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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:

  1. Does my attempt of constructing an $E_6$ make sense?
  2. Is it known in the literature?
  3. Is the resulting group anisotropic iff $D$ has index $3$ (i.e iff $A_2$ is anisotropic) ?
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    $\begingroup$ I may be blind, but I’m not seeing any claimed construction of $E_6$ in the above...? $\endgroup$ Nov 29, 2018 at 6:13
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    $\begingroup$ Are you refering to the fact that this actually constructs $E_6$ torsors? Or do really mean it?Such constructions (i.e. considering certain sub groups of a split group and applying Galois cohomology to get maps of torsors) are known as Tits constructions. There are many such constructions but some are less obvious than others (for some $E_8$'s take $F_4$ and $G_2$ for example). In case of $E_8$ one trick of Tits was to consider the map $F_4 \rightarrow F_4 \times G_2$, which sends an Albert algebra onto itself and its underlying Oction algebra to produce $E_8$'s with trivial Rost Invariant. $\endgroup$
    – nxir
    Nov 29, 2018 at 22:14
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    $\begingroup$ More likely I was blind — or to put it more mildly, not in the target audience for whom what you say in the comment should already be obvious from the question :-) $\endgroup$ Nov 29, 2018 at 22:41
  • $\begingroup$ @FrancoisZiegler that's indeed the point of Galois cohomology; roughly speaking it indexes $k$-forms of an object from a very partial information about the given object. $\endgroup$
    – YCor
    Nov 30, 2018 at 18:17

1 Answer 1

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  1. 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, $t$).

  2. 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.

  3. 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 $D$).

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  • $\begingroup$ Yes, I was being a bit imprecise on the simply connected and adjoint cases. In your paper with Semenov and Garibaldi on Shells you calculate the motives for the adjont case mod 3. So i thought it may be worth thinking about how the strongly inner $E_6$'s arise. Also I was a bit confused as Garibali has introduced a degree four invariant in his old paper on structurable algebras. But only for the simply connected case. This now makes sense to me, as you point out that theres another variable involved in that case. $\endgroup$
    – nxir
    Dec 1, 2018 at 11:23

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