Constructing groups of Type E^{66}_{7,1} having non trivial Tits algebra

Question

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

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.

Answer 1

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 $E^{66}_{7,1}$.