research31 min studyLesson 601 in Mathematics

The modular group and fundamental domains

Track fractional-linear actions, boundary identifications, and stabilizers before treating the quotient as a parameter space.

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What you’ll understand

  • Compute the generators' actions and distinguish SL₂(ℤ) from its effective quotient.
  • Reduce an explicit upper-half-plane point to the standard domain.
  • Identify elliptic stabilizers and the cusp without confusing them with ordinary interior points.

The upper half-plane parametrizes oriented lattice shapes, but changing an integral lattice basis changes that parameter. The modular group records these changes through fractional-linear maps. A fundamental domain chooses representatives while its paired edges remember when different representatives describe the same shape. We work over the complex numbers with τ=x+iy and y>0, and use determinant-one integral matrices. Keeping their central sign visible is essential when transformation laws of odd weight appear later.

The matrix action has a central kernel

A matrix γ with entries a,b,c,d and determinant one acts by (aτ+b)/(cτ+d). Its denominator cannot vanish in the upper half-plane: a zero would put τ on the real axis, unless c=0, when d is nonzero. Multiplying numerator and denominator by the conjugate denominator gives the new imaginary part y divided by |cτ+d| squared. The action therefore preserves the upper half-plane and its hyperbolic metric, although it usually changes Euclidean lengths and heights.

The matrices γ and −γ induce the same map. The kernel is precisely the two scalar matrices ±I, so the faithful action is by PSL₂(ℤ), the quotient of SL₂(ℤ) by that kernel. With S=[[0,−1],[1,0]] and T=[[1,1],[0,1]], the maps are inversion with a minus sign and unit translation. In SL₂(ℤ), S squared and (ST) cubed equal −I; their images have orders two and three in PSL₂(ℤ). Group orders must be stated in the group actually being used.

γτ=aτ+bcτ+d,Im⁡(γτ)=Im⁡τ∣cτ+d∣2,S2=(ST)3=−I\gamma\tau=\frac{a\tau+b}{c\tau+d},\qquad \operatorname{Im}(\gamma\tau)=\frac{\operatorname{Im}\tau}{|c\tau+d|^2},\qquad S^2=(ST)^3=-I
The imaginary-part identity uses real matrix entries and determinant one. The central relations become identity relations only after passage to PSL₂(ℤ).↔ Scroll wide equations horizontally.

The standard region is unique only away from its boundary

Take the closed region defined by |x|≤1/2 and |τ|≥1. Every orbit meets it, and no two distinct points in its interior are equivalent. Boundary points still require identification: translation pairs the two vertical edges, and S pairs the two halves of the unit-circle arc. Calling the closed region a set of unique representatives without these qualifications counts some lattice shapes twice. A half-open boundary convention can remove duplication, but must handle the special vertices consistently.

A practical reduction alternates translation into the vertical strip with S whenever the point lies inside the unit circle. Inversion then increases its height. Termination is not a consequence of an arbitrary increasing real sequence; discreteness of the possible denominators supplies the missing argument. The domain theorem and generation by S,T are structural results, while explicit examples can be checked directly. On the quotient, the identified region has a noncompact end extending upward, despite having finite hyperbolic area.

F‾={τ∈H:∣Re⁡τ∣≤12, ∣τ∣≥1},∫Fdx dyy2=∫−1/21/2dx1−x2=π3\overline{\mathcal F}=\{\tau\in\mathbb H: |\operatorname{Re}\tau|\le\tfrac12,\ |\tau|\ge1\},\qquad \int_{\mathcal F}\frac{dx\,dy}{y^2}=\int_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}}=\frac\pi3
Boundary sets have measure zero. The area is for one effective modular-domain region, with hyperbolic measure dx dy/y².↔ Scroll wide equations horizontally.

Fixed points and the cusp carry different local geometry

The point i is fixed by S. The point ρ=exp(2πi/3), on the left circular endpoint, is fixed by ST. Their effective stabilizers have orders two and three. In the upper-half-plane covering, a local quotient coordinate consequently behaves like the square or cube of a local coordinate. The coarse quotient is still a Riemann surface; the stabilizer information explains its ramified uniformizing map and later fractional contributions in the valence formula. It is not a singularity of the upper half-plane itself.

The rational boundary points together with infinity form the cusp set. For the full modular group every rational cusp is equivalent to infinity, because a reduced fraction can be completed to a determinant-one matrix. Infinity is an end rather than a point of the original domain. Adding that cusp compactifies the quotient, with exponential coordinate q=exp(2πiτ) near the added point. This local coordinate identifies unit translations, while its vanishing records escape to infinite height. It will impose a genuinely additional holomorphy condition on modular forms.

Reduce a point without guessing its orbit

Reduce τ=7/3+i/3 to the standard closed region and track a matrix effecting the reduction.

  1. Apply T⁻² to obtain τ₁=1/3+i/3. It is in the vertical strip but has squared modulus 2/9, so it is inside the excluded unit disk.

  2. Apply S: τ₂=−1/τ₁=−3/2+3i/2. The height changes from 1/3 to 3/2, agreeing with y/|τ₁|².

  3. Apply T once to obtain τ₃=−1/2+3i/2. Its squared modulus is 5/2, so it lies on the left vertical boundary of the closed domain. The composite is TS T⁻²; its matrix [[1,−3],[1,−2]] has determinant one.

The result

The final point is an allowed boundary representative. Its right-edge translate 1/2+3i/2 represents the same orbit, so uniqueness must include an edge convention.

Explore the model

Interactive lab

Interactive lab · matrix

Inspect an integral change of basis

Reset, then set a=1 and b=1. The matrix [[1,1],[0,1]] preserves area and shears an integral basis; predict determinant 1. Compare a fresh Reset with a=1 and b=0, also determinant 1. The instrument shows a real linear basis change, not the fractional-linear map on τ. Deriving that latter action requires taking the ratio of the transformed lattice generators.

Inspect an integral change of basisThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 integral matrix and its action on a shape parameter are related but distinct representations.

Check your understanding

What is the order of the image of S in PSL₂(ℤ)?

Try it yourself

Find the effective stabilizer orders of i and ρ=−1/2+i√3/2. Explain why a nonreal point strictly inside the domain other than i has trivial effective stabilizer.

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

Give me a hint

Solve Sτ=τ and STτ=τ, then use the fundamental-domain theorem's exceptional-point classification.

Compare with the solution

The fixed equation for S gives τ²=−1, selecting i in the upper half-plane. For ST it gives τ²+τ+1=0, selecting ρ. Their stabilizer orders in PSL₂(ℤ) are two and three. The domain theorem says these are the only nontrivial stabilizers up to the boundary equivalences. Thus an ordinary interior point has trivial effective stabilizer. In SL₂(ℤ), even an ordinary point retains the central subgroup ±I, so the effective qualifier matters.

Take it with you

  • SL₂(ℤ) acts through PSL₂(ℤ), but the central sign matters for weights.
  • The standard domain needs edge identifications and exceptional stabilizers.
  • The unique full-level cusp becomes q=0 after compactification.

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