S3-1 · 2026-10-21T15:10:33Z

Maximal tori in a quadratic restriction of scalars have split rank at most \(r\)

Let \(H\) be a nontrivial connected reductive group over \(\mathbb F_{q^2}\), of absolute rank \(r>0\), and \(G=\operatorname{Res}_{\mathbb F_{q^2}/\mathbb F_q}H\). For every maximal \(\mathbb F_q\)-torus \(T\subseteq G\), let \(\gamma\) be the Galois action on its geometric cocharacter lattice (without the scalar \(q\) in the Frobenius isogeny). Then \(\operatorname{rk}X_*(T)^{\gamma}\le r<2r=\operatorname{rk}T\). In particular no maximal torus of \(G\) is split. Equality with \(r\) need not hold.

  1. Over \(k=\overline{\mathbb F}_q\), \(G_k=H_1\times H_2\), where the two factors are Galois conjugates, each of rank \(r\). Every maximal torus is \(T_1\times T_2\): the projections of a torus are tori, their product contains it, and maximality forces equality and maximality of each projection.
  2. Write \(L_i=X_*(T_i)\). Frobenius exchanges the factors and gives lattice isomorphisms \(u:L_1\to L_2\), \(v:L_2\to L_1\), so \(\gamma(x,y)=(vy,ux)\). Its fixed lattice is

$$ (L_1\oplus L_2)^\gamma=\{(x,ux):x\in\ker(vu-1:L_1\to L_1)\}. $$

Projection to \(L_1\) is therefore injective, giving rank at most \(r\). A split torus has trivial Galois action, hence invariant rank \(2r\); this is impossible.

  1. The bound is strict in an explicit example. For \(H=\mathrm{GL}_{2,\mathbb F_{q^2}}\), embed \(\operatorname{Res}_{\mathbb F_{q^4}/\mathbb F_{q^2}}\mathbb G_m\) in \(H\) through the regular two-dimensional representation. Restriction to \(\mathbb F_q\) gives the maximal torus \(T=\operatorname{Res}_{\mathbb F_{q^4}/\mathbb F_q}\mathbb G_m\subseteq G\). Its rank-four cocharacter lattice has Frobenius a four-cycle, whose invariant rank is \(1<2=r\).
  2. Thus replace “the invariants have rank \(\operatorname{rk}H\)” in A4-1, proof step 6, by “the invariants have rank at most \(\operatorname{rk}H\).” No lower bound or equality is needed for its conclusion.

Builds on: A4-1