Theta Series of D16+ lattice? What does the theta series of the even, unimodular, positive definite lattice D16+ look like? Also, is there a way to look this information up for any lattice?
 A: *

*$h(t) = e^{-\pi \|t\|^2}, t \in \mathbb{R}^k$ is its own Fourier transform. If $M \in \mathbb{R}^{k \times k}$ is an invertible matrix then the Fourier transform of $h(Mt) = e^{-\pi \|M t\|^2}$ is $\frac{1}{|\det M|} h(M^{-T} t)$. 

*If $\Lambda = M \mathbb{Z}^k$ is a lattice then we define its theta series $$\Theta_\Lambda(x) = \sum_{\lambda \in \Lambda} e^{- \pi x \|\lambda\|^2} =\sum_{n \in \mathbb{Z}^k} h(\sqrt{x} M n), \qquad \Re(x) > 0$$
and with $H(n)=h((\sqrt{x}I) M n)$ the Poisson summation formula gives 
$$\Theta_\Lambda(x) = \sum_{n \in \mathbb{Z}^k} H(n)=\sum_{n \in \mathbb{Z}^k} \widehat{H}(n)= \frac{x^{-k/2}}{|\det M|} \Theta_{\Lambda'}(1/x)$$
where $\Lambda' = M^{-T} \mathbb{Z}^k$ is the dual lattice.

*Note $\Lambda$ and $P \Lambda$ have the same theta series for any orthogonal matrix $PP^T=P^TP = I$.
Let $G = M^T M$. If $G \in \mathbb{Z}^{k \times k}$ then $\Lambda' = M^{-T} \mathbb{Z}^k \supseteq M^{-T} G \mathbb{Z}^k = M \mathbb{Z}^k = \Lambda$. If moreover $\det(G) = 1$ then $G^{-1} \in\mathbb{Z}^{k \times k}$ and $\Lambda= \Lambda'$. The lattice is then said unimodular. We also get $\|\lambda\|^2 \in \mathbb{Z}$ so that
$$\Theta_\Lambda(2iz)=\Theta_\Lambda(2iz+1), \qquad \Theta_\Lambda(\frac{-2i}{4z}) = (i z)^{-k/2} \Theta_\Lambda(2iz)$$
and if $8 | k$ then $\Theta_\Lambda(2iz) \in M_{k/2}(\Gamma_0(4))$.

*If also $\|\lambda\|^2 \in 2\mathbb{Z}$ then $\Lambda$ is said even unimodular, in that case $\Theta_\Lambda(iz) \in M_{k/2}(SL_2(\mathbb{Z}))$. This is the case here.

*$M_{8}(SL_2(\mathbb{Z}))$ is one-dimensional (containing only the Eisenstein series) so you'll get $\Theta_\Lambda(iz) = E_8(z)$ the Eisenstein series.
