advanced29 min studyLesson 586 in Mathematics

Dirichlet characters and L-functions

Encode multiplicative residue classes while distinguishing a character’s modulus from its conductor.

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What you’ll understand

  • Extend unit-group characters by zero with the correct multiplicativity.
  • Relate an induced L-function to its primitive conductor.
  • Explain why nonprincipal periodic cancellation extends convergence.

To distinguish prime counts in different residue classes, zeta alone is too coarse: all integers receive the same coefficient. A Dirichlet character attaches phases to invertible residue classes and zero to the others. Its L-function preserves an Euler product while recording this extra arithmetic information. The modulus specifies where the function is displayed, whereas the conductor identifies the smallest modulus carrying its essential character data.

Characters live on the unit group

A Dirichlet character modulo a positive integer qq starts as a group homomorphism from (Z/qZ)×(\mathbb Z/q\mathbb Z)^\times to the nonzero complex numbers. Extend it periodically to the integers and assign zero whenever the argument is not coprime to the modulus. The nonzero values have modulus one and are roots of unity. This extension is completely multiplicative, including products involving nonunits, because a product is a unit exactly when both factors are units.

The principal character is one on units and zero elsewhere; it is not the constant-one arithmetic function unless the modulus is one. For a nonprincipal character, summing over a complete residue period gives zero. Indeed, multiplication by a unit with character value different from one permutes the unit classes but multiplies their sum by that value. The partial sums are consequently bounded in absolute value by the modulus. Their cancellation will matter analytically even though the absolute values of the nonzero coefficients never decay.

L(s,χ)=∑n≥1χ(n)ns=∏p(1−χ(p)p−s)−1(Re⁡s>1).L(s,\chi)=\sum_{n\geq1}\frac{\chi(n)}{n^s}=\prod_p(1-\chi(p)p^{-s})^{-1}\qquad(\operatorname{Re}s>1).
Both the series and product converge absolutely in this half-plane, including for imprimitive characters.↔ Scroll wide equations horizontally.

The conductor removes redundant modulus

A character modulo qq is induced from a character modulo a divisor when their values agree on all integers coprime to qq. The unique minimal such modulus is its conductor, and the inducing character at that modulus is primitive. Extending the modulus can turn some formerly nonzero values into zeros, because additional primes become nonunits. The L-function therefore changes by removing precisely those Euler factors. Primes already dividing the conductor need no additional factor: their primitive character values were already zero.

The finite product in the displayed identity is first justified to the right of one, then continues wherever the primitive L-function does. Its factors contain the primitive character values; replacing all those values by one is generally wrong. Functional equations with conductor-dependent gamma factors are naturally stated for primitive characters. An imprimitive L-function inherits the primitive equation together with these extra finite factors, so blindly substituting its larger modulus into a primitive formula can create incorrect constants and zeros.

L(s,χ)=L(s,χ∗)∏p∣qp∤f(1−χ∗(p)p−s),f=cond⁡(χ).L(s,\chi)=L(s,\chi^*)\prod_{\substack{p\mid q\\p\nmid f}}(1-\chi^*(p)p^{-s}),\qquad f=\operatorname{cond}(\chi).
The primitive inducing character has modulus f; only new prime divisors of q remove Euler factors.↔ Scroll wide equations horizontally.

Cancellation, poles, and parity

Partial summation with bounded character partial sums shows that a nonprincipal character series converges when the real part is positive, locally uniformly away from that boundary. Absolute convergence still requires real part greater than one: on units its absolute coefficients equal one. The primitive functional equation continues a nonprincipal L-function to an entire function, and the finite-factor identity gives the same entireness for an imprimitive nonprincipal character. Conditional convergence of the original series and analytic continuation beyond it remain distinct statements.

For the principal character, L(s,χ0)=ζ(s)∏p∣q(1−p−s)L(s,\chi_0)=\zeta(s)\prod_{p\mid q}(1-p^{-s}), with a simple pole at one of residue φ(q)/q\varphi(q)/q. A primitive nonprincipal character has parity a∈{0,1}a\in\{0,1\} determined by χ(−1)=(−1)a\chi(-1)=(-1)^a; its completed expression uses (f/π)(s+a)/2Γ((s+a)/2)L(s,χ)(f/\pi)^{(s+a)/2}\Gamma((s+a)/2)L(s,\chi). Its reflection relates the character to its complex conjugate, rather than always to itself. In particular, a complex character need not have its own zeros symmetric across the real axis; conjugation can transfer them to another character’s L-function.

Induce the odd character from four to twelve

Relate the L-function induced modulo twelve from the nonprincipal character modulo four to its primitive L-function, and calculate its value at one.

  1. The primitive character χ4\chi_4 is zero on even integers, one on residues one modulo four, and minus one on residues three. Its conductor is four, and L(1,χ4)=1−1/3+1/5−⋯=π/4L(1,\chi_4)=1-1/3+1/5-\cdots=\pi/4.

  2. Inducing to modulus twelve additionally removes integers divisible by three. The new prime is three, with χ4(3)=−1\chi_4(3)=-1, so the missing Euler factor is restored by multiplication with 1+3−s1+3^{-s}, not 1−3−s1-3^{-s}.

  3. Thus L(s,χ12)=(1+3−s)L(s,χ4)L(s,\chi_{12})=(1+3^{-s})L(s,\chi_4), first in the absolutely convergent region and then by continuation. At one the multiplier is four thirds, giving L(1,χ12)=π/3L(1,\chi_{12})=\pi/3. This character is nonprincipal and has no pole there.

The result

The conductor remains four, while changing the displayed modulus to twelve multiplies the L-function by one plus three to the minus s.

Explore the model

Interactive lab

Interactive lab · modular

Separate residues from character values

Pause/Reset, set modulus 12 and step 1, and use Step to enumerate residues. Assign the induced character by hand: values at 1 and 5 are 1, at 7 and 11 are -1, and at all other residues are zero. For a fresh comparison Pause/Reset, set both modulus 4 and step 1 and compare the nonzero value at residue 3.

Separate residues from character valuesAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe instrument enumerates additive residues; the character values are assigned from the unit-group homomorphism. It does not compute multiplicative powers, conductors, or L-functions.

Check your understanding

For the character induced from the odd primitive character modulo four to modulus twelve, what is the extra factor in its L-function?

Try it yourself

On the unit group modulo five, define χ(2)=i\chi(2)=i. Find its values on residues 1,2,3,41,2,3,4, its parity, and the sum over a full period including zero.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The successive powers of two modulo five are one, two, four, three. Assign the corresponding powers of i before reordering the residues.

Compare with the solution

The values in the requested order are 1,i,−i,−11,i,-i,-1, since two squared is four and two cubed is three modulo five. The value at zero is zero, so the full-period sum is 1+i−i−1=01+i-i-1=0. Because minus one is residue four, χ(−1)=−1\chi(-1)=-1 and the character is odd. It is primitive: the only proper divisor of the prime modulus five is one, whose character induces only the principal character, whereas this one is nonprincipal.

Take it with you

  • Characters assign zero to nonunits and unit complex phases to units.
  • An induced L-function differs from its primitive one by specified finite Euler factors.
  • Nonprincipal cancellation improves series convergence, while principal characters retain a pole at one.

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