Extending metrics from $M =\mathbb{T}^2 \times (-\pi , \pi)$ to $ \mathbb{T}^3$

Question

I would like to know of a similar result for the below but for the torus:

$\textbf{Cylinder to sphere rule:}$ Let $0< w \leq \infty$, and let $g$ be a metric on the topological cylinder $(-w, w) \times S^n$ of the form $$g = \phi(z)^2 dz^2 + \psi(z)^2 g_{can}$$ where $\phi,\psi : (-w, w) \to \mathbb{R_+}$ and $g_{can}$ is the canonical round metric of radius $1$ on $S^n$. Then $g$ extends to a smooth metric on $S^{n+1}$ if and only if

$$ \int^w_{-w}\phi(r)dr < \infty$$ $$ \lim_{z \to \pm w} \psi(z) = 0 $$ $$\lim_{z \to \pm w} \frac{\psi'(z)}{\phi(z)} = \mp1 $$ and $$ \lim_{z \to \pm w} \frac{d^{2k}\psi(z)}{ds^{2k}} = 0 $$

for all $k \in \mathbb{N}$, where $ds$ is the element of arc length induced by $\phi$.

Well, I tried to reformulate and it was like this:

Notation: $$S^r =\{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = r^2 \}$$ $$S^R = \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = R^2 \}$$ $$S^a = \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = a^2 \}$$ $\mathbb{T}^2 = S^R \times S^r \texttt{ with } R>r>0$ $\mathbb{T}^3 = S^R \times S^r \times S^a \texttt{ with } R>r>0 \texttt{ and } a> R + r $

$\textbf{Reformulation:}$ Let $M = \mathbb{T}^2 \times (-\pi , \pi)$ be a manifold with the metric:

$$g = r^2 dx^2 + (R + r.cos(x))^2 dy^2 + \theta(x,y,z)^2dz^2$$ and $\theta: (-\pi, \pi)^3 \to \mathbb{R_+}$ so $g$ extends to a metric in $\mathbb{T}^3$ if and only if

$$ |{\int_A\theta(x,y,z)dA}| < \infty$$ $$ \lim_{(x,y,z) \to \pm (\pi, \pi, \pi)} \theta(x,y,z) = a+r-R $$

where $A = (-\pi, \pi)^3 $.

Explanation about the formulation and some ideas: When a fix the metric

$g = r^2 dx^2 + (R + r.cos(x))^2 dy^2$ in $\mathbb{T}^2$ and make such extension, then I would like to know if the metric in $\mathbb{T}^3$ must be like this

$g = r^2 dx^2 + (R + r.cos(x))^2 dy^2 + (a + (R + r.cos(x))cos(y))^2dz^2$?

and, if so, the conditions above are sufficient?

In addition, suppose that $g$ extends to a metric in $ \mathbb{T}^3$, then for any parametrization $f$ we have $\langle \partial_x f, \partial_x f \rangle= r^2 $, $\langle \partial_y f, \partial_y f \rangle= (R + rcos(x))^2 $ and $\langle \partial_z f, \partial_z f \rangle= \theta(x,y,z) ^2 $ so $||\partial_x f|| = r$, $||\partial_y f|| = (R + rcos(x))$ and $||\partial_y f|| = \theta(x,y,z) $. Now

$$ |\int_A\theta(x,y,z)dA| \leq \int_A |\theta(x,y,z)|dA = \int_A ||\partial_y f|| < \infty $$

Is that right? thanks in advance!

Answer 1

The integral condition in the first example is just saying that the total arclength is finite- the appropriate integral condition for you is $\int_{-\pi}^\pi \theta dz$ being finite.

The other condition for a smooth closing is easier to understand in your situation than in the closing-to-$S^{n+1}$ situation that you cited, because here none of the factors in the coordinates you've written degenerate to $0$. What you need is that $\theta$ extends to a (strictly positive) function on $T^3$, which is $C^{\infty}$ if you want the metric to be $C^{\infty}$. Basically just take the conditions for a periodic function.