Setup: Let $u: (V,g) \rightarrow (M,h)$ be a map where
- $V$ is a Lorentz manifold of dimension $n+1$, $V= R \times S$ where $S$ is a $n$ dimensional smooth oriented manifold with metric $\sigma$. On $V$, we have a Robertson Walker metric $g=-dt^2+R^2(t) \sigma$
- $M$ is a Riemannian manifold.
Now if we have local coordinates $x_{\alpha}$ on $V$ and $y^A$ on $M$, we can define $du \in \Gamma(V, T^{*}V \otimes u^{*}TM)$,
$$ du= \partial_{\alpha} u^A [dx_{\alpha} \otimes u^{*} \partial_A] $$
We also use the notation
$$d_{\alpha} u= \partial_{\alpha} u^A u^{*} \partial_A$$
It then holds that $h( d_{\alpha} u, d_{\beta} u) \in C^{\infty}(V)$.
Now look at an expression of the kind:
$$\sigma^{ij}[(\nabla_i^{\sigma} (h(d_tu, d_ku) dx_k) )(\partial_j)]$$
Now it says that this is, for each $t$, a "space divergence in the metric $\sigma$" and therefore
$$\int\limits_{S_t} \sigma^{ij}[(\nabla_i^{\sigma} (h(d_tu, d_ku) dx_k) )(\partial_j)]=0$$
Here, $S_t=\{t\} \times S$ is a submanifold of $V$.
Comments on notations:
- $\partial_t$ is the basis tangent vector on $R$
- $t$ is the index used for $R$ and other latin indices are used for $S$
My question: I don't see why this is true. I was told to apply Stokes theorem but I don't see how and why I am allowed to do so.
Write In general for a one form $\alpha = \alpha_i \mathrm dx^i$, then
$$A = \sigma^{ij} \nabla_i \alpha_j$$
is the divergence of $\alpha$, which can also be written as
$$X = -*d *\alpha, $$ where $*$ is the Hodge star operator. So
$$ \int _S X \; \mathrm dV = -\int_S * ( d*\alpha \mathrm dV) = \int_S d* \alpha$$
and from there one can apply Stokes theorem.