Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 24H (Differential Geometry)

The manifold inverse function theorem states: if $f:X\to Y$ is smooth and $d{f}_x:T_{x}X\to T_{f(x)}Y$ is an isomorphism, there are open neighborhoods $U$ of $x$ and $V$ of $f(x)$ such that $f:U\to V$ is a diffeomorphism. In particular the manifolds have the same dimension at those points.

Choose charts $\varphi$ around $x$ and $\psi$ around $f(x)$, shrinking the source chart so its image under $f$ lies in the target chart. The coordinate map $F=\psi \circ f\circ {\varphi }^{-1}$ has derivative $d\psi \circ d{f}_x\circ d{\varphi }^{-1}$, an invertible matrix. Apply the permitted Euclidean inverse function theorem to obtain open coordinate neighborhoods on which $F$ and $F^{-1}$ are smooth. Pull them back through the charts. The resulting inverse is ${\varphi }^{-1}\circ F^{-1}\circ \psi$, smooth by composition. This proves the manifold statement.

Differentiate the homogeneity identity with respect to $t$ at one to obtain Euler's identityOn $X_a$, its right side is $ma\ne 0$, so $d{p}_x\ne 0$. Thus $a$ is a regular value. The preimage theorem says that the nonempty inverse image of a regular value of a smooth map $M^k\to N^l$ is a smooth submanifold of codimension $l$. Applied to $p:{\mathbb{R}}^k\to \mathbb{R}$, it gives $\dim X_a=k-1$.

For $a,b>0$, dilation by $t=(b/a{)}^{1/m}$ maps $X_a$ to $X_b$, because $p(tx)=t^{m}p(x)=b$. Its inverse is dilation by $t^{-1}$, so all nonempty positive level manifolds are diffeomorphic.

A negative level need not be diffeomorphic to a positive one, even when both are nonempty. For $p(x,y,z)=x^2+y^2-z^2$, $X_1$ is connected (parametrize it by $(\sqrt{1+z^2}\cos \theta ,\sqrt{1+z^2}\sin \theta ,z)$), whereas $X_{-1}$ has two components, according to $z=\pm \sqrt{1+x^2+y^2}$. Thus the answer for opposite signs is no. If $m$ is odd, the additional symmetry $x\mapsto -x$ does interchange opposite levels up to scaling, but the question allows even degrees.