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Fourier Analysis Fourier Analysis Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Friday 18th October, 2013 1 Fourier basics Let G be a finite abelian group. A character of G is simply a homomorphism from G to the multiplicative group of the complex numbers, ∗ 1 C : (a + b) = (a) (b), and (−a) = (a) . Since G is finite, we have that every element in the image of 1 is a root of unity, and thus (a) = (a). Characters form a group under multiplication. We define the dual group of G, G^, to be the group of all characters of G. Let 0 be the trivial character, which maps all of G to 1; this is the identity element of G^. 1.1 Examples 1. Let G = Zp. For a 2 Zp, define a : G ! C by: 2πiax=p ax a(x) = e = !p ; 2πi=p ^ (where !p is the primitive pth root of unity e ). Then G = f a j a 2 Zpg. (Here p need not be prime). n 2. Let G = Z2n . For a 2 Z2 , define a : G ! C by: Pn ha;xi aixi a(x) = (−1) = (−1) j=1 : ^ Then G = f a j a 2 Zpg. (In the above examples, it is clear that the given a are indeed characters. That these are all the characters can be verified using a minimal set of generators for G. Soon we will see a proof of the fact that in general, the size of the dual group is at most jGj). Lk For a general finite abelian group G, by the classification of finite abelian groups, we can write G ≡ i=1 Zdi . ∗ This gives us, for each a = (a1; : : : ; ak) 2 Zdi , the character a : G ! C given by: k Y (x ; : : : ; x ) = !aixi : a 1 k di i=1 Observe that these characters are distinct and form a group. Again, these are all the characters of G. It so happens that in all these cases, G^ is isomorphic to G, but this is just a coincidence that occurs only in the finite case. 1.2 Characters form an orthonormal basis We use CG to denote the C vector space of all complex valued functions on G, equipped with the inner G product hf; gi = Ex2g[f(x)g(x)]. We now show that the characters of G form an orthonormal basis for C . While studying additive questions on G, it will be terribly useful to represent functions on G in this basis. 1 Lemma 1. Let be a character of G. Then: ( 1 = 0 Ex2G[ (x)] = : 0 6= 0 Proof. If = 0, then the lemma is obvious. Suppose 6= 0. Let S = Ex2G[ (x)]. For a fixed y 2 G, as x varies over all elements of the group, then so does xy. Then: S = Ex2G[ (xy)] = (y) · S: Take y 2 G to be such that (y) 6= 1. Thus S = 0. This immediately gives us the orthonormality of characters: Lemma 2. Let ; 0 be characters of G. Then: ( 0 0 0 1 = h ; i = Ex2G[ (x) (x)] = : 0 6= 0 Proof. Apply the previous lemma to the character · 0. Thus the number of characters of G is at most jGj. We already exhibited, for every finite abelian group G, a set of jGj distinct characters of G. Thus those are all the characters, and the set of all characters forms a basis for CG. Now that we have determined the dual group G^ of every finite abelian group, we change notation somewhat Lk ^ Lk ^ to make our future expressions cleaner. For G ≡ i=1 Zdi , we will let G = i=1 Zdi , and for a 2 G, we usel a to denote the character of G corresponding to a as given in the previous section. 1.3 The Fourier transform By the previous section, every function f : G ! C can be written as a linear combination of characters of G. Lemma 3. Every f : G ! C has the following expression in terms of the characters of G: X ^ f = f(a) · a; a2G^ where: ^ f(a) = hf; ai = Ex2G[f(x) a(x)]: The function f^ : G^ ! C is called the Fourier transform of f. There is also an inverse Fourier transform: given any function h : G^ ! C, there is a unique function f : G ! C such that f^ = h. 1.4 Parseval and Plancherel The orthonormality characters also implies that the Fourier transform is an isometry. This is given by the next two lemmas1. Lemma 4. Let f : G ! C. Then: 2 X ^ 2 Ex2G[jf(x)j ] = jf(a)j : a2G^ 1Somewhere I got the impression that the first of these lemmas is called the Parseval identity, and the second is called the Plancherel identity, but just now I wasn't able to confirm that from the internet. 2 Lemma 5. Let f; g : G ! C. Then: X ^ Ex2G[f(x)g(x)] = f(a)g^(a): a2G^ The second lemma implies the first. To prove the second lemma, we expand and simplify: 0 1 0 1 X ^ X Ex2G[f(x)g(x)] = Ex2G[@ f(a1) a1 (x)A @ g^(a2) a2 (x)A] a12G^ a22G^ X ^ = Ex2G[ f(a1) a1 (x)g^(a2) a2 (x)] a1;a22G^ X ^ = Ex2G[f(a1) a1 (x)g^(a2) a2 (x)] a1;a22G^ X ^ = f(a1)g^(a2)Ex2G[ a1 (x) a2 (x)] a1;a22G^ X ^ = f(a1)g^(a2)1a1=a2 a1;a22G^ X = f^(a)g^(a); a2G^ as desired. 1.5 Convolution So far we have not seen why the Fourier transform has anything to do with the group structure on G. Now we will. Given functions f; g : G ! C, define their convolution f ∗ g : G ! C as follows: f ∗ g(x) = Ey2Gf(y)g(x − y): Convolution shows up naturally in many additive-combinatorial problems. For example, if f = 1S, 1 f ∗ f(x) = · ( the number of (y ; y ) 2 S2 s.t. y + y = x): q 1 2 1 2 Convolution is an associative operation (check this!), and thus we may write expressions such as f ∗ g ∗ h. We denote f ∗ f ∗ · · · ∗ f (k times) by f (∗k). Convolution interacts nicely with the Fourier transform. Lemma 6. If f; g : G ! C, and h = f ∗ g. Then for every a 2 G^, h^(a) = f^(a) · g^(a): We also have a relation in the other direction. Lemma 7. If f; g : G ! C, and h = f · g. Then for every a 2 G^, X h^(a) = f^(b)^g(a − b): b2G^ 3 1.6 Big and small There is a lot of useful information in the magnitudes of Fourier coefficients. To get acclimatized, we should understand how big Fourier coefficents typically are, and how big can they possibly be. There are two special scenarios we will be especially interested in. p P ^ 2 jSj ^ jSj 1. S is a set, and f = 1S. By Parseval, a2G^ jf(a)j = jGj , and thus at least one jf(a)j is at least jGj . ^ ^ jSj Furthermore, for all a 2 G, we have jf(a)j ≤ jGj . P ^ 2 2. f : G ! C is such that jf(x)j = 1 for all x 2 G. By Parseval, a2G^ jf(a)j = 1, and thus at least one jf^(a)j ≥ p1 . Furthermore, for all a 2 G^, we have jf^(a)j ≤ 1. j Gj n ^ What does it mean for a Fourier coefficient to be large? Take G = Z2 , let a 2 G be nonzero, let S ⊆ G, and ^ jSj let f = 1S. For jf(a)j to be near its largest possible value jGj , we must have the following extreme scenario: Let H0 = fx 2 G j hx; ai = 0g and H1 = fx 2 G j hx; ai = 1g be the two hyperplanes defined by a that partition G. Then the manner in which S gets partitioned by H0 and H1 must be highly skewed. Conversely, jSj the Fourier coefficent is very small (o( jGj )) if S is partitioned almost equally into two parts by H0 and H1. Thus largeness of Fourier coefficients detects \nonuniform" distribution of the set S (with respect to the hyperplanes of the group G). ^ For G = Zp, the relationship is slightly less precise. For S ⊆ G to have a large Fourier coefficient a 2 G, it ax must be the case that most of the elements of f!p j x 2 Sg are \aligned" on the unit circle. Thus the scaled set aS should be strongly concentrated in an interval, or equivalently, S should be strongly concentrated in an arithmetic progression. Again, this is a form of nonuniform distribution. What do we expect for typical f? 1. Let f be a uniformly random function from G to the unit circle of C. Then for any fixed nonzero a 2 G^, we have by the Chernoff/Hoeffding/Bernstein bounds: t 2 Pr[jf^(a)j > ] < e−t =2jGj: jGj Taking union bound over all nonzero a 2 G^, we have that: t 2 Pr[ 9 nonzero a 2 G^ s.t. jf^(a)j > ] < jGj · e−t =2jGj: jGj p ^ q log jGj Taking t jGj · log jGj, we have that with high probability, jf(a)j ≤ O( jGj ) for all a 2 G. 2. Take S to be a random multiset of size k, selected by choosing x1; : : : ; xk uniformly at random from G. Then for any fixed nonzero a 2 G^, we have by the Chernoff/Hoeffding/Bernstein bounds: t 2 Pr[jf^(a)j > ] < e−t =2k: jGj Taking union bound over all nonzero a 2 G^, we have that: t 2 Pr[ 9 nonzero a 2 G^ s.t.
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