RIEMANNIAN GEOMETRY
(Unit-III (a))
Gradient vector Field: A vector field Z on a Riemannian manifold (M, g) is said to
be gradient vector field if there exists a smooth function f ∈ C ∞ (M ), such that
g(grad f, Y ) = g(Z, Y ) = (df )Y = Y f,
∀ Y ∈ χ(M ).
In this case we write Z = grad f .
(d : ∧r −→ πr+1 ) P
∂f ∂
1
2
ij
(Note: On a local chart (U, x , x , ..., xn ) on M , grad f = i,j g ij ∂x
= (hij )−1 .)
i ∂xj ; g
Proof. Let (U, x1 , x2 , ..., xn ) be a local chart on the Riemannian manifold (M, g) and
f ∈ C ∞ (M ). Then
∂f i X l ∂ dx
x l
df (X) =
g
∂xi
∂x
i,j
l
∂ X X ∂f
l
i
X
dx
=
∂xi
∂xl
i
l
X X ∂f
X l δli
=
i
∂x
i
l
X X ∂f X
l
ij
=
X
g
g
jl
∂xi
i
j
l
∂f ∂f X X X ∂f
l ij
X
g
g
,
=
i
j ∂xl
∂x
∂x
i
j
l
XX
∂f ∂ X l ∂f =g
X
g ij i j ,
∂x
∂x
∂xl
i
j
l
X
ij ∂f ∂
⇒ df (X) = g
g
,X
∂xi ∂xj
i,j
X
∂f ∂
∀X ∈ χ(M )
⇒ g(grad f, X) = g
g ij i j , X ,
∂x ∂x
i,j
X
∂f ∂
grad f =
gij i j .
∂x ∂x
i,j
X
∴
ij
Theorem 1. Let Z be a gradient vector field on a Riemannian manifold (M, g). Then,
g(∇X Z, Y ) = g(∇Y Z, X), ∀X, Y ∈ χ(M ).
1
Proof. Since ∇ is a Levi-Civita connection on (M, g). Therefore by definition of LeviCivita connection
(∇X g)(Y, Z) = 0,
or
⇒
(∇X g)(Y, Z) − g(∇X Y, Z) − g(Y, ∇X Z) = 0
⇒
Xg(Y, Z) = g(∇X Y, Z) + g(Y, ∇X Z)
⇒
g(∇X Z, Y ) = Xg(Y, Z) − g(∇X Y, Z).
(1)
Interchanging X and Y in (1), we have
g(∇Y Z, X) = Y g(X, Z) − g(∇Y X, Z).
(2)
Subtraction (2) from (1), we get
g(∇X Z, Y ) − g(∇Y Z, X) = Xg(Y, Z) − Y g(X, Z) − g(∇X Y, Z) + g(∇Y X, Z).
Since Z is a gradient vector field on M , therefore there exists f ∈ C ∞ (M ), such that
g(Y, Z) = df (Y ) = Y f.
Thus, we have
g(∇X Z, Y ) − g(∇Y Z, X) = Xdf (Y ) − Y df (X) − df (∇X Y ) + df (∇Y X)
= X(Y f ) − Y (Xf ) − (∇X Y )(f ) + ∇Y X(f )
= [X, Y ](f ) − [X, Y ](f )
= 0.
Hence
g(∇X Z, Y) = g(∇Y Z, X).
Divergence: Let (M, g) be a Riemannian manifold of dimension n and X ∈ χ(M ).
Then the divergence of the vector field X is defined as
div X =
n
X
g(∇ei X, ei ),
i=1
where {e1 , e2 , ..., en } is local orthogonal frame of vector fields on M .
Problem. Consider R4 with usual Riemannian metric. Take X = (x2 , y 2 , z 2 , w2 ),
where x, y, z and W are usual coordinate functions on R4 . Compute div X.
Solution. By definition
div X =
n
X
g(∇ei X, ei ),
i=1
2
∂
∂
∂
∂
where e1 = ∂x
, e2 = ∂y
, e3 = ∂z
and e0 = ∂w
, g is usual Riemannian metric on R4 and ∇
is Levi-Civita connection on R4 . Then, we have
∇e1 X = e1 (x2 ), e1 (y 2 ), e1 (z 2 ), e1 (w2 ) ,
where
D
E
e1 (x ) = (1, 0, 0, 0), (2x, 0, 0, 0)
= 2x,
4
D
E R
e1 (y 2 ) = (1, 0, 0, 0), (0, 0, 0, 0)
= 0,
2
R4
2
2
and e1 (y ) = 0 = e1 (y ).
Therefore, ∇e1 X = (2x, 0, 0, 0).
Similarly,
∇e2 X = (0, 2y, 0, 0),
∇e3 X = (2x, 0, 2z, 0),
and ∇e4 X = (2x, 0, 0, 2w).
Thus,
D
E
g(∇e1 X, e1 ) = (2x, 0, 0, 0), (1, 0, 0, 0) = 2x.
div(X) =
∴
4
X
g(∇ei X, ei ),
i=1
which implies
div(X) = 2(x + y + z + w).
* Let {e1 , e2 , ..., en } be a local orthogonal frame of vector fields on a Riemannian manifold
(M, g) of dimension n then for X = X k ek , then we have,
div(X) =
n
X
g(∇ei X, ei )
i=1
=
X
g(∇ei (X k ek ), ei )
i
X =
g ei (X k )ek + X k ∇ei ek , ei
i
=
X
=
X
ei (X k )g(ek , ei ) +
X
i
i
ei (X
k
)δik
+
i
∴
or
g(X k Γjik ej , ei )
X
i
div(X) = ek (X k ) + X k Γ,
div(X) = ek (Xk ) + Xk Γ.
3
X Γjik δik
k
Theorem 2. Let f ∈ C ∞ (M ) and X ∈ χ(M ), then
div(f X) = Xf + f div(X).
Proof. Let {e1 , e2 , ..., en } be a local orthogonal frame of vector fields on a Riemannian
manifold (M, g), then
n
X
div(f X) =
g ∇ei (f X), ei
i=1
n
X
=
g ei (f )X + f ∇ei X, ei
=
i=1
n
X
ei (f )g(X, ei ) + f
n
X
i=1
g(∇ei X, ei )
i=1
= Xf + f div(X).
Result: Prove that div([X, Y]) = X div(Y) − Y div(X).
Proof.
∵
∴
∴
div(Y ) = ek (Y k ) + Γiki Y k
Xdiv(Y ) = X ek (Y k ) + X Γiki Y k
k
Xdiv(Y ) = X ek (Y ) + X Γiki Y k + Γiki X(Y k ).
In the same way
Y div(X) = Y ek (X k ) + Y Γiki X k + Γiki Y (X k ).
Therefore,
k
k
Γiki
k
XY − Y X
Xdiv(Y ) − Y div(X) = ek XY − Y X +
= ek [X, Y ]k + Γiki [X, Y ]k
= div [X, Y ] .
k
Curl: The curl of a 1-form u on a Riemannian manifold (M, g) is defined as
curl u = du
(exterior derivative of u).
To Prove that
(Curl u)(X, Y ) = ∇X U Y − ∇Y U X,
where ∇ is the levi-Civita connection on M .
4
∀X, Y ∈ χ(M ),
Proof.
∵
∴
(Curl u)(X, Y ) = du(X, Y )
= Xu(Y ) − Y u(X) − u [X, Y ]
= ∇X u Y + u ∇X Y − ∇Y u X − u ∇Y X − u [X, Y ]
= ∇X u Y − ∇Y u X + u ∇X Y − ∇Y X − u [X, Y ]
= ∇X u Y − ∇Y u X + u [X, Y ] − u [X, Y ]
(Curl u)(X, Y) = ∇X u Y − ∇Y u X, ∀X, Y ∈ χ(M)
Definition. A Riemannian manifold (M, g) is said to be a Einstein manifold if the Ricci
tensor on M is of the form
S(X, Y ) = λg(X, Y ),
∀X, Y ∈ χ(M ), λ ∈ C ∞ (M ).
Theorem 3. if a Riemannian manifold (M, g) of dim n > 3, is of constant curvature
then it is an Einstein manifold.
Proof. Let (M, g) be a Riemannian manifold of dim n > 3, which is of constant curvature.
then the curvature tensor on M is given by
R(X, Y )Z = K g(Y, Z)X − g(X, Z)Y , ∀X, Y, Z ∈ χ(M )
⇒ R(X, Y, Z, U ) = g(R(X, Y )Z, U )
= K g(Y, Z)g(X, U ) − g(X, Z)g(Y, U ) , ∀X, Y, Z ∈ χ(M ).
Let {e1 , e2 , ..., en } be a local orthogonal frame of vector fields onM . Then taking X◦U = ei
and summing over i, we get
n
X
R(ei , Y, Z, ei ) =
i=1
⇒
⇒
n
X
h
i
K g(Y, Z)g(ei , ei , g( ei, z)g(Y, ei )
i=1
S(Y, Z) = K ng(Y, Z) − g(Y, Z)
S(Y, Z) = K(n − 1)g(Y, Z), ∀Y, Z ∈ χ(M ).
Hence, (M, g) is an Einstein manifold.
Remark: The converse of theorem is not true, in general.
However, if M is a 3-dimensional Einstein manifold then it a manifold of constant
curvature.
Examples: Rn , S n (n > 3) are Einstein manifolds.
Theorem 4. The scalar curvature of an n-dimensional Einstein manifold is constant if
n > 2.
5
Covariant differentiation along a curve: Let M be a smooth manifold and σ :
I −→ M be a smooth curve (I is interval).
A map X : t −→ X(t) ∈ Tσ(t) M , is called a vector field along the curvev σ. A
vector field X along the curve σ is called is said to be smooth, if the map t −→ X(t)(g)
is smooth, ∀f ∈ C ∞ M . The set of all smooth vector fields along a smooth curve σ is
denoted by χ(σ).
Remarks:
(i) Let σ : I −→ M be a smooth curve in a smooth manifold. Suppose t is the
parameter of the curve. Then σ 0 (t) is tangent vector field along t.
(ii) Let J denotes the counterclockwise rotation by π2 . Then Jσ 0 (t) is also a smooth
vector field along σ, which is the normal vector field.
Theorem 5. Let M be a smooth manifold with an affine connection ∇ and σ : I −→ M
be a smooth curve with parameter t, where I is an interval in R. Then, there exists a
D
unique map dt
: χ(σ) −→ χ(σ), such that,
(i)
D
(X
dt
+Y)=
DX
dt
+
DY
dt
,
∀X, Y ∈ χ(M ).
(ii) If f : I −→ R is a smooth map, then
Df
DX
D
(f X) =
X +f
,
dt
dt
dt
∀X ∈ χ(M ), f ∈ C ∞ (M ).
(iii) If V is a smooth vector field along σ, which is induced by a vector field Y ∈ χ(M ),
that is, V (t) = Yσ(t) , ∀ t ∈ I. Then
DV
= ∇σ0 (t) Y = ∇ dσ Y.
dt
dt
This map
D
dt
is called the operator of covariant derivative along the curve σ.
D
Proof. Uniqueness: Let a map dt
: χ(σ) −→ χ(σ) exists, which satisfy all the properties
1
2
n
listed above. Let (U, x , x , ..., x ) be a chart on M and σ(I) ⊆ U . If σ(t) = (x1 , x2 , ..., xn ),
then we have
n
X
0
∂
0
σ (t) =
xi (t) i
.
∂x
σ(t)
i=1
Let X be a smooth vector along the curve. Then X(t) ∈ Tσ(t) M,
X(t) =
n
X
X j (t)
j=1
6
∂
∂xj
.
σ(t)
∀t ∈ I. Let
Then
n
DX
DX j
∂ =
X (t) j
dt
dt j=1
∂x σ(t)
=
=
=
n
X
D
∂
X (t) j
dt
∂x
j=1
n X
j=1
n
X
j=1
n
X
dX j ∂
dt ∂xj
dX j ∂
dt ∂xj
=
D ∂
+ X (t)
dt ∂xj
σ(t)
+
σ(t)
=
=
Hence
DX
=
dt
j=1
n
X
j=1
n
X
(X j )0 (t)
∂
∂xj
dX k ∂
dt ∂xk
k=1
n h
X
k=1
∂
∂xj
n
X
+
σ(t)
+
σ(t)
+
σ(t)
+
σ(t)
X j (t)∇σ0 (t)
j=1
j 0
(X j )0 (t)
(by (i))
σ(t)
j
∂
=
(X ) (t) j
∂x
j=1
n
X
j
n
X
j=1
n
X
(by (ii))
σ(t)
∂
∂xj
X j (t)∇(X i )0 (t)
(by (iii))
|
∂
∂xi σ(t)
(X i )0 (t)X j (t)∇
i,j=1
n
X
∂
∂xj
|
∂
∂xi σ(t)
∂
(xi )0 (t)X j (t) Γkij
∂xk
i,j=1
n
X
dxi j k ∂
X (t) Γij
dt
∂xk
i,j=1
σ(t)
σ(t)
n
X
dX k
dxi j i ∂
+
Γkij
X (t)
dt
dt
∂xk
i,j=1
∂
∂xj
(i),
σ(t)
D
which proves the uniqueness of dt
because right hand side of (i) depends only on given
connection ∇ and given curve σ.
Existence:
D
: χ(σ) −→ χ(σ) be defined as
Let dt
n
n
DX X h dX k X k dxi j i ∂
=
+
Γij
X (t)
dt
dt
dt
∂xk
i,j=1
k=1
for any X(t) =
Pn
i=1
X i (t)
∂
∂xi σ(t)
,
σ(t)
∈ χ(σ). Suppose that V ∈ χ(σ) which is induced by a
7
vector field X ∈ χ(M ). Let X = X i ∂x∂ i on U . Then,
V (t) = X(σ(t))
∂ = X i i (σ(t))
∂x
∂
= X i (σ(t)) i
∂x σ(t)
= (X i ◦ σ)(t)
Then, we let Y =
Pn
i=1
∂
∂xi
.
σ(t)
Y i ∂x∂ i on U .
(i)
n h
n
i ∂
X
D
d(X + Y )k X k dxi
j
(X + Y ) =
+
(X + Y ) (t)
Γij
dt
dt
dt
∂xk
i,j=1
k=1
=
n h
X
k=1
n
X
i ∂
i
dX k
k dx
j
+
Γij
X (t)
dt
dt
∂xk
i,j=1
+
σ(t)
n h
X
k=1
σ(t)
n
dY k X k dxi j i ∂
+
Γij
Y (t)
dt
dt
∂xk
i,j=1
D
DX DY
(X + Y ) =
+
.
dt
dt
dt
(ii)
n h
n
i ∂
X
D
d(f X k ) X k dxi
j
(f X) =
+
Γij
(f X )(t)
dt
dt
dt
∂xk
i,j=1
k=1
=
n h
X
k=1
df
=
dt
=
σ(t)
n
X
i ∂
i
dX k
df k
k dx
j
X +f
+
Γ
f (X (t))
dt
den i,j=1 ij dt
∂xk
n
X
∂
X
∂xk
k=1
k
+
σ(t)
n h
X
k=1
df
DX
X +f
,
dt
dt
i ∂
i
dX k
k dx
j
+
Γij
X (t)
dt
dt
∂xk
i,j=1
∀f ∈ C ∞ (σ).
8
σ(t)
n
X
σ(t)
σ(t)
(iii)
n
X d
∂
DV
=
(X i ◦ σ) i
dt
dt
∂x
k=1
+
σ(t)
n
X d
DV
∂
=
(X i ◦ σ) i
dt
dt
∂x
k=1
n
X
+
σ(t)
∂
=
X σ(t) ∇σ0 (t)
∂xj
j=1
= ∇σ0 (t)
n
X
i,j,k=1
n
X
i,j,k=1
i=1
∂
X
∂xi
+
σ(t)
n
X
σ 0 (t)(X i )
i=1
i
dxi ∂
dt ∂xk
(X i ◦ σ)(t)∇σ0 (t)
j
n
X
Γkij (X i ◦ σ)(t)
σ(t)
∂
∂xi
∂
∂xj
σ(t)
σ(t)
σ(t)
d
∵ ∀f ∈ C (M ), σ (t) = (f ◦ σ)(t)
dt
∞
0
= ∇σ0 (t) X.
Now, suppose that (V, ψ) is another chart on M , such that, U ∩ V 6= {0}. Define DX
by
dt
DX
equation (i) on V then definition agrees on U ∩ V by uniqueness of dt in U . Thus, by
D
D
uniqueness of dt
, various definitions of dt
agrees when chart overlap. Therefore, above
definition can be extended on whole of M , which completes the proof.
Definition. Let M be a smooth manifold with an affine connection ∇. Suppose X is a
smooth vector field along a smooth curve σ : I −→ M , I being in interval. Then X is
= 0.
said to be parallel along σ of the covariant derivative of X is zero, that is DX
dt
Example. Let σ : [a, b] −→ M be the constant curve, such that, σ(t) = p, ∀ t ∈ [a, b].
Let X ∈ χ(σ). Then
n
n
DX X h dX k X k dxi j i ∂
=
+
Γij
X (t)
dt
dt
dt
∂xk
i,j=1
k=1
(i).
σ(t)
Since, σ is a constant curve, thereforeX i are constant and
from (i)
dxi
dt
= 0, ∀i = 1, 2, ..., n. So,
n
DX X dX k ∂
=
dt
dt ∂xk
k=1
⇒
σ(t)
DX
dX
=
, f or a constant curve.
dt
dt
Question.
Let γ(t) = (t, 0, 0, ..., 0) in M = Rn , with the usual connection ∇. If
Pn
∂
∞
i=1 fi ∂xi is parallel along γ, fi ∈ C (R), then what we can say about X?
9
Solution. Since, X is parallel along γ, we have
⇒
⇒
DX
=0
dt
∇γ 0 (t) X = 0
∇ ∂1 X = 0
(by(iii))
∂x
n
X
∂fi ∂
⇒
=0
1 ∂xi
∂x
i=1
∂fi
= constant along x1 − axis, ∀i
∂x1
⇒ f i = ci
(say).
⇒
P
Therefore, X = ni=1 Ci ∂x∂ i on x1 -axis, where, Ci ∈ R.
Question: Let γ(t) = (cos(t), sin(t), 0), t ∈ R be a curve on the unit sphere S 2 , endowed
with induced Riemannian metric from R3 . If X(t) = e3 , ∀t. Determine whether X is
parallel along γ.
solution. We know that
T
DX
= ∇γ 0 (t) X ,
dt
given
γ 0 (t) = − sin(t), cos(t), 0
∂
∂
+ cos(t) ,
= −sin(t)
∂x
∂y
where x, y, z are coordinate forms on R3 . Therefore
∇γ 0 (t) X
T
X=∇
∂
∂
−sin(t) ∂x
+cos(t) ∂y
= −sin(t)
∂
∂z
∵ e3 = (0, 0, 1) ≡
∂ ∂z
∂1 ∂
∂1 ∂
+ cos(t)
∂x ∂z
∂y ∂z
= 0.
Therefore
DX
dt
= 0. Hence X is parallel along γ.
Definition. Let M be a smooth manifold with an affine connection ∇. A vector field
V along a smooth curve γ : T −→ M is said to be parallel along γ with respect to
connection ∇ if DV
= 0, ∀t ∈ I.
dt
A vector field on M is said to be parallel if it is parallel along every smooth curve in
M.
Question. Let γ : I −→ Rn be a smooth curve. show that a vector field V along γ is
10
parallel with respect to usual connection on Rn if and only if V is a constant vector field.
Solution. Let x1 , x2 , ..., xn be the coordinate forms on Rn . Assume that
V =
X
i
Vi
∂
, V i ∈ C ∞ Rn .
i
∂
Then DV
=0
Pn dt dV j ∂
P
P
i
⇔ j=1 dt ∂xj γ(t) + nk=1 ni,j=1 Γkij dx
V j (t) ∂x∂ k γ(t) = 0
dt
(∵ on Rn with usual connection ∇, Γkij = 0, ∀ i, j, k)
Pn dV j ∂
⇔
j=1 dt ∂xj γ(t) = 0
dV j
dt
⇔
∵ ∂x∂ j are linearly independent
⇔
⇔ V is a constant field.
= 0, ∀j = 1, 2, ..., n
V j = constant, ∀j = 1, 2, ..., n
11