Pullbacks, Isometries & Conformal Maps
Outline
1. Pullbacks
Let V and W be vector spaces, and let T : V → W be an injective linear transformation.
Given an inner product h−, −i on W, the pullback of h−, −i is the inner product T ∗h−, −i
on V defined by
T ∗hv, v0 i = T (v), T (v0 ) .
In terms of matrices, if T is defined by a matrix A, and h−, −i is defined by a matrix P , then
the pullback T ∗h−, −i is defined by the matrix ATP A. In particular, the pullback of the dot
product corresponds to the matrix ATA.
If σ : U → R3 is a surface patch, the first fundamental form at a point p ∈ U can be
thought of as the pullback of the dot product on R3 under the derivative Dp σ. In particular,
the matrix I for the first fundamental form is given by the formula
I = (Dσ)T (Dσ).
2. Isometric Linear Transformations
Now suppose that V and W are inner product spaces, with inner products h−, −iV and
h−, −iW . A linear transformation T : V → W is said to be isometric if
T (v), T (v0 ) W = hv, v0 iV
for all v, v0 ∈ V. Here are a few equivalent definitions of isometric:
1. T is isometric if and only if kT (v)k = kvk for all v ∈ V.
2. T is isometric if and only if the pullback T ∗h−, −iW of the inner product on W is equal
to the inner product h−, −iV on V.
3. Let PV and PW be the matrices for h−, −iV and h−, −iW with respect to a certain pair of
bases, and let A be the matrix for T . Then T is isometric if and only if
ATPW A = PV .
(This is simply the matrix form of the previous definition.)
4. Let b1 , . . . , bn be a basis for V. Then T is isometric if and only if
T (bi ), T (bj ) W = hbi , bj iV
for all i and j.
5. Let e1 , . . . , en be an orthonormal basis for V. Then T is isometric if and only if the
vectors T (e1 ), . . . , T (en ) are orthonormal in W.
3. Isometries
A smooth map f : S1 → S2 between two surfaces is called a local isometry if it preserves the
lengths of curves. That is, f is a local isometry if, for each curve γ on S1 , the image curve
f ◦ γ on S2 has the same length as γ.
A local isometry f : S1 → S2 is called an isometry if it is bijective. In this case, the
surfaces S1 and S2 are said to be isometric.
The following theorem characterizes local isometries:
Theorem. A smooth map f : S1 → S2 is a local isometry if and only if the derivative
Dp f : Tp S1 → Tf (p) S2 at each point p ∈ S1 is an isometric linear transformation.
In light of this theorem, we can use any one of the following criteria to determine whether a
map is a local isometry:
1. A map f is a local isometry if and only if
kDp f (t)k = ktk
for each point p ∈ S1 and each tangent vector t ∈ Tp S1 .
2. A map f is a local isometry if and only if
(Dp f )∗h−, −if (p) = h−, −ip
for each point p ∈ S1 . That is, f is a local isometry if and only if the pullback under Df
of the first fundamental form on S2 is the first fundamental form on S1 .
3. A map f is a local isometry if and only if
(Df )T I2 (Df ) = I1 ,
where I1 and I2 are the matrices for the first fundamental forms on S1 and S2 .
4. Let t1 , t2 be a basis of tangent vectors at each point of
S1 . Then f is a local isometry if
and only if kDf (t1 )k = kt1 k and kDf (t2 )k = kt2 k and Df (t1 ), Df (t2 ) = ht1 , t2 i.
5. In particular, suppose that t1 , t2 is a basis of orthonormal tangent vectors at each point
of S1 . Then f is a local isometry if and only if Df (t1 ) and Df (t2 ) are orthonormal.
There is a special trick that can be used to simplify condition (3) further. Given a surface
patch σ1 : U → S1 , the composition σ2 = f ◦ σ can be thought of as a surface patch U → S2 .
With respect to these surface patches, the 2 × 2 derivative matrix D f is simply the identity
matrix, so f is a local isometry if and only if matrices I1 and I2 for the first fundamental forms
are equal.
Conversely, suppose that σ1 : U → S1 and σ2 : U → S2 are surface patches with the same
domain. If the matrices I1 and I2 for the first fundamental forms are equal, then the images
σ1 (U ) and σ2 (U ) must be isometric, with the isometry being the composition σ2 ◦ σ−1
1 .
4. Conformal Linear Transformations
Let V and W be inner product spaces, with inner products h−, −iV and h−, −iW . A linear
transformation T : V → W is said to be conformal if it preserves the angles between vectors.
The following theorem characterizes conformal transformations:
Theorem. A linear transformation
T: V → W is conformal if and only if there exists a
0
scalar λ > 0 so that T (v), T (v ) = λhv, v0 i for all v, v0 ∈ V.
That is, a conformal linear transformation must be the composition of an isometric transformation and a dilation. Here are a few equivalent definitions of conformal:
1. T is conformal if and only T (v) and T (v0 ) are orthogonal for each pair of orthogonal
vectors v, v0 ∈ V.
2. T is conformal if and only if T ∗h−, −iW = λh−, −iV for some scalar λ > 0.
3. Let PV and PW be the matrices for h−, −iV and h−, −iW with respect to a certain pair of
bases, and let A be the matrix for T . Then T is isometric if and only if
ATPW A = λPV .
for some scalar λ > 0.
4. Let b1 , . . . , bn be a basis for V. Then T is isometric if and only if there exists a scalar
λ > 0 so that
T (bi ), T (bj ) W = λhbi , bj iV
for all i and j.
5. Let e1 , . . . , en be an orthonormal basis for V. Then T is isometric if and only if the
vectors T (e1 ), . . . , T (en ) are orthogonal and all have the same length.
5. Conformal Maps
A smooth map f : S1 → S2 between two surfaces is said to be conformal if the derivative
Dp f at each point p ∈ S1 is a conformal linear transformation. Such a map will preserve the
angles between curves on the surface.
As with local isometries, there are several ways to show that a map is conformal:
1. A map f is conformal if and only if Dfp (t1 ) and Dfp (t2 ) are orthogonal for each point
p ∈ S1 and each pair of orthogonal tangent vectors t1 , t2 ∈ Tp S1 .
2. A map f is a conformal if and only if there exists a scalar function λ : S1 → (0, ∞) such
that
(Dp f )∗h−, −if (p) = λ(p) h−, −ip
for each point p ∈ S1 .
3. A map f is conformal if and only if
(Df )T I2 (Df ) = λI1 ,
for some scalar function λ, where I1 and I2 are the matrices for the first fundamental
forms on S1 and S2 .
4. Let t1 , t2 be a basis of tangent vectors at each point of S1 . Then f is conformal if and
only if there exists a scalar function λ such that
√
√
kDf (t1 )k = λkt1 k, kDf (t2 )k = λkt2 k, and Df (t1 ), Df (t2 ) = λht1 , t2 i
5. In particular, suppose that t1 , t2 is a basis of orthonormal tangent vectors at each point
of S1 . Then f is conformal if and only if Df (t1 ) and Df (t2 ) are orthogonal and have the
same length.
By the way, the scalar function λ appearing in these criteria is actually the Jacobian of the
conformal transformation. (See the notes on Area & Jacobians).
As with isometries, there is a special trick that can be used to simplify condition (3) further.
Specifically, given surface patches σ1 : U → S1 and σ2 : U → S2 satisfying σ2 = f ◦ σ1 , the
map f is conformal if and only if I2 = λI1 for some scalar function λ, where I1 and I2 are the
matrices for the first fundamental forms obtained from σ1 and σ2 .
6. Special Surface Patches
It is also possible to define the notions of local isometry and conformal map for surface patches.
Given a surface patch σ : U → S,
• We say that σ is isometric if the matrix I for the first fundamental form obtained from
σ is the identity matrix.
• We say that σ is conformal if the matrix I for the first fundamental form obtained from
σ is the product of the identity matrix with a scalar function.
That is, σ is isometric if the first fundamental form on U is the same as the dot product, and
σ is conformal if the first fundamental form on U is a scalar multiple of the dot product.
If we think of U as a subset of the xy-plane in R3 , then these definitions for a surface patch
reduce to the definitions given above for a map between surfaces. This makes it possible to
use any of the criteria given before for maps between surfaces to show that a surface patch is
isometric or conformal.
Practice Problems
1. Let P be the xy-plane in R3 , let C be the cylinder x2 + y 2 = 1, and let f : P → C be the map
f (x, y, 0) = (cos x, sin x, y).
Show that f is a local isometry.
2. Let C be the cylinder x2 + y 2 = 1, and let f : C → R2 be the map
f (x, y, z) = (xez , yez ).
(a) What is the image of f ?
(b) Show that f maps the cylinder C conformally onto its image.
3. Let γ : R → R2 be a smooth unit-speed curve without self-intersections, and define a surface
patch σ : R2 → R3 by
σ(u, v) = γ1 (u), γ2 (u), v .
Show that σ is isometric.
4. Let H 2 be the hyperbolic plane, and let σ : R × (0, ∞) → H 2 be a surjective surface patch
with first fundamental form ds2 = (du2 + dv 2 )/v 2 .
(a) Is the surface patch σ conformal? Is it isometric?
(b) Find a pair of vectors {t1 , t2 } at each point (u, v) that are orthonormal with respect to
the first fundamental form.
(c) Show that the map f : H 2 → H 2 defined by f (σ(u, v)) = σ(2u, 2v) is an isometry.
5. Let S be the cone x2 + y 2 = z 2 with the origin removed, and let f : S → S be the map
f (x, y, z) = x/z 2 , y/z 2 , 1/z .
(a) Find an orthonormal basis of tangent vectors at each point (x, y, z) ∈ S.
(b) Show that f is conformal. What is the Jacobian?