We conjectures for abean varetes over fnte feds Kwun Chung Abstract Ths s an expostory paper on zeta functons of abean varetes over fnte feds. We woud ke to go through how zeta functon s defned, and dscuss the We conjectures. The man purpose of ths paper s to f n more detas to the proofs provded n Mne. Subject to ength constran, we w not ncude a detaed proof for Remann hypothess n ths paper. We w many be foowng Mne s notes [2]. 1 Introducton We conjectures are a st of conjectures for varetes over fnte fed. Let V be a smooth n-dmensona projectve varety over F q. Denote by N m the number of ponts of V over F q m. The zeta functon s defned to be ( ) ζ(v, s) = Z(V, t) := exp n=1 N m t m m, where t = q s. We conjectures have two parts: (a) Z(V, t) s a ratona functon. In fact, Z(V, t) = P 1(t)P 3 (t) P 2n 1 (t) P 0 (t)p 2 (t) P 2n (t), where P (t) = j (1 a,jt) for some a,j, and P 0 (t) = 1 t, P 2n (t) = 1 q n t. (b) (Remann hypothess) a,j = q /2. Ths mpes that the zeros of Z(V, t) are of absoute vaues q odd/2. Thus zeros of ζ(v, s) has rea part 1 2, 3 2,..., 2n 1 2, and smary poes of ζ(v, s) has rea part 0, 1, 2,..., n. It s hard to prove th We conjectures for a genera varety V over F q. In 1964, Grothendeck proved the ratonaty of zeta functons usng étae cohomoogy. The Remann hypothess was not proved unt 1974 by Degne, where he used -adc cohomoogy and Lefschetz pencs. 1
The proof for abean varetes are much smper compared to Degne s proof. We w show the proof of ratonaty of zeta functons. The proof for Remann hypothess n Mne nvokes Rosat nvouton, and s more n depth. We w sketch the outne of the proof, but w not gve the detas. Throughout the paper, A stands for an abean varety over a base fed k. F q s the fnte fed of order q. For an nteger n, we w use [n] to denote the mutpcaton by n map on A. 2 Characterstc Poynoma To state We conjectures, we need to be abe to tak about characterstc poynoma of the Frobenus endomorphsm. We w do ths for a genera endomorphsm α End(A). Let s frst ook at the case for abean varetes A over C. Then A = C g /Λ, where Λ C g s a attce of rank 2g. Then α defnes a map Λ Q Λ Q. Thus we have a characterstc poynoma of α as an endomorphsm of the vector space,.e. P α (X) = det(α X Id), α X Id vewed as an endomorphsm over Λ Q. Snce α(λ) Λ, P α (X) us a poynoma over Z. There s another descrpton of P α (X) that does not refer to the unformzaton. Reca that the attce Λ s the same as H 1 (A, Z). Thus Λ Q = H 1 (A, Q). Let be a prme not equa to the characterstc of the base fed, and defne V A := T A Q be extendng the Tate modue T A to be a vector space over Q. For abean varetes over C, we have that Thus, over C we have that V A = H 1 (A, Q) Q. P α (X) = det(α X Id), where α X Id s vewed as an endomorphsm over V A, a vector space over Q. One may try to defne characterstc poynoma for abean varetes over F q. Unfortunatey, ths s uncear that the poynoma defned ths way has ratona coeffcents, nor s t cear that the defnton s ndependent on the choce of. Thus, we are adoptng another approach done by We. To do ths, we w modfy what we have for abean varetes over C to have another descrpton of P α (X) that does not refer to the attce Λ. Proposton 2.1. For any nteger r, abean varetes A over C and α End(A), we have that P α (r) = deg(α [r]). 2
Proof Reca that for a separabe sogeny α End(A), we have deg(α) = ker α. In our case, a sogenes are separabe as char(c) = 0. We have that ker α = (α 1 Λ)/λ = Λ/αΛ. Thus, ker α = Λ/αΛ = det(α), where α s vewed as an endomorphsm of the vector space Λ Q = H 1 (A, Q). Repacng α by α [r], we see that deg(α [r]) = det(α [r]) = P α (r). We w defne the characterstc poynoma for endomorphsms of abean varetes over other feds k n a smar fashon, by the foowng theorem. Theorem 2.2. Let A be an abean varety and α End(A). There s a unque poynoma P α (X) Z[X] such that P α (n) = deg(frob [n]). Ths s caed the characterstc poynoma of α. The strategy of our proof w be gong nto End 0 (A) := End(A) Q and check that deg(frob [r]) s a poynoma of degree 2g n r (we use r here as r can now be ratona, not just nteger). Then we w go back to End(A) and show that the poynoma has ntegra coeffcents and s monc. Frst, notce that snce deg(nα) = deg([n]) deg(α) = n 2g deg(α) for a α End(A), we can extend the degree functon to End 0 (A) = End(A) Q. That s, for α End 0 (A) we defne for n Z >0 wth nα End(A). deg(α) = n 2g deg(nα) Proposton 2.3. deg : End 0 (A) Q s a homogeneous poynoma functon of degree 2g. That s, for any fnte neary ndependent set {e 1,..., e n }, deg(x 1 e 1 + + x n e n ) = P (x 1,..., x n ) where P (x 1,..., x n ) s a homogeneous poynoma. We need a emma to prove ths. Lemma 2.4. Let N 0 be an nteger, V be a vector space over a fed k wth k > N, and f : V k be a functon such that for any v, w V, f(xv + w) s a poynoma n x of degree at most N (ndependent on choce of v, w). Then f s a poynoma functon, defned smary as a homogeneous poynoma functon. 3
As a remark, the proof of Lemma 10.12 n Mne ([2]) s ncorrect. The notes supposed that f(x 1,..., x n ) s a poynoma n x n wth coeffcents as functons from k n 1 to k. But ths s apror not cear. Thus we need to bound the degree unformy n the emma. The author of ths paper has aso consdered whether the nta verson of Lemma 10.12 n Mne, wthout a unform bound on degree, s correct. Wth a smar argument, but a trck n addton, whch requres that chark = 0, we can prove that f(x 1 v 1 + + x n v n + w) = a (x 1,..., x n 1 )x n, where a k[x 1,..., x n 1 ] The key step of the above statement s to reaze that ( xn ) s a bass for the vector space k[xn ]. Ths dea comes from Yfeng Huang, a PhD student (by the tme of wrtng) n Unversty of Mchgan. For detas, readers can check the author s notes on t ([1]). But after ths, the author s not sure whether we can move on and show that a = 0 for a 0. Now et s go back to the Lemma 2.4. Proof (of Lemma 2.4) The case n = 1 s by our hypothess. Now suppose t s true for n 1. By usng the nta hypothess (not the nductve one), there are some functons a (x 1,..., x n 1 ) from k n 1 to k such that f(x 1 v 1 + + x n v n + w) = a 0 (x 1,..., x n 1 ) + a 1 (x 1,..., x n 1 )x n + + a N (x 1,..., x n 1 )x N n. Ths s done by, for b 1,..., b n 1 k, defnng a (b 1,..., b n 1 ) to be the coeffcent of x n n the poynoma f(b 1 v 1 + + b n 1 v n 1 + x n v n + w). Snce k > N, we can pck c 0,..., c N k to be a dstnct. By nducton hypothess, f(x 1 v 1 + +x n 1 v n 1 +c v n +w) are poynomas n v 1,..., v n 1 for a 0 N. We now have the foowng system of equatons n k[x 1,..., x n 1 ]: =0 f(x 1 v 1 + + x n 1 v n 1 + c 0 v n + w) 1 c 0 c 2 0 c N 0 a 0 (x 1,..., x n 1 ) f(x 1 v 1 + + x n 1 v n 1 + c 1 v n + w). = 1 c 1 c 2 1 c N 1 a 1 (x 1,..., x n 1 )...... f(x 1 v 1 + + x n 1 v n 1 + c N v n + w) 1 c N c 2 N cn N a N (x 1,..., x n 1 ) Note that the square matrx s a Vandermonde matrx. Snce c are a dstnct, the matrx has non-zero determnant. Snce t s a matrx over a fed k, t s nvertbe. Thus each a s a k-near combnaton of f(x 1 v 1 + + x n 1 v n 1 + c j v n + w). Therefore a a are poynomas, and thus f(x 1 v 1 + + x n 1 v n 1 + x n v n + w) s a poynoma. 4
Proof (of Proposton 2.3) By the emma, t suffces to show that once we fx α, β End 0 (A), deg(nα + β) s a poynoma n n of degree at most 2g. Notce that ths deg s the extended degree functon from End 0 (A) to Q. Usng that deg(nα) = n 2g deg(α),.e. degree s homogeneous of degree 2g, we can suppose α, β End(A) are actua endomorphsms of A. Fx α, β End(A), and et D be a very ampe dvsor on A, and defne D n to be the puback (nα+β) D. We w use a resut from Mne s Dvsors and Intersecton Theory notes ([3]), whch computes powers of D n for us: (D n D n ) = deg(nα + β)(d D). By ncuson-excuson prncpe (Cor 5.3 of [2]) apped to nα + β, α, α and the ne bunde L = L(D), we have that.e. D n+2 2D n+1 (2α) D + D n + 2(α D) 0, D n+2 2D n+1 + D n = D, where D = (2α) D 2(α D). By usng the resut from ntersecton theory, we can further smpfy D as deg(2α)d 2 deg(α)d = (2 2g 2) deg(α)d = ND for some constant N (ndependent on n). By nducton, t can be shown that D n = n(n 1) D + nd 1 (n 1)D 0 = 2 Now usng the resut from ntersecton theory, we have that n(n 1) ND + n deg(α)d (n 1)D. 2 ( ) g n(n 1) deg(nα + β)d g = Dn g = N g D g + O(n 2g 1 )D g. 2 Ths shows that deg(nα + β) s a poynoma n n of degree 2g. Proof (of Theorem 2.2) Frst, the unqueness s cear snce any two such poynomas agree on a ntegers, whch s an nfnte set. As for exstence, Proposton 2.4 shows that for each α End 0 (A), there s a poynoma P α (X) of degree 2g such that P α (r) = deg(α [r]) for a ratona r. It remans to show that f α End 0 (A), then we get a monc poynoma wth ntegra coeffcents. Let D be an ampe symmetrc dvsor on A (reca that by symmetrc, we mean [ 1] D D). By puttng α [1] and β α, we have that P α ( n) = deg([n] + α) = (D g n)/(d g ), where D n = ([n] + α) D. By the computaton n the proof of Proposton 2.3, we have that ( ) n(n 1) D n = D + n([1] + α) D (n 1)α D, 2 5
where D = [2] D 2D. Now we reca the foowng fact about pung back a ne bunde L aong [n], whch s, In partcuar, f L s symmetrc, then [n] L L n(n+1)/2 [ 1] L n(n 1)/2. [n] L L n2. Appy ths to L = L(D), we have that D = 4D 2D = 2D. Thus D n = n(n 1)D+n([1]+α) D (n 1)α D = n(n 1)D+n deg([1]+α)d (n 1) deg(α)d, where a coeffcents are ntegra. Therefore, D g n = n 2g D g + O(n 2g 1)D g, where a coeffcents are ntegra. Ths shows that P α ( n) s a monc ntegra poynoma of degree 2g. 3 Characterstc poynoma and the representaton T A Let a notatons be as above. As we mentoned n the setup, we attempted to defne P α (X) as the characterstc poynoma of how α s actng on the vector space V A = T A Q, where s a prme not dvdng the characterstc of the base fed and T A s the Tate modue. The goa of ths secton s to show that the P α (X) we have defned above does agree wth the characterstc poynoma of the acton. Proposton 3.1. Let α End(A) be a separabe endomorphsm. For a char(k), P α (X) s the characterstc poynoma of α actng on V A. We need two emmas nvovng some agebra. Lemma 3.2. Let P (X) = (X a ) and Q(X) = (X b ) be monc poynomas of the same degree over Q. If F (a ) = F (b ) for a poynomas F Z[T ], then P = Q. Proof By contnuty, f F Z [X] s a poynoma, there s a poynoma F Z[X] such that F (a ) = F (a ) for a. To see ths, wrte F (X) = c 0 + c 1 X + + c m X m, where c j Z. 6
By contnuty, for a ε > 0, there s d 0,..., d n Z such that (c j d j )a j < ε and (c j d j )b j < ε for a, j. Thus by non-archmedean trange nequaty, m m c j a j d j a j < ε j=0 j=0 Choose ε to be smaer than n j=0 c ja j. By non-archmedean trange nequaty, we now have m c j a j m = d j a j. j=0 We have a smar equaty for b. Hence we have that for a poynomas F Z [X], F (a ) = F (b ). We can easy extend t to a poynomas n Q [T ], smpy by cearng the denomnator. Now, et d, e be mutpcty of a 1 as a root of P and Q respectvey. We w show d = e and t s suffcent to deduce our resut (by nducton on degree of P, Q). Pck α Q such that α a 1, and α a 1 1. Let F (X) be the mnma poynoma of α over Q, and G be the Gaos group of the spttng fed of F over Q. Thus, F (a ) = (a σα). 1 n Snce G permutes a, we have that F (a ) = 1 n j=0 1 n,σ G 1 n,σ G (σa σα). Snce Q s a oca fed, σa σα = a α for a, σ. Hence, F (a ) = a α m, 1 n where m = G. Smary, 1 n F (b ) = 1 n 1 n b α m, By assumpton, the eft hand sde of the prevous two equatons are equa. Thus, a α m = b α m 1 n 1 n 1 n a α = 1 n b α a 1 α d a α = b 1 α e b α a a 1 b a 1 7
WLOG suppose d e. For α a for a, We can wrte a 1 α d e b = b 1 b α a a 1 a α snce a α are nonzero. As α s gettng cose to a 1, a 1 α can be any postve ratona number suffcenty cose to 0, but rght hand sde s constant by non-archmedean trange nequaty. Thus f d > e, a postve ratonas cose to 0 are roots of the poynoma x d e b = C = b 1 b α. a a 1 a α Ths s a contradcton snce a poynoma can ony have fntey many roots. Therefore, d = e. Lemma 3.3. Let E be an agebra over a fed K, and δ : E K be a poynoma functon on E (vewng E as a vector space over K) such that δ(αβ) = δ(α)δ(β) for a α, β E. Fx α E. If P (X) = (X a ) s the poynoma δ(α X). Then for a monc poynomas F K[X], δ(f (α)) = ± F (a ). Proof Let L be the spttng fed of F. We w extend δ to a map E K L L that s st a poynoma functon and mutpcatve. Let {e } be a K-bass for E (as a vector space). Snce δ : E K s a poynoma functon, ( ) δ x e fnte s a poynoma over K n the x s nvoved. Use the same sets of poynomas (w.r.t. each fnte sets of e s) to defne δ : E K L L. It s we-defned,.e. compatbe w.r.t. ncuson of fnte sets of e s, as the poynomas we get come from a snge functon δ : E K. It s a poynoma functon, snce {e } s a K-bass of E, thus we can wrte a eements n E K L as L-near combnatons of fntey many e s. Fnay, t s mutpcatve. The computaton to show that δ s mutpcatve s tedous. The dea s that we can wrte every fnte product en as near combnatons of fntey many e j s, and then expand both sdes of δ(αβ) = δ(α)δ(β) expcty and compare. The emma s now easy. Let b 1,..., b n be roots of F (X). Snce our extended functon δ : E K L L s mutpcatve, we have that = δ( j (α b j )) = j δ(α b j ) = j P (b j ), recang that P (X) = δ(α X). By the factorzaton of P, we have δ(f (α)) =,j (b j a ) = ± F (a ). 8
We can now prove proposton 3.1. Proof (of Proposton 3.1) We may assume k s separaby cosed (snce characterstc poynoma of an sogeny s not changed by defnton, characterstc poynoma of a map s unchanged after base changed to agebrac cosure). For a separabe endomorphsm β End(A), deg(β) = # ker β = (# ker β()) 1 where ker β() s the -prmary part of the sze of kerne of β. Equvaenty, t s the kerne of the nduced map of β from T A to T A. We have the nverse snce a = a. Reca that T A = Z 2g, and that for any Z -near map f : Z n Zn, there exsts two Z -near somorphsms Z n Zn, as the vertca maps of the foowng dagram, such that the foowng the dagram commutes, and that the bottom map s obtaned by a dagona matrx. Z n f Z Z n Z The szes of ker β() and coker T β can be computed usng the dagram and the bottom row. Suppose the dagona matrx correspondng to the bottom map has dagona entres a 1,..., a n, where a Z. The sze of ker β() s n a 1, snce n each standard 1-dmensona summand of T A, a annhates precsey Z/ ν (a ) = Z/ a 1. On the other hand, the cokerne of T β s n =1 Z /(a ), whch aso has sze n a 1. Therefore, we have that =1 deg(β) = (# ker β()) 1 = (# coker T β) 1 = det(t β. Now suppose α End(A) s a separabe sogeny, and et a 1,..., a 2g be the roots of P α (X). Let F (X) Z[X] be an arbtrary poynoma wth eadng coeffcent c. Then by Lemma 3.3, apped to the functon X deg X, 1 c F (a ) = deg F (α). By computaton above, deg F (α) = det(t F (α)). =1 9
By Lemma 3.3 agan, apped to the functon X det(t X), 1 det(t F (α)) = c F (b ), where b are the roots of the characterstc poynoma of T α, the Z -near map on T A nduced by α. Therefore, 1 F (a ) = c F (b ). By Lemma 3.2, the characterstc poynomas for α and for T α agree. 4 Zeta Functon and We Conjectures We are fnay gong to defne the zeta functon and dscuss the We conjectures. Throughout ths secton, A s an abean varety of dmenson g over F q, a fnte fed wth q eements. Let Frob : A A be the Frobenus map. For readers seeng t the frst tme, to defne Frobenus precsey, frst defne t on projectve spaces by takng q-th power on each coordnate,.e. [x 0 : : x n ] [x q 0 : : xq n]. Note that on the underyng topoogca space, the map s ndeed dentty, as the radca of (f q ) s same as the radca of (f ). Now we use the fact that abean varetes are projectve to get a map on A. Such a map s we-defned (.e. does not depend on the embeddng we pck to a projectve space), as Frobenus commutes wth reguar maps on any affne charts (by drect computaton). Note that the kerne of Frob Id on A(F q ) s A(F q ). Ths can be check on ponts by the cacuaton on projectve space P n (F q ). Now snce 0 A(F q ), we have that Frob(0) = 0. Thus Frob s an endomorphsm of A, and we can tak about ts characterstc poynoma. Let P Frob (X) be the characterstc poynoma. By defnton, deg P Frob (X) = 2g and P Frob Z[X] s monc. It s tme to state our man theorem for the paper. Theorem 4.1. (We conjectures for abean varetes over fnte feds) Let P Frob (X) = (X a ) and N m = A(F q m). Then (a) N m = 2g =1 (1 am ) for a n 1. (b) (Remann hypothess) a = q for a. The theorem we had here seems dfferent from what we had n the ntroducton. Let s frst see how ths verson mpes the verson n the ntroducton. 10
Let P r (t) = (1 a,rt), where the product goes over a a 1 a 2... a r for 1 1 < 2 < < r 2g. We w show that Z(A, t) = P 1(t)P 3 (t)... P 2g 1 (t) P 0 (t)p 2 (t)... P 2g (t) by showng that ther ogarthms are equa,.e. t m N m m = og P 1(t)+og P 3 (t)+ +og P 2g 1 (t) og P 0 (t) og P 2 (t) og P 2g (t). m=1 To see ths, reca that we have a power seres expanson x n og(1 x) = for x < 1. n By Remann hypothess, a,r = q r/2. Thus for t < q g, we have that og P r (t) = n=0 og(1 a,r t) = m=1 a m,r tm Snce the sum over s fnte and each sum over m converges absoutey for t < q g, we can exchange the order of summatons. Thus, ( ) t m og P r (t) = m. Now, we have m=1 a m,r og P 1 (t) + + og P 2g 1 (t) og P 0 (t) og P 2g (t) ( = a m,0 a m,1 + a m,2 + ) a m t m,2g m m=1 ( ) = (1 a m t m ) m. m=1 The ast step s true by expandng (1 am ). Now we see that ths equas t m N m by (a) of the Theorem 4.1. Hence Theorem 4.1 does mpy (a) of m=1 m We conjectures as stated n the ntroducton. Wth ths descrpton of P (t), we can easy see that (b) we had here mpes (b) n the ntroducton. Now we w prove (a) of Theorem 4.1. Proof Reca that our goa s to show that m N m = (1 a m ). We w frst prove t for m = 1. Reca that A(F q ) s the kerne of Frob Id : A A, and Frob Id s separabe (snce t s on projectve pane P N F q ). Hence, N 1 = A(F q ) = ker(frob Id) = deg(frob Id) = P Frob (1) = (1 a ). 11
For genera m, we w use the resut from Secton 3. Reca that for any α End(A), P α (X) agrees wth the characterstc poynoma of α actng on the vector space V A. Put α = Frob, we get that a 1,..., a 2g are the egenvaues of Frob over V A. Thus a m 1,..., a m 2g are the egenvaues of Frob m over V A. Hence the characterstc poynoma of Frob m s (X am ). Snce the kerne of Frob m Id s A(F q m), by appyng the above argument for m = 1 to a genera m, we can show that N m = (1 a m ). Despte not gvng fu detas, we w sketch the prove of Remann hypothess for abean varetes over fnte feds here. The proof n Mne ([2]) makes use of Rosat nvouton. The frst step s to show that Frob Frob s the same as [q] (q s the sze of base fed), where Frob s the Rosat nvouton of Frobenus defned by some poarzaton. The second step s to show that f α End 0 (A) such that α α = [r] for some nteger r, then a 2 = r for any root a of P α. The frst step requres unwndng the defnton of Rosat nvouton and some computatons of dvsors. The second s many commutatve agebra. In partcuar, the key nput for the second step s to reaze that Rosat nvouton essentay sends an eement to ts compex conjugate. To be precse, f α End 0 (A) s as above, and ρ : Q[α] C s a homomorphsm, then α Q[α], and ρ(α ) = ρ(α). Readers can refer to Mne ([2]) for the compete proof. References [1] Kwun Chung. An attempt to fx mnes abean varetes emma 10.12, 2017. [2] James S. Mne. Abean varetes (v2.00), 2008. Avaabe at www.jmne.org/math/. [3] James S. Mne. Dvsors and ntersecton theory, 2015. Avaabe at www.jmne.org/math/. 12