Question

The third fundamental form of a surface σ(u, v) is
||N̂u|| ^2 du^2 + 2N̂u.N̂v dudv + ||N̂v||^2 dv^2
where N̂ (u, v) is the standard unit normal to σ(u, v). Let FIII be the associated 2 × 2
symmetric matrix.
a) ) Show that FIII − 2HFII + KFI = 0, where K and H are the Gaussian and mean
curvatures, respectively, of σ.

Answer 1

$I = dS^2=a_{11}(du^1)^2+2a_{12}du^1du^2+a_{22}(du^2)^2\\ a_{ij}=(\overline{r_i}, \overline{r_j}); i,j=1,2\\ II=b_{11}(du^1)^2+2b_{12}du^1du^2+b_{22}(du^2)^2\\ b_{ij}=(\overline{r_{ij}},\overline{n}); i,j=1,2\\ III=d\overline{n}^2=(\overline{n_{u^1}})^2(du^1)^2+2\overline{n_{u^1}}\overline{n_{u^2}}du^1du^2+(\overline{n_{u^2}})^2(du^2)^2\\ \begin{pmatrix} (\overline{n_{u^1}})^2 & \overline{n_{u^1}}\overline{n_{u^2}} \\ \overline{n_{u^1}}\overline{n_{u^2}} & (\overline{n_{u^2}})^2 \end{pmatrix} -2H \begin{pmatrix} b_{11} & b_{12} \\ b_{12} & b_{22} \end{pmatrix} +K \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix} =0\\ H=\frac{1}{2}(k_1+k_2)\text{ --- mean curvature}\\ 2H=k_1+k_2=\frac{a_{11}b_{22}+a_{22}b_{11}-2a_{12}b_{12}}{a_{11}a_{22}-a_{12}^2}\\ K=k_1\cdot k_2=\frac{b_{11}b_{22}-b_{12}^2}{a_{11}a_{22}-a_{12}^2}\text{ --- Gaussian curvature}$