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We can try to generalise the above results from two variables u, v to n variables u1, ..., un for any positive integer n. A manifold M in parametric form is the locus of points (in some $ \mathbb {R}$k) obtained as the image of a vector-valued function x(u1,..., un). Moreover, to ensure that we have n ``directions of freedom'' we also require the derivatives xu1, ..., xun to be a linearly independent set at every point; we use Tp(M) to denote the linear span of these vectors which is called the tangent space to M at p. (For those who know more, this is a local manifold and not a global one).

If (u1(t),..., un(t)) is an n-tuple of functions of one variable we obtain a curve in M. The velocity and acceleration of this curve as a curve in $ \mathbb {R}$k are

v = xu1u1, t + ... + xunun, t and  
a = $\displaystyle \sum_{i,j}^{}$xuiujui, tuj, t$\displaystyle \sum_{i}^{}$xuiui, tt  

The vector v is in the tangent space to M. We need to understand the projection of acceleration into the tangent space to see whether a curve is indeed ``curved'' in M or is a geodesic.

Exercise 19   Let v1, ..., vn be linearly independent vectors. Show that the matrix whose ij-th entry is gij = vi . vj is a positive definite symmetric matrix. Let hij denote the matrix entries of its inverse. The orthogonal projection of a vector a to the linear span of v1, ..., vn is given by $ \sum_{i,j}^{}$hij(a . vi)vj.

Thus we see that the orthogonal projection of the acceleration into the tangent space of M is given by

$\displaystyle \sum_{i,j,k,l}^{}$(hkl(xuiuj . xuk)ui, tuj, t)xul + $\displaystyle \sum_{i}^{}$ui, ttxui

Where we will use gij = (xui . xuj) and hij for the matrix entries of the inverse of the matrix (gij). Now we have the following equations for the derivatives of gij

(gij)uk = (xuiuk . xuj) + (xui . xujuk)

We ``solve'' these to obtain

(xuiuj . xuk) = (1/2)((gik)uj + (gjk)ui - (gij)uk)

Finally, we can put

$\displaystyle \Gamma_{ij,l}^{}$ = $\displaystyle \sum_{k}^{}$hkl(1/2)((gik)uj + (gjk)ui - (gij)uk)

so that the components of acceleration along xuk are given by $ \sum_{i,j,k}^{}$$ \Gamma_{ij,k}^{}$ui, tuj, t + uk, tt.

First of all we note that the $ \Gamma_{ij,k}^{}$ and hence the acceleration of a curve on M can be computed one we are given the gij as functions of ui. We no longer need the ``crutch'' of the vector-valued functions x(ui). Thus, the geometry of the manifold M is entirely described by the matrix-valued function (gij) with values in positive-definite matrices. This function gives the speed of a curve (ui(t)) as $ \sum$gijui, tuj, t and the accleration is given by the formula above.

Exercise 20   Use the geodesic normal form of the surface as given above and compute the speed of a curve (u(t), v(t)) to be

ut2 + vt2 + (kl /3)(uvt - vut)2

modulo order 3 terms in u and v.

Thus the curvature can also be obtained by comparing the speed of curves with the ``usual'' speed in the (u, v) plane. We shall see that this is the key idea used by Riemann to understand curvature in many dimensions.

Secondly, we note that the equation for a curve on M with 0 accleration can easily be solved inductively given the starting point and the velocity at that point. In other words given (u1, 0,..., un, 0) and (u1, 1,..., un, 1) we consider a curve given by the functions ui(t) = ui, 0 + ui, 1t + .... By comparing the coefficients of tr in the equation $ \sum_{i,j}^{}$$ \Gamma_{ij,k}^{}$ui, tuj, t + uk, tt = 0, we can inductively solve for ui, 2, ui, 3 and so on.

Let p be a fixed point on M and choose a collection of vectors e1, ...en so that (gij) becomes the identity matrix with respect to this basis. For each n-tuple (v1,..., vn) we consider the geodesic $ \gamma_{(v_1,\dots,v_n)}^{}$(t) (upto some order in t) obtained as above, starting at p and proceeding along $ \sum_{i}^{}$viei. One easily sees that the functions

ui(v1,..., vn) = ui($\displaystyle \gamma_{(v_1,\dots,v_n)}^{}$(1))

give another parametric form (in the variables v1,...,vn) for the manfiold M (upto some order r). This form is called the geodesic normal form for M. We can re-express any function on M in terms of the new variables by substitution. Similarly, we can compute the speed of curves (v1(t),..., vn(t) in terms of a new matrix (Gij) which can easily be computed in terms of the gij and the expression of u's as functions of v's.

It is clear that this special parametric form should reflect the geometry of M. In fact, Riemman proved

Theorem 2   If the manifold M is expressed in the variables (v1, in geodesic normal form, and the expression for spped of a curve takes the form

$\displaystyle \sum_{ij}^{}$Gijvi, tvj, t = $\displaystyle \sum_{i}^{}$vi, t2 + $\displaystyle \sum_{i<j,k<l}^{}$cij, kl(vivj, t - vjvi, t)(vkvl, t - vlvk, t)

modulo order 3 terms in the variables vi.

The second term can be thought of as a quadratic form $ \sum_{i<j,k<l}^{}$cij, klzijzkl in a new set of variables zij. By diagonalising this quadratic form as earlier we obtain n(n - 1)/2 ``eigenvalues'' which are called the principal sectional curvatures of M in analogy with the two dimensional case given in the exercise above. Many interesting properties of curvature have been developed over the years. Riemann himself showed that the curvature is 0 at all points of M if and only if M is Euclidean. He also claimed that the functions on M given by the principal sectional curvatures uniquely determine the geometry of M. This claim has only been proved partially.

next up previous
Next: Suggested Reading Up: Coordinate Geometry Previous: Surfaces
Kapil H. Paranjape 2001-01-20