Constructing algebraic groups of type E6 with split Tits algebras Let us assume our base field $k$ has characteristic zero.
From a series of papers by Borel and Siebenthal it is known that there is an embedding of groups
$A_2 \times A_2 \times A_2$ into $E_6$.
This gives a map
$H^1(k, A_2 \times A_2 \times A_2) \rightarrow H^1(k, E_6)$.
Let us consider the map
$A_2 \rightarrow A_2 \times A_2 \times A_2$, which sends the CSA $D$ to $(D,D,D)$.
If we apply Galois cohomolgy again and compose this, we obtain
$H^1(k, A_2) \rightarrow H^1(k, E_6)$.
I assume the Tits algebra of every group constructed this way is
$D\otimes D\otimes D \in H^2(k,\mu_3)$ and thus trivial as the period of
$D$ is $1$ or $3$.
Also it is known, that every group of type $E_6$ (considered "mod 3"), which has trivial Tits algebras is either anisotropic or split.
Questions: 


*

*Does my attempt of constructing an $E_6$ make sense?

*Is it known in the literature?

*Is the resulting group anisotropic iff $D$ has index $3$ (i.e iff $A_2$ is anisotropic) ?

 A: *

*Yes, it definitely makes sense, but you should be slightly more accurate. The actual group sitting inside $E_6$ is indeed semisimple of type $A_2+A_2+A_2$, but it is $(SL_3\times SL_3\times SL_3)/\mu_3$ if your $E_6$ is simply connected and $(SL_3\times SL_3\times SL_3)/\mu_3^2$ if it is adjoint. So not every cocycle from $H^1(F,PGL_3\times PGL_3\times PGL_3)$ lifts to this group, and even if it does, the lifting is not unique but depends on some constants coming from the long exact sequence in cohomology. Say, in the case of $E_6^{sc}$ you have two restrictions (basically they say that you indeed have just three copies of $D$) and two constants, but the group itself (that is the image in $H^1(F,E_6^{ad})$) depends only on one constant (say, $t$).

*In the simply connected case what you get is called "the first Tits construction". Usually it is described in terms of the exceptional Jordan (Albert) algebra. However, in the adjoint case (when you have one restriction and one constant) I am not aware if this construction explicitly mentioned in the literature.

*The first Tits consruction is anisotropic iff $[D]\cup (t)$ is nonzero in $H^3(F,Z/3)$ (that is $D$ must be division and $t$ is not in the image of the reduced norm of $D$).
