# How many independent equations are contained in the Bianchi identity?

In consideration of the various symmetries of the Riemann curvature tensor, how many independent equations are contained in the Bianchi identity $$R_{rsmn|t}+R_{rsnt|m}+R_{rstm|n}=0$$ ?

Symmetries of Riemann tensor:

1. $$R_{rsmn}=-R_{srmn}$$
2. $$R_{rsmn}=-R_{rsnm}$$
3. $$R_{rsmn}=R_{mnrs}$$
4. $$R_{rsmn}+R_{rmns}+R_{rnsm}=0$$
• As an upper limit, we have ${d \choose 2}$ choices for $\{r,s\}$ and ${d \choose 3}$ for $\{m, n, t\}$, which works out to at most $\frac{1}{12} d^2 (d-1)^2 (d-2)$ equations. ($d$ denotes the number of dimensions.) But this doesn't take into account the identities (3) and (4). Mar 23, 2022 at 14:58
• Thanks, that's the hard part. Wondering if any text has the info I seek or if anyone has calculated this. Mar 23, 2022 at 16:03
• I've posted an answer below, but for the reasons mentioned in the disclaimer I would recommend you wait some time before accepting it. Mar 23, 2022 at 17:42

Disclaimer: I think the argument below is valid, but I don't often use coordinate-based proofs in GR, and it's possible I've made a conceptual error. I welcome corrections or comments telling me that I'm an idiot.

The first derivative of a tensor field with the symmetries of the Riemann tensor should have $$\frac{1}{12} d^3 (d^2-1)$$ independent components (the number of components of the Riemann tensor times $$d$$, the dimension of the manifold.) So if we can figure out how many independent components the tensor $$\nabla_a R_{bcde}$$ actually has, then the difference between these two numbers must be the number of constraints imposed by the (second) Bianchi identity.

To do this, we can use a counting argument, similar to a standard argument for the local flatness theorem (see, for example, §6.2 of Schutz's First Course in General Relativity). In general, when we apply a coordinate transformation, the nth derivatives of the metric at a point $$P$$ are modified by various combinations of the $$(n+1)$$th derivatives of the new coordinate functions $${x'}^{\mu}(x^\nu)$$. Since under diffeomorphism invariance we can specify these derivatives freely, we can use this freedom to set many (but not necessarily all) of the derivatives of the metric to zero. Any "leftover" derivatives of the metric must correspond to physically meaningful quantities.

So, for example, we can set all $$d {d +1 \choose 2}$$ ($${}=40$$ when $$d = 4$$) derivatives of the form $$\partial_\rho g_{\mu \nu}$$ to zero via a local coordinate transformation, since there are $$d {d +1 \choose 2}$$ second derivatives of the coordinate functions ($$\partial^2 {x'}^\mu/\partial x^\nu \partial x^\rho$$) that we can choose freely. This then implies that we can always choose a set of coordinates in the neighborhood of a point $$P$$ such that the Christoffels vanish at $$P$$, which is the crux of the local flatness theorem. Similarly, we can't choose coordinates such that all the second derivatives of the metric vanish; this is because there are $${d +1 \choose 2} {d +1 \choose 2}$$ second derivatives of the metric $$\partial_\rho \partial_\sigma g_{\mu \nu}$$ but only $$d {d +2 \choose 3}$$ independent third derivatives $$\partial^3 {x'}^\mu/\partial x^\nu \partial x^\rho x^\sigma$$ of our coordinate functions. So that means that there are $${d +1 \choose 2} {d +1 \choose 2} - d {d +2 \choose 3} = \frac{d^2(d^2-1)}{12}$$ second derivatives of the metric that are physically meaningful, and these correspond to the components of the Riemann tensor.

To apply this logic to your question, then: by similar logic, there must be $${d + 2 \choose 3} {d +1 \choose 2} - d {d+3 \choose 4} = \frac{d^2(d^2 - 1)(d+2)}{24}$$ independent third derivatives of the metric that we cannot eliminate via a coordinate transformation. These will manifest as independent components of the tensor $$\nabla_a R_{bcde}$$. But from above, we would expect this tensor to have $$\frac{1}{12} d^3 (d^2-1)$$ components if it was unconstrained. This means that the differential Bianchi identity $$\nabla_{[a} R_{bc]de} = 0$$ must constrain $$\frac{d^3 (d^2-1)}{12} - \frac{d^2(d^2 - 1)(d+2)}{24} = \boxed{\frac{d^2 (d^2 - 1)(d-2)}{24}}$$ combinations of these derivatives, meaning that it corresponds to that number of independent equations. When $$d$$ takes on the values 2 through 5, this means that the Bianchi identity corresponds to 0, 3, 20, and 75 independent equations respectively.

• This seems wrong. In vacuum case ($R_{ab}=0$), linearized Bianchi identity is equivalent to wave equation for spin-2 field. This is most easily seen in spinor form, where Bianchi identity is $\nabla_{AA'}\Psi^{ABCD}=0$ (See Penrose & Rindler, vol 1, 4.10). Mar 23, 2022 at 18:48
• @A.V.S. Hmm. I'm not familiar with that result, but you could be right. It does seem like the quantity $\nabla_a R_{bcde}$ should depend on third derivatives of the metric when expressed in a local coordinate system, though. Maybe all the third derivatives cancel out from $\nabla_{[a} R_{bc]de}$? Mar 23, 2022 at 19:00
• I wrote a Mathematica program that uses brute force methods to eliminate the symmetries. I'll link it here soon. The result exactly matches your boxed answer, checked up to d=7. Indeed a genius answer Mr. Seifert! Mar 23, 2022 at 20:51
• Here you go: bit.ly/bianchi-identity ;this is a Mathematica 12.0 notebook. Mar 25, 2022 at 20:07

There is a somewhat systematic way to determine the number of independent components of any tensor expression with symmetries using Young diagrams. A good reference that explains the relevant formulas and manipulations is Georgi's book "Lie algebras in particle physics." An online reference can be found here.

Young diagrams are a way to describe how to decompose tensors into irreducible components by symmetrizing and antisymmetrizing various collections of indices. For the Riemann tensor $$R_{bcde}$$, the full set of symmetries can be summarized by saying that it has Young symmetry

This diagram means the corresponding tensor should be symmetrized along the rows (indices $$bd$$ and $$ce$$), and then antisymmetrized along the columns (indices $$bc$$ and $$de$$).

For the Bianchi identity, we are interested in the tensor $$\nabla_a R_{bcde}$$, which as a tensor lives in a tensor product representation of a standard vector with a tensor with symmetries of Riemann. Decomposing such a tensor into irreducible representations involves applying the branching rules, which are simple to learn and are described in the above references. The relevant branching rule for this problem is

where the left hand side represents the tensor symmetries of $$\nabla_a R_{bcde}$$, while the right hand side gives its decomposition into irreducible representations. For the Bianchi identity, we want to examine $$\nabla_{[a}R_{bc]de}$$. Since this is antisymmetric on three indices, it must correspond to a Young diagram with a column with three boxes. This is clearly the second diagram shown above, so we see the Bianchi identity deals with one irreducible component of $$\nabla_a R_{bcde}$$.

Finally, we need a way of determining the dimension of the representation. The general formula for this dimension discussed in the above references goes as follows: beginning in the upper left corner of the Young diagram, label the first box with the dimension of the vector representation ($$d$$). As you move to the right, increase the label by $$1$$, and as you move down, decrease the label by $$1$$. So in $$d=4$$, the labeled diagram of interest would be

The dimension of the representation is then the product of all terms in the labeled diagram, divided by the product of the Hook lengths of the diagram, which for the above diagram is $$4\cdot 3\cdot 2\cdot 1\cdot 1 = 24$$. Applying this formula in arbitrary spacetime dimension yields

$$\frac{d^2(d^2-1)(d-2)}{24},$$

in agreement with the answer by @michael-seifert.

Hence, the Bianchi identity can be viewed as saying the tensor $$\nabla_a R_{bcde}$$ has Young symmetry

and using the above dimension formula, the number of independent components is $$\frac{d^2(d^2-1)(d+2)}{24}$$

• Thanks, looks like a wonderfully systematic method. I'll take some time to study this. Mar 25, 2022 at 20:18