curvature

curvature

- [Curve](https://hi-ogawa.github.io/markdown-tex/?id=e40372524f96337f1f2066ad332b4d2b&filename=curvature-00-curve)
- [Surface](https://hi-ogawa.github.io/markdown-tex/?id=e40372524f96337f1f2066ad332b4d2b&filename=curvature-01-surface)

curvature-00-curve

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## Curve


**Notations**

- $\gamma(t) : \RR \to \RR^2$ : twice differentiable curve,

- $\l< \bb{a}, \bb{b} \r> \equiv \bb{a}^T \bb{b}$,

- $\hat{\bb{a}} \equiv \dfrac{\bb{a}}{|\bb{a}|}$.


### Def. 1 (Osculating circle)

Define $\bb{v}(t \mid t_0) \in \RR^2$ s.t. 

- $\l< \bb{v}(t \mid t_0), \gamma'(t_0) \r> = 0$,

- $|\gamma(t) - (\gamma(t_0) + \bb{v}(t \mid t_0))| = |\bb{v}(t \mid t_0)|$.


**N.B.**

- We ocasionally omit $t, t_0$ and $t_0$ and write e.g. $\bb{v} = \bb{v}(t) = \bb{v}(t \mid t_0)$.

- Geometrically speaking, it gives a circle s.t.
    - center : $\gamma(t_0) + \bb{v}(t \mid t_0)$,
    - tangent line: $\gamma(s) + \gamma'(t_0) (s - t_0)$ at $\gamma(t_0)$,
    - passing points: $\gamma(t)$, $\gamma(t_0)$.

- Such circle can be still found even when $\gamma(t) \in \RR^3$.

- It's not well defined if
    - $\gamma'(t_0) = \bb{0}$ or
    - $\l< \gamma(t) - \gamma(t_0), \gamma'(t_0) \r> = 0$
    


### Prop. 2

$$ \l< \bb{v}(t_0 \mid t_0), \gamma''(t_0) \r> = |\gamma'(t_0)|^2$$


**N.B.**

- If $|\gamma'(t)|$ : constant, then we have
  $0 = \del_t |\gamma'(t)| = \dfrac{\l< \gamma'(t), \gamma''(t) \r>}{|\gamma'(t)|}$.
  
  Thus it gives:
  $$
    \bb{v}(t_0 \mid t_0)
      = \dfrac{|\gamma'(t_0)|^2}{|\gamma''(t_0)|^2} \gamma''(t_0)
      = \dfrac{|\gamma'(t_0)|^2}{|\gamma''(t_0)|} \hat{\gamma''}(t_0),
  $$
  
  and especially if $|\gamma'(t_0)| = 1$, then
  $\bb{v} = \dfrac{1}{|\gamma''(t_0)|} \hat{\gamma''}(t_0)$.
  

***Proof***


From Tayler's theorem, we note that:

$$
\Ba
\l<\bb{v}(t), \gamma(t) - \gamma(t_0) \r> 
  &= \l<\bb{v}(t), \gamma'(t_0) \r> (t - t_0)
    + \l<\bb{v}(t), \gamma''(t_0) \r> \frac{1}{2} (t - t_0)^2
    + O((t - t_0)^3 \\
  &=
    \frac{1}{2}
    (
      \l<\bb{v}(t), \gamma''(t_0) \r>
      + O(t - t_0)
    )
    (t - t_0)^2
\Ea
$$

Thus, we have:

$$
\Ba
|\bb{v}(t)| &= |\gamma(t) - (\gamma(t_0) + \bb{v}(t))| \\
  &\iff
    |\gamma(t) - \gamma(t_0)|^2 
      = 2 \l<\bb{v}(t), \gamma(t) - \gamma(t_0) \r> \\
  &\iff
    \l| \frac{\gamma(t) - \gamma(t_0)}{t - t_0} \r|^2 
      = 2 \frac{ \l<\bb{v}(t), \gamma(t) - \gamma(t_0) \r>}{(t - t_0)^2}
      = \l<\bb{v}(t), \gamma''(t_0) \r> + O(t - t_0).
\Ea
$$


### Prop. 3

Define $s : \RR \to \RR : t \mapsto s(t)$ by
$
s(t) \equiv \int^t du |\gamma'(u)|
$
and write its inverse as $t(s)$.

Also define $\eta(s) \equiv \gamma(t(s))$, then we have:

$$
\eta''(s)
  = \frac{1}{|\gamma'|^2} (I - \hat{\gamma'} \hat{\gamma'}^T) \gamma''
  = \frac{1}{|\gamma'|^4} (|\gamma'|^2 I - \gamma' \gamma'^T) \gamma''
  = \frac{\!\!-1}{\;|\gamma'|^4} \gamma' \xx (\gamma' \xx \gamma'')
$$

**N.B.**

- We denote e.g. $\gamma' \equiv \gamma'(t(s))$ and $t' \equiv t'(s)$.


***Proof***

We note that $s'(t) = |\gamma'(s(t))|$ and
$t'(s) = \dfrac{1}{s'(t(s))} = \dfrac{1}{|\gamma'(t(s))|}$.

Then we see:

$$
t''(s)
  = \del_s \frac{1}{|\gamma'|}
  = - |\gamma'|^{-2} \frac{\gamma'^T}{|\gamma'|} \gamma'' t'
  = - \frac{\l< \gamma', \gamma'' \r>}{|\gamma'|^4}.
$$

Thus:

$$
\Ba
\eta''(s)
  = \gamma'' (t')^2 +  \gamma' t''
  = \gamma'' \frac{1}{|\gamma'|^2}
    - \gamma' \frac{\l< \gamma', \gamma'' \r>}{|\gamma'|^4}
  = \frac{1}{|\gamma'|^2}
      (I - \frac{\gamma' \gamma'^T}{|\gamma'|^2}) \gamma''.
\Ea
$$


### Prop. 4

For $\gamma(x) \equiv \Bmp x \\ y(x) \Emp$ and $\eta$ from **Prop.3**, we have:

$$
\eta''
  = \frac{y''}{(1 + (y')^2)^2}
      \Bmp -y' \\ 1 \Emp.
$$

***Proof***

We have
$
\gamma' = \Bmp 1 \\ y' \Emp
$
and
$
\gamma'' = \Bmp 0 \\ y'' \Emp
$
, thus:

- $|\gamma'|^2 = 1 + (y')^2$,
- $
|\gamma'|^2 I - \gamma' \gamma'^T
  =
    \Bmp
      1 + y'^2 - 1 & - y' \\
      - y'         & 1 + y'^2 - y'^2
    \Emp
  =
    \Bmp
      y'^2 & - y' \\
      - y' &   1
    \Emp    
$.

Therefore:

$$
\Ba
\eta''
  &= \frac{1}{|\gamma'|^4} (|\gamma'|^2 I - \gamma' \gamma'^T) \gamma''
  = \frac{1}{1 + (y')^2}
    \Bmp
      y'^2 & - y' \\
      - y' &   1
    \Emp
    \Bmp 0 \\ y'' \Emp \\
  &= \frac{y''}{(1 + (y')^2)^2}
      \Bmp -y' \\ 1 \Emp.
\Ea
$$

### Prop. 5

For $f(x, y(x)) = 0$ and $\gamma$, $\eta$ from **Prop. 4**, we have:

$$
\eta''
  = \frac{f_{11} f_2^2 - 2 f_{12} f_1 f_2 + f_{22} f_1^2}{(f_1^2 + f_2^2)^2}
    \Bmp f_1 \\ f_2 \Emp
  = -
    \frac{
      \quad \l|\Bmp f_2 \\ - f_1 \Emp \r|_{\bb{D}^2f}^2
    }{
      \l|\Bmp f_1 \\ f_2 \Emp \r|^2
    }  \Bmp f_1 \\ f_2 \Emp
$$

**N.B.**

- Given $f : \RR^n \to \RR^{n - 1}$ with $\del_\bb{y} f(x, \bb{y})$ is invertible
  where $x \in \RR$ and $\bb{y} \in \RR^{n - 1}$,
  by implicit function theorem, there (locally) exists
  $x \mapsto \bb{y}(x)$ s.t. $f(x, \bb{y}(x))$.
  
- We denote $f_1 \equiv \del_1 f(x, y)$ and $f_2 \equiv \del_2 f(x, y)$.

- We denote $|v|_A^2 \equiv \l< \bb{v}, A \bb{v} \r> = \bb{v}^T A \bb{v}$.


***Proof***

Since $f(x, y(x)) = 0$, we have:

- $
0 = \del_x f(x, y(x))
  = \bb{D} f \cdot \Bmp 1 \\ y' \Emp
  = f_1 + f_2 \, y'
$,

- $
0 = (\del_x)^2 f(x, y(x))
  = \bb{D} f \cdot \Bmp 0 \\ y'' \Emp + \l| \Bmp 1 \\ y' \Emp \r|_{\bb{D}^2 f}^2
  = f_2 \, y'' + \l| \Bmp 1 \\ y' \Emp \r|_{\bb{D}^2 f}^2
$.

Therefore:

- $f_2 \, y' = - f_1$,

- $
(f_2)^3 \, y''
  = - (f_2)^2 \l| \Bmp 1 \\ y' \Emp \r|_{\bb{D}^2 f}^2
  = - \l| \Bmp f_2 \\ f_2 y' \Emp \r|_{\bb{D}^2 f}^2  
  = - \l| \Bmp f_2 \\ - f_1 \Emp \r|_{\bb{D}^2 f}^2
$

and thus:

$$
\Ba
\eta''
  &= \frac{y''}{(1 + (y')^2)^2}
      \Bmp -y' \\ 1 \Emp
   = \frac{(f_2)^3 y''}{((f_2)^2 + (f_2 \, y')^2)^2}
      \Bmp - f_2 \, y' \\ f_2 \Emp
  = -
    \frac{
      \l| \Bmp f_2 \\ - f_1 \Emp \r|_{\bb{D}^2 f}^2  
    }{((f_1)^2 + (f_2)^2)^2}
      \Bmp f_1 \\ f_2 \Emp.
\Ea
$$

curvature-01-surface

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## Surface

**Goals**

- Prove classical results using usual Jacobian and matrix manipulation
  (i.e. without differential form etc...)


**Notations**

For $F : \RR^2 \to \RR^3$, we denote:

- $F(u, v)$,
- $F_u = \del_u F$,
- $\bb{D} F = \Bmp F_u,  F_v\Emp \in \RR^{3 \xx 2}$,
- $g = (\bb{D} F)^T \bb{D} F \in \RR^{2 \xx 2}$

For quadratic $Q \in \RR^{n \xx n}$, we denote:

- $Q[\bb{x}, \bb{y}] = \l< \bb{x}, Q \bb{y} \r> = \bb{x}^T Q \bb{y}$.

**Preliminary. 1**

- $
|g| = 
  \l|\Bm
    |F_u|^2 & \l< F_u, F_v \r> \\
    \l< F_u, F_u \r> & |F_v|^2
  \Em\r|
  =
     |F_u|^2 |F_v|^2 -  \l< F_u, F_v \r>^2
  = |F_u \xx F_v|^2
$.



**Preliminary. 2**

In **Prop.3** of the [previous section](https://hi-ogawa.github.io/markdown-tex/?id=e40372524f96337f1f2066ad332b4d2b&filename=curvature-00-curve),
we found that 

$$
\eta''(s)
  = \frac{1}{|\gamma'|^4} (|\gamma'|^2 I - \gamma' \gamma'^T) \gamma''
  = \frac{\!\!-1}{\;|\gamma'|^4} \gamma' \xx (\gamma' \xx \gamma'').
$$

where

- $\gamma : \RR \to \RR^2$ : curve,
- $\eta(s) = \gamma(t(s))$ : arc-length parametrized curve.

Here we consider the surface defined by $F : \RR^2 \to \RR^3$ and
the curve $\gamma(t) = F(\mu(t)) : \RR \to \RR^2 \to \RR^3$ within the surface.
Then we deine:

- $N \equiv \widehat{F_u \xx Fv}$ : (right-handed) normal
- $k \equiv \l< N, \eta'' \r>$ : normal curvature

where arc-length parametrization $\eta(s)$ is naturally defined for
3-dimentional case $\gamma : \RR \to \RR^3$ as well.
But, note that it's not necessarily $\eta'' \parallel N$
even though it's always $\eta'' \perp \gamma'$ and $N \perp \gamma'$.


### Prop.1 

$$
k = \frac{ A[\mu', \mu'] }{ g[\mu', \mu'] }
$$

where $
A[\bb{x}, \bb{y}]
  = \l< N, \bb{D}^2 F[\bb{x}, \bb{y}] \r>
  = \l< N, \bb{D}^2 F \r> [\bb{x}, \bb{y}]
$.

**N.B.**

- More explicitly:

$$
\Ba
A 
  &= \l< N, \bb{D}^2 F \r>
  = 
    \Bmp
      \l< N, F_{uu} \r> & \l< N, F_{uv} \r> \\
      \l< N, F_{uv} \r> & \l< N, F_{vv} \r> 
    \Emp
  = 
    \frac{1}{|F_u \xx F_v|}
    \Bmp
      \l< F_u \xx F_v, F_{uu} \r> & \l< F_u \xx F_v, F_{uv} \r> \\
      \l< F_u \xx F_v, F_{uv} \r> & \l< F_u \xx F_v, F_{vv} \r> 
    \Emp \\\\
  &= 
    \frac{1}{|g|^{1/2}}
    \Bmp
      \l< F_u \xx F_v, F_{uu} \r> & \l< F_u \xx F_v, F_{uv} \r> \\
      \l< F_u \xx F_v, F_{uv} \r> & \l< F_u \xx F_v, F_{vv} \r> 
    \Emp.
\Ea
$$


***Proof***

First we note:

- $\gamma' = \bb{D}F \cdot \mu'$,
- $\gamma'' = \bb{D}^2 F [\mu', \mu'] + \bb{D}F \cdot \mu''$,
- $\l< N, \bb{D} F \r> = 0$,
- $\l< N, \gamma' \r> = 0$,
- $|\gamma'|^2 = \l< \bb{D}F \mu', \bb{D}F \mu' \r> = g[\mu', \mu]$,
- $\l< N, \gamma'' \r> = \l< N, \bb{D}^2 F [\mu', \mu'] \r>$.

Thus,

$$
\l< N, \eta'' \r>
  =
    \frac{1}{|\gamma'|^2}
    \l< N, \gamma'' \r>
    -
    \frac{\gamma'^T \gamma''}{|\gamma'|^4}
    \l< N , \gamma' \r>
  =
    \frac{\l< N, \bb{D}^2 F [\mu', \mu'] \r>}{g[\mu', \mu]}.
$$


### Prop. 2

Writing $\mu' = g^{-\frac{1}{2}} \bb{x}$ and
$B \equiv g^{-\frac{1}{2}} A g^{-\frac{1}{2}}$ , we have:

$$
k = \frac{ A[\mu', \mu'] }{ g[\mu', \mu'] }
  = \frac{ g^{-\frac{1}{2}} A g^{-\frac{1}{2}}[\bb{x}, \bb{x}] } { |\bb{x}|^2 }
  = B[\hat{\bb{x}}, \hat{\bb{x}}]
$$

Thus, denoting its diagonalization by $B = P \Bmp k_1 & 0 \\ 0 & k_2 \Emp P^{-1}$
with $P = \Bmp \bb{x}_1, \bb{x}_2 \Emp$ and $k_1 \ge k_2$, we have:

- $\t{max}(k) = k_1$ when $\mu' = \mu'_1 = g^{- \frac{1}{2}} \bb{x}_1$,
- $\t{min}(k) = k_2$ when $\mu' = \mu'_2 = g^{- \frac{1}{2}} \bb{x}_2$,
- $k = \cos^2(\phi) k_1 + \sin^2(\phi) k_2$
  when $\hat{\bb{x}} = \cos(\phi) \bb{x}_1 + \sin(\phi) \bb{x}_2$
  (aka. Euler's theorem)
- $\bb{x}_1 \perp \bb{x}_2$,
- N.B. maximality, minimality and orthogonality comes from usual spectral theorem.


Also, for $\gamma'_i = \bb{D} F \mu'_i = \bb{D} F g^{- \frac{1}{2}}\bb{x}_i$ where ($i = 1, 2$),
we have $\gamma'_1 \perp \gamma'_2$ since

$$
\l< \gamma'_1, \gamma'_2 \r>
  = \bb{x}_1^T g^{- \frac{1}{2}} g g^{- \frac{1}{2}} \bb{x}_2
  = \l< \bb{x}_1,  \bb{x}_2 \r> = 0.
$$


### Def. 3

- Gaussian curvature : $k_G \equiv k_1 k_2$,
- Mean curvature : $k_M \equiv \dfrac{k_1 + k_2}{2}$.


### Prop. 4

Noting that $|F_u \xx F_v| = |g|^{1/2}$,

$$
\Ba
2 k_M
  &= \t{tr}[B] = \t{tr}[g^{-1} \l< \widehat{F_u \xx F_v}, \bb{D}^2 F \r>] \\
  &=
    \frac{1}{|g|^{3/2}}
    \t{tr}[\alpha_g^T \l< F_u \xx F_v, \bb{D}^2 F\r> ] \\  
  &= 
    \frac{1}{|g|^{3/2}}
    \bigg(
      |F_v|^2 \l< F_u \xx F_v, F_{uu} \r>
      + |F_u|^2 \l< F_u \xx F_v, F_{vv} \r>
      - 2 \l< F_u, F_v \r> \l< F_u \xx F_v, F_{uv} \r>
    \bigg).
\Ea
$$

where we used:

- $
\alpha_g^T
  = \Bmp
      |F_v|^2            & - \l< F_u, F_v \r> \\
      - \l< F_u, F_v \r> & |F_u|^2
    \Emp.
$


### Prop. 5

$$
\Ba
k_G
  &= \t{det}[B]
  = \frac{\t{det}[\l< \widehat{F_u \xx F_v}, \bb{D}^2 F \r>]}{\t{det}[g]} 
  = \frac{1}{|g|^2} \t{det}[\l< F_u \xx F_v, \bb{D}^2 F \r>] \\
  &=
    \frac{1}{|g|^2}
    \bigg(
      \l< F_u \xx F_v, F_{uu} \r> \l< F_u \xx F_v, F_{vv} \r>     
      - \l< F_u \xx F_v, F_{uv} \r>^2
    \bigg).
\Ea
$$


### Prop. 6

As a special case of $F(u, v) = (u, v, f(u, v))$ (aka. scalar surface), we have:

- $F_u = \Bmp 1 \\ 0 \\ f_u \Emp$, $F_v = \Bmp 0 \\ 1 \\ f_v \Emp$,

- $F_u \xx F_v = \Bmp - f_u \\ - f_v \\ 1 \Emp $,

- $g = |F_u \xx F_v|^2 = 1 + f_u^2 + f_v^2$,

- $\l< F_u \xx F_v, \bb{D}^2 F \r> = \bb{D}^2 f$.

Thus,

$$
\Ba
2 k_M
  &=
    \dfrac{1}{(1 + f_u^2 + f_v^2)^{3/2}}
    \bigg(
      (1 + f_v^2) f_{uu}
      + (1 + f_u^2) f_{vv}
      - 2 f_u f_v f_{uv}
    \bigg), \\
k_G
  &=
    \dfrac{1}{(1 + f_u^2 + f_v^2)^2}
    ( f_{uu} f_{vv} - f_{uv}^2 ).    
\Ea
$$


### Def. 7 (Shape operator)

Define $S$ as unique $S \in \RR^{2 \xx 2}$ s.t.

$$
\bbD F \cdot S = \bbD N \in \RR^{3 \xx 2}.
$$

**N.B.**

Uniqueness holds since:

$$
\Ba
&\bbD F \cdot S = \bbD N \\
  &\implies
    \bbD F^T \cdot \bbD F \cdot S = \bbD F^T \cdot \bbD N \\
  &\iff
    S = (\bbD F^T \cdot \bbD F)^{-1} \cdot \bbD F^T \cdot \bbD N
      = g^{-1} \cdot \bbD F^T \cdot \bbD N
\Ea
$$

and its converse holds since $\bbD N \in \t{span}(\bbD F)$.


### Prop. 8

$$
S = - g^{-1} \l< N, \bbD^2 F \r>
$$

Thus, especially

- $\t{tr}[S] = - \t{tr}[g^{-1} \l< N, \bbD^2 F \r>] = - \t{tr}[B] = - (k_1 + k_2)$,
- $\det[S] = \det[g^{-1} \l< N, \bbD^2 F \r>] = \det[B] = k_1 k_2$.

So, this already captures information about the curvature
without going through $B = g^{-1/2} \l< N, \bbD^2 F \r> g^{-1/2}$.

***Proof***

It safices to show $\bbD F^T \cdot \bbD N = - \l< N, \bbD^2 F\r>$.
From $\bbD F^T \cdot N = 0$, we see that:

$$
\Ba
0 &= \bbD (\bbD F^T \cdot N)
  = \bbD^2 F^T \cdot N + \bbD F^T \cdot \bbD N \\
  &= \l< N, \bbD^2 F \r> + \bbD F^T \cdot \bbD N.
\Ea
$$


### Def. 9 (Laplace-Beltrami operator)

For $f : \RR^n \to \RR$, define $\Delta f : \RR^n \to \RR$ by:

$$
\Delta f
  \equiv
    \frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j} \del_j f)
$$

**N.B.**

- We can separate 1st/2nd derivatives as:

$$
\Ba
\frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j} \del_j f)
  &=
    g^{-1}_{i, j} \del_{i, j} f
    +
    \frac{1}{|g|^{1/2}} 
      \del_j f \;
        \del_i (|g|^{1/2} g^{-1}_{i, j} ) \\
  &=
    \t{tr}[g^{-1} \bbD^2 f]
    +
    \frac{1}{|g|^{1/2}} 
      \bbD f
        \cdot
        \Bmp
          \del_i |g|^{1/2} g^{-1}_{i, 1} \\
          \del_i |g|^{1/2} g^{-1}_{i, 2} \\
          \vdots
        \Emp.
\Ea
$$

- For vector-valued function (e.g. $F : \RR^2 \to \RR^3$), we define
  $\Delta F$ in a pointwise manner (i.e. $\Delta$ for each component).


### Prop. 10

$$
\Delta F = 2 k_M N
$$

**N.B.**

- Here we use a different notation for derivative by
  $F_1 \equiv = F_u = \del_u F$ and $F_2 \equiv F_v = \del_v F$ since
  I found that the subscripts $u$ and $v$ are sometimes not easy to discern.

***Proof***

It safices to show:

- (1) $\bbD F^T \cdot \Delta F = \bb{0} \in \RR^{2}$, i.e. $\Delta F$ doesn't have tangential component, and
- (2) $N^T \cdot \Delta F = 2 k_M \in \RR$.

First, we show (2).

Since $N^T \cdot \bbD F = 0$, the term with 1st derivative vanishes and we have:

$$
\l< N, \Delta F \r>
  = \l< N, \t{tr}[g^{-1} \bbD^2 F] \r>
  = \t{tr}[g^{-1} \l< N, \bbD^2 f \r>]
  = 2 k_M.
$$

Next, we show (1).

Looking at 1st/2nd derivative terms:

$$
\bbD F^T \cdot \Delta F
  =
    \bbD F^T \cdot \t{tr}[g^{-1} \bbD^2 F]
    +
    \bbD F^T \cdot \bbD F \cdot 
      \frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j}) \\
  =
    \bbD F^T \cdot \t{tr}[g^{-1} \bbD^2 F]
    +
    g \cdot 
      \frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j}).
$$

Here we denote $\beta \in \RR^{2 \xx 2}$ by $g^{-1} = \dfrac{1}{|g|} \beta$, i.e.

$$
\beta =
  \Bmp
    |F_2|^2 & - \l< F_1, F_2 \r> \\
    - \l< F_1, F_2 \r> & |F_1|^2 \\    
  \Emp.
$$

Also remember that $|g| = |F_1 \xx F_2|^2$.

Now, expanding the last term:

$$
\Ba
\del_i (|g|^{1/2} g^{-1}_{i, j})
  &= \del_i ( |g|^{-1/2} \beta_{i, j})
  =
    (\del_i |g|^{-1/2}) \; \beta_{i, j}
    +
    |g|^{-1/2} \del_i \beta_{i, j} \\
  &=
    \beta \cdot (\bbD |g|^{-1/2})^T
    +
    |g|^{-1/2} \del_i \beta_{i, j},    
\Ea
$$

$$
\bbD |g|^{-1/2}
  = - \frac{1}{2} |g|^{-3/2} \bbD |g|
  = - |g|^{-3/2} \l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>,
$$

$$
\del_i \beta_{i, j}
  = 
    \Bmp
      \del_1 |F_2|^2 - \del_2 \l< F_1, F_2 \r> \\
      - \del_1 \l< F_1, F_2 \r> + \del_2 |F_1|^2
    \Emp
  = 
    \Bmp
      \l< F_2, F_{12} \r> - \l< F_1, F_{22} \r> \\
      \l< F_1, F_{12} \r> - \l< F_{11}, F_2 \r>
    \Emp,
$$

so, we found that (note that $\; g \beta = |g| I \;$):

$$
\Ba
g \cdot \frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j})
  &=
    - \frac{1}{|g|^2}
      g \cdot \beta \cdot \l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>^T
    + \frac{1}{|g|}
      g \cdot (\del_i \beta_{i, j}) \\
  &=
    \frac{1}{|g|}
    \l(
      - \l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>^T
      + g \cdot (\del_i \beta_{i, j}) 
    \r).
\Ea
$$

Inserting this into the original formula, we obtain:

$$
\Ba
\bbD F^T \cdot \Delta F
  &=
    \bbD F^T \cdot \t{tr}[g^{-1} \bbD^2 F]
    +
    g \cdot 
      \frac{1}{|g|^{1/2}} \del_i (|g|^{1/2} g^{-1}_{i, j}) \\
  &=
    \frac{1}{|g|}
      \l(
        \bbD F^T \cdot \t{tr}[\beta \; \bbD^2 F]
        - \l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>^T
        + g \cdot (\del_i \beta_{i, j})         
      \r)
\Ea
$$

Finally, we're going to expand all terms:

$$
\Ba
\bbD F^T \cdot \t{tr}[\beta \; \bbD^2 F]
  &=
    \t{tr} \l[
      \beta \cdot
      \Bmp
        \l< F_i, F_{11} \r> & \l< F_i, F_{12} \r> \\
        \l< F_i, F_{12} \r> & \l< F_i, F_{22} \r>        
      \Emp
    \r]_i \\
  &=
    \Bmp
      |F_2|^2 \l< F_1, F_{11} \r>
        + |F_1|^2 \l< F_1, F_{22} \r> 
        - 2 \l< F_1, F_2 \r> \l< F_1, F_{12} \r> \\
      |F_2|^2 \l< F_2, F_{11} \r>
        + |F_1|^2 \l< F_2, F_{22} \r> 
        - 2 \l< F_1, F_2 \r> \l< F_2, F_{12} \r>
    \Emp, \\\\
\l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>^T
  &=
    \Bmp
      \l< F_1 \xx F_2, \del_1 (F_1 \xx F_2) \r> \\
      \l< F_1 \xx F_2, \del_2 (F_1 \xx F_2) \r>
    \Emp \\
  &=
    \Bmp
      \l< F_1 \xx F_2, F_{11} \xx F_2 + F_1 \xx F_{12} \r> \\
      \l< F_1 \xx F_2, F_1 \xx F_{22} + F_{12} \xx F_2 \r>
    \Emp \\
  &=
    \Bmp
      \l< F_1, F_{11} \r> |F_2|^2 - \l< F_1, F_2 \r> \l< F_{11}, F_2 \r> +
      \l< F_2, F_{12} \r> |F_1|^2 - \l< F_1, F_2 \r> \l< F_1, F_{12} \r> \\
      \l< F_2, F_{22} \r> |F_1|^2 - \l< F_1, F_2 \r> \l< F_1, F_{22} \r> +
      \l< F_1, F_{12} \r> |F_2|^2 - \l< F_1, F_2 \r> \l< F_2, F_{12} \r>
    \Emp, \\\\
g \cdot \del_i \beta_{i, j}
  &= 
    g \cdot
    \Bmp
      \l< F_2, F_{12} \r> - \l< F_1, F_{22} \r> \\
      \l< F_1, F_{12} \r> - \l< F_{11}, F_2 \r>
    \Emp \\
  &= 
    \Bmp
      |F_1|^2 (\l< F_2, F_{12} \r> - \l< F_1, F_{22} \r>)
      + \l< F_1, F_2 \r> ( \l< F_1, F_{12} \r> - \l< F_{11}, F_2 \r> ) \\
      \l< F_1, F_2 \r> (\l< F_2, F_{12} \r> - \l< F_1, F_{22} \r>)
      + |F_2|^2 (\l< F_1, F_{12} \r> - \l< F_{11}, F_2 \r>)
    \Emp,
\Ea
$$

and we see all terms cancels each other:

$$
|g| \; \bbD F^T \cdot \Delta F
  =
    \bbD F^T \cdot \t{tr}[\beta \; \bbD^2 F]
    - \l< F_1 \xx F_2, \bbD (F_1 \xx F_2) \r>^T
    + g \cdot (\del_i \beta_{i, j})       
  = 0.
$$