3
$\begingroup$

I am reading these notes of an excellent course by Kuznetsov on Homological Projective Duality. On page 10 there is Example 1''.

One starts with projective space, endowed with the identity embedding in itself and the slightly tweaked Lefschetz decomposition $A_0 = <\mathcal{O},\mathcal{O}(1),\mathcal{O}(2)>$, $A_{i} = <\mathcal{O}(2)>$. It is stated that the HP-dual of this is a non-commutative space, actually a fibration in non-commutative projective lines. It's not obvious to me why this space cannot be realised geometrically, say by a space $Y$, perhaps together with a sheaf of algebras over it.

So my question is: why can't we find a dual of geometric origin?

$\endgroup$

1 Answer 1

1
$\begingroup$

It seems I can't comment on your question so I'll try an answer...

Perhaps you can try to prove that $C \simeq D^b(\mathbb{P}(V^*), \mathcal{A})$, where $\mathcal{A} = p_* \mathcal{E}nd(q^* (\mathcal{O}(1)\oplus \mathcal{O}(2)))$ and $p : \mathbb{P}(V) \times \mathbb{P}(V^*) \rightarrow \mathbb{P}(V^*)$, $q : \mathbb{P}(V) \times \mathbb{P}(V^*) \rightarrow \mathbb{P}(V)$ are the projections.

I am not sure it works....

$\endgroup$
3
  • $\begingroup$ I think this is true, but with a small adjustment. Instead of $P(V) \times P(V^*)$ you should consider the incidence quadric $Q \subset P(V)\times P(V^*)$ (the canonical divisor of bidegree $(1,1)$), take for $p$ and $q$ the projections $p:Q\to P(V^*)$ and $q:Q \to P(V)$ and define $A$ by the same formula. $\endgroup$
    – Sasha
    Jul 22, 2013 at 9:53
  • $\begingroup$ @ Sasha : you're perfectly right, forgot that $C$ was defined in the $D^b$ of the incidence quadric. By the way, it is claimed at the end of these notes that a HPD for $G(3,6)$ is known to be a double cover of $\mathbb{P}^{19}$ ramified along some hyperquartic. Is there a reference for the proof? $\endgroup$
    – Libli
    Jul 22, 2013 at 21:57
  • $\begingroup$ I think that the current state of art can be found here: repository.upenn.edu/cgi/… $\endgroup$
    – Sasha
    Jul 23, 2013 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.