Let $G$ be a reductive (or just semisimple) algebraic group over an algebraically closed field $k$ of characteristic $p > 0$, let $T$ be a maximal Torus and let $f:G \rightarrow G$ be an isogeny. Suppose the induced morphism of root data $f^\star: \Psi(G,T) \rightarrow \Psi(G,T)$ is a Frobenius morphism multiplying roots by $q = p^r$ (confer 9.6.3 in Springer's book on algebraic groups). Is it true that $f$ is the geometric Frobenius of an $\mathbb{F}_{q}$-structure on $G$?

Equivalent formulation: Is the morphism $F:G \rightarrow G$ defined by $F( u_\alpha(c) ) = u_\alpha(c^q)$ for all roots $\alpha$ and $F(t) = t^q$ for all $t \in T$ the geometric Frobenius of an $\mathbb{F}_{q}$-structure on $G$?

I'm pretty sure this is true (perhaps under some additional conditions). I know that there is an easy characterization of Frobenius morphisms in terms of the comorphism of the coordinate rings but I was not able to verify this.


The answer is "yes", but not in a good way: the descent it arises from is the split form, so this does not encode an interesting $\mathbf{F}_ q$-structure. More precisely, $f$ arises from the $q$-Frobenius of a split descent down to $\mathbf{F}_ q$ (and so even to the prime field $\mathbf{F}_ p$). In particular, this is definitely not the way to encode the information of "interesting" descents of $(G,T)$ to $\mathbf{F}_ q$, if that may be your ultimate intention.

To see this, let $(G_0, T_0)$ be the split form over $\mathbf{F}_ q$, and $F_0:(G_0, T_0) \rightarrow (G_0,T_0)$ the $q$-Frobenius morphism. This induces the self-map of the root datum given by $q$-multiplication on ${\rm{X}}(T_0)$. Extending scalars to $k$ has no effect on the root datum for split groups, so $(F_ 0)_ k$ induces the endomorphism of the root datum of $((G_ 0)_ k, (T_ 0)_ k)$ given by $q$-multiplication on the root datum. Pick a $k$-isomorphism between the $k$-split pairs $(G,T)$ and $((G_ 0)_ k, (T_ 0)_ k)$ (as we may, by the isomorphism Theorem). The resulting isomorphism of root data identifies the maps from $(F_ 0)_ k$ and $f$ since $q$-multiplication is functorial with respect to all homomorphisms between abelian groups. Thus, these maps coincide up to the action of some $t \in \overline{T}(k) = (\overline{T}_ 0)(k)$, where $\overline{T} := T/Z_G$ and $\overline{T}_ 0 := T_ 0/Z_ {G_ 0}$ are the "adjoint tori''. Using an isomorphism $\overline{T}_ 0 \simeq \mathbf{G}_m^r$, $t$ goes over to some $r$-tuple $(t_i)$ with $t_i \in k^{\times}$. Since $k$ is separably closed, we can solve $t_i = y_i^{q-1}$ with $y_i \in k^{\times}$. In other words, $t = f(y)/y$ for some $y \in T_0(k)$. So if we modify the identification of $(G_0, T_0)$ as an $\mathbf{F}_ q$-descent of $(G,T)$ by composing with the action of $y$ or $y^{-1}$ (depending on direction of maps) then $f = (F_ 0)_ k$ via the new identification of $(G_ 0, T_ 0)$ as a split $\mathbf{F} _q$-descent of $(G,T)$. QED

Remark: A refinement of the argument (using some care if $k$ is not algebraic over $\mathbf{F}_ p$) proves that the split descent to $\mathbf{F} _q$ is the only one which can work, but I won't get into that here. (It comes down to the fact that the solutions to $y^{q-1} = 1$ in $k^{\times}$ lie in $\mathbf{F} _q$.)

Note that the above works over any separably closed field of characteristic $p > 0$, not necessarily algebraically closed, provided the Isomorphism Theorem is valid over such fields. The Existence, Isomorphism, and even Isogeny Theorems are valid for split connected reductive groups over any field. The paper of Steinberg mentioned in the previous answer is really nice, but unfortunately assumes from the outset that the ground field is algebraically closed. Good news, in case you may care, is that one can deduce the results over any field from that case as a "black box'' by using descent theory. See Appendix A.4 of the book "Pseudo-reductive groups''.

  • $\begingroup$ Thanks a lot! I have one additional question because I'm not entirely sure: what is "the split form over F_q"? Is this the base extension of Demazure's Z-group scheme with root datum equal that of G to F_q? If so, is there some uniqueness which justifies the "the"? $\endgroup$ – user717 Mar 6 '10 at 17:35
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    $\begingroup$ Yes, and yes. For a field k, a conn'd reductive k-group G is called split if there is a max k-torus T that is k-split. (Any max k-torus remains maximal after extension on k; not obvious, but important.) To any such pair we assign a root datum. Existence Thm says all root data arise this way (for any k). Isom Thm says for 2 k-split pairs (G,T) and (G',T'), an isom between root data arises from a k-isom between the pairs (unique up to (T/Z_G)(k)-action on G). Using a pinning removes the (T/Z_G)(k) stuff; see Theorem A.4.13 in "Pseudo-reductive groups". SGA3 does analogue over Z, etc. $\endgroup$ – BCnrd Mar 6 '10 at 17:52
  • $\begingroup$ Isn't your "yes but not in an interesting way" answer a little too negative? If you extend the notion of an isogeny of the root datum as in Steinberg's paper, then any Frobenius comes from an isogeny of root datum, so the isogeny theorem does capture $\mathbb F_q$-descents no? $\endgroup$ – Kevin McGerty Mar 7 '10 at 14:26
  • $\begingroup$ Do you mean to say an F_q-descent of (G,T) "is" a suitable isomorphism h:(G,T) ---> (G^{(q)},T^{(q)}), which we express as an isogeny f:(G,T) ---> (G,T) with kernel ker(F_{G/k})? This says R(f):R(G,T) --> R(G,T) has image q X(T) on X(T)'s. In the question, R(f) = q "on the nose" and you are suggesting to allow other R(f)'s? That's a nice idea. If k is algebraic over F_q then need "continuity" condition on f, perhaps (1/q)R(f):X(T) ---> X(T) has finite order. Is that exactly the right condition? If k not algebraic over F_q then more care is needed (but maybe it works). Please let me know. $\endgroup$ – BCnrd Mar 7 '10 at 15:06
  • $\begingroup$ One further thing: "finite order" on (1/q)R(f) isn't quite right either if k/F_q is a finite extension. Then perhaps it should have order dividing [k:F_q]. If k/F_q is algebraic of infinite degree but not an algebraic closure, then it looks even more delicate. Anyway, I think it would be very interesting if one could prove a precise statement along such lines, but it sounds a bit delicate. Perhaps the only case to expect a good answer, and maybe the most interesting case for application to finite groups, is k = Fbar_q. Let me know what you get. $\endgroup$ – BCnrd Mar 7 '10 at 15:08

The geometric Frobenius is an isogeny of the algebraic group to itself, since it clearly has finite kernel. A mild extension of the "isomorphism theorem" for algebraic groups of arbitrary characteristic classifies isogenies in terms of their action on root data, as you require. The theorem says that an "isogeny" of root data corresponds, uniquely up to conjugation, to an isogeny of groups. The cases you are interested in occur via Frobenius morphisms and hence those morphisms are determined by their action on root systems. This is all discussed in Steinberg's paper, so hopefully that's a sufficient reference.

  • $\begingroup$ Hm, by the isogeny theorem I know that to a Frobenius morphism of root data there corresponds an isogeny of the group (defined as in "equivalent formulation"), ok. But why is this isogeny the Frobenius of some F_q-structure? Am I missing something obvious? $\endgroup$ – user717 Mar 6 '10 at 17:38
  • $\begingroup$ Well it is "obvious" if you know there is a split form of the algebraic group over $\mathbb F_q$, which I was assuming, (as Brian also mentions in his answer above, the classification of split groups goes through over any field). I thought you were just asking about characterization, not the existence though? $\endgroup$ – Kevin McGerty Mar 7 '10 at 14:10

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