§5.1 Gauss’ Remarkable Theorem
Recall that, for a surface M , its Gauss curvature is defined by K = κ1 κ2 where
κ1 , κ2 are the principal curvatures of M . The principal curvatures are the eigenvalues
of the shape-operator SP . The shape-operator
SP (v) = −Dv n(P )
is defined by using the Gauss map n(P ), which depends, not only the surface itself,
but also on its position in R3 . Note that, the computational formula (4.19) which
we derived for calculating the Gauss curvature depends both on the first and second
fundamental form of M (the second fundamental form of M depends on the Gauss
map n(P ), hence it is an “extrinsic” quantity).
Gauss discovered that the Gauss curvature, in fact, depends only on the first
fundamental form of M . Hence, it is an intrinsic property(only depends on
the surface M itself). In other words, the Gauss curvature is unchanged when the
surface is bent without stretching(for the the first fundamental form, i.e. the measurement about the length, does not change, if the surface is bent without stretching).
Gauss called this result ’egregium’, and the Latin word for ’remarkable’ has remained
attached to this theorem ever since. This discovery is one of the most important discoveries about surface and is the foundation for the theory of Riemannian Geometry.
1
Review of the definition of the Gauss curvature
In this section, we review the definition of the Gauss curvature (see section 3.4 and
section 4.1). Recall that the Gauss curvature is defined by K = κ1 κ2 where κ1 , κ2 are
the principal curvatures of M . The principal curvatures are the eigenvalues of the
shape-operator SP ,
The shape-operator SP is defined by
SP (v) = −Dv n(P ).
Let σ : U → R3 be a parametrization for the surface M , then {σ u , σ v } is a basis of
the vector space TP (M ). Then, the matrix of SP with respect to the basis {σ u , σ v },
denoted by A, is (see (3.4.8))
A = FI−1 FII ,
1
where
FI =
E F
F G
!
FII =
,
e f
f g
!
,
Here {E, F, G} and {e, f, g} are the First and Second Fundamental forms of M . If
we write
!
a c
A=
,
b d
then(see (3.4.9))
f F − eG
eF − f E
, b=−
,
2
EG − F
EG − F 2
gF − f G
f F − gE
c=−
, d=−
.
2
EG − F
EG − F 2
a=−
The Gauss curvature is defined by K = κ1 κ2 where κ1 , κ2 are the eigenvalues of
SP . Hence, we have the following formula
(5.1.1)
2
K = det A = ad − bc =
eg − f 2
.
EG − F 2
The Christoffel symbols
Let σ : U → R3 be a parametrization of a surface M , and let {E, F, G}, {e, f, g} be
its first and second fundamental form. Let P ∈ M , note that {σ u , σ v } forms a basis
of the vector space TP (M ), hence {σ u , σ v , n} forms a basis of R3 , where n is the unit
normal to the surface M . Therefore, we can express σ uu , the (partial) derivatives of
the vector σ u , as the linear combination of (basis) {σ u , σ v , n}, i.e.:
σ uu = Γ111 σ u + Γ211 σ v + c3 n,
where Γ111 , Γ211 , and c3 are some real numbers which depend on the point P (hence
they are, in fact, functions on M ). Since σ u · n = 0, σ v · n = 0, we have c3 = σ uu · n.
By the definition of the Second Fundamental Form (see (3.4.4)), σ uu · n = e, where
{e, f, g} is the second fundamental form of M . Hence c3 = e. Therefore, we have
(5.1.2)
σ uu = Γ111 σ u + Γ211 σ v + en.
Similarly, we have
(5.1.3)
σ uv = Γ112 σ u + Γ212 σ v + f n,
(5.1.4)
σ vu = Γ121 σ u + Γ221 σ v + f n,
2
(5.1.5)
σ vv = Γ122 σ u + Γ222 σ v + gn.
Using σ uv = σ vu , it is easy to check that Γlik = Γlki , 1 ≤ i, l, k ≤ 2. The six functions
Γlik = Γlki , 1 ≤ i, l, k ≤ 2, are called the Christoffel symbols. The are the functions
on the surface M . Note that the Christoffel symbols Γlik are determined by writing
σ uu , σ uv and σ vv as a linear combination of {σ u , σ v , n}.
Also, write
(5.1.6)
nu = a11 σ u + a21 σ v ,
(5.1.7).
nv = a12 σ u + a22 σ v ,
Since nu = Dσ u n = −SP (σ u ), nv = Dσ v n = −SP (σ v ), we have that
a11 a21
a12 a22
!
= −A,
where A is the matrix of SP with respect to the basis {σ u , σ v }. Since A = FI−1 FII ,
we have
!
a11 a21
= −FI−1 FII .
a12 a22
So
(5.1.8)
3
a11 =
f F − eG
eF − f E
,
a
=
,
21
EG − F 2
EG − F 2
a12 =
gF − f G
f F − gE
,
a
=
.
22
EG − F 2
EG − F 2
The method of computing the Christoffel symbols
To compute the Christoffel symbols, we take the inner product in (5.1.2), (5.1.3),
(5.1.4), (5.1.5) with σ u and σ v respectively, and notice that σ u · n = 0, σ v · n = 0.
For example, taking the inner product (on both sides) in (5.1.2) with σ u , we get
σ uu · σ u = Γ111 σ u · σ u + Γ211 σ v · σ u .
Since, by the definition, E = σ u · σ u , ‘F = σ v · σ u , we have
(5.1.9)
σ uu · σ u = Γ111 E + Γ211 F.
3
From σ u · σ u = E, taking partial differentiation with respect to u, we get 2σ uu · σ u =
Eu . Substituting σ uu · σ u = 21 Eu into (5.1.9) yields
1
Γ111 E + Γ211 F = Eu .
2
(5.1.10)
Similarly, taking the inner product (on both sides) in (5.1.2) with σ v , we get
1
Γ111 F + Γ211 G =< σ uu , xv >= Fu − Ev .
2
(5.1.11)
So, we can compute the Christoffel symbol by solving the above system of linear
equations (5.1.10) and (5.1.11) jointly (with unknowns Γ111 and Γ211 ) to get
Γ111 =
(5.1.12)
GEu − 2F Fu + F Ev
,
2(EG − F 2 )
Γ211 =
2EFu − EEv − F Eu
.
2(EG − F 2 )
Other Christoffel symbols can be computed in a similar way, we have
GEv − F Gu
EGu − F Ev
2
2
,
Γ
=
Γ
=
12
21
2(EG − F 2 )
2(EG − F 2 )
2GFv − GGu − F Gv
EGv − 2F Fv + F Gu
=
, Γ222 =
.
2
2(EG − F )
2(EG − F 2 )
Γ112 = Γ121 =
Γ122
(5.1.13)
In the case that F = 0, then the above formula can be the following simple form
1 ∂ ln E
1 ∂E
,
Γ211 = −
2 ∂u
2G ∂v
1
∂
ln
E
1 ∂ ln G
= Γ121 =
, Γ212 = Γ221 =
2 ∂v
2 ∂u
1
∂
ln
G
1 ∂G
, Γ222 =
.
= −
2E ∂u
2 ∂v
Γ111 =
Γ112
Γ122
Note that the Christoffel symbols depend only on the first fundamental
form.
Example. Let
2u
2v
u2 + v 2 − 1
,
,
.
1 + u2 + v 2 1 + u2 + v 2 1 + u2 + v 2
!
σ(u, v) =
Calculate its Christoffel symbols.
Solution. Since
2(1 − u2 + v 2
−4uv
4u
,
,
,
2
2
2
2
2
2
(1 + u + v ) (1 + u + v ) (1 + u2 + v 2 )2
!
σ u (u, v) =
4
−4uv
2(1 + u2 − v 2
4v
,
,
,
2
2
2
2
2
2
(1 + u + v ) (1 + u + v ) (1 + u2 + v 2 )2
!
σ v (u, v) =
we can calculate the first fundamental form for σ is
4
E=G=
, F = 0.
2
(1 + u + v 2 )2
Hence, using (5.1.3) (or the simplified version wince F = 0), we get, by a direct
calculation,
2u
,
1 + u2 + v 2
2v
= Γ112 = Γ221 = −
,
1 + u2 + v 2
2v
2u
, Γ122 =
.
=
2
2
1+u +v
1 + u2 + v 2
Γ111 = Γ212 = Γ121 =
Γ222
Γ211
4
Gauss equations, Codazzi-Mainardi equations
We recall that, in Calculus III, we have the theorem that fxy = fyx , for any smooth
functions f , i.e. it doesn’t matter whether we take the partial with respect to x first
and then take the partial w.r.t. y or, we take the partial w.r.t. y first and then take
the partial w.r.t. x. For σ uu , σ uv , σ vv , we then have, by taking partials again,
(5.1.17)
σ uuv = σ uvu ,
(5.1.18)
σ vvu = σ vuv .
These relations give equations which will be refereed as Gauss equations and the
Codazzi-Mainardi equations. We first consider (5.1.17), i.e. σ uuv = σ uvu . Using
(5.1.2) and (5.1.3), the equation σ uuv = σ uvu gives
(Γ111 σ u + Γ211 σ v + en)v = (Γ112 σ u + Γ212 σ v + f n)u .
Hence,
(Γ111 )v σ u + Γ111 σ uv + (Γ211 )v σ v + Γ211 σ vv + ev n + env
= (Γ112 )u σ u + Γ112 σ uu + (Γ212 )u σ v + Γ212 σ uv + fu n + f nu
It yields
(Γ111 )v − (Γ112 )u σ u + (Γ211 )v − (Γ212 )u σ v + (ev − fu )n
= Γ112 σ uu + Γ212 − Γ111 σ uv − Γ211 σ vv + f nu − env .(5.1.19)
5
Using (5.1.3) to substitute σ uv , (5.1.5) to substitute σ vv , (5.1.6), (5.1.7) and (5.1.8)
to substitute nu and nv , we get(we only look at the coefficient before σ u for simplicity
reason!), from (5.1.19)
(Γ111 )v − (Γ112 )u σ u + (Γ211 )v − (Γ212 )u σ v + (ev − fu )n
= Γ112 Γ111 + (Γ212 − Γ111 )Γ112 − Γ211 Γ122 + f a11 − ea12 σ u + · · ·
=
Γ212 Γ112
−
Γ211 Γ122
(f 2 − eg)F
+
EG − F 2
!
σu + · · ·
(5.1.20)
By comparing the coefficients before σ u appearing in (5.1.20), we get
(Γ111 )v − (Γ112 )u = Γ212 Γ112 − Γ211 Γ122 +
(f 2 − eg)F
.
EG − F 2
Since, by (5.1.1),
K=
eg − f 2
.
EG − F 2
Hence, the above equation becomes
(5.1.21)
F K = Γ212 Γ112 − Γ211 Γ122 − (Γ111 )v + (Γ112 )u .
Similarly, by comparing the coefficients before σ v appearing in (5.1.20), we get
(5.1.22)
−EK = (Γ212 )u − (Γ211 )v + Γ112 Γ211 + Γ212 Γ212 − Γ211 Γ222 − Γ111 Γ212 .
The equations (5.1.21) and (5.1.22) are called Gauss equations.
Similarly, by comparing the coefficients before n appearing in (5.1.20), we obtain
(5.1.23)
ev − fu = eΓ112 + f (Γ212 − Γ111 ) − gΓ211 ,
and, using (5.1.18) and comparing the coefficients before n,
(5.1.24)
fv − gu = eΓ122 + f (Γ222 − Γ112 ) − gΓ212 .
The equations (5.1.23) and (5.1.24) are called Codazzi-Mainardi equations.
5
Gauss’s remarkable theorem
We are now ready to state and prove the following Gauss’s remarkable theorem:
6
Theorem(Gauss’ remarkable theorem) Let M be a regular surface(so EG−F 2 6=
0), then
(EG − F 2 )K =
(Γ211 )v − (Γ212 )u − Γ112 Γ211 − Γ212 Γ212 + Γ211 Γ222 + Γ111 Γ212 G
− Γ212 Γ112 − Γ211 Γ122 − (Γ111 )v + (Γ112 )u F.
(5.1.25)
In particular, the Gauss curvature depends only on the first fundamental form.
Proof. Since M is a regular surface, σ u × σ v 6= 0, in particular, EG − F 2 6= 0.
Multiplying F to (5.1.21) and G to (5.1.22) and then add them up, we can obtain
the (5.1.25). This proves the theorem.
Q.E.D.
(5.1.25) allows us to calculate the Gauss curvature in terms of the first fundamental
form only. Assume F = 0, then, from the Gauss equations above, we can derive that
(5.1.26)
−1
K= √
2 EG
E
√ v
EG
!
Gu
+ √
EG
v
! !
.
u
Example: For the surface of revolution
σ(u, v) = (f (u) cos v, f (u) sin v, g(u))
where f > 0 and f 0 2 + g 0 2 = 1. We found that E = 1, F = 0, G = f (u)2 . Hence, from
the above formula, we get
√
!
−1
f 00
Gu
1 ∂2 G
√
K= √
==
−
.
= −√
f
2 G
G u
G ∂u2
6
Isometry
Gauss’ remarkable theorem can be interpreted in a slightly different way in terms of
isometries. We discuss this point here briefly and informally.
Let M, N be two surfaces. A one-to-one map f : M → N that preserves lengths
and angles is called an isometry. (This may be understood at the level of tangent
vectors: f takes curves to curves, and hence velocity vectors to velocity vectors. f
is an isometry if and only if f preserves angle between velocity vectors and preserve
length of velocity vectors.)
Theorem Gauss curvature is a bending invariant, i.e. is invariant under isometries,
by which we mean: if f : M → N is an isometry then
KN (f (P )) = KM (P ),
7
i.e. the Gauss curvature is the same at corresponding points.
Proof: f preserves lengths and angles. Hence, with appropriate parameterizations,
the first fundamental forms for M and N are the same(at corresponding points). Since
the Gauss curvature pnly depends on the irst fundamental form, Gauss curvature will
be the same at corresponding points.
Application 1. The cylinder has the Gaussian curvature K = 0 (because a plane
has zero Gaussian curvature).
Applicaition 2. No piece of a plane can be bent to a piece of a sphere without
distorting lenghts (becuase the Gauss curvature of plane is zero, the Gauss curvature
of a sphere is 1/r2 where r =radius).
Theorem (Riemann). Let M be a surface with vanishing Gaussian curvature, i.e.
K ≡ 0, then each P ∈ M has a neighborhood which is isometric to an oepn set in the
Euclidean place.
The proof of Riemann’s theorem is omitted here.
7
Recap
• The Christoffel symbols:
2EFu − EEv − F Eu
GEu − 2F Fu + F Ev
, Γ211 =
2
2(EG − F )
2(EG − F 2 )
GEv − F Gu
EGu − F Ev
= Γ121 =
, Γ212 = Γ221 =
2
2(EG − F )
2(EG − F 2 )
2GFv − GGu − F Gv
EGv − 2F Fv + F Gu
=
, Γ222 =
.
2
2(EG − F )
2(EG − F 2 )
Γ111 =
Γ112
Γ122
• The Christoffel symbols(in the case F = 0):
1 ∂ ln E
1 ∂E
,
Γ211 = −
2 ∂u
2G ∂v
1
∂
ln
E
1 ∂ ln G
= Γ121 =
, Γ212 = Γ221 =
2 ∂v
2 ∂u
1 ∂G
1 ∂ ln G
2
= −
, Γ22 =
.
2E ∂u
2 ∂v
Γ111 =
Γ112
Γ122
8
• Gauss equations and Codazzi-Mainardi equations: Gauss equations:
F K = Γ212 Γ112 − Γ211 Γ122 − (Γ111 )v + (Γ112 )u ,
−EK = (Γ212 )u − (Γ211 )v + Γ112 Γ211 + Γ212 Γ212 − Γ211 Γ222 − Γ111 Γ212 .
Codazzi-Mainardi equations:
ev − fu = eΓ112 + f (Γ212 − Γ111 ) − gΓ211 ,
fv − gu = eΓ122 + f (Γ222 − Γ112 ) − gΓ212 .
• Formula for Gauss curvature:
(EG − F 2 )K =
(Γ211 )v − (Γ212 )u − Γ112 Γ211 − Γ212 Γ212 + Γ211 Γ222 + Γ111 Γ212 G
− Γ212 Γ112 − Γ211 Γ122 − (Γ111 )v + (Γ112 )u F.
In particular, when F = 0, then
−1
K= √
2 EG
E
√ v
EG
9
!
Gu
+ √
EG
v
! !
.
u