Definition and intuition of a tubular neighborhood of a submanifold

Question

Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$ and $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary$^1$.

Now let $$T_xM:=\left\{v\in\mathbb R^d\mid\exists\varepsilon>0,\gamma\in C^1((-\varepsilon,\varepsilon),M):\gamma(0)=x,\gamma'(0)=v\right\}$$ denote the tangent space of $M$ at $x$ and $N_xM:=(T_xM)^\perp$ for $x\in M$.

I'm trying to understand the definition of a tubular neighborhood of $M$. I was only able to find definitions of this notion in a way more general setting which built up on several concepts I'm not familiar with. So, I'd like to find a simplified, but equivalent, definition for my present setting.

What I've understood is that one considers the space $$N(M):=\{(x,v):x\in M\text{ and }v\in N_xM\}$$ and the map $$E(x,v):=x+v\;\;\;\text{for }(x,v)\in N(M).$$

Now the usual definition of a tubular neighborhood $U$ of $M$ is that it is a (open?) neighborhood of $M$ in $\mathbb R^d$ (really a neighborhood of all of $M$?) such that $U$ is the diffeomorphic image under $E$ of an open subset $V$ of $N(M)$ with $$V=\{(x,v)\in N(M):\left\|v\right\|<\delta(x)\}\tag1$$ for some continuous function $\delta:M\to(0,\infty)$.

I really got trouble to understand this. The vector $E(x,v)$ is simply a vector originating from $x$ and pointing in the direction $v$. If $M$ is a circle in $\mathbb R^3$, I guess the intuition is that for each point $x$ on the circle all the $v\in N_xM$ built a ring around $x$ by "rotating" the normals around $x$. Doing so for all $x$ on the circle, we obtain a torus. Is this correct so far? All the pictures I've found have confused me.

And how would a tubular neighborhood of a sphere in $\mathbb R^3$ look like? For a sphere, the normal spaces are 1-dimensional ...

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$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$.

If $E_i$ is a $\mathbb R$-Banach space and $B_i\subseteq E_i$, then $f:B_1\to E_2$ is called $C^1$-differentiable if $f=\left.\tilde f\right|_{B_1}$ for some $E_1$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,E_2)$ and $g:B_1\to B_2$ is called $C^1$-diffeomorphism if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.

Answer 1

Your intuition seems to be right. I like to describe it as an open disk bundle i.e. the fibre over the point $p \in M$ looks like $\{p\} \times O^n$ where $O$ is an open disk of dimension $n$ (which is also equal to the dimension of the normal bundle, or even fancier, the codimension of $M$). So you have a tube that encapsulates the submanifold.

If $M$ is a circle in $\mathbb{R}^3$ then you are right, its tubular neighbourhood is exactly an (open solid) torus i.e. $S^1 \times O^2$ so it's actually a trivial bundle.

(really a neighborhood of all of 𝑀?)

Yes, all of $M$. Now let us consider the sphere $S^2 \subset \mathbb{R}^3$. You get essentially an (open) thickened sphere, you can think of it as $S^2 \times O^1 \cong S^2 \times (0,1)$. Note the dimension of the open disk you are working with, since the sphere is codimension $1$!

The general idea is that any submanifold can be identified with the zero section in its normal bundle, and in the normal bundle you have a little "wiggle room" in the normal direction around each point and gluing all these little wiggle rooms together give you an open neighbourhood of $M$ in the ambient manifold.