Stewart-Walker lemma

The Stewart-Walker lemma provides necessary and sufficient conditions for the linear of a tensor field to be gauge-invariant. $Delta\; delta\; T\; =\; 0$ if and only if one of the following holds

1. $T\_\{0\}\; =\; 0$

2. $T\_\{0\}$ is a constant scalar field

3. $T\_\{0\}$ is a linear combination of products of delta functions $delta\_\{a\}^\{b\}$

Derivation

A 1-parameter family of manifolds denoted by $mathcal\{M\}\_\{epsilon\}$ with $mathcal\{M\}\_\{0\}\; =\; mathcal\{M\}^\{4\}$ has metric $g\_\{ik\}\; =\; eta\_\{ik\}\; +\; epsilon\; h\_\{ik\}$. These manifolds can be put together to form a 5-manifold $mathcal\{N\}$. A smooth curve $gamma$ can be constructed through $mathcal\{N\}$ with tangent 5-vector $X$, transverse to $mathcal\{M\}\_\{epsilon\}$. If $X$ is defined so that if $h\_\{t\}$ is the family of 1-parameter maps which map $mathcal\{N\}\; o\; mathcal\{N\}$ and $p\_\{0\}\; in\; mathcal\{M\}\_\{0\}$ then a point $p\_\{epsilon\}\; in\; mathcal\{M\}\_\{epsilon\}$ can be written as $h\_\{epsilon\}(p\_\{0\})$. This also defines a pull back $h\_\{epsilon\}^\{*\}$ that maps a tensor field $T\_\{epsilon\}\; in\; mathcal\{M\}\_\{epsilon\}$ back onto $mathcal\{M\}\_\{0\}$. Given sufficient smoothness a Taylor expansion can be defined

:$h\_\{epsilon\}^\{*\}(T\_\{epsilon\})\; =\; T\_\{0\}\; +\; epsilon\; ,\; h\_\{epsilon\}^\{*\}(mathcal\{L\}\_\{X\}T\_\{epsilon\})\; +\; O(epsilon^\{2\})$

$delta\; T\; =\; epsilon\; h\_\{epsilon\}^\{*\}(mathcal\{L\}\_\{X\}T\_\{epsilon\})\; equiv\; epsilon\; (mathcal\{L\}\_\{X\}T\_\{epsilon\})\_\{0\}$ is the linear perturbation of $T$. However, since the choice of $X$ is dependent on the choice of gauge another gauge can be taken. Therefore the differences in gauge become $Delta\; delta\; T\; =\; epsilon(mathcal\{L\}\_\{X\}T\_\{epsilon\})\_\{0\}\; -\; epsilon(mathcal\{L\}\_\{Y\}T\_\{epsilon\})\_\{0\}\; =\; epsilon(mathcal\{L\}\_\{X-Y\}T\_\{epsilon\})\_\{0\}$. Picking a chart where $X^\{a\}\; =\; (xi^\{mu\},1)$ and $Y^\{a\}\; =\; (0,1)$ then $X^\{a\}-Y^\{a\}\; =\; (xi^\{mu\},0)$ which is a well defined vector in any $mathcal\{M\}\_\{epsilon\}$ and gives the result

:$Delta\; delta\; T\; =\; epsilon\; mathcal\{L\}\_\{xi\}T\_\{0\}.,$

The only three possible ways this can be satisfied are those of the lemma.

Sources

* Describes derivation of result in section on Lie derivatives