Hasse diagrams of G/P_1 and G/P_2

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.

Answer

Attribution
Source : Link , Question Author : nxir , Answer Author : Victor Petrov