# Can I simply reverse the indices in a contraction?

Suppose I have something like

$$\left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu$$

Can since all the terms involving $\mu$ on the left and $\nu$ on the right are contractions, can I simply do:

$$\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu}$$

Seems like a pretty nifty trick.

• You can, except that you should keep track of what the position of the indices are when raised. So you should write $R^\nu{}_\beta$ rather than $R^\nu_\beta$. Jun 5, 2015 at 2:36
• But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via $\nabla_\mu (\nabla_\beta) V^\mu$ and not on the vector $V^\mu$. I thought for contractions, they have to act on one another? Jun 5, 2015 at 2:37
• No, that's not true. It is acting as $\nabla_\mu ( \nabla_\beta V^\mu )$. It is certainly acting on $V^\mu$. Similarly, the second term is $\nabla_\beta ( \nabla_\mu V^\mu)$. Jun 5, 2015 at 2:38
• Ok then if I have something like $F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, how what will $\nabla^\mu F_{\mu \nu}$ look like? Jun 5, 2015 at 2:55
• It will look like $\nabla^\mu ( \nabla_\mu A_\nu ) - \nabla^\mu ( \nabla_\nu A_\mu)$. It acts on everything to the right of it unless otherwise specified by explicit brackets. Did you want me to explicitly specify what this is? Jun 5, 2015 at 2:59

Typically yes, but generally no. That is, your exchange of index positions relies on something called metric compatibility. This is the assumption that the covariant derivative of the metric is zero: $$\nabla_\alpha g_{\beta \gamma} = 0.$$ This is actually a condition imposed on $\nabla$. It turns out [see, e.g., Wald p. 35] that for any metric there exists exactly one such $\nabla$ (assuming also, as Danu points out, that the covariant derivative is torsion free). So typically in GR, this assumption is made because it simplifies things so much -- as in your case.

So if you assume metric compatibility, then your expression is valid. In particular, from the fact that $g_{\beta \gamma} g^{\gamma \epsilon} = {\delta_\beta}^\epsilon$ is constant (so its derivative is zero), metric compatibility also implies $\nabla_\alpha g^{\beta \gamma} = 0$. Then, you can calculate \begin{align} \left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu &= \left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) \left( g^{\mu \tau} V_\tau \right) \\ &= \nabla_\mu \nabla_\beta\left( g^{\mu \tau} V_\tau \right) - \nabla_\beta \nabla_\mu \left( g^{\mu \tau} V_\tau \right) \\ &= \nabla_\mu \left( g^{\mu \tau} \nabla_\beta V_\tau \right) - \nabla_\beta \left( g^{\mu \tau} \nabla_\mu V_\tau \right) \\ &= g^{\mu \tau} \nabla_\mu \nabla_\beta V_\tau - \nabla_\beta \nabla^\tau V_\tau \\ &= \nabla^\tau \nabla_\beta V_\tau - \nabla_\beta \nabla^\tau V_\tau \\ &= \nabla^\mu \nabla_\beta V_\mu - \nabla_\beta \nabla^\mu V_\mu \end{align} I've used metric compatibility three times here, to move the metric from one side of a $\nabla$ to the other. I've also used the definition $\nabla^\alpha = g^{\alpha \beta} \nabla_\beta$.

That's the left-hand side, at least. As for the right-hand side, @Prahar rightly points out that you should try to be careful with the order of your indices, since a lot of tensors are not symmetric, and if you get into the habit of being sloppy with indices it can hurt. It so happens that the Ricci tensor $R_{\mu\nu}$ is symmetric, so you can get away with it in this particular case, but it's easy to be careful. Anyway, the right-hand side is straightforward: \begin{align} R_{\nu\beta} V^\nu &= R_{\nu\beta} g^{\nu \sigma} V_\sigma \\ &= g_{\nu \mu} {R^{\mu}}_{\beta} g^{\nu \sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} g_{\mu \nu} g^{\nu \sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} {\delta_{\mu}}^{\sigma} V_\sigma \\ &= {R^{\mu}}_{\beta} V_\mu \\ &= {R^{\nu}}_{\beta} V_\nu \end{align} Now, since the Ricci tensor is symmetric, you could have swapped the order of the indices on $R_{\nu\beta}$ at the beginning, and you would have come up with ${R_{\beta}}^{\nu} V_\nu$ at the end. This is the only reason you'll some times see those indices written directly above each other; if $R$ had been some asymmetric tensor, this wouldn't be true.

Finally, I'll also point out that these index swaps on the two different sides are independent of each other, so you can do either or both independently.

• In the second line, for the term on the right why did you raise $\mu$ to $\nabla^\mu$? Jun 5, 2015 at 3:10
• Because I was typing too fast...
– Mike
Jun 5, 2015 at 3:11
• From third line to fourth line, how did the term on the right go from $g^{\mu \tau} \nabla_\mu$ to $\nabla^\tau V_\tau$? Can you lower/raise indices on operators like $\nabla_\alpha$ too? Jun 5, 2015 at 3:13
• In general, you should be wary of raising and lowering indices on operators. But in this case, that's actually the definition of $\nabla^\tau$. (And because of possible metric incompatibility, it's important to remember which side the metric comes on in that definition.)
– Mike
Jun 5, 2015 at 3:17
• So as long as $\nabla_\alpha g^{\beta \gamma} = 0$ holds, I can raise/lower indices on operators using $g^{\alpha \beta}$? Jun 5, 2015 at 3:22

Sure! This comes from the way that raising and lowering of indices is defined in terms of the metric tensor. Suppose we have a rank $(0, 2)$ tensor $S_{ab}.$ Then its trace is $S^{a}{}_a.$ We may write:

$$S^{a}{}_{a} = g^{ab} S_{b a} = S_{b}{}^{b}.$$

Since $a$ and $b$ are dummy indices, we may interchange them readily. This yields the final equation

$$S^{a}{}_{a} = S_{a}{}^a.$$

So any time you have a contraction, you may switch which is the "up-index" and which is the "down-index" without changing your result.