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The theory of Severi-Brauer varieties is well-known. Let $k$ be a (perfect) field. There may exist varieties not isomorphic to $\mathbf{P}^n$ over $k$, which are isomorphic to $\mathbf{P}^n$ over $\overline{k}$. They are classified by $H^1(k,\mathrm{PGL}_n)$.

How about quadrics? Say $k$ is a (perfect) field and $X$ is a smooth, projective $k$-variety of dimension $n$. Assume that $X \otimes_k \overline{k}$ is isomorphic to a quadric. Is $X$ necessarily an $n$-dimensional quadric itself? If not, can you give some nice examples (e.g. over number fields) which show that this need not be the case?

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It is not too difficult to see that any automorphism of a smooth quadric hypersurface $$X : Q(x) = 0,$$ over a field $k$ must be a projective automorphism (see for instance the argument I give in Automorphism group of a smooth quadric $Q\subset\mathbb{P}^4$). Hence the automorphism group of any quadric $X$ is the projective orthogonal group $\mbox{PO}(Q)$. Thus twists of $X$ are parametrised by $H^1(k, \mbox{PO}(Q))$.

As to your question: yes there exist twists of quadrics which are not themselves quadrics.

Recall that $\mathbb{P}^1 \times\mathbb{P}^1$ embeds into $\mathbb{P}^3$ via the Segra embedding as a quadric surface. So let now $k$ be a field for which there exists two conics $C_1$ and $C_2$ over $k$ without rational points, and take

$$X = C_1 \times C_2.$$

This is a twist of a quadric surface, but is not isomorphic to a quadric surface. To see this, note that any quadric surface contains an effective divisor $D$ of self-intersection $D^2 = 2$. However it is not too difficult to see that for any effective divisor $D$ on $X$ we have $D^2 = 0$ or $D^2 \geq 8$, hence $X$ is not isomorphic to a quadric surface, as required.

To see what is happening in general, note that those twists of a quadric hypersurface $X$ which are themselves quadrics are parametrised by $H^1(k,\mbox{O}(Q))$. Thus, there are twists of a quadric which are not quadrics whenever the map $$H^1(k,\mbox{O}(Q)) \to H^1(k,\mbox{PO}(Q)),$$ is not surjective. I believe this is the case exactly for quadrics hypersurfaces in $\mathbb{P}^{2n+1}$ over a field $k$ for which the Brauer group $\mbox{Br}(k)$ of $k$ has non-trivial $2$-torsion, but I did not check all the details.

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  • $\begingroup$ I realized that $H^1(k,\mathrm{PO}(Q))$ would classify these twists, but I didn't want to try computing this group -- it might be painful? :-) Very nice example! $\endgroup$ – Wanderer Sep 23 '14 at 15:29
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    $\begingroup$ In fact it is not too difficult to compute, at least over number fields (note that it is a pointed set, not a group). This is more or less the classification of quadratic forms in terms of the discriminant, Hasse-Witt invariant and signature, plus taking care of the cokernel of the above map I give, which should not be too difficult. These are all cohomological invariants which arise from considering various exact sequence of groups related to $\mbox{O},\mbox{SO}, \mbox{PO}$ and $\mbox{Spin}$. $\endgroup$ – Daniel Loughran Sep 23 '14 at 15:36
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This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$).

As discussed in the Wikipedia article, there is an extension of algebraic groups
$$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$.

This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. (A possible reference for these would be the book on central simple algebras by P. Gille and T. Szamuely; alternatively, check out survey papers on the Milnor conjecture.) So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.

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    $\begingroup$ Thanks for working this out. Note that you don't need to invoke any deep theorems in K-theory to show that $H^2(k, \mathbb{Z}/2) \cong {}_2\mbox{Br}(k)$; this follows easily from Kummer theory and Hilbert's Theorem 90. $\endgroup$ – Daniel Loughran Sep 23 '14 at 21:08
  • $\begingroup$ Also I'm not sure your interpretation of my construction is correct; conics over $k$ are classified by $H^1(k,\mbox{PGL}_2)$, not $H^1(k,\mathbb{Z}/2)$. $\endgroup$ – Daniel Loughran Sep 23 '14 at 21:25
  • $\begingroup$ @DanielLoughran: true, both comments. Sorry for bringing up the Milnor conjecture, but the easier ways of proving this (Hilbert 90 etc.) are also explained in Gille-Szamuely. I removed the remark with the cup product; but I still wonder how to interpret the construction. The conics correspond to quaternion algebras, hence also to elements in $H^2(k,\mathbb{Z}/2)\cong {}_2Br(k)$. The only operation I see would be the additive structure of ${}_2Br(k)$... Sorry, I did not have the time yet to sit down and actually compute the obstruction class of your example. $\endgroup$ – Matthias Wendt Sep 24 '14 at 18:48
  • $\begingroup$ @Mattias: Its ok, I have thought about it a bit as well, and I could not find an "elementary" way to describe the obstruction class in my case either. $\endgroup$ – Daniel Loughran Sep 24 '14 at 19:42

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