Identity for the norm of the second fundamental form I am reading the paper on the proof of Lawson's conjecture of Simon Brendle.
During his calculations, he claims, that for 2 dimensional minimal surfaces, the following identity follows from the Codazzi equations:
$$ \lvert \nabla A \rvert^2 = 2 \lvert \nabla \lvert A \rvert \rvert^2,$$
where $\lvert A \rvert^2 = h_{ik}h_{ik}$ is the norm of the second fundamental form.
I am not sure how to understand $\lvert \nabla A \rvert$, i would guess
$$\lvert \nabla A \rvert^2 = \sum_{j=1}^2 \frac{\partial h_{ik}}{\partial x_j} \frac{\partial h_{ik}}{\partial x_j}$$
and i don't really see how this follows from the Codazzi equations.
Does anyone have a reference for this identity?
 A: Putting together the Codazzi equation $\nabla_i h_{jk} - \nabla_j h_{ik} = R_{ij}{}^\nu{}_k$ with $R_{ijkl} = g_{ik}g_{jl} - g_{il}g_{kj}$ yields $\nabla_i h_{jk} = \nabla_j h_{ik}$, since the slots $i,j$ of $h$ come from the tangent space to $\Sigma$, which is orthogonal to $\nu$. Thus $b := \nabla A$ is totally symmetric. Additionally, the fact that $\Sigma$ is minimal implies $\nabla_ih_{jj} = 0$, so $b_{i11} + b_{i22} = 0$. Thus we have the following relations amongst the $8$ components of $b$: $$b_{122} = b_{212} = b_{221} = -b_{111}\\ b_{211} = b_{121} = b_{112} = -b_{222}$$
Now, fix a point $p$ on $\Sigma$ and choose an orthonormal basis aligned to the principal vectors at $p$, so that at $p$, $A$ is diagonal. The minimal surface equation in this situation implies $h_{22} = -h_{11}$. Now we can easily compute $|\nabla A|^2$ - it's just the sum of squares of components of $b$, which (reducing everything to $b_{111}$ and $b_{222}$ using the relations above) is
$$ |\nabla A|^2 = 4(b_{111}^2 + b_{222}^2). $$
For $|\nabla|A||^2$, first compute
$$ \nabla_i|A| =\frac 1{2|A|}\nabla_i(h_{jk}h_{jk})=\frac1{|A|}b_{ijk}h_{jk} = \frac{h_{11}}{|A|}(b_{i11}-b_{i22}),$$
and thus (noting $|A|^2 = 2h_{11}^2$ and again putting everything in terms of $b_{111},b_{222}$) we get 
$$ |\nabla|A||^2 = \frac12 \left((b_{111} - b_{122})^2 + (b_{211} - b_{222})^2\right) = 2\left(b_{111}^2 + b_{222}^2\right);$$
so $|\nabla A|^2 = 2 |\nabla|A||^2$ as desired.
