## Lecture 6: Riemannian Manifolds

In the last lecture, we saw how the scalar product in Es gave rise to a type (0, 2) covariant tensor field gij. Here, we generalize this concept.

Definition 6.1 A smooth inner product on a manifold M is a function -,- that associates to each pair of smooth contravariant vector fields X and Y a scalar (field) X, Y , satisfying the following properties.

 symmetry: X, Y = Y, X for all X and Y, bilinearity: aX, bY = ab X, Y for all X and Y, and scalars a and b X, Y+Z = X, Y + X, Z X+Y, Z = X, Z + Y, Z . non-degeneracy: If X, Y = 0 for every Y, then X = 0.
We also call such a gizmo a symmetric bilinear form. A manifold endowed with a smooth inner product is called a Riemannian manifold.

Before we look at some examples, let us see how these things can be specified. First, notice that, if x is any chart, and p is any point in the domain of x, then

 X, Y = XiYj xi , xj

This gives us smooth functions

 gij = xi , xj

such that

X, Y = gijXiYj

and which, by Proposition 5.3, constitute the coefficients of a type (0, 2) symmetric tensor. We call this tensor the fundamental tensor or metric tensor of the Riemannian manifold.

Examples 6.2

(a) Take M = En, with the usual inner product; we find gij = ij.

(b) (Minkowski Metric) M = E4, with gij given by the matrix

G=
 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -c2
,

where c is the speed of light.

Question How does this effect the length of vectors?
Answer We saw in Lecture 3 that, in En, we could think of tangent vectors in the usual way; as directed line segments starting at the origin. The role that the metric plays is that it tells you the length of a vector; in other words, it gives you a new distance formula:

Euclidean 3- space: d(x, y) = [(y1 - x1)2 + (y2 - x2)2 + (y3 - x3)2]1/2
Minkowski 4-space: d(x, y) = [(y1 - x1)2 + (y2 - x2)2 + (y3 - x3)2 - c2(y4 - x4)2]1/2

Geometrically, the set of all points in Euclidean 3-space at a distance r from the origin (or any other point) is a sphere of radius r. In Minkowski space, it is a hyperbolic surface. In Euclidean space, the set of all points a distance of 0 from the origin is just a single point; in M, it is a cone, called the light cone. (See the figure.)

Minkowski 4-Space

(c) If M is any manifold embedded in Es, then we have seen above that M inherits the structure of a Riemannian metric from a given inner product on Es. In particular, if M is any 3-dimensional manifold embedded in E4 with the metric shown above, then M inherits such a inner product.

(d) As a particular example of (c), let us calculate the metric of the two-sphere M = S2, with radius r, using polar coordinates x1 = , x2 = . To find the coordinates of g** we need to calculate the inner product of the basis vectors /x1, /x2 in the ambient space Es. We saw in Section 3 that the ambient coordinates of /xi are given by

 j th coordinate = yjxi

where

y1 = r sin(x1) cos(x2)
y2 = r sin(x1) sin(x2)
y3 = r cos(x1)

Thus,

 x1 = r(cos(x1)cos(x2), cos(x1)sin(x2), -sin(x1)) x2 = r(-sin(x1)sin(x2), sin(x1)cos(x2), 0)

This gives

g11 = /x1, /x1 = r2
g22 = /x2, /x2 = r2 sin2(x1)
g12 = /x1,/x2 = 0,

so that

g**=
 r2 0 0 r2sin2 (x1)
.

(e) The n-Dimensional Sphere Let M be the n-sphere of radius r with generalized polar coordinates.

y1 = r cos x1
y2 = r sin x1 cos x2
y3 = r sin x1 sin x2 cos x3
...
yn-1 = r sin x1 sin x2 sin x3 sin x4 ... cos xn-1
yn = r sin x1 sin x2 sin x3 sin x4 ... sin xn-1 cos xn
yn+1 = r sin x1 sin x2 sin x3 sin x4 ... sin xn-1 sin xn.

(Notice that x1 is playing the role of and the x2, x3, . . . , xn-1 the role of .) Following the line of reasoning in the previous example, we have

 x1 = (-r sin x1, r cos x1 cos x2, r cos x1 sin x2 cos x3 , ... , r cos x1 sin x2... sin xn-1 cos xn, r cos x1 sin x2 ... sin xn-1 sin xn) x2 = (0, -r sin x1 sin x2, . . . , r sin x1 cos x2 sin x3... sin xn-1 cos xn, r sin x1 cos x2 sin x3 ... sin xn-1 sin xn) x3 = 0, 0, -r sin x1 sin x2 sin x3, r sin x1 sin x2 cos x3 cos x4 . . . , r sin x1 sin x2 cos x3 sin x4... sin xn-1 cos xn, r sin x1 sin x2 cos x3 sin x4 ... sin xn-1 sin xn)

and so on. This gives

g11 = /x1, /x1 = r2
g22 = /x2, /x2 = r2sin2x1
g33 = /x3, /x3 = r2sin2x1 sin2 x2
...
gnn = /xn, /xn = r2sin2x1 sin2 x2 ... sin2 xn-1
gij = 0 if i j

so that

g**=
 r2 0 0 ... 0 0 r2sin2x1 0 ... 0 0 0 r2sin2x1 sin2 x2 ... 0 ... ... ... ... ... 0 0 0 ... r2sin2x1 sin2 x2 ... sin2 xn-1
.

(f) Diagonalizing the Metric Let G be the matrix of g** in some local coordinate system, evaluated at some point p on a Riemannian manifold. Since G is symmetric, it follows from linear algebra that there is an invertible matrix P = (Pji) such that

PGPT=
 ±1 0 0 0 0 ±1 0 0 ... ... ... ... 0 0 0 ±1

at the point p. Let us call the sequence (±1,±1, . . . , ±1) the signature of the metric at p. (Thus, in particular, the Minkowski metric has signature (1, 1, 1, -1).) If we now define new coordinates j by

xi = Pjij,

(so that we are using the inverse of P for this) then xi/j = Pji, and so

 ij = xai gab xbj = PiagabPjb = Piagab(PT)bj = (PGPT)ij

showing that, at the point p,

**=
 ±1 0 0 0 0 ±1 0 0 ... ... ... ... 0 0 0 ±1
.

Thus, in the eyes of the metric, the unit basis vectors ei = /i are orthogonal; that is,

ei, ej = ±ij.

Note The non-degeneracy condition in Definition 6.1 is equivalent to the requirement that the locally defined quantities

g = det(gij)

are nowhere zero.

Here are some things we can do with a Riemannian manifold.

 Definition 6.3 If X is a contravariant vector field on M, then define the square norm norm of X by ||X||2 = X, X = gijXiXj. Note that ||X||2 may be negative. If ||X||2 < 0, we call X timelike; if ||X||2 > 0, we call X spacelike, and if ||X||2 = 0, we call X null. If X is not spacelike, then we can define ||X|| = (||X||2)1/2= (gijXiXj)1/2

In the exercise set you will show that null need not imply zero.

Note Since X, X is a scalar field, so is ||X|| is a scalar field, if it exists, and satisfies ||úX|| = |ú|á||X|| for every contravariant vector field X and every scalar field ú. The expected inequality

||X + Y|| ||X|| + ||Y||

need not hold. (See the exercises.)

Arc Length One of the things we can do with a metric is the following. A path C given by xi = xi(t) is non-null if ||dxi/dt||2 0. It follows that ||dxi/dt||2 is either always positive ("spacelike") or negative ("timelike").

Definition 6.4 If C is a non-null path in M, then define its length as follows: Break the path into segments S each of which lie in some coordinate neighborhood, and define the length of S by

L(a, b)=
b

a
 ± gij dxidt . dxjdt
1/2

dt

where the sign ±1 is chosen as +1 if the curve is space-like and -1 if it is time-like. In other words, we are defining the arc-length differential form by

ds2 = ±gijdxidxj.

To show (as we must) that this definition is independent of the choice of chart x, all we need observe is that the quantity under the square root sign, being a contraction product of a type (0, 2) tensor with a type (2, 0) tensor, is a scalar.

 Proposition 6.5 (Paramaterization by Arc Length) Let C be a non-null path xi = xi(t) in M. Fix a point t = a on this path, and define a new function s (arc length) by s(t) = L(a, t) = length of path from t = a to t. Then s is an invertible function of t, and, using s as a parameter, ||dxi/ds||2 is constant, and equals 1 if C is space-like and -1 if it is time-like. Conversely, if t is any parameter with the property that ||dxi/dt||2 = ±1, then, choosing any parameter value t = a in the above definition of arc-length s, we have t = ±s + C for some constant C. (In other words, t must be, up to a constant, arc length. Physicists call the parameter = s/c, where c is the speed of light, proper time for reasons we shall see below.)

Exercise Set 6

1. Give an example of a Riemannian metric on E2 such that the corresponding metric tensor gij is not constant.

2. Let aij be the components of any symmetric tensor of type (0, 2) such that det(aij) is never zero. Define

X, Y a = aijXiYj.

Show that this is a smooth inner product on M.

3. Give an example to show that the ||X|| + ||Y|| is not always true on a Riemannian manifold.

4. Give an example of a Riemannian manifold M and a nowhere zero vector field X on M with the property that ||X|| = 0. We call such a field a null field.

5. Show that if g is any smooth type (0, 2) tensor field, and if g = det(gij) 0 for some chart x, then = det(ij) 0 for every other chart (at points where the change-of-coordinates is defined). [Use the property that, if A and B are matrices, then det(AB) = det(A)det(B).]

6. Suppose that gij is a type (0, 2) tensor with the property that g = det(gij) is nowhere zero. Show that the resulting inverse (of matrices) gij is a type (2, 0) tensor. (Note that it must satisfy gijgkl = kilj.)

7. (Index lowering and raising) Show that, if Rabc is a type (0, 3) tensor, then Raic given by

Raic = gibRabc,
is a type (1, 2) tensor. (Here, g** is the inverse of g**.) What is the inverse operation?

8. A type (1, 1) tensor field T is orthogonal in the Riemannian manifold M if, for all pairs of contravariant vector fields X and Y on M, one has

TX, TY = X, Y,
where (TX)i = TikXk. What can be said about the columns of T in a given coordinate system x? (Note that the ith column of T is the local vector field given by T(/xi).)

Last Updated: January, 2002