Dedekind Sums with Equal Values · 2020. 4. 24. · values of the function. But doesn’t tell us much about the equality. Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat

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)


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


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.


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


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.


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.


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.


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.


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.


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.


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.


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


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.


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


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.


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


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)


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)


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.


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.


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)

)


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


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.


20 40 60 80

500

1000

1500

2000

2500

3000

inv values vs. b


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.


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 .


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


Questions?


Documents

Dedekind Sums with Equal Values · 2020. 4. 24. · values of the function. But doesn’t tell us much about the equality. Yiwang Chen, Nicholas Dunn, Campbell Hewett and Shashwat