dedekind sums with equal values · 2020. 4. 24. · values of the function. but doesn’t tell us...
TRANSCRIPT
Dedekind Sums with Equal Values
Yiwang Chen, Nicholas Dunn,Campbell Hewett and Shashwat Silas
Brown University
August 6, 2014
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 1 / 29
Overview
The Dedekind sum s(a, b) for (a, b) = 1 is defined by,
s(a, b) =b−1∑i=1
((i
b
))((ai
b
))where (()) denotes the sawtooth function,
((x)) =
{x − bxc − 1
2 if x 6∈ Z0 if x ∈ Z
We would like to be able to characterize a1, a2 such that s(a1, b) = s(a2, b)
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 2 / 29
Overview
1 Conjectures for when b is a prime power in s(a, b)
2 Generating Functions I
3 Generating Functions II
4 On the Number of Inversions
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 3 / 29
Preliminary Results and Facts
1 s(a, b) = s(a + kb, b) for k ∈ Z.
s(a + kb, b) =b−1∑i=1
((i
b
))(((a + kb)i
b
))=
b−1∑i=1
((i
b
))((ai
b+
kbi
b
))
=b−1∑i=1
((i
b
))((ai
b
))= s(a, b)
2 s(a1, b) = s(a2, b) if a1a2 ≡ 1 (modb).
s(a1, b) =b−1∑i=1
((i
b
))((a1i
b
))=
b−1∑i=1
((a2i
b
))((a2a1i
b
))= s(a2, b)
3
s(a, b) + s(b, a) = −1
4+
1
12
(a
b+
1
ab+
b
a
)Remark: This allows us to compute Dedekind Sums fast.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 4 / 29
Visualization Strategies
Here’s a simple graph of what s(a, 37) looks like,
5 10 15 20 25 30 35
-3
-2
-1
1
2
3sHa,37L
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 5 / 29
Visualization Strategies
Here’s a graph of many values of b (divided by b on both axes),
0.2 0.4 0.6 0.8 1.0
-0.05
0.00
0.05
This gives a few insights especially relating to the highest/n–th highestvalues of the function. But doesn’t tell us much about the equality.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 6 / 29
Visualization Strategies (b is prime)
But then we came up with another visualization which was somewhatmore fruitful for a while,
1 2 3 4 5 6 7
1
2
3
4
5
6
7Pairs for b=7
2 4 6 8 10 12
2
4
6
8
10
12
Pairs for b=13
5 10 15
5
10
15
Pairs for b=19
Each of the axes range up to b.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 7 / 29
A Theorem by Jabuka, Robins and Wang
Let b be a positive integer and a1, a2 any two integers that are relativelyprime to b. If s(a1, b) = s(a2, b), then b | (a1a2 − 1)(a1 − a2).
This is gives us the corollary that when b is prime,s(a1, b) = s(a2.b) ⇐⇒ a1 ≡ a2 (mod b) or if a1a2 ≡ 1 (mod b).
So all equal Dedekind Sums can be explained by the previous two one lineproofs when b is prime. Let’s take a look at our old picture again.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 8 / 29
Prime Case Explained
1 2 3 4 5 6 7
1
2
3
4
5
6
7Pairs for b=7
2 4 6 8 10 12
2
4
6
8
10
12
Pairs for b=13
5 10 15
5
10
15
Pairs for b=19
The red dots correspond to pairs which are inverses mod b and the blueones obviously correspond to ones that are the same mod b.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 9 / 29
What About the b = p2?
We looked at the same kinds of graphs again,
0 10 20 30 40
10
20
30
40
Pairs for b=72
0 50 100 1500
50
100
150
Pairs for b=132
0 50 100 150 200 250 300 3500
50
100
150
200
250
300
350
Pairs for b=192
The green dots correspond to pairs which are inverses mod b and the blueones correspond to ones that are the same mod b. And the red dots arethe ones which are aren’t in either of those cases.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 10 / 29
Red Dots
They look suspiciously well organized, so we were able to guess what linesthey lie on.
0 10 20 30 40
10
20
30
40
Pairs for b=72
0 50 100 1500
50
100
150
Pairs for b=132
0 50 100 150 200 250 300 3500
50
100
150
200
250
300
350
Pairs for b=192
The lines in each graph are kp ± 1 for 0 < k < p and k ∈ Z.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 11 / 29
Coming Up With Formulae
This led us to the conjecture:s(a1, p
2) = s(a2, p2) if and only if one of the following is true:
(i) a1 = a2 (mod p2),(ii) a1 = a−1
2 (mod p2),(iii) a1 = a2 = ±1 (mod p).
That is, the non–trivial pairs can be given by,
s(kp − 1, p2) =(p − 1)(p + 1)
6p2=
1
6− 1
6p2,
and
s(kp + 1, p2) = −(p − 1)(p + 1)
6p2= −1
6+
1
6p2.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 12 / 29
Coming Up With Formulae
From pn|(a1 − a2)(a1a2 − 1), we know
a1 = a2 (mod pk) and a1 = a−12 (mod pn−k)
for some k .
0 10 20 30 40
10
20
30
40
Pairs for b=72
0 50 100 150 200 250 3000
50
100
150
200
250
300
Pairs for b=73
0 500 1000 1500 20000
500
1000
1500
2000
Pairs for b=74
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 13 / 29
For b = pn, some nontrivial pairs are given by
s(kpn−m − 1, pn) = −p2n−2m − 3pn + 2
12pn=
1
4− pn−2m
12− 1
6pn,
and
s(kpn−m + 1, pn) =p2n−2m − 3pn + 2
12pn= −1
4+
pn−2m
12+
1
6pn
for 1 ≤ m ≤ n/2 and (k , p) = 1.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 14 / 29
For the b = pq case, there were no clear patterns.
0 5 10 15 20 25 30 35
5
10
15
20
25
30
35Pairs for p=5*7
0 20 40 60 80 100 120 1400
20
40
60
80
100
120
140
Pairs for p=11*13
0 20 40 60 80 100 1200
20
40
60
80
100
120
Pairs for p=7*19
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 15 / 29
Some roots of the polynomial on the complex plane for various b:
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
Roots of f11HxL
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
Roots of f14HxL
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
Roots of f21HxL
Some are on the unit circle, and some are roots of unity.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 16 / 29
A plot of sizes of roots (blue dots) for several values of b:
0 10 20 30 400.0
0.5
1.0
1.5
2.0Sizes of roots vs. b
They are bounded by
e− 8 logϕ(b)
b2−1 < x < e8 logϕ(b)
b2−1
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 17 / 29
Some factorizations of the polynomial:
b fb(x)2 13 1 + x
4 (1 + x)(1 − x + x2)
5 (1 + x)2(1 − x + x2)2
6 (1 + x2)(1 − x2 + x4 − x6 + x8)
7 (1 + x)(1 − x + x2)(1 − x3 + 3x6 − x9 + x12)
8 (1 + x)(1 − x + x2)(1 + x4)(1 − x3 + x6)(1 − x4 + x8)
9 1 + 2x10 + 2x18 + x28
10 (1 + x2)2(1 − x2 + x4)2(1 − x6 + x12)2
11 (1 + x)(1 − x + x2)(1 − x3 + x6 − x9 + x12 + x15 + x18 − x21 + x24 + x27 + x30 − x33 + x36 − x39 + x42)
12 (1 + x)(1 − x + x2)(1 + x4)(1 − x3 + x6)(1 − x9 + x18)(1 − x4 + x8 − x12 + x16 − x20 + x24)
13 (1 + x)2(1 − x + x2)2(1 − 2x3 + 3x6 − 4x9 + 5x12 − 6x15 + 7x18 − 6x21 + 5x24 − 4x27 + 5x30 − 4x33 + 5x36
−6x39 + 7x42 − 6x45 + 5x48 − 4x51 + 3x54 − 2x57 + x60)
14 (1 + x2)(1 − x2 + x4)(1 − x6 + x12 − x18 + x24 + x30 − x36 + x42 + x48 − x54 + x60 − x66 + x72)
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 18 / 29
Proposition. (x + 1)|fb(x) precisely when b is odd and not a square orwhen 4|b.Proof.If b is odd, then (−1)inv(a,b) =
(ab
). Hence,
fb(−1) =∑
(a,b)=1
(−1)inv(a,b) =∑
(a,b)=1
(ab
)=
{0 if b is not square
ϕ(b) if b is square
When b is even, it turns out
(−1)inv(a,b) =
{(−1)
a−12 if b = 0 (mod 4)
1 if b = 2 (mod 4)
from which it follows that
fb(−1) =
{0 if b = 0 (mod 4)
ϕ(b) if b = 2 (mod 4)
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 19 / 29
Other facts:
1 If b is odd, not a square, and not divisible by 3, then (x3 + 1)|fb(x).
2 If b is not a square and is 1 (mod 4), then (x + 1)2|fb(x). If b is alsonot divisible by 3, then (x3 + 1)2|fb(x).
3 If b = ck is odd, c is not a square, and (c, k) = 1, then fb(ζ2k) = 0.If b is also not divisible by 3, then fb(ζ6k) = 0.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 20 / 29
Conjecture. For b = ck , ζ2k and ζ6k are roots of fb(x) precisely under thefollowing conditions:(i) If c = 2 (mod 4), then ζ2k and ζ6k are roots of fb(x) precisely whenk = 4 (mod 8).(ii) If c = 0 (mod 4), then ζ2k and ζ6k are roots of fb(x) precisely whenk 6= 2 (mod 4) and 8 - k.(iii) If c = 3mn, m ≥ 1, and (n, 6) = 1, then only ζ2k is a root of fb(x)precisely when 3n - k and c is not a square.(iv) If (c , 6) = 1, then ζ2k and ζ6k are roots of fb(x) precisely when c - kand c is not a square.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 21 / 29
In trying to understand this conjecture, we discovered that the polynomialtakes values of Kloosterman sums:
Km(x , y) =∑
(a,m)=1
e2πim
(xa+ya−1)
If 4k|b, then
fb
(eπik
)=
1
2eπi2k K2b
(b
4k,b
4k
)If 2k|b but 4k - b, then
fb
(eπik
)=
1
4ieπi2k K4b
(b
2k,b
2k(1− b)
)
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 22 / 29
The conjecture seems to cover all cases when ζ2k or ζ6k is a root for k |b.There are rarer cases when k - b which are more mysterious.
b Unexplained roots
8 ζ1818 ζ1622 ζ20, ζ6026 ζ20, ζ6029 ζ1840 ζ18, ζ9045 ζ8, ζ4046 ζ3656 ζ1857 ζ5470 ζ18, ζ9074 ζ3680 ζ14, ζ4283 ζ18
114 ζ108117 ζ8
b Unexplained roots
117 ζ8136 ζ18, ζ306138 ζ108148 ζ18, ζ36173 ζ18186 ζ20198 ζ20200 ζ18204 ζ54296 ζ18317 ζ18325 ζ18, ζ90332 ζ72345 ζ54362 ζ36398 ζ18
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 23 / 29
If we have the following sequence
r−1 = b, r0 = a, r1, r2, r2, . . . , rn, rn+1 = 1
where rj+2 is rj reduced modulo rj+1 for all −1 ≤ j ≤ n − 1, then
inv(a, b) =a − 1
4ab2 +
(a − 1)(a − 2)
4a+
1
4
n∑j=1
(−1)j
rj rj−1
(rj − 1)(rj−1 − 1)(rj + rj−1 − 1)
b −(a − 1)2
4a
This means that every inv value lies on a parabola in b.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 24 / 29
20 40 60 80
500
1000
1500
2000
2500
3000
inv values vs. b
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 25 / 29
Proposition. If 2 ≤ a ≤ b − 2, then
b2 − 1
8≤ inv(a, b) ≤ 3b2 − 12b + 9
8.
Proof. The proof is by induction on b. There are three cases:
1 If b 6= ±1 (mod a), then use the induction hypothesis:
inv(a, b) ≥ −1
8(b + 2)a +
1
4(b2 + 3b + 2)− 1
8(2b2 + 5b + 2)
1
a.
2 If b = 1 (mod a), then by reciprocity,
inv(a, b) =1
4(b − 1)a +
1
4(b2 − 3b + 2)− 1
4(b2 − 2b + 1)
1
a.
3 If b = −1 (mod a), then again by reciprocity,
inv(a, b) = −1
4(b + 1)a +
1
4(b2 + 3b + 2)− 1
4(b2 + 2b + 1)
1
a.
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 26 / 29
More generally, it appears that inv(k , b) is the kth smallest value ofinv(a, b) for 1 ≤ k ≤ 6.
More specifically, the kth smallest value occurs with the remaindersequence
b, k , k − 1, 1
The longer the remainder sequence b, a, ..., 1, the closer inv(a, b) is to(b−1)(b−2)
4 .
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 27 / 29
5 10 15 20
1´109
2´109
3´109
4´109
5´109
Values of invHa,bL vs. Length of remainder sequence for b=100000
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 28 / 29
Questions?
Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat Silas (ICERM)Dedekind Sums with Equal Values August 6, 2014 29 / 29