Local Isometry in Dilatation

From: Narasimham (mathma18_at_hotmail.com)
Date: 01/19/05 

Date: 19 Jan 2005 07:56:21 -0500

Requesting your help in understanding a long-standing, seemingly
elementary doubt in differential geometry.

A square patch between points [A(pi,0), B(0,pi), C(-pi,0), D(0,-pi)] in
x-y plane undergoes dilatation, represented as as mapping from (x,y,0)
to ( x,y, cos(x)+ cos(y) ).The four boundary lines at planes x+/- y =
+/- pi do not change position and planes parallel to z-axis and passing
through join of mid-points of AB,BC,CD,DA do not undergo any change in
length (no dilatation) as they preserve Gauss Curvature at initial zero
value during dilatation.

Local isometry seems to apply here, we have inextensional deformation
along these lines. One principal normal curvature is always zero on
these lines during dilatation.

Egregium theorem requires that distance between any pair of mid-points
where local isometry applies,say between (-pi/2,pi/2),(pi/2,pi/2)
should not change.

However, distance between any two points on 2D surface is always more
than the straight Euclidean distance in 3D. How then is the
contradiction resolved?

Is it a case of converse of Egregium theorem (if Gauss Curvature is
preserved, then it may not always preserve lengths) not being true?

Are there any other lines that preserve length between any two points
in this dilatation?