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in the Paper http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.30.5052&rep=rep1&type=pdf at the end, we can see Hasse diagrams for several projective, homogeneous $G$-varieties for $G$ being a exceptional linear algebraic group.

Note that $D_4/P_1$ is isomorphic to a six dimensional quadric, that i will denote as $Q^6$. In an unfinished book by Gille, Petrov,N. Semenov and Zainoulline, which can be found on the last authors page, we can learn that:

$G_2/P_1$ $\simeq Q^5$, while $G_2/P_2$ is isomorphic to a Fano variety.

Note also, that some authors use "reversed index" notation for denoting parabolic subgroups, but in the case of $G_2$ there cant be too much confusion.

Checking the Hasse-diagrams in the first reference the case $G_2/P_1$ has seven vertexes. This would mean that $G_2/P_1$ can be isomorphic to $Q^6$ and not to $Q^5$.

This is contradicting. Where is the mistake?

The diagram for $G_2/P_2$ is obviously not representing a quadric, so messing indexes, as i feared, cant be the problem.

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    $\begingroup$ $G_2/B$ is 6-dimensional. That could be the source of confusion. The other 2 $G_2$-varieties are 5-dimensional. $\endgroup$
    – Ben McKay
    Jul 17, 2015 at 21:05
  • $\begingroup$ Thats the problem! By the way. Do you have a source for your claim or some tables with chow rings of G/P_i for several G? $\endgroup$
    – nxir
    Jul 17, 2015 at 21:43
  • $\begingroup$ You could look at page 20 of my paper arxiv.org/abs/0704.2555 $\endgroup$
    – Ben McKay
    Jul 18, 2015 at 7:59
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    $\begingroup$ This being the Internet, there's no need to be coy about where books are located; you can just link to them. The Gille–Petrov–Semenov–Zainoulline book is Introduction to motives and algebraic cycles on projective homogeneous varieties. $\endgroup$
    – LSpice
    Apr 28, 2018 at 23:29

1 Answer 1

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"Hasse diagrams" may refer to several different things. They are definitely coincide in a microweight case (say, for $E_6/P_6$ and $E_7/P_7$), however, $G_2$ has no microweight representations, and you need to settle zero weights (one zero weight in the case of $G_2/P_1$) somehow.

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