Derivation of shape operator.

[Ming-Yeung-Alfred-Meng]

Could you please explain how you derived the second and third equality above?

[alecjacobson]

S_p(v) = ∇n⋅v

∇n can be realized as a matrix A ∈ ℝ²ˣ² in the 2D tangent plane spanned by basis vectors e₁ and e₂, that is:

S_p(v) = A v

What are the entries of A? Well, we can pick them out by multiplying against basis vectors e₁ = [1 0] and e₂ = [0 1]:

Aᵢⱼ = eⱼᵀ A eᵢ = eⱼᵀ ( S_p(eᵢ) ) = S_p(eᵢ) ⋅ eⱼ