2
$\begingroup$

Let $A$ be a central, simple algebra of degree $2n$ with orthogonal involution $\sigma$.

Let SB$(A)$ the Severi-Brauer variety, which is the variety of left ideals of reduced dimension one in $A$.

The involution variety Inv$(A,\sigma)= \{ I \in $SB$(A)$ $|$ $\sigma(I)I = 0\}$ is known to be a hyperplane section of SB$(A)$.

We define a generalized SB$(A)_{l,n}$ as the variety of left ideals in $A$ of reduced dimension $l$, when $deg(A) = n$. In an analogue way one can define a generalized involution variety Inv$(A,\sigma)_{l,n}$.

I asked myself, if for $l$ other than $1$, Inv$(A,\sigma)_{l,n}$ is a hyperplane section of SB$(A)_{l,n}$. Calculating the dimensions i think i found that this is not possible. However we still can ask

Question: For central, simple algebras $A,B$ of degrees $n,2m$, and for $\sigma$ beeing an orthogonal involution on $B$. Can there be $l,k$ such that Inv$(B,\sigma)_{k,2m}$ is a hyperplane section of SB$(A)_{l,n}$ ?

What inspired me to think about such things is a result from

D. Krashen - Zero cycles on homogeneous varieties,

stating that for $A$ of degree $4$ and $B$ of degree $6$ there is an isomorphism between Inv$(B,\sigma)_{1,6}$ and SB$(A)_{2,4}$.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ If you had an isomorphism of $\text{Inv}(B,\sigma)_{k,2m}$ and $\text{SB}(A)_{\ell,n}$, resp. a hyperplane section thereof, then after an etale extension of your original base, you would have an isomorphism between the orthogonal Grassmannian of isotropic $k$-dimensional subspaces in a vector space of dimension $2m$ and the Grassmannian of $\ell$-dimensional subspaces of an $n$-dimensional vector space, resp. a hyperplane section thereof. As Krashen points out, this is actually the case for $(k,2m)=(1,6)$ and $(\ell,n)=(2,4)$, i.e., $\text{Grass}(2,4)$ is a quadric fourfold . . . $\endgroup$
    – Jason Starr
    Commented Nov 27, 2017 at 1:42
  • $\begingroup$ . . . For the dimensions to work out, you need that $k(4m-3k-1)/2$ equals $\ell(n-\ell)$, resp. equals $\ell(n-\ell) - 1$. Then you can also compute the dimension of the Lie algebra of infinitesimal automorphisms of both sides. For the orthogonal Grassmannian, it is the dimension of the special orthogonal group, $m(2m-1)$. For the Grassmannian, it is the dimension of the special linear group $n^2-1$. For a hyperplane section that happens to be homogeneous (so that $h^1$ of the tangent bundle equals $0$), the dimension is $n^2-\binom{n}{\ell}$. $\endgroup$
    – Jason Starr
    Commented Nov 27, 2017 at 2:27
  • $\begingroup$ I tried to find other solutions of the two Diophantine equations from my comment above, and it looks to me like there are very few solutions (maybe none, beyond the case $(k,2m)=(1,6)$ and $(\ell,n)=(2,4)$). So that might settle your question. $\endgroup$
    – Jason Starr
    Commented Nov 28, 2017 at 10:38
  • $\begingroup$ I didnt have the time yet to do any calculations using computers. What did you use to figure that out? $\endgroup$
    – nxir
    Commented Nov 28, 2017 at 23:40
  • $\begingroup$ I do not know understand your last comment. $\endgroup$
    – Jason Starr
    Commented Nov 29, 2017 at 0:22

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .