# Extending $f: X\subset \mathbb{R}^m\to\mathbb{R}^n$ an isometric immersion.

Let $X\subset \mathbb{R}^m$ not empty and $f: X\to\mathbb{R}^n$ an isometric immersion. Prove that there exists an isometric immersion $\varphi: \mathbb{R}^m \to\mathbb{R}^n$ such that $\varphi|_X=f$. When $X$ generates (in the sense of linear algebra) $\mathbb{R}^m$, $\varphi$ is unique. Metrics are euclidean metrics (in particular, $\mathbb{R}^m$ and $\mathbb{R}^n$ metrics comes from usual inner product).

Part of proof:

Without loss of generality we can say that $f(0)=0$ (up translations). Then $f$ satisfies $f(x+y)=f(x)+f(y)$ and $f(\lambda x)=\lambda f(x) \ \forall \lambda\in\mathbb{R}$ because $f$ preserve distances and the fact that $\mathbb{R}^n$ euclidean metric comes from an inner product, is not so hard to prove.

I need to prove the existence of $T$ linear transformation such that $T|_X=f$ (assumming $f(0)=0$). I've tried extending $f$ using values of a basis of $X$, but is not factible to prove existence of $T$.

Also, is question possible if $m>n$? or exercise is just incomplete?.

I think that if $X$ generates $\mathbb{R}^m$ the uniqueness is obvious if $T$ is linear extension.

Added: I was opened a bounty for +100 because I need to prove existence of $T$. This is the only fact that I can't be formal. If you want to explain what happend if $m>n$ better. But the priority is a formal proof of the fact of existence of $T$.

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"generate" in the sense of linear algebra? Is $f$ supposed to be linear? Upon reading your question, I imagined a piece of paper immersed in $\mathbb{R}^3$ as a cilinder, but I guess this is not at all what you mean by your question? – Daan Michiels May 22 '12 at 22:07
@DaanMichiels in the sense of linear algebra. $f$ is not supposed linear. $f$ is given isometric immersion not necesarly coincide with R^2->R^3 cilinder inmersion. – Gastón Burrull May 22 '12 at 22:21
Indeed, the statement seems to be false when $m>n$: you can immerse a point in $\mathbb R^2$ into $\mathbb R$, but there is no immersion $f\colon \mathbb R^2\to\mathbb R$. To prevent misunderstanding, could you define exactly what you mean by isometric immersion? – user31373 May 22 '12 at 22:49
If you know the result holds for a subset $S \subset X$ of your domain, it should be possible to construct an extension that holds for your subset plus another point, $S \cup p$, $p \in S \cap X$. Then well-order the points in your domain, and use transfinite induction to prove your result. I will try to write this up in an answer later. – Nick Alger May 25 '12 at 0:51
Ewan's answer is a more reasonable approach to the problem than what I wrote above. – Nick Alger May 28 '12 at 10:38

As noted in the OP, we may assume $0\in X$, and $f(0)=0$. The first thing to note is that $f$ preserves the scalar product, since for any $x,x'$ in $X$ we have

$$<f(x),f(x')>=\frac{d(f(x),0)^2+d(f(x'),0)^2-d(f(x),f(x'))^2}{2}= \frac{d(x,0)^2+d(x',0)^2-d(x,x')^2}{2}=<x,x'>$$

Let $\lbrace y_1,y_2, \ldots ,y_r \rbrace \subseteq f(X)$ be a basis of ${\sf span}(f(X))$. For each $i$ there as an $x_i$ such that $f(x_i)=y_i$. Let us put $H={\sf span}(x_1,x_2, \ldots ,x_r)$, and denote by $H^{\perp}$ its orthogonal complement in ${\mathbb R}^m$. Let $\phi$ be the linear mapping ${\sf span}(f(X)) \to H$, defined by $\phi(y_i)=x_i$. Then $\phi$ preserves scalar products by the equality above, so $\phi$ is an isometry.

Let $x\in X$ and $y=f(x)$. We can decompose $x$ as $x=h+k$, with $h \in H, k \in H^{\perp}$. Then $< h,k >=0$ , and we have Pythagoras' equality $||x||^2=||h||^2+||k||^2$. For any $i$ we have $$<h,x_i>=<x,x_i>=<f(x),f(x_i)>=<y,y_i>=<\phi(y),\phi(y_i)>=<\phi(y),x_i>$$

So we see that $h$ and $\phi(y)$ are both in $H$ and have equal coordinates in the basis $(x_1, \ldots ,x_n)$. So $h=\phi(y)$. Then

$$||h||^2=<h,h>=<\phi(y),\phi(y)>=<y,y>=d(y,0)^2=d(x,0)^2=||x||^2$$

We deduce $||k||^2=||x||^2-||h||^2=0$, so $k=0$, and $x=h=\phi(y)$. Thus $f$ coincides with $\phi^{-1}$ where it is defined, so $f$ is linear and we are done.

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Thanks. You have expanded $f$ isometric inmersion to $\phi^{-1}$ linear application with domain $H=\operatorname{span}(X)$. But, for expanding to whole $\mathbb{R^m}$ what do you do? Expanding $\phi^{-1}$ in $H^{\perp}$ or expanding $\phi$ in $\operatorname{span}(f(X))^{\perp}$? – Gastón Burrull May 26 '12 at 4:11
Oops sorry have not sense expanding $\phi^{-1}$ to whole $\mathbb{R}^n$. I was full understand now. – Gastón Burrull May 26 '12 at 4:18