\begin{align*}
\mathrm{\mathbf{(a)}} \quad U_{\pm} \cdot U_{\pm} &= \dfrac{1}{2} (n_a n^a \pm 2 n_a m^a + m_a m^a) = \pm n_a m^a = 0 \\
U_+ \cdot U_- &= \dfrac{1}{2} (n_a n^a - m_a m^a) = \dfrac{1}{2} (-1-1) = -1 \\ \\
\mathrm{\mathbf{(b)}} \quad P^a_b &= \delta^a_b + U_{\mp}^a (U_{\pm})_b + U_{\pm}^a (U_{\mp})_b \\
&= \delta^a_b + \dfrac{1}{2} (n^a \mp m^a)(n_b \pm m_b) + \dfrac{1}{2} (n^a \pm m^a)(n_b \mp m_b) \\
&= \delta^a_b + n^a n_b - m^a m_b \\
&= h^a_b - m^a m_b \\ \\
\mathrm{\mathbf{(c)}} \quad \theta_{\pm} &= P^{ab} \nabla_a (U_{\pm})_b \\
&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) \nabla_a (n_b \pm m_b) \\
&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (K_{ab} - n_a n^c \nabla_c n _b) \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b) \\
&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) K_{ab} - \dfrac{1}{\sqrt{2}} \underbrace{h^{ab} n_a}_{= \, 0} n^c \nabla_c n_b + \dfrac{1}{\sqrt{2}} m^b \underbrace{m^a n_a}_{= \, 0} n^c \nabla_c n_b \\
&\hspace{45pt} \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b) \\ \\
&= \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) K_{ab} \pm \dfrac{1}{\sqrt{2}} (h^{ab} - m^a m^b) (k_{ab} - m_a m^c \nabla_c m _b)
\end{align*}Why is there a factor of ##\dfrac{1}{\sqrt{2}}##, and how do you re-write the second term?
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