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 D4/P1 is isomorphic to a six dimensional quadric, that i will denote as Q6.
In an unfinished book by Gille, Petrov,N. Semenov and Zainoulline, which can be found on the last authors page, we can learn that:G2/P1 ≃Q5, while G2/P2 is isomorphic to a Fano variety.
Note also, that some authors use “reversed index” notation for denoting parabolic subgroups, but in the case of G2 there cant be too much confusion.
Checking the Hasse-diagrams in the first reference the case G2/P1 has seven vertexes. This would mean that G2/P1 can be isomorphic to Q6 and not to Q5.
This is contradicting. Where is the mistake?
The diagram for G2/P2 is obviously not representing a quadric, so messing indexes, as i feared, cant be the problem.
Answer
“Hasse diagrams” may refer to several different things. They are definitely coincide in a microweight case (say, for E6/P6 and E7/P7), however, G2 has no microweight representations, and you need to settle zero weights (one zero weight in the case of G2/P1) somehow.
Attribution
Source : Link , Question Author : nxir , Answer Author : Victor Petrov