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?



What is the automorphism group of the tensor square of the Leech lattice?
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](https://projecteuclid.org/euclid.nmj/1118796476), 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 $\Lambda$](https://en.m.wikipedia.org/wiki/Leech_lattice#Geometry), 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.


