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In the last lecture, we saw how the scalar product in E_{s} gave rise to a type (0, 2) covariant tensor field g_{ij}. 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.

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  =  X^{i}Y^{j}  x^{i} 
,  x^{j} 
This gives us smooth functions
g_{ij}  =  x^{i} 
,  x^{j} 
such that
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 = E_{n}, with the usual inner product; we find g_{ij} = _{ij}.
(b) (Minkowski Metric) M = E_{4}, with g_{ij} given by the matrix
G  = 

where c is the speed of light.
Question How does this effect the length of vectors?
Answer We saw in Lecture 3 that, in E_{n}, 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:
Geometrically, the set of all points in Euclidean 3space 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.)
(c) If M is any manifold embedded in E_{s}, then we have seen above that M inherits the structure of a Riemannian metric from a given inner product on E_{s}. In particular, if M is any 3dimensional manifold embedded in E_{4} 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 twosphere M = S^{2}, with radius r, using polar coordinates x^{1} = , x^{2} = . To find the coordinates of g_{**} we need to calculate the inner product of the basis vectors /x^{1}, /x^{2} in the ambient space E_{s}. We saw in Section 3 that the ambient coordinates of /x^{i} are given by
j th coordinate  =  x^{i} 
where
Thus,
x^{1}  =  r(cos(x^{1})cos(x^{2}), cos(x^{1})sin(x^{2}), sin(x^{1})) 
x^{2}  =  r(sin(x^{1})sin(x^{2}), sin(x^{1})cos(x^{2}), 0) 
This gives
so that
g_{**}  = 

(e) The nDimensional Sphere Let M be the nsphere of radius r with generalized polar coordinates.
(Notice that x^{1} is playing the role of and the x^{2}, x^{3}, . . . , x^{n1} the role of .) Following the line of reasoning in the previous example, we have
x^{1}  =  (r sin x^{1}, r cos x^{1} cos x^{2}, r cos x^{1} sin x^{2} cos x^{3 }, ... , r cos x^{1} sin x^{2}... sin x^{n1} cos x^{n}, r cos x^{1} sin x^{2} ... sin x^{n1} sin x^{n}) 
x^{2}  =  (0, r sin x^{1} sin x^{2}, . . . , r sin x^{1} cos x^{2} sin x^{3}... sin x^{n1} cos x^{n}, r sin x^{1} cos x^{2} sin x^{3} ... sin x^{n1} sin x^{n}) 
x^{3}  =  0, 0, r sin x^{1} sin x^{2} sin x^{3}, r sin x^{1} sin x^{2} cos x^{3} cos x^{4} . . . , r sin x^{1} sin x^{2} cos x^{3 }sin x^{4}... sin x^{n1} cos x^{n}, r sin x^{1} sin x^{2} cos x^{3 }sin x^{4} ... sin x^{n1} sin x^{n}) 
and so on. This gives
so that
g_{**}  = 

(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 = (P_{ji}) such that
PGP^{T}  = 

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
_{ij}  =  ^{i} 
g_{ab}  ^{j} 
=  P_{ia}g_{ab}P_{jb}  =  P_{ia}g_{ab}(P^{T})_{bj} = (PGP^{T})_{ij} 
showing that, at the point p,
_{**}  = 

Thus, in the eyes of the metric, the unit basis vectors e_{i} = /^{i} are orthogonal; that is,
Note The nondegeneracy condition in Definition 6.1 is equivalent to the requirement that the locally defined quantities
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
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

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 = úáX for every contravariant vector field X and every scalar field ú. The expected inequality
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 x^{i} = x^{i}(t) is nonnull if dx^{i}/dt^{2} 0. It follows that dx^{i}/dt^{2} is either always positive ("spacelike") or negative ("timelike").
Definition 6.4 If C is a nonnull 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
where the sign ±1 is chosen as +1 if the curve is spacelike and 1 if it is timelike. In other words, we are defining the arclength differential form by

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 nonnull path x^{i} = x^{i}(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, dx^{i}/ds^{2} is constant, and equals 1 if C is spacelike and 1 if it is timelike. Conversely, if t is any parameter with the property that dx^{i}/dt^{2} = ±1, then, choosing any parameter value t = a in the above definition of arclength s, we have
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 E_{2} such that the corresponding metric tensor g_{ij} is not constant.
2. Let a_{ij} be the components of any symmetric tensor of type (0, 2) such that det(a_{ij}) is never zero. Define
Show that this is a smooth inner product on M.
3. Give an example to show that the
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(g_{ij}) 0 for some chart x, then = det(_{ij}) 0 for every other chart (at points where the changeofcoordinates is defined). [Use the property that, if A and B are matrices, then det(AB) = det(A)det(B).]
6. Suppose that g_{ij} is a type (0, 2) tensor with the property that g = det(g_{ij}) is nowhere zero. Show that the resulting inverse (of matrices) g^{ij} is a type (2, 0) tensor. (Note that it must satisfy g_{ij}g^{kl} = ^{k}_{i}^{l}_{j}.)
7. (Index lowering and raising) Show that, if R_{abc} is a type (0, 3) tensor, then R_{a}^{i}_{c} given by
R_{a}^{i}_{c} = g^{ib}R_{abc},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} = T^{i}_{k}X^{k}. What can be said about the columns of T in a given coordinate system x? (Note that the i^{th} column of T is the local vector field given by T(/x^{i}).)
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