# why is the homological projective dual of this Lefschetz decomposition non-commutative?

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?

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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 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. –  Sasha Jul 22 '13 at 9:53
@ 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? –  Libli Jul 22 '13 at 21:57