Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 50 / 1 / b

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.

A type $(0,s)$ covariant tensor is a multilinear form on $s$ vectors. For arbitrary $V_1,\dots ,V_s\in T_p\mathcal{M}$, the pullback of a covariant tensor isProjection of all covariant slots means $(P\omega )(X_1,\dots ,X_s)=\omega (P{X}_1,\dots ,P{X}_s)$. Since every pushed-forward vector is already tangential,Equality on all arguments proves ${\varphi }^{\ast }\omega ={\varphi }^{\ast }(P\omega )$. No antisymmetry is needed: this holds for every covariant tensor, not only for differential forms.

The pushforward of a contravariant tensor applies $d{\varphi }_p$ to each of its $r$ vector slots. Choose a basis $e_i$ of $T_p\mathcal{M}$ and expandIts pushforward is $T^{i_1\cdots i_r}(d\varphi e_{i_1})\otimes \cdots \otimes (d\varphi e_{i_r})$. Each factor is tangent to the hypersurface and fixed by $P$. ConsequentlyThe basis expansion proves the result for arbitrary tensors, rather than only a single decomposable tensor product.