Academia Sinica, Taipei October 30, 2003
Hopf algebras and
Diophantine problems
Michel Waldschmidt
miw@math.jussieu.fr http://www.math.jussieu.fr/∼miw/
Hopf algebras (commutative, cocommutative, of finite type) Algebraic groups (commutative, linear, over Q)
Exponential polynomials
Transcendence of values of exponential polynomials
Algebra of multizeta values
Algebras (over k = C or k = Q)
A k -algebra (A, m, η ) is a k-vector space A with a product m : A ⊗ A → A and a unit η : k −→ A which are k -linear maps such that the following diagrams commute:
(Associativity)
A ⊗ A ⊗ A
m⊗Id−−→ A ⊗ A
Id⊗m
↓ ↓
mA ⊗ A − →
m
A
(Unit)
k ⊗ A
η⊗Id−−→ A ⊗ A
Id⊗η←−− A ⊗ k
↓ ↓
m↓
A = A = A
Commutative algebras A k -algebra is commutative if the diagram
A ⊗ A − →
τA ⊗ A
m
↓ ↓
mA = A
commutes. Here τ (x ⊗ y) = y ⊗ x.
Coalgebras
A k -coalgebra (A, ∆, ) is a k -vector space A with a coproduct
∆ : A → A ⊗ A and a counit : A −→ k which are k-linear maps such that the following diagrams commute:
(Coassociativity)
A − →
∆A ⊗ A
∆
↓ ↓
∆⊗IdA ⊗ A −−→
Id⊗∆
A ⊗ A ⊗ A
(Counit)
A = A = A
↓ ↓
∆↓
k ⊗ A ←−−
⊗Id
A ⊗ A −−→
Id⊗
A ⊗ k
Commutative coalgebras A k -coalgebra is commutative if the diagram
A = A
∆
↓ ↓
∆A ⊗ A ← −
τ
A ⊗ A
commutes.
Bialgebras
A bialgebra (A, m, η, ∆, ) is a k -algebra (A, m, η ) together with a coalgebra structure (A, ∆, ) which is compatible: ∆ and are algebra morphisms
∆(xy ) = ∆(x)∆(y), (xy ) = (x)(y).
Hopf Algebras
A Hopf algebra (H, m, η, ∆, , S ) is a bialgebra (H, m, η, ∆, ) with an antipode S : H → H which is a k-linear map such that the following diagram commutes:
H ⊗ H ←
∆− H − →
∆H ⊗ H
Id⊗S
↓
η◦↓ ↓
S⊗IdH ⊗ H − →
m
H ← −
m
H ⊗ H
In a Hopf Algebra the primitive elements
∆(x) = x ⊗ 1 + 1 ⊗ x
satisfy (x) = 0 and S (x) = −x; they form a Lie algebra for the bracket
[x, y] = xy − yx.
In a Hopf Algebra the primitive elements
∆(x) = x ⊗ 1 + 1 ⊗ x
satisfy (x) = 0 and S (x) = −x; they form a Lie algebra for the bracket
[x, y] = xy − yx.
The group-like elements
∆(x) = x ⊗ x, x 6= 0
are invertible, they satisfy (x) = 1, S (x) = x
−1and form a
Example 1.
Let G be a finite multiplicative group, kG the algebra of G over k which is a k vector-space with basis G. The mapping
m : kG ⊗ kG → kG extends the product
(x, y) 7→ xy of G by linearity. The unit
η : k → kG
maps 1 to 1
G.
Define a coproduct and a counit
∆ : kG → kG ⊗ kG and : kG → k
by extending
∆(x) = x ⊗ x and (x) = 1 for x ∈ G
by linearity. The antipode
S : kG → kG is defined by
S (x) = x
−1for x ∈ G.
Since ∆(x) = x ⊗ x for x ∈ G this Hopf algebra kG is cocommutative.
It is a commutative algebra if and only if G is commutative.
The set of group like elements is G: one recovers G from kG.
Example 2.
Again let G be a finite multiplicative group. Consider the k- algebra k
Gof mappings G → k, with basis δ
g(g ∈ G), where
δ
g(g
0) =
1 for g
0= g, 0 for g
06= g.
Define m by
m(δ
g⊗ δ
g0) = δ
gδ
g0.
Hence m is commutative and m(δ
g⊗ δ
g) = δ
gfor g ∈ G.
The unit η : k → k
Gmaps 1 to P
g∈G
δ
g.
Define a coproduct ∆ : k
G→ k
G⊗ k
Gand a counit : k
G→ k by
∆(δ
g) = X
g0g00=g
δ
g0⊗ δ
g00and (δ
g) = δ
g(1
G).
The coproduct ∆ is cocommutative if and only if the group G is commutative.
Define an antipode S by
S (δ
g) = δ
g−1.
Remark. One may identify k
G⊗ k
Gand k
G×Gwith
δ
g⊗ δ
g0= δ
g,g0.
Duality of Hopf Algebras
The Hopf algebras kG from example 1 and k
Gfrom example 2 are dual from each other:
kG × k
G−→ k (g
1, δ
g2) 7−→ δ
g2(g
1)
The basis G of kG is dual to the basis (δ
g)
g∈Gof k
G.
Example 3.
Let G be a topological compact group over C. Denote by R (G) the set of continuous functions f : G → C such that the translates f
t: x 7→ f (tx), for t ∈ G, span a finite dimensional vector space.
Define a coproduct ∆, a counit and an antipode S on R (G) by
∆f (x, y) = f (xy ), (f ) = f (1), Sf (x) = f (x
−1) for x, y ∈ G.
Hence R (G) is a commutative Hopf algebra.
Example 4.
Let g be a Lie algebra, U ( g ) its universal envelopping algebra, namely T ( g )/ I where T ( g ) is the tensor algebra of g and I the two sided ideal generated by XY − Y X − [X, Y ].
Define a coproduct ∆, a counit and an antipode S on U ( g ) by
∆(x) = x ⊗ 1 + 1 ⊗ x, (x) = 0, S (x) = −x for x ∈ g .
Hence U ( g ) is a cocommutative Hopf algebra.
The set of primitive elements is g : one recovers g from U ( g ).
Duality of Hopf Algebras (again)
Let G be a compact connected Lie group with Lie algebra g .
Then the two Hopf algebras R (G) and U ( g ) are dual from each
other.
Bicommutative Hopf algebras of finite type 1.
H = k[X ], ∆(X ) = X ⊗ 1 + 1 ⊗ X , (X ) = 0, S (X ) = −X .
Bicommutative Hopf algebras of finite type 1.
H = k[X ], ∆(X ) = X ⊗ 1 + 1 ⊗ X , (X ) = 0, S (X ) = −X .
k[X ] ⊗ k [X ] ' k[T
1, T
2], X ⊗ 1 7→ T
1, 1 ⊗ X 7→ T
2∆P (X ) = P (T
1+ T
2), P (X ) = P (0), SP (X ) = P (−X ).
Bicommutative Hopf algebras of finite type 1.
H = k[X ], ∆(X ) = X ⊗ 1 + 1 ⊗ X , (X ) = 0, S (X ) = −X .
k[X ] ⊗ k [X ] ' k[T
1, T
2], X ⊗ 1 7→ T
1, 1 ⊗ X 7→ T
2∆P (X ) = P (T
1+ T
2), P (X ) = P (0), SP (X ) = P (−X ).
G
a(K ) = Hom
k(k[X ], K ), k[G
a] = k[X ]
k [G ] is a bicommutative Hopf algebra of finite type.
Bicommutative Hopf algebras of finite type 2.
H = k[Y, Y
−1], ∆(Y ) = Y ⊗ Y , (Y ) = 1, S(Y ) = Y
−1.
Bicommutative Hopf algebras of finite type 2.
H = k[Y, Y
−1], ∆(Y ) = Y ⊗ Y , (Y ) = 1, S(Y ) = Y
−1.
H ⊗ H ' k [T
1, T
1−1, T
2, T
2−1], Y ⊗ 1 7→ T
1, 1 ⊗ Y 7→ T
2∆P (Y ) = P (T
1T
2), P (Y ) = P (1), SP (Y ) = P (Y
−1).
Bicommutative Hopf algebras of finite type 2.
H = k[Y, Y
−1], ∆(Y ) = Y ⊗ Y , (Y ) = 1, S(Y ) = Y
−1.
H ⊗ H ' k [T
1, T
1−1, T
2, T
2−1], Y ⊗ 1 7→ T
1, 1 ⊗ Y 7→ T
2∆P (Y ) = P (T
1T
2), P (Y ) = P (1), SP (Y ) = P (Y
−1).
G
m(K ) = Hom
k(k[Y, Y
−1], K ), k [G
m] = k [Y, Y
−1],
k [G
m] is a bicommutative Hopf algebra of finite type.
Bicommutative Hopf algebras of finite type 3.
H = k[X
1, . . . , X
d0, Y
1, Y
1−1, . . . , Y
d1, Y
d−11
] ' k[X ]
⊗d0⊗ k[Y, Y
−1]
⊗d1Primitive elements: k -space kX
1+ · · · + kX
d0, dimension d
0.
Group-like elements: multiplicative group hY
1, . . . , Y
d1i, rank d
1.
G = G
da0× G
dm1k [G] = H, G(K ) = Hom (H, K ).
Bicommutative Hopf algebras of finite type 3.
H = k[X
1, . . . , X
d0, Y
1, Y
1−1, . . . , Y
d1, Y
d−11
] ' k[X ]
⊗d0⊗ k[Y, Y
−1]
⊗d1The category of commutative linear algebraic groups over k G = G
da0× G
dm1is anti-equivalent to the category of Hopf algebras of finite type which are bicommutative (commutative and cocomutative)
H = k [G].
The vector space of primitive elements has dimension d
0while
the rank of the group-like elements is d
1.
Other examples
If W is a k-vector space of dimension `
0, Sym(W ) is a bicommutative Hopf algebra of finite type, anti-isomorphic to k [G
`a0]:
For a basis ∂
1, . . . , ∂
`0of W , Sym(W ) ' k[∂
1, . . . , ∂
`0].
Other examples
If W is a k-vector space of dimension `
0, Sym(W ) is a bicommutative Hopf algebra of finite type, anti-isomorphic to k [G
`a0].
If Γ is a torsion free finitely generated Z-module of rank `
1, then the group algebra k Γ is again a bicommutative Hopf algebra of finite type, anti-isomorphic to k[G
`m1]:
For a basis γ
1, . . . , γ
`1of Γ, kΓ ' k [γ
1, γ
1−1, . . . , γ
`1, γ
`−11
].
Other examples
If W is a k-vector space of dimension `
0, Sym(W ) is a bicommutative Hopf algebra of finite type, anti-isomorphic to k [G
`a0].
If Γ is a torsion free finitely generated Z-module of rank `
1, then the group algebra k Γ is again a bicommutative Hopf algebra of finite type, anti-isomorphic to k[G
`m1].
The category of bicommutative Hopf algebras of finite type is equivalent to the category of pairs (W, Γ) where W is a k -vector space and Γ is a finitely generated Z-module:
H = Sym(W ) ⊗ kΓ.
Commutative linear algebraic groups over Q G = G
da0× G
dm1d = d
0+ d
1G(Q) = Q
d0× (Q
×)
d1(β
1, . . . , β
d0, α
1, . . . , α
d1)
Commutative linear algebraic groups over Q G = G
da0× G
dm1d = d
0+ d
1G(Q) = Q
d0× (Q
×)
d1exp
G: T
e(G) = C
d−→ G(C) = C
d0× (C
×)
d1(z
1, . . . , z
d) 7−→ (z
1, . . . , z
d0, e
zd0+1, . . . , e
zd)
Commutative linear algebraic groups over Q G = G
da0× G
dm1d = d
0+ d
1G(Q) = Q
d0× (Q
×)
d1exp
G: T
e(G) = C
d−→ G(C) = C
d0× (C
×)
d1(z
1, . . . , z
d) 7−→ (z
1, . . . , z
d0, e
zd0+1, . . . , e
zd) For α
jand β
iin Q,
exp
G(β
1, . . . , β
d0, log α
1, . . . , log α
d1) ∈ G(Q)
Baker’s Theorem. If
β
0+ β
1log α
1+ · · · + β
nlog α
n= 0 with algebraic β
iand α
j, then
1. β
0= 0
2. If (β
1, . . . , β
n) 6= (0, . . . , 0), then log α
1, . . . , log α
nare Q- linearly dependent.
3. If (log α
1, . . . , log α
n) 6= (0, . . . , 0), then β
1, . . . , β
nare Q-
linearly dependent.
Example: (3 − 2 √
5) log 3 + √
5 log 9 − log 27 = 0.
Example: (3 − 2 √
5) log 3 + √
5 log 9 − log 27 = 0.
Corollaries.
1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e
√2
.
Example: (3 − 2 √
5) log 3 + √
5 log 9 − log 27 = 0.
Corollaries.
1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e
√2
.
2. Gel’fond-Schneider (n = 2, β
0= 0): transcendence of 2
√2
, log 2/ log 3, e
π.
Example: (3 − 2 √
5) log 3 + √
5 log 9 − log 27 = 0.
Corollaries.
1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e
√2
.
2. Gel’fond-Schneider (n = 2, β
0= 0): transcendence of 2
√2
, log 2/ log 3, e
π.
3. Example with n = 2, β
06= 0: transcendence of Z
1dx = 1
log 2 + π
√ ·
Values of exponential polynomials Proof of Baker’s Theorem. Assume
β
0+ β
1log α
1+ · · · + β
n−1log α
n−1= log α
n(B
1) (Gel’fond–Baker’s Method)
Functions: z
0, e
z1, . . . , e
zn−1, e
β0z0+β1z1+···+βn−1zn−1Points: Z(1, log α
1, . . . , log α
n−1) ∈ C
nDerivatives: ∂/∂z
i, (0 ≤ i ≤ n − 1).
Values of exponential polynomials Proof of Baker’s Theorem. Assume
β
0+ β
1log α
1+ · · · + β
n−1log α
n−1= log α
n(B
1) (Gel’fond–Baker’s Method)
Functions: z
0, e
z1, . . . , e
zn−1, e
β0z0+β1z1+···+βn−1zn−1Points: Z(1, log α
1, . . . , log α
n−1) ∈ C
nDerivatives: ∂/∂z
i, (0 ≤ i ≤ n − 1).
n + 1 functions, n variables, 1 point, n derivatives
Another proof of Baker’s Theorem. Assume again β
0+ β
1log α
1+ · · · + β
n−1log α
n−1= log α
n(B
2) (Generalization of Schneider’s method)
Functions: z
0, z
1, . . . , z
n−1, e
z0α
z11· · · α
zn−1n−1=
exp{z
0+ z
1log α
1+ · · · + z
n−1log α
n−1}
Points: {0} × Z
n−1+ Z(β
0, , . . . , β
n−1) ∈ C
nDerivative: ∂/∂z
0.
Another proof of Baker’s Theorem. Assume again β
0+ β
1log α
1+ · · · + β
n−1log α
n−1= log α
n(B
2) (Generalization of Schneider’s method)
Functions: z
0, z
1, . . . , z
n−1, e
z0α
z11· · · α
zn−1n−1=
exp{z
0+ z
1log α
1+ · · · + z
n−1log α
n−1} Points: {0} × Z
n−1+ Z(β
0, , . . . , β
n−1) ∈ C
nDerivative: ∂/∂z
0.
Six Exponentials Theorem. If x
1, x
2are two complex numbers which are Q-linearly independent and if y
1, y
2, y
3are three complex numbers which are Q-linearly independent, then one at least of the six numbers
e
xiyj(i = 1, 2, j = 1, 2, 3)
is transcendental.
Proof of the six exponentials Theorem
Assume x
1, . . . , x
aare Q-linearly independent numbers and y
1, . . . , y
bare Q-linearly independent numbers such that
e
xiyj∈ Q for i = 1, . . . , a, , j = 1, . . . , b with ab > a + b.
Functions: e
xiz(1 ≤ i ≤ a) Points: y
j∈ C (1 ≤ j ≤ b)
a functions, 1 variable, b points, 0 derivative
Linear Subgroup Theorem G = G
da0× G
dm1, d = d
0+ d
1.
W ⊂ T
e(G) a C-subspace which is rational over Q. Let `
0be its dimension.
Y ⊂ T
e(G) a finitely generated subgroup with Γ = exp(Y ) contained in G(Q) = Q
d0× (Q
×)
d1. Let `
1be the Z-rank of Γ.
V ⊂ T
e(G) a C-subspace containing both W and Y . Let n be the dimension of V .
Hypothesis:
n(`
1+ d
1) < `
1d
1+ `
0d
1+ `
1d
0n(`
1+ d
1) < `
1d
1+ `
0d
1+ `
1d
0d
0+ d
1is the number of functions
d
0are linear
d
1are exponential
n is the number of variables
`
0is the number of derivatives
`
1is the number of points
d
0d
1`
0`
1n
Baker B
11 n n 1 n
Baker B
2n 1 1 n n
Six exponentials 0 a 0 b 1
d
0d
1`
0`
1n
Baker B
11 n n 1 n
Baker B
2n 1 1 n n
Six exponentials 0 a 0 b 1 Baker:
n(`
1+ d
1) = n
2+ n
`
1d
1+ `
0d
1+ `
1d
0= n
2+ n + 1
Six exponentials: a + b < ab
n(`
1+ d
1) = a + b
duality:
(d
0, d
1, `
0, `
1) ←→ (`
0, `
1, d
0, d
1)
d dz
sz
te
xzz=y
=
d dz
tz
se
yzz=x
.
Fourier-Borel duality:
(d
0, d
1, `
0, `
1) ←→ (`
0, `
1, d
0, d
1)
d dz
sz
te
xzz=y
=
d dz
tz
se
yzz=x
. L
sy: f 7−→
d dz
sf (y ).
f
ζ(z ) = e
zζ, L
sy(f
ζ) = ζ
se
yζ. L
sy(z
tf
ζ) =
d dζ
tL
sy(f
ζ).
For v = (v
1, . . . , v
n) ∈ C
n, set D
v= v
1∂
∂z
1+ · · · + v
n∂
∂z
n·
Let w
1, . . . , w
`0
, u
1, . . . , u
d0
, x and y in C
n, t ∈ N
d0and s ∈ N
`0. For z ∈ C
n, write
(uz )
t= (u
1z )
t1· · · (u
d0
z )
td0and D
ws= D
ws11
· · · D
ws``00