Isometry of surfaces We know isometry of euclidean spaces is composition of an orthogonal transformation and a translation. Also differential map of an isometry of regular surfaces is an orthogonal transformation.
 I want to know that, is there any such composition or some expression for  isometry of surfaces? 
Thank you.
 A: There is no particular form of the map $u$, no. First of all, consider an isometry from the $xy$-plane to itself, one that fixes the plane. Then $u$ has the form 
$$
u(x, y, 0) = (x, y, 0)
$$
for any $x, y$. But off the plane, it can be almost anything. Letting $f$ be any (continuous) function of $x, y, z$, we can write something that's the "same" as $u$ (from the point of view of isometry) but looks rather different:
$$
v(x, y, z) = (x, y, z f(x, y, z))
$$
Indeed, we could write
$$
v(x, y, z) = (x, y, 0) + z(f(x, y, z), g(x, y, z), h(x, y, z))
$$
and still have an isometry, as long as $f, g, h$ are continuous functions. 
Of course, all this depends on $u$ being "messy" away from the surface $F$. Another possibility is that on the surface $F$ itself, the function must be "nice", but even that is pretty much hopeless. For once again letting $F$ be the $xy$-plane, we can write
$$
u(x, y, z) = (x, y, zk(z) + h(x))
$$
where $k$ and $h$ are any continuous functions, and get an isometry (indeed, a special isometry that's the identity on the surface $F$).
In short: isometry, for surfaces in 3-space, just isn't very "rigid". 
If you require that the surfaces be compact surfaces without boundary, you might find some more rigidity, but even then I have my doubts, for if your compact surface happens to have a 'flat region' (some open disk that lies entirely in a plane), then within this flat region, you can do the tricks that I did with the plane above. 
Maybe for compact surfaces with everywhere (or almost-everywhere) nonzero gaussian curvature, there's a rigidity theorem -- I suspect there is, but cannot recall it -- but in general, things can be pretty ugly. 
