# Trace and divergence of the Bach tensor

Let $$(M, g)$$ be a compact Riemannian manifold of dimension $$n \geq 4$$.

In Besse's book, the Bach tensor is defined as the gradient of the functional

$$SW(g) = \int_M |W(g)|_g^2 d\mu^g$$

meaning that, for any symmetric 2-tensor field,

$$SW(g + t h) = SW(g) + t \int_M \langle B, h\rangle_g d\mu^g + O(t^2)$$

As a consequence, this symmetric 2-tensor is

• Divergence-free as one has $$SW(\phi_t^* g) = SW(g)$$ for any 1-parameter family of diffeomorphisms $$\phi_t$$. So choosing for $$\phi_t$$ the flow of some vector field $$X$$, we get $$\phi_t^* g = g + t L_X g + O(t^2)$$. As $$SW(\phi_t^* g) = SW(g) + t \int_M \langle B, L_X g\rangle_g d\mu^g + O(t^2),$$ we conclude that $$\int_M \langle \mathrm{div} B, X\rangle_g d\mu^g = \int_M \langle B, L_X g\rangle_g d\mu^g = 0$$ for any vector field $$X$$.
• Trace-free in dimension 4.This comes from the fact that $$SW$$ is conformally invariant in dimension $$4$$ so one can repeat the previous argument with the 1-parameter family of metric $$g_t = e^{\lambda u} g$$ (for an arbitrary function $$u$$).

Now the formula given for the Bach tensor (see e.g. the Wikipedia page of the Bach tensor, but this is very similar to the formula in Besse's book) is

$$B_{ab} = P^{cd} W_{acbd} + \nabla^c (\nabla_c P_{ab} - \nabla_a P_{cb})$$ (here $$P$$ is the Schouten tensor)

This formula for $$B$$ gives a trace-free tensor in any dimension. This follows immediately as the Weyl tensor is trace-free and as the second term is the divergence of the Cotton tensor, so, up to some coefficient, this is the double divergence of the Weyl tensor. This is independent of the dimension.

The divergence, however, can be computed explicitely as $$\nabla^a B_{ab} = P^{cd} (\nabla_c P_{db} - \nabla_b P_{cd})$$ which is a priori non-zero.

I do not understand where my mistake is. Any help would be appreciated.

## 1 Answer

The Bach tensor is only divergence-free in dimension four. In general dimension, it holds that $$\nabla^j B_{ij} = (n-4)P^{jk} ( \nabla_i P_{jk} - \nabla_j P_{ik} )$$. I pulled this formula from the bottom of Page 1966 in Robin Graham's article "Extended obstruction tensors and renormalized volume coefficients".

Without seeing your computation, it is impossible to say where your mistake is. The basic idea of the computation is that:

(a) the divergence of $$P^{kl}W_{ikjl}$$ gives $$(n-3)P^{kl}C_{kli} + \frac{1}{2}C^{klj}W_{jkil}$$, where $$C_{kli}=\nabla_i P_{kl} - \nabla_lP_{kl}$$ is the Cotton tensor; this is just using the relation between the divergence of the Weyl tensor and the Cotton tensor that you noted; and

(b) the divergence of the other term gives $$-P^{kl}C_{kli} - \frac{1}{2}C^{klj}W_{jkil}$$, as follows from commuting derivatives (which you seem to be doing in your computation).

• Thank you for pointing this paper! I guess the divergence depends on the normalization you use: my calculation gives the same result up to some coefficient. As calculations are fairly boring I did not include them. Feb 3, 2021 at 12:13
• This then means that the Bach tensor is not the gradient of the integral of the squared norm of the Weyl tensor, right? Feb 3, 2021 at 12:22
• Correct; the realization of the Bach tensor as the gradient only works in dimension four. As far as I can tell, the generalization of the Bach tensor to other dimensions is motivated by the Fefferman--Graham ambient metric. Feb 3, 2021 at 13:05