In the class I am teaching I tried to count number of independent components of the Riemann curvature tensor accounting for all the symmetries. We’ll call it RCT in this note.

It turned out to be not so straightforward, so I decided to write it down here.

First of all, here are the three symmetries/skew-symmetries of :

,

,

.

The first two expressions tell us that is skew-symmetric in the first two and the last two indices. And form the last we can deduce that is symmetric in the pairs and .

There is an interesting fact behind these identities. Listen to it carefully, is a **symmetric bilinear form** on wedge-products and . In some sense it is a bilinear form on a pair of planes. Each defines a plane spanned by . When we swap and the wedge product changes sign ( is skew-symmetric). Thus we have a minus in the first identity above. Similar situation is with the second identity. The third identity corresponds to swapping the planes and . It is a symmetric form on planes, so the sign doesn’t change.

Another set of symmetries is the Bianchi identity, involving cyclic permutations:

**(*)** .

Here the curvature tensor is with the raised index. We are using the definition .

I have found two ways to compute number of independent components of RCT. Let us consider the first one.

In -dimensional space there are possible values for not counting the symmetries.

Consider , where and are each a pair of indices: .

The number of different ‘s that correspond to different values of RCT is (pairs and will define the same value, upto a sign; a pair will make ). The same for ‘s. So far we took care of the first two identities. Denote by $\latex D:=N(N-1)/2$. The third identity tells us that is symmetric in , thus we have distinct entries of RCT. Or, in terms of ,

.

Now we need to subtract the number of distinct equations **(*)**, since each equation will reduce the number of independent values of RCT by one.

This part is not so trivial. It turns out, by lowering the index and using all the anti-/symmetries one can write equation **(*)** as

, which means the sum over all permutations of indices with a sign, determined by a parity of a permutation. The number of equations of the type is equal to the number of ways one can choose 4 **distinct** indices from . It is equal to .

The final answer to the number of independent components of the RCT is

.

Which is equal to (after some algebra)

.

Ooof, that was long. I will describe the second method of obtaining the same number briefly.

It basically consists of evaluating how many ways you can choose four indices out of , and then, based on identities, how many different ‘s that corresponds to.

If all the indices are equal than . Not interesting. If there are only two distinct indices , they correspond to only one value, say . All others are zeros or equal to . So there are $latex\binom{N}{2}$ ways of choosing 2 distinct indices out of , thus there are $latex\binom{N}{2}$ distinct values of RCT, so far. For 3 distinct indices , there are $latex\binom{N}{3}$ possible values. And the fourth index should be a repeated index, it is either , or or . So values of RCT so far.

Now, there are ways of choosing 4 distinct indices. The three symmetries reduce this number by 8 (each one halves). And the Bianchi identity is a famous example of “*buy 2 get third free!*“. In other words, if we know two of the values of RCT, then the third one is determined by the Bianchi identity. So we need to multiply the last number by 2/3, i.e. we have .

Finally,

.

Which yields exactly the same value.

P.S. I became a bit dizzy after typing so many times unusual plural of index. Should I have used indexes? WordPress underlines word *indices* as wrong.

## 4 comments

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April 17, 2011 at 3:45 pm

TinosThanks! I couldn’t work out how to handle the Bianchi identity. The best I managed was to times by 2/3, which takes away too much. NC4 is genius!

April 18, 2011 at 2:02 am

sergeyskThank you for your reply, Tinos. I don’t know who NC4 is, so he might as well be as clever as you described 🙂

August 13, 2011 at 2:36 pm

PSHello Sergeysk

The final formula for the number of independent components is fine. My problem is this. Suppose n=2., i.e. we have 2-dimensional space-time. nc4 = n!/(n-4)!4! makes no sense, i.e. what is (-2)!? The first derivation/proof surely only applies to the case n>=4.

Similarly, the second derivation/proff is only valid if n>=3 due to the presence of the nc3 = n!/(n-3)!3! term.

Please advise. Your opinion is apprecated and welcome.

PS

October 16, 2011 at 1:55 pm

David A. de Wolf, (emeritus professor ECE)The part about including the Bianchi identities is not clear. Here is an improvement:

R(ijkl) + R(iljk) + R(iklj) = 0 Now exchange j k to get

R(ikjl) + R(ilkj) + R(ijlk) = 0 Subtract

R(ijkl) + R(iljk) + R(iklj) – R(kjl) – R(ilkj) – R(ijlk) = 0

This equation contains i and then all 6 distinct permutations of jkl. Even permutations with + sign, odd ones with – sign. But i (not equal to j,k,l) is arbitrarily one of the N values so each of these equations with 6 terms occurs for each distinct permutation of N values in ijkl. And thus there are N(N-1)(N-2)(N-3).4! extra constraints due to the Bianchi identities.

–DAdW–