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A non-null hypersurface projection annihilates the unit normal vector and fixes tangent vectors to the hypersurface. It obeys $P^2=P$. Apply it to every tensor index to remove normal components, with the metric used to lower or raise the index pair for covariant slots. For a timelike unit normal in mostly-plus signature it is the spatial projection tensor. A null normal cannot be normalized this way, and a null hypersurface requires additional choices for a corresponding projection.

This question uses proper cosmological time $t$. The future-directed unit normal isThe non-null hypersurface projection removes all normal components. The only relevant Christoffel symbols are ${\Gamma }_{ij}^0=a\dot{a}{\delta }_{ij}$. Since the components of $n_{\mu }$ are constant,ConsequentlyWith the printed positive extrinsic curvature convention, this is the expansion of the normal comoving congruence. A fixed coordinate cell has proper volume proportional to $a^3$, so $K=d\log (a^3)/dt={\nabla }_{\mu }{n}^{\mu }$: the fractional proper-volume expansion rate.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

Write $\epsilon =n_a{n}^a\in \{-1,+1\}$. The non-null hypersurface projection is $P^a{}_b={\delta }^a{}_b-\epsilon n^a{n}_b$, reproducing both signs in the question. It annihilates the normal and is the identity on tangent vectors to the hypersurface.

Represent $V$ by the tangent to a curve $\lambda$ through $p$. The curve $\varphi \circ \lambda$ lies in $\Sigma$, so its tangent $d{\varphi }_{p}V$ is tangent to $\Sigma$. Orthogonality therefore gives $n_a({\varphi }_{\ast }V{)}^a=0$, andThis uses the stated smooth hypersurface assumption; the dimension relation alone would not make the image of an arbitrary smooth map a regular hypersurface.