The Weyl Kac Character Formula — Examples

The rank-one presentation in F2 has the three generators $e,h,f$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$, so it is ${\mathfrak{s}\mathfrak{l}}_2$. There is one positive root $\alpha =2\omega$, and its reflection sends $\omega$ to $-\omega$, giving $W=\{1,s\}$ and $\rho =\omega$. Inserting these in F1 gives the displayed quotient.

Set $x=e^{-2\omega }$. After cancelling a monomial, the quotient in step 1.1 is $e^{m\omega }(1-x^{m+1})\mathrm{/}(1-x)$. The polynomial identity $(1-x){\sum }_{j=0}^m{x}^j=1-x^{m+1}$ proves the finite expansion, since $1-x$ is a formal unit. Distinct $j$ give distinct weights, each with coefficient one, so the total dimension is $m+1$. At $m=0$ the sum is $1$, and at $m=1$ it is $e^{\omega }+e^{-\omega }$. No numerical division at $x=1$ is used; the dimension is read from the already finite polynomial.

Here $\mathrm{\ell }=1$ and ${\mathrm{\Phi }}_0^{+}=\{\alpha \}$. The roots $n\delta$ give $q^n$ for $n\ge 1$; $\alpha +n\delta$ give $z{q}^n$ for $n\ge 0$; $-\alpha +n\delta$ give $z^{-1}{q}^n$ for $n\ge 1$. F1 says all these factors have exponent one in this rank. Substitution gives the stated product.

In particular its degree-zero factor is $1-z$. To first degree in $q$, the remaining factors are $(1-q)(1-zq)(1-z^{-1}q)$; factors with index at least two contribute only at degree at least two. Thus $P=(1-z)(1-(1+z+z^{-1})q+O(q^2))\mathrm{.}$ This calculation checks both index endpoints: omitting $n=0$ from the positive family loses $1-z$, while including it in the negative family adds the nonexistent root $-\alpha$. The grouping is coefficientwise formal as in F1, with no analytic Jacobi identity asserted.

Put $\lambda ={\mathrm{\Lambda }}_0+\rho$. Then $\lambda (h)=1$ and $\lambda (c)=3$, by F2 and $\rho (h_0)=\rho (h)=1$ in F1. F4's translation formula, with $\nu (h)=\alpha$ and $B(mh,mh)\mathrm{/}2=m^2$, gives $t_{mh}\lambda -\lambda =3m\alpha -(3m^2+m)\delta ,t_{mh}s\lambda -\lambda =(3m-1)\alpha -(3m^2-m)\delta \mathrm{.}$ Translations have sign plus, because $t_h=s_{0}s$ and powers have even sign; the second family has sign minus. Thus the normalized alternant in F1 is $U={\sum }_{m\in \mathbb{Z}}(z^{-3m}{q}^{3m^2+m}-z^{1-3m}{q}^{3m^2-m})$. Only $m=0$ and the degree-two terms from $m=-1$ in the first family and $m=1$ in the second contribute below degree three. Indeed both quadratic expressions are at least four for the other nonzero choices. Hence $U=(1-z)+(z^3-z^{-2})q^2+O(q^3)$.

By F3 the product is $P=(1-z){\prod }_{n\ge 1}(1-q^n)(1-z{q}^n)(1-z^{-1}{q}^n)$. Put $S=1+z+z^{-1}$. The $n=1$ triple is $1-Sq+S{q}^2-q^3$ and the $n=2$ triple is $1-S{q}^2+O(q^4)$; all later triples begin at degree three or more. Thus $P=(1-z)(1-Sq+O(q^3))$, with zero coefficient at degree two inside the parentheses.

Polynomial division gives $(z^3-z^{-2})\mathrm{/}(1-z)=-(z^{-2}+z^{-1}+1+z+z^2)=-T$. By step 1.1, $U\mathrm{/}(1-z)=1-T{q}^2+O(q^3)$. By step 1.2, the inverse of $P\mathrm{/}(1-z)$ through degree two is $1+Sq+S^2{q}^2+O(q^3)$. Consequently F1 gives $U\mathrm{/}P=1+Sq+(S^2-T)q^2+O(q^3)$. Direct multiplication gives $S^2=z^2+z^{-2}+3+2z+2z^{-1}$ and $S^2-T=2+z+z^{-1}$, proving the statement. Cancellation of $1-z=1-e^{-\alpha }$ is valid in the downward completion by its geometric inverse; division never lowers $q$ degree. At each fixed degree the numerator has finitely many terms by the quadratic bounds and the denominator has finitely many relevant positive-degree factors. This also justifies every displayed truncation without an analytic identity or AC.

By F2 the factor for $\delta$ in F1 is $(1-e^{-\delta }{)}^2=1-2e^{-\delta }+e^{-2\delta }$; the proposed replacement is $1-e^{-\delta }$. A positive root below $\delta$ that is imaginary must equal $\delta$: in F2's computed loop list the other positive roots have nonzero finite-root part and squared length two, hence are real, while positive imaginary roots are $n\delta$. Terms from $n\ge 2$ cannot contribute at degree $\delta$ because their simple coordinates exceed those of $\delta$.

Let $R$ be the common product of all factors relevant at or below $\delta$ other than the $\delta$ factor. Its constant coefficient is one by F1. If $r$ is its coefficient at $e^{-\delta }$, the true product has coefficient $r-2$ and the modified product has coefficient $r-1$. Cross products with the nonconstant part of the $\delta$ factor require the zero coefficient of $R$, already one; no other terms can reach this degree. Thus the modified coefficient exceeds the true one by exactly one, refuting the claim. The zero-degree coefficients agree, so that agreement cannot detect the error. Rank-one imaginary multiplicity one would give no such witness; the rank-two diagonal space in F2 is essential. The calculation is finite and choice-free.