The Weyl Kac Character Formula

A positive root other than ${\alpha }_i$ has a positive simple coordinate at an index different from $i$: the only roots on the $i$th axis are $\pm {\alpha }_i$ by F1's root conventions. Reflection changes only coordinate $i$, and its image is a root by F2, so the one-sign property makes it positive. Applying the involution twice proves that $s_i$ permutes ${\mathrm{\Delta }}^{+}\setminus \{{\alpha }_i\}$, preserving every multiplicity. Also $s_i\rho =\rho -{\alpha }_i$ since $\rho (h_i)=1$.

Transform all exponents of the defining product. Step 1.1 gives $s_{i}D=e^{\rho -{\alpha }_i}(1-e^{{\alpha }_i}){\prod }_{\alpha >0,\alpha \mathrm{\ne }{\alpha }_i}(1-e^{-\alpha }{)}^{m_{\alpha }}=-D\mathrm{.}$ The positive-root product after reindexing is coefficientwise finite by F1; the single exceptional factor is a polynomial with two terms. Thus this manipulation really equals the transformed coefficient array, rather than assuming an action on the entire completion. A finite word of reflections now transforms this particular array repeatedly, producing one minus sign per reflection. Each simple reflection fixes a hyperplane and negates its complementary root line, so has determinant $-1$; the accumulated sign is $\det (w)$ independent of the word. The identity word has sign one. Every transformation used only finite coefficient computations, with no AC.

Fix a target weight $\tau$. For a subquotient $X$ of $V$, let $p(X)={\sum }_{\eta \ge \tau }\dim X_{\eta }$, a finite nonnegative integer by F1. If nonzero, choose a maximal weight $\mu \ge \tau$ in its finite support window and $0\mathrm{\ne }v\in X_{\mu }$. All positive root operators kill $v$, since a nonzero image would have larger weight in that same window. Thus $v$ is a highest vector, hence primitive in F6's terminology. Its cyclic submodule $H$ is a quotient of $M(\mu )$ by F3 and has simple quotient $L(\mu )$ by F4: the kernel of $M(\mu )\to H$ is proper and is contained in the unique maximal proper submodule. Write $K$ for the kernel of $H\to L(\mu )$. Both $p(K)<p(X)$ and $p(X\mathrm{/}H)<p(X)$, because the nonzero top weight of $H$ survives in $L(\mu )$.

The two exact sequences give $\mathrm{ch}X=\mathrm{ch}K+\mathrm{ch}L(\mu )+\mathrm{ch}(X\mathrm{/}H)$. Induction on $p(X)$ in step 1.1 expresses the character above $\tau$ as a finite sum of characters of simple subquotients, with nonnegative integral multiplicities, plus terms zero on that window. The same applies to every Verma module by F3. These coefficients are unique where needed: order the finite interval $[\tau ,\mathrm{\Lambda }]$ by decreasing height relative to its top; the character of $L(\mu )$ has top coefficient one and all other weights strictly lower. Successively subtracting top coefficients determines each multiplicity. Enlarging the window therefore gives the same answers on the old window. This defines a coefficientwise locally finite simple-character expansion without assuming finite length of $V$.

F5 acts on $V$ as $a=(\mathrm{\Lambda }+2\rho ,\mathrm{\Lambda })$. Its pointwise finite defining operator commutes with passage to a submodule or quotient, so every simple subquotient in step 2.1 also has scalar $a$. On $L(\mu )$, F5 gives scalar $(\mu +2\rho ,\mu )$. Their equality is exactly $\mathrm{\parallel }\mu +\rho {\mathrm{\parallel }}^2=\mathrm{\parallel }\mathrm{\Lambda }+\rho {\mathrm{\parallel }}^2$, where the squares denote the bilinear form, not a positive norm assumption. Applying the same reasoning to $M(\nu )$ shows that its simple-character expansion has coefficient one at $\nu$ and no coefficient across distinct shifted-square values.

On every finite interval, the matrix expressing Verma characters in simple characters is integral unitriangular. Its inverse is integral: if the strictly triangular part is $N$, then $I-N+N^2-\cdots$ stops after the number of interval weights. By step 3.1 it is block diagonal by shifted-square value, and each power and the inverse preserve those same blocks. Inverting the simple expansion of $V$ consequently gives the asserted constrained Verma expansion. The coefficients agree across windows by the same top-down recursion; equivalently they are the coefficients of the uniquely defined series $P\mathrm{ch}V$, where $\mathrm{ch}M(\mu )=e^{\mu }{P}^{-1}$ by F3. The top coefficient is one because a nonzero highest module is a quotient of $M(\mathrm{\Lambda })$ and its top survives. All windows and vector selections above are finite; the unique coefficient recursion defines the global series without AC.

For a reduced word $w=s_{i_1}\cdots s_{i_l}$, each prefix is reduced and its next root $s_{i_1}\cdots s_{i_{r-1}}{\alpha }_{i_r}$ is positive by F3. Telescoping therefore gives $\rho -w\rho ={\sum }_{r=1}^l{s}_{i_1}\cdots s_{i_{r-1}}{\alpha }_{i_r}\in Q^{+},\mathrm{ht}(\rho -w\rho )\ge l\mathrm{.}$ There are finitely many words of length at most any fixed integer. Thus only finitely many $w$ contribute at a bounded depth, and $w\rho =\rho$ forces $l=0$. The alternant has top coefficient one, and its normalization $a$ is invertible by F4. Reindexing the orbit sum by left multiplication gives $s_i{A}_{\rho }=-A_{\rho }$ as an array; finite multiplicities justify every coefficient.

We justify division of these skew arrays without presuming a Weyl action on the whole completion. Fix $i$ and write $x=e^{-{\alpha }_i}$ and $y^{\mathbf{b}}={\prod }_{j\mathrm{\ne }i}{e}^{-b_j{\alpha }_j}$ for $\mathbf{b}\ge 0$. Let $m(\mathbf{b})=-{\sum }_{j\mathrm{\ne }i}{a}_{ij}{b}_j\ge 0$, where nonnegativity is the off-diagonal sign axiom for a GCM. By F7, reflection sends $x$ to $x^{-1}$ and $y^{\mathbf{b}}$ to $x^{m(\mathbf{b})}{y}^{\mathbf{b}}$. Put $p=e^{-\rho }D=P$ by F6. The skewness of $D,A_{\rho }$ implies for either $f=p$ or $f=a$ that its fixed transverse coefficient satisfies $f_{\mathbf{b}}(x)=-x^{m(\mathbf{b})+1}{f}_{\mathbf{b}}(x^{-1})\mathrm{.}$ This is an equality of coefficient arrays. Since its original exponents in $x$ are nonnegative, the equality forces them to be at most $m(\mathbf{b})+1$; hence each $f_{\mathbf{b}}$ is a polynomial. Moreover $p_{\mathbf{0}}=1-x$ by F6. Also $a_{\mathbf{0}}=1-x$: in step 1.1, a reduced word whose first letter is not $i$ already contributes the off-axis root ${\alpha }_j$; if its first letter is $i$ and it has a second letter $j$, reducedness gives $j\mathrm{\ne }i$, while F7 gives $s_i{\alpha }_j={\alpha }_j-a_{ij}{\alpha }_i$, whose ${\alpha }_j$-coordinate is one because the simple roots are independent. All associated roots are positive, so later summands cannot cancel that off-axis coordinate. Thus only $1,s_i$ occur on the axis.

Form $C=p\mathrm{/}a$ by F4. We show by induction on ${\sum }_{j\mathrm{\ne }i}{b}_j$ that every $C_{\mathbf{b}}(x)$ is a polynomial satisfying $C_{\mathbf{b}}(x)=x^{m(\mathbf{b})}{C}_{\mathbf{b}}(x^{-1})$. The base is $C_{\mathbf{0}}=1$. At a nonzero transverse index, the product equation gives $(1-x)C_{\mathbf{b}}=R_{\mathbf{b}}:=p_{\mathbf{b}}-{\sum }_{\begin{array}{c}\mathbf{u}+\mathbf{v}=\mathbf{b}\\ \mathbf{u}\mathrm{\ne }\mathbf{0}\end{array}}{a}_{\mathbf{u}}{C}_{\mathbf{v}}\mathrm{.}$ Every $\mathbf{v}$ in the finite sum has smaller total transverse degree. Step 2.1 and the induction hypothesis make $R_{\mathbf{b}}$ a polynomial satisfying $R_{\mathbf{b}}(x)=-x^{m(\mathbf{b})+1}{R}_{\mathbf{b}}(x^{-1})$, since $m$ is additive. At $x=1$ this gives $R_{\mathbf{b}}(1)=0$, so polynomial division yields $R_{\mathbf{b}}=(1-x)q$ with $q\in \mathbb{Z}[x]$. Its expansion equals $C_{\mathbf{b}}$ by uniqueness of inversion in formal power series. Substitution and cancellation of $1-x$ then give $q(x)=x^{m(\mathbf{b})}q(x^{-1})$. This completes the induction, including rank one where only the base transverse index exists.

Step 3.1 proves exactly $s_{i}C=C$ as an array for every $i$, with each transverse slice finite. Repeating over a finite word gives $wC=C$. If the coefficient of $e^{-\beta }$ is nonzero, all coefficients of $e^{-w\beta }$ are the same nonzero value. Since the original support lies in $-Q^{+}$, this implies $w\beta \in Q^{+}$ for every $w$, proving the asserted cone support. It does not say that $\beta$ is a root at all, so makes no identification with F5's imaginary-root set.

Suppose nonconstant support exists and take a nonzero $\beta$ of least positive height in it. If $\beta (h_i)>0$, then $s_i\beta =\beta -\beta (h_i){\alpha }_i$ is a nonzero element of $Q^{+}$ by step 4.1 and invertibility of the reflection. Its coefficient is the same and its height is smaller, a contradiction. Thus every pairing is nonpositive. Least height exists in the positive integers; finitely many lattice points have that height, so choosing one uses no AC.

Apply F2 to the one-dimensional trivial module, whose highest weight is zero and character one. Multiplying its Verma expansion by $e^{\rho }P$ using F3 and identifying this factor as $D$ by F6 gives $D={\sum }_{\mu }{c}_{\mu }{e}^{\mu +\rho }$. Therefore a nonzero coefficient of $e^{\rho -\beta }$ in $D$ must satisfy $(\rho -\beta {)}^2={\rho }^2$, or ${\beta }^2=2(\rho ,\beta )$. All multiplications are coefficientwise finite by F1 and F2.

If $C$ has nonconstant support, take a least-height $\beta ={\sum }_i{k}_i{\alpha }_i\mathrm{\ne }0$ in it. F1 gives $k_i\ge 0$ and $\beta (h_i)\le 0$ for every $i$. F4 gives $({\alpha }_i,\xi )=d_i\xi (h_i)$ with $d_i>0$. Thus ${\beta }^2={\sum }_i{k}_i{d}_i\beta (h_i)\le 0,2(\rho ,\beta )=2{\sum }_i{k}_i{d}_i>0.$ This contradicts the equality required in step 1.1, if that coefficient of $D$ is nonzero.

It is nonzero. Write $a=e^{-\rho }{A}_{\rho }$ and $p=e^{-\rho }D=aC$. In the coefficient at $e^{-\beta }$ every mixed product with a nonzero degree of $C$ strictly smaller than $\beta$ vanishes by minimality. Hence $p_{\beta }=C_{\beta }+a_{\beta }$. A nonzero $a_{\beta }$ would require $\beta =\rho -w\rho$ for some $w$. The reflection formula and F4 show the form is Weyl invariant: expansion of a reflected pairing cancels its two cross terms against $({\alpha }_i,{\alpha }_i)=2d_i$. Thus such a $\beta$ would satisfy $(\rho -\beta {)}^2={\rho }^2$, contradicting the two inequalities of step 2.1. So $a_{\beta }=0$ and $p_{\beta }=C_{\beta }\mathrm{\ne }0$, contradicting step 1.1 after all. There is no nonconstant support, and F1's unit constant coefficient gives $C=1$. The choice of a least-height term is finite, so no AC is used.

By F1 the normalized alternant $a=e^{-\rho }{A}_{\rho }$ has constant coefficient one and an inverse, and the normalized product $p=e^{-\rho }D$ satisfies $p=aC$. The product is coefficientwise defined.

F2 gives $C=1$, hence $p=a$ by step 1.1. Multiplying by the monomial $e^{\rho }$ gives exactly the asserted equality, with the original exponents $m_{\alpha }$ on every factor. All products used are coefficientwise finite by F1.

Fix $i$ and write $E=e_i,F=f_i,H=h_i$. By F2 and F5 the module has the local nilpotence in F3, while F8 and F9 give $V\in \mathcal{O}$ and hence well-defined finite-dimensional weight spaces. Thus $T=\exp (E)\exp (-F)\exp (E)$ is defined on each vector by finite sums. Its inverse is $\exp (-E)\exp (F)\exp (-E)$: each adjacent exponential cancellation is the finite binomial identity on a vector. F4 permits all computations involving $E,F,H$ on a finite-dimensional invariant subspace containing the vector under consideration.

The brackets in F7 give conjugations $\exp (E)H\exp (-E)=H-2E$, $\exp (-F)H\exp (F)=H-2F$ and $\exp (-F)E\exp (F)=E+H-F$. These follow by expanding the finite exponentials, or by differentiating their polynomial conjugations and using the first two commutators. Consequently the successive conjugations of $H$ in $TH{T}^{-1}$ are $H-2E$, then $-H-2E$, then $-H$. For an arbitrary $h\in \mathfrak{h}$, put $h^{\perp }=h-{\alpha }_i(h)H\mathrm{/}2$. F7 makes $h^{\perp }$ commute with $E,F$, so $Th{T}^{-1}=h-{\alpha }_i(h)H=s_{i}h$. The polynomial identities hold on every vector by step 1.1; commutation with $h^{\perp }$ requires no finite-dimensional invariant space for the whole Cartan.

If $v\in V_{\mu }$, step 2.1 and $s_i^2=1$ give $hTv=T(s_{i}h)v=\mu (s_{i}h)Tv$. Thus $T:V_{\mu }\to V_{s_i\mu }$ is an isomorphism, with the inverse from step 1.1. The two finite dimensions in F8 are equal, including when both spaces are zero. Hence $s_i\mathrm{ch}V=\mathrm{ch}V$ as coefficient arrays.

Transform the coefficient sum for $D\mathrm{ch}V$ by $s_i$. Relabelling its pairs of exponents bijectively gives the coefficient sum for $(s_{i}D)(s_i\mathrm{ch}V)$; it is finite because its preimage is a coefficient sum of the original product, finite by F8. F1 and step 3.1 therefore give $s_{i}N=-N$. Iterating a finite word gives the claimed determinant sign. No assertion of an action on arbitrary completion elements is needed. All exponential identities were finite on each vector and no bases of infinitely many spaces were selected.

Let $\eta =\lambda -\beta$ have a nonzero coefficient, so $\beta \in Q^{+}$ by F2. All its simple labels are integers by F3 and the integral Cartan matrix. If $\eta (h_i)=0$, reflection fixes $\eta$ and F1 makes its coefficient its negative, impossible over $\mathbb{Z}$. If $\eta (h_i)<0$, reflect: $s_i\eta =\eta +k{\alpha }_i$ for the positive integer $k=-\eta (h_i)$. Its coefficient is still nonzero by F1, so F2 implies $\beta -k{\alpha }_i\in Q^{+}$. Its height is smaller by $k$. Repeatedly choosing the least negative index terminates in finitely many steps, at a support weight ${\eta }^{\mathrm{\prime }}$ with every label strictly positive. This termination uses the actual cone bound on the orbit's support, not a claim that every integral weight can be moved to the dominant chamber.

Write ${\eta }^{\mathrm{\prime }}=\lambda -\gamma$, $\gamma ={\sum }_i{k}_i{\alpha }_i\in Q^{+}$. F2 says ${\eta }^{\mathrm{\prime }2}={\lambda }^2$. But F6 gives ${\lambda }^2-{\eta }^{\mathrm{\prime }2}=(\gamma ,\lambda +{\eta }^{\mathrm{\prime }})={\sum }_i{k}_i{d}_i(\lambda (h_i)+{\eta }^{\mathrm{\prime }}(h_i))\mathrm{.}$ Since $\lambda (h_i)=\mathrm{\Lambda }(h_i)+1\ge 1$ and ${\eta }^{\mathrm{\prime }}(h_i)>0$, this is strictly positive unless every $k_i=0$. Hence ${\eta }^{\mathrm{\prime }}=\lambda$. Step 1.1 now puts every support weight on the orbit of $\lambda$. The comparison is between real sums of labels even if complementary Cartan coordinates are complex. No positive-definiteness assumption was made.

For a reduced word $w=s_{i_1}\cdots s_{i_t}$, telescoping gives $\lambda -w\lambda ={\sum }_{j=1}^t\lambda (h_{i_j})s_{i_1}\cdots s_{i_{j-1}}{\alpha }_{i_j}\mathrm{.}$ Every prefix is reduced, and its next root is positive by F5. Every scalar is an integer at least one by F3. Thus the difference belongs to $Q^{+}$ and has height at least $t=\mathrm{\ell }(w)$, using F4's length convention. If $w\lambda =\lambda$, this forces $t=0$, so the stabilizer is trivial. A fixed height bound allows only finitely many words in the finite alphabet, proving local finiteness. F2 gives coefficient one at $\lambda$; F1 then gives exactly $\det (w)$ at $w\lambda$. Together with step 2.1 this proves all assertions. Empty words and the empty simple system give the sole term of coefficient one. No infinite choices occur.

F3 permits application of F2. Its complete coefficient description gives the first displayed equality. In particular no additional exponents or cancellations from a stabilizer are omitted. Its length bound proves that at any fixed depth only finitely many Weyl terms occur. F1 ensures that the denominator here is the actual root-multiplicity product, not a product with artificial imaginary-root multiplicities.

The series $P$ has constant coefficient one and other support in $-Q^{+}\setminus \{0\}$. Its inverse is the formal geometric series in $1-P$: at depth $n$ only powers at most $n$ contribute, and finite telescoping verifies both inverse identities. Multiplying step 1.1 by $e^{-\rho }{P}^{-1}$ gives the second formula, with finite coefficient sums because each simple-root coordinate is bounded by the target depth. Conversely multiplication by $e^{\rho }P$ recovers the first. At $\mathrm{\Lambda }=0$ this is F1, since the trivial module is the simple highest-weight module of weight zero. Empty root data give a single monomial. All operations are formal and choice-free.

Put $\lambda =\mathrm{\Lambda }+\rho$ and $\beta =\mathrm{\Lambda }-\mu$. A contributing $w$ must satisfy $\gamma =w\lambda -(\mu +\rho )\in Q^{+}$ by F2. F1's height bound gives ${\delta }_w=\lambda -w\lambda \in Q^{+}$ and $\mathrm{ht}({\delta }_w)\ge \mathrm{\ell }(w)$. Since $\beta ={\delta }_w+\gamma$, no $w$ contributes unless $\beta \in Q^{+}$. If it does, $\mathrm{\ell }(w)\le \mathrm{ht}(\beta )$, so only finitely many words in the finite simple-reflection alphabet, and hence finitely many elements, can contribute.

Expand F1's inverse product using F2. The term indexed by $(w,\gamma )$ has exponent $w(\mathrm{\Lambda }+\rho )-\rho -\gamma$ and coefficient $\det (w)K(\gamma )$. Equating this exponent to $\mu$ forces exactly the argument of $K$ in the statement. The coefficient extraction is finite by step 1.1, so gives the displayed identity. For $\mu =\mathrm{\Lambda }$, step 1.1 forces $\mathrm{\ell }(w)=0$, and the value is $K(0)=1$. Outside the cone every summand and the corresponding weight space are zero. There is no analytic summation or choice of an infinite family.

By F4 the root set is finite and all its roots are real, so F3 gives multiplicity one. Weyl transformations permute the real roots. This permutation action is faithful: the finite-type Cartan matrix is nonsingular, so in its minimal realization the independent simple roots are a basis of the dual Cartan, and a linear map fixing every root fixes that basis. Thus $W$ embeds in the permutation group of a finite set and is finite.

Substitute the multiplicities of step 1.1 into F1. Its infinite-index notation is now the finite sum and finite product displayed above. The semisimple algebra, Cartan, simple coroot labels and positive generators coincide under F2's generator identification, so the simple highest-weight module and dominance convention are the same. This is the finite Weyl character formula with that positive Borel. The quotient denotes the equality after multiplication by its denominator, or the formal inverse used in F1; no division by a numerically zero specialization is made. Disconnected finite types are included by F2, and empty data give one monomial. No choice beyond finite linear algebra enters.

F3 and F4 give real roots $\alpha +n\delta$ for every finite root $\alpha$ and integer $n$, with multiplicity one. Their positive members are those with $n>0$, together with $n=0$ and $\alpha \in {\mathrm{\Phi }}_0^{+}$. For $n>0$ and $\alpha \in {\mathrm{\Phi }}_0^{+}$, F5 gives $\alpha +n\delta =n{\alpha }_0+(n\theta +\alpha )\in Q_{+}\mathrm{.}$ For $\alpha =-\beta$ with $\beta \in {\mathrm{\Phi }}_0^{+}$, it gives $-\beta +n\delta =n{\alpha }_0+(n-1)\theta +(\theta -\beta )\in Q_{+}\mathrm{.}$ These exhaust the two finite-root signs by F6. Negation handles $n<0$, while $n=0$ has the finite-root sign. Imaginary positive roots are exactly $n\delta$, $n\ge 1$, because $\delta ={\alpha }_0+\theta \in Q_{+}$; F3 gives their multiplicity $\mathrm{\ell }$. The value $n=0$ gives Cartan weight zero and is not a root.

Split the finite roots in step 1.1 into $\alpha \in {\mathrm{\Phi }}_0^{+}$ and their negatives. Their exponentials are respectively $e^{-\alpha }{q}^n$ for $n\ge 0$ and $e^{\alpha }{q}^n$ for $n\ge 1$. Imaginary roots contribute $q^n$ for $n\ge 1$. Inserting these disjoint exhaustive families, with their multiplicities, in F1's product gives the displayed expression. No root is lost or repeated, and regrouping is permitted because each coefficient has only finitely many contributing root factors, as in F1. Rank one gives exponent one on the imaginary product, whereas larger rank retains $\mathrm{\ell }$. This is a formal reindexing with no analytic convergence or AC assumption.