Serre Duality on Curves

Home , Duality (mathematics), Serre duality

Serre duality on curves

Justin Campbell

1 Let k be a ﬁeld and X and smooth and proper variety of dimension d over k. Write ΩX for the rank d vector bundle of 1-forms on X, and

d ^ 1 ωX := ΩX for the canonical line bundle.

d Theorem 1.1 (Serre duality). There exists a canonical linear map trX : R Γ(X, ωX ) → k such that for any coherent sheaf F on X, the induced pairing

trX RΓ(X, F) ⊗ R HomOX (F, ωX )[d] −→ RΓ(X, ωX )[d] −−→ k is perfect.

Here the pairing is constructed using the identiﬁcation RΓ(X, F) = R HomOX (OX , F) and the composition map

RΓ(X, F) ⊗ R HomOX (F, ωX ) = R HomOX (OX , F) ⊗ R HomOX (F, ωX )

−→ R HomOX (OX , ωX ) = RΓ(X, ωX ).

i Since X is d-dimensional we have R Γ(X, ωX ) = 0 for i > d, so there is a canonical map

d trX RΓ(X, ωX )[d] −→ R Γ(X, ωX ) −−→ k which we also denote by trX . The assertion that the pairing is perfect means that the induced map

∨ R HomOX (F, ωX )[1] −→ RΓ(X, F)

∨ is a quasi-isomorphism, where for any complex of vector spaces V we write V := R Homk(V, k) for the dual complex. At the level of cohomology, this means that for any i ∈ Z, the pairing of vector spaces tr RiΓ(X, F) ⊗ Extd−i(F, ω ) −→ RdΓ(X, ω ) −−→X k OX X X is perfect. Let us explore some consequences of the theorem. If X is geometrically connected then 0 R Γ(X, OX ) = k, so taking F = OX , in this case the theorem says that there is a canonical isomorphism d trX : R Γ(X, ωX )−→ ˜ k.

1 Recall that if E is a vector bundle on X and we put

∨ E := HomOX (E, OX ), then for any quasicoherent sheaves F and G we have a canonical isomorphism

∨ R HomOX (F ⊗ E, G)−→ ˜ R HomOX (F, G ⊗ E ).

Thus the theorem in this case yields a canonical quasi-isomorphism

∨ ∨ RΓ(X, ωX ⊗ E )[d]−→ ˜ RΓ(X, E) , and hence for all i ∈ Z an isomorphism

d−i ∨ i ∨ R Γ(X, ωX ⊗ E )−→ ˜ R Γ(X, E) .

2 The Riemann-Roch theorem

For F a coherent sheaf on the proper smooth variety X, we will use the notation

i i h (F) := dimk R Γ(X, F) for i ∈ Z, as well as X χ(F) := (−1)ihi(F). i∈Z Recall that given a short exact sequence of coherent sheaves

0 −→ F1 −→ F2 −→ F3 −→ 0, we have χ(F1) + χ(F3) = χ(F2). From now on we assume that X is a smooth, proper, and geometrically connected curve. Recall that for a curve, proper implies projective. The hypothesis that X be geometrically connected is 0 1 equivalent to h (OX ) = 1. Observe that since X is 1-dimensional, we have ωX = ΩX . Since X is smooth and in particular normal, the notions of Weil divisor and Cartier divisor on X coincide. Let Div(X) denote the group of divisors on X and K the function field of X. The homomorphism Div(X) → Pic(X) which sends a divisor Z to the line bundle OX (Z) fits into an exact sequence 1 −→ k× −→ K× −−→div Div(X) −→ Pic(X) −→ 1. Given a closed point x in X, its degree is defined by

deg(x) := dimk kx, where kx denotes the residue ﬁeld of x. This extends Z-linearly to a homomorphism

deg : Div(X) −→ Z.

Proposition 2.1. For any f ∈ K×, we have deg(div(f)) = 0.

2 Proof. By the valuative criterion of properness, the rational function f extends uniquely to a map 1 f : X → P . If f is constant then div(f) = 0, so we are done. Otherwise we have by deﬁnition

div(f) = f −1(0) − f −1(∞)

(here we take the scheme-theoretic preimage). Note that f is ﬁnite ﬂat, being a nonconstant morphism between smooth, proper, and connected curves. But this implies that deg(f −1(0)) and −1 deg(f (∞)) both equal the rank of the vector bundle f∗OX , whence the claim.

Thus the degree homomorphism factors through Div(X) → Pic(X), and we obtain

deg : Pic(X) −→ Z.

In fact, the degree of a line bundle can also be deﬁned cohomologically.

Proposition 2.2. For any line bundle L on X, we have

deg L = χ(L) − χ(OX ). ∼ Proof. First suppose that L = OX (−Z) where Z is an eﬀective divisor on X. Then we have the short exact sequence 0 −→ OX (−Z) −→ OX −→ i∗OZ −→ 0, where i : Z → X is the inclusion. Thus

0 χ(OX (−Z)) − χ(OX ) = −χ(i∗OZ ) = −h (OZ ) = − deg Z, where the second to last equality holds because Z is ﬁnite, therefore aﬃne, and hence

RΓ(X, i∗OZ ) = RΓ(Z, OZ ) = Γ(Z, OZ ). ∼ Note that L ⊗ i∗OZ = i∗OZ noncanonically for any line bundle L on X. Thus tensoring the above exact sequence with OX (Z) yields

0 −→ OX −→ OX (Z) −→ i∗OZ −→ 0, and we deduce that χ(OX (Z)) − χ(OX ) = χ(i∗OZ ) = deg Z.

Finally, if W is another eﬀective divisor then we can tensor the ﬁrst exact sequence with OX (W ) to obtain 0 −→ OX (W − Z) −→ OX (W ) −→ i∗OZ −→ 0, and hence χ(OX (W − Z)) = χ(OX (W )) − χ(i∗OZ ) = deg W + χ(OX ) − deg Z as desired.

Corollary 2.2.1. If L1 → L2 is a nonzero morphism of line bundles on X, then deg L1 ≤ deg L2.

3 Proof. First observe that any nonzero morphism ϕ : L1 → L2 is a monomorphism. Namely, if ker ϕ 6= 0 then im ϕ must be torsion, but since it is a subsheaf of the line bundle L2 this would imply that im ϕ = 0. Thus we have an exact sequence

ϕ 0 −→ L1 −→ L2 −→ coker ϕ −→ 0, with coker ϕ having ﬁnite support and hence χ(coker ϕ) = h0(coker ϕ). We deduce that

0 χ(L1) ≤ χ(L1) + h (coker ϕ) = χ(L1) + χ(coker ϕ) = χ(L2), and we obtain the desired inequality by subtracting χ(OX ) from both sides.

Let us now assume Serre duality for curves has been proved and deduce some consequences. 0 Recall that the geometric genus of X is deﬁned to be h (ωX ), while the arithmetic genus is 1 h (OX ). Serre duality supplies an isomorphism

0 1 ∨ R Γ(X, ωX )−→ ˜ R Γ(X, OX ) , so these numbers are equal and we refer to

0 1 g := h (ωX ) = h (OX ) simply as the genus of X.

Theorem 2.3 (Riemann-Roch). For any line bundle L on X, we have

0 0 ∨ h (L) − h (ωX ⊗ L ) = deg L + 1 − g.

0 ∨ 1 Proof. Since h (ωX ⊗ L ) = h (L) by Serre duality, the left side of the equation equals χ(L). By Proposition 2.2, we have χ(L) = deg L + χ(OX ), and χ(OX ) = 1 − g by the deﬁnition of g and the hypothesis that X is geometrically connected.

Corollary 2.3.1. We have deg ωX = 2g − 2. Proof. Applying the theorem, we have

0 0 deg ωX = h (ωX ) − h (OX ) − 1 + g = 2g − 2.

Corollary 2.3.2. If deg L > 2g − 2 then h1(L) = 0.

1 1 Proof. Suppose that h (L) 6= 0, i.e. R Γ(X, L) 6= 0. Thus by Serre duality HomOX (L, ωX ) 6= 0, which implies that deg L ≤ deg ωX = 2g − 2 by Corollaries 2.2.1 and 2.3.1.

4 3 The Riemann-Hurwitz formula

Let Y be another smooth, projective, and geometrically connected curve over k. Suppose that we are given a finite separable morphism f : X → Y , i.e. the associated extension of function fields is finite separable. Then we have the canonical short exact sequence ∗ 1 1 1 0 −→ f ΩY −→ ΩX −→ ΩX/Y −→ 0, (3.1) 1 and the sheaf of relative differentials ΩX/Y is a torsion sheaf of finite length. In particular we have a canonical direct sum decomposition

1 M 1 ΩX/Y = ΩX/Y,x x where the sum runs over closed points of X, and only ﬁnitely many summands are nonzero. For any closed point x in X, we write OX,x for the completed local ring at x. Consider the restriction homomorphism ] fx : OY,f(x) −→ OX,x.

Let t be a coordinate at x, i.e. a generator of the maximal ideal mx ⊂ OX,x, and similarly let s ] ex × be a coordinate at f(x). Then we have fx(s) = ut where u ∈ OX,x and ex ≥ 1 is a uniquely determined integer called the ramiﬁcation degree of f at x. Letting p denote the characteristic of k, we can divide the behavior of f at x into three cases: we say that f is

• unramiﬁed at x if ex = 1;

• tamely ramiﬁed at x if ex > 1 and p - ex;

• wildly ramified at x if ex > 1 and p|ex. 1 Some information about the ramification behavior of f can be extracted from the sheaf ΩX/Y . 1 Proposition 3.1. The morphism f is ramified at x if and only if ΩX/Y,x 6= 0. We have 1 ex ≤ 1 + dimkx Γ(X, ΩX/Y,x), with equality if and only if f is unramified or tamely ramified at x. ∗ 1 1 Proof. For the first assertion, observe that the fiber of f ΩY → ΩX at a closed point x is ∗ ∗ kx ⊗kf(x) Tf(x)(Y ) −→ Tx (X). ∗ 2 Here Tx (X) := mx/mx denotes the cotangent line. By definition, this map is an isomorphism if and only if f is unramified at x. 1 1 Denote by ΩX,x the OX,x-module attached to ΩX , obtained by completion at x. There is a 1 natural differential d : OX,x → ΩX,x, and moreover f determines an OY,f(x)-linear map 1 1 dfx :ΩY,f(x) −→ ΩX,x.

] ex × Choose coordinates s and t at f(x) and x respectively, so that fx(s) = ut for some u ∈ OX,x. Then we have ex−1 ex dfx(ds) = exut dt + t du. 1 If p - ex then it follows that Γ(X, ΩX/Y,x) = coker dfx has kx-dimension ex − 1, and otherwise this dimension is strictly greater than ex − 1.

5 Lemma 3.2. If L is a line bundle on Y and n denotes the degree of f, then deg(f ∗L) = n · deg L. ∼ Proof. Since deg is a homomorphism, it suﬃces to consider the case L = OY (−Z) where Z ⊂ Y is an eﬀective Cartier divisor. Then we have the short exact sequence

0 −→ OY (−Z) −→ OY −→ i∗OZ −→ 0 where i : Z → Y is the inclusion. Applying f ∗, we obtain

∗ ∗ 0 −→ f OY (−Z) −→ OX −→ f i∗OZ −→ 0 and hence ∗ ∗ ∗ deg(f OY (−Z)) = χ(f OY (−Z)) − χ(OX ) = −χ(f i∗OZ ). Since f is ﬂat, base change yields

∗ −1 −1 RΓ(X, f i∗OZ ) = RΓ(f (Z), Of −1(Z)) = Γ(f (Z), Of −1(Z)).

But then we have

∗ −1 χ(f i∗OZ ) = dimk Γ(f (Z), Of −1(Z)) = n · deg Z = −n · deg L as needed.

Now we give an application of the Riemann-Roch theorem and hence, indirectly, of Serre duality.

Theorem 3.3 (Riemann-Hurwitz). Let n be the degree of f. Then we have

0 1 2gX − 2 = n · (2gY − 2) + h (ΩX/Y ).

If f only has tame ramiﬁcation, then

0 1 X h (ΩX/Y ) = |kx : k| · (ex − 1). x

1 Proof. Consider the exact sequence (3.1) and take Euler characteristics. Since ΩX/Y has ﬁnite 1 1 support we have h (ΩX/Y ) = 0 and hence

∗ 0 1 χ(ωX ) = χ(f ωY ) + h (ΩX/Y ).

Applying Proposition 2.2 yields

∗ 0 1 deg ωX = deg(f ωY ) + h (ΩX/Y ).

Then we obtain the ﬁrst formula by Corollary 2.3.1 and Lemma 3.2. The second formula follows immediately from Proposition 3.1.

Corollary 3.3.1. We have gX ≥ gY .

0 1 Proof. This is immediate from the theorem, since n ≥ 1 and h (ΩX/Y ) ≥ 0.

6 4 Tate vector spaces

In complex analysis, the residue of a meromorphic function at a pole depends only on a small punctured disk centered there. In algebraic geometry, analytic disks and punctured disks are replaced by formal disks and punctured disks. The algebras of functions on these spaces are power series and Laurent series respectively, which carry natural complete topologies. Although not very interesting geometrically, these topologies play an essential bookkeeping role, and in particular are fundamental to the algebraic theory of residues. In what follows, the ﬁeld of scalars k is always equipped with the discrete topology. Proposition 4.1. Let V be a topological vector space over k. The following are equivalent: (i) V is Hausdorﬀ, complete, and 0 ∈ V admits a base of open neighborhoods consisting of linear subspaces;

(ii) the canonical map V → limU V/U is an isomorphism, where the limit runs over open linear subspaces of V ;

(iii) there exists an isomorphism V →˜ limα Vα, where α 7→ Vα is a coﬁltered diagram of discrete vector spaces with surjective transition maps. Proof. Left as an exercise.

Deﬁnition 4.2. A linearly complete vector space is a topological vector space satisfying the conditions of Proposition 4.1. Clearly a ﬁnite direct sum of linearly complete vector spaces is linearly complete. We warn the reader that like topological vector spaces as a whole, the category of linearly complete vector spaces is not abelian. For example, if V is a linearly complete vector space, then we can consider the underlying discrete vector space V disc. Then we have a continuous linear map V disc → V which is a monomorphism and an epimorphism, but not an isomorphism unless V is discrete. We call a monomorphism, respectively an epimorphism, admissible if it is a closed embedding, respectively an open surjection. We call a sequence

0 −→ V −→ W −→ U −→ 0 exact if the underlying sequence of discrete vector spaces is exact and V → W and W → U are admissible. ∨ Given a linearly complete vector space V , we can consider its continuous dual V := Homcts(V, k). Note that a functional V → k is continuous if and only if its kernel is open. In some cases V ∨ also admits a natural linearly complete topology. Example 4.3. If V is a discrete vector space, then its dual V ∨ has a natural structure of linearly complete vector space. Namely, we have V = colimW W where the colimit runs over all ﬁnite- ∨ ∨ dimensional subspaces of V ordered by inclusion, and thus V = limW W . Proposition 4.4. The following are equivalent for a topological vector space V : (i) V is linearly complete and V/U is ﬁnite-dimensional for any open linear subspace U ⊂ V ;

(ii) there exists an isomorphism V →˜ limα Vα, where α 7→ Vα is a coﬁltered diagram of ﬁnite- dimensional vector spaces with surjective transition maps;

7 (iii) there exists an isomorphism V →˜ W ∨ where W is a discrete vector space and W ∨ is topologized as in Example 4.3.

Proof. If (i) holds, then by Proposition 4.1 we have V →˜ limU V/U where U runs over open linear subspaces in V . Thus (ii) follows because V/U is finite-dimensional for any U. Conversely, suppose that (ii) holds and let U ⊂ V be an open linear subspace. Then there exists α such that U contains the kernel of the given map ϕα : V → Vα. It follows that V/U→˜ Vα/ϕα(U), and the latter is finite-dimensional, so (i) holds. Evidently (iii) implies (ii). To see that (i) implies (iii), note that since V →˜ limU V/U, we have ∨ ∨ ∨∨ ∨∨ V →˜ colimU (V/U) and hence V →˜ limU (V/U) . But V/U is finite-dimensional for any U, so that V/U→˜ (V/U)∨∨ and V −→˜ lim V/U−→˜ lim(V/U)∨∨−→˜ V ∨∨. U U Thus V is dual to the discrete vector space V ∨.

Deﬁnition 4.5. A topological vector space satisfying the conditions of Proposition 4.4 is called linearly compact.A lattice in a linearly complete vector space is an open linearly compact subspace. A Tate vector space is a linearly complete vector space which admits a lattice.

Equivalently, a Tate vector space is a topological vector space isomorphic to V ⊕ W ∨, where V and W are discrete. Linearly compact vector spaces are sometimes called proﬁnite-dimensional because of condition (ii) in Proposition 4.4. Note that a linearly compact vector space is discrete if and only if it is ﬁnite-dimensional.

Example 4.6. The power series algebra k[[t]] with its t-adic topology is linearly compact. The Laurent series ﬁeld k((t)) is Tate, since it contains the lattice k[[t]] ⊂ k((t)).

Example 4.7. A linearly complete vector space V is compact (in the usual topological sense) if and only if k is ﬁnite and V is linearly compact, and it is locally compact if and only if k is ﬁnite and V is Tate.

Proposition 4.8. If V is a Tate vector space, then V ∨ admits a natural structure of Tate vector space, and the canonical map V → V ∨∨ is an isomorphism. The duality V 7→ V ∨ on Tate vector spaces interchanges linearly compact and discrete vector spaces.

Proof. We endow V ∨ with the topology generated by subspaces of the form

L⊥ := {λ ∈ V ∨ | λ(v) = 0 for all v ∈ L}

∼ ∨ ∨ ∼ ∨ where L ⊂ V is a lattice. Writing V = V1 ⊕V2 where V1 and V2 are discrete, we have V = V1 ⊕V2, ∨∨ ∨ since we saw that V2→˜ V2 in the proof of Proposition 4.4. Thus V is Tate, and the rest of the proposition is now clear.

The category of linearly complete vector spaces admits limits: one takes the limit of the underlying vector spaces and equips it with the limit topology. In particular, the forgetful functor to discrete vector spaces preserves limits. Colimits of linearly complete vector spaces also exist, and can be computed by taking the colimit of the underlying vector spaces equipped with the colimit topology, then completing. So the forgetful functor to discrete vector spaces generally does not preserve colimits. However, in some cases the colimit topology is already complete.

8 Proposition 4.9. Suppose α 7→ Vα is a ﬁltered diagram of linearly complete vector spaces with all transition maps Vα → Vβ being open embeddings. Then the colimit V := colimα Vα in the category of (not necessarily complete) topological vector spaces is linearly complete.

Proof. For a ﬁxed α the image of Vα → V is an open linearly complete subspace. Thus we have ∼ V = Vα ⊕ V/Vα, so V is a direct sum of two linearly complete vector spaces and hence linearly complete.

∼ Corollary 4.9.1. Any Tate vector space V can be written as a ﬁltered colimit V = colimα Vα where the Vα are linearly compact and all transition maps Vα → Vβ are open embeddings. The colimit can be taken in either the category of all topological vector spaces or in linearly complete vector spaces. ∼ Proof. Choose a lattice L ⊂ V , so V = L ⊕ V/L. Since V/L is discrete we have V/L = colimW W , ∼ where W runs over all ﬁnite-dimensional subspaces of V/L. Thus V = colimW L ⊕ W , and clearly each L ⊕ W is linearly complete and all the transition maps are open embeddings.

It follows that for any Tate vector space V we have V = colimL L, where the colimit runs over lattices L ⊂ V and can be taken in the category of all topological vector spaces. This corollary gives a diﬀerent perspective on the completeness of V ∨ for V Tate. Namely, we ∼ can write V = colimL L as above, and then

V ∨ =∼ lim L∨. L

∨ Since each L is linearly compact L is discrete, and each transition map L1 → L2 is an open ∨ ∨ embedding, which implies that L2 → L1 is surjective.

5 Completed tensor products

Given two linearly complete vector spaces V and W , it is natural to ask whether the tensor product V ⊗W of their underlying discrete vector spaces can be endowed with a linear topology. This would allow us to construct a completed tensor product and therefore stay in the category of linearly complete vector spaces. In fact there are several such topologies, which interact in an interesting way. The only case in which the answer is unambiguous is when either V or W is finite-dimensional, in which case V ⊗ W is already complete for the finite product topology. Define V ⊗! W to be the completion of V ⊗ W with respect to the topology generated by linear subspaces of the form V0 ⊗ W + V ⊗ W0, where V0 ⊂ V and W0 ⊂ W are open linear subspaces. Next, let V ⊗∗ W denote the completion of V ⊗ W with respect to the topology generated by linear subspaces U ⊂ V ⊗ W satisfying the following three conditions:

• there exist open subspaces V0 ⊂ V and W0 ⊂ W such that V0 ⊗ W0 ⊂ U;

• for any v ∈ V there exists an open subspace W1 ⊂ W such that v ⊗ W1 ⊂ U;

• for any w ∈ W there exists an open subspace V1 ⊂ V such that V1 ⊗ w ⊂ U. → Finally, we deﬁne V ⊗ W to be the completion of V ⊗ W with respect to the topology generated by linear subspaces U ⊂ V ⊗ W satisfying

9 • there exists an open subspace W0 ⊂ W such that V ⊗ W0 ⊂ U;

• for any w ∈ W there exists an open subspace V0 ⊂ V such that V0 ⊗ w ⊂ U. → ← → Note that the definition of V ⊗ W is not symmetric in V and W . We put V ⊗ W := W ⊗ V . One can verify that ⊗! and ⊗∗ define symmetric monoidal structures on linearly complete → vector spaces, while ⊗ defines a non-symmetric monoidal structure. In addition to the associativity constraints on each of the tensor products, we also have a natural continuous map

! ∗ ! ∗ (V ⊗ W ) ⊗ U −→ V ⊗ (W ⊗ U) (5.1) for any linearly complete vector spaces V , W , and U. Proposition 5.1. All three tensor products preserve exact sequences in either variable. Proof. Left as an exercise.

It is easy to see that the ∗-topology is ﬁner than the →-topology, and that the →-topology is stronger than the !-topology. It follows that we have natural continuous linear maps

∗ → ! V ⊗ W −→ V ⊗ W −→ V ⊗ W. (5.2)

Observe that we also have natural maps

∗ ∗ ← ! ! V ⊗ W −→˜ W ⊗ V −→ V ⊗ W −→ W ⊗ V −→˜ V ⊗ W.

Consider the resulting sequence

∗ → ← ! 0 −→ V ⊗ W −→ V ⊗ W ⊕ V ⊗ W −→ V ⊗ W −→ 0, (5.3) where the first map is the canonical one and the second map is the difference of the natural maps on the summands. Proposition 5.2. The sequence (5.3) is exact. Proof. Let us show that the second map is an open surjection, which will be used later. For this it suffices to show that the map is open, since it has dense image. The rest of the assertion is left as an exercise to the reader. To see that the map is open, we claim that a linear subspace U ⊂ V ⊗ W is open in the !- topology if it is open in the →- and ←-topologies. Then there exist open subspaces V0 ⊂ V and W0 ⊂ W such that V0 ⊗ W, V ⊗ W0 ⊂ U and hence

V0 ⊗ W + V ⊗ W0 ⊂ U.

Next, we investigate under what conditions the various tensor products coincide. → Proposition 5.3. If V is discrete or W is linearly compact, then V ⊗ W → V ⊗! W is an isomor- → phism. If V is linearly compact or W is discrete, then V ⊗∗ W → V ⊗ W is an isomorphism. Thus all three tensor products coincide in any of the following cases:

10 • V and W are both discrete;

• V and W are both linearly compact;

• either V or W is ﬁnite-dimensional (in which case they also coincide with V ⊗W , topologized as a ﬁnite product).

→ Proof. Suppose that V is discrete. To show that V ⊗W → V ⊗!W is an isomorphism, let U ⊂ V ⊗W be open in the →-topology, so we must prove it is open in the !-topology. Then there exists an open subspace W0 ⊂ V such that V ⊗ W0 ⊂ U. But {0} ⊂ V is open because V is discrete, so we have {0} ⊗ W + V ⊗ W0 = V ⊗ W0 ⊂ U, whence U is open in the !-topology. Now suppose that W is linearly compact, and again we ﬁx a linear subspace U ⊂ V ⊗ W which is open in the →-topology. As before there exists an open subspace W0 ⊂ V such that V ⊗W0 ⊂ U, and now W/W0 is ﬁnite-dimensional because W is linearly compact. Choose w1, ··· , wn ∈ W which span W/W0, so for each 1 ≤ i ≤ n there exists an open subspace Vi ⊂ V such that Vi ⊗ wi ⊂ U. Now put V0 := V1 ∩ · · · ∩ Vn, so we have

n X V0 ⊗ W + V ⊗ W0 = V0 ⊗ W0 + V0 ⊗ wi + V ⊗ W0 ⊂ U i=1 as needed. The remaining cases are similar and are left as an exercise to the reader.

It is helpful to know how to compute these tensor products as limits and colimits. The following result can also be made into an alternate deﬁnition with some additional care to ensure naturality.

Proposition 5.4. Let V and W be linearly complete vector spaces. All limits (respectively colimits) below are assumed cofiltered (respectively filtered), and colimits are taken in the category of linearly complete vector spaces (i.e. in general they must be completed). ∼ ∼ (i) Choose isomorphisms V = limα Vα and W = limβ Wβ where Vα and Wβ are discrete. Then the resulting map ! V ⊗ W −→ lim lim Vα ⊗ Wβ α β is an isomorphism. ∼ (ii) Fix an isomorphism W = limα colimβ Wαβ where the Wαβ are finite-dimensional. Then we have an isomophism → V ⊗ W −→˜ lim colim V ⊗ Wαβ. α β ∼ (iii) Suppose that V is Tate, so by Corollary 4.9.1 we have V = colimL L where L runs over lattices in V . Then we have an isomorphism

∗ → V ⊗ W −→˜ colim L ⊗ W. L

11 ∼ (iv) If W is Tate, so that W = colimL L where L runs over lattices in W , then we have an isomorphism ∗ ← V ⊗ W −→˜ colim V ⊗ L. L Proof. Left as an exercise.

n Example 5.5. Writing k((t)) = limn k((t))/t k[[t]], we obtain ! k((s)) ⊗ k((t)) = lim lim(k((s))/smk[[s]]) ⊗ (k((t))/tmk[[t]]) = k[[s, t, s−1, t−1]], m n where the right side is the space of doubly infinite series X m n amns t m,n∈Z with amn ∈ k. −m n Using the expression k((t)) = limn colimm t k[[t]]/t k[[t]], we can compute → k((s)) ⊗ k((t)) = lim colim k((s)) ⊗ (t−mk[[t]]/tnk[[t]]) = k((s))((t)), n m and similarly we have ← k((s)) ⊗ k((t)) = k((t))((s)). −n Finally, since k((s)) = colimn s k[[s]], we find that ∗ k((s)) ⊗ k((t)) = colim (s−nk[[s]]) ⊗ k((t)) = k[[s, t]][s−1, t−1]. n One can observe directly in this case that the sequence (5.3) 0 −→ k[[s, t]][s−1, t−1] −→ k((s))((t)) ⊕ k((t))((s)) −→ k[[s, t, s−1, t−1]] −→ 0 is exact. Note that none of the tensor products are Tate. The description of ⊗∗ in the previous proposition only applies to Tate vector spaces, but the next result helps to explain its significance in general. Proposition 5.6. Let V , W , and U be linearly complete vector spaces. Restriction along the natural map V × W → V ⊗∗ W induces an isomorphism from the space of continuous linear maps V ⊗∗ W → U to the space of continuous bilinear maps V × W → U.

Proof. First suppose that λ : V ⊗ W → U is such that λ|V ×W is continuous, so we must show that λ is continuous in the ∗-topology. Since U is the limit of its discrete quotients, we can assume that U is discrete, and thus it suﬃces to show that ker λ is open. Now (V ×W )∩ker λ is open in V ×W , so there exist open linear subspaces V0 ⊂ V and W0 ⊂ W such that V0 × W0 ⊂ ker λ and hence V0 ⊗ W0 ⊂ ker λ. Given v ∈ V , we obtain a continuous linear map W → U given by w 7→ λ(v, w), so there exists and open subspace W1 ⊂ W such that {v} × W0 ⊂ ker λ and hence v ⊗ W0 ⊂ ker λ. Similarly, for any w ∈ W we can ﬁnd an open subspace V0 ⊂ V such that V0 ⊗ w ⊂ ker λ. Conversely, assume that λ : V ⊗ W → U is continuous in the ∗-topology. Again we can assume that U is discrete, so we must show that any (v, w) ∈ V ×W has an open neighborhood on which λ is constant. There exist open linear subspaces V0 ⊂ V and W0 ⊂ W such that V0 ⊗ w, v ⊗ W0 ⊂ ker λ, and by shrinking V0 and W0 if necessary we may assume that V0 ⊗ W0 ⊂ ker λ. Then we have

(v + V0) ⊗ (w + W0) = v ⊗ w + V0 ⊗ w + v ⊗ W0 + V0 ⊗ W0 ⊂ v ⊗ w + ker λ, which implies that λ is constant on (v + V0) × (w + W0).

12 6 The trace on Tate vector spaces

Let V and W be Tate vector spaces. Denote by Homcts(V,W ) the k-vector space of continuous linear maps. We write Homd(V,W ) ⊂ Homcts(V,W ) for the subspace consisting of maps with open kernel. A subset of a Tate vector space is said to be bounded if it is contained in a lattice, and we deﬁne the subspace

Homc(V,W ) ⊂ Homcts(V,W ) to consist of maps with bounded image. We call

Homtr(V,W ) := Homc(V,W ) ∩ Homd(V,W ) the space of trace-class maps. Proposition 6.1. We have canonical isomorphisms

! ∨ V ⊗ W −→˜ Homcts(V,W ), → ∨ V ⊗ W −→˜ Homc(V,W ), ← ∨ V ⊗ W −→˜ Homd(V,W ), and ∗ ∨ V ⊗ W −→˜ Homtr(V,W ).

Proof. Write V = colimL L where L runs over the lattices in V , and W = limU W/U where U runs over open linear subspaces in W . Then we have

! ∨ ∨ Homcts(V,W )−→ ˜ lim lim Homcts(L, W/U)−→ ˜ lim lim L ⊗ (W/U)−→ ˜ V ⊗ W. L U L U For the next isomorphism, observe that → ∨ ∨ Homc(V,W )−→ ˜ lim Homc(V, W/U)−→ ˜ lim V ⊗ (W/U)−→ ˜ V ⊗ W. U U Here the ﬁrst isomorphism is because V → W has bounded image if and only V → W/U has bounded image for every U, and the second isomorphism is because V → W/U has bounded image if and only if it has ﬁnite rank. Note that a continuous linear map V → W has open kernel if and only if the transpose W ∨ → V ∨ has bounded image. Thus → ← ∨ ∨ ∨ ∨ Homd(V,W )−→ ˜ Homc(W ,V )−→ ˜ W ⊗ V = V ⊗ W. We could deduce the last isomorphism from Proposition 5.2, but let us give a direct argument. Writing W = colimL L where L runs over the lattices in W , we have

Homtr(V,W )−→ ˜ colim Homtr(V,L) L because trace-class maps are bounded. But Homtr(V,L) = Homd(V,L) because L is linearly compact, so we obtain ← ∗ ∗ ∨ ∨ ∨ Homtr(V,W )−→ ˜ colim Homd(V,L)−→ ˜ colim V ⊗ L−→˜ colim V ⊗ L−→˜ V ⊗ W. L L L

13 Corollary 6.1.1. Each of the spaces Homcts(V,W ), Homc(V,W ), Homd(V,W ), and Homtr(V,W ) admits a natural linearly complete topology. The sequence

0 −→ Homtr(V,W ) −→ Homc(V,W ) ⊕ Homd(V,W ) −→ Homcts(V,W ) −→ 0 is exact, where the ﬁrst map is given by the two inclusions and the second is the diﬀerence of the inclusions. In particular we have

Homcts(V,W ) = Homc(V,W ) + Homd(V,W ).

Proof. Apply Proposition 5.2.

Recall that for any vector space V , there is a canonical trace functional deﬁned on ﬁnite rank linear endomorphisms of V . Namely, such endomorphisms identify with V ∨ ⊗ V , and trace is just the evaluation pairing.

Proposition 6.2. If V is a Tate vector space, then the trace pairing V ∨ ⊗ V → k extends uniquely to a continuous functional ∗ ∨ ∼ tr : V ⊗ V = Endtr(V ) −→ k. Proof. By Proposition 5.6, it suﬃces to show that the trace is continuous as a map V ∨ × V → k. Since k is discrete, this means we must show that for any λ ∈ V ∨ and v ∈ V , the trace is constant in a neighborhood of (λ, v). For this, choose a lattice L1 ⊂ ker λ and let L2 := L1 + k · v. Then for ⊥ any µ ∈ L2 and w ∈ L1, we have

(λ + µ)(v + w) = λ(v) + µ(v) + λ(w) + µ(w) = λ(v),

⊥ i.e. the trace is constant on (λ + L2 ) × (v + L1).

Let us give a concrete construction of the trace, which involves some choices but is more suitable for computation. If V is a Tate vector space and ϕ : V → V is a trace-class operator, we can choose lattices L1 ⊂ ker ϕ and L2 ⊃ im ϕ since ϕ has open kernel and bounded image. Then L1 + L2, L1 ∩ L2 are lattices and that (L1 + L2)/(L1 ∩ L2) is ﬁnite-dimensional. Since ϕ sends L1 + L2 into L2 ⊂ L1 + L2 and vanishes on L1 ⊃ L1 ∩ L2, it induces an endomorphism of (L1 + L2)/(L1 ∩ L2). One can check that the trace of ϕ on (L1 + L2)/(L1 ∩ L2) does not depend on the choice of L1,L2. Proposition 6.3. The two deﬁnitions of trace agree.

Proof. First suppose that ϕ : V → V has ﬁnite rank. Then we can moreover assume that ϕ = λ⊗v ∨ for some λ ∈ V and v ∈ V . Choose L1 ⊂ ker ϕ = ker λ and put L2 := L1 + k · v. Then ∼ ∼ (L1 + L2)/(L1 ∩ L2) = L2/L1 = k · v and ϕ acts on this line by the scalar λ(v). Thus the trace of ϕ in either sense is λ(v). Since the ﬁnite rank operators are dense, it remains to show that the second trace map (involving ∨ ∗ the choice of lattices) is continuous on V ⊗ V . Write V = limL1 V/L1 = colimL2 L2, where both the limit and the colimit run over lattices in V . Then we have

! ∨ ∗ ∨ V ⊗ V = colim (V/L1) ⊗ L2, L1,L2

14 where the latter is equipped with the colimit topology. On each term the trace map in question factors through a ﬁnite-dimensional vector space:

! ∨ ∨ (V/L1) ⊗ L2 −→ ((L1 + L2)/(L1 ∩ L2)) ⊗ (L1 + L2)/(L1 ∩ L2) −→ k. But the kernel of the ﬁrst map is the open subspace

! ! ∨ ∨ (V/(L1 + L2)) ⊗ L2 + (V/L1) ⊗ (L1 ∩ L2).

Example 6.4. Let V , L, and W be ﬁnite-dimensional, linearly compact, and discrete respectively. Suppose we have an endomorphism ϕ0 : V → V and an injection ψ : W → L. Then the map ϕ : V ⊕ L ⊕ W −→ V ⊕ L ⊕ W given by the matrix   ϕ0 0 0  0 0 ψ 0 0 0 is trace-class, and tr ϕ = tr ϕ0. The following observation will be useful in the sequel.

Proposition 6.5. If ϕ ∈ Endc(V ) and ψ ∈ Endd(V ), then [ϕ, ψ] ∈ Endtr(V ) and tr([ϕ, ψ]) = 0. Proof. Left as an exercise.

7 The Tate extension

For a Tate vector space V , we can view the continuous endomorphisms Endcts(V ) as an associative algebra under composition. One checks that Endc(V ) and Endd(V ) are two-sided ideals, hence so is Endtr(V ). Note that Endcts(V ) is a topological algebra with respect to the linearly complete topology coming from the isomorphism with V ∨⊗!V . One way to see this is by interpreting the multiplication as the continuous map

! ∗ ! ! ∗ ! (V ∨ ⊗ V ) ⊗ (V ∨ ⊗ V )−→ ˜ (V ∨ ⊗ V ) ⊗ (V ∨ ⊗ V ) ! ∗ ! −→ V ∨ ⊗ (V ⊗ (V ∨ ⊗ V )) ! ∗ ! −→ V ∨ ⊗ (V ⊗ V ∨) ⊗ V ! −→ V ∨ ⊗ V, where the ﬁrst map is the symmetry constraint for ⊗∗, the second and third are induced by (5.1), and the last map is given by the trace functional. Recall that any associative algebra A acquires the structure of Lie algebra under the commutator bracket [x, y] := xy − yx (x, y ∈ A). Namely, the commutator bracket is bilinear and we have

15 • (antisymmetry) [x, y] = −[y, x], and

• (Jacobi identity) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for x, y, z ∈ A. Any two-sided ideal I ⊂ A is then a Lie ideal, i.e. we have [x, y] ∈ I for x ∈ A and y ∈ I. If A is a linearly complete topological algebra, one checks that the commutator bracket makes A into a topological Lie algebra. Thus Endcts(V ) is a topological Lie algebra under the commutator bracket, and the two-sided ideals Endc(V ), Endd(V ), and Endtr(V ) are Lie ideals with respect to this bracket. Given Tate vector spaces V and W , we deﬁne the space of asymptotic maps V → W by ∼ Hom∞(V,W ) := Homcts(V,W )/ Homc(V,W ) = Homd(V,W )/ Homtr(V,W ).

If V = W then End∞(V ) is naturally a topological associative algebra, being the quotient of Endcts(V ) by the closed two-sided ideal Endc(V ). The projection Endd(V ) → End∞(V ) is a morphism of Lie algebras, but not of unital associative algebras because idV ∈/ Endd(V ). We deﬁne the Tate extension of End∞(V ) to be the pushout

Endtr(V ) Endd(V )

[ k End∞(V ) , or more explicitly the quotient

[ End∞(V ) := (Endd(V ) ⊕ k)/{(ϕ, − tr(ϕ)) | ϕ ∈ Endtr(V )}.

One can check that for ϕ ∈ Endcts(V ) and ψ ∈ Endtr(V ) we have tr([ϕ, ψ]) = 0, which implies that [ End∞(V ) is the quotient of Endd(V ) by a Lie ideal and hence inherits a Lie bracket. By construction we have an extension of topological Lie algebras

[ 0 −→ k −→ End∞(V ) −→ End∞(V ) −→ 0, which is moreover a central extension in the sense that k is contained in the center of the Lie [ algebra End∞(V ) . Next, we deﬁne the Tate extension of Endcts(V ) to be the ﬁber product

[ Endcts(V ) Endcts(V )

[ End∞(V ) End∞(V ).

It is a central extension of topological Lie algebras

[ 0 −→ k −→ Endcts(V ) −→ Endcts(V ) −→ 0. We will write 1 for the generator of the central copy of k when necessary to disambiguate the notation. It is immediate from the construction that this extension admits canonical splittings σc and σd over Endc(V ) and Endd(V ) respectively, and that σc(ϕ) − σd(ϕ) = tr(ϕ)1 for ϕ ∈ Endtr(V ). Moreover, the splittings σc and σd commute with the adjoint action of Endcts(V ), and in particular [ σc(Endc(V )) and σd(Endd(V )) are Lie ideals in Endcts(V ) .

16 [ Proposition 7.1. If V is either discrete or linearly compact, then the Tate extension Endcts(V ) splits canonically.

Proof. If V is discrete then we have Endcts(V ) = Endd(V ), and if V is linearly compact then [ we have Endcts(V ) = Endc(V ). By the above description of Endcts(V ) , in either case it splits canonically.

[ Note that the Lie bracket on Endcts(V ) factors as

∗ ∗ [ [ [ [ , ] [ Endcts(V ) ⊗ Endcts(V ) −→ Endcts(V ) ⊗ Endcts(V ) −−−→ Endcts(V ) , with the composition

∗ [ [ , ] [ Endcts(V ) ⊗ Endcts(V ) −−−→ Endcts(V ) −→ Endcts(V ) being the commutator bracket [ , ]. In order to write an explicit formula for the bracket [ , ][, it is necessary to ﬁrst make an auxiliary choice. Suppose we are given a central extension of Lie algebras

0 −→ k −→ g[ −→ g −→ 0 with the bracket denoted by [ , ][ : g ⊗ g −→ g[. Then a choice of k-linear splitting σ : g → g[ determines a skew-symmetric pairing

[ h , iσ : g ⊗ g −→ k x ⊗ y 7→ [x, y][ − σ([x, y]).

[ [ Conversely, the pairing h , iσ determines the bracket [ , ] because [ [ 1 [x, y] = σ([x, y]) + hx, yiσ . In our situation, the Tate extension can be split by a choice of decomposition V =∼ L⊕W where L ⊂ V is a lattice and W is discrete. Denote by πL ∈ Endc(V ) and πW ∈ Endd(V ) the operators of projection onto L and W respectively, so in particular πL + πW = idV . Then we obtain a splitting

[ σL : Endcts(V ) −→ Endcts(V )

ϕ 7→ σc(ϕπL) + σd(ϕπW )

(note the abuse of notation: the section σL depends on W as well as L). We simply write ∗ [ h , iL : Endcts(V ) ⊗ Endcts(V ) −→ k for the associated continuous pairing. ∼ Proposition 7.2. Fix a decomposition V = L ⊕ W as above. For any ϕ, ψ ∈ Endcts(V ), the operator [ϕπL, ψπL] − [ϕ, ψ]πL is trace-class and we have

[ hϕ, ψiL = tr([ϕπL, ψπL] − [ϕ, ψ]πL), or equivalently [ [ϕ, ψ] = σL([ϕ, ψ]) + tr([ϕπL, ψπL] − [ϕ, ψ]πL)1.

17 Proof. Certainly [ϕπL, ψπL] − [ϕ, ψ]πL ∈ Endc(V ), since πL ∈ Endc(V ). Now observe that

[ϕπL, ψπL] − [ϕ, ψ]πL = [ϕ, ψ] − [ϕπL, ψπW ] − [ϕπW , ψπL] − [ϕπW , ψπW ] − [ϕ, ψ]πL

= [ϕ, ψ]πW − [ϕπL, ψπW ] − [ϕπW , ψπL] − [ϕπW , ψπW ] ∈ Endd(V ) because πW ∈ Endd(V ). Thus [ϕπL, ψπL] − [ϕ, ψ]πL is trace-class as desired. The second formula can be rewritten as

[ [ϕ, ψ] = σc([ϕ, ψ]πL) + σd([ϕ, ψ]πW ) + (σc − σd)([ϕπL, ψπL] − [ϕ, ψ]πL).

Expanding and simplifying the right side, we obtain

σc([ϕπL, ψπL]) + σd([ϕ, ψ] − [ϕπL, ψπL])

= σc([ϕπL, ψπL]) + σd([ϕπL, ψπW ] + [ϕπW , ψπL] + [ϕπW , ψπW ]) [ [ [ [ = [ϕπL, ψπL] + [ϕπL, ψπW ] + [ϕπW , ψπL] + [ϕπW , ψπW ] = [ϕ, ψ][.

The following identity will be needed in what follows.

Proposition 7.3. For any ϕ, ψ, ρ ∈ Endcts(V ) we have

[ϕψ, ρ][ + [ρϕ, ψ][ + [ψρ, ϕ][ = 0.

Proof. An easy calculation shows that

[ϕψ, ρ] + [ρϕ, ψ] + [ψρ, ϕ] = 0.

Choosing a decomposition V =∼ L ⊕ W , it therefore suﬃces to show that

[ [ [ hϕψ, ρiL + hρϕ, ψiL + hψρ, ϕiL = 0.

Applying Proposition 7.2 and using the identity for [ , ] again, the left side can be written as

tr([ϕψπL, ρπL] + [ρϕπL, ψπL] + [ψρπL, ϕπL]).

Now observe that [ϕψπL, ρπL] = [ϕπLψπL, ρπL] + [ϕπW ψπL, ρπL], and that the second term on the right side is trace-class with vanishing trace by Proposition 6.5. Applying similar reasoning to the other terms, the expression in question becomes

tr([ϕπLψπL, ρπL] + [ρπLϕπL, ψπL] + [ψπLρπL, ϕπL]).

Using the identity for [ , ] once more, we see that this operator vanishes even before taking its trace.

18 8 Abstract residues

Let A be a commutative topological algebra whose underlying topological vector space is linearly complete (the reader should keep the example A = k((t)) in mind). Suppose that M is a linearly complete A-module, i.e. a linearly complete vector space equipped with an associative and unital action ∗ A ⊗ M −→ M. A continuous derivation with values in M is a continuous k-linear map ∂ : A → M satisfying

∂(xy) = x∂(y) + y∂(x)(x, y ∈ A).

1 Proposition 8.1. There exists a topological A-module ΩA equipped with a continuous derivation 1 d : A → ΩA which is universal in the sense that that for any A-module M, any continuous derivation 1 A → M factors uniquely through a continuous A-linear map ΩA → M. ∗ ∗ Proof. Let I∆ ⊂ A ⊗ A be the kernel of the multiplication A ⊗ A → A. It is a closed ideal in the topological algebra A ⊗∗ A. We put

1 ∼ 2 ΩA := I∆ ⊗ A = I∆/I∆, A⊗∗A

1 which is equipped with the quotient topology via the projection I∆ → ΩA. The universal continuous derivation is deﬁned by the formula

1 d : A −→ ΩA x 7→ x ⊗ 1 − 1 ⊗ x.

We leave the veriﬁcation of the universal property to the reader.

Note that a continuous homomorphism A → B of linearly complete commutative algebras 1 1 induces a continuous A-linear map ΩA → ΩB. Namely, the composition

d 1 A −→ B −→ ΩB

1 is a continuous derivation, and therefore factors through A → ΩA to give the desired map. 1 Proposition 8.2. The functor A 7→ ΩA preserves limits, as well as ﬁltered colimits along open embeddings.

1 Proof. It is easy to see that A 7→ ΩA preserves products, so let α 7→ Aα be a coﬁltered diagram of linearly complete commutative algebras and put A := limα Aα. Passing to the limit of the derivations A → Ω1 yields a continuous derivation A → lim Ω1 , which we claim is universal. α Aα α Aα Suppose we are given a linearly complete A-module M and a continuous derivation ∂ : A → M. Without loss of generality, we can assume that M is discrete. Then ∂ factors through A → Aα for some α, and hence through a continuous A -linear map Ω1 → M. The resulting continuous α Aα A-linear map lim Ω1 −→ Ω1 −→ M α Aα Aα yields the desired factorization of ∂. Now suppose that β 7→ Bβ is a ﬁltered diagram of linearly complete commutative algebras with the transition maps being open embeddings, and put B := colimβ Bβ. The forgetful functor

19 from linearly complete commutative algebras to linearly complete vector spaces preserves ﬁltered colimits, as is the case for commutative algebras in any category. Thus Proposition 4.9 implies that the formula B = colimβ Bβ also holds in the category of all topological vector spaces, i.e. colimβ Bβ is already complete in the colimit topology. Thus the derivations 1 1 Bβ −→ ΩB −→ colim ΩB β β β

1 give rise to a continuous derivation B → colimβ Ωβ, which we claim is universal. Suppose we are given a linearly complete B-module M and a continuous derivation ∂ : B → M. Then each of the derivations ∂ Bβ −→ B −→ M 1 factors uniquely through a continuous Bβ-linear map Ωβ → M. Passing to the colimit over β, we 1 obtain the required B-linear continuous map colimβ Ωβ → M.

If the topological k-vector space underlying M is Tate, we will call M a Tate A-module. In that case the action of A can be expressed as a continuous homomorphism of topological associative algebras A → Endcts(M). We deﬁne the Heisenberg extension of A attached to M to be the ﬁber product [ AM A

[ Endcts(M) Endcts(M). It is a central extension of topological Lie algebras [ 0 −→ k −→ AM −→ A −→ 0, where A is given the zero Lie bracket. Such an extension is determined by a continuous pairing ∗ [ h , iM : A ⊗ A −→ k. Explicitly, the composition

∗ ∗ [ [ , ] [ A ⊗ A −→ Endcts(M) ⊗ Endcts(M) −−−→ Endcts(M) [ lands in the central copy of k because A is commutative, and this deﬁnes the pairing h , iM . If we ∼ [ choose a decomposition M = L ⊕ W with L ⊂ M a lattice and W discrete, then the pairing h , iM [ agrees with the restriction to A of the previously deﬁned pairing h , iL. Proposition 8.3. If the Tate vector space underlying M is discrete or linearly compact, then the [ pairing h , iM vanishes identically. Proof. This follows immediately from Proposition 7.1.

Suppose that the topological vector space underlying A is Tate. We will call such a topological algebra a Tate commutative algebra. Then A∨ is a Tate vector space, and we can consider the continuous k-linear map A −→ A∨ [ g 7→ (f 7→ hf, giM ).

20 Proposition 7.3 implies that for any f, g, h ∈ A we have

[ [ [ hfg, hiM = hf, ghiM + hg, fhiM , which says precisely that A → A∨ is a derivation. Thus there is a unique continuous A-linear map

1 ∨ ΩA −→ A

ﬁtting into the commutative triangle d 1 A ΩA

A∨. In particular, we obtain a continuous functional

1 ∨ resM :ΩA −→ A −→ k, where the second map is evaluation at 1 ∈ A. We will be especially interested in this functional in the case M = A with the action of A on itself by multiplication. Notice also that resM determines the continuous pairing

∗ 1 A ⊗ ΩA −→ k f ⊗ η 7→ resM (fη).

Proposition 8.4. Choose a decomposition M =∼ L ⊕ W as above. For any f, g ∈ A, we have [fπL, g] ∈ Endtr(M) and resM (f dg) = tr([fπL, g]). Proof. By Proposition 7.2, the operator

[fπL, gπL] − [f, g]πL = [fπL, gπL] is trace-class and we have

[ resM (f dg) = hf, giL = tr([fπL, gπL]).

The right side can be rewritten as

tr([fπL, gπL]) = tr([fπL, g] − [fπL, gπW ]) = tr([fπL, g]), where the second equality used Proposition 6.5.

We record two more basic properties of abstract residues for later use.

Proposition 8.5. For any short exact sequence of Tate A-modules

0 −→ M −→ N −→ P −→ 0, we have resN = resM + resP .

21 Proof. Let Endcts(N,M) ⊂ Endcts(N) be the subalgebra consisting of endomorphisms which preserve M, so we have a natural homomorphism

Endcts(N,M) −→ Endcts(M) × Endcts(P ).

The proposition will follow if we can show that the pullback along this map of the central extension [ [ [ (Endcts(M) ×Endcts(P ) )/(1M −1P ) is canonically isomorphic to the pullback of Endcts(N) along the inclusion Endcts(N,M) → Endcts(N). By the construction of these central extensions it suﬃces to check the commutativity of the square

Endtr(N,M) Endtr(M) × Endtr(P )

trM + trP

trN Endtr(N) k, which is easily deduced from Proposition 6.3 and the additivity of the usual trace on ﬁnite- dimensional vector spaces.

Proposition 8.6. Let A → B be a continuous homomorphism of Tate commutative algebras and M a Tate B-module. Then the following triangle commutes:

1 1 ΩA ΩB

resM resM k.

[ Proof. It suﬃces to show that the pairing h , iM on B coincides with the same-named pairing when restricted to A. This follows immediately from the existence of the cartesian square

[ AM A

[ BM B.

9 Local Serre duality

For now X will denote a smooth curve over k. For simplicity we will assume that k is perfect. Given a closed point x in X, we denote by Ox the completed local ring at x. Write mx ⊂ Ox for the maximal ideal and kx := Ox/mx for the residue field, which is a finite extension of k. By n completeness we have Ox = limn Ox/mx, and in particular Ox is naturally equipped with a linearly compact topology. Recall that Ox is a discrete valuation ring, and that a coordinate at x is a choice 2 of t ∈ mx \ mx, i.e. a generator for mx. Writing Kx for the fraction field of Ox, for any coordinate −1 t we have Kx = Ox[t ]. In particular Kx is is naturally a Tate vector space. We think of Ox and Kx as the rings of functions on the formal disk and formal punctured disk centered at x, respectively. This suggests Spec Ox and Spec Kx as candidates for the the disk and

22 punctured disk. However, these objects are insensitive to the topologies on Ox and Kx, and for this reason are insuﬃcient for our purposes. A better candidate for the disk is the formal spectrum

n Spf Ox = colim Spec(Ox/m ), n x from which we can recover Ox as a topological algebra. Unfortunately this construction does not make sense for Kx, since its topology is not induced by any ideal in this way. Moreover, it is not difficult to see that (Spf Ox) \{x} = ∅, i.e. nothing is left over after puncturing this version of the disk. For the same reason, there is no nonconstant map Spec Kx → Spf Ox. In what follows we generally work with Ox and especially Kx simply as topological algebras, without attempting to view them as functions on an algebro- geometric object. The next result tells us that the formal disk at x can be identified with the formal disk at the origin in 1 . Akx Proposition 9.1. A choice of coordinate t at x determines a continuous isomorphism of k-algebras kx[[t]]→ ˜ Ox, and hence an isomorphism kx((t))→ ˜ Kx. n Proof. We claim that the projection Ox/mx → kx splits uniquely as a k-algebra homomorphism for any n ≥ 1. Namely, this map splits uniquely over k, and since the extension kx/k is separable the map Spec kx → Spec k is étale.Since the kernel of the projection is nilpotent, the claim follows from the infinitesimal lifting property of étalemaps. Having obtained a k-algebra map kx → Ox, a choice of coordinate yields an extension to a n n homomorphism kx[t] → Ox. By the definition of a coordinate, the induced map kx[t]/(t ) → Ox/mx is an isomorphism for any n ≥ 1. Passing to the limit, we obtain the desired continuous isomorphism kx[[t]]→ ˜ Ox. The other isomorphism follows immediately by inverting t.

Letting k0/k be any ﬁnite extension of k, consider the maps

k0[[t]] −→ k0[[t]]dt and k0((t)) −→ k0((t))dt, both given by the formula f(t) 7→ f 0(t)dt (here dt is just a formal symbol). These are clearly continuous derivations over k, and hence uniquely factor through the universal continuous derivations

0 1 0 1 k [[t]] −→ Ωk0[[t]] and k ((t)) −→ Ωk0((t)).

1 0 1 0 Proposition 9.2. The resulting maps Ωk0[[t]] → k [[t]]dt and Ωk0((t)) → k ((t))dt are isomorphisms. In particular Ω1 and Ω1 are free modules of rank one over O and K respectively, and both are Ox Kx x x generated by the element dt for any coordinate t at x. Proof. Suppose we are given a continuous k-linear derivation ∂ : k0[[t]] → M, where M is a Tate 0 0 1 k [[t]]-module over k. Since k /k is separable we have Ωk0/k = 0, from which it follows that ∂ is k0-linear. Then it is easy to check that

k0[[t]]dt −→ M f(t)dt 7→ f(t)∂t is the unique continuous k0[[t]]-linear factorization of ∂ through the derivation k0[[t]] → k0[[t]]dt de- 1 0 ﬁned above. Thus the latter is the universal continuous derivation, which implies that Ωk0[[t]]→˜ k [[t]]dt.

23 Any continuous k-linear derivation k0((t)) → M, where M is a Tate k0((t))-module over k, can be restricted to obtain a derivation k0[[t]] → M. By the previous step this extends uniquely to a continuous k0[[t]]-linear map k0[[t]]dt → M. Since t acts invertibly on M, this map extends uniquely to a k0((t))-linear map k0((t))dt → M, which is easily seen to be continuous. This shows that k0((t))dt has the required universal property. The second statement follows immediately by Proposition 9.1.

Motivated by complex analysis, we consider the explicit continuous functional

kx((t))dt −→ k X i ait dt 7→ trkx/k(a−1). i∈Z Choosing a coordinate at t, we obtain

1 ΩKx −→˜ kx((t))dt −→ k. (9.1) When k has characteristic zero it is not diﬃcult to prove that this functional is coordinate- independent, but in positive characteristic this is far from clear.

Theorem 9.3. The functional (9.1) agrees with resKx , and in particular does not depend on the choice of coordinate at x. The continuous pairing of Tate vector spaces

∗ 1 Kx ⊗ ΩKx −→ k

f ⊗ η 7→ resKx (fη) is perfect, i.e. it induces an isomorphism

1 ∨ ΩKx −→˜ Kx . Proof. Since the construction of abstract residues is manifestly invariant under isomorphism, for the ﬁrst claim it suﬃces to show that ! X i reskx((t)) ait dt = trkx/k(a−1). i∈Z

When k = kx, this can be veriﬁed by a direct calculation which is left as an instructive exercise for the reader. For the general case, recall that if V is a ﬁnite-dimensional vector space over kx and ϕ : V → V is a kx-linear endomorphism, then

trV/k(ϕ) = trkx/k(trV/kx (ϕ)). Proposition 6.3 implies that the same formula holds when V is Tate and ϕ is trace-class, and then the claim follows from Proposition 8.4. ∨ Next, we claim that Kx is a one-dimensional Kx-vector space. Fix a nonzero k-linear map λ : kx → k and a coordinate t at x, and for any n ∈ Z consider the continuous functional

λn : Kx −→ k X i ait 7→ λ(an). i∈Z

24 m ∨ Note that for m ∈ Z we have t ·λn = λn−m, and clearly λ spans kx as a kx-vector space. It follows m ∨ that the functionals at · λ0 span a dense subspace in Kx . It is straightforward to check that the subspace topology on Kx · λ0 agrees with the topology on Kx, which implies that Kx · λ0 is closed ∨ because Kx is complete. Thus Kx is one-dimensional over Kx as desired. According to Proposition 9.2, the K -vector space Ω1 is also one-dimensional. Thus it suﬃces x Kx to show that the map Ω1 → K∨ induced by the residue pairing is nonzero. Choosing a coordinate Kx x t at x, this map sends dt to the functional

kx((t)) −→ k X i ait 7→ trkx/k(a−1). i∈Z

The trace map trkx/k is nonzero because kx/k is separable, whence the claim.

Corollary 9.3.1. We have O⊥ = Ω1 under the pairing in the theorem, i.e. it induces isomor- x Ox phisms 1 ∨ 1 1 ∨ ΩOx →˜ (Kx/Ox) and Ox−→˜ (ΩKx /ΩOx ) . Proof. It suﬃces to show that O⊥ = Ω1 under the residue pairing, i.e. for η ∈ Ω1 , we have x Ox Ox res (fη) = 0 for all f ∈ O if and only if η ∈ Ω1 . Choosing a coordinate t at x, we can write Kx x Ox

X i η = ait dt. i∈Z

n Then we have resKx (t η) = trkx (a−n−1) for n ∈ Z, which immediately implies the claim.

Note that any ﬁnite-dimensional vector space V over Kx naturally acquires the structure of Tate vector space over k. Namely, a choice of Kx-basis for V allows us to transport the product ∼ r topology along V = Kx, and this topology is easily seen to be independent of the basis. We think of V as the space of sections of a vector bundle on the punctured disk at x, and we will see below that any vector bundle on X gives rise to such a V .

Corollary 9.3.2 (Local Serre duality). For any ﬁnite-dimensional vector space V over Kx, the pairing of Tate vector spaces over k

∗ 1 1 resKx V ⊗ HomKx (V, ΩKx ) −→ ΩKx −−−→ k is perfect. If L ⊂ V is an Ox-stable lattice such that Kx ⊗Ox L→˜ V , then under the above pairing we have ⊥ 1 1 1 L = HomOx (L, ΩOx ) ⊂ HomOx (L, ΩKx ) = HomKx (V, ΩKx ).

Proof. After choosing a Kx-basis of V , the ﬁrst statement follows immediately from the theorem. For the second assertion, we claim that if V is r-dimensional over Kx, then any such L is free of rank r over Ox. Namely, the isomorphism Kx ⊗Ox L→˜ V implies that L has rank r, and since L is torsion-free over the PID Ox, it is therefore free. It is easy to see that any Ox-basis for L is also a Kx-basis for V , and after choosing such a basis, we are done by the previous corollary.

25 10 The ring of ad`eles

Our main tool for passing from local to global is the ring of adèles Y Y AX := colim Kx × Ox, I x∈I x/∈I where the colimit runs over all finite sets I of closed points in X. In what follows we usually omit the subscript and simply write A when it will not cause confusion. Explicitly, an element of A is a tuple (fx), where fx ∈ Kx for all closed points x in X and fx ∈ Ox for all but finitely many x. We equip A with the colimit topology, and since the transition maps are open embeddings, Proposition 4.9 implies that A is linearly complete. In fact A is a Tate commutative algebra containing the canonical lattice Y OX := Ox, x which is also a subalgebra. Again, we will often omit the subscript and write O. Proposition 10.1. We have canonical continuous isomorphisms Y Ω1 −→˜ Ω1 O Ox x and 1 Y 1 Y 1 Ω −→˜ colim ΩK × ΩO A I x x x∈I x/∈I of O-modules and A-modules respectively. Proof. This follows immediately from Proposition 8.2, since the transition maps in the colimit which defines A are open embeddings.

Corollary 10.1.1. The modules Ω1 and Ω1 are free of rank one over and respectively. O A O A Proof. Combine the previous proposition and Proposition 9.2.

We now compute the residue map for the action of A on itself, which unsurprisingly turns out to be the sum of residues over all the closed points of X. Proposition 10.2. In terms of the isomorphism of Proposition 10.1, the residue map res :Ω1 → k A A is given by the formula X (ηx) 7→ resKx (ηx). x Proof. For any (η ) ∈ Ω1 there exists a ﬁnite set of closed points I such that η ∈ Ω1 for x∈ / I. x A x Ox In particular resKx (ηx) = 0 for x∈ / I, so the sum makes sense. Observe that Y AX = AX\I × Kx x∈I as a k-algebra, so by Proposition 8.5 we have X res = res + resQ = res + res . AX AX\I x∈I Kx AX\I Kx x∈I

26 Proposition 8.6 implies that the triangle

Ω1 Ω1 AX Kx

resKx resKx k

commutes for any x, so it suﬃces to show that resAX\I ((ηx)) = 0. Using commutativity of the triangle Ω1 Ω1 AX AX\I res res AX\I AX\I k

1 and the fact that (ηx) ∈ Ω , replacing X by X \ I we reduce to proving that res ((ηx)) = 0 x/∈I OX\I AX 1 for (ηx) ∈ Ω . Applying Proposition 8.5 once more, we obtain OX res ((η )) = res ((η )) + res ((η )). AX x OX x AX /OX x

Since OX is linearly compact and AX /OX is discrete, Proposition 8.3 implies that the right side vanishes.

Finally, we prove the ad`elicversion of local Serre duality as a rather formal consequence of the pointwise version.

Proposition 10.3. The continuous pairing of Tate vector spaces

∗ 1 1 resA A ⊗ ΩA −→ ΩA −−→ k is perfect, and ⊥ = Ω1 . That is, it induces an isomorphism O O

1 ∨ ΩA−→˜ A , as well as 1 ∨ 1 1 ∨ ΩO−→˜ (A/O) and O−→˜ (ΩA/ΩO) . Proof. Note that by Proposition 10.2, this pairing is given by X (fx) ⊗ (ηx) 7→ resKx (fxηx). x

Since is a lattice in , we see that res :Ω1 → k vanishes by combining Propositions 8.3 and 8.5. O A A O It follows that Ω1 → ∨ sends Ω1 into ( / )∨. Then Corollary 9.3.1 implies that the resulting A A O A O map !∨ Y M Y Ω1 = Ω1 −→ ( / )∨ = K /O = (K /O )∨ Ox O A O x x x x x x x is an isomorphism. This is equivalent to ⊥ = Ω1 and →˜ (Ω1 /Ω1 )∨. O O O A O

27 For any set of closed points I, we therefore have !∨ Y Y Y × K −→˜ ( )∨ × K∨−→˜ Ω1 /Ω1 × Ω1 . OX\I x OX\I x AX\I OX\I Kx x∈I x∈I x∈I Passing to the limit over I and taking note of the isomorphism M Ω1 /Ω1 −→˜ Ω1 /Ω1 , A O Kx Ox x we obtain an isomorphism ∨ 1 A −→˜ ΩA, which by construction is inverse to the map induced by resA.

If V is a free A-module of ﬁnite rank then it acquires a canonical structure of Tate vector space over k, similarly to a Kx-vector space. Corollary 10.3.1. For any free A-module V of ﬁnite rank, the pairing of Tate vector spaces ∗ 1 1 resA V ⊗ HomA(V, ΩA) −→ ΩA −−→ k is perfect. If L ⊂ V is a lattice which is also a free O-submodule and A ⊗O L→˜ V , then under the above pairing we have

⊥ 1 1 1 L = HomO(L, ΩO) ⊂ HomO(L, ΩA) = HomA(V, ΩA). Proof. This is deduced from the previous proposition in exactly the same way that Corollary 9.3.2 is deduced from Proposition 9.3.

11 Local-to-global methods Q Q For any ﬁnite set I of closed points in X, we will write OI := x∈I Ox and KI := x∈I Kx. Thus we have a Spec OI = Spec Ox x∈I and a Spec KI = Spec Kx. x∈I The idea of our approach is to think of the natural maps

jX\I : X \ I −→ X and

jOI : Spec OI −→ X as a covering of X, with the overlap being

jKI : Spec KI −→ X.

Of these maps, only jX\I is a Zariski open embedding. Nonethess, we still have a version of descent for this covering.

28 Proposition 11.1. For any quasicoherent sheaf F on X, the natural sequence 0 −→ F −→ j j∗ F ⊕ j j∗ F −→ j j∗ F −→ 0 X\I,∗ X\I OI ,∗ OI KI ,∗ KI is exact. Proof. The statement is local on X, so without loss of generality we can assume that X = Spec A is aﬃne. Shrinking X further if necessary, we can assume that (as a set) I is the vanishing locus n of some f ∈ A. Then OI = Ab := limn A/f A and KI = Abf . Letting M := Γ(X, F), we are trying to show exactness of the sequence

0 −→ M −→ Mf ⊕ (M ⊗A Ab) −→ M ⊗A Abf −→ 0. This is the conclusion of Lemma 15.83.11 in Part 1 of [1], which applies here by Remark 15.83.12(3) in loc. cit. because A is Noetherian.

In what follows, we write F := j∗ F and F := j∗ F. OI OI KI KI Corollary 11.1.1. If X \I is aﬃne, then applying RΓ to the short exact sequence in the proposition yields an exact triangle

RΓ(X, F) −→ Γ(X \ I, F) ⊕ FOI −→ FKI .

Proof. Since X is separated and the maps jX\I , jOI , and jKI have aﬃne source, the functors jX\I,∗, jOI ,∗, and jKI ,∗ are exact and therefore agree with their right derived functors. Moreover, each of these functors preserves injective objects because all three maps are ﬂat. Thus we have

∗ ∗ RΓ(X, jX\I,∗jX\I F) = RΓ(X, RjX\I,∗jX\I F) = RΓ(X \ I, F) = Γ(X \ I, F), and similarly for jOI and jKI . The conclusion follows immediately.

It will be helpful to remove the dependence on I. We put Y F := FO = lim FO and F := colim F ⊕ FK , O x I I A I OX\I I x which are respectively an O-module and an A-module. For simplicity, we will assume for the rest of these notes that X is connected. Letting K denote the ﬁeld of rational functions on X and ν : Spec K → X the canonical morphism, we will write

∗ FK := ν F for the generic ﬁber of F. Corollary 11.1.2. We have a natural exact triangle

RΓ(X, F) −→ FK ⊕ FO −→ FA. Proof. Starting with the exact triangle in the previous corollary, we immediately obtain an exact triangle

RΓ(X, F) −→ Γ(X \ I, F) ⊕ FO −→ FOX\I ⊕ FKI for any I. Since ﬁltered colimits in the category of vector spaces are exact, passing to the colimit over I yields the desired exact triangle.

29 Now suppose that F is coherent. Then by Lemma 10.96.1 in Part 1 of [1], for any closed point x the Ox-module FOx is mx-adically complete, i.e.

n FO −→˜ lim FO /m FO . x n x x x

n In particular FOx is naturally a linearly complete k-vector space. Moreover, since FOx /mxFOx is

ﬁnite-dimensional for all n, the limit FOx is even linearly compact. It follows that

FKx = Kx ⊗Ox FOx has a canonical structure of Tate vector space over k for which the action of Kx is continuous. We also deduce that FO and FA are naturally linearly compact and Tate over k respectively, with their respective actions of O and A being continuous. Let Ftor ⊂ F denote the torsion subsheaf. Since X is a smooth curve, for any coherent sheaf F the quotient F/Ftor is a vector bundle. If F = Ftor then clearly we have

FK = FKx = FA = 0. tor It follows that in general FK , FKx , and FA depend only on F/F . Moreover, the map FOx → FKx tor is injective if and only if x does not belong to the support of F , and hence FO → FA is injective if tor and only if F = 0, i.e. F is a vector bundle. In any case the images of FOx → FKx and FO → FA are lattices. tor Suppose that F has rank r, i.e. F/F is a vector bundle of rank r, or equivalently FK is an r-dimensional vector space over K. Then FKx is an r-dimensional vector space over Kx, from which it follows that FA is a free A-module of rank r. The Ox-module FOx has rank r, but it is free if and tor only if x does not belong to the support of F . Thus FO is an O-module of rank r, which is free if and only if F is a vector bundle.

Proposition 11.2. The topological modules (ωX )Ox and (ωX )Kx for Ox and Kx respectively canonically identify with the modules of diﬀerentials Ω1 and Ω1 introduced in Proposition 8.1. Ox Kx Proof. The statement only depends on an open neighborhood of x, so without loss of generality we 1 can assume that X = Spec A is aﬃne. Then we have Γ(X, ωX ) = ΩA. The composition

d 1 A −→ Ox −→ ΩOx is a derivation, hence factors uniquely through an A-linear map Ω1 → Ω1 . By extension of scalars A Ox we obtain the desired map 1 1 (ωX )Ox = Ox ⊗A ΩA −→ ΩOx .

Note that this map is automatically continuous, since it is Ox-linear and both sides are complete 1 with respect to the mx-adic topology. To see that it is an isomorphism, choose a function t : X → Ak which is ´etaleat x, so the image of t under A → Ox is a coordinate at x. By Nakayama’s lemma dt freely generates (ωX )Ox over Ox, and by Proposition 9.2 the map in question sends it to a free generator for the O -module Ω1 . x Ox We immediately obtain the isomorphism (ω ) →˜ Ω1 by inverting a coordinate t. X Kx Kx

Corollary 11.2.1. We have canonical continuous isomorphisms (ω ) →˜ Ω1 and (ω ) →˜ Ω1 . X O O X A A Proof. This is immediate from the proposition and Corollary 10.1.

30 12 Global Serre duality

For the rest of these notes, we assume that X is a projective curve (in addition to being smooth 1 and connected). Our ﬁrst goal is to construct the trace functional on R Γ(X, ωX ).

Lemma 12.1. For any coherent sheaf F on X, the natural map FK → FA embeds FK as a discrete closed subspace of FA, and the quotient FA/FK is linearly compact. Proof. The statement only depends on F/Ftor, so we can assume that F is a vector bundle. It follows that FO maps isomorphically onto a lattice in FA. Corollary 11.1.2 implies that we have an exact sequence

0 1 0 −→ R Γ(X, F) −→ FK ⊕ FO −→ FA −→ R Γ(X, F) −→ 0. ∼ 0 Thus FK ∩ FO = R Γ(X, F) is ﬁnite-dimensional, which implies that FK is closed and discrete in ∼ 1 FA, since FO is a lattice. Similarly, the fact that FA/(FK + FO) = R Γ(X, F) is ﬁnite-dimensional implies that FA/FK is linearly compact.

Proposition 12.2. The composition

1 1 resA ΩK −→ ΩA −−→ k vanishes.

Proof. By Proposition 8.6, this composition equals resA. Consider the short exact sequence of Tate K-modules 0 −→ K −→ A −→ A/K −→ 0

(the ﬁrst map is admissible by the lemma), so by Proposition 8.5 we have resA = resK + resA/K . Since K is discrete, and A/K is linearly compact by the lemma, the claim follows from Proposition 8.3.

Corollary 12.2.1 (Sum of residues formula). For any meromorphic 1-form η on X, we have X resx(η) = 0. x Proof. By Proposition 10.2, this is a restatement of the previous proposition.

To construct the trace map, we combine Corollaries 11.1.2 and 11.2.1 to obtain a canonical exact sequence

0 1 1 1 1 0 −→ R Γ(X, ωX ) −→ ΩK ⊕ ΩO −→ ΩA −→ R Γ(X, ωX ) −→ 0 1 1 (we have (ωX )K = ΩK since the functor A 7→ ΩA commutes with localization). As observed in the proof of Proposition 10.3, the composition

1 1 resA ΩO −→ ΩA −−→ k vanishes, so combining with Proposition 12.2 we obtained the desired trace map

1 1 1 1 resA trX : R Γ(X, ωX ) = coker(ΩK ⊕ ΩO → ΩA) −−→ k.

31 Theorem 12.3 (Serre duality for curves). With this deﬁnition of trX , for any coherent sheaf F on X the pairing trX RΓ(X, F) ⊗ R HomOX (F, ωX )[1] −→ RΓ(X, ωX )[1] −−→ k from Theorem 1.1 is perfect.

The remainder of these notes will be devoted to the proof of this theorem. Note that the pairing in the theorem is natural in the sense that for any morphism F → G of coherent sheaves, the following square commutes:

R HomOX (G, ωX )[1] R HomOX (F, ωX )[1]

RΓ(X, G)∨ RΓ(X, F)∨.

The key global input in the proof of Theorem 12.3 is the following result, a strengthening of Proposition 12.2.

Proposition 12.4. Under the pairing

∗ 1 1 resA A ⊗ ΩA −→ ΩA −−→ k ⊥ 1 we have K = ΩK , i.e. the pairing induces isomorphisms

1 1 ∨ 1 ∨ K−→˜ (ΩA/ΩK ) and ΩK −→˜ (A/K) . 1 ∨ Proof. Proposition 12.2 implies that the pairing induces a map ΩK → (A/K) , which is evidently 1 ∨ K-linear. Since ΩK = (ωX )K is 1-dimensional as a K-vector space, it suffices to show that (A/K) is also 1-dimensional over K and that the map is nonzero. 1 ∨ 1 To see that ΩK → (A/K) is nonzero, fix a closed point x and let η ∈ ΩK be a 1-form with a pole at x, say of order n. Choose a coordinate t at x, and observe that the functional A → k n−1 n−1 induced by η sends the element t ∈ Kx ⊂ A to resx(t η) 6= 0. ∨ It remains to show that (A/K) is 1-dimensional over K. First, we will need the following description of A/K. For any effective divisor Z on X, the map OX (−Z) → OX is an isomorphism away from Z and therefore induces isomorphisms OX (−Z)K →˜ K and OX (−Z)A→˜ A. Corollary 11.1.2 therefore yields an isomorphism

1 A/(OX (−Z)O + K)−→ ˜ R Γ(X, OX (−Z)).

P Q nx Writing Z = x nxx, the image of the embedding OX (−Z)O → O is the ideal x mx , so in particular limZ OX (−Z)O = 0. Combining this with the previous isomorphism, we obtain a continuous isomorphism 1 A/K−→˜ lim R Γ(X, OX (−Z)). Z Passing to the continuous duals, we have

∨ 1 ∨ (A/K) −→˜ colim R Γ(X, OX (−Z)) . Z To see the action of K on the right side, ﬁrst observe that

K = colim Γ(X, OX (Z)). Z

32 1 ∨ Given f ∈ Γ(X, OX (Z)) and λ ∈ R Γ(X, OX (−W )) , we obtain

1 f· 1 λ fλ : R Γ(X, OX (−Z − W )) −→ R Γ(X, OX (−W )) −→ k.

∨ Fix λ, µ ∈ (A/K) , which we will now show are dependent over K. Let Z be an eﬀective divisor 1 ∨ such that λ, µ ∈ R Γ(X, OX (−Z)) . Fix a closed point x, and for any n ≥ 0 consider the map

1 ∨ Γ(X, OX (nx)) ⊕ Γ(X, OX (nx)) −→ R Γ(X, OX (−Z − nx)) (12.1) (f, g) 7→ fλ + gµ.

0 Since Z + nx is eﬀective and X is connected, we have h (OX (−Z − nx)) = 0. Thus, by Proposition 2.2 the dimension of the right side is

1 h (OX (−Z − nx)) = −χ(OX (−Z − nx)) = deg Z + n − χ(OX ).

1 On the other hand, by Serre’s theorem we have h (OX (nx)) = 0 for n 0, so that in that range

0 h (OX (nx)) = χ(OX (nx)) = n + χ(OX ).

0 1 Thus for n 0 we have 2h (OX (nx)) ≥ h (OX (−Z − nx)), which implies that in that range (12.1) fails to be injective. In particular there exist f, g ∈ K such that fλ + gµ = 0, as desired.

Corollary 12.4.1. Let V be a ﬁnite-dimensional vector space over K. Under the pairing

∗ 1 1 resA (A ⊗K V ) ⊗ HomA(A ⊗K V, ΩA) −→ ΩA −−→ k of Corollary 10.3.1, we have

⊥ 1 1 1 V = HomK (V, ΩK ) ⊂ HomK (V, ΩA) = HomA(A ⊗K V, ΩA). Proof. This is immediate from the proposition after choosing a K-basis of V .

We are now in a position to prove the theorem.

Proof of Theorem 12.3. First, let us reduce to the case that F is a vector bundle. Since X is projective (in particular quasi-projective), for an arbitrary coherent sheaf F there exists a vector bundle E0 and an epimorphism E0 → F. Put

E1 := ker(E0 → F), which is a vector bundle because it is a subsheaf of the vector bundle E0, hence coherent and torsion-free. Considering the Serre duality pairing for each term in the short exact sequence

0 −→ E1 −→ E0 −→ F −→ 0, we obtain a morphism of exact triangles

R HomOX (F, ωX )[1] R HomOX (E0, ωX )[1] R HomOX (E1, ωX )[1]

∨ ∨ ∨ RΓ(X, F) RΓ(X, E0) RΓ(X, E1) .

33 If the middle and right vertical arrows are isomorphisms then so is the left vertical arrow, so we have performed the desired reduction. It is immediate from the deﬁnitions that

(F ⊗ G)A−→˜ FA ⊗A GA for any coherent sheaves F and G on X. Similarly, we have

(F ⊗ G)O−→˜ FO ⊗O GO and (F ⊗ G)K −→˜ FK ⊗K GK . Recall that for F a vector bundle, we write

∨ F := HomOX (F, OX ) for the dual vector bundle. The A-linear map

∨ ∨ (F )A ⊗A FA−→˜ (F ⊗ F)A −→ (OX )A = A induces a map ∨ (F )A −→ HomA(FA, A), which is easily seen to be an isomorphism by choosing a local trivialization of F. Similarly, we obtain isomorphisms

∨ ∨ (F )O−→˜ HomO(FO, O) and (F )K −→˜ HomK (FK ,K). Combining the above isomorphisms with Proposition 11.2, we obtain an isomorphism

∨ 1 (F ⊗ ωX )A−→˜ HomA(FA, ΩA), and similarly

∨ 1 ∨ 1 (F ⊗ ωX )O−→˜ HomO(FO, ΩO) and (F ⊗ ωX )K −→˜ HomK (FK , ΩK ). For any vector bundle F we have a canonical isomorphism

∨ R HomOX (F, ωX )−→ ˜ RΓ(X, F ⊗ ωX ). Applying Corollary 11.1.2 and the above identiﬁcations, we obtain an exact triangle

R Hom (F, ω ) −→ Hom (F , Ω1 ) ⊕ Hom (F , Ω1 ) −→ Hom (F , Ω1 ). (12.2) OX X K K K O O O A A A On the other hand, note that the morphism of two-step complexes

0 FK ⊕ FO FA 0

0 FA FA/FK ⊕ FA/FO 0 is a quasi-isomorphism using the observation that both kernels identify with FK ∩ FO, and both cokernels with FA/(FK + FO). Combine this with Corollary 11.1.2 and dualize to obtain an exact triangle ∨ ∨ ∨ ∨ (FA/FK ) ⊕ (FA/FO) −→ (FA) −→ RΓ(X, F) .

34 Consider the diagram

Hom (F , Ω1 ) ⊕ Hom (F , Ω1 ) Hom (F , Ω1 ) R Hom (F, ω )[1] K K K O O O A A A OX X ∼ ∼

∨ ∨ ∨ ∨ (FA/FK ) ⊕ (FA/FO) (FA) RΓ(X, F) , where the ﬁrst row is the rotation of the exact triangle (12.2), and the ﬁrst and second vertical maps are the isomorphisms of Corollaries 10.3.1 and 12.4.1. It is straightforward to check that this diagram commutes, as all of the vertical maps arise from the pairing ⊗∗ Ω1 → k induced by res . A A A The theorem now follows.

References

[1] The Stacks Project, available at https://stacks.math.columbia.edu.