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Suppose $G$ is a connected, reductive algebraic group over a nonarchimedian local field $F$, which splits over a finite extension $E/F$.

I frequently see a result stating that "all maximal $F$-tori are conjugate over $E$", by which I understand the following: Let $G(E)$ denote the $E$-points of the algebraic group $G$; then for any maximal $F$-tori $T, T'$ of $G$, there exists $x \in G(E)$ such that $T(E) = xT'(E)x^{-1}$.

Furthermore, it is clear from the definitions that if $T, T'$ are any maximal $F$-tori of $G$, then there exists an isomorphism of $T(F)$ onto $T'(F)$ which is defined over $E$.

My question is this: can the isomorphism in the second statement be assumed to be conjugation (as in the first statement)? That is to say: does it follow from these results that if $T, T'$ are any maximal $F$-tori in $G$, then there exists $x \in G(E)$ such that $T(F) = xT'(F)x^{-1}$?

Any help (including a proof of the first statement) is greatly appreciated!

itsricobitches
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    I might be totally off, but the second paragraph feels wrong. Some maximal $F$-tori might not split over $E$, and I doubt that split and non-split tori can be conjugate. As an example, take $E\vert F$ the unramified quadratic extension of some $p$-adic field, $G$ the multiplicative group of the quaternion division algebra over $F$. Then the maximal $F$-tori in $G$ should be (isomorphic to) multiplicative groups of quadratic extensions of $F$ (which are all contained in $G$), but e.g. the one of a ramified extension remains non-split over $E$. Cf. https://math.stackexchange.com/q/44154/96384. – Torsten Schoeneberg Feb 25 '20 at 07:03
  • The discussion in the linked post seems to be working over the field of definition $F$, and not the splitting field $E$ of the group $G$ - see theorem 6.4.1 of Springer's "Linear algebraic groups;" in the example you give, the maximal torus has rank 2, while the maximal split torus has rank one; the point of this theorem, as I see it, is that over $E$, all tori are split. This is easily seen if we assume $F$ is algebraically (and separably) closed, for instance in $GL_n$, where it reduces to the statement that every invertible matrix is diagonalizable. – itsricobitches Feb 25 '20 at 21:45
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    I should have clarified that my example is a bit different from the one in the linked post. However, I think both are valid (the latter one just has $E=F$). I think what needs to be pointed out that a group is called split over $F$ if there exists a split maximal $F$-torus, but this does not imply that all maximal $F$-tori of that group are split. (Of course, if $F$ happens to be algebraically closed, then every torus is split anyway, but basically over every non-algebraically closed field there are counterexamples.) – Torsten Schoeneberg Feb 26 '20 at 04:39
  • Here's maybe the easiest example, similar to the link: Let $G=GL_2$ and $F$ any field which has a proper quadratic extension $F(\sqrt{x})$. Then $G$ is already split over $E:=F$ (there is the obvious split maximal ($2$-dim.) torus of diagonal matrices); but there is also the maximal ($2$-dim.) torus $\lbrace \pmatrix{a &bx \b &a}:a,b \in F \rbrace$, which is non-split and not conjugate to the diagonal matrices (it's isomorphic to the multiplicative group of $K:=F(\sqrt{x})$). Note this torus would split over $K$, but as said, $G$ is already split over $E=F$, giving another counterexample. – Torsten Schoeneberg Feb 26 '20 at 04:47
  • Thank you - your italicized comment is illuminating! :) However, I am struggling to reconcile your example (which seems to give infinitely many non-$F$-isomorphic) tori with the following: the set of $F$-isomorphism classes of maximal tori in $G$ can be identified with a subset of $H^1(\mathrm{Gal}(F_s/F), N_G(T)(F_s))$; by theorem 6.14 of Platonov-Rapinchuck, this set is finite, so there should only be finitely many such $F$-isomorphism classes. Am I missing something? – itsricobitches Feb 26 '20 at 20:26
  • Well, in a fixed separable closure, a local field most often has only finitely many extensions of given degree $n$, so the above constructions give only finitely many isomorphism classes of maximal tori. The only exception is actually if $char(F)$ divides $n$, https://math.stackexchange.com/q/353928/96384. I don't know if Platonov-Rapinchuck's theorem covers this case, which then might need more reconciliation efforts, but e.g. for $p$-adic fields we're fine. – Torsten Schoeneberg Feb 27 '20 at 03:51

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