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.



Hasse diagrams of G/P_1 and G/P_2
"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.


