# Isometric isomorphism between $R^2$ and $R$

Can someone help me solving the following problems?

1. $(\mathbb R^2,d_2)$ and $(\mathbb R, d_1)$, $d_2, d_1$ being the respective euclidean norms, are not isometric isomorphic, i.e. there is no distance preserving isomorphism between them.
2. Let $d$ be the restriction of the euclidean metric on $(-1,1) \times(-1,1).$ Find a metric $d':\mathbb R \times \mathbb R \to \mathbb R$ such that the spaces $((-1,1),d)$ and $(\mathbb R,d')$ are isometric isomorphic.

For 1.)
I thought it suffices to prove that if there was a isomorphism between those two, it wouldn't be distance preserving since it then also must be norm preserving which can't be injective since $||f(x)||_2 = ||x||_2 = ||f(-x)||_2$.

I don't know if that suffices or if it is even true. I would be glad if someone could prove me why they can't be isomorphic in the first place.

For 2.) I don't know what to do.

• Do you know what it means for a space to be "connected"? Jun 11, 2016 at 19:00
• And no, it is not true that distance preserving maps are necessarily norm-preserving. Jun 11, 2016 at 19:01
• @Omnomnomnom A norm is a stronger condition then a metric,there are lots of metric spaces which do not have corresponding induced norms. Jun 11, 2016 at 19:04
• I know what connectedness means, however I must solve this exercise without the notion of connectedness. Jun 11, 2016 at 19:06

Hint: How many elements of $\Bbb R$ are of distance $1$ from $0$?

• Good hint,wish I'd had it in baby topology.......lol +1 Jun 11, 2016 at 19:05
• So it suffices to say there are more Elements of distance 1 from 0 in $\mathbb R^2$ than in $\mathbb R$ and hence it can't be surjective? And for 2.)? Jun 11, 2016 at 19:11

Here's an elementary way to solve 1. First the idea: The euclidean norms are quite special, since they come from inner products, and isometries in inner product spaces are very well-behaved.

Namely, we have the following result: We denote $\langle,\rangle$ inner products and $\Vert\cdot\Vert$ the respective norm.

Proposition: If $H$ and $K$ are real vector spaces with inner products, and $T:H\to K$ is an isometry, then $T$ is of the form $L=T-T(0)$ is linear and $\langle Lx,Ly\rangle=\langle x,y\rangle$ for all $x,y\in H$.

Proof: Note that $L$ is isometric and $L(0)=0$. In particular, $$\Vert L(x)-L(y)\Vert^2=\Vert x-y\Vert^2$$ We can rewrite this as $$\Vert Lx\Vert^2-2\langle Lx,Ly\rangle+\Vert L(y)\Vert^2=\Vert x\Vert^2-2\langle x,y\rangle+\Vert y\Vert^2$$ Now again since $L$ is isometric and $L(0)=0$, we obtain $\Vert Lx\Vert=\Vert x\Vert$, and similarly for $y$, so $\langle L(x),L(y)\rangle=\langle x,y\rangle$.

Now we calculate \begin{align*} \Vert L(x+y)-L(x)-L(y)\Vert^2&=\Vert L(x+y)\Vert^2+\Vert L(x)\Vert^2+\Vert L(y)\Vert^2\\ &\phantom{=} -2\langle L(x+y),L(x)\rangle-2 L(x+y),L(y)\rangle+2\langle Lx,Ly\rangle \end{align*} and again using the fact that $L$ preserves norm and inner product, \begin{align*} \Vert L(x+y)-L(x)-L(y)\Vert^2&=\Vert x+y\Vert^2+\Vert x\Vert^2+\Vert y\Vert^2\\ &\phantom{=} -2\langle x+y,x\rangle-2\langle x+y,y\rangle+2\langle x,y\rangle\\ &=0 \end{align*} So $L(x+y)=L(x)+L(y)$. I'll leave it to you to show $L(\lambda x)=\lambda L(x)$ for $\lambda\in\mathbb{R}$ and $x\in H$. QED

Therefore, if $\mathbb{R}$ and $\mathbb{R}^2$ were isometric with their euclidean norms, there would be a linear isometry $L:\mathbb{R}^2\to\mathbb{R}$, which would imply $2=\dim\mathbb{R}^2=\dim T(\mathbb{R}^2)=\dim\mathbb{R}=1$, an absurd.

This in fact works to show that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not isometric with their euclidean norms for $n\neq m$. There are several results which are a lot stronger, but they require more advanced stuff.

For $2.$ here's a hint: If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces and $f:X\to Y$ is an isometry, then $d_Y(x,y)=d_X(f^{-1}(x),f^{-1}(y))$. This formula shows, in particular, that $d=d_Y$ is the only metric on $Y$ which makes $f$ an isometry (with $d_X$ on $X$).

Now if you have a bijection $f:(-1,1)^2\to\mathbb{R}^2$, use the same formula to induce a (unique) metric on $\mathbb{R}^2$ which makes $f$ an isometry.

• Thank you for your in-depth proof. Jun 11, 2016 at 22:39

In the line there is no regular triangle.

• Can you elaborate? Jun 11, 2016 at 20:10
• @bavor42 By a regular triangle, he means three distinct points $x,y,z$ with $d(x,y)=d(y,z)=d(x,z)$. We can find such points in $\mathbb{R}^2$ but not in $\mathbb{R}$. Jun 11, 2016 at 22:28
• Okay. Does that suffice to prove that there can't be an isomorphism since it can never be surjective? Jun 11, 2016 at 22:29