All Questions
7 questions
10
votes
1
answer
1k
views
Characterization of schemes whose dualizing complex is perfect
I'm wondering if there is a characterization of schemes over a a field $k$ whose dualizing complex is a perfect complex in terms of singularities. E.g. on a proper Cohen-Macauley scheme over a field, ...
9
votes
1
answer
909
views
Grothendieck-Verdier duality without the noetherian condition
The Grothendieck-Verdier duality:
$$
Rf_*\big(R\mathcal{H}\textit{om}_X^\bullet(\mathcal{E}^\bullet,f^!\mathcal{F}^\bullet)\big) \cong R\mathcal{H}\textit{om}^\bullet_Y(Rf_*\mathcal{E}^\bullet,\...
5
votes
0
answers
383
views
Serre functors for non-proper categories
One usually defines a Serre functor to be a functor on a $k$-linear category $\mathcal{C}$ which has finite dimensional $Hom$s over $k$. In that case, the standard definition is that a Serre functor $...
4
votes
0
answers
264
views
Does GKZ's reflexivity theorem imply the Plucker formula?
Let $S\subset\mathbb{P}^n$, Gelfand-Kapranov-Zelevinsky defined its dual variety $S^\vee\subset\mathbb{P}^{n^\ast}$. In this paper (http://arxiv.org/pdf/math/0111179v1.pdf), the author obtained the ...
3
votes
1
answer
296
views
Confusion about the (Grothendieck–Poincaré) double dual of reflexive differentials vs usual differentials on a normal Cohen–Macaulay scheme
$\DeclareMathOperator\Hom{Hom}$Let $\mathcal{A}$ be an abelian category, my question is about the case when $\mathcal{A}$ is the category of quasi-coherent sheaves on a scheme $X$. There is a fully ...
2
votes
1
answer
512
views
Grothendieck duality for resolution of singularities
I would like to know a reference for Grothendieck duality in a resolution of singularities. More precisely, let $Y$ be a normal, Gorenstein variety with finite quotient singularities, and suppose that ...
0
votes
1
answer
600
views
Grothendieck-Verdier duality for affine morphisms
Suppose $X,Y$ are varieties over $\mathbb{C}$, $Y$ is smooth and $X$ is Gorenstein ($X$ is not smooth in my case). Let $f: X \to Y$ be an affine morphism, and each fibre of $f$ has the same dimension $...