How can I show that if the second fundamental form of a surface is identically equal to zero, then the surface is a plane?

Solution 1:

Proposition 4.2 If the second fundamental form of a surface vanishes, it is part of a plane.

Proof: If the second fundamental form vanishes,

$0 = r_u \cdot n_u = r_v \cdot n_u = r_u \cdot n_v = r_v \cdot n_v$

so that

$n_u = n_v = 0$

since $n_u$, $n_v$ are orthogonal to $n$ and hence linear combinations of $r_u$, $r_v$.

Thus $n$ is constant.

This means

$(r \cdot n)_u = r_u \cdot n = 0$,

$(r \cdot n)_v = r_v \cdot n = 0$

and so

$r \cdot n$ = const

which is the equation of a plane.

Hint: If the second fundamental form is always $0$, what does this tell you about $\vec n_u$ and $\vec n_v$?