Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 2

Take $n^a{n}_a=-1$. The first fundamental form, or induced metric, is the spatial projection tensorWith the convention used in the question, define the second fundamental form byLocally $n_a$ is proportional to the gradient of a defining function for the hypersurface, so $h_a{}^c{h}_b{}^d{\nabla }_{[c}{n}_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

Let $D$ be the Levi-Civita connection of $h$. Apply $(D_c{D}_d-D_d{D}_c)$ to a tangent vector, replace each $D$ by the tangential projection of $\nabla$, and decompose $\nabla n$ using $K_{ab}$. Comparison with the intrinsic Riemann curvature tensor gives the Gauss equationChanging the sign convention for $K_{ab}$ does not alter these quadratic terms.

Contract the Gauss equation with $h^{ac}{h}^{bd}$. Since $h^{ab}=g^{ab}+n^a{n}^b$ and the antisymmetries of the Riemann curvature tensor eliminate the term containing four normals,The extrinsic terms contract to $K^2$ and $K_{ab}{K}^{ab}$. Hence the intrinsic Ricci scalar is

Contract the Einstein field equations $G_{ab}=8\pi T_{ab}$ twice with $n^a$. Since $n^a{n}_a=-1$ and $\rho =T_{ab}{n}^a{n}^b$,If $K_{ab}$ is proportional to $h_{ab}$ in three dimensions, tracing gives $K_{ab}=(K/3)h_{ab}$ and $K_{ab}{K}^{ab}=K^2/3$. The Scalar Gauss equation gives the Hamiltonian constraintTherefore