# Tensor trace

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.3.

The trace of a rank-2 tensor is given by the contraction ${F_{\;\; i}^{i}}$. In matrix terminology, it is the sum of the diagonal elements. If we start with a contravariant tensor ${F^{ij}}$, then we can calculate the trace as follows:

 $\displaystyle F_{\;\; j}^{i}$ $\displaystyle =$ $\displaystyle g_{jk}F^{ik}\ \ \ \ \ (1)$ $\displaystyle F_{\;\; i}^{i}$ $\displaystyle =$ $\displaystyle g_{ik}F^{ik} \ \ \ \ \ (2)$

That is, we first lower the second index, then contract the top and bottom indices.

Since the trace contains no free index, it should be a scalar, which means it should be invariant. We can prove this by doing the transformation.

 $\displaystyle g_{ik}^{\prime}F^{\prime ik}$ $\displaystyle =$ $\displaystyle \frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{m}}{\partial x^{\prime k}}g_{lm}\frac{\partial x^{\prime i}}{\partial x^{a}}\frac{\partial x^{\prime k}}{\partial x^{b}}F^{ab}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\partial x^{l}}{\partial x^{\prime i}}\frac{\partial x^{\prime i}}{\partial x^{a}}\right)\left(\frac{\partial x^{m}}{\partial x^{\prime k}}\frac{\partial x^{\prime k}}{\partial x^{b}}\right)g_{lm}F^{ab}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{\;\; a}^{l}\delta_{\;\; b}^{m}g_{lm}F^{ab}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ab}F^{ab} \ \ \ \ \ (6)$