There is a nice 4-dimensional Weyl identity that I can never seem to remember off the top of my head; so I decided I need to write this note so I don’t have to remember it in the future. The identity is:

This identity comes in handy for simplifying the equations of motion of Einstein-dilaton-Gauss-Bonnet gravity.

So, how do we work out this identity so we don’t have to remember it? I gave a hint when I said this is a dimension-dependent identity: it’s specific to dimension 4. Most¹ dimension-dependent identities come from antisymmetrizing some expression over more than $d$ indices, where $d$ is the dimension. Because there are only $d$ coordinates, antisymmetrizing over more than $d$ slots will automatically vanish (i.e. more than $d$ vectors must be linearly-dependent in a $d$-dimensional vector space).

This particular identity comes from antisymmetrizing over $[abcdf]$ in the expression:

Obviously this will produce $5!=120$ terms in a sum, but the vast majority of them vanish because Weyl is tracefree on every pair of indices.

In fact, up to index order, there are only 5 combinations that are not obviously vanishing. The first Weyl tensor vanishes unless the lower indices are two of $\{c,d,f\}$; similarly, the second Weyl tensor vanishes unless the lower indices are two of $\{a,b,f\}$. This allows us to enumerate all of the possibilities in short order. After applying the antisymmetry of Weyl on the latter two indices, there are exactly 5 index permutations of $\{abcdf\}$ that are allowed, and they are (lexicographically):

All that remains to do is to contract the $\delta$ indices and figure out the signature of each permutation to get the sign correct. Doing so gives the identity \eqref{eq:theIdentity} (with an index raised and indices renamed).

Of course, all of this is much easier with the xAct/xTensor package. I highly recommend this to find e.g. the Riemann form of the above identity. It comes from the same expression as \eqref{eq:antisymmetrize} but replacing $C$ with $R$, and now there are many more combinations that do not vanish, but instead produce Ricci terms. Explicitly, we get the unwieldy

While I was hunting for my lost identity, I came across a nice paper² on the more general topic of dimension-dependent identities. This included, for example, the fact that the Cayley-Hamilton theorem can be derived from a dimension-dependent identity. Specifically, consider an $n$-dimensional vector space $V$ and the matrix $T^a{}_b$ with indices in $V$ and $V^*$. Then the Cayley-Hamilton theorem for $T$ can be written as

where there are $n$ copies of $T$, and an antisymmetrization over $n+1$ indices.

This doesn’t obviously look like the Cayley-Hamilton theorem, but: it is a linear combination of various matrix powers of $(T^k)^a{}_b$, with coefficients determined by various traces of other powers of $T$ in a very specific way (they turn out to be elementary symmetric polynomials of the eigenvalues of $T$).

An example makes this a bit easier to see. Let’s expand this for $n=4$ dimensions. Again, I recommend using xTensor. For shorthand, let me write the matrix power

and denote the trace with $[T] \equiv T^a{}_a$. Then the four-dimensional Cayley-Hamilton theorem says

What are all those strange combinations of traces? Well, use the identity that $[T^k] = \sum_{i=1}^n \lambda_i^k$ (easiest proved for diagonalizable matrices by going into the diagonal basis) where $\lambda_i$ is the i’th eigenvalue. Then you will readily verify that these are in fact the elementary symmetric polynomials of the eigenvalues of $T$, explicitly:

So this is indeed the characteristic polynomial of $T$! (Aside: equations such as these are examples of Newton-Girard identities).

Footnotes