# Dualizing sheaf on a Cohen-Macaulay variety

Let $k$ be an algebraically closed field and let $X$ be a Cohen-Macaulay variety over $k$, i.e. all local rings are Cohen-Macaulay (perhaps this can later be generalized). What is the dualizing sheaf on $X$, what is its central/characterizing property and can one write it down explicitly?

I'm asking this in such a naive way (knowing that there is a vast literature on this) because I have the following problem: I'm (obviously) not an expert in algebraic geometry and came across a definition stating that the dualizing sheaf of $X$ is $\omega_X := j_* \omega_{X_s}$, where $j:X_s \rightarrow X$ is the inclusion of the smooth locus $X_s$ of $X$ and $\omega_{X_s} := \mathrm{det}(\Omega_{X_s/k}^1)$. I can take this of course as a definition (it is pretty easy, and I also know something about dualizing sheaves on smooth varieties) but I don't know what the central properties of this sheaf in this setting are. I know that there is Grothendieck-Verdier-Neeman(-more names) duality and while browsing through Hartshorne's book Residues and Duality I tried to deduce this definition from the abstract "nonsense" but I failed. I know that there exists this very general duality and that there happens something in the Cohen-Macaulay case but this was over my head!

Amnon Neeman defines in his article Derived categories and Grothendieck duality a dualizing complex of a noetherian and separated scheme $X$ to be an object $\mathcal{J} \in \mathbf{D}^b( \mathrm{Coh}(X))$ such that $\mathbb{R}\mathcal{H}om(-,\mathcal{J}):\mathbf{D}^b(\mathrm{Coh}(X))^{op} \rightarrow \mathbf{D}^b(\mathrm{Coh}(X))$ is an (triangulated) equivalence. Is it correct that if $X$ is a (separated?) noetherian Cohen-Macaulay scheme the sheaf defined above is a dualizing complex in this sense and that this is the characterizing property I was asking for? This is the only idea I have so far...

• I think $X$ would have to be normal in addition to Cohen-Macaulay for this formula to work. There are many experts who can explain this better than I could. In one sentence that the dualizing sheaf (or complex) is the thing which makes the Grothendieck-Serre duality theorem work; it would be unique up to isomorphism if it exists. – Donu Arapura Jul 19 '10 at 19:18
• Uniqueness of $\omega$ up to shift & twist by line bundle when $X$ conn'd. For smooth $X$, any line bundle is dualizing; top-diff'tls makes duality "explicit" via residues. CM equiv. to $\omega$ being sheaf (in some degree). For CM $X$ and $j:U \rightarrow X$ open with complement of codim $\ge 2$, want $\omega \rightarrow j_{\ast}(\omega|_U)$ an isom. By SGA2, III, Cor. 3.5, need depth$(\omega_x) > 1$ for $x \in X-U$. Crux: $\omega_x$ satisfies local duality, so by Ext depth criterion same as $H^i_x(k(x))=0$ for $i \ge {\rm{dim}}O_x-1$ ($> 0$!). Verify via excision (cf. SGA2, II, Cor. 4)! – BCnrd Jul 19 '10 at 20:16
• I was hoping you'd jump in, but now I'm confused. $\omega=f^!\mathcal{O}_{spec k}$ is only unique up to shift & twist? – Donu Arapura Jul 19 '10 at 20:28
• @Donu: the formula you give is a dualizing object, but not the only one. The property that defines a dualizing object $R^{\bullet}$ in $D^b_c(X)$ is finite injective dimension and on $D^b_c(X)$ the functor $\mathbf{D} = {\mathbf{R}}\mathcal{H}om^{\bullet}(\cdot,R^{\bullet})$ satisfies the "double duality" map ${\rm{id}} \rightarrow \mathbf{D} \circ \mathbf{D}$ (which makes sense due to the finite injective dimension hypothesis) is an isom. By inspection, applying shift and line bundle twist has no effect! Remarkable fact is that this is the only ambiguity, for connected $X$. – BCnrd Jul 19 '10 at 20:49
• The pushforward formula is only correct when the open set has complement with codimension $\ge 2$, also used in the vanishing aspect of the local duality argument in my initial comment. In general I don't like to say the formulas (pushforward, Ext on an affine space, etc.) are the "definition" of a dualizing sheaf, but rather a way to compute one (coupled with a suitable trace map, which can be quite hard to make "explicit" away from simple examples). A dualizing object yields a "codimension function", and demanding it match usual codimension (can always be done) eliminates shift ambiguity. – BCnrd Jul 19 '10 at 22:40