Why is the shape operator self adjoint regarding the first fundamental form?

323 Views Asked by At

Why is the shape operator $A: \mathbb R^2 \to \mathbb R^2 $self adjoint regarding the first fundamental form? I see it is written as a fact everywhere but cannot find a proof for it.

1

There are 1 best solutions below

0
On BEST ANSWER

I already found out myself that it is equivalent to the symmetry of the 2nd fundamental form.