Dirichlet L-functions & Characters Flashcards
Define the zeta function.
Show this representation of ζ converges for Re(s)>1.
State and prove eulers product formula.
Prove that if we can show the limit as Re(s) tends to 1 of the ζ function is infinite then there are infinitely many primes.
ζ(s) = Σn^(-s), sum over all n>=1
s € C, Re(s)>1.
Take the modulus of the sum up to N and then evaluate using an integral.
Eulers product formula:
ζ(s)=Π(1-p^-s)^-1 summing over all primes p.
to price, just sum up to N and take the product up to primes less than N. Then this product hits everything in the sum exactly once, with error term you can bound and show tends to zero.
Now, taking the log of the zeta function using eulers product, we see:
ΣpΣn (1/(n•p^ns)) =log(ζ(s))
Now the sum can be split into the sum over all primes and n=1, and the sum over all primes and n>1.
Computing the second sum, we see this is finite as Re(s) tends to 1. So that if the limit of ζ is infinite, then the first sum if infinite. So there must be an infinite number of primes.
Define the Set of Dirichlet characters modulo q.
Show they satisfy an orthogonality condition.
Show that: Σχ(b)= 0 for some non trivial character, where the sum is over all b mod q.
Define χ over all Z
Bound the sum of χ(b) over all b in Z up to x, where χ is a non trivial character.
Let G=(Z/qZ)*
Let G’= set of homomorphisms of G into C*, ie:
χ : (Ζ/qZ)* –> {z€C : |z|=1}
With χ(1)=1
χ(ab)=χ(a)•χ(b).
CLAIM: Σχ(a)•x(b)* =
φ(q) if aΞb mod q
0 otherwise
PROOF: obvious in first case, so if a isn’t congruent to b mod q then there exists so character ψ such that ψ(a•b^-1)~=0. Then multiply the above sum by this and show that this is in fact the same as the original sum, and so the sum must be zero.
As χ is non trivial, pick some a such that χ(a) is NOT 1. Then:
χ(a)•Σχ(b)= Σχ(a)•χ(b)
= Σχ(ab)
= Σχ(b)
So Σχ(b)= 0
Define χ on Z as follows:
•if (m, q) =1 then χ(m)=χ(b) where bΞm mod q and b<q.
• if (m, q) =0 then χ(m)=0
The sum can be easily bounded by q (and in fact φ(q) by using the last two parts.
Define the Dirichlet L function:
L(s,χ)
Find a similar product expression for it as we found for the ζ function.
L(s,ζ)= Σχ(n)/n^(s) Re(s)>1
Then: Σχ(n)/n^(s)
= Π(1+χ(p)/p^s+χ(p^2)/p^2s+..)
= Π(1-χ(p)/p^s)^-1
Where we take the product over all primes.
Using the product formula for the Dirichlet L-function, how can we show that there are infinitely many primes such that pΞa mod q for any a, q st (a,q)=1?
Firstly, by taking the log of the L-function we see:
log(L(s,χ))= ΣpΣn (χ(p^n))/n•p^ns
Now if we fix a with (a,q)=1 then sum over all χ in G’:
1/φ(q) • Σχ(a)*•log(L(s,χ)) =ΣpΣn(1/n•p^ns)[Σχ (χ(a)*•χ(p^n)/φq]
And the expression in [ • ] is 0 unless pΞa mod q, and is 1 when this happens, so the sum becomes:
ΣnΣp (1/n•p^ns)
Where we only sum over primes which are equivalent to a mod q, ie primes which appear in the arithmetic series a, a+q, a+2q,…
But again, for the sum over all n>=2 we can bound uniformly for Re(s)>=1. So if we can show that this sum is infinite we have shown the infinitude of primes in this arithmetic progression.
Essentially, this boils down to showing that for non trivial characters, L(1,χ) is not zero.
From basics, find an explicit expression for any character mod q.
[hint: use the CRT to reduce to the case where q is a prime power. Then treat the cases p odd and p=2 separately. Also use the fact that (Z/2^eZ)* is given by the elements:
(-1)^ε • 5^ν For ε=0 or 1,
0<2^e-2]
Reducing to prime powers is easy. So look at q=p^e
CASE 1 - p is odd:
Then (Z/p^eZ)* is cyclic, generated by α=αp say.
Then as χ(α)^φ(p^e)
=Χ(α^φ(p^e))
=Χ(1) =1
we see that:
χ((αp)^v)= exp(2πi•mv/φq) where m is some integer 0=/φ(pj^ej)]
If q’|q and χ’ is a character mod q’ define the induced character mod q.
Define the primitive character inducing χ.
What is an imprimitive character?
How can we classify primitive characters easily?
Set:
χ(a) = χ’(a) if (a,q)=1
= 0 otherwise
Given an character χ mod q, there is some minimal q’’==q then χ is called primitive. Else χ is called imprimitive
A character is primitive IFF it is primitive in each prime factorisation, which happens obviously when our indexing element vi is such that (vi,pi)=1
Describe the square free kernel of an integer
This is the product of all the distinct prime which divide n.
Prove that, for a non-trivial character χ mod q, L(s,χ) has a holomorphic continuation to Re(s)>0.
Find a bound for |L(s,χ)|
Use the stieljes integral with continuous function f(x)=1/x^s and then a function of bounded variation g(x) = Σχ(n) for n=<q•|s|/σ
Prove that ζ(s) has a meromorphic continuation to Re(s)>0. Use this to prove that for the trivial character ψ mod q, L(s,ψ) has a meromorphic continuation into this strip and satisfies a bound, once we subtract the right pole at s=1.
By using the stieljes integral again with a continuous function 1/x^s and the function g(x) =[x], ie the round down function, we immediately get a meromorphic continuation with the only pole at s=1.
Then observe the equation:
L(s,ψ)=ζ(s)•Π(1-p^-s)
Where the product is finite, over all p|q.
Putting this with our analytic continuation we know immediately what to subtract from the L-function and get a nice bound, not spoilt here.
How do you show that, for a complex character, L(1,χ) is not zero?
Well by our bound for L when the real part of s is positive, we see L has not got a pole at 1, and L is holomorphic in this half plane.
If χ is complex then it’s conjugate, called χ’ is a distinct character. Now, if we consider the product of all characters mod q. By taking the log of this product, we see that as the real part of s tends to 1, this value is larger than or equal to 1. But then as the L function for the principal character has a simple pole at 1, and all others are holomorphic, if some complex character L is zero at s=1 then so is its conjugate, and so this product would have to be zero here, which is a contradiction.
What is a quadratic character?
A quadratic character mod q is χ such that χ^2 is the principal character mod q. Thus χ^2(a)=1 for all a coprime to q, and so χ(a)=+/-1 for all such a. Hence χ is a real character.
For an odd prime power p^a show that there are exactly two real characters mod p^a.
Show that if a>=3, there are exactly four real characters mod 2^a.
Well, if χ is a character mod p^a then it is a homomorphism from (Z/p^a•Z)* which is a cyclic group of order φ(p)=(p-1)p^(a-1). Then whatever the character is at a generator for this group, this decides the whole character. But as χ is real, it’s value on a generator is +/-1, which gives exactly two distinct real characters.
Characters mod 2^a are given by how they act on: (-1)^ε•5^v, which generates all elements in the group. Now, we index the characters by ε’, v’, and by inspection, the characters are real IFF 2v•v’/2^(a-2) is an integer. But this must be true for v=1 and so this gives two choices of v’, by looking at the range of v’. Combining this with the two choices of ε gives all real characters.
Characterise the primitive characters mod an odd prime power p^a.
How about characters mod 2^a?
The characters mod p^a are indexed by 0=<p^a. Then a character χm is primitive IFF p does not divided m.
Proof:
Show that if p|m then χm is induced by some character mod p^(a-1).
Show that if p does not divide m then p^(a-1) is not an induced modulus, and so conclude that χm is primitive.
Characterise all real primitive characters mod prime powers (consider even too). Can you do it in terms of the Legendre symbol?
Aaa
What is a quadratic character?
A quadratic character mod q is χ such that χ^2 is the principal character mod q. Thus χ^2(a)=1 for all a coprime to q, and so χ(a)=+/-1 for all such a. Hence χ is a real character.
