This can be considered as a continuation of my last useful question:


[Constructing groups of Type E7 with certain Tits Index](https://mathoverflow.net/questions/242664/constructing-groups-of-type-e7-with-certain-tits-index)


It is known that a quadratic form $q$ of dimension $12$, having splitting pattern $(2,4)$ (Vishik notation), gives an algebraic group with Tits index $E^{66}
\_{7,1}$ (anisotropic kernel is of type $D\_6$) and trivial Tits algebra, by a Galois cohomological construction (see my old question for that.) 


It is known that there are also groups with such Tits index, having Tits algebras with index $2$ or $4$.


For quite some time, i was under the impression that these are realized in the same manner, by taking some dimension $12$ quadratic form $q\in I^2$ (since we want none trivial Tits algebras), check back with an appropriate splitting pattern (all splitting patterns for dimension $12$ forms are classified) and we are done. I never really checked the details.


Now i did and it does not look like things work that way this time. /#/


**Question: Can you give a method for constructing algebraic groups with Tits index $E^{66}\_{7,1}$ and Tits algebras with index $2$ or $4$ ?**


/#/ The index $2$ case might be easier. Writing 6 = n = 2\*r and d = 2 (these are the parameters from Tits famous classification of indexes, meaning there are $r=3$ orbits) one can see that there should be three "splitting steps", giving the sequence $D\_6, D\_4$ x $A\_1, A\_1$ x $A\_1$ x $A\_1, \emptyset$ of anisotropic kernels. This would correspond to the splitting pattern $(1,2,3)$, which does not occur.



Constructing groups of Type E^{66}_{7,1} having non trivial Tits algebra
This question was posed by Jacques Tits on page 215 of his 1971 paper "Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque". (He emphasizes: "It would be interesting to know if the case of index 4 can be presented effectively.") Here is an outline of a construction that produces all of the desired groups, and a proof that such groups occur.


As Tits points out there, by his Witt-type Theorem, isogeny classes of groups of type $E\_7$ whose semisimple anisotropic kernel has Dynkin diagram contained in the $D\_6$ subdiagram (i.e., whose Tits index is $E^{66}\_{7,1}$ or has more circles) are in bijection with isogeny classes of certain groups of inner type $D\_6$. Which groups? In the notation of the Book of Involutions, it is a half-spin group isogenous to SO$(A, \sigma)$ where (1) $A$ is a central simple algebra of degree 12, (2) $\sigma$ is an orthogonal involution with trivial discriminant and (3) the Clifford algebra $C(A, \sigma)$ --- which is a direct product of two central simple algebras of degree 32 --- has one of its two components a split matrix algebra. (Necessarily therefore the other is Brauer-equivalent to $A$.) Conversely, such a group determines $(A, \sigma)$ up to isomorphism. Here and below I assume characteristic different from 2 for convenience.


Such groups of type $D\_6$ were studied in my paper with Quéguiner "Pfister's Theorem for orthogonal involutions of degree 12" <http://dx.doi.org/10.1090/S0002-9939-08-09674-3> Using the existence of an open orbit in the projectivization of the natural representation of a half-spin group of order 12, it is shown in Theorem 3.1 that such an $(A, \sigma)$ is constructed from a central simple algebra of degree 6 with unitary involution $(B, \tau)$ in a such a way that $A$ is Brauer-equivalent to the discriminant algebra $D(B,\tau)$ of $(B, \tau)$ as defined in the Book of Involutions. (Note that $(B, \tau)$ determines $(A, \sigma)$ but not necessarily vice versa.)


So far: each $(B, \tau)$ where $B$ has degree 6 and exponent 1 or 2 produces an $(A, \sigma)$ and in turn one of the desired groups of type $E\_7$, and all of the desired groups of type $E\_7$ are obtained in this way.


Now it remains to prove that such $E\_7$ exist by producing the desired $(B, \tau)$. That is, to construct $(B, \tau)$ such that $D(B, \tau)$ has index 2 or 4 (so that that resulting $E\_7$ has Tits algebra of index 2 or 4) and the resulting $\sigma$ is anisotropic (so that the Tits index of the $E\_7$ is precisely $E^{66}\_{7,1}$).


In case $D(B, \tau)$ has index 4, then the resulting $\sigma$ is necessarily anisotropic. Indeed, if the resulting $\sigma$ were isotropic, then the Tits index of the resulting $E\_7$ would have more than the 1 circle we started by assuming, yet examining the list of possible indexes for $E\_7$ and comparing with Tits algebras for smaller groups we find that the Tits algebra of such an $E\_7$ has index at most 2.


On page 148 of the Book of Involutions, Exercise 13, you find outlined a concrete construction of a $(B, \tau)$ where $D(B,\tau)$ has Tits index 4. This proves existence for the index 4 case of groups of type $E^{66}\_{7,1}$.


For index 2, one can construct the desired $(B, \tau)$ using Example 2.3 of the paper "Pfister's Theorem..." referenced above by picking a 1-Pfister quadratic form $q$ (norm of a quadratic extension $K$), a 6-dimensional trivial discriminant quadratic form $\psi$, and a quaternion algebra $Q$ split by $K$. The referenced example takes $B = M\_6(K)$ and $\tau$ is adjoint to the $K/F$-hermitian form constructed from $\psi$, and one gets $(A, \sigma)$ with $A$ Brauer-equivalent to $Q$. If you pick $q$, $\psi$, $Q$ generically (using indeterminates for the parameters), then after base change to the function field of the Severi-Brauer variety of $Q$, the algebra $Q$ is split and $\sigma$ is adjoint to $\psi \otimes q$, which remains anisotropic, and therefore the half-spin group isogenous to SO$(A, \sigma)$ must have been anisotropic over the original field. This proves existence for the index 2 case of groups with Tits index $E^{66}\_{7,1}$.


