2

So I'm revising for an exam in differential geometry, and have got stuck on a question regarding isometries.

We define a local isometry as a map $f:X\rightarrow Y$, (between 2 regular m-surfaces X,Y) where, for all $p\in X$ and all $u \in T_{p}X, |df_{p}(u)|=|u|$

The example in question involves finding a local isometry between 2 surfaces:

Let $X=\mathbb{R}^2${($0,0$)} and $Y$={$x$ $\in$ $\mathbb{R}^3$ $8x_{1}^2 +8x_{2}^2=x_{3}^2$ and $x_{3} >0$}.

We can use polar coordinates $q(r,\theta)=(rcos\theta, rsin\theta)$ on X to define a map $f:X\rightarrow Y$,

Now apparently, this map is $f(r,\theta)=(\frac{r}{3}cos\theta,\frac{r}{3}sin\theta,\frac{\sqrt{8}}{3}r)$

We then go on to prove this is well defined etc, but I really don't see where f has come from. Has anyone got any ideas? Many thanks.

Ma Ming
  • 7,482
Olivia77989
  • 289

1 Answers1

1

Without loss of generality, note that any point in $Y$ can be written in the form $$x_1=\rho\cos \theta,x_2=\rho \sin\theta,x_3=\pm \sqrt{8x_1^2+8x_2^2}=+ \sqrt 8 \times \rho$$ just by using plane polars for $(x_1,x_2)$.

Then they set $\rho(r)=r/3$ (no idea why they make that choice, but it'll do) and their map $f$ follows by identifying $\theta$.

not all wrong
  • 16,178
  • 2
  • 35
  • 57