Weil s Conjecture on Tamagawa Numbers (Lecture 1)

Weil s Conjecture on Tamagawa Numbers (Lecture ) January 30, 204 Let R be a commutative ring and let V be an R-module. A quadratic form on V is a ma q : V R satisfying the following conditions: (a) The construction (v, w) q(v + w) q(v) q(w) determines an R-bilinear ma V V R. (b) For every element λ R and every v V, we have q(λv) = λ 2 q(v). A quadratic sace over R is a air (V, q), where V is a finitely generated rojective R-module and q is a quadratic form on V. One of the basic roblems in the theory of quadratic forms can be formulated as follows: Question. Let R be a commutative ring. Can one classify quadratic saces over R (u to isomorhism)? Examle 2. Let V be a vector sace over the field R of real numbers. Then any quadratic form q : V R can be diagonalized: that is, we can choose a basis e,..., e n for V such that q(λ e + + λ n e n ) = λ 2 + + λ 2 a λ 2 a+ λ 2 a+b for some air of nonnegative integers a, b with a + b n. Moreover, the integers a and b deend only on the isomorhism class of the air (V, q) (a theorem of Sylvester). In articular, if we assume that q is nondegenerate (in other words, that a + b = n), then the isomorhism class (V, q) is comletely determined by the dimension n of the vector sace V and the difference a b, which is called the signature of the quadratic form q. Examle 3. Let Q denote the field of rational numbers. There is a comlete classification of quadratic saces over Q, due to Minkowski (later generalized by Hasse to the case of an arbitrary number field). Minkowski s result is highly nontrivial, and reresents one of the great triumhs in the arithmetic theory of quadratic forms: we refer the reader to [5] for a detailed and readable account. Let us now secialize to the case R = Z. We will refer to quadratic saces (V, q) over Z as even lattices. The classification of even lattices u to isomorhism is generally regarded as an intractable roblem (see Remark 7 below). Let us therefore focus on the following variant of Question : Question 4. Let (V, q) and (V, q ) be even lattices. Is there an effective rocedure for determining whether or not (V, q) and (V, q ) are isomorhic? Let (V, q) be a quadratic sace over a commutative ring R, and suose we are given a ring homomorhism φ : R S. We let V S denote the tensor roduct S R V. An elementary calculation shows that there is a unique quadratic form q S : V S S for which the diagram V q R V S q S S φ

is commutative. The construction (V, q) (V S, q S ) carries quadratic saces over R to quadratic saces over S; we refer to (V S, q S ) as the extension of scalars of (V, q). If (V, q) and (V, q ) are isomorhic quadratic saces over R, then extension of scalars yields isomorhic quadratic saces (V S, q S ) and (V S, q S ) over S. Consequently, if we understand the classification of quadratic saces over S and can tell that (V S, q S ) and (V S, q S ) are not isomorhic, it follows that (V, q) and (V, q ) are not isomorhic. Examle 5. Let q : Z Z be the quadratic form given by q(n) = n 2. Then the even lattices (Z, q) and (Z, q) cannot be isomorhic, because they are not isomorhic after extension of scalars to R: the quadratic sace (R, q R ) has signature, while (R, q R ) has signature. Examle 6. Let q, q : Z 2 Z be the quadratic forms given by q(m, n) = m 2 + n 2 q (m, n) = m 2 + 3n 2. Then (Z 2, q) and (Z 2, q ) become isomorhic after extension of scalars to R (since both quadratic forms are ositive-definite). However, the quadratic saces (Z 2, q) and (Z 2, q ) are not isomorhic, since they are not isomorhic after extension of scalars to the field F 3 = Z/3Z (the quadratic form q F3 is nondegenerate, but is degenerate). q F 3 Using variants of the arguments rovided in Examles 5 and 6, one can roduce many examles of even lattices (V, q) and (V, q ) that cannot be isomorhic: for examle, by arranging that q and q have different signatures (after extension of scalars to R) or have nonisomorhic reductions modulo n for some integer n > 0 (which can be tested by a finite calculation). This motivates the following definition: Definition 7. Let (V, q) and (V, q ) be ositive-definite even lattices. We say that (V, q) and (V, q ) of the same genus if (V, q) and (V, q ) are isomorhic after extension of scalars to Z/nZ, for every ositive integer n (in articular, this imlies that V and V are free abelian grous of the same rank). Remark 8. One can also define study genera of lattices which are neither even nor ositive definite, but we will restrict our attention to the situation of Definition 7 to simly the exosition. More informally, we say that two even lattices (V, q) and (V, q ) are of the same genus if we cannot distinguish between them using variations on Examle 5 or 6. It is clear that isomorhic even lattices are of the same genus, but the converse is generally false. However, the roblem of classifying even lattices within a genus has a great deal of structure. One can show that there are only finitely many isomorhism classes of even lattices in the same genus as (V, q). Moreover, the celebrated Smith-Minkowski-Siegel mass formula allows us to say exactly how many (at least when counted with multilicity). Notation 9. Let (V, q) be a quadratic sace over a commutative ring R. We let O q (R) denote the automorhism grou of (V, q): that is, the grou of R-module isomorhisms α : V V such that q = q α. We will refer to O q (R) as the orthogonal grou of the quadratic sace (V, q). More generally, if φ : R S is a ma of commutative rings, we let O q (S) denote the automorhism grou of the quadratic sace (V S, q S ) over S obtained from (V, q) by extension of scalars to S. Examle 0. Suose q is a ositive-definite quadratic form on an real vector sace V of dimension n. Then O q (R) can be identified with the usual orthogonal grou O(n). In articular, O q (R) is a comact Lie grou of dimension n2 n 2. Examle. Let (V, q) be a ositive-definite even lattice. For every integer d, the grou O q (Z) acts by ermutations on the set V d = {v V : q(v) d}. Since q is ositive-definite, each of the sets V d is finite. Moreover, for d 0, the action of O q (Z) on V d is faithful (since an automorhism of V is determined by its action on a finite generating set for V ). It follows that O q (Z) is a finite grou. Let (V, q) be a ositive-definite even lattice. The mass formula gives an exlicit formula for the sum (V,q ) O q (Z), where the sum is taken over all isomorhism classes of even lattices (V, q ) in the genus of (V, q). The exlicit formula is rather comlicated in general, deending on the reduction of (V, q) modulo for various rimes. For simlicity, we will restrict our attention to the simlest ossible case. 2

Definition 2. Let (V, q) be an even lattice. We will say that (V, q) is unimodular if the bilinear form b(v, w) = q(v + w) q(v) q(w) induces an isomorhism of V with its dual Hom(V, Z). Remark 3. Let (V, q) be a ositive-definite even lattice. The condition that (V, q) be unimodular deends only on the reduction of q modulo for all rimes. In articular, if (V, q) is unimodular and (V, q ) is in the genus of (V, q), then (V, q ) is also unimodular. In fact, the converse also holds: any two unimodular even lattices of the same rank are of the same genus (though this is not obvious from the definitions). Remark 4. The condition that an even lattice (V, q) be unimodular is very strong: for examle, if q is ositive-definite, it imlies that the rank of V is divisible by 8. Theorem 5 (Mass Formula: Unimodular Case). Let n be an integer which is a ositive multile of 8. Then (V,q) O q (Z) = Γ( 2 )Γ( 2 2 ) Γ( n 2 )ζ(2)ζ(4) ζ(n 4)ζ(n 2)ζ( n 2 ) 2 n π n(n+)/4 = B n/4 n j<n/2 B j 4j. Here ζ denotes the Riemann zeta function, B j denotes the jth Bernoulli number, and the sum is taken over all isomorhism classes of ositive-definite even unimodular lattices (V, q) of rank n. Examle 6. Let n = 8. The right hand side of the mass formula evaluates to 696729600. The integer 696729600 = 2 4 3 5 5 2 7 is the order of the Weyl grou of the excetional Lie grou E 8, which is also the automorhism grou of the root lattice of E 8 (which is an even unimodular lattice). Consequently, the fraction 696729600 also aears as one of the summands on the left hand side of the mass formula. It follows from Theorem 5 that no other terms aear on the left hand side: that is, the root lattice of E 8 is the unique ositive-definite even unimodular lattice of rank 8, u to isomorhism. Remark 7. Theorem 5 allows us to count the number of ositive-definite even unimodular lattices of a given rank with multilicity, where a lattice (V, q) is counted with multilicity O q(z). If the rank of V is ositive, then O q (Z) has order at least 2 (since O q (Z) contains the grou ± ), so that the left hand side of Theorem 5 is at most C 2, where C is the number of isomorhism classes of ositive-definite even unimodular lattices. In articular, Theorem 5 gives an inequality C Γ( 2 )Γ( 2 2 ) Γ( n 2 )ζ(2)ζ(4) ζ(n 4)ζ(n 2)ζ( n 2 ) 2 n 2 π n(n+)/4. The right hand side of this inequality grows very quickly with n. For examle, when n = 32, we can deduce the existence of more than eighty million airwise nonisomorhic (ositive-definite) even unimodular lattices of rank n. We now describe a reformulation of Theorem 5, following ideas of Tamagawa and Weil. Suose we are given a ositive-definite even lattice (V, q), and that we wish to classify other even lattices of the same genus. If (V, q ) is a lattice in the genus of (V, q), then for every ositive integer n we can choose an isomorhism α n : V/nV V /nv which is comatible with the quadratic forms q and q. Using a comactness argument (or some variant of Hensel s lemma) we can assume without loss of generality that the isomorhisms {α n } n>0 are comatible with one another: that is, that the diagrams V/nV α n V /nv V/mV αm V /mv 3

commute whenever m divides n. In this case, the data of the family {α n } is equivalent to the data of a single isomorhism α : Ẑ Z V Ẑ Z V, where Ẑ lim Z/nZ denotes the rofinite comletion of the ring Z. n>0 By virtue of the Chinese remainder theorem, the ring Ẑ can be identified with the roduct Z, where ranges over all rime numbers and Z denotes the ring lim Z/ k Z of -adic integers. It follows that (V, q) and (V, q ) become isomorhic after extension of scalars to Z, and therefore also after extension of scalars to the field Q = Z [ ] of -adic rational numbers. Since the lattices (V, q) and (V, q ) are ositive-definite and have the same rank, they also become isomorhic after extending scalars to the field of real numbers. It follows from Minkowski s classification that the quadratic saces (V Q, q Q ) and (V Q, q Q ) are isomorhic (this is known as the Hasse rincile: to show that quadratic saces over Q are isomorhic, it suffices to show that they are isomorhic over every comletion of Q; see 3.3 of [5]). We may therefore choose an isomorhism β : V Q V Q which is comatible with the quadratic forms q and q. Let A f denote the ring of finite adeles: that is, the tensor roduct Ẑ Z Q. The isomorhism Ẑ Z induces an injective ma A f Ẑ Z Q (Z Z Q) Q, whose image is the restricted roduct res Q Q : that is, the subset consisting of those elements {x } of the roduct Q such that x Z for all but finitely many rime numbers. The quadratic saces (V, q) and (V, q ) become isomorhic after extension of scalars to A f in two different ways: via the isomorhism α which is defined over Ẑ, and via the isomorhism β which is defined over Q. Consequently, the comosition β α can be regarded as an element of O q (A f ). This element deends not only the quadratic sace (V, q ), but also on our chosen isomorhisms α and β. However, any other isomorhism between (VẐ, qẑ) and (V, Ẑ q ) can be written in the form α, where O Ẑ q(ẑ). Similarly, the isomorhism β is well-defined u to right multilication by elements of O q (Q). Consequently, the comosition β α is really well-defined as an element of the set of double cosets O q (Q)\ O q (A f )/ O q (Ẑ). Let us denote this double coset by [V, q ]. It is not difficult to show that the construction (V, q ) [V, q ] induces a bijection from the set of isomorhism classes of even lattices (V, q ) in the genus of (V, q) with the set O q (Q)\ O q (A f )/ O q (Ẑ) (the inverse of this construction is given by the rocedure of regluing, which we will discuss in the next lecture). Moreover, if O q (A f ) is a reresentative of the double coset [V, q ], then the grou O q (Z) is isomorhic to the intersection O q (Ẑ) O q (Q). Consequently, the left hand side of the mass formula can be written as a sum O q (Ẑ) O q (Q), () where ranges over a set of double coset reresentatives. At this oint, it will be technically convenient to introduce two modifications of the calculation we are carrying out. For every commutative ring R, let SO q (R) denote the subgrou of O q (R) consisting of those automorhisms of (V R, q R ) which have determinant (if R is an integral domain, this is a subgrou of index at most 2). Let us instead attemt to comute the sum SO q (Ẑ) SO q (Q), (2) where runs over a set of reresentatives for the collection of double cosets X = SO q (Q)\ SO q (A f )/ SO q (Ẑ). 4

If q is unimodular, exression (2) differs from the exression () by an overall factor of 2 (in general, the exressions differ by a ower of 2). Remark 8. Fix an orientation of the Z-module V (that is, a generator of the to exterior ower of V ). Quantity (2) can be written as a sum SO q (Z), where the sum is indexed by all isomorhism classes of oriented even unimodular ositive-definite lattices (V, q ) which are isomorhic to (V, q) as oriented quadratic saces after extension of scalars to Z/nZ, for every integer n > 0. Let A denote the ring of adeles: that is, the ring A f R. Then we can identify X with the collection of double cosets SO q (Q)\ SO q (A)/ SO q (Ẑ R). The virtue of this maneuver is that A has the structure of a locally comact commutative ring containing Q as a discrete subring. Consequently, SO q (A) is a locally comact toological grou which contains SO q (Q) as a discrete subgrou and SO q (Ẑ R) as a comact oen subgrou. Let µ be a Haar measure on the grou SO q (A). One can show that the grou SO q (A) is unimodular: that is, the measure µ is invariant under both right and left translations. In articular, µ determines a measure on the quotient SO q (Q)\ SO q (A), which is invariant under the right action of SO q (Ẑ R). We will abuse notation by denoting this measure also by µ. Write SO q (Q)\ SO q (A) as a union of orbits x X O x for the action of the grou SO q (Ẑ R). If x X is a double coset reresented by an element SO q(a), then we can identify the orbit O x with the quotient of SO q (Ẑ R) by the finite subgrou SOq(Ẑ R) SO q (Q). We therefore have SO q (Ẑ R) SO q (Q) = x X µ(o x ) µ(so q (Ẑ R)) (3) = µ(so q(q)\ SO q (A)) µ(so q (Ẑ R)). (4) Of course, the Haar measure µ on SO q (A) is only well-defined u to scalar multilication. Rescaling the measure µ has no effect on the left hand side of the receding equation, since µ aears in both the numerator and the denominator of the right hand side. However, it is ossible to say more: it turns out that there is a canonical normalization of the Haar measure, known as Tamagawa measure. Construction 9. Let G be a connected semisimle algebraic grou of dimension d over the field Q of rational numbers. Let Ω denote the collection of all left invariant d-forms on G, so that Ω is a -dimensional vector sace over Q. Choose a nonzero element ω Ω. The vector ω determines a left-invariant differential form of to degree on the smooth manifold G(R), which in turn determines a Haar measure µ R,ω on G(R). For every rime number, an analogous construction yields a measure µ Q,ω on the -adic analytic manifold G(Q ). One can show that the roduct of these measures determines a measure µ Tam on the restricted roduct res G(R) G(Q ) G(A). Let λ be a nonzero rational number. Then an elementary calculation gives µ R,λω = λ µ R,ω µ Q,λω = λ µ Q,ω; here λ denotes the -adic absolute value of λ. Combining this with the roduct formula λ = λ, we deduce that µ Tam is indeendent of the choice of nonzero element ω Ω. We will refer to µ Tam as the Tamagawa measure of the algebraic grou G. If (Λ, q) is a ositive-definite even lattice, then the restriction of the functor R SO q (R) to Q-algebras can be regarded as a semisimle algebraic grou over Q. We may therefore aly Construction 9 to obtain 5

a canonical measure µ Tam on the grou SO q (A). We may therefore secialize equation (4) to obtain an equality SO q (Z) = µ Tam(SO q (Q)\ SO q (A)) µ Tam (SO q (Ẑ R)), (5) where it makes sense to evaluate the numerator and the denominator of the right hand side indeendently. Remark 20. The construction R O q (R) also determines a semisimle algebraic grou over Q. However, this grou is not connected, and the infinite roduct µ Q,ω does not converge to a measure on the restricted roduct res O q(q ) = O q (A f ). This is the reason for referring to work with the grou SO q in lace of O q. Remark 2. The numerator on the right hand side of (5) is called the Tamagawa number of the algebraic grou SO q. More generally, if G is a connected semisimle algebraic grou over Q, we define the Tamagawa number of G to be the measure µ(g(q)\g(a)). The denominator on the right hand side of (5) is comutable: if we choose a differential form ω as in Construction 9, it is given by the roduct µ R,ω (SO q (R)) µ Q,ω(SO q (Z )). The first term is the volume of a comact Lie grou, and the second term is a roduct of local factors which are related to counting roblems over finite rings. Carrying out these calculations leads to a very retty reformulation of Theorem 5: Theorem 22 (Mass Formula, Adelic Formulation). Let (V, q) be a nondegenerate quadratic sace over Q. Then µ Tam (SO q (Q)\ SO q (A)) = 2. The aearance of the number 2 in the statement of Theorem 22 results from the fact that the algebraic grou SO q is not simly-connected. Let Sin q denote the (2-fold) universal cover of SO q, so that Sin q is a simly-connected semisimle algebraic grou over Q. We then have the following more basic statement: Theorem 23. Let (V, q) be a ositive-definite quadratic sace over Q. Then µ Tam (Sin q (Q)\ Sin q (A)) =. Remark 24. For a deduction of Theorem 22 from Theorem 23, see [4]. Theorem 23 motivates the following: Conjecture 25 (Weil s Conjecture on Tamagawa Numbers). Let G be a simly-connected semisimle algebraic grou over Q. Then µ Tam (G(Q)\G(A)) =. Conjecture 25 was roven by Weil in a number of secial cases. The general case was roven by Langlands in the case of a slit grou ([3]), by Lai in the case of a quasi-slit grou ([2]), and by Kottwitz for arbitrary simly connected algebraic grous satisfying the Hasse rincile ([]; this is now known to be all simlyconnected semisimle algebraic grous over Q, by work of Chernousov). Our goal in this course is to rove an analogue of Conjecture 25 in which Q has been relaced by global field of ositive characterstic. We will review the formulation (and interretation) of this conjecture in the next lecture. 6

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