It is well known to differential geometers that there are many symmetries of the Riemannian curvature tensor , namely:

- , and
- (Bianchi identity)

In yesterday’s Riemannian geometry lesson, professor wanted to prove (2) above from (1) and (3). Of course this is more on algebra than the geometry itself. But then Bean Ng and I found ourselves having fun playing with those symmetries, and we discovered the following diagram (which certainly is not original, perhaps is due to Milnor) which I shall call B-diagram (B for Bianchi, or Bean):

Before explaining the B-diagram, let’s explain what we want to do. Basically we want to prove that , using the two other identities (1) and (3). For simplicity we will denote by etc. i.e we want to show

With some hindsight, the strategy is, basically, that if we want to show , which is difficult to prove directly, we then try to “symmetrize” it, using the known symmetries as much as possible. More precisely, if we want to show , and we know that

perhaps it’s a good idea to show instead that which, by our symmetry that,

Back to our case, we want to show that , but by (1), we know that and , so perhaps we could try to show that

We haven’t used the Bianchi identity, this is used in the last step by adding zero terms (by Bianchi) to the equation (1) above to see if there’s any cancelation. This is explained in the B-diagram above.

We want to show that the entries at the 4 vertices (the boxed terms) add up to zero. We know that the 3 terms in the upper left triangle cancels, due to the Bianchi identity (3), similarly for the other 3 triangles. We then look for cancelation in the “unboxed terms”. Thanks to the symmetries in (1), we see that the terms connected by the “broken arrows” cancel each others, e.g. . Thus all terms in the B-diagram sum up to zero, and by the above remark, that the four “boxed terms” also sum up to zero, this is (1), which is what we want to prove.

Good. I understand.