The Riemann-Roch Theorem: a Proof, an Extension, and an Application MATH 780, Spring 2019

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1 The Riemann-Roch Theorem: a Proof, an Extension, and an Application MATH 780, Spring 2019 This handout continues the notational conentions of the preious one on the Riemann-Roch Theorem, with one slight addition We will denote ideles ie, elements of K A = GL 1 with upper-case Latin letters: A, B, The associated diisors will be denoted using the corresponding fraktur font: A, B, In other words, if A = A is an idele, then the corresponding diisor is Recall that for an idele A = A, A = di A = MK ord A ΛA = {b : b A 1 O for all places MK} We first proe the following Theorem 1: Let K be a function field with field of constants F q Then for all ideles A GL 1 with corresponding diisor A, la = dega + 1 g + dim Fq ΛA + K, where g := dim Fq ΛI + K Proof: We claim that for ideles A and B satisfying ΛA ΛB, dim Fq ΛA/ΛB = dega degb To see why, it suffices to consider the case where A 1 O = B 1 O for all but one place But this case is just the definition of the degree of a place and the order: dim Fq A 1 O /B 1 O = [A 1 = ord B 1 O : B 1 O ] deg ord A 1 deg = ord A ord B deg Next, still assuming ΛA ΛB, from the standard isomorphism theorems and Therefore ΛA/ΛB ΛB = + ΛA K /ΛB ΛB + ΛA K ΛB = ΛA ΛA + K = ΛB + ΛA K ΛB + K ΛA K ΛB ΛA K = ΛA K ΛB K = LA LB 1 dega degb = la lb + dim Fq ΛA + K ΛB + K 1

2 By Lemma 2 of the second handout on the adele ring, there is an idele C such that ΛC + K = Let B be an arbitrary idele and choose an idele A with ΛA = ΛB + ΛC Then ΛA ΛB and by 1 dega degb = la lb + dim Fq ΛB + K In particular, the last summand aboe is finite, so that we may rewrite 1 in the form 2 dega degb = la lb + dim Fq ΛB + K dim Fq ΛA + K Now let A and B be arbitrary ideles and choose an idele D satisfying both ΛD ΛA and ΛD ΛB Using 2 twice yields dega dim Fq ΛA + K + la = degd dim Fq This shows that for any idele A, the quantity dega dim Fq ΛD + K + ld = degb dim Fq ΛA + K + la ΛB + K + lb is the same Now considering the case of the identity idele I with corresponding diisor 0, we get dega dim Fq Rearranging yields Theorem 1 ΛA + K + la = deg0 dim Fq = 1 g ΛI + K + l0 Theorem 2 Riemann s Theorem: The genus g may be characterized as the maximum of dega la + 1 oer all diisors A DiK Further, there is an integer z, depending only on K, such that dega la + 1 = g for all diisors A with dega z Proof: By Theorem 1 we hae dega la + 1 = g dim Fq ΛA + K for all diisors A The first part of Theorem 2 is thus tatamount to the existence of a diisor A with = ΛA + K, which was a preious lemma Now choose a diisor A 0 with = ΛA 0 +K, ie, one which satisfies dega 0 la 0 +1 = g, and set z = dega 0 + g Then for any diisor A with dega z we hae la A 0 dega A g z dega g 1 2

3 Specifically, there is a non-zero α LA A 0 Consider B = A + diα; this diisor is linearly equialent to A and satisfies B A 0 We now hae by 2 and the containment ΛB + K ΛA 0 + K dega la = degb lb = dega 0 la 0 + dim Fq dim Fq ΛA 0 + K ΛB + K dega 0 la 0 = g 1 Recall the notion of the algebraic dual space of a ector space: if V is a ector space oer a field F, then the dual space V consists of the linear transformations from V into F For a finite-dimensional ector space, it s well-known that the space and its dual are isomorphic, whence hae the same dimension The Riemann-Roch Theorem is a consequence of Theorem 1 together with an appropriate realization of / ΛA + K as a dual space We ll take the most direct approach gien what we e already proen possible, but do note that there are many approaches, each with their own attributes Definition: A Weil differential is an F q -linear transformation ω : F q whose kernel contains a subset of the form ΛA + K for some idele A For a gien idele A we denote the collection of Weil differentials anishing on ΛA + K by Ω, and denote the set of all Weil differentials by Ω K It s a triial matter to confirm that Ω K is a subspace of the dual space of iewed as a ector space oer F q in the obious manner In fact, it s clear that Ω is the dual space of / ΛA + K, so that 3 dim Fq Ω = dim Fq ΛA + K Directly from the definition we see that Ω Ω K B wheneer the associated diisors satisfy A B, implying ΛA ΛB We may also iew Ω K as a ector space oer K as follows For α K and ω Ω K, set αω a = ω αa for all a In particular, note that if α 0 and ω anishes on ΛA + K, then αω anishes on ΛαA + K Proposition 1: As a ector space oer K, Ω K has dimension 1 Proof: We first note that Ω K {0} Indeed, by Theorem 1 we must hae a non-zero element of Ω wheneer the associated diisor A has degree less than 1, say Fix a non-zero ω Ω K ; we must show that Ω K = ωk Towards that end, suppose ω Ω K If ω = 0, then ω = 0ω and we re done, so we will now assume ω 0 Since ω, ω Ω K, there are ideles A and A with ω Ω and ω Ω These ideles are certainly not unique Indeed, if ω Ω, then ω Ω K B wheneer B A Neertheless, we may certainly fix/choose such ideles For a gien idele B consider the F q -linear and one-toone maps φ : LA + B Ω K B 1 and φ : LA + B Ω K B 1 gien by φα = αω and φ α = αω Exercise 1: Proe that these maps are, indeed, F q -linear and one-to-one 3

4 We claim that one can choose B such that the images of φ and φ hae non-triial intersection Note that, gien this claim, we then hae α, α K with αω = α ω, so that ω ωω K As for the claim, let z be the quantity in Riemann s Theorem and let B be such that degb max{z dega, z dega, 1, 3g 1 dega dega } Then both degb + A, degb + A z, so by Riemann s Theorem 4 la + B = dega + 1 g + degb, la + B = dega + 1 g + degb On the other hand, by Theorem 1 and 3, Ω K B 1 has dimension 5 dim Fq ΩK B 1 = l B deg B 1 + g = degb 1 + g since deg B = degb 1 Now 4 tells us the dimensions as F q -ector spaces of the respectie images of φ and φ, which when added are larger than the dimension of the codomain Ω K B 1 by 5 and construction Thus these images hae non-triial intersection As remarked in the proof aboe, for any non-zero ω Ω K there are infinitely many ideles A such that ω Ω Howeer, there is a least upper bound of sorts among such ideles/diisors the ordering coming from the inherent partial ordering on the diisor group Proposition 2: Gien any non-zero Weil differential ω, there is a unique diisor, called the diisor of ω and denoted diω, such that ω Ω K diω and for all ideles A with ω Ω we hae A diω For any α K and non-zero ω Ω K, diαω = diα + diω Proof: Let z be the constant in Riemann s Theorem Then a direct consequence of that result and 3 is that Ω = {0} wheneer dega z Thus, there is a diisor diω not necessarily proen to be unique at this point! of maximal degree such that ω Ω K diω We must proe that this diisor is uniquely determined Suppose to the contrary that there is a diisor A diω such that ω Ω Since A diω there is a place 0 MK such that ord 0 A > ord 0 diω Now consider the diisors B := 0 + diω and B = diω Let α ΛB Write α = β + γ where { α if 0, β = 0 if = 0 Then one readily erifies that β ΛB and γ ΛA We now hae ω α = ω β + ω γ = 0, so that ω Ω K B But degb = deg 0 + deg diω > deg diω, contradicting the hypothesis on diω Exercise 2: Complete the proof of Proposition 2 We remark that the definitions imply immediately that Ω consists of those Weil differentials ω satisfying diω A together with 0 Also, Propositions 1 and 2 together show that the following definition makes sense Definition: The canonical class is the equialence class of diisors containing diω for all non-zero ω Ω K 4

5 The last part of our proof of the Riemann-Roch Theorem is the following Theorem 3: Let A DiK and W = diω be a canonical diisor Then the mapping φ: LW A Ω gien by φα = αω is an isomorphism of F q -ector spaces Proof: Let α be a non-zero element of LW A assuming there is one Then by definition diα A W, so that diαω A This clearly implies that αω Ω Now let ω be a non-zero element of Ω By Proposition 1 we must hae ω = αω for some α K This implies that diαω A, so that diα A W, ie, α LW A Exercise 3: Show that φ aboe is one-to-one and F q -linear One can go about things in a similar but somewhat different manner than aboe Consider the topological dual of the locally compact group Gien any locally compact abelian group G, the dual group G consists of continuous homomorphisms called characters of G from G to the unit circle in the complex plane The dual group G inherits a topology from G by using the topology of uniform conergence on compact sets; G is then called the topological dual of G Moreoer, if H is any closed subgroup, the characters of G which induce the triial character on H make up a closed subgroup H of G ; this is called the subgroup associated with H by duality and is isomorphic to the dual of G/H Weil s approach is ia the following Theorem 4: Let V be a finite-dimensional ector space oer a number field or function field K Let χ be a character of, triial on K, and denote the adele ring of V by V A For any V A we hae a unique V A gien by = χ for all V A This determines an isomorphism between VA and V A Moreoer, it maps V to the subgroup of VA associated with the discrete subgroup V We now see how to extend the Riemann-Roch Theorem Suppose K is a function field and consider an element A = A GL n Exercise 4: Proe that A GL n K for all places MK and A O n = O n for almost all places Moreoer, gien such A for all places, A = A GL n Note that the determinant map takes GL n to GL 1, as usual Similar to the onedimensional case, we define and set LA = ΛA K n ΛA = {b KA n : b A 1 O n for all places MK} Exercise 5: Proe that LA is a finite-dimensional ector space oer F q for all A GL n With Exercise 5 in hand, we denote the dimension of LA by la exactly as before The proof of Theorem 1 applies almost word-for-word to get the following Theorem 1+: Let K be a function field with field of constants F q Then for all A GL n la = deg di deta + n1 g + dim Fq K n A ΛA + K n As for the term on the right aboe, ia the isomorphism from K+ΛI 1 to K ΛωI 1, where ω is any idele whose diisor is in the canonical class, one sees that the dual to 5 K n A K n +ΛA is

6 isomorphic to K n ΛωA, where A is the inerse transpose of A GL n We thus arrie at the following extension of the Riemann-Roch Theorem Theorem 5: Let K be a function field and let ω K A canonical class Then for all A GL n we hae la = deg di deta + n1 g + lωa be an idele such that diω is in the Of course, the case n = 1 here is just the usual Riemann-Roch Theorem Recall that we may iew K n A as our analog for Euclidean n-space, and then Kn is the analog of the integer lattice Z n We get a generic lattice ia an element of the general linear group: AK n Here the adelic absolute alue called the adelic module of an idele a GL 1 is a A = q deg dia Thus the determinant of lattice AK n would be q deg di deta Our analog of the unit ball is B A = On Thus LA is precisely the set of lattice points in the unit ball, so that q la is the number of such lattice points This is where the analogy would break down entirely if the field of constants were infinite Theorem 5 tells us exactly how many such lattice points there are Sticking with this identification, the analog of the length of a lattice point would be as follows Gien any A GL n and x K n \ {0}, the adelic length of Ax is gien by Ax A = min { a A : Ax ab A } a GL 1 = min a GL 1 { a A : a 1 Ax B A }, = min a GL 1 { a A : x Λa 1 A} Of course, 0 A is defined to be 0 It s sometimes simpler to deal with the logarithm rather than the length itself, so we will gie it a name: the logarithmic height of the lattice point Ax is h Ax = log q Ax A We can now try our hand at some classic geometry of numbers but in the function field setting The successie minima λ 1 A λ n A are gien by λ i A = min{m : there are i linearly independent x 1,, x i K n with h Ax i m} Minkowski s First Theorem: For any A GL n we hae nλ 1 A ng deg di deta Proof: We ll cheat just a little bit here and assume the image of the degree function on DiK is all of Z which it is Let a be an idele with deg dia = λ 1 A 1 Then by definition laa = 0 Since the far righthand term in Theorem 5 is necessarily non-negatie it s a dimension, after all, we see that 0 = laa deg di detaa + n1 g = deg di deta + n deg dia + 1 g = deg di deta + nλ 1 A ng 6

7 Compare this with the classical result: for any lattice Λ R n, the first minima satisfies λ 1 Λ n VolB 2 n detλ, where VolB is the olume of the unit ball in R n and the successie minima are the actual lengths rather than logarithms of lengths - whence the multiplicatie nature of the statement aboe rather than the additie statement of the function field ersion The 2 n term arises from conexity: B + B = 2B In our case B A + B A = B A, so we don t expect to see this term So what is the olume of B A? Note that we set the determinant of AK n to be q deg di deta In other words, we set K n to hae determinant 1 This means we hae some sort of measure technically a haar measure on KA n where the discete subgroup Kn has coolume 1 This turns out to imply that B A has olume q n1 g Indeed, if we take the haar measure on KA n obtained by setting the measure of each O n to be 1 a reasonable enough choice, then we need to normalize multiply by q n1 g in order for K n to hae coolume 1 Therefore, our function field Minkowski s First Theorem is just like the classical case, but with the 2 n term replaced by q n That term is actually a result of the discrete nature of our lengths as compared to the Euclidean case Indeed, we hae the following Minkowski s First Theorem Alternate Version: We hae AK n B A {0} for all A GL n with deta A < VolB A This is exactly the classical ersion sans the 2 n term, which we don t expect in the function field case as explained aboe Note that AK n B A {0} is equialent to λ 1 A 0 Exercise 6: Assuming the image of deg : DiK Z is all of Z, show that the function field Minkowski s First Theorem is equialent to the Alternate Version 7