# Tubular Neighborhood in local coordinates

I am trying to verify if I understand the concept of a tubular neighborhood correctly. Any comments on my line of thoughts are very much appreciated.

Let $$\mathcal{M}$$ be a $$d-$$dimensional manifold embedded in $$\mathbb{R}^D$$ through $$f:\mathcal{U}\subset \mathbb{R}^d \to \mathcal{M}$$. Then, for every point $$n$$ in the normal space of $$\mathcal{M}$$ there exists a $$u\in \mathcal{U}$$ and $$v\in \mathbb{R}^{D-d}$$ such that $$n = f(u) + \bot J_f(u)v =:\varphi(u,v)$$ where $$\bot J_f(u)$$ denotes the matrix consisting of column vectors forming a basis of the normal space $$N_x$$ in $$x$$. These column vectors, together with the column vectors of the Jacobian $$J_f(u)$$, form a basis of $$\mathbb{R}^D$$ (justifying my notation).

Now, a tubular neighborhood of $$\mathcal{M}$$ is the image of the mapping $$\varphi$$ if $$v$$ is restricted to be sufficiently small, i.e. $$||v||<\varepsilon$$ for some $$\varepsilon >0$$, such that $$\varphi$$ is bijective.

Question: Assuming that this is correct, under which conditions is the tubular neighborhood diffeomorphic to $$\mathbb{R}^D$$?

Intuition: This is true whenever $$\varphi$$ is sufficiently smooth and $$\varepsilon$$ is sufficiently small.

• Have you tried drawing some examples? Commented Sep 30, 2021 at 16:29
• Yes. Try using the implicit or inverse function theorem to confirm your intuition. Commented Sep 30, 2021 at 18:43
• Thanks for the hint Deane. I followed your advice and added an answer. Would you mind having a look at it? Commented Oct 1, 2021 at 10:06

Following the advice of Deane, I use the inverse function theorem to show that indeed the tubular neighborhood is diffeomorphic to $$\mathbb{R}^D$$ under certain conditions.

Let $$\mathcal{B}_{\varepsilon}^{D-d}(0)$$ be the open ball in $$\mathbb{R}^{D-d}$$. Let $$\varepsilon$$ be sufficiently small such that the image of

$$\mathcal{U} \times \mathcal{B}_{\varepsilon}^{D-d}(0)$$

under $$\varphi$$ is a tubular neighborhood of $$\mathcal{M}$$.

From the inverse function theorem, we know the following: If the Jacobian determinant of $$\varphi$$ is non-zero at $$(u,0)$$, then there exists an open neighborhood $$U_u$$ of $$(u,0)$$ such that

$$\varphi:U_u\to \varphi(U_u)$$

is a diffeomorphism. Without loss of generality, $$U_u$$ is diffeomorphic to $$\mathbb{R}^D$$, and thus $$\varphi(U_u)$$ is diffeomorphic to $$\mathbb{R}^D$$.

If we can show that this is indeed true for all points $$(u,0)$$ then we can simply unify all neighborhoods $$U_u$$, intersect this union with $$\mathcal{U} \times \mathcal{B}_{\varepsilon}^{D-d}(0)$$, and $$\varphi$$ restricted to this intersection will yield a tubular neighborhood which is diffeomorphic to $$\mathbb{R}^D$$

What remains to show is: The Jacobian determinant of $$\varphi$$ is non-zero at every point (u,0).

Proof: The Jacobian of $$\varphi$$, $$J_{\varphi}$$ is a square matrix. Thus, $$\det J_{\varphi}(u,0) \neq 0$$ if and only if $$\det J_{\varphi}(u,0)^T J_{\varphi}(u,0) \neq 0$$. The Gram determinant of $$\varphi$$ is given by:

\begin{align}\label{eq:} \det \left( J_{\varphi}(u,0)^T J_{\varphi}(u,0) \right) &= \det \left[ \begin{array} JJ_f(u)^T J_f(u) & J_f(u)^T \cdot \bot J_f(u)\\ \bot J_f(u)^T \cdot J _f(u)^T & \bot J_f(u)^T \bot J_f(u) \end{array} \right] \\ &= \det \left[ \begin{array} JJ_f(u)^T J_f(u) & 0_{d\times {D-d}}\\ 0_{{D-d}\times d} & \bot J_f(u)^T \bot J_f(u) \end{array} \right] \\ &= \det J_f(u)^T J_f(u) \cdot \det \bot J_f(u)^T \bot J_f(u) \end{align}

where we have exploited the fact that the column vectors of $$J_f(u)$$ and $$\bot J_f(u)$$ are orthogonal. Since $$f$$ is an embedding for $$\mathcal{M}$$, $$\det J_f(u)^T J_f(u) \neq 0$$. As the column vectors of $$\bot J_f(u)^T$$ span the normal space, we may assume without loss of generality that $$\det \bot J_f(u)^T \bot J_f(u) =1$$. This ends the proof.

Remark: Note that the inverse function theorem requires $$\varphi$$ to be in $$C^1$$. Therefore, $$f$$ needs to be in $$C^2$$.