Isometry of surfaces

Question

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.

Answer 1

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.