# Twists of cubic threefolds

Let k be a field of characteristic $0$.

Let $X$ be a variety over k which is isomorphic to a smooth cubic threefold over $\bar{k}$. Then is $X$ isomorphic to a smooth cubic threefold over $k$?

For motivation, let's consider some other cases.

• For cubic curves, the analogous question has a a negative answer in general, as elliptic curves can have quite complicated twists.
• For cubic surfaces however the answer is yes. This is because cubic surfaces are embedded by the anticanonical divisor, which is already defined over $k$.

This latter argument does not work however for cubic threefolds, as for a cubic threefold, the hyperplane section is half the anticanonical divisor. So the question is whether the anticanonical divisor is always divisible by $2$ in the Picard group of $X$.

Note that the Hochschild-Serre spectral sequence yields the exact sequence $$0 \to \mathrm{Pic}(X) \to \mathrm{Pic}(X_\bar{k})^{\mathrm{Gal}(\bar{k}/k)}=\mathbb{Z} \to \mathrm{Br}(k),$$ so the obstruction to $X$ being isomorphic to a cubic threefold over $k$ is given by an element of $\mathrm{Br}(k)[2]$.

• Just a comment, which I'm sure you already know: the case of plane conics is very similar, and in that case the obstruction class in $\mathrm{Br}(k)$ is given by the conic itself (as Severi-Brauer variety). So I would wonder whether there is a natural way to associate a Brauer class to a smooth cubic threefold. – Martin Bright Sep 11 '15 at 12:58
• The Brauer class in question is that of the Severi-Brauer variety parametrising effective divisors in the class $-K_X/2$ (see e.g. Serre, Local Fields, X.6). Looking at the case of a smooth cubic threefold, it seems to have dimension 4, since then it's just the dual of the ambient $(\mathbf{P}^4)$. – Martin Bright Sep 11 '15 at 15:09
• @MartinBright doesn't your description of the S-B variety answer the question with "Yes, the Brauer obstruction is trivial because a twist of ${\bf P}^4$ is in ${\rm Br}[5]$, and thus trivial if it's also 2-torsion"? – Noam D. Elkies Sep 20 '15 at 0:21
• Where does $n \geq 43$ come from? That condition certainly doesn't hold here. Nor, for that matter, for an elliptic curve, with $n = 2$; how does a curve such as $y^2 + x^4 + 1 = 0$ over $\bf R$ escape this argument? – Noam D. Elkies Sep 21 '15 at 3:08
• I think Will meant to say that $n \geq 4$. In this case the Lefschetz hyperplane section theorem tells you that the Picard group of such a hypersurface is cyclic and generated by the hyperplane class. – Daniel Loughran Sep 21 '15 at 15:28

As pointed out by Noam Elkies, my comment above can be turned into a positive answer to the question.

In the exact sequence

$\DeclareMathOperator{\Pic}{Pic}\DeclareMathOperator{\Br}{Br}0 \to \Pic X \to (\Pic \bar{X})^{G_k} \to \Br k$,

the arrow $\delta \colon (\Pic \bar{X})^{G_k} \to \Br k$ sends the class of an effective Cartier divisor $D$ to the class of the Severi--Brauer variety whose $\bar{k}$-points are isomorphic to the projectivisation of $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$. See Serre, Local Fields, X.6 for a passing mention of this fact, and Grothendieck, Le groupe de Brauer, III, 5.4 for more details (though he doesn't actually prove it, saying "on (plus précisément, J. Giraud) vérifie que la classe de cet élément dans $\Br(Y)$ est bien $\delta(\xi)$.")

In the particular example here, $\Pic X$ is of index $2$ in $(\Pic \bar{X})^{G_k}$, so the image of $\delta$ is killed by $2$. On the other hand, taking $D=-K_X/2$, the vector space $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$ has dimension $5$, since we are assuming that $\bar{X}$ is isomorphic to a cubic threefold, on which $D$ is the class of a hyperplane section. It follows (for example Serre, loc. cit.) that $\delta(D) \in \Br k$ is killed by $5$. So $\delta(D)=0$, and $D$ lies in $\Pic X$.

As noted above the obstruction is given by some $A \in Br(k)[2]$, and this obstruction vanishes if $X$ has a $k$-rational point. Since $X$ has a rational point over some odd degree extension, $A$ must become trivial over an odd degree extension. This means that $A$ is already trivial over $k$.

• I agree that a cubic threefold over $k$ has a point over some odd degree extension. But why does a twisted cubic threefold have such a point? Is there a natural odd degree morphism to projective $3$-space? – Ariyan Javanpeykar Sep 12 '15 at 19:21