# Rost Invariant of $E_7$

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 $(a) \cup (b) \cup (c)$.

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 $D_4$).
• That is interessting to hear (i mean the statment: $r(G)$ is symbol -> kernel of $G$ is $D_4$. I hoped that in case someone posts a proof (disproof), that proof would contain something about $r(G)$ in the case when $G$ has kernel $D_6$ and is not strongly inner.
• Ok, i knew there something unusal about your post: The Dynkin index of $E_7$ is 12. For $E_8$ it is 60, but I did not realize it either. But still you are right about the $\mu_4$ coefficients.
• Actually, the anisotropic kernel is $D_6$ if and only if $r(G)$ is a sum of two symbols with a common slot. This is proved in my paper with Garibaldi and Semenov "Shells of twisted flag variteits and the Rost invariant" Oops, but this is only for strongly inners forms, sorry for that. Commented May 14, 2018 at 17:52