What is the automorphism group of the tensor square of the Leech lattice? The tensor square of the Leech lattice is an even unimodular lattice of dimension 576 which, unless I am very mistaken, has no roots. Its automorphism group contains a group of shape $2 \cdot \mathrm{Co}_1^{\times 2} : 2$, but I expect it is larger than that. I would like to understand what I can about this group, e.g. how it is built from simple groups, what its maximal subs are, things like that. Has anything about this automorphism group been calculated?
 A: This post provides details for the answer outlined by Noam Elkies in the comments. It is Community Wiki, so anyone can edit it to improve it.
Answer: Let $\Lambda$ denote the Leech lattice. The full automorphism group of the lattice $\Lambda^{\otimes 2}$ is the group $2\cdot \mathrm{Co}_1^{\times 2}:2 = (\mathrm{Co}_0^{\times 2}:2) / 2$, where $2\cdot \mathrm{Co}_1 = \mathrm{Co}_0 = \mathrm{Aut}(\Lambda)$.
(The action is: the wreath product $\mathrm{Aut}(\Lambda)^{\times 2} : 2$ obviously acts on $\Lambda^{\otimes 2}$, but the diagonal central $\mathbb Z/2$ acts trivially. As in the Atlas, $2$ denotes the group of order $2$, $:$ denotes semidirect product, and $\cdot$ denotes an extension that does not split.)
By Kitaoka, Scalar extensions of quadratic lattices II, minimal vectors in $\Lambda^{\otimes 2}$ are all of the form $v \otimes w$ for $v,w \in \Lambda$ minimal.
By inspecting the shapes of minimal vectors in $\Lambda$, one sees that if $v,v'\in \Lambda$ are minimal, then $|\langle v,v'\rangle| \leq 4$, with equality only if $v = \pm v'$. It follows that $\langle v\otimes w,v'\otimes w'\rangle = 8$ iff either $v = \pm v'$ and $\langle w,w'\rangle = \pm 2$ or $\langle v,v'\rangle = \pm 2$ and $w = \pm w'$.
Suppose now that $v_1\otimes w_1$, $v_2 \otimes w_2$, and $v_3 \otimes w_3$ are minimal vectors in $\Lambda^{\otimes 2}$ which pairwise pair to $8$. Then by pigeonhole at least two of the $v$s or at least two of the $w$s are equal up to sign, and so all of the $v$s or all of the $w$s are equal up to sign. (The sign of $v$ is not determined by the tensor product $v\otimes w$, of course.) If all the $w$s are equal, then all the $v$s pair pairwise to $+2$. Let us call such a triple of $v$s a "trio".
Choose a basis $e_1,\dots,e_{24}$ for $\Lambda$ consisting entirely of minimal vectors with the following property: the graph whose vertices are trios and whose edges are when trios intersect in sets of two is connected. It is easy to find such bases.
Let $\phi\in \mathrm{Aut}(\Lambda^{\otimes 2})$. Consider the set of 24 vectors $\phi(e_1\otimes e_1), \phi(e_2\otimes e_1),\dots,\phi(e_{24}\otimes e_1)$. These are all minimal vectors in $\Lambda^{\otimes 2}$ with many triples that pair pairwise to $8$ (one for each trio), and so up to perhaps multiplying $\phi$ by the "switch" $v\otimes w \mapsto w \otimes v$, we have $\phi(e_i \otimes e_1) = v_i \otimes w_1$. Since $\phi$ is metric-preserving, so is $\varphi_L : e_i \mapsto v_i$.
(Note that the sign of $w$, and hence of the $v_i$s, is ambiguous, but that is all.)
Compare now the set $\phi(e_1\otimes e_1),\dots,\phi(e_1\otimes e_{24})$. Either $\phi(e_1\otimes e_j) = v_1 \otimes w_j$ for some vectors or $v_j' \otimes w_1$ (since we already know that $\phi(e_1\otimes e_1) = v_1\otimes w_1$). But in the latter case the set of 47 linearly independent vectors $e_{24}\otimes e_1,\dots, e_1\otimes e_1,\dots,e_1,\otimes e_{24}$ is mapped under $\phi$ to the 24-dimensional space $\Lambda \otimes w_1$, and so is ruled out. Let $\varphi_R : e_j \mapsto w_j$.
Similarly, $\phi(e_2 \otimes e_j) = v_2 \otimes w_j'$, where at first all we know is that $w_1' = w_1$. But considering $\phi(e_i\otimes e_2)$ shows that $w_2 = w_2'$.
All together we find that $\phi(e_i \otimes e_j) = \varphi_L(e_i) \otimes \varphi_R(e_j)$ for $(\varphi_L,\varphi_R) \in \mathrm{Co}_0^{\times 2}$.
This description is redundant because we can switch the signs of both $\varphi_L,\varphi_R$ simultaneously without changing $\phi$. Also we allowed up to one multiplication by the "switch" earlier. All together this shows that $\phi \in 2 \cdot \mathrm{Co}_1^{\times 2} : 2$ as claimed.
