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Is every (adjoint) Chevalley group over the field with two elements $G_{\mathbb{F}_2}$ isomorphic to a subgroup of its counterpart over the rationals $G_\mathbb{Q}$?

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  • $\begingroup$ I'm suspecting you mean specifically with structure coefficients just 0 or 1, or something similar? $\endgroup$ Sep 3, 2021 at 1:16
  • $\begingroup$ Note that Will's proof even shows it doesn't, in general, embed into $G_\mathbf{C}$. $\endgroup$
    – YCor
    Sep 3, 2021 at 11:26
  • $\begingroup$ @‍WillSawin's proof mentioned by @YCor. $\endgroup$
    – LSpice
    Sep 3, 2021 at 11:44
  • $\begingroup$ The smallest non-trivial representation of $E_8(2)$ has dimension at least $402653184$. $\endgroup$ Sep 15, 2021 at 13:10

1 Answer 1

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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\PGL{PGL}$No.

Here's a cheap argument. Let $G = \PGL_n $ for $n>8$ even. Inside $G_{\mathbb F_2}$, we have the group of upper-triangular matrices which differ from the identity matrix only in the upper-right quadrant of the matrix. This group is isomorphic to $(\mathbb Z/2)^{n^2/4}$.

It suffices to prove that $(\mathbb Z/2)^{n^2/4}$ does not embed into $\PGL_n$ in characteristic $0$.

Let's first lift from $\PGL_n$ to $\GL_n$. The commutator map from $(\mathbb Z/2)^{n^2/4} \times (\mathbb Z/2)^{n^2/4}$ to the $n$th roots of unity clearly lies in the $2$nd root of unity, and is symplectic, so it must have an isotropic subspace of dimension $n^2/8$. The inverse image of that group is abelian, thus also contains a subgroup isomorphic to $(\mathbb Z/2)^{n^2/8}$.

But any finite abelian subgroup of $\GL_n$ may be simultaneously diagonalized, so $(\mathbb Z/2)^m$ only embeds in $\GL_n$ if $m \leq n$. Since $n>8$ we have $n^2/8 > n$ and this is impossible.

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