Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to learn a little about Grothendieck duality. One version of the theorem states that if $f: X \to Y$ is a proper morphism of schemes, then the induced functor on derived categories $f_*: D^+(\mathrm{QCoh}(X)) \to D^+(\mathrm{QCoh}(Y))$ has a right adjoint $f^!$ (and under nice hypotheses, these will preserve the subcategories with coherent cohomology). The existence of an adjoint can be proved via adjoint functor arguments, even without assuming $f$ proper; this was done, e.g., by Neeman. (The point is that a triangulated functor between nice triangulated categories (or, stable $\infty$-categories) which preserves coproducts is a left adjoint.)

However, in trying to identify $f^{!}$, we might want to be able to localize on $X$ and $Y$, and thus deal with the non-proper case. My understanding is that the upper-shriek functor $f^{!}$ there is not supposed to be the right adjoint to $f_{\ast}$: for example, for an open immersion it should be the upper-star $f^*$.

In the topological version, one can define a $f_{!}$ functor for sheaves (push-forward with compact support) and $f^!$ is the right adjoint to $f_{!}$. Is there any "functorial" way to interpret $f^{!}$ when $f$ is not proper?

share|improve this question
    
I'm afraid there's not really more to say than you've said. You have descriptions for open embeddings and proper morphisms and that's enough to describe it in general (say, for schemes finite type over a field). But $j^*=j^!$ doesn't commute with products (for $j$ an open embedding) and therefore doesn't have a left adjoint. Do you have a more precise question in mind? –  Moosbrugger Aug 3 '12 at 1:58
    
Not really. I was just curious if there was any way of making the formalism clearer (e.g., a way of defining this functor which would generalize nicely to a derived setting). –  Akhil Mathew Aug 3 '12 at 2:26
2  
Doing it right is a non-trivial issue, discussed in some detail e.g. in arxiv.org/abs/1105.4857. But note that just defining it naively is easy: Nagata compactification exists for derived schemes as well. The analysis that this (homotopy) doesn't depend on the compactification essentially follows from the analysis in Section 6.2.2 of loc. cit. –  Moosbrugger Aug 3 '12 at 3:07
add comment

1 Answer

I don't have an answer, but maybe these notes by Lipman help.

From what I understand the upper pling functor $f^!$ is a sort of Frankenstein, definitely for etale maps it's given by ordinary pullback. More generally for smooth maps you have to tensor with the relative canonical bundle.

The only description I know of is via dualising complexes, but that's perhaps not categorical enough for what you want.

share|improve this answer
    
The ! is probably the gadget with more pronounciations in all of current notation :-) –  Mariano Suárez-Alvarez Aug 2 '12 at 23:36
1  
Isn't that symbol used to represent the "click" sound in click languages? –  Lee Mosher Aug 2 '12 at 23:55
    
These notes look nice. Thanks. –  Akhil Mathew Aug 3 '12 at 2:29
    
Tyler Lawson suggested the pronunciation "surprise", because you should be (pleasantly) surprised that the functor exists. –  S. Carnahan Aug 3 '12 at 8:54
1  
When I first heard someone say "shriek" for "exclamation point" I broke out laughing (certainly interrupting an explanation they were giving). It's such an evocative word, which one easily forgets as a mathematician repeating it frequently (e.g., the person explaining it to me had never quite thought about the word choice). I also appreciated that in our science we have a much more poetical term for this punctuation mark than in poetry. –  Moosbrugger Aug 3 '12 at 12:50
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.