# Regular Surface + Isometry

• Dec 21st 2010, 04:02 PM
slevvio
Regular Surface + Isometry
Let$\displaystyle S_1$ and $\displaystyle S_2$ be regular surfaces in $\displaystyle \mathbb{R}^3$. Let $\displaystyle \phi: S_1 \rightarrow S_2$ be an isometry, that is to say

$\displaystyle I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y)$ for all $\displaystyle x,y \in T_p S_1$, the tangent plane at$\displaystyle p\in S_1$, where I is the first fundamental form on the appropriate tangent plane, and $\displaystyle d_p$ is the differential map.

Let $\displaystyle (U_1,F_1,V_1)$ and $\displaystyle (U_2,F_2,V_2)$ be local parametrisations of $\displaystyle S_1$ at $\displaystyle p$ and $\displaystyle S_2$ at $\displaystyle \phi(p)$ respectively.
Let $\displaystyle u = F^{1}(p).$

I am trying to show that the gaussian curvature is the same at $\displaystyle p$ and $\displaystyle \phi(p).$I guess that this means I am trying to show that the first fundamental forms at $\displaystyle p$ and $\displaystyle \phi(p)$ are the same when i put the appropriate basis vectors in, i.e.

$\displaystyle I_p^{S_1} (X_i, X_j)= I_{\phi(p)}^{S_2} (X_i ', X_j ')$, where $\displaystyle X_i = DF_1(u) e_i$ and $\displaystyle X_i = DF_2 (F_2^{-1} \circ \phi \circ F_1 (u)) e_i$. I figured showing this would be sufficient because then the gaussian curvature is completely determined by the first fundamental form (the notation I've used means you seperate the two columns of the jacobian and use them as basis vectors).

So if I try to show this, then I end up with

$\displaystyle I_p^{S_1} (X_i, X_j) = I_{\phi(p)}^{S_2} (D(\phi \circ F_1)(u)e_i, D(\phi \circ F_1)(u)e_j )$and then I am totally stuck. Can anyone offer any advice? I understand this question is a bit long winded.
• Dec 21st 2010, 09:13 PM
xxp9
since isometries keep the first fundamental form I and I determines the gaussian curvature, this is already done. what are you trying to prove? Re-writing an abstract expresion in specific coordinates does not make mush sense.
• Dec 21st 2010, 11:05 PM
slevvio
why are the first fundamental forms the same? this is what I am trying to show, i.e.

$\displaystyle X_i \cdot X_j = X_i ' \cdot X_j '$

or am I fundamentally misunderstanding something here? The only definition of an isometry I have is $\displaystyle I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y)$. Why does this mean the first fundamental forms have the same matrix?
• Dec 22nd 2010, 03:43 AM
xxp9
do you understand the expression I_p(x,y) = I_f(p)( df(x), df(y) ) ? ( I'm using f instead of \phi)
This expression tells: the first fundamental form I of S1 at p, corresponds to the first fundamental form I of S2 at f(p).
"Corresponds to" here means: Given any two vectors x and y of S1 at p, their inner product I_p(x,y), or you can write as x . y, equals to the inner product of their images under the differential map df. This is exactly you Xi . Xj = Xi' . Xj'.

Using any choice of coordinates won't change that. You need to understand that coordinates are only used for simplifying computation. The important thing is the underlying geometry.

For your question, consider two vector spaces V and W, and a linear isomorphism f: V -> W. Also their are two extra structures, their inner products, are defined for each. Suppose that f is further an isometry, that is, f keeps inner products, <f(x), f(y)> = <x,y>. Then their two inner products have the same matrix, in any choice of basis v1,..., vn in V and f(v1),...,f(vn) in W.
• Dec 22nd 2010, 07:02 AM
slevvio
so if $\displaystyle \phi:S_1 \rightarrow S_2$ is an isometry is $\displaystyle d_p \phi$ necessarily an isomorphism?
• Dec 22nd 2010, 07:21 AM
slevvio
YES it is!

Every tangent vector $\displaystyle \alpha'(0) \in T_{\phi(p)} S_2$ looks like $\displaystyle d_p \phi (\phi^{-1} \circ \alpha) ' (0),$ so we have surjectivity $\displaystyle \iff$ we have injectivity $\displaystyle \iff$ bijectivity

since the linear map is on a finite dimensional vector space. I guess there is a more general way of doing this