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Let $F$ be a local p-adic field and $G$ a semisimple simply connected group over $F$, $\mathfrak{g}$ its Lie algebra. Let $T$ a maximal anisotropic torus of $G$, split over an etale extension of $F$ and $\gamma\in T(F)$, can we write $\gamma=exp(g)$ for some element $g\in\mathfrak{g}(F)$?

We can assume at first that $F$ is of characteristic zero and then for general characteristics is $p$ good sufficient?.

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    $\begingroup$ Given that the question involves the exponential map, the statement at the end that we can "assume at first" that $F$ has characteristic 0 seems redundant (setting aside that the phrase "$p$-adic field" is generally understood to include characteristic 0 as well). Also, probably "maximal anisotropic torus" should be "anisotropic maximal torus" (in English the properties build up from right to left). Also, tori always split over a separable finite extension, so the condition "split over an etale extension of $F$" seems redundant as well. $\endgroup$
    – user27920
    Commented Nov 13, 2014 at 8:36
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    $\begingroup$ Since $\mathfrak{g}(F)$ is totally disconnected, please make more precise what you mean by ${\rm{exp}}(g)$. Do you mean that you are considering ${\rm{exp}}$ as a map of connected rigid-analytic spaces, from a maximal connected admissible open domain around 0 inside $\mathfrak{g}$ (viewed as an analytic affine space over $F$) into $G^{\rm{an}}$? $\endgroup$
    – user27920
    Commented Nov 13, 2014 at 8:42
  • $\begingroup$ we can have exponential map in characteristic $p$ as well and if you consider the torus $Res_{E/F}\mathbb{G}_{m}$ for an inseparable extension $E$, it is a counter-example to what you say. $\endgroup$
    – prochet
    Commented Nov 13, 2014 at 17:42
  • $\begingroup$ What do you mean intrinsically by "exponential map in characteristic $p$"? It would help to see a definition in the question, as this is not a standard notion (away from certain unipotent groups), as far as I am aware. Also, the Weil restriction ${\rm{R}}_{E/k}(\mathbf{G}_m)$ for non-separable $E/k$ is actually never a torus (the geometric fiber has a non-trivial unipotent radical corresponding to the 1-units in $E \otimes_k \overline{k}$). What I said about tori is really correct, and is proved in the usual books on algebraic groups. $\endgroup$
    – user27920
    Commented Nov 13, 2014 at 18:24

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