Is the Shape Operator Always Symmetrical?

[Alteran]

Please, help me to solve that task related to Shape Operator.

We have surface $$S$$ and its normal $$N$$. Alse we have surface patch $$r : U -> S$$ in local coordinates $$r_1, r_2, ..., r_n$$. Shape operator (Weingarten Linear Operator) is defined as follow:
$$L_p : T_{r(p)}S -> T_{r(p)}S$$, where $$T_{r(p)}S$$ - tangent plane to surface.
It is known that $$L_p(w) = -D_vN(p), w \in T_{r(p)}S$$.
It is necessary to proof, that shape operator is symmetrical.

There is theorem that shows that shape operator is symmetrical $$L_p : T_pS -> T_pS, L_p(v)*w = v*L_p(w)$$, but on the surface. For patch we need to prof that matrix is symmetrical or something like that..

Can anyone lead me to right direction?
Thank you.

 

[Jhero]

Thank you for your question. The proof that the shape operator is symmetrical can be done using the definition of the shape operator and the properties of the derivative operator. First, we need to understand the definition of a symmetrical operator.

A linear operator L: V -> V on a vector space V is called symmetrical if, for all vectors v and w in V, we have L(v)*w = v*L(w). In other words, the operator L is symmetrical if it satisfies the property that the inner product of L(v) and w is equal to the inner product of v and L(w).

Now, let's consider the shape operator L_p : T_{r(p)}S -> T_{r(p)}S. We know that L_p(w) = -D_vN(p), where D_vN(p) is the directional derivative of the normal vector N at the point p in the direction of the tangent vector v. Since the shape operator is defined as a linear operator, we can write L_p(w) as L_p(v*w), where v*w is the inner product of v and w.

Using the definition of the shape operator, we have L_p(v*w) = -D_vN(p). Now, let's consider the right-hand side of the equation. By the properties of the derivative operator, we know that D_vN(p) = v*(D_N(p)). Therefore, L_p(v*w) = -v*(D_N(p)).

Now, substituting L_p(v*w) and -v*(D_N(p)) in the original equation, we have v*(L_p(w)) = -v*(D_N(p)). This shows that the inner product of L_p(w) and v is equal to the inner product of v and L_p(w). Therefore, the shape operator L_p is symmetrical.

I hope this helps to guide you in the right direction. Please let me know if you need further clarification or assistance with the proof. Good luck!