# Quick one-line on Tensor Contraction

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1. Feb 28, 2015

### unscientific

What do they mean by 'Contract $\mu$ with $\alpha$'? I thought only top-bottom indices that are the same can contract? For example $A_\mu g^{\mu v} = A^v$.

2. Feb 28, 2015

### strangerep

It means "replace $\mu$ by $\alpha$ -- or vice versa".

3. Feb 28, 2015

### unscientific

How can we simply do that?

4. Feb 28, 2015

### strangerep

Suppose you have an ordinary matrix. If I told you to "take the trace of that matrix", what would that mean to you?

5. Feb 28, 2015

### unscientific

Summing up the diagonal, i.e. $\sum M_{ii}$.

6. Feb 28, 2015

### strangerep

Correct.

So if I have a matrix equation like $M_{ij} = 0$, then it is also true that $\sum_i M_{ii} = 0$, or in summation convention notation, $M_{ii} = 0$. One says that we have contracted $i$ with $j$, though a slightly more helpful phrase might be to say that we have contracted over $i,j$.

Similarly in your OP, except that it deals with contraction over 2 indices of a 4th rank tensor instead of a 2nd rank matrix.

7. Feb 28, 2015

### unscientific

Do you mind showing the working how they 'contracted the indices'? I think it's not so simple as cancelling them.

8. Feb 28, 2015

### samalkhaiat

$$R^{\alpha}{}_{\beta \alpha \nu} = \delta^{\mu}_{\alpha} R^{\alpha}{}_{\beta \mu \nu} = g_{\alpha \gamma} g^{\mu \gamma} R^{\alpha}{}_{\beta \mu \nu}$$

9. Feb 28, 2015

### strangerep

Indeed, it is not index "cancellation".

Multiply both sides of the original equation by the 2 factors of $g$ indicated on the rhs of Samalkhaiat's post #8. Then the only remaining "trick" is realizing that you can pass them through the covariant derivative operators -- since the metric is assumed to covariantly constant in GR, e.g., $\nabla_\lambda g_{\alpha\gamma} = 0$, etc.

10. Mar 1, 2015

### unscientific

So, the working is

$$\nabla_\gamma R^\mu _{\nu \alpha \beta} + \nabla_\beta R^\mu _{\nu \gamma \alpha} + \nabla_\alpha R^\mu _{\nu \beta \gamma} = 0$$

$$\delta_\mu ^\alpha \nabla_\gamma R^\mu _{\nu \alpha \beta} + \delta_\mu ^\alpha \nabla_\beta R^\mu _{\nu \gamma \alpha} + \delta_\mu ^\alpha \nabla_\alpha R^\mu _{\nu \beta \gamma} = 0$$

$$\nabla_\gamma g_{\mu \gamma} g^{\alpha \gamma} R^\mu _{\nu \alpha \beta} + \nabla_\beta g_{\mu \gamma} g^{\alpha \gamma} R^\mu _{\nu \gamma \alpha} + \nabla_\alpha g_{\mu \gamma} g^{\alpha \gamma} R^\mu _{\nu \beta \gamma} = 0$$

$$\nabla_\gamma R^\alpha _{\nu \alpha \beta} + \nabla_\beta R^\alpha _{\nu \gamma \alpha} + \nabla_\alpha R^\alpha _{\nu \beta \gamma} = 0$$

$$\nabla_\gamma g^{\nu \gamma} R^\alpha _{\nu \alpha \beta} + \nabla_\beta g^{\nu \gamma} R^\alpha _{\nu \gamma \alpha} + \nabla_\alpha g^{\nu \gamma} R^\alpha _{\nu \beta \gamma} = 0$$

$$2 \nabla_\nu R^\nu _\beta + \nabla_\beta g^{\nu \gamma} R^\alpha _{\nu \gamma \alpha} = 0$$

Shouldn't it be a $+$ on the second term: $2 \nabla_\nu R^\nu _\beta + \nabla_\beta R = 0$?

11. Mar 1, 2015

### Bartolius

No, because R is defined to be
$$R^\alpha _\alpha = g^{\nu \gamma} R^\alpha _{\nu \alpha \gamma}$$
and you can obtain that form by exchange of the last two indices, hence the minus sign