Copasetic Flow

What's Gone Before and What Will Ensue :  Yesterday, he first step of an exercise regarding the rotation of cross products was worked out.  Today, the identity proved yesterday will be used to show that when the two vectors in a cross product are rotated by the same rotation matrix, the resulting vector of the cross product is rotated by the same rotation matrix.  In the end, yet another property will be proven using tensor index notation. The identity from yesterday is: $\epsilon_{ijk}W_{iq}W_{jl}W_{km} = det\left(W\right)\epsilon_{qlm}$ We'll also need the definition of the cross product in index notation $\vec{A} \times \vec{B} = \epsilon_{ijk}A_jB_k$ and the rotation matrix transpose identity $M^TM = 1$ also known as $M_{iq}M_{in} = \delta_{qn}$ We want to prove that if $\vec{A}$ and $\vec{B}$ are both rotated by the same rotation matrix, $M_{in}$, then so is the result of the cross product, $\vec{V}$ First, rotate the two input vectors $\vec{A^{\prime}...

Summary of what's gone on before   The use of index notation to indicate a transpose was explained and shown with a concrete example.  Today, work on old homeworks begin.  The big issue of the day was figuring out an elegant way of showing that a contracted product was antisymmetric. The insights about commuting terms in index based products and how the Levi-Civita symbol works were worth the effort, but it was in fact a lot of effort!  There's a heap of broken attempts to get the short answer at the bottom of the post. The first part of the rotated cross product problem is to show that $W_{il}W_{jm}W_{kn}\epsilon_{lmn} = det\left(W\right)\epsilon_{lmn}$ We're to do this by first showing that the left hand side is antisymmetric with respect to the $i$, $j$, and $k$ indices and therefore proportional to $\epsilon_{lmn}$ and then by showing that for a concreted example, $i=1$, $j=2$, $k=3$, the left hand side is equal to $det\left(W\right)$ There's a long way ...