# Note on a (dimension-dependent) Weyl identity

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:

\begin{align} \label{eq:theIdentity} C_{a}{}^{cde}C_{bcde} = \frac{1}{4}g_{ab} C^{cdef}C_{cdef} \,. \end{align}

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:

\begin{align} \label{eq:antisymmetrize} 0 = C^{ab}{}_{[ab}C^{cd}{}_{cd}\delta^e{}_{f]} \,. \end{align}

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):

\begin{align*} C^{ab}{}_{cd}C^{cd}{}_{ab}\delta^e{}_{f} \,, C^{ab}{}_{cd}C^{cd}{}_{af}\delta^e{}_{b} \,, C^{ab}{}_{cd}C^{cd}{}_{bf}\delta^e{}_{a} \,,\\ C^{ab}{}_{cf}C^{cd}{}_{ab}\delta^e{}_{d} \,, C^{ab}{}_{df}C^{cd}{}_{ab}\delta^e{}_{c} \,. \end{align*}

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

\begin{align} R_{ebcd} R_{f}{}^{bcd} = \frac{1}{4} g_{ef} R_{abcd}R^{abcd} - g_{ef} R_{ab}R^{ab} + 2 R^{bc} R_{ebfc} + 2 R_{eb} R_f{}^b + \frac{1}{4} g_{ef} R^2 - R R_{ef} \,. \end{align}

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

\begin{align} T^{i_1}{}_{[i_1} T^{i_2}{}_{i_2} \cdots T^{i_n}{}_{i_n} \delta^{a}{}_{b]} = 0 \,, \end{align}

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

\begin{align} (T^k)^a{}_b \equiv T^a{}_{i_1} T^{i_1}{}_{i_2} \cdots T^{i_{k-1}}{}_b \end{align}

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

\begin{align} 0 ={}& 5 T^{i_1}{}_{[i_1} T^{i_2}{}_{i_2} T^{i_3}{}_{i_3} T^{i_4}{}_{i_4} \delta^{a}{}_{b]} \\ 0={}& (T^4)^a{}_b - (T^3)^a{}_b [T] + (T^2)^a{}_b \tfrac{1}{2} ([T]^2 - [T^2]) \\ &{}+ T^a{}_b \tfrac{1}{6} (-[T]^3 + 3 [T] [T^2] - 2 [T^3]) \nonumber\\ &{}+ \delta^a{}_b \tfrac{1}{24} ([T]^4 - 6 [T]^2 [T^2] + 3 [T^2]^2 + 8 [T] [T^3] - 6 [T^4]) \,. \nonumber \end{align}

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:

\begin{align*} \tfrac{1}{24} ([T]^4 - 6 [T]^2 [T^2] + 3 [T^2]^2 + 8 [T] [T^3] - 6 [T^4]) &= \lambda_1 \lambda_2 \lambda_3 \lambda_4 \\ \tfrac{1}{6} ([T]^3 - 3 [T] [T^2] + 2 [T^3])&= \lambda_1 \lambda_2 \lambda_3 + \lambda_1 \lambda_2 \lambda_4 + \lambda_1 \lambda_3 \lambda_4 + \lambda_2 \lambda_3 \lambda_4 \\ \tfrac{1}{2} ([T]^2 - [T^2]) &= \lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_1 \lambda_4 + \lambda_2 \lambda_3 + \lambda_2 \lambda_4 + \lambda_3 \lambda_4 \\ [T] &= \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 \end{align*}

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

# Footnotes

1. I think there may be more complicated ones which come from Garnir relations, but I know nothing about these.

2. Edgar and Höglund, J.Math.Phys. 43 (2002) 659-677 arXiv:gr-qc/0105066

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