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 Exercise [14.30] 
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Joined: 12 Jul 2010, 07:44
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Post Exercise [14.30]
Consider the Riemann tensor Rabcd in a particular coordinate system, on a Riemannian manifold of dimension n.
In this case, the total set of components is given by specifying all possible values for each of a, b, c, d.
Each of these can be a number from 1 to n, representing the different dimensions or coordinate axes.
(Alternatively you can number them from 0 to n-1, if you're counting from 0 - it makes no difference).

However, due to the symmetries:

Rabcd = -Rbacd = -Rabdc = Rcdab (antisymmetries, interchange symmetry)

and

Rabcd + Rcabd + Rbcad = 0 (Bianchi symmetry)

these components are not independent of each other.


One approach to finding the number of independent components of R is to first partition all its components according to 4 separate cases, and then find the number of independent components in each case.

The four cases I will consider divide the components of R up as follows:

  • Components with all four indices taking different values. (e.g. R1243)
  • Components with the four indices each taking one of three distinct values. (e.g. R1223)
  • Components with the four indices taking only two distinct values between them. (e.g. R3133, R1212)
  • Components with all four indices taking the same value. (e.g. R4444)


Consider first the case, where each of the a, b, c, d are different.

Due to the antisymmetry and interchange symmetries, the following sets of permutations of abcd yeild components that are all equal to each other up to a sign change:

[1] abcd, abdc, bacd, badc, cdab, cdba, dcab, dcab
[2] acbd, acdb, cabd, cadb, bdac, bdca, dbac, dbac
[3] adbc, adcb, dabc, dacb, bcad, bcda, cbad, cbda

Thus, of the 24 total possible permutations of indices, only 3 of the corresponding components are independent due to these symmetries. But now consider the Bianchi symmetry also (see above). The three terms on its LHS are from the above sets [1], [2], and [3] respectively. Since it allows any one of these terms to be expressed in terms of the other two, that eliminates one of these 3 components, leaving only two independent components remaining.

Thus, in the first case, for each possible choice of 4 distinct index values, there are exactly two independent components.

In the next case, two of the indices are equal, and I'll use an "x" to indicate the positions of the two equal indices. (e.g. Raxbx)

The possible permutations divide into two sets in this case:

[1] abxx, baxx, xxab, xxba
[2] axbx, axxb, xabx, xaxb, bxax, bxxa, xbax, xbxa

Due to antisymmetry, all components corresponding to the first set of permutations are zero. All components corresponding to the second set are equal up to a sign change, leaving only one independent component for each possible choice of a, b, and x. In this case, the Bianchi symmetry degenerates into (a combination of) the other symmetries, and so has no additional effect.

There is a slight subtlety here, of which note must be taken: The role of the x index is different from that of the a and b indices, and it is thus not interchangeable with them. That means that for each choice of 3 distinct index values together with a choice of which of them is the duplicated 'x' index value, there is one independent component. So for each choice of 3 distinct index values, there are actually three independent components.

Similarly in the next case, where we choose only two distinct index values (denoting the indices as "a" and "b" where their roles are interchangeable, and as "a" and "x" where not), the possible permutations divide into three sets:

[1] axxx, xaxx, xxax, xxxa
[2] aabb, bbaa
[3] abab, abba, baab, baba

Similarly to the previous case, all components corresponding to the first two sets are zero, and components corresponding to the third set are equal up to a sign change. Thus there's only one independent component for each choice of two distinct index values.

Finally, components of the form Raaaa are all zero, by antisymmetry, so there are no independent components in the final case.


Summary so far:
  • 2 independent components are associated with each possible choice of four distinct index values.
  • 3 independent components are associated with each possible choice of three distinct index values.
  • 1 independent component is associated with each possible choice of two distinct index values.
  • There are no other independent components.


Now as the manifold is n-dimensional, the number of ways of choosing k distinct index values is nCk, where:


nCk = n! / k!(n-k)!


Thus, for a given dimension n, the number of independent components of the Riemann tensor

= 2(nC4) + 3(nC3) + nC2


...which some simple algebra will show

= n^2 * (n^2 - 1) / 12


as given in the book.

This evaluates to 20 when n=4, as required.


QED


Last edited by deant on 14 Nov 2010, 17:42, edited 1 time in total.

14 Nov 2010, 15:46
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