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Definition and intuition of a tubular neighborhood of a submanifold

Mathematics Asked on November 9, 2021

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

Now let $$T_xM:=left{vinmathbb R^dmidexistsvarepsilon>0,gammain C^1((-varepsilon,varepsilon),M):gamma(0)=x,gamma'(0)=vright}$$ denote the tangent space of $M$ at $x$ and $N_xM:=(T_xM)^perp$ for $xin M$.

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 $$N(M):={(x,v):xin Mtext{ and }vin N_xM}$$ 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 $$V={(x,v)in N(M):left|vright|<delta(x)}tag1$$ for some continuous function $delta:Mto(0,infty)$.

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 $vin N_xM$ 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 …


$^1$ 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 $B_isubseteq E_i$, then $f:B_1to E_2$ is called $C^1$-differentiable if $f=left.tilde fright|_{B_1}$ for some $E_1$-open neighborhood $Omega_1$ of $B_1$ and some $tilde fin C^1(Omega_1,E_2)$ and $g:B_1to B_2$ 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.

One Answer

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 ??)

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.

Answered by Osama Ghani on November 9, 2021

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