- #1

- 14

- 0

## Homework Statement

Two graphs defined for a two dimensional torus:

[tex]f_1(t) = (\frac{1}{\sqrt{2}}(a\ +\ b\ sin\ t),\frac{1}{\sqrt{2}}(a\ +\ b\ sin\ t),b\ cos\ t),\

t \in (-\frac{\pi}{2},\frac{\pi}{2})[/tex]

[tex]f_2(t) = (a\ cos(t+\frac{\pi}{4}),a\ sin(t+\frac{\pi}{4}),b),\

t \in (-\frac{\pi}{4},\frac{\pi}{4})[/tex]

Then [tex]f_1(0)=f_2(0)=(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}},b)[/tex]

Are the graphs equivalent at a point [tex](\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}},b)[/tex]?

## Homework Equations

Graphs f

_{1}and f

_{2}on manifold M are equivalent at a point m (m is in an open group U), if for some chart f

_{c}(q

^{1},...,q

^{n}): U ->

**R**

^{n}of manifold M

[tex]\frac{d}{dt}q^i(f_1(t))=\frac{d}{dt}q^i(f_2(t))\ |_{t=0}[/tex]

I also know that q is a coordinate function:

[tex]q^i:=pr^i \circ f_c[/tex] where the projection [tex]pr^i:R^n \rightarrow R, (x^1,...,x^n) \mapsto x^i[/tex]

I'm also told that if [tex]q^i:=pr^i \circ f_c[/tex] then f can be written as: [tex]f_c=(q^1,...,q^n)[/tex]

## The Attempt at a Solution

I can't actually give anything of my own to go with this. Last week I posted a question relating to this problem, but as anyone hasn't been able to comprehend my output, I'm hoping that someone could help me with the problem directly.

So the problem is I really don't understand the behaviour of the coordinate function in this, but I'll happily read any kind of suggestions (using the method mentioned in 2. or not) for solving the problem.

Last edited: