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

For a smooth map between manifolds $\varphi :\mathcal{M}\to \mathcal{N}$, the four constructions go in the directions dictated by composition and the differential of a smooth map.

The pullback of a smooth function $f\in C^{\infty }(\mathcal{N})$ is ${\varphi }^{\ast }f=f\circ \varphi$, a function on $\mathcal{M}$. The pushforward of a curve $\lambda :I\to \mathcal{M}$ is ${\varphi }_{\ast }\lambda =\varphi \circ \lambda$, a curve in $\mathcal{N}$ with the same parameter interval.

For a tangent vector $V\in T_p\mathcal{M}$, its pushforward is ${\varphi }_{\ast }V=d{\varphi }_p(V)\in T_{\varphi (p)}\mathcal{N}$. Intrinsically, treating a tangent vector as a derivation on functions,For a covector $\eta \in T_{\varphi (p)}^{\ast }\mathcal{N}$, the pullback of a covector is the dual linear map:No inverse map is required. A pushed-forward field for a general map is a field along that map; it need not assign a unique vector to an image point with several preimages.

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.

The two coordinate tangent vectors of the embedding areTheir Euclidean inner products give the pullback of a Riemannian metric, equivalently the induced metric:Locally set $u=\rho \phi$. The line element becomes $d{u}^2+d{\hat{z}}^2$, so the induced Riemann curvature tensor vanishes identically. This is the intrinsic flatness of a circular cylinder. Its bending in the ambient space is extrinsic curvature, not intrinsic curvature; periodicity of the angular coordinate does not change the local flatness.

Choose the outward normal vector $n=(\cos \phi ,\sin \phi ,0)$ and the positive conventionThe ambient connection vanishes in Cartesian coordinates. Since ${\partial }_{\phi }n=e_{\phi }/\rho$ and ${\partial }_{\hat{z}}n=0$, the extrinsic curvature components are $K_{\phi \phi }=\rho$, $K_{\phi \hat{z}}=K_{\hat{z}\hat{z}}=0$. Taking the trace using the induced metric,Reversing the normal or using the negative extrinsic-curvature convention gives $-1/\rho$. The two principal curvatures are $1/\rho$ and zero in the chosen convention; the averaged mean curvature would be $K/2$, so it must not be confused with the requested trace.