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Introduction to Characters

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Introduction to Characters Introduction to Characters Travis Scholl April 23, 2015 Abstract *These notes were taken from a course by Ralph Greenburg in Spring 2015 [Gre15] on counting points on varieties over finite fields. It also overlaps with [Ser12, Ch. VI] and [Was97]. Any mistakes should be due to me. 1 Characters This section will define the basic objects required for studying characters. Much of this theory can be generalized but for simplicity we will stick to the \hands on" approach. Definition 1.1. Let A be a finite abelian group. The dual of A to be the group A “ HompA; C˚q, that is the group of homomorphisms A Ñ C˚. This is also called the Pontryagin dual. Elements χ P A are examples of characters. In this group the trivial character χ0 is the mapp sending A to 1. Exercisep 1.2. Prove that (a) Let χ P A. Show the image χpAq is contained in the set of roots of unity: n ` p χpAq Ď tz P C | z “ 1 for some n P Z u: (b) For any fixed A, there exists an isomorphism A – A. (c) There is a canonical isomorphism A – A, i.e. thep \double dual" functor is naturally isomorphic to the identity functor on the category of finite abelian groups. x Hint. Use the structure theorem for finite abelian groups. If A is a product of cyclic groups Z{nZ, then maps out of A are uniquely determined by where a generator in each component go. Notice that there is a unique cyclic subgroup of order n in C˚. Definition 1.3. Let FA denote the vector space of C-valued functions on A. Note that A Ă FA. We give an inner product to FA by defining p 1 xf; gy “ fpaqgpaq |A| aPA ¸ Remark 1.4. The inner product defined in Definition 1.3 can also be interpreted as an integral. We could instead write A f1f2dµA where µA is a normalized Haar measure on A. In this case this just means any singleton tau has measure 1 so that the measure of A is normalized to 1. ³ |A| 1 Exercise 1.5. Show dimC FA “ |A|. Hint. Note that FA is arbitrary set-theoretic functions. So they are uniquely defined only by their value on each point in A. Theorem 1.6 (Fourier Series). The elements χ P A form an orthogonal basis for FA. If f P FA then f “ cχχ. χPA p ř Remark 1.7. This shouldp remind of you Fourier series you might have seen in analysis. Just note 2πinx that the map R Ñ C˚ given by x ÞÑ e for some fixed n P Z is a character. Use this to compare with the standard Fourier series. Before proving Theorem 1.6, we first prove a very helpful lemma. Lemma 1.8. If χ P A then 1 1 if χ “ χ0 χpaq “ p |A| aPA #0 if χ ‰ χ0. ¸ d th Proof. Let d be the order of χ so that χ “ χ0. Then χpAq is exactly the d roots of unity (each with multiplicity |A|{d). The sum of the dth roots of unity is 0 unless d “ 1, in which case χ “ χ0 and we have aPA χpaq “ |A|. Proof of Theoremř 1.6. By Exercise 1.5 it is sufficient to show that the χ are orthogonal since then they will form an independent set of the same size as the dimension. Let χ1; χ2 P A. By Exercise 1.2 we know the image of any χ P A is contained in the roots of unity. Hence χpaq “ χpaq´1 for any a P A. Using this and the previous lemma we can write p 1 p xχ1; χ2y “ χ1paqχ2paq |A| aPA ¸ 1 ´1 “ χ1paq pχ2paqq |A| aPA ¸ 1 ´1 “ χ1χ paq |A| 2 aPA ¸ ` ˘ 1 if χ1 “ χ2 “ #0 if χ1 ‰ χ2. Theorem 1.6 gives an expansion of any element f P FA into a linear combination χPA cχχ. But the coefficients cχ in the theorem can be calculated via the inner product. Fix P A and consider ř p the inner product with as follows p xf; y “ cχχ, C G χ¸PA “ cχ p xχ, y χ¸PA “ c p which shows 1 c “ fpaq ´1paq: (1) |A| aPA ¸ 2 2 Gauss Sums In the previous section we considered characters as group homomorphisms into C˚. In this section we expand our objects from finite abelian groups A to finite fields Fq where q is some prime power. The extra structure allows us to talk about additive characters (homomorphisms on Fq with ˚ the additive structure) and multiplicative characters (homomorphisms on Fq ). There is a natural action of Fq on Fq. Given b P Fq let mb : Fq Ñ Fq be the multiplication by b map. This is a group homomorphism with respect to the additive structure. Then for P Fq define b ¨ “ b “ ˝ mb. It's clear this is againx a character. x Proposition 2.1. Let be a non-trival character in Fq. For any b1; b2 P Fq, if b1 ‰ b2 then b1 ‰ b2 . x Proof. b1 “ b2 ô pb1aq “ pb2aq @a P Fq ô ppb1 ´ b2qaq “ 1 @a P Fq Now pb1 ´ b2qa varies over all of Fq since b1 ‰ b2. Since is non-trivial it follows that b1 ‰ b2 . Corollary 2.2. For any non-trivial P Fq we have Fq “ t b | b P Fqu. Exercise 2.3. Prove Corollary 2.2. x x ˚ Next we want to consider multiplicative characters, i.e. Fq , and relate them to additive ones. ˚ Given χ P Fq we extend it to a map χ : Fq Ñ C as follows. If χ ‰ χ0 then x x χpaq if a ‰ 0 rχpaq “ #0 if a “ 0 and if χ “ χ0 then we extend it by r χ0paq “ 1 @a P Fq: Warning 2.4. Note that χ is extended differently! This will make things easier later. Also it's 0 Ă nice that it is still a constant function. Also be careful because χ may not necessarily lie in Fq because it is not additive.Ă However, χ is still multiplicative so χpabq “ χpaqχpbq for all a; b P Fq. r x Note that the extension χ is a C-valued function on Fq so that χ P F q . So we can apply r r r rF Theorem 1.6 which says χ can be written as a linear combination of the P Fq. So by Corollary 2.2 we have r r r χ “ cb b x bP ¸Fq r for some fixed non-trivial P Fq. ˚ Exercise 2.5. Let χ P and fix some non-trivial P q. By above we can write χ “ c . Fq x F bPFq b b (a) In the notation above, show ř x x r 0 if χ ‰ χ0 c0 “ #1 if χ “ χ0 3 Hint. See Warning 2.4 and compute xχ, 0y. The proof of Lemma 1.8 may be helpful. (b) Show xχ, χy “ |c |2. b r Hint. Notice x¨ř; ¨y is a Hermitian inner product on FFq . r r Definition 2.6. A Gauss sum is the sum of a multiplicative character times an additive one. ˚ Specifically, given χ P Fq and P Fq, then define the Gauss sum to be x x γpχ, q “ χpaq paq: aP ¸Fq There are many cool things one can do with Gaussr sums. Here is a small application where we compute the absolute value. ˚ ? Theorem 2.7. Let χ P Fq and P Fq. If and χ are non-trivial then |γpχ, q| “ q. Proof. From Theorem 1.6 and Corollary 2.2 we can write χ “ c for some coefficients x x bPFq b b ca P C. ř First we will show |cb1 | “ |cb2 | for any non-zero b1; b2 P Fq.p Recall that χ is multiplicative, so for any b ‰ 0 in Fq we can write r cb “ xχ, by “ χpaq bpaq aPr ¸Fq ´1 “ χrpabqχpb q 1pbaq aP ¸Fq ´1 “ χrpabqχrpb q 1pbaq aP ¸Fq ´1 “ χpb rq rχpabq 1pbaq aP ¸Fq ´1 “ χrpb q xχ, r1y ´1 “ χpb qc1 r r Since χ b´1 is a root of unity it follows c c . p q r| b| “ | 1| ?1 Now we can show |cb| “ q . Using Exercise 2.5 we have r 2 2 xχ, χy “ |cb| “ pq ´ 1q|c1| bP ¸Fq ˚ r r and as χ is a character for Fq we have 1 1 χ, χ χ a χ a “ x y “ ˚ p q p q |Fq | ˚ a¸PFq 1 “ χpaqχpaq q ´ 1 aPFq q ¸ “ xχ, χry r q ´ 1 r r 4 Note: the two inner products above are for different character groups! Together these inequalities show 1 |c1| “ ? q Now unwinding definitions we have γpχ, q “ cb bpaq paq ¨ ˛ aP bP ¸Fq ¸Fq ˝ ‚ “ cb bpaq paq bP aP ¸Fq ¸Fq “ cb|Fq| x b; y bP ¸Fq “ c1q which with above finishes the proof. Exercise 2.8. Find the absolute value of the Gauss sum γpχ, q in the following cases: ‚ χ “ χ0, ‰ 0 ‚ χ ‰ χ0, “ 0 ‚ χ “ χ0, “ 0 Hint. ‚ 0 ‚ 0 ‚ q 3 A Theorem of Gauss Another example of Gaussian sums comes from a theorem of Gauss. Definition 3.1. Let p be an odd prime. Define a map Z Ñ t´1; 0; 1u by 1 if p - a and a is a square mod p a a ÞÑ “ ´1 if p a and a is not a square mod p p $ - ˆ ˙ &'0 if p | a ˚ It's not hard to see this is a non-trivial%' character on Fp extended to Fp.
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