Return to Answer

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.

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.

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 $D^2 \geq 4$, 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 $k$ with non-trivial $2$-torsion in the Brauer group, but I did not check all the details.

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.

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 $D^2 \geq 4$, as required.

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 $D^2 \geq 4$, 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 $k$ with non-trivial $2$-torsion in the Brauer group, but I did not check all the details.