# Differential geometry – Double frame in Riemannian metrics.

Suppose we have a Riemannian metric. $$ds ^ 2 = Edu ^ 2 + 2Fdudv + Gdv ^ 2$$ in a coordinated local neighborhood $$(U; (u, v))$$ Prove that for the following vector fields:

$$e_ {1} = frac {1} { sqrt {E}} frac { partial} { partial u}, e_ {2} = frac {-1} { sqrt {EG- F ^ 2}} left ( frac {F} { sqrt {E}} frac { partial} { partial} – sqrt {E} frac { partial} { partial v} right )$$

the $$1-$$shapes:
$$omega_1 = sqrt {E} left (du + frac {F} {E} dv right), omega_2 = sqrt { frac {EG-F ^ 2} {E}} dv PS$$
satisfy
$$omega_i (e_k) = delta_ {ik}$$

My work: Let $$p (u, v)$$ A differentiable function, I believe that I must show fervently that $$omega_1 (e_1 (p)) = p$$, that is to say $$omega_1 (e_1) = 1$$ The identity function.

$$omega_1 left ( tfrac {1} { sqrt {E}} tfrac { partial} { partial u} (p) right) = tfrac {1} { sqrt {E}} tfrac { partial} { partial u} ( omega_1 (p))$$
I do this because a $$1-form$$ $$alpha$$ It is such that $$alpha (fX) = f alpha (X)$$. Then it is correct that
$$omega_1 (p) = sqrt {E} left (pdu + frac {F} {E} pdv right)?$$
and then apply the partial derivative, my problem is that I do not know how to operate the $$1$$-to form. Anyone can guide me on how I can achieve the result or an explicit way to operate $$omega_1$$ Y $$omega_2$$?

I am using the book "Umehara, differential geometry of surfaces".