Syntomic regulators and special values of p-adic
L-functions
Amnon Besser
p-adic Geometry and Homotopy theory
Loen, Norway
August 5, 2009
Amnon Besser
Syntomic regulators
Plan
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Deligne cohomology
I
Syntomic cohomology
I
Regulators
I
Beilinson conjecture
I
p-adic Beilinson conjecture
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The Karoubi regulator
Amnon Besser
Syntomic regulators
Motivation - complex theory
X /C - smooth proper irreducible of dimension d.
Cohomology of complex varieties
M = H i (X /Z)/torsion - a Hodge structure
I
M/Z - free finite rank
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MC - a finite dimensional complex vector space with a Hodge
decomposition MC = ⊕k+l=i M k,l ⇒ descending Hodge
filtration F n = ⊕k≥n M k,l
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ι : M ⊗ C → MC an isomorphism
Amnon Besser
Syntomic regulators
Abel-Jacobi map
CH0 (X ) - free abelian group on points of X / rational equivalence
deg : CH0 (X ) → Z = H0 (X , Z)
Abel-Jacobi map
α : Ker(deg) → (Ω1 (X ))∗ /image of H1 (X , Z)
X
X Z xi
α(
ni xi )(ω) =
ni
ω
P
P ∈ X arbitrary. Integral taken on some path.
Amnon Besser
Syntomic regulators
Cohomological interpretation
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CH0 (X ) = CHd (X ) = codimension d cycles.
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H0 (X , Z) = H 2d (X , Z)
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(Ω1 (X ))∗ = H 2d−1 (X , C)/F d (Poincaré duality)
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H1 (X , Z) = H 2d−1 (X , Z)
deg : CHd (X ) → H 2d (X , Z)
h
i
α : Ker(deg) → H 2d−1 (X , C)/ F d + H 2d−1 (X , Z)
Amnon Besser
Syntomic regulators
Extensions interpretation
Define
I
M0 = H 2d (X ), M1 = H 2d−1 (X )
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1 = Z with F 0 1C = 1C , F 1 1C = 0
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Twists M(n) = M, F j M(n)C = F j+n MC , ι multiplied by
(2πi)n .
Then
I
I
H 2d (X , Z) = M0 = M0 ∩ F d = M0 (d) ∩ F 0 = Hom(1, M0 (d))
H 2d−1 (X ,C)/ F d + H2d−1 (X , Z) =
M1 (d)C / F 0 + M1 (d) = Ext1 (1, M1 (d))
Higher dimensional analogues
c : CHk (X ) → Hom(1, H 2k (X )(k))
α : Ker(c) → Ext1 (1, H 2k−1 (X )(k))
Amnon Besser
Syntomic regulators
Abel-Jacobi and extensions
Z - a cyle of codimension k on X , c(Z ) = 0
s(Z ) - support of Z
2k
2k
0 → H 2k−1 (X ) → H 2k−1 (X −s(Z )) → Ker(Hs(Z
) (X ) → H (X )) → 0
2k (X ) → H 2k (X )))(k))
c(Z ) ∈ Hom(1, (Ker(Hs(Z
)
Pullback gives
0 → H 2k−1 (X )(k) → H 2k−1 (X − s(Z ))(k) → 1 → 0
whose extension class is α(Z )
Amnon Besser
Syntomic regulators
Deligne cohomology
Idea of Deligne: Both c and α should come from
2k (X , Z(k)), with
CHk (X ) → HD
i
0 → Ext1 (1, H i−1 (X )(n)) → HD
(X , Z(n)) → Hom(1, H i (X )(n)) → 0(∗)
(for i = 2k, k = n).
How to define?
First step: find functorial complexes computing all cohomologies
involved:
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RΓB (X , Z) computing integral cohomology.
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RΓdR (X /C) - filtered complex computing de Rham
cohomology with its filtrations (If X is non-proper take care to
introduce log singularities).
Second step: Take R Hom(1, •) on the resulting object.
Resulting spectral sequence gives (∗)
Amnon Besser
Syntomic regulators
Comments
I
I
To get RΓdR (X /C) take RΓ(X̄ , Ω•X̄ /C (log X̄ − X )) for a
compactification with normal crossings divisor as complement.
To make it functorial take a limit over all possible
compactifications.
˜ C • B • = MF (A• ⊕ B • → C • ). Then
Set A• ×
˜ RΓdR (X /C) RΓ(X , Z)
RΓD (X , Z(n)) = F n RΓdR (X /C)×
Amnon Besser
Syntomic regulators
p-adic theory
K /Qp finite, OK ring of integers, k-residue field
X /OK smooth.
Similar to complex case: V = H i (X ), an object of a category of
“filtered Frobenius modules” consisting of
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V /K0 finite dimensional vector space
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φ : V → V a semi-linear Frobenius map
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VK a K -vector space with desending filtration F n .
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ι : VK → V ⊗ K - comparison map
Amnon Besser
Syntomic regulators
p-adic theory
Here
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i (X /K ) is the rigid cohomology of the special fiber.
V = Hrig
0
k
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φ is the natural Frobenius
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i (X /K ) is the de Rham cohomology of the generic
VK = HdR
K
fiber
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ι is the “specialization map”
This category has
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1 - object where V = K0 with φ = σ, F 0 = VK = K , F 1 = 0
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Twists - filtrations shift as in Hodge, φ divided by p n .
Amnon Besser
Syntomic regulators
Syntomic cohomology
By analogy want
i
0 → Ext1 (1, H i−1 (X )(n)) → Hsyn
(X , n) → Hom(1, H i (X )(n)) → 0
How to define?
1. Define RΓrig (Xk /K0 ) with a Frobenius φ, RΓrig (Xk /K ) and a
base change map
2. Define a specialization map on the level of complexes
RΓdR (XK /K ) → RΓrig (Xk /K )
3. Define
˜ RΓrig (Xk /K ) F n RΓdR (XK /K )
RΓsyn (X , n) = MF (1 − p −n φ)×
n
Hom(1, V (n)) = V φ=p ∩ F n VK
Ext1 (1, V (n)) = V /(1 − φ/p n )F n (when K = K0 )
Amnon Besser
Syntomic regulators
Other versions (often usefull)
Gros style - without log singularities
Niziol style
without log singularities and using convergent rather than rigid
cohomology.
Same as above for proper X , hopeless otherwise.
Connection with étale cohomology (Niziol, Inventioness 127)
Y /K , K any field.
i (Y , Q (n)) - continuous étale cohomology (Jannsen).
Hét
p
Spectral sequence
j
i+j
E2i,j = H i (K , Hét
(Ȳ , Qp (n))) ⇒ Hét
(Y , Qp (n))
Amnon Besser
Syntomic regulators
Niziol’s Theorem
For X /OK proper smooth there is a functorial map
i
i
Hsyn
(X , n) → Hét
(XK , Qp (n))
compatible with Chern classes. It is compatible with the spectral
sequences, e.g. with
i−1
Ext1 (1, H i−1 (X )(n)) → H 1 (K , Hét
(XK̄ , Qp (n)))
which turns out to be the Bloch-Kato exponential map.
Amnon Besser
Syntomic regulators
Modified versions
Replace RΓrig (Xk /K0 ) with RΓrig (Xk /K ) and 1 − φ/p n by
1 − φr /p rn where φr is linear.
Benefit - Can use X to compute the rigid cohomology.
Obvious problem - It’s a different cohomology (but see later).
Finite polynomial cohomology and Coleman integration
Replace 1 − φr /p rn with more general polynomials ⇒ (X proper
smooth)
i−1
i
i
0 → HdR
(XK /K )/F n → Hfp
(X , n) → F n HdR
(XK /K ) → 0
E.g. i = n = 1
1
0 → K → Hfp
(X , 1) → Ω1 (XK /K ) → 0
1 (X , 1) unique up to constant =
ω ∈ Ω1 (XK /K ) ⇒ ω̃ ∈ Hfp
Coleman integral of ω.
Amnon Besser
Syntomic regulators
Regulators (aka Chern classes)
2n−s
cns : Ks (X ) → Hsyn
(X , n)
Main interest: explicitly compute these
For X proper and s > 0 the target is very simple
n
2n−s
1. Hrig
(Xk /K0 )φ=p = 0 for weight reasons
2n−s−1
2. 1 − φ/p n is invertible on Hrig
(Xk /K0 ) ⇒
2n−s−1
1
2n−s−1
∼
Ext (1, H
(X )(n)) = Hrig
(Xk /K )/F n
(Modified versions are the same) So usually
2n−s−1
(Xk /K )/F n
cns : Ks (X ) → Hrig
Amnon Besser
Syntomic regulators
Construction when X = Spec(A)
2n (BGL /O , n), i >> 0
First compute Hsyn
i
K
I BGLi has cohomology only in even degrees
I The standard classes xn ∈ H 2n (BGLi ) exist in de Rham and
rigid cohomology and are in
2n (BGL /O , n)
Hom(1, H 2n (BGLi )(n)) = Hsyn
i
K
I These classes are compatible so give classes in
2n (BGL/O , n)
Hsyn
K
Chern classes now follow from standard machinary:
∼ Hs (BGL(A), Z)
1. Hurewich Ks (X ) → Hs (BGL(A)+ , Z) =
2. Hs (BGL(A), Z) is computed with a complex having in degree
s maps from X = Spec(A) to the degree s part of BGL(A).
3. using pullbacks get
Hs (BGL(A), Z) × H k (BGL, n) → H k−s (X , n)
4. Compose the Hurevich map with the pairing applied to xn .
Amnon Besser
Syntomic regulators
p-adic Abel Jacobi
For K0 and X proper Chern classes induce a p-adic Abel-Jacobi
map
αp : Ker(deg) → Ω1 (XK /K )∗
This has the same formula as in the classical case, only with
ordinary integration replaced by Coleman integration.
Amnon Besser
Syntomic regulators
Beilinson’s conjecture: Motivic cohomology
i
HM
(X , Q(n)) = (K2n−i (X ) ⊗ Q)(n)
(n) refers to the part where the Adams operation ψ k acts by k n .
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An integral version can be defined using Bloch’s higher Chow
groups.
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The indexing is such that the Chern character defines
i (X , Q(n)) → H i (X , Q(n)) and
HM
D
i
i (X , n) in the appropriate situations.
HM (X , Q(n)) → Hsyn
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For Beilinson conjectures need to introduce subspace
i (X , Q(n)) (Scholl).
HM
Z
Amnon Besser
Syntomic regulators
Deligne cohomology over R
X /R ⇒ X (C) has an action of complex conjugation c.
i (X /R, R(n)) = H i (X /C, R(n))c=1 where c acts
Definition: HD
D
both on space and on coefficients.
For s > 0
2n−s
2n−s−1
HD
(X /C, R(n)) = HdR
(X /C)/ F n + H 2n−s−1 (X (C), (2πi)n R)
Taking c invariants
2n−s−1
0 → H 2n−s−1 (X (C), (2πi)n R)c=1 → HdR
(X /R)/F n
2n−s
→ HD
(X /R, R(n)) → 0
Suppose X /Q smooth and projective. H 2n−s−1 (X (C), (2πi)n R)c=1
2n−s−1
and HdR
(X /R)/F n have obvious Q structures on them, hence
2n−s
HD (X /R, R(n)) has one, denoted D.
Amnon Besser
Syntomic regulators
Beilinson’s conjecture
∼
I
2n−s
2n−s
HM
(X , Q(n))Z ⊗ R −
→ HD
(X /R, R(n)) hence provides a
2n−s
Q structure M on HD (X /R, R(n)).
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det M = L(H 2n−s−1 (X ), n) det D.
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More concretely, if bases {Mi } and {Di } are chosen
^
^
α · Mi = L(H 2n−s−1 (X ), n) · Di
with α ∈ Q×
Amnon Besser
Syntomic regulators
Example: number fields
X = Spec(L), L a number field with r1 real embeddings and 2r2
complex embeddings.
1 (Spec L/R, R(n))?
What is HD
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X (C) = set of embeddings, permuted by c
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c acts on (2πi)n as (−1)n .
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dim H 0 (X (C), (2πi)n R)c=1 = r2 (+r1 if 2|n)
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1 (X /R, R(n)) = r (+r if 2|n + 1)
dim HD
2
1
In this case, Beilinson’s conjecture is Borel’s Theorem.
Important observation: K-theory knows information at infinity.
Amnon Besser
Syntomic regulators
p-adic Beilinson conjecture
X /Q smooth proper, s > 0
Xp - an integral model for X ⊗ Qp , still assumed smooth and
proper.
2n−s
2n−s−1
2n−s
HM
(X , Q(n)) → Hsyn
(Xp , n) = (HdR
(X /Q)/F n ) ⊗ Qp
Problem: dimensions don’t match !!!
Missing the H 2n−s−1 (X (C), (2πi)n R)c=1 piece.
A possible solution: import this part using comparison maps.
Perrin-Riou’s solution: Make everything depend on the choice of a
basis for a complimentary subspace of dimension
d± = dim H 2n−s−1 (X (C), (2πi)n R)c=1 !!!
Amnon Besser
Syntomic regulators
The conjecture
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Fixing the bases as for Beilinson’s conjecture.
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Adding the complelementary basis to the image of the
regulator.
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Replacing Beilinson’s regulator by the syntomic regulator.
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Replacing the L-function by the p-adic L-function (for the
complementary basis).
It is the same conjecture!!! (with the same rational factor up to
trivial terms).
Amnon Besser
Syntomic regulators
What are p-adic L-functions?
3 answers
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The precise - An element of an Iwasawa algebra coming via a
logrithm map from an Euler system...
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The easy - It is defined by the above conjecture: Rephrase to
say: There is a p-adic continuous (or even analytic) function
that interpolates the numbers arising in the conjecture above
(with L-factor at p removed)
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The practical - For some (so called critical) n no motivic
cohomology is needed. The p-adic L-function interpolates
these values.
Better yet: It is a measure on Zp such that the integral of x n χ, for
a Dirichlet character χ, gives α corresponding to
L(H 2n−s−1 (X ) ⊗ χ, n) for critical n, χ.
Amnon Besser
Syntomic regulators
Evidence for p-adic Beilinson
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X an elliptic curve with CM over Q, s = n = 2 (Coleman-de
Shalit, B.)
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Certain numerical evidence for totally real fields and totally
real Artin motives (B. Buckingham, de Jeu, Roblot)
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Modular curves (Gealy)
Amnon Besser
Syntomic regulators
Relation with the Karoubi regulator
Problem: The syntomic regulator depends on an integral model
Should expect dependence on the generic fiber only: It’s étale
cohomology determines the filtered Frobenius module via
Fontaine’s functor.
Karoubi’s regulator may provide the solution
Amnon Besser
Syntomic regulators
Karoubi’s regulator
A - non archimedean Banach ring
create a (rigid) homotopy invariant K -theory
Make a simplicial ring A•
X
An = A{x0 , . . . , xn }/(
xi − 1)
Definition: Ktop (A) = BGL(A• ).
Map K (A) → Ktop (A)
Krel (A) = MF(K (A) → Ktop (A))
de Rham Chern classes
K (A) → F n Ω•+2n (A)
By integrating on simplices
Ktop (A) → Ω•+2n (A)
Krel (A) → MF(F n → Ω• (A))
2n−s−1
Ksrel (A) → HdR
(A)/F n
just like the syntomic regulator
Amnon Besser
Syntomic regulators
Relations with the syntomic regulator
Kstop (A) is torsion in some cases. so Krel is like K .
I
Karoubi conjectures a relation with p-adic polylogarithms for
p-adic fields
I
B. conjectured a relation with the syntomic regulator
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Hamida - formulas for Karoubi’s regulator for p-adic fields
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Tamme - relation with the syntomic regulator for fields
(maybe in general)
Expectation: MF (Ktop ⊕ F n → Ω•+2n ) should be a replacement
for syntomic cohomology.
Amnon Besser
Syntomic regulators