Answer to Question #148406 in Differential Geometry | Topology for anjali

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.
Show that FIII = FIIF^−1I FII , where FI and FII are the 2 × 2 symmetric matrices
associated with the first and the second fundamental forms, respectively

Answer 1

$Solution$

From the introductions to First and second fundamental forms (https://warwick.ac.uk/fac/sci/maths/people/staff/weiyi_zhang/ma3d92018fall.pdf)

We prefer $E,F, and\ G$ to the first fundamental form and $L,M,N$ to the second fundamental form, hence;

$I=Edu^2+ 2Fdudv+Gdv^2,\ II=Ldu^2+ 2Mdudv+Ndv^2.$

We denote

$\digamma_{I}=\begin{pmatrix} E & F \\ F & G \end{pmatrix}, \digamma_{II}=\begin{pmatrix} L & M \\ M & N \end{pmatrix}$

From the above concept, we deduced the first, second and third fundamental forms replacing $Edu^2+ 2Fdudv+Gdv^2$ with $a_{11}(du^1)^2+2a_{12}du^1du^2+a_{22}(du^2)^2$ , $II=Ldu^2+ 2Mdudv+Ndv^2$ with $b_{11}(du^1)^2+2b_{12}du^1du^2+b_{22}(du^2)^2$ and used $d\overline{n}^2=(\overline{n_{u^1}})^2(du^1)^2+2\overline{n_{u^1}}\overline{n_{u^2}}du^1du^2+(\overline{n_{u^2}})^2(du^2)^2$ for the third fundamental form

$I = dS^2=a_{11}(du^1)^2+2a_{12}du^1du^2+a_{22}(du^2)^2\\ a_{ij}=(\overline{r_i}, \overline{r_j}); i,j=1,2\\ II=b_{11}(du^1)^2+2b_{12}du^1du^2+b_{22}(du^2)^2\\ b_{ij}=(\overline{r_{ij}},\overline{n}); i,j=1,2\\ III=d\overline{n}^2=(\overline{n_{u^1}})^2(du^1)^2+2\overline{n_{u^1}}\overline{n_{u^2}}du^1du^2+(\overline{n_{u^2}})^2(du^2)^2\\ \begin{pmatrix} (\overline{n_{u^1}})^2 & \overline{n_{u^1}}\overline{n_{u^2}} \\ \overline{n_{u^1}}\overline{n_{u^2}} & (\overline{n_{u^2}})^2 \end{pmatrix} = \begin{pmatrix} b_{11} & b_{12} \\ b_{12} & b_{22} \end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix}^{-1} \cdot \begin{pmatrix} b_{11} & b_{12} \\ b_{12} & b_{22} \end{pmatrix}\\ A=\begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix}\\ A^{-1}=\begin{pmatrix} \frac{a_{22}}{\Delta} & -\frac{a_{12}}{\Delta} \\ -\frac{a_{12}}{\Delta} & \frac{a_{11}}{\Delta} \end{pmatrix}\\ \Delta=|A|=a_{11}a_{22}-a_{12}^2$