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 indices, where is the dimension. Because there are only coordinates, antisymmetrizing over more than slots will automatically vanish (i.e. more than vectors must be linearly-dependent in a -dimensional vector space).

This particular identity comes from antisymmetrizing over in the expression:

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

Obviously this will produce 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 ; similarly, the second Weyl tensor vanishes unless the lower indices are two of . 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 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 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 with , 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 -dimensional vector space and the matrix with indices in and . Then the Cayley-Hamilton theorem for 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 copies of , and an antisymmetrization over indices.

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

An example makes this a bit easier to see. Let’s expand this for 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 . 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 (easiest proved for diagonalizable matrices by going into the diagonal basis) where is the i’th eigenvalue. Then you will readily verify that these are in fact the elementary symmetric polynomials of the eigenvalues of , 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 ! (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