The embedding $\Uu\gg \mono \Ooo(G)^*$ is given by the map $u \mapsto \lambda_u$ in \cref{LieAlgAldLemma2}, and is an algebra homomorphism. \end{prop} \begin{defn} Let $G$ be any group; then we define the \emindexing{group algebra} $\CC[G]$ of $G$ to be the free vector space on the set $G$, with the multiplication given on the basis by the multiplication in $G$. The unit $e\in G$ becomes the unit $1 \cdot e \in \CC[G]$. \end{defn} \begin{lemma} If $G$ is an algebraic group, then $\CC[G] \mono \Ooo(G)^*$ is an algebra homomorphism given on the basis $g \mapsto \ev_g$. \end{lemma} \subsection{Constructing $G$ from $\gg$} \cite[Lectures 42 and 43]{notes} A Lie algebra $\gg$ does not determine the group $G$ with $\gg = \Lie(G)$. We will see that the correct extra data consists of prescribed representation theory. Throughout the discussion, we gloss the details, merely waving at the proofs of various statements. \begin{lemmadef} Let $G$ be an algebraic group. A finite-dimensional module $G\acts V$ is {\em algebraic}\index{algebraic module}\index{module!algebraic} if the map $G \to \GL(V)$ is a morphism of affine varieties. Any finite-dimensional algebraic (left) action $G\acts V$ of an algebraic group $G$ gives rise to a (left) \emindexing{coaction} $V^* \to \Ooo(G) \otimes V^*$: \eqn{ \begin{tikzpicture}[auto,baseline=(X.base)] \coordinate (X) at (0,1); \node (v) at (0,2) {$V^*$}; \node (ov1) at (5,2) {$\Oo(G) \otimes V^*$}; \node (ov2) at (0,0) {$\Oo(G) \otimes V^*$}; \node (oov) at (5,0) {$\Oo(G) \otimes \Oo(G) \otimes V^*$}; \\ \draw[->] (v) to node {\scriptsize \rm coact} (ov1); \draw[->] (v) to node {\scriptsize \rm coact} (ov2); \draw[->] (ov2) to node {\scriptsize \rm id $\otimes$ coact} (oov); \draw[->] (ov1) to node {\scriptsize \rm comult $\otimes$ id} (oov); \end{tikzpicture} } This in turn gives rise to a (left) action $\Ooo(G)^* \acts V$, which specializes to the actions $G \acts V$ and $\Uu\gg \acts V$ under $G \mono \CC[G] \mono \Ooo(G)^*$ and $\Uu\gg \mono \Ooo(G)^*$. \end{lemmadef} We will take the following definition, referring the reader to \cite{ES} for the connections between rigid categories and Hopf algebras, and \cite{BK} and references therein for a thorough category-theoretic discussion. \begin{defn} A \emindexing{rigid category} is an abelian category $\Mm$ with a (unital) monoidal product and duals. We will write the monoidal product as $\otimes$. A {\em rigid subcategory} of $\Mm$ is a full subcategory that is a tensor category with the induced abelian and tensor structures. I.e.\ it is a full subcategory containing the zero object and the monoidal unit, and closed under extensions, tensor products, and duals. \end{defn} \begin{defn} A rigid category $\Mm$ is \emindexing{finitely generated} if for some finite set of objects $V_1,\dots,V_n \in \Mm$, any object is a subquotient of some tensor product of $V_i$s (possibly with multiplicities). Of course, by letting $V_0 = V_1 \oplus \dots \oplus V_n$, we see that any finitely generated rigid category is in fact generated by a single object. \end{defn} \begin{eg} For any Lie algebra $\gg$, the category \cat{$\gg$-mod} of finite-dimensional representations of $\gg$ is a tensor category; indeed, if $U$ is any Hopf algebra, then \cat{$U$-mod} is a tensor category. \end{eg} \begin{defn} Let $\gg$ be a finite-dimensional Lie algebra over $\CC$, and let $\Mm$ be a rigid subcategory of \cat{$\gg$-mod}. By definition, for each $V \in \Mm$, we have a linear map $\Uu\gg \to \End V$. Thus for each linear map $\phi: \End V \to \CC$ we can construct a map $\{\Uu\gg \to \End V \overset{\phi}\to \CC\} \in \Uu\gg^*$; we let $A_\Mm \subseteq \Uu\gg^*$ be the set of all such maps. Then $A_\Mm$ is the set of {\em matrix coefficients}\index{matrix coefficient} of $\Mm$. Indeed, for each $V$, the maps $\Uu\gg \to \End V \to \CC$ are the matrix coefficients of the action $\gg \acts V$. In particular, for each $V\in \Mm$, the space $(\End V)^*$ is naturally a subspace of $A_\Mm$, and $A_\Mm$ is the union of such subspaces. \end{defn} \begin{lemma} If $\Mm$ is a rigid subcategory of \cat{$\gg$-mod}, then $A_\Mm$ is a subalgebra of the commutative algebra $\Uu\gg^*$. Moreover, $A_\Mm$ is a Hopf algebra, with comultiplication dual to the multiplication in $\Uu\gg \otimes \Uu\gg \to \Uu\gg$. \end{lemma} \begin{proof} The algebra structure on $A = A_\Mm$ is straightforward: the multiplication and addition stem from the rigidity of $\Mm$, the unit is $\epsilon: \Uu\gg \to \CC \isom \Cc$, and the subtraction is not obvious but is straightforward; it relies on the fact that $\Mm$ is abelian, and so contains all subquotients. We will explain where the Hopf structure on $A$ comes from --- since $\Uu\gg$ is infinite-dimensional, $\Uu\gg^*$ does not have a comultiplication in general. But $\Mm$ consists of finite-dimensional representations; if $V \in \Mm$, then we send $\{\Uu\gg \to \End V \overset{\phi}\to \CC\} \in A$ to $\{(\End V \otimes \End V) \overset{\textrm{multiply}}{\longto} \End V \overset{\phi} \to \CC\} \in (\End V)^* \otimes (\End V)^*\subseteq A \otimes A$. That this is dual to the multiplication in $\Uu\gg$ comes from the fact that $\Uu\gg \to \End V$ is an algebra homomorphism. \end{proof} \begin{cor} The map $\Uu\gg \to A^*$ dual to $A \mono \Uu\gg^*$ is an algebra homomorphism. \end{cor} \begin{prop} \label{thepropabove} Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod}. Then $A_\Mm = \Ooo(G)$ for some algebraic group $G$. \end{prop} \begin{proof} If $\Mm$ is finitely generated, then there is some finite-dimensional representation $V_0 \in \Mm$ so that $(\End V_0)^*$ generates $A_\Mm$. Then $A_\Mm$ is a finitely generated commutative Hopf algebra, and so $\Ooo(G)$ for some algebraic group $G$. \end{proof} \begin{lemma} \label{thepreviouslemma} Let $\gg$ be a finite-dimensional Lie algebra, $\Mm$ a finitely-generated rigid subcategory of \cat{$\gg$-mod} , and $G$ the algebraic group corresponding to the algebra $A_\Mm$ of matrix coefficients of $\Mm$. We will henceforth write $\Ooo(G)$ for $A_\Mm$. Then $G$ acts naturally on each $V \in \Mm$. \end{lemma} \begin{proof} Let $\{v^1,\dots,v^n\}$ be a basis of $V$ and $\{\xi_1,\dots,\xi_n\}$ the dual basis of $V^*$. For each $i$, we define $\lambda_i: V \to \Ooo(G)$ by $v \mapsto \{ u \mapsto \<\xi_i,uv\>\}$ where $v\in V$ and $u\in \End V$. \ Then we define $\sigma: V \to V \otimes \Ooo(G)$ a right coaction of $\Ooo(G)$ on $V$ by $v \mapsto \sum_{i=1}^n v^i \otimes \lambda_i(v)$. It is a coaction because $uv = \sum_{i=1}^n v^i\lambda_i(v)(u)$ by construction. In particular, it induces an action $G \acts V$. \end{proof} \begin{prop} Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod} that contains a faithful representation of $\gg$. Then the map $\Uu\gg \to A_\Mm^*$ is an injection. \end{prop} \begin{proof} Let $\sigma: G \acts V$ as in the proof of \cref{thepreviouslemma}. Then the induced representation $\Lie(G) \acts V$ is by contracting $\sigma$ with point derivations. But $\gg \acts V$ and the map $\Uu\gg \to \Ooo(G)^*$ maps $x\in \gg$ to a point derivation since $x\in \gg$ is primitive. Thus the following diagram commutes for each $V \in \Mm$: \eqn{ \begin{tikzpicture}[baseline=(X.base), scale=1.5] \path coordinate (X) +(0,.5) node (g) {$\gg$} +(1.5,.5) node (G) {$\Ooo(G)^*$} +(0,-.5) node (L) {$\Lie(G)$} +(1.5,-.5) node (V) {$\gl(V)$}; \draw[->] (g) -- (G); \draw[->] (g) -- (L); \draw[->] (L) -- (V); \draw[->] (G) -- (V); \end{tikzpicture}} The map $\gg \to \Lie(G)$ does not depend on $V$. Thus, if $\Mm$ contains a faithful $\gg$-module, then $\Uu\gg \mono \Uu\Lie(G) \mono \Ooo(G)^*$. \end{proof} \begin{eg} Let $\gg = \CC$ be one-dimensional, and let $\Mm$ be generated by one-dimensional representations $V_\alpha$ and $V_\beta$, where the generator $x\in \gg$ acts on $V_\alpha$ by multiplication by $\alpha$, and on $V_\beta$ by $\beta$. Then $\Mm$ is generated by $V_\alpha \oplus V_\beta$, and $x$ acts as the diagonal matrix $\left[ \begin{matrix} \alpha & \\ & \beta \end{matrix}\right]$. Let $\alpha,\beta \neq 0$, and let $\alpha \not\in \QQ\beta$. Then $\Lie(G)$ will contain all diagonal matrices, since $\alpha / \beta \not\in \QQ$, but $\gg \mono \Lie(G)$ as a one-dimensional subalgebra. The group $G$ is the complex torus, and the subgroup corresponding to $\gg \subseteq \Lie(G)$ is the irrational line. \end{eg} \begin{prop} Let $V_0$ be the generator of $\Mm$ satisfying the conditions of \cref{thepropabove}, and let $W$ be a neighborhood of $0\in \gg$. Then the image of $\exp(W)$ is Zariski dense in $G$. \end{prop} \begin{proof} Assume that $\Mm$ contains a faithful representation of $\gg$; otherwise, mod out $\gg$ by the kernel of the map $\gg \to \Lie(G)$. Thus, we may consider $\gg \subseteq \Lie(G)$, and let $H \subseteq G$ be a Lie subgroup with $\gg = \Lie(H)$. Let $f \in \Ooo(G)$ and $u \in \Uu\gg$; then the pairing $\Uu\gg \otimes \Ooo(G) \to \CC$ sends $u\otimes f \mapsto u(f|_H)(e)$. In particular, the pairing depends only on a neighborhood of $e\in H$, and hence only on a neighborhood $W \ni 0$ in $\gg$. But the pairing is nondegenerate; if the Zariski closure of $\exp W$ in $G$ were not all of $G$, then we could find $f,g \in \Ooo(G)$ that agree on $\exp W$ but that have different behaviors under the pairing. \end{proof} \begin{defn} Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod}, and let $G$ be the corresponding algebraic group as in \cref{thepropabove}. Then $\gg$ is \emindexing{algebraically integrable} with respect to $\Mm$ if the map $\gg \to \Lie(G)$ is an isomorphism. In particular, $\Mm$ must contain a faithful representation of $\gg$. \end{defn} \begin{eg} \label{abelianhenceintegrable} Let $\gg$ be a finite-dimensional abelian Lie algebra over $\CC$, and let $X \subseteq \gg^*$ be a lattice of full rank, so that $X \otimes_\ZZ \CC = \gg^*$. Let $\{\xi_1,\dots,\xi_n\}$ be a $\ZZ$-basis of $X$ and hence a $\CC$-basis of $\gg^*$, and let $\Mm = \{\bigoplus \CC_\lambda \st \lambda \in X\}$, where $\gg \acts \CC_\lambda$ by $z \mapsto \lambda(z)\times$. Then $V_0 = \bigoplus \CC_{\xi_i}$ is a faithful representation of $\gg$ in $\Mm$ and generates $\Mm$. Then $G \subseteq \GL(V_0)$ is the Zariski closure if $\exp \gg$, and for $z\in \gg$, $\exp(z_1,\dots,z_n)$ is the diagonal matrix whose $(i,i)$th entry is $e^{\xi_i(z)}$. Thus $G$ is a torus $T \cong (\CC^\times)^n$, with $\Oo(T) = \CC[t_1^{\pm 1},\dots,t_n^{\pm 1}]$. In particular, $\gg$ is algebraically integrable with respect to $\Mm$, since $X$ is a lattice. \end{eg} \begin{prop} \label{sumhenceintegrable} Let $\gg$ be a finite-dimensional Lie algebra and $\Mm$ a finitely generated rigid subcategory of \cat{$\gg$-mod} containing a faithful representation. Suppose that $\gg = \bigoplus_{i=1}^r \gg_i$ as a vector space, where each $\gg_i$ is a Lie subalgebra of $\gg$; then $\Mm$ embeds in \cat{$\gg_i$-mod} for each $i$. If each $\gg_i$ is algebraically integrable with respect to (the image of) $\Mm$, then so is $\gg$. \end{prop} \begin{proof} Let $G$, $G_i$ be the algebraic groups corresponding to $\gg \acts \Mm$ and to $\gg_i \acts \Mm$. Then for each $i$ we have a map $G_i \to G$. Let $H\subseteq G$ be the subgroup of $G$ corresponding to $\gg \subseteq \Lie(G)$. Consider the map $m: G_1 \times \cdots \times G_r \to G$ be the function that multiplies in the given order; it is not a group homomorphism, but it is a morphism of affine varieties. Since each $G_i \to G$ factors through $H$, and since $H$ is a subgroup of $G$, the map $m$ factors through $H$. Indeed, the differential of $m$ at the identity is the sum map $\bigoplus \gg_i \to \gg$. Thus we have an algebraic map $m$, with Zariski dense image. But it is a general fact that any such map (a \emindexing{dominant morphism}\index{morphism!dominant}) is dimension non-increasing. Therefore $\dim G \leq \dim (G_1 \times \cdots \times G_r) = \dim \gg$, and so $\gg = \Lie(G)$. \end{proof} \begin{thm}[Semisimple Lie algebras are algebraically integrable] \label{algint1thm} Let $\gg$ be a semisimple finite-dimensional Lie algebra over $\CC$, and let $\hh \subseteq \gg$ be its Cartan subalgebra and $Q$ and $P$ the root and weight lattices. Let $X$ be any lattice between these: $Q \subseteq X \subseteq P$. Let $\Mm$ be the category of finite-dimensional $\gg$-modules with highest weights in $X$. Then $\Mm$ is finitely generated rigid and contains a faithful representation of $\gg$, and $\gg$ is algebraically integrable with respect to $\Mm$. \end{thm} \begin{proof} Let $V \in \Mm$; then its highest weights are all in $X$, and so all its weights are in $X$ since $X \supseteq Q$. Moreover, the decomposition of $V$ into irreducible $\gg$-modules writes $V = \bigoplus L_\lambda$, where each $\lambda \in P_+ \cap X$. This shows that $\Mm$ is rigid. It contains a faithful representation because the representation of $\gg$ corresponding to the highest root is the adjoint representation, and the highest root is an element of $Q$ and hence of $X$. Moreover, $\Mm = \{ \bigoplus V_\lambda \st \lambda \in P_+ \cap X\}$ is finitely generated: $P_+ \cap X$ is $\ZZ_{\geq 0}$-generated by finitely many weights. We recall the triangular decomposition (c.f.\ \cref{triangulardecompprop}) of $\gg$: $\gg = \nn_- \oplus \hh \oplus \nn_+$. Then $\hh$ is abelian and acts on modules in $\Mm$ diagonally; in particular, $\hh$ is algebraically integrable by \cref{abelianhenceintegrable}. On the other hand, on any $\gg$-module, $\nn_+$ and $\nn_-$ act by strict upper- and strict lower-triangular matrices, and the matrix exponential restricted to strict upper- (lower-) triangular matrices is a polynomial. In particular, by finding a faithful generator of $\Mm$ (for example, the sum of the generators plus the adjoint representation), we see that $\nn_\pm$ are algebraically integrable. The conclusion follows by \cref{sumhenceintegrable}. \end{proof} \begin{thm}[Classification of Semisimple Lie Groups over $\CC$] Let $\gg$ be a finite-dimensional semisimple Lie algebra over $\CC$. Any connected Lie group $G$ with $\Lie(G)$ is semisimple; in particular, the algebraic groups constructed in \cref{algint1thm} comprise all integrals of $\gg$. \end{thm} \begin{proof} Let $\tilde G$ be the connected and simply connected Lie group with $\Lie(G) = \gg$; then any integral of $\gg$ is a quotient of $\tilde G$ be a discrete and hence central subgroup of $\tilde G$, and the integrals are classified by the kernels of these quotients and hence by the subgroups of the center $Z(\tilde G)$. Let $G_X$ be the algebraic group corresponding to $X$. Since $Z(G_P) = P/Q$, it suffices to show that $G_P$ is connected and simply connected. We show first that $G_X$ is connected. It is an affine variety; $G_X$ is connected if and only if $\Ooo(G_X)$ is an integral domain. Since $G_X$ is the Zariski closure of $\exp W$ for a neighborhood $W$ of $0\in \gg$, and $\exp W$ is connected, so is $G_X$. Let $U_\pm$ be the image of $\exp(\nn_\pm)$ in $G_X$, and let $T = \exp(\hh)$. But $\exp: \nn_\pm \to U_\pm$ is the matrix exponential on strict triangular matrices, and hence polynomial with polynomial inverse; thus $U_\pm$ are simply connected. We quote a fact from algebraic geometry: the image of an algebraic map contains a set Zariski open in its Zariski closure. In particular, since the image of $U_- \times T \times U_+$ is Zariski dense, it contains a Zariski open set, and so the complement of the image must live inside some closed subvariety of $G_X$ with complex codimension at least 1, and hence real codimension at least 2, since locally this subvariety is the vanishing set of some polynomials in $\CC^n$. So in any one-complex-dimensional slice transverse to this subvariety, the subvariety consists of just some points. Therefore any path in $G_X$ can be moved off this subvariety and hence into the image of $U_- \times T \times U_+$. It suffices to consider paths in $G_X$ from $e$ to $e$, and by choosing for each such path a nearby path in $U_-TU_+$, we get a map $\pi_1(U_-TU_+) \onto \pi_1(G_X)$. On the other hand, by the LU decomposition (see any standard Linear Algebra textbook, e.g.\ \cite{WikiLU}), the map $U_- \times T \times U_+ \to U_-TU_+$ is an isomorphism. Since $U_\pm$ are isomorphic as affine varieties to $\nn_\pm$, we have: \eqn{ \pi_1(U_-TU_+) = \pi_1(U_- \times T \times U_+) = \pi_1(T) } And $\pi_1(T) = X^*$, the co-lattice to $X$, i.e.\ the points in $\gg$ on which all of $X$ takes integral values. Thus, it suffices to show that the map $\pi_1(T) \onto \pi_1(G_P)$ collapses loops in $T$ when $X = P$. But then $\pi_1(T) = P^* = Q^\vee$ is generated by the simple coroots $\alpha_i^\vee$. For each generator $\alpha_i^\vee = h_i$, we take $\sl(2)_i \subseteq \gg$ and exponentiate to a map $\SL(2,\CC) \to G$. Then the loops in $\exp(\RR h_i)$, which generate $\pi_1(T)$, go to loops in $\SL(2,\CC)$ before going to $G$. But $\SL(2,\CC)$ is simply connected. This shows that the map $\pi_1(T) \onto \pi_1(G_P)$ collapses all such loops, and $G_P$ is simply connected. \end{proof} \section{Conclusion} \cite[Lecture 44]{notes} In Chapter 5, we put semisimple Lie algebras over $\CC$ into bijection with possibly-disconnected Dynkin diagrams, and classified all such diagrams. In the current chapter, we described how to compute the representation theory of any such diagram: the finite-dimensional irreducible representations correspond to elements of the weight lattice of the diagram. Then we showed that the integrals of any semisimple Lie algebra are algebraic, and correspond to lattices between the weight and root lattice: the simply connected group corresponds to the weight lattice, and the adjoint group corresponds to the root lattice. The category of representations of the Lie group corresponding to a given lattice is precisely the category of representations whose highest (and hence all) weights lie in the lattice. We remark that the index of the weight lattice on the root lattice --- the size of the quotient --- is precisely the determinant of the Cartan matrix. The story we have told can be generalized. We will not justify it, but only sketch how it goes. We recall \cref{reductivedefn}. We say that a complex Lie group $G$ is \emindexing{reductive} if it is linear and if its finite-dimensional representations are completely reducible. We demand linearity to assure that the group have finite-dimensional representations: for example, an elliptic curve is a complex Lie group but it is compact, so any holomorphic function is constant, so its only finite-dimensional representations are trivial. Let $G$ be a reductive Lie group and $\gg = \Lie(G)$ its Lie algebra. Then we have a Levi decomposition $\gg = \ss \oplus \rr$, and if $G$ is reductive, then $\rr = \zz$ is abelian and is the center of $\gg$, so $\gg$ is reductive as a Lie algebra. Let $\gg$ be a reductive Lie algebra. Then the Cartan subalgebra $\hh$ of $\gg$ is $\hh_{\ss} \oplus \zz$, where $\hh_\ss$ is the Cartan subalgebra $\ss$. We have a coroot lattice $Q^\vee \subseteq \hh_\ss$, and we pick a lattice $X^* \supseteq Q^\vee$ such that $X \supseteq Q$ and $X^*$ spans $\hh$. Then let $\Mm$ be the category of finite-dimensional representations of $\gg$ with weights in $X$ and on which $\hh$ acts diagonally --- this is an extra condition, because $\zz$ need not act diagonally on a representation of $\gg$. Then the entire story of algebraic integrals applies. For any \emindexing{root datum} --- a choice of roots $\alpha$ and coroots $\alpha^\vee$ satisfying natural conditions --- we get a Cartan matrix, the Cartan matrix of $\ss$. In particular, as we will not explain here, the root data $(X,\alpha_i\in X,\alpha_i^\vee\in X^*)$ classify: \begin{enumerate} \item reductive Lie groups over $\CC$ \item reductive Lie algebras over $\CC$ \item compact real Lie groups \item reductive algebraic groups over any algebraically closed field in any characteristic \item {\em group schemes}\index{group scheme}, or ``algebraic groups over $\ZZ$'', and therefore {\em Chevalley groups}\index{Chevalley group}\index{group!Chevalley}, the tensor products of group schemes with finite fields \item finite groups of Lie type, essentially the source of the finite simple groups \end{enumerate} Some of this story is outlined but mostly not proved in \cite{anton}. It would make a good second semester for this one-semester course, but in fact the second semester will tell a different story of quantum groups \cite{Qnotes}. \Section{Exercises} \begin{enumerate} \item Show that the simple complex Lie algebra $\gg$ with root system $G_2$ has a $7$-dimensional matrix representation with the generators shown below. \eqn[hw61gen]{ \begin{array}{ccc} e_1 = \left[ \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] & & f_1 = \left[ \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \\ \\ e_2 = \left[ \begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] & & f_2 = \left[ \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right] \end{array} } \item \begin{enumerate} \item Show that there is a unique Lie group $G$ over $\CC$ with Lie algebra of type $G_2$. \item Find explicit equations of $G$ realized as the algebraic subgroup of $\GL(7, \CC)$ whose Lie algebra is the image of the matrix representation in Problem 1. \end{enumerate} \item Show that the simply connected complex Lie group with Lie algebra $\so(2n, \CC)$ is a double cover $\Spin(2n, \CC)$ of $\SO(2n, \CC)$, whose center $Z$ has order four. Show that if $n$ is odd, then $Z$ is cyclic, and there are three connected Lie groups with this Lie algebra: $\Spin(2n, \CC)$, $\SO(2n, \CC)$ and $\SO(2n, \CC)/ \{\pm I \}$. If $n$ is even, then $Z \cong (\ZZ/2\ZZ)^2$, and there are two more Lie groups with the same Lie algebra. \item If $G$ is an affine algebraic group, and $\gg$ its Lie algebra, show that the canonical algebra homormorphism $\Uu\gg \to \Ooo(G)^*$ identiŽes $\Uu\gg$ with the set of linear functionals on $\Ooo(G)$ whose kernel contains a power of the maximal ideal $\mm = \ker(\ev_e)$. \item Show that there is a unique Lie group over $\CC$ with Lie algebra of type $E_8$. Find the dimension of its smallest matrix representation. \item Construct a finite dimensional Lie algebra over $\CC$ which is not the Lie algebra of any algebraic group over $\CC$. [Hint: the adjoint representation of an algebraic group on its Lie algebra is algebraic.] \end{enumerate} \printindex % For whatever reason, TeX, or at least this document style, does not want to include the Index in the Table of Contents. I have included it be hand: in my Intro file, I included an index command, with some added instruction: %% \addcontentsline{toc}{chapter}{Bibliography} \begin{thebibliography}{99} \bibitem{BK} B. Bakalov and A. Kirillov, Jr. {\em Lectures on Tensor Categories and Modular Functors}. American Mathematical Society, Providence, RI, 2001. An earlier version is available at \url{http://www.math.sunysb.edu/~kirillov/tensor/tensor.ps.gz}. \bibitem{Borel} A. Borel. {\em Linear Algebraic Groups}. Second enlarged edition. Graduate Texts in Mathematics; 126. Springer-Verlag New York Inc., New York, NY, 1991. \bibitem{Hom1} H. Cartan and S. Eilenberg. {\em Homological Algebra}. Princeton Mathematics Series; 19. Princeton University Press, Princeton, NJ, 1999. \bibitem{ES} P. Etingof and O. Schiffmann. {\em Lectures on quantum groups}. Second edition. Lectures in Mathematical Physics. International Press, Somerville, MA, 2002. \bibitem{FH} W. Fulton and J. Harris. {\em Representation Theory}. Graduate Texts in Mathematics; 129. Springer-Verlag New York Inc., New York, NY, 1991. \bibitem{Hom2} S.I. Gelfand and Y.I. Manin. {\em Methods of Homological Algebra}. Second edition. Springer Monographs in Mathematics. Springer-Verlag Berlin Heidelberg, 2003. \bibitem{GW} K.R. Goodearl and R.B. Warfield, Jr. {\em An Introduction to Noncommutative Noetherian Rings}. London Mathematics Society; Student Texts 16. Cambridge University Press, Cambridge, UK, 1989. \bibitem{notes} M. Haiman. {\em Math 261A: Lie Groups}, UC Berkeley, Fall 2008. Lecture notes by T. Johnson-Freyd available at \url{http://math.berkeley.edu/~theojf/LieGroups.pdf}. \bibitem{HarishChandra} Harish-Chandra. {\em Faithful Representations of Lie Algebras}, Annals of Mathematics, vol. 50 (1) pp. 68-76, 1949. \href{http://www.ams.org/mathscinet-getitem?mr=30945}{MR0030945}. \bibitem{Iwasawa} K. Iwasawa. {\em On the representation of Lie algebras}, Japanese Journal of Mathematics, vol. 19 pp. 405--426, 1948. \href{http://www.ams.org/mathscinet-getitem?mr=32613}{MR0032613}. \bibitem{Knapp} A.W. Knapp. {\em Lie Groups Beyond an Introduction}. Second edition. Progress in Mathematics, 140. Birkh{\"a}user Boston Inc., Boston, MA, 2002. \bibitem{Knutson} A. Knutson. {\em Math 261: Lie Groups and Lie Algebras}, UC Berkeley, 2001-02. \url{http://math.berkeley.edu/~allenk/courses/fall01/261/} and \url{http://math.berkeley.edu/~allenk/courses/spr02/261b/}. \bibitem{Lam} T.Y. Lam. {\em A First Course in Noncommutative Rings}. Graduate Texts in Mathematics; 131. Springer-Verlag New York Inc., New York, NY, 2001. \bibitem{Lang} S. Lang. {\em $\SL_2({\bf R})$}. Graduate Texts in Mathematics; 105. Springer-Verlag New York Inc., New York, NY, 1975. \bibitem{WikiLU} ``LU decomposition''. {\em Wikipedia, the free encyclopedia}. Wikimedia Foundation, Inc.,2003. Available at \url{http://en.wikipedia.org/wiki/LU_decomposition}. \bibitem{McCR} J.C. McConnell and J.C. Robson. {\em Noncommutative Noetherian Rings}. Pure and Applied Mathematics. Wiley Interscience, John Wiley and Sons, New York, 1987. \bibitem{Qnotes} N. Reshetikhin. {\em Math 261B: Quantum Groups}, UC Berkeley, Spring 2009. Lecture notes by T. Johnson-Freyd available at \url{http://math.berkeley.edu/~theojf/QuantumGroups.pdf}. \bibitem{anton} N. Reshetikhin, V. Serganova, and R. Borcherds. {\em Math 261A: Lie Groups and Lie Algebras}, UC Berkeley, Spring 2006. Lecture notes by A. Geraschenko et al. available at \url{http://math.berkeley.edu/~anton/written/LieGroups/LieGroups.pdf}. \bibitem{Hom3} C.A. Weibel. {\em An Introduction to Homological Algebra}. Cambridge Studies in Advanced Mathematics; 38. Cambridge University Press, Cambridge, UK, 1997. \end{thebibliography} \end{document}