Proving F is an Isometry for C^1 Functions in Elementary Geometry

[MathematicalPhysicist]

I might have forgotten about it cause I took a similar course two years ago.

So I have this assertion:
Let F be a C^1 function from R^n to R^n, show that if F is injective, and for each curve \gamma : I\rightarrow R^n Length(\gamma)=Length(F o \gamma) then F is an isometry.

So I thought basically if I pick \gamma to be a straight line, and if the metric is the euclidean metric then basically I have Length(\gamma)=d(\gamma(a),\gamma(b))
and from what is given I get that the metric is conserved as well under F, so it's an isometry.
But I am not sure I should use here a specific metric.

Any other thought of this question?
Thanks.

 

[Landau]

Well, give your definitions! I assume that the length \ell(\gamma) is defined by

\ell(\gamma):=\int_I |\gamma'(t)|dt?

Then

\ell(F\circ\gamma):=\int_I |F'(\gamma(t))||\gamma'(t)|dt

by the chain rule.

Also, "isometry" could mean precisely

d(F(a),F(b))=d(a,b)

for all a,b in R^n, with d some fixed metric, probably the Euclidean metric. But it might mean something else?

 

[lavinia]

Take \gamma to be parameterized by arc length. Then

0 = \int^{t}_{0}(|F'(\gamma(s))|- 1)ds

Differentiating with respect to t gives

0 = (|F'(\gamma(t))|- 1) d/dt|F'(\gamma(t))|

I think by the Chain Rule. Is this right?

If |F'(\gamma(t))| does not equal one in some interval then

d/dt|F'(\gamma(t))| = 0 in the interval and so |F'(\gamma(t))| must be constant. But the only possible constant is 1.

Something like this must work. Thie idea of this is to write the length of the image curve in a Taylor approximation with respect to the length of the original curve

 

[MathematicalPhysicist]

Well, give your definitions! I assume that the length \ell(\gamma) is defined by

\ell(\gamma):=\int_I |\gamma'(t)|dt?

Then

\ell(F\circ\gamma):=\int_I |F'(\gamma(t))||\gamma'(t)|dt

by the chain rule.

Also, "isometry" could mean precisely

d(F(a),F(b))=d(a,b)

for all a,b in R^n, with d some fixed metric, probably the Euclidean metric. But it might mean something else?

Yes these are the definitions I am using (F should also be injective in order to be an isometry), but I am not sure I can use here the Euclidean metric.

Is there a general way to show this without assuming what type of a metric I have here?

 

[lavinia]

But I am not sure I should use here a specific metric.

You can use any metric on the domain and any other metric on the range.

Parametrize the curve by arc length in the first metric.

(<dF(\gamma'(s),dF(\gamma'(s)>^{1/2} - 1) integrates to zero over any interval.

The derivative of the integral also equals zero.

So 0 ={ (<dF(\gamma'(s),dF(\gamma'(s)>^{1/2} - 1)/<dF(\gamma'(s),dF(\gamma'(s)>^{1/2} }d/dt<dF(\gamma'(s),dF(\gamma'(s)>

This means that <dF(\gamma'(s),dF(\gamma'(s)> is constant and so must equal 1.

dF carries vectors of length 1 in the first metric into vectors of length 1 in the second.

There are technicalities that make this argument wrong for all cases in that even though F is injective dF might have a non-trivial kernel at some points and also since F is only C^{1} the last derivative may not exist.

 

[MathematicalPhysicist]

Well obviously we know that F is onto its image (so I guess that we assume its surjective by default).

Anyway, I think I solved with some help from my TA.

If we take \gamma to be a striaght line then from the fact that
d(\gamma(a),\gamma(b)) \le Length(\gamma) (in general, for a straight line they are equal)
and from the equality that d(x,y) > = d(F(x) , F(y))

Now if I pick gamma to be a curve such that the composition of F with gamma equals a straight line then I get with the same reasoning that
d(F(x),F(y)) >= d(x,y).