The Weyl-Kac denominator formula states
$$\prod_{\alpha\in\Phi^+}(1-e(-\alpha))^{mult(\alpha)}=\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho))$$
Where the product is taken over the positive roots and the sum is taken over the Weyl group. Here $\epsilon(w)=(-1)^{l(w)}$ where $l(w)$ denotes the length of $w$ and $e(\alpha)$ is the formal exponential.
I'm trying to apply this to the affinization of $\mathfrak{sl}_2$ with simple roots $\{\alpha_0,\alpha_1\}$. In this case the $\rho=\frac{\alpha_1}{2}+2\Lambda_0$ seems to work. Here $\Lambda_0$ is defined on the coroots st. $\Lambda_0(\alpha_i^\vee)=\delta_{i0}$.
The Weyl group of the affinization of $\mathfrak{sl}_2$ is generated by $s_0,s_1$ st. $s_0^2=s_1^2=1$ (it's the infinite dihedral group). The generators works on $\rho$ in the following way
$$s_0(\rho)=\rho-\rho(\alpha_0^\vee)\alpha_0=\rho+\alpha_0=\rho-\alpha_1+\delta$$
$$s_1(\rho)=\rho-\rho(\alpha_1^\vee)\alpha_1=\rho-\alpha_1$$
Now for the formula. The positive roots are exactly $\alpha_1+k\delta$ for $k\geq 0$, $-\alpha_1+k\delta$ for $k\in\mathbb{N}$ and $k\delta$ for $k\in\mathbb{N}$. Here $\delta=\alpha_0+\alpha_1$. Writing $e(-\delta)=q$ and $e(-\alpha_1)=r$ then the left side is
$$\prod_{k>0}(1-q^k)(1-q^{k-1}r)(1-q^kr^{-1})$$
Turning to the right side, is where I get in trouble. I've seen it should equal $\sum_{k\in\mathbb{Z}}(-1)^kr^kq^{\frac{k(k-1)}{2}}$, so working my way backwards then $\sum_{w\in W}\epsilon(w)e(w(\rho)-\rho)$ should equal
\begin{align*}
\sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1)e(-\frac{k(k-1)}{2}\delta) &= \sum_{k\in\mathbb{Z}}(-1)^ke(-k\alpha_1-\frac{k(k-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-\frac{2m(2m-1)}{2}\delta)\\
&-e(-(2m+1)\alpha_1-\frac{(2m+1)((2m+1)-1)}{2}\delta) \\
&=\sum_{m\in\mathbb{Z}}e(-2m\alpha_1-m(2m-1)\delta)\\
&-e(-(2m+1)\alpha_1-(2m+1)m\delta)
\end{align*}
Which is where I'm stuck.
Best Answer
Let $(\delta, \varpi_1, \Lambda_0)$ be the basis of $\mathfrak{h}^*$ which is dual to the basis $(d, \alpha_1^\vee, c)$ of $\mathfrak{h}$, where $c$ is the canonical central element and $d$ is the derivation. The element $\varpi_1$ is a lift of the fundamental weight of $\mathfrak{sl}_2$ to the affine algebra, but is not a fundamental weight of the affine algebra: instead the element $\Lambda_1 = \varpi_1 + \Lambda_0$ is. In this basis the missing roots and coroots are $$ \alpha_1 = 2 \varpi_1, \quad \alpha_0 = \delta - 2 \varpi_1, \quad \alpha_0^\vee = c - \alpha_1^\vee.$$ The reason we use this basis is because it shows the structure of the Weyl group a bit better: abstractly, the affine Weyl group (for untwisted affine Lie algebras) is the semidirect product of the finite coroot lattice and the finite Weyl group. In terms of the basis $(\delta, \varpi_1, \Lambda_0)$ of $\mathfrak{h}^*$, the matrices (with zeros omitted) of the simple reflections are $$ s_0 = \begin{pmatrix} 1 & 1 & -1 \\ & -1 & 2 \\ & & 1 \end{pmatrix}, \quad s_1 = \begin{pmatrix} 1 & & \\ & -1 & \\ & & 1\end{pmatrix}. $$ We can see the semidirect product structure by defining $t = s_0 s_1$, then we have $s_1 t^k = t^{-k} s_1$, and the Weyl group is the semidirect product of $\mathbb{Z}$ (generated by $t$) and $\mathbb{Z}_2$ (generated by $s_1$). In terms of matrices, we have $$ t^k = \begin{pmatrix} 1 & -k & -k^2 \\ & 1 & 2k \\ & & 1 \end{pmatrix}.$$
A little computation shows that $$ t^k \rho - \rho = (-k - 2k^2)\delta + 4k \varpi_1, \quad s_1 t^k \rho - \rho = (-k -2k^2) \delta + (-4k - 2) \varpi_1.$$ Therefore letting $q = e^{-\delta}$ and $r = e^{\alpha_1} = e^{2 \varpi_1}$ we get $$ \begin{aligned} \sum_{w \in W} (-1)^w e^{w \rho - \rho} &= \sum_{k \in \mathbb{Z}}(e^{t^k \rho - \rho} - e^{s_1 t^k \rho - \rho}) \\ &= \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{2k} - \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{-2k-1} \\ &= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 - l)/2} r^{-l} \\ &= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 + l)/2} r^{l} \\ &= \sum_{l \in \mathbb{Z}} (-1)^l q^{(l^2 + l)/2} r^l. \end{aligned}$$
The first equality is using the semidirect product fact, the third equality is reindexing the first sum over even integers $l = 2k$ and the second sum over odd integers $l = 2k + 1$, and the fourth equality is using the $l \mapsto -l$ symmetry in the second sum. The final answer is equivalent to yours because I used $r = e^{\alpha_1}$ rather than $r = e^{- \alpha_1}$.