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# On Serre Duality

Serre Duality is the statement, for $X$ a smooth projective (integral) variety and $\mathcal{E}$ a locally-free sheaf on $X$,

$H^i(X,\mathcal{E}) \cong H^{n-i}(X, \mathcal{E}^* \otimes \omega_X)^*$,

where $\omega_X$ is the canonical bundle and $n = \dim X$. This isomorphism is almost canonical: it depends on the choice of an isomorphism

$t: H^n(X,\omega_X) \to k$,

called a trace map. I’m going to sketch out my understanding of what’s going on with this duality statement and how it comes up (non-rigorously).

Derived Functors of $(H^n)^*$

The starting point for sheaf cohomology on $X$ is that the global sections functor is only left-exact: for a short exact sequence of sheaves

$0 \to A \to B \to C \to 0,$

taking global sections gives

$0 \to H^0(A) \to H^0(B) \to H^0(C),$

not, in general, exact on the right. So, since the category of sheaves on $X$ has enough injectives, we can define derived functors $H^i(-)$, which continue the above sequence to the right. If $X$ has dimension $n$, the sequence necessarily terminates at the $n$-th level:

$\cdots \to H^n(A) \to H^n(B) \to H^n(C) \to 0,$

since $H^i(\mathcal{F}) = 0$ for any sheaf $\mathcal{F}$, for $i > \dim X$. This is the Grothendieck Vanishing Theorem.

Consider dualizing this sequence (of vector spaces):

$0 \to H^n(C)^* \to H^n(B)^* \to H^n(A)^* \to \cdots.$

This is kind of neat: we see that $H^n(-)^*$ is again a left-exact (contravariant) functor. And, evidently, its higher derived functors are the duals of sheaf cohomology, in reverse order, $H^{n-i}(-)^*$.

So we can think of the two ends of the familiar long exact sequence in sheaf cohomology as being interchangeable — either one could be the “starting point”. (I’m glossing over the fact that our category doesn’t have enough projectives, so technically these are “universal $\delta$-functors,” not strictly speaking derived functors.)

Representing $H^0$ and $(H^n)^*$

Another standard fact about sheaves on $X$ is that a map of sheaves $\mathcal{O}_X \to \mathcal{F}$ is the same as a global section of $\mathcal{F}$:

$\text{Hom}(\mathcal{O}_X,-) \cong H^0(-).$

In other words, the sheaf $\mathcal{O}_X$ represents the global-sections functor for sheaves on $X$. This ties in nicely with higher cohomology as well: taking derived functors “of both sides” of the equation above, we get

$\text{Ext}^i(\mathcal{O}_X,-) \cong H^i(-).$

(To establish this, we observe/show that both sides are universal $\delta$-functors that agree in degree 0. Or we observe, informally, that the functors are the same (naturally isomorphic) for $n=0$, hence must have naturally isomorphic derived functors. This is the main tool for justifying this kind of manipulation, and I’ll gloss over it from now on.)

Now let’s make the following observation. Let $\omega$ be any sheaf whose top cohomology is $1$-dimensional, and fix a trace map $t : H^n(\omega) \to k$. Then, for any map $\mathcal{F} \to \omega$, we have an induced map

$H^n(F) \to H^n(\omega) \cong k,$

that is, we get an element of $H^n(F)^*$. If every element arises naturally and uniquely in this way — this is a big if — then we can say

$\text{Hom}(-,\omega) \cong H^n(-)^*,$

that is, $\omega$, combined with the choice of trace map $t$, (co)represents the dual-of-top-cohomology functor $H^n(-)^*$. We call such a pair $(\omega,t)$ a dualizing sheaf.

And, thinking optimistically, we might hope for the derived functors to agree as well — namely,

$\text{Ext}^i(-,\omega) \cong H^{n-i}(-)^*.$

Note that this is Ext in the second argument, not the first, so there are some technical hiccups here. In any case, plugging in a locally free sheaf $\mathcal{E}$, we can get rid of the Ext part and get a statement entirely in terms of sheaf cohomology. Specifically, we know that

$\text{Hom}(\mathcal{E},\omega) \cong \text{Hom}(\mathcal{O}_X,\mathcal{E}^* \otimes \omega) \cong H^0(\mathcal{E}^* \otimes \omega)$

holds by hom-tensor adjunction, so “taking derived functors of both sides” gives

$\text{Ext}^i(\mathcal{E}, \omega) \cong H^i(\mathcal{E}^* \otimes \omega).$

Note that here, we’re letting $\mathcal{E}$ be fixed in the equation, taking derived functors in the other argument, and then plugging back in $\omega$. And to even talk about $\mathcal{E}^*$, we need $\mathcal{E}$ to be locally free. Luckily, that’s enough to make this justifiable and correct (you can prove this, for example, with the spectral sequence relating “Ext in the first argument” to “Ext in the second argument”.)

Anyway, combining the two equations gives the usual statement of Serre duality, in terms of the dualizing sheaf $\omega$:

$H^i(\mathcal{E}^* \otimes \omega) \cong H^{n-i}(\mathcal{E})^*.$

Finding a Good Dualizing Sheaf

The construction above relied, crucially, on the existence of a dualizing sheaf to (co)represent the $H^n(-)^*$ functor. So a few remarks are in order to say where this comes from. Personally, I find the exposition in Hartshorne to be a bit frustrating, since the method is as follows:

1. (Projective Space) Observe that it just so happens to work for $X = \mathbb{P}^n$, with $\omega = \mathcal{O}(-n-1)$, by inspection.
2. (Duality for Closed Embeddings) Use sheaf Ext to construct explicitly a dualizing sheaf on $Y$, where $Y \to X$ is an embedding, $X,Y$ are smooth (or, more generally, Cohen-Macaulay), and $X$ has a dualizing sheaf $\omega_X$. Explicitly, the dualizing sheaf on $Y$ is $\omega_Y = \mathscr{E}xt^{\dim X - \dim Y}(\mathcal{O}_Y,\omega_X)$. This gains the desired functor-representation property from the Cohen-Macaulay formalism, using regular sequences and/or Koszul complexes and/or vanishing of certain Ext functors.
3. (Canonical Bundle). Observe that the adjunction formula for the canonical bundle, $\omega_Y = \omega_X \otimes \bigwedge^{\dim X - \dim Y} \mathcal{I} / \mathcal{I}^2$, where $\mathcal{I}$ is the ideal sheaf of $Y$, is compatible with the transition functions when building the dualizing sheaf out of Koszul complexes. In other words, the canonical bundle on $Y$ is, itself, a dualizing sheaf. This allows us to replace the more abstract definition, using sheaf Ext, with more familiar (not to mention geometric) differential forms.

An alternate approach is to replace “closed embedding” with “finite cover”, but I’m not familiar with it. Ravi Vakil’s notes do it this way, so refer to those for the exposition.

While the approach of “pulling back” Serre Duality across embeddings or finite maps is elegant and (I think) fairly theoretically satisfying, it still leaves that magical step 1 above, where Serre Duality is practically coincidental, and unenlightening, when $X = \mathbb{P}^n$.

I remember thinking this when reading Serre’s GAGA paper as well: the approach for general smooth projective varieties is to reduce to the case of projective space, then to observe that all the statements of GAGA (particularly the Betti numbers) just happen to be true for projective space, whether considered algebraically or analytically. It’s a little weird that there’s no “deeper” reason, or that the proof isn’t more intrinsic for arbitrary smooth varieties. If anyone reading this has any thoughts, I’d be interested to hear them.

* I actually have no idea how one actually does this with derived categories. Someone should explain that to me…

1. alexyoucis says:

Hey Jake! I saw when you commented on my blog, that you yourself had a blog–so I thought I’d check it out.

I think there is a somewhat satisfying answer to your complaint about the reduction to $\mathbb{P}^n$. Namely, I believe (although, math history is not my strong suit 🙂 )that this treatment of Serre duality is a bit anachronistic.

People had versions of Serre duality on complex compact manifolds before Serre in ’55 gave the modern statement of the theorem. They roughly thought about Serre duality as giving a sort of pairing

$H^i(X,\mathcal{E})\times H^{n-i}(X,\mathcal{E}^\vee\otimes\omega_X)\to\mathbb{C}$

which is, very roughly, wedging forms together and integrating. This can be made more rigorous, and if you’re interested, you should look at Chapter 5 of Voisin’s book.

So, with our GAGA caps on, and a healthy dose of the Lefschetz principle, we know that not only Serre duality should hold, but that the dualizing sheaf must be the canonical sheaf. In fact, if you squint your eyes hard, and believe (or, alternatively, look at some sources by Serre and Tate on the proof of Serre duality) the proof for $\mathbb{P}^n$ is also “pairing and integrating”. In fact, the trace map $t:H^n(\omega)\to k$ is nothing but the ‘residue map’.

So, as you said, once you do the abstract statement of “there exists a unique (up to isomorphism) sheaf which satisfies BLAH and respects closed embeddings”, then you immediately knew you were reduced to $\mathbb{P}^n$. There you could try and imitate (and, as above, I maintain that is roughly what is done) the proof from complex manifold land, and then conclude.

• alexyoucis says:

PS, this was a nice post! Your informal derivation of why something representing $H^\ast$ should be the dualizing sheaf was quite enlightening!

• Hi Alex, thanks for the comments. It’s a good point: the trace map does seem more motivated when we think of it as ‘pairing and integrating’. Have you seen (or, were you referring to) Serre’s proof of duality / Riemann-Roch for curves in those terms? It is explicitly as you say — in local coordinates, the trace map picks out the coefficient of the $1/x$ term of the Laurent expansion (where $x$ is a uniformizer), i.e. the residue. It’s a good point that things are not so mysterious after all if we stop thinking purely in terms of homological algebra in the abstract.

Lastly, this is probably obvious to you but I thought I’d mention it anyway, I could have written out the construction in terms of perfect pairings, just in terms of Ext and Hom (for whatever reason, I didn’t do this when I wrote the post). Namely, we start with

$\mathrm{Ext}^n(\mathcal{O}_X,-) \otimes \mathrm{Hom}(-,\omega) \to \mathrm{Ext}^n(\mathcal{O}_X,\omega) \cong k,$

and its derived analogue,

$\mathrm{Ext}^{n-i}(\mathcal{O}_X,-) \otimes \mathrm{Ext}^i(-,\omega) \to \mathrm{Ext}^n(\mathcal{O}_X,\omega) \cong k,$

which becomes your pairing (with indices swapped) under the identification of the Ext functor with (twisted) sheaf cohomology.

2. alexyoucis says:

Jake, for some reason it won’t let me respond to your actual response, so I’m just posting a new comment. 🙂

Which proof are you referring to? The proof in Serre’s book ‘Algebraic Groups and Class Fields’? If so, yes, it is of the form I am referencing. Literally thinking of the trace map as being the residue map. That said, some of the details in the general sort of cases are VERY messy. The only rigorous reference I know for the definition of the full residue map in general is this very terrible paper of Tate: ‘Residues of Differentials on Curves’.

Just as a comment, if you’re interested in going even further down the abstraction rabbit hole, it is (as you may already know) ‘best’ to view Serre duality in the more general context of ‘duality theory’. This is the subject of coming up with the right notions of dualizing complexes for (possibly non-smooth) maps of schemes. If you’re interested in how this might look for the special case of Serre duality, you should look at this paper by Matt Baker: http://csirik.net/serre-duality-long.pdf. Or, the entire book on duality theory by Hartshorne.

Yes, the perfect pairing method is the one followed by Hartshorne, and is really Harkening more towards the general duality theory as referenced in the last paragraph.

• Yes, that’s the proof I was thinking of. Thanks for the suggestions — I’ll definitely give those a look.