# Simplifying the trace of the extrinsic curvature tensor

I am following a derivation of the Gibbons-Hawking action term, and in it the trace of the extrinsic curvature $$K_{ab}$$ is found by

\begin{align} K & = {n^\alpha}_{; \alpha} \\ & = g^{\alpha \beta} n_{\alpha ; \beta} \\ & = \left (\epsilon n^\alpha n^\beta + h^{\alpha \beta} \right ) n_{\alpha ; \beta} \\ & = h^{\alpha \beta} n_{\alpha ; \beta} \\ & = h^{\alpha \beta} (n_{\alpha, \beta} - \Gamma^\gamma_{\alpha \beta} n_\gamma). \end{align}

Why did we say that $$n^\alpha n^\beta n_{\alpha;\beta}=0$$? I could now show this no matter how I tried to go about it. Could somebody please help me, I’m sure it’s a silly thing for I can’t think of it for some reason?

Why did we say that $$n^\alpha n^\beta n_{\alpha;\beta}=0$$?
Because you can write \begin{align} n^\alpha n^\beta n_{\alpha;\beta} &= n^\beta\frac{1}{2}(n^\alpha n_{\alpha;\beta}+n^\alpha n_{\alpha;\beta})\\ & = n^\beta\frac{1}{2}(n^\alpha n_{\alpha;\beta}+n_\alpha {n^\alpha}_{;\beta})\\ & = n^\beta\frac{1}{2}(n_\alpha n^\alpha)_{;\beta}\\ & = n^\beta\frac{1}{2}\epsilon_{,\beta}=0 \end{align} And the last term is $$0$$ because $$\epsilon=\pm1$$, therefore $$\partial_\beta\epsilon=0$$.