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Let $\Sigma$ be a $n-1$ dimensional space-like submanifold of a $n+1$ dimensional space-time $(V,g)$ and let $x \in \Sigma$. Then $(T_x \Sigma)^\perp$ is of dimension $2$ and is time-like. Such a $2$-plane as $(T_x \Sigma)^\perp$ cuts the future lightcone at $x$ along two linearly independent future null directions. Let $l ^+$ and $l^-$ be two future null vectors along these directions.

Let $e_a$ with $a \in {2,3,4,5,...}$ be a reference frame on $\Sigma$ and hence orthogonal to $l^+$ and $l^-$. Then the following sequence of identities hold,

$K^+_{ab} = K^+(e_a,e_b) = g(l^+,\nabla _{e_a} e_b) = g(e_b,\nabla _ {e_a} l^+) = \nabla _b l^+_a$

  • By metric compatibility and the fact that $l^+$ and $e_b$ are orthogonal I would have expected a minus between the last but two steps. Why is there no negative sign?

and similarly,

$K^-_{ab} = \nabla _b l^-_a$

If ${h_{ab}}$ is the induced metric on the submanifold $\Sigma$ then the {\bf Null Mean Curvature} of $\Sigma$ is defined as,

$\chi ^+ = h^{ab}K^+_{ab}$

$\chi ^- = h^{ab}K^-_{ab}$

Now suppose that $\Sigma$ is an oriented $n-1$ submanifold of a space-like n-submanifold $M$ of $V$. Because of the orientability assumption one can make a continuous choice of unit normal $\nu$ to $\Sigma$ (and tangent to $M$) distinguishing the interior and exterior of $\Sigma$ in $M$. Also define $n$ as the future unit normal to $M$. Choose $\nu$ to be pointing outwards with respect to $\Sigma$ and $n$ to be pointing outwards with respect to $M$. Then one can choose $l^+$ and $l^-$ such that $l^+ = n + \nu$ and $l^- = n - \nu$ ensuring $l^+_\mu l^{-\mu} = -2$.

The metric ${h_{ab}}$ defined on $\Sigma$ can be lifted to a metric $h$ (calling it the same by abuse of notation) on $V$ by extending it trivially along the two orthogonal directions.

  • Then apparently the following equation is true which I do not understand,

$h^{\mu \nu} = g^{\mu \nu} + \frac{1}{2}(l^{+^\mu} l^{-\nu} + l^{-^\mu} l^{+^\nu})$

(where $\mu$ and $\nu$ run over coordinates of $V$ unlike $a$ and $b$ which run over coordinates of $\Sigma$)

How does one see that the lift to $V$ ($h_{\mu \nu}$) of the pulled back metric $h_{ab}$ and the original metric on $V$ ($g_{\mu \nu}$) are related thus?

Now by parallel translation extend $l^+$ and $l^-$ to be tangent field to geodesic congruences. Then the following equation will be true for any $\nu$, $l^{+-\mu}\nabla _{\mu} l^{+-}_\nu = 0$. Further since these vector fields are also of constant norm one has for any $\nu$, $l^{+-\mu}\nabla {\nu} l^{+-}\mu = 0$.

Using the above one gets the following identity for the expansion $\theta = \nabla _\mu l^{+\mu}$ of the geodesic congruence to which $l^+$ is tangent,

$\nabla_\mu l^{+\mu} = g^{\mu \nu}\nabla_\mu l^{+}\nu = h^{\mu \nu} \nabla\mu l^{+}_\nu$

  • Now I don't understand why $h^{ab}\nabla_a l^+b = h^{\mu \nu} \nabla\mu l^+_\nu$
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