Let $E\_7$ denote the split group of type $E\_7$. Assume $G := \xi\overline{G}$ is a semisimple algebraic group over a field $k$ with characteristic zero for some $\xi \in H^1(k,E\_7)$. Let $r(G)$ $\in$ $H^3(k,\mu\_2)$ be the even part of the Rost Invariant. Assume further that the Tits Index of $G$ has all but three points circled (these are necessarily number $2,5,7$ in Bourbaki notation).


Note that the classical Rost invariant is defined for simply connected groups.
In the adjoint case one needs to consider some generalized version.


Question: Is $r(G)$ a pure symbol in $H^3(k,\mu\_2)$?


For me pure symbol in is an element of the form $(a) \cup (b) \cup (c)$.



Rost Invariant of $E_7$
Well, you can't obtain a group with such Tits index using cocycle from simply connected split $E\_7$, because its Tits algebras are non-trivial (they are Brauer equivalent to a quaternion algebra). And for non simply connected groups the Rost invariant is not defined.


By the way, even when it is defined, its 2-part is from $H^3(k,\mu\_4)$, not $H^3(k,\mu\_2)$ in general. That is because the Dynkin index for $E\_7$ is 60. It is known that if it is, nevertheless, a pure symbol from $H^3(k,\mu\_2)$, then your twisted group is indeed isotropic but with circled vertices 1, 6, 7 (so the anisotropic kernel is of type $D\_4$).


