 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014-resit/12/vii/solution

Every nonzero component of the n-dimensional Levi-Civita symbol is

\pm 1

, so its square is one. There are exactly

n!

n

distinct indices; all remaining index lists have a repeat and contribute zero. Hence

i_{1}, \dots, i_{n} = 1 \sum n ϵ_{i_{1} \dots i_{n}}^{2} = n! .

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