# Quick question on Riemannian geometry

Hey.

I got a quick question on Riemannian geometry. I'm not quite sure whether this is the right place to ask this question, since it might be a rather elementary one from a research point of view. Nontheless I think it might be too advanced for a typical homework question (which by the way it isn't)

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 want 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 characterization 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?

Thanks in advance.

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This appears to be a fragment of a question arising from either a course or textbook in differential geometry. I suggest posting this question on math.stackexchange.com. But the key tool here is "implicit function theorem", which turns any local question about a manifold or submanifold into a local question in Euclidean space. – Deane Yang Dec 25 '10 at 17:50
Okay. I posted the question on math.stackexchange.com However if anyone wants to give a (maybe more detailed) answer, I'd be very happy. Thanks anyway – user11823 Dec 25 '10 at 18:01
By the way, you should clarify whether you are restricting the curves to lie in $M$ or not. – Deane Yang Dec 25 '10 at 18:04
Note that what is being said here is that every submanifold locally looks like the graph of a function (the function f). If you think about it that way perhaps you may see how to prove it---as Deane says, using the implicit function theorem. – Dick Palais Dec 25 '10 at 18:10
Okay. Thanks for all the remarks. If I just may ask another question. In my course we've introduced a quite strange looking form of the implicit function theorem, namely: $M \subset R^k$ smooth m-manifold and $N \subset R^l$ smooth n-manifold. f:M-->N , q $\in$N regular value of f Then P:=f^{-1} is a m-n dimensional manifold. I don't really see how this theorem is related to the ordinary implicit function theorem which one usually learn in an introductory analysis course. (I used the classical version to proof my 'problem' above). I'd be really happy if anyone could help me. Thanks. – user11823 Dec 25 '10 at 21:15