Can these two terms cancel out? In trying to prove that $$\Gamma_{\mu\nu\lambda} = \eta_{ab}J^a_bJ^b_{\nu\lambda}.$$
The author canceled out while expanding the first equation $$J^a_{\mu\lambda}J^b_\nu$$ with $$J^b_{\mu\lambda}J^a_\nu$$ since they were carrying opposite signs, is this eligible? If so, why? 
Note: $$J^a_{\mu\lambda} := \frac{\partial^2\phi^a}{\partial x^\mu\partial x^\lambda}.$$
 A: I'm pretty sure I know the answer to this question, even though this question provides very little context for what the various tensors are, it gives a couple expressions without including the important surrounding equation for context, and it includes an equation with a typo.
First of all, as it stands, the first equation has unbalanced indices.  I assume the first equation is actually supposed to be
$$\Gamma_{\mu\nu\lambda} = \eta_{ab}{J^a}_{\mu}{J^b}_{\nu\lambda}\ .$$
Although I don't understand what the $J$'s are, all that matters that I understand correctly is that $\eta_{ab}$ is the metric tensor (presumably the Minkowski metric, because $\eta$ is commonly used to denote the Minkowski metric, although the Latin indices instead of Greek indices is unusual).
The first equation is a type (0, 3) tensor on both sides, but ${J^a}_{\mu\lambda}{J^b}_\nu$ and ${J^b}_{\mu\lambda}{J^a}_\nu$ are type (2, 3) tensors, so clearly they are getting contracted with something in the whole equation that they are in.  Given the form of the RHS of the first equation, my presumption is that the expression in which the cancellation occurs is something like
$$\eta_{ab}({J^a}_{\mu\lambda}{J^b}_\nu - {J^b}_{\mu\lambda}{J^a}_\nu + …)\ .$$
In that context, it's valid for those two terms to cancel out, but the reason that they cancel out is because the metric tensor is always symmetric.  We have
$$\eta_{ab}{J^a}_{\mu\lambda}{J^b}_\nu=\eta_{ba}{J^b}_{\mu\lambda}{J^a}_\nu=\eta_{ab}{J^b}_{\mu\lambda}{J^a}_\nu\ ,$$
where the first step is just renaming the indices, and the second step is due to the metric tensor being symmetric.  So
$$\eta_{ab}({J^a}_{\mu\lambda}{J^b}_\nu - {J^b}_{\mu\lambda}{J^a}_\nu) = 0\ ,$$
no matter what the $J$'s are denoting.
