I've been trying to understand some of the idiosyncrasies associated to algebraic groups over non-algebraically closed fields $K$ of characteristic $p > 0$.

Let $G$ is a connected almost absolutely simple adjoint algebraic group over $K$ and $\widetilde{G}$ is the simply connected cover of $G$ also defined over $K$ with central isogeny $\pi : \widetilde{G} \rightarrow G$. Let $\widetilde{T} \subseteq \widetilde{G}$ be a maximal $K$-torus and let $T= \pi(\widetilde{T}) \subseteq G$.

Is it possible for the $\mathbb{Z}$-rank of the $K$-defined character groups $X_K(\widetilde{T})$ and $X_K(T)$ to be different? In other words, is the $K$-split part of $\widetilde{T}$ always the same rank as the $K$-split part of $T$?


2 Answers 2


The map $\widetilde{T} \rightarrow T$ is an isogeny, and an isogeny between tori over any field $K$ induces a finite-index inclusion between the associated geometric character groups as ${\rm{Gal}}(K_s/K)$-lattices, hence an isomorphism on associated rational vector space, so same dimension for spaces of Galois-invariants; i.e., same rank for $K$-rational character groups. However, this is a central isogeny, and that is really the key to the good behavior.

For isogenies $f:G \rightarrow G'$ between connected semisimple $K$-groups such that $\ker f$ is not central in $G$ (which can only occur in positive characteristic $p$, such as Frobenius isogenies $F_{G/K}:G \rightarrow G^{(p)}$) it can happen that $G'$ has larger $K$-rank than $G$. This might sound paradoxical if you are not familiar with it, since you might reason that if $T' \subset G'$ is a maximal $K$-torus (say containing a maximal split $K$-torus of $G'$) then the identity component of the underlying reduced scheme of $f^{-1}(T')$ seems to be a smooth $K$-subgroup scheme of $G$, hence a torus mapping onto $T'$ via an isogeny, so it has the same $K$-rank as $T'$ by the first paragraph above. That reasoning is valid provided that $f^{-1}(T')_{\rm{red}}$ really is a smooth $K$-subgroup scheme of $G$. For perfect $K$ such conditions hold (since the underlying reduced scheme of a finite type $K$-group scheme is a smooth $K$-subgroup scheme for such $K$). But it can fail when $K$ is imperfect.

For example, if $K$ is a local function field of characteristic $p$ and $A$ is a central division algebra of dimension $p^2$ over $K$ then $G := {\rm{SL}}_1(A)$ is a $K$-anisotropic form of ${\rm{SL}}_p$ but $G^{(p)}$ is $K$-split since $A^{(p)}$ is $K$-split (by local class field theory). So $F_{G/K}:G \rightarrow G^{(p)}$ is an isogeny from a $K$-anisotropic absolutely simple semisimple $K$-group onto a $K$-split one. And of course this is a non-central isogeny. I suspect that this is the kind of phenomenon you were trying to find when formulating the OP.

[EDIT: I should probably have explained why in the case of central isogenies there are no surprises. That is, if $f:G \rightarrow G'$ is a central isogeny between connected reductive $K$-groups for a field $K$ (i.e., the scheme-theoretic kernel $\ker f$ centralizes $G$ in the functorial sense) and if $T' \subset G'$ is a maximal $K$-torus then the scheme-theoretic preimage $T := f^{-1}(T')$ is a maximal $K$-torus of $G$ (in particular, smooth and connected, even if $f$ is inseparable or has disconnected kernel). The reason is that to prove the $K$-group scheme $T$ is a maximal $K$-torus it suffices to do so after a ground field extension, since by Grothendieck's theorem on the "geometric maximality" of maximal tori over the ground field in a smooth connected affine $K$-group we do not lose the maximality hypothesis on $T'$ after a ground field extension. Hence, we may assume $K$ is algebraically closed.

With $K = \overline{K}$, all choices of $T'$ are $G'(K)$-conjugate and the map $G(K) \rightarrow G'(K)$ is surjective, so it suffices to treat the case of a single $T'$. Ah, but then we simply choose a maximal $K$-torus $S$ in $G$, so $T' := f(S)$ is a maximal $K$-torus in $G' = f(G)$, and thus the problem is to show that the inclusion $S \subset f^{-1}(f(S))$ of $K$-group schemes is an equality. Since $f$ is necessarily faithfully flat, so $G' = G/(\ker f)$ as fppf group sheaves, it suffices to show that $\ker f \subset S$ as subfunctors of $G$. Since $\ker f$ is central in $G$ by hypothesis, so it centralizes $S$, and hence it suffices to show that the scheme-theoretic centralizer $Z_G(S)$ of $S$ is equal to $S$. We know equality on $K$-points by the classical theory, so one just has to show that the group scheme $Z_G(S)$ is smooth, which is to say that it has the "expected" tangent space (i.e., that of $S$). This is a problem on dual numbers, and is proved in section 9 of Chapter III of Borel's textbook in a more classical language.]


Since the question is formulated in the language of the papers by Borel-Tits, I'd recommend $\S 22$ of Borel's GTM 126 Linear Algebraic Groups (on central isogenies) as a reasonable reference. See especially 22.6-22.7. Borel and Tits developed their language for a connected reductive group defined over an arbitrary field, avoiding the greqter generality of scheme theory but also being well aware of the advantages which scheme language can provide when the field of definition is imperfect. (SGA3 and the more recent writings by Milne, Conrad-Gabber-Prasad, provide an alternative route into the complexities of working over function fields or over more general rings.)

As long as your isogeny is central, there is not much difficulty in comparing an adjoint group with any covering group. Otherwise, as pointed out by user36938, life is more complciated. It's certainly essential to make appropriate choices of language and source materials when dealing with these kinds of questions. In any case, the basic facts have been written down pretty clearly a long time ago. Applying these facts to groups over function fields and the like remains a challenge.


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