2
$\begingroup$

Let $X$ be a smooth projective variety of dimension $d$ over $\mathbb C$ and let $E$ be a zero-dimensional coherent sheaf on $X$. The dual sheaf $E^D=\mathscr Ext_X^d(E,\omega_X)$ is again zero-dimensional.

Question. Do $E$ and $E^D$ have support of the same length?

In other words, if $Z,Z^D\subset X$ are the supports of $E,E^D$ respectively, is it true that $h^0(X,\mathscr O_Z)=h^0(X,\mathscr O_{Z^D})$? A more ambitious question would be to ask if the supporting points are the same. Maybe these questions can be reduced to a commutative algebra problem involving canonical modules, but I do not see how to do that. Any reference (I feel this is well-known) is very much appreciated. Thanks!

Improve this question
$\endgroup$
3
  • $\begingroup$ This is a small part of the local duality in local cohomology. One reference is Theorem A4.2, p. 694, of Eisenbud's "Commutative Algebra with a View Toward Algebraic Geometry". Take $i$ equals $0$ in the statement there. $\endgroup$
    – Jason Starr
    Commented Dec 4, 2015 at 16:53
  • 4
    $\begingroup$ In case nfdc23 is out there, the reference in "Residues and Duality" is Theorem 6.2, p. 278. Of course the proof is in Section 4 of "Local Cohomology". $\endgroup$
    – Jason Starr
    Commented Dec 4, 2015 at 17:04
  • $\begingroup$ I misread your question: the comments above are about the lengths of $E$ and $E^D$, not the lengths of the supports. $\endgroup$
    – Jason Starr
    Commented Dec 4, 2015 at 17:46

1 Answer 1

Reset to default
4
$\begingroup$

I misread your question, so the comments above are answering a different question than you asked. I thought you were asking about the lengths of the sheaf and its dual sheaf, not the lengths of the supports. At any rate, this does follow from duality, since the dual of the dual equals the original module (Proposition 5.1, p. 275 of "Residues and Duality").

Let us work locally: so $R$ is a local, Noetherian, Gorenstein $k$-algebra of dimension $d$ with residue field $k$ (so $k$ is a coefficient field). Denote a dualizing module by $\omega_R \cong R$. Let $E$ be a finite length $R$-module. Let $f\in R$ be any element in the annihilator of $M$. Then since the multiplication by $f$ homomorphism, $$L_{E,f}:E \to E, \ m \mapsto fm,$$ is the zero homomorphism, the same is true on $E^D = \text{Ext}^d_R(E,\omega_R)$, since Ext is an additive functor. Thus, the annihilator ideal of $E$ is contained in the annihilator ideal of $E^D$. On the other hand, $(E^D)^D$ is congruent isomorphic to $E$ as a finite length $R$-module. Thus, also the annihilator ideal of $E^D$ is contained in the annihilator ideal of $E$. So the two annihilator ideals are equal, i.e., the scheme-theoretic support of $E$ equals the scheme-theoretic support of $E^D$ (although $E$ and $E^D$ need not be isomorphic as $R$-modules).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much! Also for the reference in Eisenbud's book. $\endgroup$
    – Brenin
    Commented Dec 7, 2015 at 14:51
  • 1
    $\begingroup$ Note that the assumptions are too strong: the only relevant condition is that $E$ is Cohen-Macaulay (so that $Ext^i(E,\omega_R)=0$ for all but one value of $i$) and the same argument shows that the support of $E$ and the support of $E^D$ are equal as subschemes. (Here $E^D$ is the non-zero $Ext^i(E,\omega_R)$.) $\endgroup$
    – t3suji
    Commented Dec 7, 2015 at 17:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .