The root lattice $\Gamma_8$ of the exceptional Lie algebra $E_8$ is an eight-dimensional lattice which consists of lattice points in $\mathbb{R}^8$ which with respect to an orthonormal basis $e_1, \ldots, e_8$ have coordinates that are either all integers or all half integers and such that the sum of coordinates is an even integer. Thus, for example, $e_1 - e_2$ and $(1/2) * (e_1 + e_2 + \ldots + e_8)$ are both vectors in the $E_8$ root lattice.
One can associate a theta function to any lattice. The theta function for the $E_8$ root lattice is$$\theta_{E_8}(\tau) = \sum_{\tau \in \Gamma_8} q^{|\lambda|^2/2}, \quad q = e^{2\pi i\tau}.$$How does the $E_8$ theta function transform under the modular transformations $\tau \to \tau + 1$ and $\tau \to -1/\tau$?
Thoughts. We perhaps need to figure out something about the dual lattice of the $E_8$ root lattice if we want to use Poisson summation?
We want to find the transformation of the theta function$$\theta_{E_8}(\tau) = \sum_{\tau \in \Gamma_8} q^{\lambda^2/2}, \quad q = e^{2\pi i\tau},$$under modular transformations. Under $\tau \to \tau + 1$, $q^{\lambda^2/2} \to q^{\lambda^2/2} e^{\pi i\lambda^2}$. Let us classify the possible $\lambda^2$ for $\lambda \in \Gamma_8$, by writing $\lambda_i = \alpha_i/2$ with all $\alpha_i$ either odd or even:$$\lambda^2 = \sum_{i = 1}^8 \lambda_i^2 = {1\over4}\sum_{i = 1}^8 \alpha_i^2.$$If all $\alpha_i$ are odd, $\alpha_i^2 \equiv 1 \text{ mod }8$, so $\sum_i \alpha_i^2 \equiv 0 \text{ mod }8$ and $\lambda^2 \equiv 0 \text{ mod }2$. If all $\alpha_i$ are even, we have$$\alpha_i \equiv 2 \text{ mod }4 \implies \alpha_i^2 \equiv 4 \text{ mod }8,$$$$\alpha_i \equiv 0 \text{ mod }4 \implies \alpha_i^2 \equiv 0 \text{ mod }8.$$Since the sum is an even integer, we have an even number of $\alpha_i \equiv 2 \text{ mod }4$, so again $\sum_i \alpha_i^2 \equiv 0 \text{ mod }8$ and $\lambda^2 \equiv 0 \text{ mod }2$. Hence, $e^{\pi i\lambda^2} = 1$ and $q^{\lambda^2/2} \to q^{\lambda^2/2}$, so $\theta_{E_8}$ is invariant.
Let us now find the dual lattice to $\Lambda_8$. For this, it is helpful to write a basis for $\Lambda_8$. We will use as a basis$$B = \left\{e_1 - e_2, e_2 - e_3, e_3 - e_4, e_4 - e_5, e_5 - e_6, e_6 - e_7, 2e_1, {1\over2}(e_1 + \ldots + e_8)\right\}.$$We might be tempted to include $e_7 - e_8$, but we can check that we can express it as a linear combination of elements from this basis. Then by going through each element in the basis, $\kappa \in \Gamma_8^*$ must have:\begin{align*} \kappa_i - \kappa_j \in \mathbb{Z} & \implies \kappa_i \equiv \kappa_j \text{ mod }1,\\ 2\kappa_1 \in \mathbb{Z} & \implies \kappa_i \equiv 0,\,{1\over2}\text{ mod }1, \\ {1\over2} \sum_i \kappa_i \in \mathbb{Z} & \implies \sum_i \kappa_i \in 2\mathbb{Z}.\end{align*}These are the defining conditions for $\Gamma_8$, so $\Gamma_8$, so $\Gamma_8^* = \Gamma_8$, i.e. $\Gamma_8$ is self-dual. A self-dual lattice also must have a unit volume primitive cell. We can check this explicitly from the basis above.
Now, we can find the transformation under $\tau \to -1/\tau$. In order to use the Poisson summation formula, we want to find the Fourier transform of the summand: defining $f(\lambda) = q^{\lambda^2/2}$ so $\theta_{E_8}(\tau) = \sum_{\lambda \in \Gamma_8} f(\lambda)$, it is$$\tilde{f}(\kappa) = \int d^8\lambda\,\text{exp}(\pi i\tau \lambda^2 - 2\pi i\kappa \cdot \lambda) = \left({1\over{-i\tau}}\right)^4 \exp\left(-{{\pi i\kappa^2}\over\tau}\right),$$so that$$\sum_{\kappa \in \Gamma_8} \tilde{f}(\kappa) = (-i\tau)^{-4}\theta_{E_8}(-1/\tau).$$For a self-dual lattice, the Poisson summation formula gives$$\sum_{\lambda \in \Gamma_8} f(\lambda) = \sum_{\kappa \in \Gamma_8} \tilde{f}(\kappa),$$so we find$$\theta_{E_8}(-1/\tau) = (-i\tau)^4 \theta_{E_8}(\tau) = \tau^4 \theta_{E_8}(\tau).$$