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You have an anistropic semisimple algebraic group $G$ defined over a non-archimedean local field $k$. When can you say that the $k$-rational part of the conjugacy class of a $k$-rational point is compact?

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The answer is always. If $G$ is an anisotropic group over a non-archimedean local field $k$, then $G(k)$ is compact. The elements of $G(k)$ are semi-simple (if $char k =0$) and hence their conjugacy classes are closed in the Zariski topology and therefore also in the $k$-topology.

[Edit] To round it off, the question may easily be reduced to the case when $G$ is semi-simple and simply connected; hence $G$ is a product of simply connected $k$-simple groups $G_i$, and by a restriction of scalars argument, $G$ may be assumed to be absolutely almost simple simply connected. In this case, the classification (see Tits' article in the AMS symposia series on algebraic groups, discrete subgroups...) says that $G$ is $SL_1(D)$ for a central division algebra over $k$ or $SU_1(D)$ where $D$ is a central division algebra with an involution of the second kind. The semi-siple conjugacy classes may now be described explicitly.

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  • $\begingroup$ It's probably helpful to give a reference for the first sentence. At least in characteristic 0, this is made explicit in Cor. 9.4 of Groupes reductifs (IHES Publ. Math. 1965) by Borel-Tits, available online at numdam.org. (Note too the spelling "Zariski".) $\endgroup$ Commented Jan 10, 2014 at 14:31
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    $\begingroup$ I find it quite interesting that you say that the elements of $G(k)$ are always semisimple; are you able to give me a reference for that? $\endgroup$
    – Rupert
    Commented Jan 10, 2014 at 14:53
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    $\begingroup$ It is easy to give a proof (in char 0). If $g\in G(k)$ then write $g=g_sg_u$ (Jordan decomposition). If $g_u\neq 1$, then every unipotent element is char zero lies in the unipotent radical of a parabolic subgroup $P$ defined over $k$, and hence the group $G$ is isotropic over $k$ (you can also use Jacobson-Morozov, to get a split torus). $\endgroup$ Commented Jan 10, 2014 at 15:28
  • $\begingroup$ Okay, great, thanks, what happens in positive characteristic? Is it still true then that elements of $G(k)$ are always semisimple? $\endgroup$
    – Rupert
    Commented Jan 13, 2014 at 8:25
  • $\begingroup$ Jim Humphreys' comments contain an answer and references to your new question $\endgroup$ Commented Jan 14, 2014 at 3:05
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Since the literature is rather complicated (and scattered) when $k$ fails to be perfect, it's worthwhile to add to what Aakumadula says. Over a perfect field, say in characteristic 0, the answer to your questions is straightforward based on older work of Borel and Tits. As I commented, Cor. 9.4 in their foundational paper on reductive groups here shows that (over a local field of characteristic 0) $G_k$ is compact in the ultrametric topology iff $G$ is reductive and $k$-anisotropic. Moreover, their Cor. 8.5 shows in this anisotropic situation that $G_k$ consists of semisimple elements. (As in their $\S1$ and the earlier Chevalley seminar, it's true quite generally that in a reductive group the semisimple classes are precisely the closed ones.)

There are many more relevant papers by Tits, sometimes in collaboration with Borel or Bruhat, in which he explores the structure and classification of reductive groups over local fields. An interesting early paper in a Brussels conference volume includes some remarks on the terms $k$-isotropic and $k$-anisotropic, which arise in part from the study of quadratic forms and orthogonal groups but tend to contradict existing notions of "isotropic". In any case, these terms have persisted. This paper is probably hard to track down (and unfortunately there is no collection of his papers): Groupes semi-simples isotropes. 1962 Colloq. Th´eorie des Groupes Alg´ebriques (Bruxelles, 1962) pp. 137–147. Librairie Universitaire, Louvain; Gauthier-Villars, Paris One result stated here by Tits shows how delicate the cas of imperfect fields is: here a simple algebraic group $G$ defined over $k$ is $k$-anisotropic if and only if $G_k$ contains no "good" unipotent element other than the identity (in a certain sense of "good" explored further in a paper with Borel in Invent. Math. 12 (1971), which can be found online at GDZ.

For a more modern treatment of structure theory of reductive groups over arbitrary (especially local) fields, you should probably look at the monograph Pseudo-reductive Groups by Conrad-Gabber-Prasad (Cambridge Univ. Press, 2010).

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