5
$\begingroup$

Let $X$ be a smooth projective variety over $\mathbb{C}$, and $H \subset X$ be a smooth hypersurface.

Many properties of an ambient variety $X$ could somehow inherit to the hypersurface $H$, I was wondering if there is any know result dealing with the relation between the derived categories of (bounded) complexes of quasi-coherent sheaves (or coherent sheaves) on $X$ and $H$?

$\endgroup$
3
  • 2
    $\begingroup$ You cannot expect much: think of a hypersurface $H\subset\mathbb{P}^n$, say of degree $>n$. The derived category of $\mathbb{P}^n$ is quite simple, while that of $H$ is extremely complicated. $\endgroup$
    – abx
    Apr 22, 2014 at 14:56
  • $\begingroup$ Kuznetsov has a whole theory about that called homological projective duality. The catch is that you have to start with an appropriate exceptional collection on X. Here are some handy notes. cf.ac.uk/maths/subsites/logvinenko/talks/… $\endgroup$ Apr 22, 2014 at 15:56
  • $\begingroup$ Thanks! I will try to learn something about homological projective duality! $\endgroup$
    – Li Yutong
    Apr 23, 2014 at 1:43

2 Answers 2

7
$\begingroup$

In terms of semiorthogonal decompositions, there is an answer to this question, as it has been mentioned above.

More precisely, if one has a so called Lefschetz decomposition of the bounded derived category of coherent sheaves on a smooth projective variety (such a decomposition depends also on the choice of a projective embedding) then one has an induced semiorthogonal decomposition of the derived category of a hyperplane section, with pieces coming from the ambient variety and then an additional component. One can think of this as some sort of a Lefschetz hyperplane theorem result. More details here: Kuznetsov`s Main HPD Paper

Then, if you have a degree d hypersurface H in P^n, this hypersurface can be seen as a linear section of P^n with respect to the degree d Veronese embedding.

In this direction, the paper mentioned above (by Ballard, Deliu, Favero, Isik and Katzarkov) gives an answer to what the extra information is, either in terms of LG-models (matrix factorization categories) or in terms of the derived category of sheaves of modules over a sheaf of A_infty algebras (there is a nice, simple formula describing the sheaf of A_infty algebras).

$\endgroup$
6
$\begingroup$

Maybe it's worth spelling out the easy case of projective space (and Veronese embedding). Take the standard exceptional collection $<O,O(1),...,O(n)>.$

The derived category of a hypersurface X of degree d has then a semi-orthogonal decomposition $<A_X, O, O(1), ..., O(n-d)>$. We might think of $A_X$ as being the "interesting" part of the derived category of X. We can also intersect further and $D(X_1 \cap X_2) = <A_{X_1 \cap X_2}, O,\ldots, O(n-2d)>$.

Take the universal hypersurface of degree d, call it $H$. This sits inside $P^n \times P^N$, where $N$ is the ambient dimension of the degree d Veronese embedding.

The category $H$ (being itself a hypersurface) has a decomposition where we call the interesting part just $A$.

For each point $p \in P^N$ we have a hypersurface $H_p$ and thus an interesting category $A_{H_p}$.

The "homological dual" variety is a $Y$, mapping to the dual $P^N$, such that $D(Y)$ is the total $A$. In some sense, $Y$ is parameterezing all the categories $A_{H_p}$.

The point of HPD is then the compatibility between passing to base loci and base changing Y.

If $L < P^N$ is some linear system, then the derived category of the corresponding base locus has a decomposition and the interesting part is the interesting part of $D(Y \times_{P^N} L$.

(this is perhaps too sloppy to be useful but I tried)

The Viennese school of Katzarkov, Ballard, Favero etc. has shown that in this context (of degree d Veronese embedding) there is a (sort of tautological) dual. Unfortunately this dual is a highly non-commutative variety, in fact it's the dual $P^N$ with a sheaf of $A_\infty$-algebras over it.

$\endgroup$

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.