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?

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 $H^1(k, \mbox{PO}(Q))$.

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.

• 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! – Wanderer Sep 23 '14 at 15:29
• 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 $\mbox{Spin}$. – Daniel Loughran Sep 23 '14 at 15:36

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.

• 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. – Daniel Loughran Sep 23 '14 at 21:08
• 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 $H^1(k,\mathbb{Z}/2)$. – Daniel Loughran Sep 23 '14 at 21:25
• @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. – Matthias Wendt Sep 24 '14 at 18:48
• @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. – Daniel Loughran Sep 24 '14 at 19:42