# How to Prove contracted Bianchi Identity

1. May 18, 2007

### jojoo

How to prove
$g^{im} \nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}R_{lk}$.

of cause $g^{im}R_{ilkm}=R_{lk}$, but I don't know how the contraction can pass through the covariant derivative?

Last edited: May 18, 2007
2. May 18, 2007

### Hurkyl

Staff Emeritus
Because derivatives and (finite) sums commute. e.g.

$$\frac{d}{dx} \sum_{i = a}^b f_i(x) = \sum_{i = a}^b \frac{df_i}{dx}(x)$$

3. May 18, 2007

### robphy

The derivative operator is usually taken to be "metric compatible", i.e.,
$$\nabla_a g_{bc}=0$$.

4. May 18, 2007

### jojoo

Yes, I know that. But $$\nabla_{\partial_j}R_{ilkm}=(\nabla_{\partial_j}R)(\partial_i,\partial_l,\partial_k,\partial_m)$$ How can the contraction really happen?(since $$\nabla_a g_{bc}=0$$ means $$(\nabla_a g)(\partial_b,\partial_c)=0$$)
Would you like to give me more detail? Thank you!

Last edited: May 18, 2007
5. May 18, 2007

### Chris Hillman

This is OT, but George Jones just pointed out that today's issue of the daily paper in Toronto featured a picture of the uncontracted Bianchi identities. For some reason a politician is in the foreground.

Industrious students can look for papers pointing out that Bianchi himself credited these identities to someone else.

6. Jun 8, 2007

### neorayner

$g^{im}\nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}(g^{im}R_{ilkm}) - R_{ilkm}\nabla_{\partial_j}g^{im}=\nabla_{\partial_j}R_{ lk}$

because $$\nabla_a g_{bc}=0$$

it respects the Leibniz derivation

Last edited: Jun 8, 2007