Is [imath]f(x^1, x^2, x^3)=(2x^1, x^2, x^3)[/imath] an isometry of the canonical metric on [imath]\mathbb{R}^3[/imath]?

[MathNugget]

I'd like to prove it isn't (unless I am missing something, it isn't).

The first thing I have to do is to express the canonical metric. I believe it is the dot/scalar product, where we have [imath]\langle e_i, e_j \rangle= \delta_{ij}[/imath] (Kronecker symbol, [imath]\delta_{ij}[/imath] =1 if i=j, 0 otherwise). Given that the basis for [imath]\mathbb{R}^3[/imath] is the same as the basis the tangent space to it, I suppose it works.

So we have [imath]g(\frac{\partial }{\partial x^i}, \frac{\partial}{\partial x^j})=g_{ij}=\delta_{ij}[/imath].

Now I also have this formula: for [imath]f(x^1, x^2, x^3)=(\overline{x^1}, \overline{x^2}, \overline{x^3})[/imath], the metric changes by this rule:

[imath]\overline{g_{ij}}=\frac{\partial x^k}{\partial \overline{x^i}}\frac{\partial x^l}{\partial \overline{x^j}}g_{kl}[/imath].

I suspect the einstein summation convention applies, so this is actually double summed, from k=1 to k=3 and from l=1 to l=3, but it's easy to notice the terms are non 0 only when k=i and l=j, and k=l, leaving exactly 1 term (in the situation of [imath]f(x^1, x^2, x^3)=(2x^1, x^2, x^3)[/imath] I am trying to solve here).

So now I have to compute them. [imath]g_{11}[/imath] would be [imath]\frac{\partial x^i}{\partial 2x^i}\frac{\partial x^i}{\partial 2x^i}g_{ii}=\frac{1}{2}\frac{1}{2}=\frac{1}{4}[/imath] (I believe the 2 in [imath]\frac{\partial x^i}{\partial 2x^i}[/imath] can get out, using a sort of continuity argument: [imath]\frac{\partial 2x^i}{\partial x^i}=2[/imath] and [imath]\frac{\partial 2x^i}{\partial 2x^i}=1[/imath], maybe [imath]\frac{\partial x^i}{\partial 2x^i}=\frac{1}{2}[/imath]).

Since [imath]g_{ii}=1[/imath] and [imath]\overline{g_{ii}}=\frac{1}{4}[/imath], it's not an isometry...or is it?

 

[blamocur]

I agree.

 

[MathNugget]

I agree.

Think it's correct? I did make a mistake, I actually calculated for i=1, not a general i.

Also I guess to complete the proof, I'd have to say the length of [imath]e_1=1[/imath] with the first metric, and [imath]\frac{1}{2}[/imath] with the 2nd metric..

 

[blamocur]

Think it's correct? I did make a mistake, I actually calculated for i=1, not a general i.

True. I'd also avoid notations like [imath]\frac{\partial x_1}{\partial 2x_1}[/imath], but I know what you mean by it.

Also I guess to complete the proof, I'd have to say the length of e1=1e_1=1e1=1 with the first metric, and 12\frac{1}{2}21 with the 2nd metric..

Isn't it another way around?

 

[MathNugget]

Isn't it another way around?

I don't know. I figured I'd have to find a vector that has different length depending on metric. but now I am very confused...
do I check [imath]g(e_1, e_1)^\frac{1}{2}[/imath] (length of [imath]e_1[/imath] with first metric) and compare it to [imath]\overline{g}(e_1, e_1)^\frac{1}{2}[/imath], or to [imath]\overline{g}(\overline{e_1}, \overline{e_1})^\frac{1}{2}[/imath].

Or switching to numbers: [imath]g((1, 0, 0), (1, 0, 0))^\frac{1}{2}[/imath] has to be equal to [imath]\overline{g}((1, 0, 0), (1, 0, 0))^\frac{1}{2}[/imath]
or to [imath]\overline{g}((2, 0, 0), (2, 0, 0))^\frac{1}{2}[/imath]?

 

[MathNugget]

After a bit more consideration, maybe the 'isomorphism' (bijection, for now) deforms both the space and the metric. Then, it wouldn't change lengths along a basis vector; but it would change lengths of vectors going 'sideways', like [imath]e_1+e_2[/imath]...