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I was wondering such a question: for a semisimple complex Lie group $G$, whether it is true that the maximal subtorus of $\mathrm{Res}_{\mathbb{C}/\mathbb{R}}(G)$ is $\mathrm{Res}_{\mathbb{C}/\mathbb{R}}(T)$, where $T$ is the maximal subtorus in $G$.

I think this is true because the maximal torus corresponds to the Cartan subalgebra in $\mathrm{Lie}(G)$, which is generated by semisimple elements for semisimple groups. As $\mathrm{Lie}(\mathrm{Res}_{\mathbb{C}/\mathbb{R}}(G))$ is just $\mathrm{Lie}(G)$ viewed as a real Lie algebra, and that Jordan decomposition is complex linear, the Cartan subalgebra in $\mathrm{Lie}(\mathrm{Res}_{\mathbb{C}/\mathbb{R}}(G))$ is the same as the Cartan subalgebra in $\mathrm{Lie}(G)$. My first question is whether this argument is right.

If the argument is right, there I didn't use the fact that $T$ is a split torus, so I think the above argument generalizes to any finite field extension of characteristic 0, and also generalizes to the case where $G$ is reductive. My second question is that if there is some additional detail here I need to pay attention to since I need to deal with algebraic group theory.

Thanks in advance for any help.

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    $\begingroup$ In general a maximal torus is not unique up to (real) conjugation in a real form (same at the Lie algebra level). So it is quite confusing to call it "the" maximal torus. $\endgroup$
    – YCor
    Commented Jun 6 at 21:31
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    $\begingroup$ However over a ground field $k$ it is true that maximal $k$-tori are maximal tori. $\endgroup$
    – YCor
    Commented Jun 6 at 21:33
  • $\begingroup$ Since there's an art to interpreting language involving rationality, I'll mention that @YCor's second comment means that, if $G$ is a smooth group over a field $k$ and $T$ is a maximal torus in $G$, then $T_E$ is a maximal torus in $G_E$ for every field extension $E/k$. (This is Lemma C.4.4 in Conrad, Gabber, and Prasad - Pseudo-reductive groups, although that's just a convenient reference, not the original source.) $\endgroup$
    – LSpice
    Commented Jun 6 at 21:36
  • $\begingroup$ @LSpice Indeed I'm using the language where objects without reference to the field are over the given algebraically closed extension (so "maximal torus" is not the same in my comment as in yours). $\endgroup$
    – YCor
    Commented Jun 6 at 22:12
  • $\begingroup$ @YCor, re, right. This is the language one sees in many references, such as the excellent Borel - Linear algebraic groups. A nice compromise, which I learned from Brian, between "all unmodified constructions are rational" (the language in my comment) and "all unmodified constructions are over the algebraic closure" is to use "geometric", e.g., "all maximal $k$-tori are geometrically maximal tori." $\endgroup$
    – LSpice
    Commented Jun 6 at 22:16

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One can test whether a subgroup is a maximal torus (as @YCor commented, not the maximal torus, unless you're in a commutative group) after base change, so it suffices to observe that the isomorphism $\operatorname{Res}_{\mathbb C/\mathbb R}(G)_{\mathbb C} \to G_{\mathbb C}^{\operatorname{Gal}(\mathbb C/\mathbb R)}$ (a product of copies of $G_{\mathbb C}$ indexed by the (two) elements of the absolute Galois group) carries $\operatorname{Res}_{\mathbb C/\mathbb R}(T)_{\mathbb C}$ to $T_{\mathbb C}^{\operatorname{Gal}(\mathbb C/\mathbb R)}$, which is a maximal torus in $G_{\mathbb C}^{\operatorname{Gal}(\mathbb C/\mathbb R)}$. This is fine for any separable extension of any field (not necessarily characteristic $0$), and any (smooth, linear, of finite type) group, not necessarily reductive. It fails over inseparable extensions because the inseparable Weil restriction of a non-trivial torus is not a torus (not even reductive!); this idea lies at the heart of the basic construction of pseudo-reductive groups in Conrad, Gabber, and Prasad - Pseudo-reductive groups.

If you prefer something citeable, this is Corollaire 6.19 of Borel and Tits - Groupes réductifs.

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