Let $$X: \quad Q_1(x)=Q_2(x) = 0 \quad \subset \mathbb{P}^{2n+1},$$ be a smooth complete intersection of two quadrics of odd dimension over a field $k$, not of characteristic $2$. Let $J(X)$ denote the intermediate Jacobian of $X$. When $k$ is algebraically closed, I know that $J(X)$ is isomorphic to the Jacobian of the hyperelliptic curve $$y^2 = \mathrm{det}(xQ_1 + Q_2). \quad (*)$$

Is there a similar description when $k$ is non-algebraically closed?

The reason I ask is that when $k$ is non-algebraically closed, it seems quite possible that we may need to take a quadratic twist of $(*)$. However the only "canonical" twists are either $(*)$ itself or the multiplication by $-1$. So my guess is that $J(X)$ should be isomorphic to the Jacobian of

$$\pm y^2 = \mathrm{det}(xQ_1 + Q_2),$$

however I don't know which sign to take, and it seems quite possible that the choice of sign could even depend on $n$.

Edit: As pointed out by Sasha, I should have explained which definition of the intermediate Jacobian I'm using. Firstly, in

Deligne - Les intersections complètes de niveau de Hodge un.

Deligne showed how to construct the intermediate Jacobian for complete intersections of Hodge level $1$ over fields of characteristic $0$, using a clever trick coming from certain universal properties (in particular, it is not explicit).

In the special case of complete intersections of two quadrics $X$ as above, I believe that one can define $J(X)$ to be the albanese variety of the Fano variety of $(n-1)$-planes inside $X$. This construction should work over fields of characteristic not equal to $2$, and moreover recover Deligne's construction when $\mathrm{char}(k)=0$, though unfortunately I know neither a proof nor a reference for this fact.

I would be happy with an answer which uses either definition, restricting to characteristic $0$ if necessary.

  • $\begingroup$ How do you define intermediate Jacobian over an arbitrary field? As for a twist, I can say that at the level of derived category the new feature is a sheaf of Azumaya algebras appearing on the curve. So, if one can twist the Jacobian of a curve by an element of its Brauer group, then probably this is the twist you need. $\endgroup$
    – Sasha
    Oct 4, 2014 at 18:32
  • $\begingroup$ @Sasha: Good point; I added my definition. Also I don't understand the rest of your comment, could you please elaborate further? $\endgroup$ Oct 4, 2014 at 19:59

1 Answer 1


This is not an answer, but this is too long for a comment.

The hyperelliptic curve (let me call its $C$) associated to an intersection of quadrics $X$ has an intrinsic meaning as the moduli space of some vector bundles (the restrictions of spinor bundles from the quadrics in the pencil to $X$). Over complex numbers it is a fine moduli space and the universal bundle gives a fully faithful functor $D^b(C) \to D^b(X)$ extending to a semiorthogonal decomposition $$ D^b(X) = \langle D^b(C), O, O(1), \dots, O(2n-3) \rangle. $$ Over other fields the difference is that the moduli space is not fine, the curve $C$ comes with an Azymaya algebra $A$ (given by components of even parts of Clifford algebras associated to quadrics in the pencil) and the decomposition one has is $$ D^b(X) = \langle D^b(C,A), O, O(1), \dots, O(2n-3) \rangle. $$ So, I believe that if one can twist the Jacobian by a Brauer class, then I think it might be the answer.

To prove this one could consider the functor $D^b(X) \to D^b(C,A)$ given by the above decomposition and apply it to the universal family of $(n-1)$-planes. This would give a morphism from the Fano variety $F(X)$ of $(n-1)$-planes to some moduli space of objects in $D^b(C,A)$ which probably is the appropriate twisting of $J(C)$. At least over an algebraically closed field of characteristic $0$ this is an isomorphism $F(X) \cong Pic^0(C)$.


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