Need assistance(Gussian curvature and differentiable vector fields)

[SuperLouisa90]

Need urgent assistance(Gussian curvature and differentiable vector fields)

Hi I have a very difficult problem where I know some of the dots but can't connect them :(

So therefore I hope that there is someone who can assist me (hopefully

Homework Statement

Let S be a surface with orientation N. Let $$V \subset S$$ be an open set in S and let $$f: V\subset S \rightarrow \mathbb{R}$$ be any nowhere zero differentiable function in V. Let $$v_1$$ and $$v_2$$ be two differentiable (tangent) vector fields in V such that at each point of V, $$v_1$$ and $$v_2$$ and that $$v_1 \land v_2 = N$$

Then prove that $$K = \frac{<d(fN)(V_1) \land d(fN)(V_2), fN>}{f^3}$$

p.s. there is also a question two but since this is so difficult I live that out for the time being hoping we can get to that later.

Homework Equations

The Attempt at a Solution

Here is what I know

Since S has the orientation N that according to do Carmo Geometry book means that it can be covered with a neighbourhood N.

From what I get is that

dN(v1) = cv1 + dv2 and dN(v2) = ev1 + fv2 but how do Carmo goes from that the above is a mystery to me. So therefore I hope there is someone who would help me understand what I am missing ?

Cheers
Louisa

 

[mighty2000]

Hi Louisa,

I understand that you are facing difficulties with the problem and I am happy to assist you. Firstly, let's try to understand the notation used in the problem.

V_1 and V_2 are differentiable vector fields, which means that they are functions that assign a vector to each point in V. The notation d(fN) represents the differential of the function fN, which is a linear transformation between tangent spaces.

Now, to prove the given statement, we can start by using the definition of Gaussian curvature K = det(dN). Since we know that v_1 \land v_2 = N, we can substitute dN(v_1) and dN(v_2) in terms of v_1 and v_2. This will give us dN(v_1) \land dN(v_2) = (cv_1 + dv_2) \land (ev_1 + fv_2).

Next, we can use the properties of the exterior product to expand this expression as (cv_1 \land ev_1) + (cv_1 \land fv_2) + (dv_2 \land ev_1) + (dv_2 \land fv_2).

Now, since v_1 and v_2 are differentiable vector fields, we can use the definition of the exterior product to show that this expression is equal to (cv_1 \land ev_1) + (dv_2 \land fv_2).

Finally, we can use the fact that v_1 \land v_2 = N to simplify this expression further as (cv_1 \land ev_1) + (dv_2 \land fv_2) = (cv_1 \land ev_1) + (dv_2 \land fv_2) \land N.

Substituting this in the original statement K = det(dN), we get K = det(dN) = <d(fN)(v_1) \land d(fN)(v_2), N>.

Since we are given that f is a nowhere zero differentiable function, we can use the chain rule to express d(fN) in terms of df and dN. This will give us d(fN)(v_i) = df(v_i)N + f(v_i)dN, where i = 1,2.

Substituting