Page 170 - Intro to Tensor Calculus

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I 20. Find the number of independent components associated with the Riemann Christoffel tensor
4
R ijkm , i,j,k,m =1, 2,... ,N. There are N components to examine in an N−dimensional space. Many of
these components are zero and many of the nonzero components are related to one another by symmetries
or the cyclic properties. Verify the following cases:
CASE I We examine components of the form R inin , i 6= n with no summation of i or n. The first index
can be chosen in N ways and therefore with i 6= n the second index can be chosen in N − 1 ways. Observe
that R inin = R nini , (no summation on i or n) and so one half of the total combinations are repeated. This
leaves M 1 = 1 N(N − 1) components of the form R inin . The quantity M 1 can also be thought of as the
2
number of distinct pairs of indices (i, n).
CASE II We next examine components of the form R inji , i 6= n 6= j where there is no summation on
the index i. We have previously shown that the first pair of indices can be chosen in M 1 ways. Therefore,
1
the third index can be selected in N − 2 ways and consequently there are M 2 = N(N − 1)(N − 2) distinct
2
components of the form R inji with i 6= n 6= j.
CASE III Next examine components of the form R injk where i 6= n 6= j 6= k. From CASE I the first pairs
of indices (i, n) can be chosen in M 1 ways. Taking into account symmetries, it can be shown that the second
1
1
pair of indices can be chosen in (N −2)(N −3) ways. This implies that there are N(N −1)(N −2)(N −3)
2 4
ways of choosing the indices i, n, j and k with i 6= n 6= j 6= k. By symmetry the pairs (i, n)and (j, k)can be
interchanged and therefore only one half of these combinations are distinct. This leaves
1
N(N − 1)(N − 2)(N − 3)
8
distinct pairs of indices. Also from the cyclic relations we find that only two thirds of the above components
are distinct. This produces
N(N − 1)(N − 2)(N − 3)
M 3 =
12
distinct components of the form R injk with i 6= n 6= j 6= k.
Adding the above components from each case we find there are

2
2
N (N − 1)
M 4 = M 1 + M 2 + M 3 =
12
distinct and independent components.
Verify the entries in the following table:

Dimension of space N 1 2 3 4 5
Number of components N 4 1 16 81 256 625
M 4 = Independent components of R ijkm 0 1 6 20 50
Note 1: A one dimensional space can not be curved and all one dimensional spaces are Euclidean. (i.e. if we have
2
2
an element of arc length squared given by ds = f(x)(dx) , we can make the coordinate transformation
p 2 2
f(x)dx = du and reduce the arc length squared to the form ds = du .)
Note 2: In a two dimensional space, the indices can only take on the values 1 and 2. In this special case there
are 16 possible components. It can be shown that the only nonvanishing components are:

R 1212 = −R 1221 = −R 2112 = R 2121 .