We know that a general finite rotation by an angle $\phi$ about an axis with unit normal $n$ (such that $n_kn_k=1$) is given by the transformation $\bar{x_i}=A_{ij}x_j$ where the matrix $A$ is represented by
$A_{ij}=\cos(\phi)\delta_{ij}+(1-\cos(\phi))n_in_j+\sin(\phi)\epsilon_{ijk}n_k$
so what I am trying to do is show that the matrix $A$ is orthogonal in the sense that $A_{ij}A_{il}=\delta_{jl}$. Clearly the first step is to expand the following:
$(\cos(\phi)\delta_{ij}+(1-\cos(\phi))n_in_j+\sin(\phi)\epsilon_{ijk}n_k)(\cos(\phi)\delta_{il}+(1-\cos(\phi))n_in_l+\sin(\phi)\epsilon_{ilp}n_p)$
which gives nine terms:
$A_{ij}A_{il}=\cos^2(\phi)\delta_{ij}\delta_{il}+\cos(\phi)(1-\cos(\phi))n_in_l\delta_{ij}+\cos(\phi)\delta_{ij}\sin(\phi)\epsilon_{ilp}n_p+(1-\cos(\phi))n_in_j\cos(\phi)\delta_{il}+(1-\cos(\phi))^2n_in_jn_in_l+(1-\cos(\phi))n_in_j\sin(\phi)\epsilon_{ilp}n_p+\sin(\phi)\epsilon_{ijk}n_k\cos(\phi)\delta_{il}+\sin(\phi)\epsilon_{ijk}n_k(1-\cos(\phi))n_in_l+\sin^2(\phi)\epsilon_{ijk}n_k\epsilon_{ilp}n_p$
and after some very tedious algebra I have reduced this to
$A_{ij}A_{il}=\delta_{jl}+\sin(\phi)(1-\cos(\phi))(n_in_jn_k\epsilon_{ilk}+n_in_ln_k\epsilon_{ijk})$
which is where my confusion lies. I have no idea how that last term becomes zero. Is there some reindexing trick we can implement that I have just missed?
I am quite new to dealing with tensors and would like to understand how to complete this proof. Any help would be greatly appreciated.
Thanks.
Consider the tensor $n_in_jn_k\epsilon_{i\ell k}$. Try interchanging indices $i\leftrightarrow k$. Since they are dummy indices, it should not change the value of the tensor. But since these indices are part of the anti-symmetric Levi-Civita, the value of the tensor should also flip: $n_in_jn_k\epsilon_{i\ell k}=n_kn_jn_i\epsilon_{k\ell i}=-n_kn_jn_i\epsilon_{k\ell i}$. But if something is equal to its own negative, then it must necessarily be $0$. There we go. Same idea applies to the other, similar term there.
In general, contracting a symmetric tensor with an anti-symmetric tensor will result in $0$.