advanced29 min studyLesson 368 in Mathematics

Conformal maps and the Schwarz lemma

Separate local angle preservation from global one-to-one behavior and derive the rigidity of normalized disk maps.

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What you’ll understand

  • Explain why a nonzero holomorphic derivative preserves oriented angles locally.
  • Verify an explicit half-plane-to-disk map and its inverse.
  • Use the Schwarz lemma and its equality case with the normalization stated.

A holomorphic map with nonzero derivative looks infinitesimally like one rotation followed by one uniform scaling. That is the source of local conformality. It does not guarantee that distant points remain distinct, and a derivative vanishing at one point can change angles there. Global maps require separate injectivity and range arguments. On the unit disk, the Schwarz lemma supplies another kind of control: a holomorphic self-map fixing the center cannot expand distance from that center, with equality forcing a rotation.

Local geometry comes from a nonzero derivative

For a holomorphic f near z₀ with f′(z0)≠0f'(z_0)\ne0, the first-order approximation is multiplication by that nonzero complex number. Two smooth curves meeting with nonzero tangent vectors therefore have their oriented angle preserved after mapping. The real Jacobian determinant equals ∣f′(z0)∣2>0|f'(z_0)|^2>0, so orientation is preserved and the inverse function theorem supplies a local inverse. These are local conclusions at a regular point, with no promise of global one-to-one behavior.

The exponential has nonzero derivative everywhere but repeats values after a vertical displacement of 2πi, so it is not injective on the whole plane. The map z² has zero derivative at the origin and doubles arguments there, while being locally conformal at every nonzero point. A conformal equivalence between domains requires a bijective holomorphic map with a holomorphic inverse. Specifying the domains is as necessary as writing an attractive formula for the transformation.

f(z0+h)=f(z0)+f′(z0)h+o(∣h∣),det⁡Df(z0)=∣f′(z0)∣2>0(f′(z0)≠0)f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|),\qquad \det Df(z_0)=|f'(z_0)|^2>0\quad(f'(z_0)\ne0)
Nonzero holomorphic derivative gives local angle and orientation preservation. It does not by itself establish a global bijection.↔ Scroll wide equations horizontally.

Normalized disk maps cannot expand radially

Let f map the open unit disk holomorphically into itself and satisfy f(0)=0. Divide by z and fill the removable value at zero with f′(0). On a smaller circle of radius r, the quotient's modulus is at most 1/r. The maximum modulus principle bounds it on that smaller disk. Letting r tend to one shows that the quotient has modulus at most one everywhere, yielding the radial and derivative inequalities of the Schwarz lemma.

If equality in the radial inequality holds at a nonzero interior point, the quotient attains its maximum modulus inside the disk and must be a constant of modulus one. Equality in the derivative bound gives the same result. Thus f is a rotation of z. Equality at the center in the radial inequality is automatic and gives no rigidity by itself. The normalization f(0)=0 is also essential; general disk automorphisms moving the center require a transformed inequality instead.

f:D→D holomorphic, f(0)=0⟹∣f(z)∣≤∣z∣,∣f′(0)∣≤1f:\mathbb D\to\mathbb D\text{ holomorphic},\ f(0)=0\quad\Longrightarrow\quad |f(z)|\le|z|,\quad |f'(0)|\le1
Equality at some nonzero interior z, or equality in |f′(0)|≤1, forces f(z)=e^{iθ}z for a real θ. Equality at z=0 alone is not informative.↔ Scroll wide equations horizontally.

Verify the range and the inverse of a fractional map

Linear fractional transformations are useful because their inverse can be solved algebraically and they map generalized circles to generalized circles. To prove that one maps the upper half-plane to the disk, compare the squared distances to i and −i. For z=x+iy with y positive, the distance to i is smaller than the distance to −i. The corresponding ratio has modulus less than one. This inequality proves the image inclusion, while the explicit inverse proves surjectivity and injectivity.

Such maps can transfer harmonic boundary problems or simplify a contour, but boundary points and poles must be tracked. A pole outside the source domain is harmless for holomorphicity inside it; a missing boundary point may correspond to infinity. Angle preservation concerns interior regular points and does not automatically give every boundary regularity statement. Domain, inverse, derivative, and boundary correspondence are separate checks that together turn a formal substitution into a usable conformal equivalence.

Map an upper-half-plane point into the disk

Verify T(z)=(z−i)/(z+i)T(z)=(z-i)/(z+i) as a conformal equivalence from the upper half-plane to the unit disk. Calculate T(2i) and T′(2i).

  1. For z=x+iy with y>0, ∣z−i∣2=x2+(y−1)2|z-i|^2=x^2+(y-1)^2 is smaller than ∣z+i∣2=x2+(y+1)2|z+i|^2=x^2+(y+1)^2. Thus |T(z)|<1, and its only pole −i lies outside the source domain.

  2. Solving w=T(z) gives z=i(1+w)/(1−w)z=i(1+w)/(1-w). For |w|<1 this has imaginary part (1−∣w∣2)/∣1−w∣2>0(1-|w|^2)/|1-w|^2>0, proving that every disk point has exactly one preimage.

  3. At z=2i, T(2i)=1/3. Differentiation gives T′(z)=2i/(z+i)2T'(z)=2i/(z+i)^2, so T′(2i)=−2i/9T'(2i)=-2i/9, a nonzero multiplier with modulus 2/9 and argument −π/2.

The result

The map is a global conformal equivalence on the stated domains. At 2i it locally rotates by −90 degrees and scales by 2/9, while mapping that point to 1/3.

Explore the model

Interactive lab

Interactive lab · matrix

Separate uniform scale from anisotropic stretching

Pause and Reset; set a=1,b=0 and predict equal singular values for the identity. Start a fresh Reset with a=2,b=0 and predict singular values 2 and 1. The second map stretches horizontal and vertical directions differently, so it cannot preserve all angles, despite keeping the coordinate axes perpendicular. This triangular real family is not a general complex-multiplication family or a disk-map solver.

Separate uniform scale from anisotropic stretchingThe 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 forConformality requires one local scale in every direction, not merely a positive determinant or one preserved right angle.

Check your understanding

A holomorphic f maps the unit disk into itself, fixes zero, and satisfies |f(1/2)|=1/2. What follows?

Try it yourself

A holomorphic disk self-map satisfies f(0)=0 and f(1/2)=i/2. Determine f everywhere and its derivative at zero. Explain why the nonzero matching point matters.

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

Give me a hint

First compare the moduli at 1/2, then determine the constant rotation from the stated complex value.

Compare with the solution

The modulus equality ∣f(1/2)∣=1/2|f(1/2)|=1/2 triggers Schwarz rigidity because 1/2 is nonzero and interior. Hence f(z)=eiθzf(z)=e^{i\theta}z. Substituting the value gives eiθ=ie^{i\theta}=i, so f(z)=iz and f′(0)=i. Equality only at zero would occur for every normalized map and would not determine f.

Take it with you

  • A nonzero holomorphic derivative gives local oriented-angle preservation.
  • Global conformal equivalence needs a bijection on stated domains.
  • The normalized Schwarz lemma has a rigid equality case at nonzero interior points.

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