Contents

Introduction
Denote by q an odd prime power, and let X ( F x ) := Hom ( F x , C x ) be the group of multiplicative characters on the finite field F q with q elements. We denote by μ q ∈ X ( F x ) the quadratic character on F q , i.e. the unique character of order 2, and we denote by ε ∈ X ( F å ) the trivial character. We extend the domain of a multiplicative character from F å to F q by setting χ (0) = 0 if χ ∈ X ( F å ) is nontrivial and ε (0) = 1. In this paper we consider Legendre sums, which are character sums of the form
where b ∈ F q , χ ∈ X (F q ), and Q b = T 2 + 2bT + 1 ∈ F q [T]. In Proposition 1.2.2, we show that Sq (χ; b) is real. In Theorem 2.0.1 we prove a "Hasse-Davenport" relation for Legendre sums, and as a corollary we obtain the bound | Sq (χ; b)| ≤ 2√ q for b ∈ Fq \ {±1} and nontrivial χ ∈ X (F x ).  This allows us to define a unique angle θ ∈ [0,π] associated to the pair (χ, b) by the equation Sq (χ; b) = 2√q cos (θ) . We attempt to study the distribution of these angles as q → ∞ for fixed b and varying χ. In particular, we prove the following result. Theorem 0.0.1. Let {qm}m>1 be a sequence of odd prime powers which tends to infin- ity. For a given m and all χ ∈ X (F q m ) let θ χ be the angle associated to the Legendre sum Sqm (χ; 0).  Then the numbers {θχ}χεχ( ε qm ) become equidistributed with respect to the measure which is the average of the Haar measure on [0,π] and the Dirac delta measure at π/2 as m → ∞, i.e. for all continuous maps f : [0,π] → R,
We motivate this theorem with the following graph. The angles corresponding to the Legendre sums S 114 (χ; 0) for χ varying over all χ ∈ X (F 114 ) were computed, and a histogram of them was plotted. Furthermore, overlaid on the graph is the constant line θ → 1/ (2π). Conjecture 0.0.2. Let { qm} m ≥ 1 be a sequence of odd prime powers which tends to infinity.  For all m ≥ 1 take b ∈ Fqm \ {0,1, − 1}, and for all χ ∈ X ( F ¥ m the angle associated to the Legendre sum Sqm (χ; bm). Then the numbers {θx}χ∈χ( r; m ) become equidistributed with respect to the Sato-Tate measure on [0,π] as m → ∞, i.e. for all continuous maps f : [0, π] → R
We motivate this conjecture with the following set of graphs.  The angles correspond- ing to the Legendre sums S q (χ; b) for q = 11 3 , 11 4 , 11 5 , fixing b = 2 and varying χ over all χ ∈ X (F ), were computed, and histograms of them were plotted. Furthermore, overlaid on each graph is the curve of the function θ →  sin 2 (θ).
Legendre Character Sums
1
In this section we discuss some properties of multiplicative characters and derive some elementary properties of Legendre sums.
Characters and Character Lifting
1.1
The group X ( F å ) of multiplicative characters on F q is a cyclic group of order q − 1 and is hence isomorphic to F å . Recall the following orthogonality relations. For a ∈ F å , we have the identity
and for χ ∈ X (F q ), we have the identity
Also recall that for a ∈ Fq, if d | q − 1, then

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and in particular for the case d = 2,
For proofs of these identities see for instance Chapter 8.1 of [1]. Proposition 1.1.5. Consider χ ∈ X ( F q ). If χ is square, then χ (−1) = 1, and if χ is non-square, then χ (−1) = −1. Proof. Let g be the generator of F q . We note that the character χ 0 ∈ X (F q ) deter- mined by χ 0 ( g ) = e 2πi/( q −1) has order q − 1 and hence generates X ( F z ). Furthermore, χo (−1) = χo (g(g−1)/2) = χo (g)(g−1)/2 = eπi = −1. If χ is square, then χ = χ ξ for some even integer k . Then χ (−1) = χ 0 (−1) k = (−1) k = 1. If χ is non-square, then χ = χ 0 for some odd integer l. Then χ (−1) = Xo (−1) l = (−1) l = −1. ● Let n be a positive integer. Let N Fqn/Fq be the norm of the field extension Fqn/Fq, given by
for α ∈ F q n . We recall the following properties of the norm. Proposition 1.1.6. The norm N = q n/F q maps F q n onto F q .  Furthermore, if we take α,β ∈ Fqn and x ∈ Fq, then
Similarly, let Tr F gn / F g  be the trace of the field extension F gn / F q , given by
We recall the following properties of the trace. Proposition 1.1.7. The trace Tr gn/F g maps F gn onto F g . Furthermore, if we take α,β ∈ Fqn and x ∈ Fq, then
For proofs of the properties of the norm and trace see Chapter 11.2 of [1]. Let χ be a multiplicative character on F q . Proposition 1.1.6 allows us to lift χ to a character on F qn by composing with the norm: χ ◦ N f qn/F q ∈ X (F qn).
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Lemma 1.1.8. Take n,d ∈ Z ≥ 1 with d | q − 1. For all χ ∈ X (F å ) with order dividing d, the lifted character χ o Nfqn/fq ∈ X (Fqn) also has order dividing d.  Furthermore, the map
is a bijection. Proof. Consider χ ∈ X (F ξ ) with χ d = ε. Then for all α ∈ F ξ n ,
and so (χ 0 N Fqn/Fq) d = e. Suppose χ, ρ ∈ X (F α ) are distinct. Then there exists a ∈ F q such that χ(α)̸= ρ(α). Because the norm is surjective, there exists α ∈ F q n such that N F q n / F q (α) = α. Then X ( Nfgn/Fq (α))̸= ρ ( Nfgn/Fq (α)). Thus φ is injective. Note that #X ( F å ) = q − 1 and #X ( F å n ) = q n − 1. Because
we have that d | q − 1 | q n − 1. Hence, because X ( F å ) and X ( F å n ) are cyclic, the number of elements of order dividing d in both groups is d. Because φ is an injective map between finite sets of the same cardinality, we have that ϕ is also surjective. ●
1.2
Legendre Sums
Definition 1.2.1. For any b ∈ F q and χ ∈ X (F x ), the corresponding Legendre sum is the sum
where Q b = T 2 + 2bT + 1 ∈ F g [T]. Proposition 1.2.2. Take b ∈ Fq and χ ∈ X (F x ). Then,
Proof. (a) Note that for x E F x ,

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Using this fact, the complex conjugate of Sq (χ; b) is given by
(b)
(c) Let △ = b² – 1. Then, using (1.1.4), we have that
□
The Hasse-Davenport Relation
2
In this section we prove what we will call the Hasse-Davenport relation for Legendre sums. Theorem 2.0.1 (Hasse-Davenport Relation for Legendre Sums). Consider nontrivial χ ∈ X (F g ) and b ∈ F g \ {±1}. There exist α,β ∈ C such that αβ = q and such that for all n E Z>1, −Sqn (X 0 Nfqn/Rq;b) = αn + βn. Here α and β implicitly depend on χ and b , and we will throughout this paper fix b and write α χ , β χ for the α , β corresponding to a particular character χ . We also prove the following. Theorem 2.0.2 (Riemann Hypothesis for Legendre Sums). The α and β that arise in Theorem 2.0.1 have |α| = |β| = √ 9. An immediate consequence of these theorems and Proposition 1.2.2 is the following.
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Theorem 2.0.3. For b ∈ F q \ {±1} and χ ∈ X (F q ), we have that | S q (χ; b)| ≤ 2 √ q. We break up the proof of Theorem 2.0.1 into two parts.  First, we find α,β ∈ C such that αβ = q and α + β = − S q (χ; b). Then, we relate α n and β n to S q n (χ o N f q n / r q; b). After proving Theorem 2.0.1, we show that α and β have the desired magnitude.
2.1
Constructing α and β
Consider b ∈ F q \ {±1} and nontrivial χ ∈ X ( F ξ ). Let z 1 , z 2 ∈ F q be the roots of Q b . Writing Q b = ( T − z 1 ) ( T − z 2 ), we immediately obtain the relations z 1 + z 2 = − 2 b and z 1 Z 2 = 1. Furthermore, because b̸= ±1, the discriminant 4 ( b 2 − 1) of Q b is nonzero, and z 1 and z 2 are distinct. Denote by F m [ T ] the set of monic polynomials over F q . We define the map
and we note that λ is well-defined, since if we write f = [ deg f ajTi , then
Furthermore, it is easy to see that λ is multiplicative, i.e. that for two monic polynomials f and g, λ(fg) = λ(f) λ(g). Lemma 2.1.1. For k ∈ Z ≥0, let
Then σ (0) = 1, σ (1) = Sq (χ; b), σ (2) = q, and σ (k) = 0 for k ≥ 3. Proof. There is exactly one monic polynomial of degree 0, namely 1, and furthermore λ (1) = 1. Hence σ (0) = 1. For polynomials of degree 1, we have:
For polynomials f of degree 2, we consider the Euclidean division f = Q b + B of f by Q b , where deg B < 2. Here the coefficient before the Q b term is 1 because f and Q b are monic of the same degree. This gives
where △y = (b² − 1) y². Then, for y ∈ Fq,
Here we used (1.1.4) between the second and third lines. Because b̸= ±1, ∆ y = 0 if and only if y = 0. This gives us,
For polynomials f of degree k ≥ 3, we consider the Euclidean division f = AQ b + B of f by Q b , where A is monic of degree k − 2 and deg B < 2. As f varies over all monic polynomials of degree k , A varies over all monic polynomials of degree k – 2, and b
Now define the formal power series
By the previous lemma, we see that L ( z ) is actually a quadratic polynomial. Namely,
We thus have for some α,β ∈ C.  Furthermore,αβ = q and α + β = − Sq (χ; b). constructed the α and β that we wanted.
2.2
Taking Powers of α and β
Lemma 2.2.1. Let L (z) be as in (2.1.2). We have that
Proof. Because F q [T], of which F q [T] is a subset, is a unique factorization domain whose prime elements are the irreducible polynomials, we may write
→ where in the last step we used the fact that the summation is a geometric series.
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Lemma 2.2.2. For all n E Z ≥1,
Proof. By (2.1.3) and Lemma 2.2.1,
Taking the logarithmic derivative of both sides and multiplying by − z gives
Expanding the left side in geometric series gives
Expanding the right side gives
Substituting n = k deg f , we obtain
We thus have that
Equating like coefficients gives the desired result. ● Lemma 2.2.3. Consider n ∈ Z ≥ 1 and irreducible f ∈ F q [T] such that deg f | n. Then for all roots ζ ∈ Fq of f,
Proof. Because f is monic and irreducible, f is the minimum polynomial of ζ, and the roots of f are ζ, ζ q , ζ q ², . . . , ζ q deg f − 1 (see for instance Chapter 6.2 of [2]). We have that
For the χ term,
For the μ q term,
This gives the desired result. ● Applying Lemmas 2.2.2 and 2.2.3 now completes the proof of Theorem 2.0.1:

The Magnitude of α and β
2.3
In order to show that |α| = |β| = √ q , we construct a smooth projective algebraic curve whose zeta function has among its roots α−¹ and β−¹ , and then we apply the Riemann hypothesis for curves over finite fields. See Chapter 4 and in particular Theorem 4.2.3 of [3] for more information on the zeta function of a curve and the corresponding Riemann hypothesis. Consider the polynomial f = Y 2 − Q b (X d ) ∈ F q [X,Y] for d | q − 1. Then f is irreducible in F g [X,Y], since b̸= ±1 implies that Q b (X d ) is not a square in F g [X]. Let C 0 be the affine variety over F q defined by f ( x, y ) = 0. We have that
both vanish only at (x,y) = (0,0) and (x,y) = d-Vb,0), since d and q are coprime. Because (0,0), d-Vb, 0) & C 0 , C 0 has no singular points and is thus smooth. We now count the number of F gn -rational points of C 0 for all n ∈ Z >1. Applying (1.1.3), we have that
By Proposition 1.2.2, Sqn (ε; b) = −1, and hence
By Lemma 1.1.8,
For all nontrivial χ ∈ X ( F q ) we can take α χ , β χ ∈ C as in Theorem 2.0.1. Then, The projective closure C 0 of C 0 is the set of solutions [ x : y : z ] ∈ P 2 ( F q ) to the poly- nomial equation f ( x, y, z ) = 0 where f ( x, y, z ) = y 2 2 ( d −1 ) − x 2 d − 2 bx d − z 2 d − z 2 d . The points at infinity of C 0 are then points on C 0 ∩ { z = 0}, i.e. the single solution [0 : 1 : 0] to the equation x 2 d = 0. Unfortunately, this point is singular, so we cannot apply the Riemann hypothesis for curves over finite fields to C 0 . In order to rectify this problem, we first define the affine variety C to be the set of points (μ 1 ,μ 2 ,··· ,μ d ,μ d+1 ) ∈ A d+1 (F q ) satisfying
Then for all 1 ≤ i ≤ d, μ i = μ i , and we can see that C is isomorphic to C 0 , e.g.  by the polynomial map ( u 1 , . . , u +1 ) → ( u 1 , u d +1 ) and its inverse ( u, v ) → ( u, u ², u ³, . . . , u đ, v ).  Hence  #C ( F gn )  =  #C o ( F g )  for  all  n  ∈  Z ≥ 1 .  The  projective closure of C is the projective curve  C ⊆ Pd+1 defined by the 2d − 2 polynomial equa- tions
The points at infinity are the points on C ∩ { μ 0 = 0}. This gives two points at infinity, namely the points [0 : ･ ･ ･ : 0 : ±1 : 1], both of which are F g -rational.  To show that these points are nonsingular, we consider the affine part C ∩ { μ d +1 = 1} of C containing both points at infinity. Its Jacobian matrix is the (2d − 2) × (d + 1) matrix
where the rows are ordered F 1 , F 2 , . . . , F d – 1 , G 0 , G 1 , . . . , G d – 3 , H .  Evaluated at the
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points at infinity, this becomes the matrix 0 0 0 0 0 0 0 ±1 0
() () () () 0
0 0 0土2 which has rank d. Thus the dimension of the tangent spaces at the points at infinity is dim A d+1 − d = (d + 1) − d = 1, and the points at infinity are nonsingular. Hence smooth, and furthermore # C ( F q n ) = 2 + C 0 ( F q n ) for all n ∈ Z ≥ 1. We now calculate the zeta function for Č
and so
By the Riemann hypothesis for curves over finite fields, the inverses of the roots of Z ( Č / F q ; T ) have magnitude √ q . If we choose d = q − 1, then the product is over all nontrivial multiplicative characters on F q , and we have completed the proof of Theorem 2.0.2.
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Equidistribution Theorems
3
Take b ∈ F q \ {±1}. By Theorem 2.0.1, for all nontrivial χ ∈ X (F x ) |Ṣq (χ; b)| = |αχ + βχ| ≤ 2√q, for some α χ , β χ ∈ C with α χβ χ = q . Because S q (χ; b) is a real number we obtain a unique angle θ x ∈ [0, π] from the equation
Equivalently,

3.1
Character Sums
Let p be the characteristic of Fq , and let ψ : Fq → C be the additive character given by ψ ( x ) = exp ( ² π i Τ r f q / f ρ ( x ) ) . For multiplicative characters χ, ρ ∈ X ( F x ), denote by G (χ) and J (χ,ρ) the respective Gauss and Jacobi sums:
Recall the following properties of Gauss and Jacobi sums. Proposition 3.1.1. We have that
Let χ ∈ X (F å ) be a nontrivial multiplicative character. Then,
In particular for the case χ = μα,
Proposition 3.1.2. If χ,ρ ∈ X (F ξ ) are such that χ,ρ,αnd χρ are all nontrivial, then
See Chapters 8.2 and 8.3 of [1] for proofs of these properties. Consider collections ( χ i ) =1 and ( ρ i ) j =1 of multiplicative characters on F q , and con- sider t ∈ F α .  In [4], Katz defines the hypergeometric sum corresponding to this data to be the sum
Lemma 3.1.3. Suppose that { χ i : 1 ≤ i ≤ m} ∩ { ρ j : 1 ≤ j ≤ n } = ∅.  Then we have the bound
We do not prove this here. See [4] for a full discussion of hypergeometric sums and their corresponding hypergeometric sheaves. In particular, this bound is a consequence of Deligne's proof of the Riemann hypothesis for varieties in [5], along with Katz's construction. Note that in the case t = 1, m = 0, and χ i = ε for i = 1,2,...,n, the hypergeometric sums reduce to the well-known hyper-Kloosterman sums which satisfy the bound
(see for instance Theorem 6 of [6]).
3.2
The Case b = 0
Proposition 3.2.1. For all odd prime powers q and all χ ∈ X (F q ), we have that
Proof. Suppose χ is non-square. Then χ (−1) = −1 by Proposition 1.1.5, and
so S q (χ; 0) = 0. Suppose χ = η² for some η ∈ X (F x ). Then, using (1.1.4), we have that
where between the second and third lines we made the substitution y → − y. ● We require the following equidistribution result for Jacobi sums. Let Proposition 3.2.2. Write Y (F å ) = X (F å ) \ {є,μ q} for all prime powers q. {9m}m>1 be a sequence of odd prime powers which tends to infinity. For η ∈ Y (F ∗ ), let φη ∈ [0, 2π) be such that η (−1) J (η,μ q ) = √ q exp (iφη). Then the numbers {Φη} η } η ∈ γ (F i ) become equidistributed with respect to the Haar measure, i.e. for all continuous maps f : [0, 2π] → R, we have that
Proof. Denote by C ([0, π]) the Banach algebra of continuous real-valued functions on [0, π] equipped with the supremum norm. By the Stone-Weierstrass theorem (see for instance Theorem 8.15 of [7]), the trigonometric polynomials are dense in C ([0,π]). Because each trigonometric polynomial can be written as a linear combination of the maps θ → cos (nθ), n ≥ 0, it suffices to prove (3.2.3) for f equal to those maps. The case n = 0 is trivial, so consider n ≥ 1. Then a simple computation shows that
and that for η ∈ Υ ( F ξ ).
We note that ∑ η ε γ ( ξ ; ) ( η ( − 1 ) J ( η , μ q )) n is real, since
Hence, we need to show that
for all n ≥ 1, where
denotes the n -th normalized moment of the character sum given by η (−1) J (η, μ q ) for η ∈ Y (F å ) over the field F q . Fix some q = q m . For η ∈ Y (F å ) we can write J (η,μ q ) as a product of Gauss sums by applying Proposition 3.1.2 as follows:
Then,
where μ q (−1) G (μ q ) / √ q has magnitude 1. Because G ( ε ) = 0, we have that
and so
Expanding each Gauss sum gives
Exchanging sums, we see that this equals
By (1.1.1), the rightmost term satisfies
Hence,
Substituting z → −z , this equals
Contained in this expression is the hypergeometric sum
Combining Lemma 3.1.3 with (3.2.4), we get that
and so we can see that
□ Theorem 3.2.5 (Restatement of Theorem 0.0.1). Let {qm}m>1 be a sequence of odd prime powers which tends to infinity. For a given m and all χ ∈ X (F* m ) let θ x be the angle associated to the Legendre sum Sqm (χ; 0). Then the numbers {θ x}χ EX( F σm become equidistributed with respect to the measure which is the average of the Haar measure on [0,π] and the Dirac delta measure at π/2 as m → ∞, i.e. for all continuous maps f : [0, π] → R,
Proof. Just as in the proof of Proposition 3.2.2, it suffices to prove (3.2.6) for f (θ) = cos ( nθ ), n ≥ 0. The case n = 0 is straightforward:
For n ≥ 1, we have that
and so we need to show that
where α χ and β χ are obtained from Theorem 2.0.1. Here we may omit the trivial character ε because its contribution to the summation is negligible in the limit as m → ∞. Fix some q = q m . We will consider first the non-square characters in X ( F å ) and then the square characters. For non-square χ ∈ X (F å ), we know that α x + β x = 0 and that αχβχ = q, which gives
Then,
This gives that
For square χ ∈ X (F x ) \ {ε}, we can write χ = η 2 . Because η̸= ε and μ qη̸= ε, we have by Proposition 3.1.2 that
Furthermore, we have that

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Thus Proposition 3.2.1 implies that
Note that η and μ gη are exactly the two "square roots" of χ . We now obtain
Applying Proposition 3.2.2 for f (φ) = cos ( nφ) , we obtain that
Summing (3.2.8) and (3.2.9) gives (3.2.7), which completes the proof. ●
The Case b ≠ 0
3.3
In this section we attempt to adapt the work of Adolphson in [8] to prove an equidis- tribution result for Legendre sums for the case b̸= 0, ±1. If χ ∈ X ( F q ), then χ q = χ . Hence for all n ∈ Z ≥ 1 and all χ ∈ X ( F q n ) we may define a “degree” of χ by
Then, let
Note that for χ ∈ X | Fq , the α x and β x that we obtain from Theorem 2.0.1 (for any b̸= ±1) satisfy |α x | = |β x | = √ q d × . Define for n ∈ Z ≥ 1 the infinite product L q,n (T) by
Conjecture 3.3.2. (a) The infinite product L q,n (T) is actually a polynomial in Z [T]. (b)  The degree of Lq,n (T) is less than or equal to c n , where c n is a constant independent of q. (c) If we write
then for all i ∈ {1,2,...,cn}, | γi| ≤ √qn+1 Theorem 3.3.4. Conjecture 3.3.2 implies Conjecture 0.0.2. Proof.  Just as in the proof of Proposition 3.2.2, it suffices to prove
for f (θ) = cos (nθ), n ∈ Z ≥ 0 . The case n = 0 is straightforward:
For n ≥ 1, a simple computation shows that
For all χ ∈ X (F q m ) \ {ε} we obtain from Theorem 2.0.1 numbers α χ ,β χ ∈ C, and by (3.0.2),
Hence, we need to show that
Fix some q = qm.  We will now apply Conjecture 3.3.2. Taking a logarithmic derivative of both sides of (3.3.1) gives
Expanding in geometric series, this equals
Note that jd x − 1 = 0 if and only if j = 1 and χ ∈ X ( F x ).  Taking a logarithmic derivative of both sides of (3.3.3) and expanding in geometric series gives
Comparing the constant terms on the right-hand sides of the previous two expressions gives for all n ≥ 1 that
and so by Conjecture 3.3.2(c),
for all n ≥ 1. This immediately implies (3.3.5) for n = 1. For n ≥ 2, note that for χ ∈ X (F x ) \ {ϵ} we have because αχβχ = q that
For n = 2, summing the right-hand side of (3.3.7) over χ ∈ X ( F q ) \ { c } and using (3.3.6) gives
which implies (3.3.5) for n = 2. Lastly, for n ≥ 3, we use (3.3.6) and (3.3.7) to obtain
which gives (3.3.5) for n ≥ 3. ●
A
Simulation Code
1 #!/usr/bin/sage c import matplotlib . pyplot as plt s import time 9 start_time = time.time() 10 11 q = 11**4 12 b = 2 #0 13 g = GF(q).multiplicative_generator() 14 15 angles = [] 16 # converts Legendre sum to its corresponding angle, after 17 # correcting for rounding errors 18 def to_angle(s): s = OC(s).real() 19 20 25 26 # generate array of muq(Q_b(x)) for all x 27 Mu = [1] * q 28 def mu(x): 29 34 for k in range(0,q-1): 35 36 37 # chi(g) where chi is the generator of X(Fq^\ast) and g is the 38 # generator of F-q^\ast 39 gen_chi_g = CC(exp(2*pi*I / (q-1))) 40 41 # array of chi(g) for all g in Fq^\ast 42 NO = np.array([0,1] + [gen_chi_g**j for j in range(1,q-1)]) 43 44 angles .append(to_angle(np.dot(NO,Mu))) 45 46 # add angles corresponding to chi(g^i) for all g in Fq^\ast and all i 47 # this corresponds to exponentiating every element of NO 48 for i in range(2,q): Ni = [x**i for x in NO] 49 angles .append(to_angle(np.dot(Ni,Mu))) 80 51 s2 print('Calculation took ' + str(time.time() - start_time) + 's.') 53 54 t = np.arange(0.0, np.pi, 0.001) s = (2 / np.pi) * np.sin(t)**2 #1 / (2*np.pi) + t * 0 56 plt . plot (t ,s ,lw=2) 57 plt.hist(angles, bins=200, normed=True) 58 plt .xlim (0,np.pi) 59 plt .ylim (0,1.25)