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Sato-Tate conjecture for a CM Picard curve
Sato-Tate conjecture for a CM Picard curve
Joan-C. Lario(joint w. Anna Somoza)
NCTS HsinchuJuly 30 – August 3, 2014
| NCTS Hsinchu July 30 – August 3, 2014 1 / 43
Sato-Tate conjecture for a CM Picard curve
Introduction
The Sato-Tate conjecture is another major example of the imapct ofcomputation on Number Theory.
Hitachi Hipac 103: economic power load equipment distribution (1961).
| NCTS Hsinchu July 30 – August 3, 2014 2 / 43
Sato-Tate conjecture for a CM Picard curve
Introduction
Start with an arithmetic object X over Z, and want to study thedistribution of a sequence xp of elements that depend on X mod p.
Attached to X one has its Sato-Tate group ST(X ). For instance,
X ST(X ) xpf polynomial Gal(f ) FrobpA abelian variety ST(A) Lp(A,T )
| NCTS Hsinchu July 30 – August 3, 2014 3 / 43
Sato-Tate conjecture for a CM Picard curve
Introduction
Once we have the Sato-Tate group ST(X ), we can:
• Theoretical approach: Show that for every nontrivial irreduciblerepresentation
φ : ST(X )→ GLn(C) ,
the associated L-function is invertible; L(φ ,xp,s) is meromorphicfor Re(s)≥ 1 and L(φ ,1) 6= 0.
• Computational approach: Check if the moment sequence arising fromthe Haar measure of ST(X ) and the limiting numerical momentsequence of xp (for p ≤ N) match as N → ∞.
| NCTS Hsinchu July 30 – August 3, 2014 4 / 43
Sato-Tate conjecture for a CM Picard curve
Sato-Tate and abelian varieties
Let A/Q be an abelian variety of dimension g . The normalized local factors
LSTp (A,T ) = Lp(A,T/
√p) =
2g
∏i=1
(1−Tαi√p
) ∈ R[T ]
can be thought as characteristic poynomials of (conjugacy classes) ofmatrices of a certain closed subgroup ST(A)⊆ USp(2g).
dim(A) #ST(A)
1 2(3)2 34(52)3 hundreds < ∞
Some specific cases in higher dimension:
Quotients of the Fermat curves: Fite-Gonzalez-L. 2013Abelian varieties with CM: Johansson 2013
| NCTS Hsinchu July 30 – August 3, 2014 5 / 43
Sato-Tate conjecture for a CM Picard curve
Sato-Tate and abelian varieties
Let A/Q be an abelian variety of dimension g . The normalized local factors
LSTp (A,T ) = Lp(A,T/
√p) =
2g
∏i=1
(1−Tαi√p
) ∈ R[T ]
can be thought as characteristic poynomials of (conjugacy classes) ofmatrices of a certain closed subgroup ST(A)⊆ USp(2g).
dim(A) #ST(A)
1 2(3)2 34(52)3 hundreds
Some specific cases in higher dimension:
Quotients of the Fermat curves: Fite-Gonzalez-L. 2013Abelian varieties with CM: Johansson 2013
| NCTS Hsinchu July 30 – August 3, 2014 6 / 43
Sato-Tate conjecture for a CM Picard curve
A CM Picard curve
A cyclic trigonal curve of genus g = 3 is called a Picard curve.
A Picard curve defined over Q is given by an affine equation
y3 = f (x) , f ∈Q[x ] , deg(f ) = 4 , f separable .
Up to isomorphism, there are five Picard curves defined over Q withcomplex multiplication by a CM-field of class number one (Koike-Weng2004):
C : y3 = f (x) CM-field
x4−x Q(ζ9)
x4−98x2 + 392x−343 Q( 3√
7 + 21ζ3)
x4−1274x2 + 24440x−130975 Q( 3√
13−39ζ3)
x4−31682x2 + 2983936x−77986111 Q( 3√
31 + 186ζ3)
x4−5772578x2 + 6297605248x−1737328844383 Q( 3√
43−258ζ3)
| NCTS Hsinchu July 30 – August 3, 2014 7 / 43
Sato-Tate conjecture for a CM Picard curve
A CM Picard curve
Our choice today is:
C : y3 = x4−x .
• it has complex multiplication by the cyclotomic field K = Q(ζ9)
• the Jacobian variety Jac(C ) is simple
• [0 : 1 : 0] is the unique point of C at infinity
• C has good reduction at all primes different from 3
| NCTS Hsinchu July 30 – August 3, 2014 8 / 43
Sato-Tate conjecture for a CM Picard curve
The local factors
For every prime p 6= 3, we want to compute the local factor
Lp(C ,T ) =6
∑i=0
biTi =
6
∏i=1
(1−αiT ) ∈ Z[T ]
It provides all NC (pr ) = |C (Fpr )| and it is determined by the first three:
b0 = 1
b1 = NC (p)− (p+ 1)
b2 = (NC (p2)− (p2 + 1) +b21)/2
b3 = (NC (p3)− (p3 + 1)−b31 + 3b2b1)/3
b4 = pb2
b5 = p2b1
b6 = p3 .
| NCTS Hsinchu July 30 – August 3, 2014 9 / 43
Sato-Tate conjecture for a CM Picard curve
The local factors
For small primes one gets:
p Lp(C ,T )
2 (1 + 2T 2)(1−2T 2 + 4T 4)5 (1 + 5T 2)(1−5T 2 + 25T 4)7 1 + 7T 3 + 343T 6
11 (1 + 11T 2)(1−11T 2 + 121T 4)13 1−65T 3 + 2197T 6
17 (1 + 17T 2)3
19 1−6T −12T 2 + 169T 3−228T 4−2166T 5 + 6859T 6
We need to compute Lp(C ,T ) for large values of p.
| NCTS Hsinchu July 30 – August 3, 2014 10 / 43
Sato-Tate conjecture for a CM Picard curve
The Grossencharakter of C
Let O be the ring of integers of K = Q(ζ9), and σi (ζ9) = ζ i9.
Consider the ideal m = (1 + ζ9 + ζ 49 )4.
Denote by IK (m) the group of fractional ideals coprime to m.
Lemma
There exists a Grossencharakter ψ : IK (m)→ C∗,
ψ(αO) = ∏σ∈Φ∗
σα if α ≡ 1 mod ∗m
of conductor m and infinite type Φ∗ = σ1,σ5,σ7 such that
Lp(C ,T ) = T 6 Irr(ψ(P),1/T f ;Q)6/(fd) ,
where N(P) = pf and d = [Q(ψ(P)) : Q].
| NCTS Hsinchu July 30 – August 3, 2014 11 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Lemma (sketch)
For every prime P of K coprime to 3, the number of points of C definedover the residue field FP is given by
|C (FP)|=
1 +N(P) if N(P) 6≡ 1 (mod 9),
1 +N(P) + TrK/Q(J(6,1)(P)) if N(P)≡ 1 (mod 9).
(Holzapfel, Nicolae): Arithmetic on a family of Picard curves.
(Fite): There is an isomorphism over Q between C and C ′ : v9 = u(u+ 1)6.
| NCTS Hsinchu July 30 – August 3, 2014 12 / 43
Sato-Tate conjecture for a CM Picard curve
The Sato-Tate group ST(C )
Proposition
Up to conjugation in USp(6,C), the Sato-Tate group of C is
⟨
u1
u1
u2
u2
u3
u3
,
0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 10 −1 0 0 0 01 0 0 0 0 0
, |ui |= 1
⟩
In particular, there is an isomorphism
ST(C )' U(1)3 o (Z/9Z)∗.
| NCTS Hsinchu July 30 – August 3, 2014 13 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Proposition
Sketch. Serre’s recipe: Fix an auxiliary prime ` of good reduction, and fixan embedding ι : Q` → C.
Let G be the Zariski closure of the image of
ρ` : Gal(Q/Q)→ GL(V`(Jac(C )))' GL(6,Q`) ,
and let G1 be the Zariski closure of G ∩Sp6(Q`).
By definition, ST(C ) is a maximal compact subgroup of G1⊗ι C.
| NCTS Hsinchu July 30 – August 3, 2014 14 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Proposition
Since the CM-type of Jac(C ) is non-degenerate, then G1 = TL(C )⊗Q`
where the twisted Lefschetz group is defined as
TL(C ) =⋃
τ∈Gal(Q/Q)
L(C )(τ) ,
where L(C )(τ) = γ ∈ Sp6(Q) : γαγ−1 = τ(α) for all α ∈ End0(Jac(C )Q).
Indeed, the CM-type of Jac(C ) is non-degenerate due to the fact that Φ∗
is simple and dimJac(C ) = 3; alternatively, check that the Z-linear map
Z[Gal(K/Q)]→ Z[Gal(K/Q)], σa 7→ ∑σb∈Φ
σ−1b σa
has maximal rank 1 + dim(Jac(C )) = 4.
| NCTS Hsinchu July 30 – August 3, 2014 15 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Proposition
Borrowing the results of Banaszak, Gadja, Krasorı [2003] and Fite,Kedlaya, Rotger, Sutherland [2012], we obtain that the identitycomponent satisfies
G 01 = TL(C )0⊗Q` = diag(x1,y1,x2,y2,x3,y3) | xi ,yi ∈Q∗` ,xiyi = 1 .
Hence,
ST(C )0 = diag(u1,u1,u2,u2,u3,u3) : ui ∈ U(1) ' U(1)3 .
Also it follows that the group of components of ST(C ) is isomorphic toGal(K/Q).
| NCTS Hsinchu July 30 – August 3, 2014 16 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Proposition
We claim that ST(C ) = ST(C )0 o 〈γ〉, where
γ =
0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 10 −1 0 0 0 01 0 0 0 0 0
.
To this end, we consider α ∈ Aut(C ) determined by α(x ,y) = (ζ 6x ,ζ 2 y).Under the basis of regular differentials of Ω1(C ):
ω1 =dx
y2, ω2 =
dx
y, ω3 =
xdx
y2,
the induced action is α∗(ω1) = ζ 2ω1, α∗(ω2) = ζ 4ω2, α∗(ω3) = ζ 8ω3.
| NCTS Hsinchu July 30 – August 3, 2014 17 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Proposition
By taking the symplectic basis of H1(Jac(C )C,C) corresponding to theabove basis (with respect to the skew-symmetric matrix J), we get
α =
ζ 2 0 0 0 0 0
0 ζ2
0 0 0 00 0 ζ 4 0 0 0
0 0 0 ζ4
0 00 0 0 0 ζ 8 0
0 0 0 0 0 ζ8
.
One checks that the matrix γ satisfies
γαγ−1 = σ2α ,
which implies γ ∈ TL(σ2). Hence, γ belongs to ST(C ); finally, one checksthat γ6 =− Id ∈ ST(C )0, but γ i is not in ST(C )0 for 1≤ i < 6.
| NCTS Hsinchu July 30 – August 3, 2014 18 / 43
Sato-Tate conjecture for a CM Picard curve
The Sato-Tate distribution
Theorem
The Picard curve C : y3 = x4−x satisfies the generalized Sato-Tateconjecture. More explicitly, the sequence
xp =
(σ2ψ(P)√N(P)
,σ4ψ(P)√N(P)
,σ8ψ(P)√N(P)
, p
)p 6=3
is equidistributed over ST(C )' U(1)3 o (Z/9Z)∗ with respect to theHaar measure.
| NCTS Hsinchu July 30 – August 3, 2014 19 / 43
Sato-Tate conjecture for a CM Picard curve
Proof of the Theorem (sketch)
To show that the distribution of LSTp (C ,T ) matches the distribution of
charcteristic polynomials of random matrices in ST(C ) given by its Haarmeasure, one proceeds as follows:
• Classify all irreducible characters φ of U(1)3 o (Z/9Z)∗.
• Rewrite the L-function
L(φ ,xp,s) = ∏p 6=3
det(1−φ(xp)p−s)−1 = L(Ψ,s)
as the L-function of a certain normalized Grossencharakter Ψ.
• Using that the CM-type of C is non-degenerate, one shows that Ψ istrivial if and only if φ is trivial. Then, invoque a classical result ofHecke.
| NCTS Hsinchu July 30 – August 3, 2014 20 / 43
Sato-Tate conjecture for a CM Picard curve
Computational approach: the moment sequence
Let us denote a1(p), a2(p), a3(p) the higher traces according to
LSTp (C ,T ) = 1 +a1(p)T +a2(p)T 2 +a3(p)T 3 +a2(p)T 4 +a1(p)T 5 +T 6 .
Due to the Weil conjectures, we know that
a1(p) ∈ [−6,6] , a2(p) ∈ [−15,15] , a3(p) ∈ [−20,20] .
We are interested in the limiting distribution of a1(p), a2(p), a3(p) over allprimes p ≤ N of good reduction, as N → ∞.
| NCTS Hsinchu July 30 – August 3, 2014 21 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
For the characteristic polynomial of a random matrix of ST(C )⊆ USp(6)we will similarly denote
P(T ) = 1 +a1T +a2T2 +a3T
3 +a2T4 +a1T
5 +T 6 ,
and
ai ∈ Ii := [−(
6
i
),+
(6
i
)] .
Let µi be the projection on the interval Ii of the Haar mesure of theSato-Tate group ST(C)' U(1)3 o (Z/9Z)∗.
The measure µi is uniquely determined by its moment sequence:
Mn[µi ] =∫Iiznµi (z) .
We obtain explicitly the moments of the distributions µi .
| NCTS Hsinchu July 30 – August 3, 2014 22 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
We split each measure µi as a sum of its restrictions kµi to eachcomponent
ST(C )0 · γk
where 0≤ k ≤ 5.
That is
µi =1
6 ∑0≤k≤5
kµi , Mn[µi ] =
1
6 ∑0≤k≤5
Mn[kµi ] .
Recall ST(C )0 ' U(1)3, and γ =
0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 10 −1 0 0 0 01 0 0 0 0 0
.
| NCTS Hsinchu July 30 – August 3, 2014 23 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
The shape of the characteristic polynomials in each component of theSato-Tate group is as follows:
ST(C )0 · Id : ∏3i=1(T −ui )(T −ui )
ST(C )0 · γ : T 6 + 1
ST(C )0 · γ2 : T 6 + (u1u2u3 +u1u2u3)T 3 + 1
ST(C )0 · γ3 :(T 2 + 1
)3
ST(C )0 · γ4 : T 6− (u1u2u3 +u1u2u3)T 3 + 1
ST(C )0 · γ5 : T 6 + 1 .
| NCTS Hsinchu July 30 – August 3, 2014 24 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
The first moments of the higher traces are as follows.
(i) First trace: M[µ1] = (1,0,1,0,15,0,310, . . .)
M[kµ1] =
(1,0,0, . . .) if k = 1, . . . ,5;
(1,0,6,0,90,0,1860, . . .) if k = 0 .
(ii) Second trace: M[µ2] = (1,1,5,35,321, . . .)
(iii) Third trace: M[µ3] = (1,0,6,0,822,0,184860,0 . . .)
| NCTS Hsinchu July 30 – August 3, 2014 25 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
The first moments of the higher traces are as follows.
(i) First trace: M[µ1] = (1,0,1,0,15,0,310, . . .)
(ii) Second trace: M[µ2] = (1,1,5,35,321, . . .)
M[kµ2] =
(1,0,0, . . .) if k = 1,2,4,5;
(1,3,9,27, . . .) if k = 3;
(1,3,21,183,1845, . . .) if k = 0 .
(iii) Third trace: M[µ3] = (1,0,6,0,822,0,184860,0 . . .)
| NCTS Hsinchu July 30 – August 3, 2014 25 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
The first moments of the higher traces are as follows.
(i) First trace: M[µ1] = (1,0,1,0,15,0,310, . . .)
(ii) Second trace: M[µ2] = (1,1,5,35,321, . . .)
(iii) Third trace: M[µ3] = (1,0,6,0,822,0,184860,0 . . .)
M[kµ3] =
(1,0,0, . . .) if k = 1,3,5;
(1,0,2,0,6,0,20,0, . . .) if k = 2,4;
(1,0,32,0,4920,0,1109120, . . .) if k = 0 .
| NCTS Hsinchu July 30 – August 3, 2014 25 / 43
Sato-Tate conjecture for a CM Picard curve
The distribution of ST(C )
The first moments of the higher traces are as follows.
(i) First trace: M[µ1] = (1,0,1,0,15,0,310, . . .)
(ii) Second trace: M[µ2] = (1,1,5,35,321, . . .)
(iii) Third trace: M[µ3] = (1,0,6,0,822,0,184860,0 . . .)
| NCTS Hsinchu July 30 – August 3, 2014 25 / 43
Sato-Tate conjecture for a CM Picard curve
The numerical moment sequence
Since the sequences ai (p)p are µi -equidistributed, the following equalitymust hold:
Mn[µi ] = limx→∞
1
π(x) ∑p≤x
ai (p)n .
To check this limit we need to compute a big amount of traces ai (p). Asmentioned before, we use the Grossencharakter atttached to C :
Lp(C ,T ) = T 6 Irr(ψ(P),1/T f ;Q)6/(fd) .
Lp(C ,T/√p) = 1+a1(p)T +a2(p)T 2 +a3(p)T 3 +a2(p)T 4 +a1(p)T 5 +T 6 .
| NCTS Hsinchu July 30 – August 3, 2014 26 / 43
Sato-Tate conjecture for a CM Picard curve
The numerical moment sequence
LSTp (C ,T ) = 1 +a1(p)T +a2(p)T 2 +a3(p)T 3 +a2(p)T 4 +a1(p)T 5 +T 6
a1 a2 a3
n Mn[µ1] Mn[µ1]≤226 Mn[µ2] Mn[µ2]≤226 Mn[µ3] Mn[µ3]≤226
0 1 1 1 1 1 11 0 −0.000 1 0.999 0 −0.0002 1 0.998 5 4.991 6 5.9843 0 −0.005 35 34.868 0 −0.1474 15 14.946 321 319.058 822 815.9375 0 −0.1516 310 308.160
| NCTS Hsinchu July 30 – August 3, 2014 27 / 43
−6 −4 −2 0 2 4 60
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
a1 histogram of y3 = x4 − x for p ≤ 212 for p ≡ 1
93 data points in 9 buckets.
Moments: 1.000 −0.102 5.013 −1.696 59.422 −57.527 987.958 −1793.535 20053.205
−6 −4 −2 0 2 4 60
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
a1 histogram of y3 = x4 − x for p ≤ 217 for p ≡ 1
2034 data points in 45 buckets.
Moments: 1.000 0.006 5.846 −0.084 85.253 −7.055 1710.579 −260.730 39879.280
−6 −4 −2 0 2 4 60
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
a1 histogram of y3 = x4 − x for p ≤ 222 for p ≡ 1
49236 data points in 221 buckets.
Moments: 1.000 0.003 5.974 −0.022 89.177 −1.556 1830.285 −46.824 43629.302
−6 −4 −2 0 2 4 60
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
a1 histogram of y3 = x4 − x for p ≤ 226 for p ≡ 1
659423 data points in 812 buckets.
Moments: 1.000 −0.002 5.990 −0.035 89.677 −0.909 1848.961 −25.241 44348.111
0 5 10 150
0.05
0.1
0.15
0.2
0.25
0.3
a2 histogram of y3 = x4 − x for p ≤ 212 for p ≡ 1
93 data points in 9 buckets.
Moments: 1.000 2.598 15.017 108.332 930.463 8965.148 94120.531 1053338.268 12350445.305
0 5 10 150
0.05
0.1
0.15
0.2
0.25
0.3
a2 histogram of y3 = x4 − x for p ≤ 217 for p ≡ 1
2034 data points in 45 buckets.
Moments: 1.000 2.921 20.099 171.042 1683.802 17991.765 202574.258 2363856.246 28311843.193
0 5 10 150
0.05
0.1
0.15
0.2
0.25
0.3
a2 histogram of y3 = x4 − x for p ≤ 222 for p ≡ 1
49236 data points in 221 buckets.
Moments: 1.000 2.992 20.861 180.835 1811.755 19714.913 226422.037 2700518.216 33123391.030
0 5 10 150
0.05
0.1
0.15
0.2
0.25
0.3
a2 histogram of y3 = x4 − x for p ≤ 226 for p ≡ 1
659423 data points in 812 buckets.
Moments: 1.000 2.996 20.947 182.211 1833.349 20052.002 231626.340 2779930.900 34322818.825
−20 −15 −10 −5 0 5 10 15 200
0.02
0.04
0.06
0.08
0.1
0.12
a3 histogram of y3 = x4 − x for p ≤ 212 for p ≡ 1
93 data points in 9 buckets.
Moments: 1.000 −0.107 22.042 −57.128 2318.032 −17077.275 420714.532 −4838070.854 99401274.285
−20 −15 −10 −5 0 5 10 15 200
0.02
0.04
0.06
0.08
0.1
0.12
a3 histogram of y3 = x4 − x for p ≤ 217 for p ≡ 1
2034 data points in 45 buckets.
Moments: 1.000 0.036 30.481 −6.383 4429.652 −2440.327 936612.201 −700317.729 229552619.137
−20 −15 −10 −5 0 5 10 15 200
0.02
0.04
0.06
0.08
0.1
0.12
a3 histogram of y3 = x4 − x for p ≤ 222 for p ≡ 1
49236 data points in 221 buckets.
Moments: 1.000 0.013 31.746 −1.283 4812.906 −357.711 1065495.433 −59937.656 276083513.105
−20 −15 −10 −5 0 5 10 15 200
0.02
0.04
0.06
0.08
0.1
0.12
a3 histogram of y3 = x4 − x for p ≤ 226 for p ≡ 1
659423 data points in 812 buckets.
Moments: 1.000 −0.002 31.905 −0.880 4883.623 −237.290 1095343.185 −68575.171 288098434.216
−2 −1.5 −1 −0.5 0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
a3 histogram of y3 = x4 − x for p ≤ 212 for p ≡ 4,7
184 data points in 13 buckets.
Moments: 1.000 −0.042 1.923 −0.185 5.602 −0.623 18.310 −2.179 63.220
−2 −1.5 −1 −0.5 0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
a3 histogram of y3 = x4 − x for p ≤ 217 for p ≡ 4,7
4070 data points in 63 buckets.
Moments: 1.000 −0.009 1.998 −0.033 5.967 −0.113 19.809 −0.404 69.125
−2 −1.5 −1 −0.5 0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
a3 histogram of y3 = x4 − x for p ≤ 222 for p ≡ 4,7
98631 data points in 314 buckets.
Moments: 1.000 −0.002 1.999 −0.007 5.994 −0.023 19.971 −0.082 69.870
−2 −1.5 −1 −0.5 0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
a3 histogram of y3 = x4 − x for p ≤ 226 for p ≡ 4,7
1319084 data points in 1148 buckets.
Moments: 1.000 −0.000 2.000 −0.001 6.000 −0.005 19.998 −0.018 69.991