# Quick question on Riemannian geometry

So let $M^m$ be a manifold embedded into euclidean space $R^n$. Let L($\gamma$) be the the lenght of a smooth curve $\gamma$: [0,1] --> $R^n$ which is the number $\int_0^1 \ |d/dt \gamma(t)| \ \mathrm{d}t$ . Define the distance function d: M x M --> [0,$\infty$) by d(p,q):=inf L ($\gamma$). The infimum is taken over all smooth paths connecting p and q. This formula defines a metric on M.

I wan't to show that $\forall$ $p_0 \in M^m \exists U \subset M^m$ open neighborhood of $p_0$ such that (1-$\epsilon$) |p-q| < d(p,q) < (1+$\epsilon$) |p-q| . Note that |.| denotes the usual euclidean distance on $M^m$.

Now we fix an arbitrary $p_0 \in M^m$. Without loss of generality (Translations and rotations preserves the length of a curve) we assume that $p_0$ = 0 and $T_{p_{0}}M$ = $R^m$ x {0} .

Now comes the step that I don't understand.

Now there's a smooth function f: $\Omega$ --> $R^{n-m}$ defined on an open neighborhood $\Omega \subset R^m$ of the origin such that { (x,y) $\in R^m x R^{n-m}$ | x $\in \Omega$ , y=f(x) } $\subset M^m$ , f(0)=0, df(0)=0

Now I don't really see where this is coming from. It seems to me that it might be some combination of a characterisation of manifolds and tangent spaces we introduced, which is the following:

Let $M^m \subset R^n$, p $\in M^m$ "smooth manifold" , then $\exists U \subset R^n$ and f: U --> $R^{n-m}$ smooth such that p $\in$ U , df(q) surjective for all q $\in$ U $\cap$ M and U $\cap$ M = $f^{-1}$(0).

Can anyone help me?

What is the relation between $U$ and $\varepsilon$ in your assertion? – Ronaldo Dec 26 '10 at 0:31
I presume the inclusion $M\subset \mathbb{R}^n$ essentially means "embedded submanifold." Now try to think of an easier example. Consider a smooth 2-surface in $\mathbb{R}^3$. Call it $S$. Let $p\in S$ and $\Pi$ the tangent plane at $p$ to $S$. One can construct a smooth chart around $p$ by projecting a small neighborhood $U$ of $p$ onto $\Pi$. This essentially gives $f$. That is, every smooth surface is locally a graph of a smooth function $f: \mathbb{R}^2\rightarrow \mathbb{R}$. Generalize this. – William Dec 29 '10 at 8:47
By the way, the result that you're trying to prove is true. Again, try to first prove the easier version, say for surfaces in $\mathbb{R}^3$, and then generalize. – William Dec 29 '10 at 8:50
This is the implicit function theorem, as Jesse Madnick pointed out. Essentially it's saying that, near $p_0=0$, the manifold $M^m$ "looks like" the graph of a function from $\mathbb{R}^m$ to $\mathbb{R}^{n-m}$, this function giving the remaining coordinates in $\mathbb{R}^n$ in terms of the first $m$ coordinates. For example, if we take the unit sphere in $\mathbb{R}^3$, then near the north pole, this looks like a graph of the function $f(x,y)=\sqrt{1-x^2-y^2}$.