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Let $k$ be a real closed field. Let $G$ be a anisotropic algebraic group of type $F_4$ over $k$. Consider the projective, homogenous $F_4$-variety $X_4$ (Bourbaki enumeration).

To avoid confusion: Note that the possible Tits indexes of any $F_4$ are anisotropic, split and $X_4$ circled (which means the semisimple anisotropic kernel of $G$ is of type $B_3$).

Question: Is $G$ split over the function field $k(X_4)$ ?

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    $\begingroup$ I am not familiar with the Bourbaki enumeration. Could you please give a reference for the Bourbaki enumeration (i.e. where in Bourbaki)? $\endgroup$
    – Ben McKay
    Mar 31, 2018 at 6:16
  • $\begingroup$ Just look here on page number 5: math.univ-paris13.fr/~declercq/pindexes.pdf $\endgroup$
    – nxir
    Mar 31, 2018 at 14:42
  • $\begingroup$ I see a table on page 5 of the simple root systems, with an ordering of the roots, following Bourbaki. But I don't understand how that gives an enumeration of the homogeneous spaces, and in particular a meaning to the $4$ in $X_4$. Is there a standard enumeration of the projective homogeneous varieties of each simple algebraic group? $\endgroup$
    – Ben McKay
    Mar 31, 2018 at 16:37
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    $\begingroup$ Yes, actually the nodes in the diagramm correspond to parabolic subgroups of $G$. Lets define $P_i$ by the ones corresponding to every node but node number $i$. Then the quotient $G/P_i$ is called $X_i$. It is a projective homogenous variety as $P$ is parabolic and has the nice property that the the node $i$ is circled in the Dynkin diagramm iff $X_i$ has a rational point over $k$. $\endgroup$
    – nxir
    Mar 31, 2018 at 22:41

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A complete answer for questions of this kind is in my paper with Nikita Semenov "Generically split projective homogeneous varieties" in Duke (or refined version in J. K-Theory). In your situation this is the case if and only if G splits by a cubic field extension.

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  • $\begingroup$ The question was under the hypothesis that $k$ is a real-closed field. So the conclusion is that $G$ does not split over $k(X_4)$, is that right? $\endgroup$
    – Gro-Tsen
    May 3, 2018 at 21:35
  • $\begingroup$ @ Gro-Tsen That was exactly my point. $\endgroup$
    – nxir
    May 5, 2018 at 12:38
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    $\begingroup$ Oh, I didn't notice that. Yes, over real closed fields obviously there are no cubic extensions, so G does not split over that function field. $\endgroup$
    – Victor Petrov
    May 14, 2018 at 17:48

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