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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{Levi-Cevita Tensor Homework 2} \\ (January 25, 2013) \\
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\noindent 1. Use the identity
\begin{eqnarray} \label{epsdel}
  \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} =
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}
\end{eqnarray} 
to eliminate the vector products from the expression
\begin{eqnarray} 
  \vec{a}\times\left(\vec{b}\times\vec{c}\right)
\end{eqnarray} 
Solution (compare book p.33):
\begin{eqnarray} \nonumber
  &~&\vec{a}\times\left(\vec{b}\times\vec{c}\right)  \\ \nonumber
  &=& \sum_i\sum_j\sum_k\sum_l\sum_m \epsilon_{ijk} \hat{x}_i a_j
      \epsilon_{klm} b_l c_m \\ \nonumber
  &=& \sum_i\sum_j\sum_k\sum_l\sum_m \epsilon_{kij} \epsilon_{klm} 
      \hat{x}_i a_j b_l c_m \\ \nonumber
  &=& \sum_i\sum_j\sum_l\sum_m \left(\delta_{il}\delta_{jm} -
      \delta_{im}\delta_{jl}\right)\hat{x}_ia_jb_lc_m \\ \nonumber
  &=& \vec{b}\,\left(\vec{a}\cdot\vec{c}\right) -
      \vec{c}\,\left(\vec{a}\cdot\vec{b}\right)\,. 
\end{eqnarray} 

\noindent 2. Use the definition
\begin{eqnarray} 
  \left(\vec{b}\times\vec{c}\right)_i\ =\
  \sum_j\sum_k \epsilon_{ijk} b_j c_k 
\end{eqnarray} 
of the $i^{th}$ component of the vector product $\vec{b}\times\vec{c}$
to proof
\begin{eqnarray} 
  \vec{a}\cdot\left(\vec{b}\times\vec{c}\right) =
  \vec{b}\cdot\left(\vec{c}\times\vec{a}\right) =
  \vec{c}\cdot\left(\vec{a}\times\vec{b}\right)\,.
\end{eqnarray} 
Solution (compare book p.30):
\begin{eqnarray} \nonumber
 &~&\vec{a}\cdot\left(\vec{b}\times\vec{c}\right) =
    \sum_i\sum_j\sum_k \epsilon_{ijk}a_ib_jc_k \\ \nonumber
 &=&\sum_j\sum_k\sum_i \epsilon_{jki}b_jc_ka_i =
    \vec{b}\cdot\left(\vec{c}\times\vec{a}\right) \\ \nonumber
 &=&\sum_k\sum_i\sum_j \epsilon_{kij}c_ka_ib_j =
    \vec{c}\cdot\left(\vec{a}\times\vec{b}\right)\,.
\end{eqnarray} 

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