← Representations of Semi-simple Lie-Algebras → Topological Groups
What should you be acquainted with? 1. Linear Algebra, in particular inner product spaces both over the real and the complex numbers. This chapter is essentially taken from Brian Hall, Lie Groups, Lie Algebras, and Representations, Chapter 10

Properties of Representations of Semi-simple Lie-Algebras

Throughout this chapter $A$ will denote a complexified semi-simple Lie-algebra, $H$ a Cartan sub-algebra, $R\sbe H$ the root system and $B$ a base for $R$.

Structure of Weights

In this section we will prove generalizations of results obtained in subsection. We start with the generalization of lemma with identical proof:
Let $\mu$ be a weight of a representation $\psi:A\rar\Hom(E)$, such that for some root $r$: $n\colon=\la\mu,r^\vee\ra > 0$. Then $\mu,\mu-r,\ldots,\mu-nr$ are weights for $\psi$.
$\proof$ Let $F$ be the sub-space of $E$ generated by all weight vectors in $E$ whose weights lie in $\mu+\Z r$. These weights are shifted by $\psi(x_r)$ and $\psi(y_r)$ by $\pm r$ and therefore the restrictions $\psi|\lhull{r,x_r,y_r}$ gives a representation $\vp$ of $\sla(2,\C)$ in $F$ - with isomorphism $r^\vee\to H$, $x_r\to X$ and $y_r\to Y$. Since $R_r(\mu)=\mu-\la\mu,r^\vee\ra r=\mu-nr$ is also a weight for $\vp$ the operator $\vp(r^\vee)\in\Hom(F)$ has eigen-values $\la r^\vee,\mu\ra=n$ and $\la r^\vee,\mu-nr\ra=-n$. By proposition $\vp(r^\vee)\in\Hom(F)$ must have eigen-values $n,n-2,\ldots,-n$ (even in the infinite dimensional case), which coincides with the set $\{\la r^\vee,\nu\ra: \nu\in\{\mu,\mu-r,\ldots,\mu-nr\}\}$. Thus for any such $\nu=\mu-kr$ there must be an eigen-vector $x\in F\sm\{0\}$ of $\vp(r^\vee)$ such that: $$ \psi(r^\vee)x=\vp(r^\vee)x=\la r^\vee,\nu\ra x $$ and this eigen-vector $x$ can be obtained by starting with a normed weight vector $x_0$ for the weight $\mu$ and applying $\psi(y_r)$ $k$ times; as $x\neq0$, it`s a weight vector for $\psi$. $\eofproof$
Suppose $\psi:A\rar\Hom(E)$ is a highest weight $\l$ representation, $B$ a base for the roots of $A$ and $\mu\in H$ is another dominant element such that $\l-\mu\in\sum_{b\in B}\N_0b$. Then $\mu$ is a weight for $\psi$.
$\proof$ Assume $\l-\mu=\sum_bn(b)b$, $n(b)\in\N_0$. Let $P\sbe H$ be the set $$ P\colon=\mu+\sum_b\{0,1,\ldots,n(b)\}b~. $$ This is the set of `possible weights $p$ between $\mu$ and $\l$`: $\mu\preceq p\preceq\l$. Define for each $p=\mu+\sum k(b)b\in P$: $d(p)\colon=\sum k(b)$, which we think of as a sort of distance from $p$ to $\mu$. We claim that if $p\in P$ is a weight and $d(p) > 0$, then there is $b_0\in B$ such that $\la p,b_0\ra > 0$: As $p-\mu\neq0$ we have $$ 0 < \norm{p-\mu}^2 =\sum_{b\in B}k(b)\la p-\mu,b\ra $$ and thus there is some $b_0\in B$ such that $k(b_0) > 0$ and $\la p-\mu,b_0\ra > 0$. Since $\mu$ is dominant we conclude that $\la p,b_0\ra > \la\mu,b_0\ra > 0$.
From lemma we infer that $p-b_0$ must be a weight. Hence for every weight $p\in P$ satisfying $d(p) > 0$ there is a weight $q\in P$ (e.g. $q\colon=p-b_0$) satisfying $d(q) < d(p)$, which eventually shows that $\mu$ must be a weight. $\eofproof$
Finally we can state the generalization of theorem to semi-simple Lie-algebras:
Suppose $\psi:A\rar\Hom(E)$ is a finite dimensional highest weight $\l$ cyclic representation and $B$ a base for the roots of $A$. An element $\mu$ of the Cartan-algebra $H$ is a weight for $\psi$ if and only if.
  1. $\mu$ is integral, i.e. $\mu\in\sum_{b\in B}\Z b^{\vee*}$,
  2. $\mu\in\l-\sum_{b\in B}\N_0b$,
  3. $\mu\in C\colon=\convex{w(\l):w\in W}$.
$\proof$ As $\psi$ is equivalent to $\g_\l:A\rar\Hom(W_\l/V_\l)$, each weight must be integral and $\mu\in\l-\sum_{b\in B}\N_0b$. Finally for all $w$ in the Weyl group $W$: $w(\mu)\preceq\l$ and since $\l$ is dominant we conclude by theorem: $\mu\in\convex{W(\l)}$.
Conversely, assume all three conditions hold. By corollary we can find some $w\in W$ such that $w(\mu)$ is dominant. By proposition $w(\mu)$ is a weight for $\psi$ (and thus $\mu$ is a weight) if $\l-w(\mu)\in\sum_{\b\in B}\N_0b$. But as $\mu\in C$ this follows from theorem provided $\l-w(\mu)\in\sum_{b\in B}\Z b$.
Now for each $c\in B$: $R_c(\mu)-\mu=\la\mu,c^\vee\ra c\in\Z c$ is integral, which implies by proposition that for all $w\in W$: $w(\mu)-\mu\in\sum_{b\in B}\Z b$ and therefore $\l-w(\mu)=\l-\mu+\mu-w(\mu)$ and since both $\l-\mu$ and $\mu-w(\mu)$ lie in $\sum_{b\in B}\Z b$, we are done. $\eofproof$
By exam we get for $\sla(2,\C)$ and $\sla(3,\C)$ for a highest weight $\l$ cyclic representation:
1. For $\sla(2,\C)$ $\mu$ is a weight of the highest weight $\l$ cyclic representation iff $\mu\in\frac12\Z H$, $\l-\mu\in\N_0 H$ and $C=\convex{\pm\l}$.
2. For $\sla(3,\C)$ $\mu$ is a weight iff $\mu\in\Z Q+\Z Y$; $\l-\mu\in\N_0 H_1+\N_0H_2$ and $C=\convex{w(\l): w\in C_{3v}}$.
More generally we get for $\sla(n,\C)$:
  1. $\mu$ is integral iff $\mu\in\sum_{j=1}^{n-1}\Z\l_j$, for $$ \l_j=(1-\tfrac jn)\sum_{l=1}^jE^{ll}-\tfrac jn\sum_{l=j+1}^nE^{ll} =diag\{1-\tfrac jn,\ldots,1-\tfrac jn,\tfrac jn,\ldots,\tfrac jn\} $$ with $j$ entries $1-\frac jn$ and $n-j$ entries $\frac jn$.
  2. $\l-\mu\in\sum_{j=1}^{n-1}\Z H_j$, $H_j\colon=E^{jj}-E^{j+1,j+1}$.
  3. If $\l=diag\{a_1,\ldots,a_n\}$, then $C=\convex{diag\{a_{\pi(1)},\ldots,a_{\pi(n)}\}:\pi\in S(n)}$.

The Casimir Element

 We assume that $A$ is a real Lie-algebra with Euclidean product $\la.,.\ra$ such that all adjoint maps $\ad(x)\in\Hom(A)$ are skew symmetric and $A\cap iA=\{0\}$. Then $A^\C=A+iA$ is its complexification, $A=\{x\in A^\C: x=-x^*\}$ and if $H_0$ denotes the Cartan algebra of $A$ the Cartan algebra of $A^\C$ is given by: $H=H_0+iH_0$; moreover for the root system $R$ of $A^\C$ we have: $R\sbe iH_0$. We will further assume that $A$ is a subset if its universal enveloping algebra $({\cal U}(A),+.)$, in particular for all $x,y\in A^\C$: $[x,y]=xy-yx$.
For an orthonormal basis $e_1,\ldots,e_n$ of the real Lie-algebra $A$ we define the Casimir element $C\in{\cal U}(A)$ by $$ C\colon=-\sum_j e_j^2~. $$
1. The Casimir element doesn`t depend on the orthonormal basis: if $b_1,\ldots,b_n$ denotes another orthonormal basis for $A$, then there is an orthonormal matrix $(u_{jk})\in\OO(n)$ such that for all $k$: $b_k=\sum_l u_{lk}e_l$. Hence $$ \sum_k b_k^2 =\sum_{k,l,m}u_{lk}u_{mk}e_le_m =\sum_{l,m}\d_{lm}e_le_m =\sum_l e_l^2~. $$ We remark that it has been important that we are only dealing with real orthonormal basis; the verification above doesn`t work for unitary matrices $(u_{jk})\in\UU(n)$!
2. The Casimir element lies in the center of ${\cal U}(A)$: Since $A$ generates ${\cal U}(A)$ we simply need to check that for all basis vectors $e_1,\ldots,e_n$: $[C,e_j]\colon=Ce_j-e_jC=0$. \begin{eqnarray*} [C,e_j] &=&\sum_k[e_j,e_k^2] =\sum_k e_je_k^2-e_k^2e_j\\ &=&\sum_k e_je_k^2-e_ke_je_k+e_ke_je_k-e_k^2e_j =\sum_k[e_j,e_k]e_k+e_k[e_j,e_k]\\ &=&\sum_{k,l}c_{lk}^j(e_le_k+e_ke_l) =\sum_{k,l}(c_{lk}^j+c_{kl}^j)e_le_k \end{eqnarray*} As $\ad(e_j)$ is skew symmetric, the structure constants must satisfy the relations $c_{lk}^j+c_{kl}^j=0$.
Define for any ONB $e_1,\ldots,e_n$ of the complexified algebra $A^\C$: $S\colon=\sum e_je_j^*$ and compute for all $j$: $[S,e_j]$ and $[S,e_j^*]$.
3. Suppose we want to compute the Casimir element starting with an ONB $h_1,\ldots,h_m$ for the real Cartan algebra $H_0$ of $A$ and normalized vectors $z_r\in A_r\sbe A^\C$, $r\in R^+$, i.e. $z_r,z_r^*$ are normalized root vectors: $[h,z_r]=\la h,r\ra z_r$ and $[h,z_r^*]=-\la h,r\ra z_r^*$. Then the collection $$ h_1,\ldots,h_m,z_r,z_r^*,\quad r\in R^+ $$ is an orthonormal basis for $A^\C$. Now the unit vectors $$ x_r\colon=\frac{z_r+z_r^*}{\sqrt2\,i} \quad\mbox{and}\quad y_r\colon=\frac{z_r-z_r^*}{\sqrt2} $$ satisfy $x_r^*=-x_r$ and $y_r^*=-y_r$ and thus all of these vectors lie in $A$. Finally they are pairwise orthogonal and therefore the set $$ h_1,\ldots,h_m,x_r,y_r,\quad r\in R^+ $$ forms an orthonormal basis for $A$. Hence we get \begin{equation}\label{tceeq1}\tag{TCE1} C+\sum_{j=1}^m h_j^2 =-\sum_{r\in R^+}(x_r^2+y_r^2) =\sum_{r\in R^+}(z_rz_r^*+z_r^*z_r) =\sum_{r\in R^+}(2z_r^*z_r+[z_r,z_r^*]) \end{equation} Now suppose we are given an arbitrary representation $\psi:A\rar\Hom(E)$. By exam there is a unique extension to an associative algebra homomorphism $\Psi:{\cal U}(A)\rar\Hom(E)$ such that for all $x\in A$: $\Psi(x)=\psi(x)$. Thus we may define $\Psi(C)$.
If $\psi:A\rar\Hom(E)$ is a finite dimensional irreducible representation or a possibly infinite dimensional highest weight $\l$ cyclic representation, then there is a constant $c_\l\in\C$ such that $$ \Psi(C)\colon=-\sum\psi(e_j)^2=c_\l id_E, $$ and the constant is given by $c_\l=\norm{\l+r_0}^2-\norm{r_0}^2\in\R_0^+$, where $r_0\colon=\frac12\sum_{r\in R^+}r$ - cf. section. Moreover $c_\l=0$ if and only if $\l=0$ i.e. iff $\psi=0$. $\Psi(C)$ is said to be the Casimir element of the representation $\psi$.
$\proof$ 1. Let $v_0$ be the weight vector with highest weight $\l$. We will first compute $\Psi(C)v_0$. As $\psi(z_r)v_0=0$ we conclude from \eqref{tceeq1} and lemma: $$ \Psi(C)v_0 =\sum_{r\in R^+}(2\psi(z_r^*)\psi(z_r)v_0+\psi([z_r,z_r^*])v_0) -\sum_{j=1}^m\psi(h_j)^2v_0 =\sum_{r\in R^+}\psi(r)v_0-\sum_{j=1}^m\psi(h_j)^2v_0 $$ As $\l=i\l_0\in iH_0$ we have: $\psi(h_j)v_0=-i\la h_j,\l_0\ra v_0$ and therefore $$ \Psi(C)v_0 =2\la r_0,\l\ra v_0+\sum_{j=1}^m\la h,\l_0\ra^2v_0\\ =(\norm{\l_0}^2+2\la r_0,\l\ra)v_0 =(\norm{\l}^2+2\la r_0,\l\ra)v_0 =\colon c_\l v_0~. $$ Finally the fact that $\l$ is dominant implies that $\la r_0,\l\ra\geq0$, i.e. $c_\l=0$ if and only if $\l=0$.
2. Since $C$ is in the center of ${\cal U}(A)$ the operator $\Psi(C)$ commutes with all operators $\psi(x)$, $x\in A$. If $\psi$ is irreducible and finite dimensional, we conclude by Schur`s lemma that $\Psi(C)=c_\l.id_E$. If $\psi$ is a highest weight $\l$ cyclic representation we put $F\colon=\{v\in E:\Psi(C)v=c_\l v\}$. As $\Psi(C)$ commutes with all operators $\psi(x)$, $x\in A$, we see that $F$ is $\psi$-invariant. On the other hand $v_0\in F$ and $v_0$ is a cyclic vector, which shows that $F=E$. $\eofproof$
By theorem this applies in particular to the representations $\g_\l:A\rar\Hom(W_\l)$ in the Verma module $W_\l$.
In exam we computed the Casimir element of the standard representation of $\so(3)$ with respect to the inner product $\tr(XY^*)$ and it turned out to be the identity.
1. The spin $l/2$ representation of $\sla(2,\C)$ has $\l=\frac12l H$ and since $r_0=\frac12H$ we get: $$ c_\l =(\tfrac12l+\tfrac12)^2-(\tfrac12)^2)\norm H^2 =\tfrac12l(l+2)~. $$ 2. For $\sla(3,\C)$ we have $R^+=\{H_1,H_2,H_1+H_2\}$ and thus $r_0=H_1+H_2$. The standard representation has highest weight $\l\colon=H_1+H_2$. Hence we get for this representation $$ c_\l =\norm{\l+r_0}^2-\norm{r_0}^2 =3\norm\l^2 =6~. $$

Complete Reducibility

In this section we will show that each finite dimensional representation $\psi:A\rar\Hom(E)$ of a semi-simple Lie-algebra $A$ is the direct sum of irreducible representations. In other words : $\psi:A\rar\Hom(E)$ is completely reducible. If for example $E$ carries a Euclidean product and all maps $\psi(x)$ are skew symmetric, then we already know that $\psi$ is completely reducible. However in the general case there is no other assumption on the linear mappings $\psi(x)$, $x\in A$ and hence it`s not at all obvious how to achieve such a decomposition. The Casimir element will play a pivotal role in the proof.
By proposition we know that $A$ contains a copy of $\sla(2,\C)$, in particular $A$ is of dimension at least $3$. Moreover, by proposition $A$ is the Lie-algebra direct sum of simple (pairwise orthogonal) Lie-algebras $A_j$ and by the same argument all of these sub-algebras must contain a copy of $\sla(2,\C)$, i.e. $\dim A_j\geq3$.
Suppose $\psi:A\rar\Hom(E)$ is a one dimensional representation of a semi-simple Lie-algebra. Then $\psi=0$.
$\proof$ Since $\psi$ is one dimensional, we may assume that $E=\C$ and thus $\Hom(E)=(\C,+,.)$. We claim that for all $j$: $\psi|A_j=0$. Assume to the contrary that e.g. $\vp\colon=\psi|A_1$ is different from $0$, then $I\colon=\ker\vp$ is an two sided ideal in $A_1$. As $A_1$ is simple we infer that either $I=A$ - which implies $\vp=0$, or $I=0$ - which says that $\vp$ is injective. Hence $\vp:A_1\rar(\C,+..)$ is injective, which is impossible, because $\dim A_1\geq2$. $\eofproof$
Suppose $\psi:A\rar\Hom(E)$ is a non trivial finite dimensional representation of a semi-simple Lie-algebra and $F$ an invariant sub-space of co-dimension $1$. Then there is a complementary invariant sub-space $G$, i.e. $E=F\oplus G$.
$\proof$ 1. Assume $\psi:A\rar\Hom(F)$ is irreducible. Since $F$ is invariant and of co-dimension $1$, we get a one dimensional representation $\vp:A\rar\Hom(E/F)$, which is trivial by lemma, i.e. for all $x\in A$: $\psi(x)(E)\sbe F$. Let $C$ be the Casimir element in ${\cal U}(A)$, then $\Psi(C)(E)\sbe F$ and by proposition: $\Psi(C)|F=c.id_F$.
If $c=0$, then by proposition we have for all $x\in A$: $\psi(x)|F=0$. Since $\psi(x)(E)\sbe F$ we conclude that for all $x,y\in A$: $\psi(x)\psi(y)=0$, in particular all operators $\psi(x)$, $x\in A$, commute. Hence for any simple sub-algebra $A_j$ containing a copy of $\sla(2,\C)$ the restriction $\psi|A_j:A_j\rar\Hom(E)$ cannot be injective, i.e. $\ker\psi|A_j=A_j$, which shows by proposition that $\psi=0$ and we may take for $G$ any complementary sub-space.
If $c\neq0$, then we choose $G=\ker\Psi(C)$. Obviously $G\cap F=\{0\}$ and $\dim G=1$; as $\Psi(C)$ commutes with all $\psi(x)$, $G$ is invariant under $\psi$.
2. If $\psi:A\rar\Hom(F)$ is not irreducible, then there is a non trivial $\psi$-invariant sub-space $F_1$ of $F$. We will prove the assertion by induction on the dimension of $F$. The case $\dim F=1$ is included in 1. (the case $c=0$). $F/F_1$ is a sub-space of $E/F_1$ of co-dimension $1$ and by induction hypothesis we may assume that there is some $\psi$-invariant one dimensional subspace $E_1/F_1$ of $E/F_1$ such that $$ E/F_1=F/F_1\oplus E_1/F_1~. $$ Moreover $E_1$ is $\psi$-invariant. Thus $F_1$ is an invariant sub-space of $E_1$ of co-dimension $1$. Since $\dim F_1 < \dim F$ we can find by induction hypothesis a one dimensional $\psi$-invariant sub-space $G$ of $E_1$ such that $$ E_1=F_1\oplus G \quad\mbox{and thus}\quad E/F_1=F/F_1\oplus 0\oplus G/F_1~. $$ This proves that $E=F+G$ and since $\dim E=\dim F+\dim G$: $E=F\oplus G$. $\eofproof$
The operator $\Psi(C)\in\Hom(E)$ was injective on $F$ and its image was a sub-space of $F$. Hence its image is $F$ and its kernel is a complementary sub-space for $F$. The invariance of $\ker\Psi(C)$ is a consequence of the commutation property: $\psi(x)\Psi(C)=\Psi(C)\psi(x)$ for all $x\in A$.
Suppose $F$ is an invariant sub-space of $E$ and $u\in\Hom(E,F)$ an intertwining map for $\psi$, i.e. for all $x\in A$: $\psi(x)u=u\psi(x)$. If $u|F$ is one-one then the kernel of $u$ is a $\psi$-invariant sub-space such that $E=F\oplus\ker u$. Moreover all eigen-spaces of $u$ are $\psi$-invariant.
If $\psi:A\rar\Hom(E)$ is a representation of the Lie-algebra $A$ in $E$, then $\d:A\rar\Hom(\Hom(E))$, $\d(x)u\colon=[\psi(x),u]$ is a representation in $\Hom(E)$ and $\bigcap\ker\d(x)$ is the space of intertwining operators $u:E\rar E$, i.e. for all $x\in A$: $\psi(x)u=u\psi(x)$.
This follows from Jacobi`s identity in $\Hom(E)$ and the fact that $\psi$ is a Lie-algebra homomorphism: \begin{eqnarray*} \d(x)\d(y)u-\d(y)\d(x)u &=&\d(x)([\psi(y),u])-\d(y)([\psi(x),u])\\ &=&[\psi(x),[\psi(y),u]]+[\psi(y),[u,\psi(x)]]\\ &=&-[u,[\psi(x),\psi(y)]] =[\psi([x,y]),u] =\d([x,y])u, \end{eqnarray*} i.e. $\d([x,y])=[\d(x),\d(y)]$. In generalizing lemma to sub-spaces of any co-dimension, we consider a representation $\d$ of the Lie-algebra $A$ in the space $\Hom(E,F)$, which relates to the given representation $\psi$ as follows: $$ \forall x\in A\ \forall u\in\Hom(E,F):\quad \d(x)u=\psi(x)u-u\psi(x)~. $$ It is well defined because $F$ is $\psi$-invariant. The verification that this is indeed a representation is similar to the proof of exam.
Every finite dimensional representation $\psi:A\rar\Hom(E)$ of a semi-simple Lie-algebra is completely reducible.
$\proof$ The essential part will be to show by means of exam that every invariant sub-space $F$ admits an invariant complementary sub-space. As we are looking for an operator $u$, which is injective on $F$, we define the sub-space $V$ of $\Hom(E,F)$ to be $$ \{u\in\Hom(E,F):u|F=c.id_F,c\in\C\}~. $$ The linear functional $c^*:V\rar\C$, $c^*(u)=c$, has kernel $W$ of co-dimension $1$ in $V$.
We claim that $V$ is $\d$-invariant, i.e. for all $x\in A$, all $u\in V$ and all $f\in F$: $\d(x)u(f)=kf$ for some $k\in \C$: as $F$ is $\psi$-invariant and $u|F=c^*(u).id_F$, we get $$ \d(x)u(f) =\psi(x)u(f)-u\psi(x)(f) =\psi(x)(c^*(u)f)-c^*(u)\psi(x)(f) =0, $$ i.e. $\d(x)V\sbe W\sbe V$, which shows that both $W$ and $V$ are $\d$-invariant.
Since $W$ is a sub-space of co-dimension $1$ in $V$ we conclude from lemma that there is a $u\in V$ such that $\lhull{u}$ is $\d$-invariant, $V=W\oplus\lhull{u}$ and w.l.o.g $u|F=id_F$. Finally by lemma: $\d(x)u=0$, i.e. $u$ commutes with all $\psi(x)$, $x\in A$. Thus we`ve found an intertwining operator $u\in\Hom(E,F)$. By exam $\ker u$ is a $\psi$-invariant sub-space of $E$ such that $E=F\oplus\ker u$.
The assertion of the theorem now follows by induction on the dimension of $E$. $\eofproof$

Weyl`s Character Formula

The character of a group representation $\Psi:G\rar\Gl(E)$ was defined as $\chi:G\rar\C$: $\chi(g)=\tr\Psi(g)$. If $G$ is a matrix Lie group with Lie-algebra ${\cal G}$, then by section: $\Psi(\exp(tX))=\exp(t\psi(X)$. Hence we define the character of a Lie-algebra representation $\psi:A\rar\Hom(E)$ in a finite dimensional space $E$ by $$ \chi_\psi:A\rar\C, \chi_\psi(x)\colon=\tr(e^{\psi(x)})~. $$ As $\psi:A\rar\Hom(E)$ is completely reducible by theorem and each irreducible representation decomposes as direct sum of its weight spaces - cf. proposition, the space $E$ is the direct sum of weight spaces $E_\mu$. For $h\in H$ the linear operator $\psi(h)$ acts on $E_\mu$ by multiplication with a factor of $\la h,\mu\ra$. Denoting by $m(\mu)\colon=\dim E_\mu$ the multiplicity of the weight $\mu$ we get \begin{equation}\label{wcfeq1}\tag{WCF1} \forall h\in H\quad \chi_\psi(h)\colon=\sum_{\mu}m(\mu)e^{\la h,\mu\ra} \end{equation} If in addition $\psi$ is irreducible, then by theorem it`s a highest weight $\l$ cyclic representation and any weight $\mu$ is integral, $\mu\in\l-\sum_{r\in R^+}\N_0r$ and $\mu\in C\colon=\convex{w(\l):w\in W}$. Hence $$ \chi_\psi(h) =\sum_{\mu\in C\cap(\l-\sum_{r\in R^+}\N_0r)}m(\mu)e^{\la h,\mu\ra}~. $$ Moreover for any $w$ in the Weyl group $W$ we have \begin{equation}\label{wcfeq2}\tag{WCF2} \forall h\in H\quad \chi_\psi(w(h))=\chi_\psi(h) \end{equation} because by proposition for each $h\in H$ the mapping $\psi(w(h))$ is similar to $\psi(h)$. Hence $e^{\psi(w(h))}$ is similar to $e^{\psi(h)}$ 1. In theorem we got for the character $\chi_l\colon=\chi_{\psi_l}$ of the spin $l/2$ representation $\psi_l:\sla(2,\C)\rar\Ma(l+1,\C)$: $$ \chi_l(e^{itH})=\frac{\sin((l+1)t)}{\sin t}~. $$ Which follows immediately from $\l\in\{\frac12lH,(\frac12l-1)H,\ldots,-\frac12lH\}$ and $m(\l)=1$.
2. The standard representation of $\sla(3,\C)$: the weights are (cf. exam): $Q,-Q+Y,-Y$ or in old notation: $(1,0),(-1,1),(0,-1)$: $$ \chi_\psi(H_1)=e+e^{-1}+1,\quad\chi_\psi(H_2)=1+e^{-1}+e $$ 3. Let`s compute the character of the $\mathbf{6}$-representation - cf. subsection. The weights are (cf. exam): $2Q,Y,-2Q+2Y,-Q,-2Y,Q-Y$ or in old notation $(2,0),(0,1),(-2,2),(-1,0),(0,-2),(1,-1)$. \begin{eqnarray*} \chi_\psi(zH_1)&=&e^{2z}+1+e^{-2z}+e^{-z}+1+e^{z}=2(1+\cosh(z)+\cosh(2z))\\ \chi_\psi(zH_2)&=&1+e^{z}+e^{2z}+1+e^{-2z}+e^{-z}=2(1+\cosh(z)+\cosh(2z)) \end{eqnarray*} 4. Finally the highest weight $Q+2Y$ representation: weights with multiplicity $1$: $Q+2Y,-Q+3Y,-2Q+2Y,-3Q+Y,-2Q-Y,-2Y,2Q-3Y,3Q-2Y,2Q$ and three weight with multiplicity $2$: $Y,-Q,Q-Y$: \begin{eqnarray*} \chi_\psi(zH_1) &=&(e^{z}+e^{-z}+e^{-2z}+e^{-3z}+e^{-2z}+1+e^{2z}+e^{3z}+e^{2z})+2(1+e^{-z}+e^{z})\\ &=&3+6\cosh(z)+4\cosh(2z)+2\cosh(3z) \end{eqnarray*} The Weyl Character Formula reduces the summation considerably: for irreducible highest weight $\l$ representations we don`t have to sum over all weights but only over the weights $w(\l)$, for $w$ in the Weyl group $W$ of the semi-simple Lie-algebra $A$
If $\psi:A\rar\Hom(E)$ is a finite dimensional irreducible representation with highest weight $\l$, then $$ \forall h\in H:\quad \chi_\psi(h) =\frac{\sum_{w\in W}\det(w)e^{\la h,w(\l+r_0)\ra}}{\sum_{w\in W}\det(w)e^{\la h,w(r_0)\ra}}~. $$ where $W$ denotes the Weyl group of $A$.
First we will establish the Weyl dimension formula, the Kostant multiplicity formula and the character formula for Verma modules as consequences of the Weyl character formula.

Weyl`s Dimension Formula

 The denominator in Weyl`s character formula \begin{equation}\label{wdfeq1}\tag{WDF1} q(h)\colon=\sum_{w\in W}\det(w)e^{\la h,w(r_0)\ra} \end{equation} is called Weyl denominator. For each $u$ in the Weyl group $W$ we get $$ q(u(h)) =\sum_{w\in W}\det(w)e^{\la u(h),w(r_0)\ra} =\sum_{w\in W}\det(u^{-1}w)\det(u)e^{\la h,u^{-1}w(r_0)\ra} =\det(u)q(h)~. $$
A function $f:H\rar\C$ is called a Weyl alternating function if $$ \forall w\in W\,\forall h\in H:\quad f(w(h))=\det(w)f(h)~. $$
Also the nominator in the Weyl formula is Weyl alternating. Hence for all $w\in W$: $\chi_\psi(w(h))=\chi_\psi(h)$. Moreover for each function $e:\C\rar\C$ the function $$ h\mapsto\sum_{w\in W}\det(w)e(\la h,w(\l)\ra) $$ is Weyl alternating. $\eofproof$
The polynomial function $$ P(h)\colon=\prod_{r\in R^+}\la h,r\ra $$ is Weyl alternating.
$\proof$ The set $$ u^{-1}(R^+) =(R^+\cap u^{-1}(R^+))\cup(-R^+\cap u^{-1}(R^+)) =\colon R_1\cup R_2 $$ is the disjoint union of the set $R_1$ of positive roots and the set $R_2$ of negative roots in $u^{-1}(R^+)$. Moreover $R_1\cap-R_2=\emptyset$, because if $r\in R_1\cap-R_2$, then $r\in u^{-1}(R^+)$ and $-r\in u^{-1}(R^+)$, i.e. $u(r)\in R^+\cap-R^+$, which is impossible by the definition of $R^+$. As $R_1\cup-R_2\sbe R^+$ and $|R_1|+|-R_2|=|u^{-1}(R^+)|=|R^+|$ we conclude that $R_1\cup-R_2=R^+$. Hence \begin{eqnarray*} P(u(h)) &=&\prod_{r\in R^+}\la u(h),r\ra =\prod_{r\in u^{-1}(R^+)}\la h,r\ra\\ &=&\prod_{r\in R_1}\la h,r\ra\prod_{r\in R_2}\la h,r\ra =\prod_{r\in R_1}\la h,r\ra(-1)^{|R_2|}\prod_{r\in-R_2}\la h,r\ra\\ &=&(-1)^{|R_2|}\prod_{r\in R_1\cup-R_2}\la h,r\ra =(-1)^{|R_2|}P(h) \end{eqnarray*} Let $B$ be the base contained in $R^+$; by proposition the Weyl group is generated by the set of reflections $R_b$, $b\in B$, and by exam $R_b$ permutes $R^+\sm\{b\}$ and sends $b$ to $-b$. Now if $B_1\sbe B$ and $b\in B$ then $$ R_b((R^+\sm B_1)\cup-B_1) =\left\{\begin{array}{cl} (R^+\sm(B_1\cup\{b\}))\cup-(B_1\cup\{b\})&\mbox{if $b\notin B_1$}\\ (R^+\sm(B_1\sm\{b\}))\cup-(B_1\sm\{b\})&\mbox{if $b\in B_1$} \end{array}\right. $$ Hence the number $k$ of points sent by $u^{-1}\colon=R_{b_1}\cdots R_{b_m}$ from $R^+$ to $-R^+$ differs from $m$ by a multiple of $2$ and we conclude that $\det(u)=\det(u^{-1})=(-1)^m=(-1)^k$. $\eofproof$
Suppose $f$ is a polynomial in $H$ (or an entire function), which vanishes at $b^\perp$, then there is a polynomial $g$ (or an entire function $g$) such that $f(h)=\la h,b\ra g(h)$.
For any basis $b,b_2,\ldots,b_n$ we put $x\colon=\la h,b\ra$ and $x_j\colon=\la h,b_j\ra$, then for fixed $x_2,\ldots,x_n$: $p(x)\colon=f(x,x_2,\ldots,x_n)$ is a polynomial in $x$ (or an entire function) and $p(0)=0$, i.e. for some $k$-homogeneous polynomials $c_k$: $$ p(x)=\sum_{k\geq0}c_k(x,x_2,\ldots,x_n)x \quad\mbox{and}\quad g(x)=\sum_{k\geq0}c_k(x,x_2,\ldots,x_n)~. $$
If $f:H\rar\C$ is a Weyl alternating polynomial (or entire function), then there is a polynomial (or entire function) $g:H\rar\C$ such that $f(h)=P(h)g(h)$. In other words: any Weyl alternating polynomial (entire function) is divisible by $P$.
$\proof$ Suppose $b_1\in B$ and $h\in b_1^\perp$, then $f(h)=f(R_{b_1}(h))=-f(h)$, i.e. $f$ vanishes at $b^\perp$ and by exam this means that $f$ is divisible by the polynomial $h\mapsto\la h,b_1\ra$. Hence for some polynomial $f_1$: $f(h)=\la h,b_1\ra f_1(h)$. If $b_2\in B$ is different from $b_1$, then $f_1$ must vanish on $b_2^\perp\cap[\la.,b_1\ra\neq0]$ and as the latter is a dense subset of $b_2^\perp$, $f_1$ must vanish on $b_2^\perp$; implying that $f_1(h)=\la h,b_2\ra f_2(h)$ for some polynomial $f_2$. Continuing this way we finally get a polynomial (or entire function) $g$ such that $f=P.g$. $\eofproof$
Suppose $f$ is an entire function on $\R^n$ and $P_k$ are $k$-homogeneous polynomials such that $$ f(x)=\sum_{k=0}^\infty P_k(x)~. $$ Prove that $f$ is Weyl alternating iff all polynomials $P_k$ are Weyl alternating.
Since $f(w(x))=\det(w)f(x)$, we conclude that $Df(w(x_0))(w(x))=\det(w)Df(x_0)x$ and by induction: $$ D^kf(w(x_0))(w(x),\ldots,w(x))=\det(w)D^kf(x_0)(x,\ldots,x) $$ Hence if $f$ is Weyl alternating, then all polynomials $P_k(x)=D^kf(0)(x,\ldots,x)/k!$ are Weyl alternating.
For any $\l\in H$ the function $$ f_\l(h)\colon=\sum_w\det(w)e^{\la h,w(\l)\ra} $$ is an entire Weyl alternating function on $H$ and $f_\l(h)=c(\l)P(h)+$ homogeneous polynomials of degree at least $\deg(P)+1$.
$\proof$ Put $m\colon=\deg(P)$. $$ f_\l(h)=\sum_{k=0}^\infty\frac1{k!}\sum_w\det(w)\la h,w(\l)\ra^k $$ As $f_\l$ is Weyl alternating all the polynomials $$ P_k(h)\colon=\frac1{k!}\sum_w\det(w)\la h,w(\l)\ra^k $$ are Weyl alternating by exam. By lemma this implies that $P_0=\cdots=P_{m-1}=0$ and $P_m=c(\l)P$. $\eofproof$
As both the nominator and the denominator in Weyl`s character formula in theorem are entire Weyl alternating functions we cannot simply evaluate both functions at $h=0$ to get the dimension $n\colon=\chi_\psi(0)$ of the representation $\psi$, because both evalute to $0$. However, we have by the previous lemma: $$ \chi_\psi(0) =\frac{c(\l+r_0)}{c(r_0)}~. $$ In order to derive a formula for the dimension we need to caculate the constant $c(\l)$ in lemma: Since $P$ is a homogeneous polynomial of degree $m\colon=|R^+|$, we take some $m$-th derivative: for any smooth function $f:H\rar\C$ we put $$ D^Rf(0) \colon=\frac{\pa^m}{\pa_{t_1}\cdots\pa_{t_m}}\Big|_{t_1=\cdots=t_m=0}f(t_1r_1+\cdots+t_mr_m) =D^mf(0)(r_1,\ldots,r_m), $$ where $R^+=\{r_1,\ldots,r_m\}$. Applying this to the function $f_\l$ gives $$ D^Rf_\l(0) =\sum_{w\in W}\det(w)\prod_{r\in R^+}\la r,w(\l)\ra =\sum_{w\in W}\det(w)\cl{P(w(\l))} =|W|\bar P(\l), $$ where we use the fact that $P$ is Weyl alternating, cf. lemma. Up to this point we could have taken any entire function $e:\C\rar\C$ in place of the exponential! On the other hand $D^Rf(0)=0$ for each homogeneous polynomial $f$ with $\deg(f)\neq m$. Hence $$ |W|\bar P(\l) =D^Rf_\l(0) =c(\l)D^RP(0) \quad\mbox{and thus}\quad \frac{c(\l+r_0)}{c(r_0)}=\frac{\bar P(\l+r_0)}{\bar P(r_0)}~. $$ Summarizing we derived the following
If $\psi:A\rar\Hom(E)$ is a finite dimensional irreducible representation with highest weight $\l$, then $$ \dim E =\frac{\prod_{r\in R^+}\la\l+r_0,r\ra}{\prod_{r\in R^+}\la r_0,r\ra} =\frac{P(\l+r_0)}{P(r_0)} =\chi_\psi(0)~. $$
For $A=\sla(3,\C)$ we have $R^+=\{H_1,H_2,H_1+H_2\}$ and $r_0=H_1+H_2$. Hence for the highest weight $\l=mQ+nY$ cyclic representation $\psi:\sla(3,\C)\rar\Hom(E)$: $$ \dim E=\tfrac12(m+1)(n+1)(m+n+2)~. $$
This follows from $\la\l+r_0,H_1\ra=m+1$, $\la\l+r_0,H_2\ra=n+1$ and $\la\l+r_0,H_1+H_2\ra=m+n+2$.

B. Kostant`s Multiplicity Formula

 This is a formula for the dimension of any weight space $\mu$ of a finite dimensional highest weight $\l$ cyclic representation $\psi:A\rar\Hom(E)$ of a semi-simple Lie-algebra $A$.

Weyl`s denominator formula

We will only need the first assertion of the following
Let $I$ be an open interval and $f_1,\ldots,f_n\in C^\infty(I)$. 1. If these functions are linearly dependent, then the Wronskian $W(f_1,\ldots,f_n)$ of $f_1,\ldots,f_n$ vanishes everywhere. 2. Conversely, if the Wronskian of $f_1,\ldots,f_n$ vanishes everywhere, then these functions need not be linearly dependent. 3. Let $A(x)$ be the matrix $(f_j^{(k-1)}(x))_{j,k=1}^n$, i.e. $W(f_1,\ldots,f_n)=\det A$. Assume that there is some $k$ and some $x_0\in I$ such that $\adj(A)_{nk}(x_0)\neq0$, then the functions $f_1,\ldots,f_n$ are linearly dependent in a neighborhood of $x_0$.
1. This is obvious!
2. Let $f_1,\ldots,f_n$ be smooth functions such that $\bigcap_j\supp(f_j)=\emptyset$, then the Wronskian of these two functions vanishes identically.
3. As $A\cdot\adj(A)=\det A.1_n$ we have for all $j,k$: $$ \sum_l f_j^{(l-1)}\adj(A)_{lk}=0~. $$ Hence for some neighborhood $U$ of $x_0$ in $I$: $$ \l_1f_j+\cdots+\l_nf_j^{(n-1)}=0 \quad\mbox{where}\quad \l_l\colon=\adj(A)_{lk}/\adj(A)_{nk} $$ are smooth functions on $U$ and $f_1,\ldots,f_n$ solve the linear differential equation $$ \l_1u+\cdots+\l_{n-1}u^{(n-2)}+u^{(n-1)}=0 $$ on $U$. However this linear differential equation has exactly $n-1$ linearly independent solutions and thus $f_1,\ldots,f_n$ are linearly dependent on $U$.
Denote by $B^{\vee*}=\{\l_1,\ldots,\l_n\}$ a fundamental weight system for the roots $R$ of $A$ - i.e. the set $\sum\Z\l_j$ is the set of integral elements. The sub-space of all finite linear combinations of functions $$ h\mapsto\exp(\la h,\mu\ra),\quad \mu\in\sum\Z\l_j $$ is a sub-algebra of the space $C(H)$ of continuous functions on $H$ containing the constant function $1$. By the Stone-Weierstraß Theorem we infer that for all compact subsets $K$ of $H$ it is dense in the Banach space $C(K)$. Moreover if $\l_1,\ldots,\l_n\in H$ are pairwise distinct, then the functions $h\mapsto\exp(\la h,\l_j\ra)$ are linearly independent: Choose $h_0\in H$ such that the numbers $a_j\colon=\la h_0,\l_j\ra$ are pairwise distinct, then the Wronskian of the functions $f_j(t)\colon=\exp(a_jt)$ is given by $$ \det\left( \begin{array}{cccc} e^{a_1t}&e^{a_2t}&\cdots&e^{a_nt}\\ a_1e^{a_1t}&a_2e^{a_2t}&\cdots&a_ne^{a_nt}\\ \vdots&\vdots&\ddots&\vdots\\ a_1^{n-1}e^{a_1t}&a_2^{n-1}e^{a_2t}&\cdots&a_n^{n-1}e^{a_nt} \end{array}\right) =e^{t\sum a_j}\prod_{k > j}(a_k-a_j) \neq0~. $$ Hence the functions $f_j$ are linearly independent.
Suppose $\L$ is a finite subset of $\sum\Z\l_j$ and $$ f(h)\colon=\sum_{\l\in\L}c_\l e^{\la h,\l\ra} \quad c_\l\neq0~. $$ If $f$ is Weyl alternating, then for all $w$ in the Weyl-group $W$: $w(\L)=\L$ and $c_{w(\l)}=\det(w)c_\l$.
$\proof$ Since all transformations $w\in W$ are linear isometries on $H$ we have $$ f(w(h)) =\sum_{\l\in\L}c_\l e^{\la w(h),\l\ra} =\sum_{\l\in\L}c_\l e^{\la h,w^{-1}(\l)\ra} $$ Now if $f$ Weyl is alternating, then $$ 0 =\sum_{\l\in\L}c_\l e^{\la h,w^{-1}(\l)\ra}-\det(w)c_\l e^{\la h,\l\ra} =\sum_{\l\in\L}(c_{w(\l)}-\det(w)c_\l)e^{\la h,\l\ra}~. $$ As the functions $h\mapsto\exp(\la h,\l\ra)$, $\l\in\L$, are linearly independent this necessarily means that $w(\L)=\L$ and $c_{w(\l)}=\det(w)c_\l$. $\eofproof$
Compute the Weyl denominator for $A=\sla(n,\C)$, $H=\sum x_je_j$ and for any $\s\in S_n$: $w(H)=(x_{\s^{-1}(1)},\ldots,x_{\s^{-1}(n)})$.
$r_0$ is given by $\frac12\sum_{j < k}(e_j-e_k)$, i.e. $$ r_0=\tfrac12((n-1)e_1+(n-3)e_2+\cdots+(1-n)e_n) $$ and thus $$ \la r_0,w^{-1}(H)\ra=\tfrac12((n-1)x_{\s(1)}+(n-3)x_{\s(2)}+\cdots+(1-n)x_{\s(n)})~. $$ Put $y_j\colon=e^{x_j}$; multiplying $q(H)$ by $(y_1\cdots y_n)^{(n-1)/2}$ we get by \eqref{wdfeq1}: $$ q(H)(y_1\cdots y_n)^{(n-1)/2} =\sum_{\s\in S_n}\sign(\s)y_{\s(1)}^{n-1}y_{\s(2)}^{n-2}\cdots y_{\s(n)}^{0}, $$ which is the Vandermonde determinant $$ \det\left(\begin{array}{cccc} y_1^{n-1}&y_2^{n-1}&\cdots&y_n^{n-1}\\ y_1^{n-2}&y_2^{n-2}&\cdots&y_n^{n-2}\\ \vdots&\vdots&\ddots&\vdots\\ y_1^{0}&y_2^{0}&\cdots&y_n^{0} \end{array}\right) =\prod_{j < k}(y_j-y_k)~. $$ Eventually we get: \begin{eqnarray*} q(H) &=&\frac{\prod_{j < k}(e^{x_j}-e^{x_k})}{\prod_k e^{x_k(n-1)/2}}\\ &=&\frac{\prod_{j < k} e^{(x_j+x_k)/2}(e^{(x_j-x_k)/2}-e^{-(x_j-x_k)/2})} {\prod_k e^{x_k(n-1)/2}}\\ &=&\prod_{j < k}(e^{(x_j-x_k)/2}-e^{-(x_j-x_k)/2})~. \end{eqnarray*} We are going to prove that this is a particular case of the Weyl product formula for the Weyl denominator:
The Weyl denominator can be written as a product over $R^+$: $$ \forall h\in H:\quad q(h)=\prod_{r\in R^+}(e^{\la h,r\ra/2}-e^{-\la h,r\ra/2})~. $$
$\proof$ 1. Expanding the right hand side we get a sum $Q(h)$ of exponential functions $e^{\la h,\l\ra}$ with coefficients $\pm1$ and $\l$ must be in the set $r_0-\sum_{r\in R^+}\N_0r$. There is no $1/2$ because each time we take $e^{-\la h,r\ra/2}$ instead of $e^{\la h,r\ra/2}$ we lower the exponent by $\la h,r\ra$, i.e. for some finite subset $\L$ of $r_0-\sum_{r\in R^+}\N_0r$: $$ Q(h)=\sum_{\l\in\L}c(\l)e^{\la h,\l\ra} $$ 2. The function $Q$ is Weyl alternating: for any $w$ in the Weyl group $W$ we have $$ Q(w(h))=\prod_{r\in R^+}(e^{\la h,w^{-1}(r)\ra/2}-e^{-\la h,w^{-1}(r)\ra/2}) $$ We proceed as in the proof of lemma and split $w^{-1}(R^+)$ into the set $R_1$ of positive roots and the set $R_2$ of negative roots. Since $R^+$ is the disjoint union of $R_1$ and $-R_2$ and $\det w$ equals $(-1)^{|R_2|}$: \begin{eqnarray*} Q(w(h)) &=&\prod_{r\in R_1}(e^{\la h,w^{-1}(r)\ra/2}-e^{-\la h,w^{-1}(r)\ra/2}) \prod_{r\in-R_2}(e^{-\la h,w^{-1}(r)\ra/2}-e^{\la h,w^{-1}(r)\ra/2})\\ &=&(-1)^{|R_2|}\prod_{r\in R_1\cup-R_2}(e^{\la h,w^{-1}(r)\ra/2}-e^{-\la h,w^{-1}(r)\ra/2})\\ &=&\det(w)\prod_{r\in R_1^+}(e^{\la h,r\ra/2}-e^{-\la h,r\ra/2}) =\det(w)Q(h) \end{eqnarray*} 3. As $r_0\in\L$, $c(r_0)=1$ and by lemma $w(\L)=\L$, we have $$ Q(h)=\sum_{w\in W}\det(w)e^{\la h,w(r_0)\ra}+\sum_{\l\in\L\sm W(r_0)}c(\l)e^{\la h,\l\ra}~. $$ We simply need to verify that $\L\sm W(r_0)=\emptyset$. So assume $\l\in\L\sm W(r_0)$. Since the $W$-orbit of $\l$ contains a dominant element (cf. corollary), we may assume that $\l$ is dominant. On the other hand $\l$ is smaller than $r_0$ and $r_0$ is the smallest strictly dominant integral element. Hence $\l$ must be orthogonal to one $b\in R^+$ at least. So we must have $R_b(\l)=\l$ and therefore by lemma: $$ c(\l)=c(R_b(\l))=\det(R_b)c(\l)=-c(\l)~. $$ $\eofproof$

Multiplicity formula

With the product formula for the Weyl denominator at hand we can easily expand its reciprocal function by means of the geometric series (provided $\la h,r\ra > 0$ for all $r\in R^+$): $$ \frac1{q(h)} =\prod_{r\in R^+}\frac1{e^{\la h,r\ra/2}-e^{-\la h,r\ra/2}} =\prod_{r\in R^+}\frac{e^{-\la h,r\ra/2}}{1-e^{-\la h,r\ra}} =e^{-\la h,r_0\ra}\prod_{r\in R^+}\sum_{k=0}^\infty e^{-k\la h,r\ra} $$ The product of the sums is the sum of exponential functions of the form $\exp(-\la h,\mu\ra)$, where $\mu\in\sum_{r\in R^+}\N_0r$.
For any integral element $\mu$, i.e. $\mu\in\sum\Z\l_j$, let $p(\mu)$ be the number of ways $\mu$ can be expressed as a non-negative integer linear combination of positive roots $\{r_1,\ldots,r_N\}=R^+$, i.e. $$ p(\mu)=\Big|\Big\{(n_1,\ldots,n_N)\in\N_0^N:\sum n_jr_j=\mu\Big\}\Big|~. $$ $p$ is called the Kostant partition function. Cf. B. Kostant. In particular we have for $\mu\notin\sum_{r\in R^+}\R_0^+r$: $p(\mu)=0$. Hence if $p(\mu)\neq0$, then $0\preceq\eta$.
For the reciprocal Weyl denominator we get: \begin{equation}\label{kmfeq1}\tag{KMF1} \frac1{q(h)} =e^{-\la h,r_0\ra}\sum_{0\preceq\mu}p(\mu)e^{-\la h,\mu\ra} \end{equation} which definitely holds for $h\in B^{p\circ}$.
Suppose $\psi:A\rar\Hom(E)$ is a finite dimensional highest weight $\l$ cyclic representation and $\mu$ is a weight of $\psi$. Then the multiplicity of the weight $\mu$, i.e. the dimension of the weight space $E_\mu$ is given by $$ m(\mu)\colon=\dim(E_\mu)=\sum_{w\in W}\det(w)p(w(\l+r_0)-(\mu+r_0))~. $$
$\proof$ By Weyl`s character formula theorem and \eqref{kmfeq1} we have for all $h\in B^{p\circ}$: $$ \chi_\psi(h) =e^{-\la h,r_0\ra}\sum_{0\preceq\eta}p(\eta)e^{-\la h,\eta\ra} \sum_{w\in W}\det(w)e^{\la h,w(\l+r_0)\ra} $$ By \eqref{wcfeq1} this equals $\sum_{\mu}m(\mu)e^{\la h,\mu\ra}$, where the sum ranges over all weights $\mu$ of $\psi$. The coefficient $m(\mu)$ of $e^{\la h,\mu\ra}$ in the product of the sums above is thus $$ \sum_{\eta\in\L}p(\eta)\det(w) $$ where the set $\L$ is given by the set of all weights $\eta$ satisfying $$ -r_0-\eta+w(\l+r_0)=\mu \quad\mbox{i.e.}\quad \eta=w(\l+r_0)-(\mu+r_0)~. $$ $\eofproof$

Characters of Verma Modules

By theorem and exam for any $\l\in H$ the representation $\g_\l:A\rar\Hom(W_\l)$ in the Verma module $W_\l$ is a highest weight $\l$ cyclic representation with cyclic vector $v_0\colon=q_\l(1)$ and for each weight $\mu\in\l-\sum_{r\in R^+}\N_0r$ the multiplicity of $\mu$ is $p(\l-\mu)$. Thus we may define the character of this representation by a formal exponential series: \begin{equation}\label{kmfeq2}\tag{KMF2} \chi_{\g_\l}(h) \colon=\sum_{\mu\preceq\l}p(\l-\mu)e^{-\la h,\mu\ra} =\sum_{0\preceq\eta}p(\eta)e^{-\la h,\l-\eta\ra}, \end{equation} where the sum is to be extended over all integral elements. The space of formal series of exponentials is also an associative algebra: the product of $\sum c_1(\eta)\exp(-\la h,\eta\ra)$ and $\sum c_2(\eta)\exp(-\la h,\eta\ra)$ is $\sum c(\mu)\exp(-\la h,\mu\ra)$, where $$ c(\mu)\colon=\sum_{\eta}c_1(\eta)c_2(\mu-\eta)~. $$ Comparing \eqref{kmfeq2} with the formal exponential series \eqref{kmfeq1} for the reciprocal Weyl denominator we get the character formula for Verma modules: \begin{equation}\label{kmfeq3}\tag{KMF3} \forall h\in H:\quad \chi_{\g_\l}(h) =e^{\la h,\l\ra}\sum_{0\preceq\mu}p(\mu)e^{-\la h,\mu\ra} =e^{\la h,\l+r_0\ra}\frac1{q(h)}~. \end{equation}

Proof of Weyl`s Character Formula

If $\psi:A\rar\Hom(E)$ is a finite dimensional irreducible representation, then by theorem it`s a highest weight $\l$ cyclic representation for some integral weight $\l$. Let us denote its character by $\chi_\l$. The function $q(h)$ is Weyl alternating by \eqref{wdfeq1} and $\chi_\l(h)$ is Weyl invariant by \eqref{wcfeq2}, which implies that $q(h)\chi_\l(h)$ is a Weyl alternating finite linear combination of exponentials $e^{\la h,\eta\ra}$. The subsequent lemma restricts the set of $\eta$`s showing up in this linear combination. A second restriction will be given in \eqref{pcfeq3}!
Let $R$ be the root system of $A$ with Weyl group $W$. If an exponential $e^{\la h,\eta\ra}$ occurs with non zero coefficient in the product $q(h)\chi_\l(h)$, then $\eta\in\convex{\{w(\l+r_0):w\in W\}}$ and $\l+r_0-\eta\in\sum_{r\in R^+}\N_0 r$.
$\proof$ For each exponential $e^{\la h,\eta\ra}$ in $q(h)$ we have by \eqref{wdfeq1}: $\eta\in W(r_0)$ and for each exponential $e^{\la h,\eta\ra}$ in $\chi_\l$ we have by theorem: $\eta\in\convex{\{w(\l):w\in W\}}$. Thus for all exponentials $e^{\la h,\eta\ra}$ occurring in $q(h)\chi_\l(h)$ we must have: $\eta=\mu+w(r_0)$ for some $w\in W$ and some weight $\mu\in\convex{\{w(\l):w\in W\}}$. Now all these weights $\eta$ are dominated by $\l+r_0$ and thus $W(\eta)\preceq\l+r_0$, which by theorem holds if and only if $\eta\in\convex{W(\l+r_0)}$.
Finally assume $\eta\colon=\mu+w(r_0)$ for some $w\in W$ and some weight $\mu\in\convex{\{w(\l):w\in W\}}$. By subsection $r_0$ is integral and dominant and all $w\in W$ $w(r_0)$ is integral and $r_0-w(r_0)\in\sum_{r\in R^+}\N_0 r$ - cf. e.g. exam. Also $\l$ is integral and by theorem we have $\l-\mu\in\sum_{r\in R^+}\N_0 r$. Hence $$ \l+r_0-\eta =\l+r_0-\mu-w(r_0)\in\sum_{r\in R^+}\N_0r+\sum_{r\in R^+}\N_0 r =\sum_{r\in R^+}\N_0 r~. $$ $\eofproof$
As for the proof of the following proposition the Casimir element will be pivotal in the following sense: given a representation $\psi:A\rar\Hom(E)$ such that its Casimir element is a multiple $c_\l$ of the identity, then the Casimir element of any sub-representation is the same multiple $c_\l$ of the identity. Hence, if $\psi$ happens to be a highest weight $\l$ cyclic representation and the sub-representation happens to be a highest weight $\mu$ cyclic representation, then by proposition: $$ \norm{\mu+r_0}=\norm{\l+r_0}~. $$ So let $\L\sbe H$ denote the set of integral elements, i.e. $\L\colon=\sum_{b\in B}\Z b^{\vee*}$ for some base $B$ of the set of roots $R$. For any dominant $\l\in\L$ put \begin{equation}\label{pcfeq1}\tag{PCF1} S_\l\colon=\{\mu\in\L:\norm{\mu+r_0}=\norm{\l+r_0}\}, \end{equation} i.e. $S_\l$ is the sphere in $\L$ with radius $\norm{\l+r_0}$ centered at $-r_0$.
Suppose $\l\in S_{\l_0}$ and $\psi:A\rar\Hom(E)$ is a (possibly infinite dimensional) highest weight $\l$ cyclic representation and $E$ is the direct sum of finite dimensional weight spaces. Then the formal character $\chi_\psi$ of $\psi$ is of the form $$ \chi_\psi =\sum_{\mu\in S_\l,\mu\preceq\l}a(\mu,\l)\chi_\mu, \quad\mbox{where}\quad a(\mu,\l)\in\N_0 \quad\mbox{and}\quad a(\l,\l)=1~. $$
$\proof$ 1. If $\psi$ is irreducible, then we may choose $\a(\l,\l)=1$ and for $\mu\neq\l$: $a(\mu,\l)=0$.
2. Let $v$ be the weight vector with weight $\l$. By proposition the Casimir element $\Psi(C)$ acts on $E$ by multiplication by $\norm{\l+r_0}^2-\norm{r_0}^2$. Suppose $\psi$ is not irreducible, then there is a $\psi$-invariant subspace $F$ of $E$ not containing $v$. By exam the weights of $\psi(x)|F$ must be contained in the set $(\l-\sum_{r\in B}\N_0r)\sm\{\l\}$. Hence it must have a maximal (with respect to $\preceq$) weight $\mu$ which is strictly smaller than $\l$. This in turn implies that there is a subspace $G$ of $F$ such that $\psi(x)|G$ is a highest weight $\mu$ cyclic representation and by proposition $\Psi(C)$ acts on $G$ by multiplication by $\norm{\mu+r_0}^2-\norm{r_0}^2$. Thus $\norm{\mu+r_0}^2=\norm{\l+r_0}^2$, i.e. $\mu\in S_\l$ and as $\mu\neq\l$ we conclude that $$ M\colon=\sum_{\mu\in S_{\l_0},\mu\preceq\l}m(\mu)\geq2, $$ where $m(\mu)$ denotes the multiplicity of the weight $\mu$ in the representation $\psi:A\rar\Hom(E)$. $M$ is finite because all weight spaces are finite dimensional!
3. Thus for $M=1$ the assertion holds and we proceed by induction on $M$: Assume $M\geq1$ and $$ \sum_{\mu\in S_{\l_0},\mu\preceq\l}m(\mu)=M+1~. $$ If $\psi$ is irreducible, then the assertion holds by 1. If $\psi$ is not irreducible then we may argue as in 2. and find a sub-space $G$ of $E$ and $\mu\in S_\l$, $\mu\preceq\l$, $\mu\neq\l$, such that $\psi(x)|G$ is a highest weight $\mu$ cyclic sub-representation of $\psi$. Hence we have another highest weight $\l$ cyclic representation $\wh\psi:A\rar\Hom(E/G)$ - the quotient representation - and by exam the multiplicity of any weight $\eta$ of $\psi$ is the sum of the multiplicities of this weight in the sub-representation and the quotient representation. Therefore the character of $\psi:A\rar\Hom(E)$ is the sum of the characters of $\psi(x)|G$ and of $\wh\psi$. Now $S_\mu=S_\l=S_{\l_0}$ and the multiplicity of $\l$ in the sub-representation equals $0$. Hence the sum of the multiplicities for this representation is smaller or equal than $M$. On the other hand the multiplicity of $\mu$ in the quotient representation also equals $0$, which implies that the sum of the multiplicities for the quotient representation is again smaller or equal than $M$. As both of these representations satisfy the premises of the proposition we conclude by induction hypotheses that the character $\chi_\l$ occurs exactly once in the character of the quotient representation. On the other hand it doesn`t occur in the character of the sub-representation. Hence $a(\l,\l)=1$. $\eofproof$
In a second satege we verify that the relation in proposition can be inverted: As $(S_\l,\preceq)$ is a finite ordered set, every subset has a minimal element. Hence we may enumerate the elements of $S_\l$ in such a way that $\l_j\preceq\l_k$ implies $j\leq k$: just choose $\l_1$ to be a minimal element of $S_\l$ and for any $j\geq2$ choose $\l_j$ to be a minimal element of $S_\l\sm\{\l_1,\ldots,\l_{j-1}\}$. Thus we get a matrix $a_{jk}\colon=a(\l_j,\l_k)$ such that for $\l=\l_k$ - thus $\psi_k$ is any highest weight $\l_k$ cyclic representation satisfying the premises of proposition: $$ \chi_{\psi_k}=\sum_{j\leq k}a_{jk}\chi_{\l_j}~. $$ As $a_{jj}=1$ the matrix $(a_{jk})\in\Ma(|S_\l|,\Q)$ is invertible, i.e. $$ \chi_{\l_k}=\sum_{j\leq k}b_{jk}\chi_{\psi_j}, \quad b_{jk}\in\Q,b_{jj}=1~. $$ Forgetting about the enumeration we conclude that \begin{equation}\label{pcfeq2}\tag{PCF2} \chi_{\l}=\sum_{\mu\in S_\l}b(\mu,\l)\chi_{\psi_\mu}, \quad b(\mu,\l)\in\Q,b(\l,\l)=1~. \end{equation} Our final goal is to show that if an exponential $e^{\la h,\mu+r_0\ra}$ occurs with non zero coefficient in the product $q(h)\chi_\l(h)$, then $\mu\in S_\l$: in order to accomplish this we choose for each $\mu\in S_\l$, $\mu\preceq\l$: $\psi_\mu=\g_\mu$, i.e. $\psi_\mu$ is the representation $\g_{\mu}:A\rar\Hom(W_{\mu})$ in the Verma module $W_{\mu}$ (cf. theorem). Notably, by exam the premises of proposition apply to $\g_\mu$! From \eqref{pcfeq2} we have for every $\l\in S_{\l_0}$ by \eqref{kmfeq3}: $$ q(h)\chi_\l(h) =\sum_{\mu\in S_\l}b(\mu,\l)q(h)\chi_{\g_\mu}(h) =\sum_{\mu\in S_\l}b(\mu,\l)e^{\la h,\mu+r_0\ra} =e^{\la h,\l+r_0\ra}+\sum_{\mu\in S_\l\sm\{\l\}}b(\mu,\l)e^{\la h,\mu+r_0\ra}~. $$ Therefore, if an exponential $e^{\la h,\mu+r_0\ra}$ occurs with non zero coefficient in the product $q(h)\chi_\l(h)$, then \begin{equation}\label{pcfeq3}\tag{PCF3} \mu\in S_\l~. \end{equation} Now we can finish the proof of the Weyl Character Formula: By lemma: $\mu+r_0\in\convex{\{w(\l+r_0):w\in W\}}$ and by \eqref{pcfeq3}: $\norm{\mu+r_0}=\norm{\l+r_0}$. As the Euclidean norm is strictly convex we infer that $\mu+r_0\in\{w(\l+r_0):w\in W\}$ and the coefficient of $e^{\la h,\l+r_0\ra}$ equals $1$. Now the function $q(h)\chi_\l(h)$ is Weyl alternating and by lemma this implies that the coefficient of $e^{\la h,w(\l+r_0)\ra}$ must equal $\det(w)$, i.e. $$ q(h)\chi_\l(h) =\sum_{w\in W}\det(w)e^{\la h,w(\l+r_0)\ra}~. $$ This completes the proof of the Weyl Character Formula.
Suppose $x_1,\ldots,x_n$ are pairwise distinct vectors in the Hilbert-space $E$, $\norm{x_j}=r > 0$, $\a_j\in[0,1]$, $\sum\a_j=1$ and $$ \norm{\sum\a_jx_j}=r~. $$ Then there is exactly on $k$ such that $\a_k=1$. This is called strict convexity of the Euclidean norm!
W.l.o.g. we may assume $r=1$. $$ 1-\norm{\sum\a_jx_j}^2 =1-\sum\a_j^2-2\sum_{j < l}\a_j\a_l\la x_j,x_l\ra =1-(\sum\a_j)^2-2\sum_{j < l}\a_j\a_l(\la x_j,x_l\ra-1)\ra =2\sum_{j < l}\a_j\a_l(1-\la x_j,x_l\ra)\ra~. $$ As $\a_j\geq0$ it follows that for all $j < l$: $\a_j\a_l(\la x_j,x_l\ra-1)=0$. By the equality case in the Cauchy-Schwarz inequality we conclude that if $\a_j,\a_l > 0$, then $x_j=x_l$.
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Last modified: Wed Jun 16 13:31:42 CEST 2021