Regular Surface + Isometry

**slevvio** · December 21st 2010, 04:02 PM

Let $S_1$ and $S_2$ be regular surfaces in $\mathbb{R}^3$ . Let $\phi: S_1 \rightarrow S_2$ be an isometry, that is to say

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

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

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

$I_p^{S_1} (X_i, X_j)= I_{\phi(p)}^{S_2} (X_i ', X_j ')$ , where $X_i = DF_1(u) e_i$ and $X_i = DF_2 (F_2^{-1} \circ \phi \circ F_1 (u)) e_i$ . 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

$<br /> I_p^{S_1} (X_i, X_j) = I_{\phi(p)}^{S_2} (D(\phi \circ F_1)(u)e_i, D(\phi \circ F_1)(u)e_j ) <br />$ and then I am totally stuck. Can anyone offer any advice? I understand this question is a bit long winded.

**xxp9** · December 21st 2010, 09:13 PM

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.

**slevvio** · December 21st 2010, 11:05 PM

why are the first fundamental forms the same? this is what I am trying to show, i.e.

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

or am I fundamentally misunderstanding something here? The only definition of an isometry I have is $I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y)$ . Why does this mean the first fundamental forms have the same matrix?

**xxp9** · December 22nd 2010, 03:43 AM

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.

**slevvio** · December 22nd 2010, 07:02 AM

so if $\phi:S_1 \rightarrow S_2$ is an isometry is $d_p \phi$ necessarily an isomorphism?

**slevvio** · December 22nd 2010, 07:21 AM

YES it is!

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

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

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