Let $F$ be a field and $A$ be an $F$-central simple algebra of degree $n$. Let $0< k< n$ and let $SB\_k(A)$ denote the generalized Severi-Brauer variety: if $E/F$ is a field extension, $SB\_k(A)(E)$ consists of the right ideals of dimension $kn$ of $A\_E=A\otimes\_F E$.


If $A$ is split, i.e. $A\simeq M\_n(F)$, then $SB\_k(A)=Gr(k,n)$, the Grassmannian.



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The result is true if $k=1$, since $SB\_k(A)$ has a rational point over $F$ iff the index of $A$ divides $k$, and Grassmannians have rational points over $F$.


This is cross-posted from Math.SE, where I posted it some days ago, with a few upvotes but no answers or comments. I hope it is ok for MO.
<https://math.stackexchange.com/questions/618616/when-are-generalized-severi-brauer-varieties-trivial>



When are generalized Severi-Brauer varieties Grassmannians?
Let's first phrase this in terms of non-abelian cohomology. You have a map $$H^1(F, PGL\_n)\to H^1(F, \operatorname{Aut}(Gr(k, n)))$$ given by sending a central simple algebra to the associated generalized Severi-Brauer. You'd like to know if this map has trivial kernel (it's just a map of pointed sets, so by this I just mean that the preimage of the distinguished element is a singleton). If $n\not=2k$, a theorem of Chow tells us that the obvious map $PGL\_n\to \operatorname{Aut}(Gr(k, n))$ is an isomorphism; it's not hard to see that the map on $H^1$ is induced by this map, hence it is also an isomorphism.


If $n=2k$, the map $PGL\_n\to \operatorname{Aut}(Gr(k, n))$ is a closed embedding with image a normal subgroup of index $2$ (by the same theorem of Chow). So there is an exact sequence (of pointed sets) $$H^0(F, \operatorname{Aut}(Gr(k, n)))\to H^0(F, \operatorname{Aut}(Gr(k, n))/PGL\_n)\to H^1(F, PGL\_n)\to H^1(F,\operatorname{Aut}(Gr(k, n))).$$


But the first map is surjective (it's just the component map, since both components of $\operatorname{Aut}(Gr(k, n))$ have rational points) so again the map in question has trivial kernel.


So it looks like the converse is indeed true, if I've remembered Chow's description of $\operatorname{Aut}(Gr(k, n))$ correctly.


