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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.

Is the converse true? If not, can you provide a counterexample?

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

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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.

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  • $\begingroup$ Actually, now that I think about it, we don't need to know what $\operatorname{Aut}(Gr(k, n))$ is--we just need to know that each connected component has an $F$-rational point, which I don't think is too hard to see. $\endgroup$ – Daniel Litt Dec 30 '13 at 20:00
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Daniel Litt's answer is completely correct, of course. You can also see the result using projective geometry (perhaps this is how Chow proves the result). For every $1$-dimensional subspace $L$ of $F^{\oplus n}$, there is an induced embedding, $$\phi_L:\text{Grass}(k-1,F^{\oplus n}/L) \hookrightarrow \text{Grass}(k,F^{\oplus n}),$$ obtained by associating to every subspace $\overline{V}$ of $F^{\oplus n}/L$ the inverse image $V$ in $F^{\oplus n}$. Denote by $P(t)$ the Hilbert polynomial of $\text{Image}(\phi_L)$ with respect to the ample invertible sheaf $\omega^\vee$, where $\omega$ is the dualizing sheaf of $\text{Grass}(k,F^{\oplus n})$. Then there is a well-defined morphism, $$\Phi:\text{Grass}(1,F^{\oplus n}) \to \text{Hilb}^{P(t)}_{\text{Grass}(k,F^{\oplus n})}, \ \ [L]\mapsto [\text{Image}(\phi_L)].$$ I claim that $\Phi$ gives an isomorphism of the domain with one of the connected components of the target (when $n\neq 2k$, there is only one connected component, so that $\Phi$ is an $F$-isomorphism). Moreover, it is equivariant for the natural actions of $\textbf{PGL}_n$. Therefore, for every central simple algebra $A$ of "degree" $n$, there is an induced isomorphism of $F$-schemes, $$ SB_1(A) \to \text{Hilb}^{P(t)}_{SB_k(A)},$$ where again the Hilbert polynomial is with respect to the very ample invertible sheaf $\omega^\vee$ on $SB_k(A)$.

In particular, if $SB_k(A)$ is isomorphic to $\text{Grass}(k,F^{\oplus n})$, then every connected component of $\text{Hilb}^{P(t)}_{SB_k(A)}$ has an $F$-rational point. Thus, also $SB_1(A)$ has an $F$-rational point. That implies that $A$ is split by the result quoted by the OP.

Of course one still needs to verify that $\Phi$ is an isomorphism. However, since this is a purely geometric result, I imagine that most of us are happier with this claim than the original problem. Also it is fairly easy to prove using some "homogeneity" and cohomology vanishing.

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