Revolving a $k$-manifold around an axis gives a $(k+1)$-manifold

Question

I want to solve the following problem from M. Spivak's Calculus on Manifolds:

Let $\mathbb{K}^n=\{x \in \mathbb{R}^n:x^1=0 \text{ and }x^2>0,\dots,x^{n-1}>0\}$. If $M \subseteq \mathbb{K}^n$ is a $k$-dimensional manifold, and $N$ is obtained by revolving $M$ around the axis $x^1=\cdots=x^{n-1}=0$, show that $N$ is a $(k+1)$ dimensional manifold. Example: the Torus (Figure 5-4).

My attempt:

At first I considered the case $n=3$, where $\mathbb{K}^3=\{(x,y,z) \in \mathbb{R}^3:x=0,y>0\}$.

for a point $\mathbf{x}$ not on the $z$ axis consider the angle $\theta(\mathbf{x})$ which is the one between the vector projection of $\mathbf{x}$ to the $[xy]$ plane and the positive $x$-axis (this is the angle $\theta$ from polar coordinates).

Now, since $M$ is $1$-dimensional manifold, for each $p \in M$ there exists open sets $U_p \ni p,V_p \subseteq \mathbb{R}^3$ and a diffeomorphism $h_p:U_p \to V_p$ such that $h_p(U_p \cap M)=V_p \cap (\mathbb{R}^1 \times \{0\}^2 )$.

Let $q \in N$ be some point. Define $$k_q(\mathbf{x}):=[R_z(\theta(\mathbf{x})) \circ h \circ R_z(-\theta(\mathbf{x}))](\mathbf{x})$$

where $R_z$ is the rotation matrix around the $z$-axis, and a branch of $\theta$ is chosen so that it is smooth around $q$. Say $p$ is the (unique) point in $\mathbb{K}^3$ such that $q$ is the result of some rotation of it around the $z$-axis. If $U_p$ is taken

to be a sufficiently small ball around $p$ (so that it doesn't intersect the $z$-axis), I claim that $k_q$ is a diffeomorphism with domain $\overline{U}_q:=R_z(\theta(q))[U_p]$ and codomain $\overline{V}_q:=k_q(\overline{U}_q)$. I also claim that

$$k_q(\overline{U}_q \cap N)=\overline{V}_q \cap (\mathbb{R}^2 \times \{0\}) .$$

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Now, trying to generalize this to arbitrary $n$ seems difficult to me, as I have no idea how rotations in $\geq 4$ dimensions work.

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My questions are:

1. Is the proof for $n=3$ valid? If not, please help me correct it.

2. How can one prove the general case?

Thank you!

Answer 1

I am about 99.9% sure that the problem statement is wrong.

1. The question makes no sense for $n = 2$.
2. The orbits of the "revolution" around a one dimensional axis in $\mathbb{R}^n$ are $n-2$ dimensional spheres. This indicates that for a $0$-dimensional manifold $M$, its rotation $N$ becomes $n-2$ dimensional which for $n > 3$ is strictly larger than $1 = k+1$.

I am about 99.9% sure that the correct problem statement is the following:

Let $n \geq 2$ and $\mathbb{K}^n = \{ x\in \mathbb{R}^n| x^1 = 0, x^2 > 0\}$ and let $M\subset \mathbb{K}^n$ be an $k$-dimensional submanifold. Show that the set $N$ obtained by revolving $M$ around the axis $x^1 = x^2 = 0$ is a $k+1$ dimensional manifold.

Note that when $n = 3$ the two formulations of the problems agree. Note further that this version makes sense also for $n = 2$. The notion of revolution is clear once you restrict to a co-dimension 2 axis: it is the action by the matrices

$$ R_{12}(\theta) = \begin{pmatrix} \cos \theta & \sin\theta & 0 & 0 & \cdots & 0 \\ - \sin\theta & \cos\theta & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & & 0 \\ \vdots & \vdots & & &\ddots & 0 \\ 0 & & \cdots & & 0 & 1\end{pmatrix} $$

and the proof in 3D basically carries over to arbitrary dimensions with very little modification. (In fact, you can start with the proof in 2D and carry it up to arbitrary dimensions.)

(One particularly simple proof uses the the "cylindrical" coordinates on $\mathbb{R}^n \setminus \{x^1 = x^2 = 0\}$ which is the same as the usual Euclidean coordinate system except $(x^1,x^2)$ is replaced by $(r = \sqrt{(x^1)^2 + (x^2)^2}, \theta = \arctan x^2/x^1)$.)