Explanation of a proof in Colding-Minicozzi's "A Course in Minimal Surfaces".

Question

I have just started reading "A Course in Minimal Surfaces" by Colding-Minicozzi on my own. I have to clarify some points in the proof of a lemma given in the book. On page $30$ of the book, they prove the following lemma (Lemma $1.19$)

Now I understand that since the image of the Gauss map $N$, in this case is the upper hemisphere, which is contractible and exterior derivative commutes with pullback, hence

$N^*\omega =N^*(d\alpha)=d(N^*\alpha)$

What I don't understand are the following two points :-

1) Why is $|A|^2d$Area=$-2Kd$Area=$2N^*\omega$ ?

Here $A$ is the Second Fundamental form and $K$ is the Gaussian curvature. I know that for a minimal graph $|A|^2=-2K$ so the first equality is fine. But why the second equality ?

2) It's written "since $\alpha$ is a one form, hence $\exists$ a constant $C_{\alpha}$ so that $|N^*\alpha|\leq C_{\alpha}|dN|$" ? Why is this true ? Also, on what quantities does this constant $C_{\alpha}$ depends ?

Thank you for your help.

Answer 1

1. The equation $K dA = (-)N^* \omega$ is often used as the definition of the Gaussian curvature. If you're instead using $K = \det dN$, note that this is essentially the same equation, since the determinant tells us how pullbacks effect volume forms and $dA, \omega$ are the natural Riemannian area forms of ${\rm graph}\ u$, $S^2$ respectively.

2. In coordinates we have $(N^* \alpha)_i = (dN)_i^j \alpha_j$, so by Cauchy-Schwarz we have the pointwise inequality $|N^* \alpha| \le |dN| |\alpha|$. To get an upper bound on $|\alpha|$ I guess you need to do a little work - I'd just construct $\alpha$ on a subset of the sphere slightly larger than the hemisphere, so that the restriction to the closed hemisphere (which is compact) is bounded.