As part of an exercise in Do Carmo's Differential Forms and Applications, it says that if we have a regular surface $M^2$ parametrized by $g(u,v)$ then we have
$$dy\wedge dz + dz\wedge dx + dx \wedge dy = \big(\sum_i(g_u\wedge g_v)_i\big)du\wedge dv$$
I think I've interpreted the question correctly but I'd like to have a second opinion on it. I've done some calculations using the machinery of differential forms but I don't know if I'm right or not.
So, if we write down $g(u,v)= g^i(u,v)e_i$ using Einstein's summation notation, chain rule gives us that
$$dg^i=\partial_u g^i\, du + \partial_v g^i\,dv$$
But $g^1(u,v)=x$,$g^2(u,v)=y$ and $g^3(u,v)=z$. Right? Therefore $dx=dg^1$,$dy=dg^2$, $dz=dg^3$. Correct?
Now that we know $dx,dy,dg$ with respect to $g^i$'s, we can calculate $$dy\wedge dz + dz\wedge dx + dx \wedge dy = (g^1_ug^2_v-g^1_vg^2_u+g^2_ug^3_v-g^2_vg^3_u+g^3_ug^1_v-g^3_vg^1_u)du\wedge dv$$
where $g^i_u=\partial_u g^i$ and $g^i_v=\partial_v g^i$.
Now my interpretation is that $g_u\wedge g_v$ is the wedge of these two (co-)vectors
$$g_u = \frac{dg}{du} = \begin{bmatrix}\frac{\partial g_1}{\partial u} & \frac{\partial g_2}{\partial u} & \frac{\partial g_3}{\partial u}\end{bmatrix}$$ $$g_v = \frac{dg}{dv} = \begin{bmatrix}\frac{\partial g_1}{\partial v}& \frac{\partial g_2}{\partial v} & \frac{\partial g_3}{\partial v}\end{bmatrix}$$
And $(g_u \wedge g_v)_i$ is the $i$-th coordinate of wedging the above vectors. The result will of course be a $2$-form. But then I can apply the $\star$-operation to it to get a $1$-form and because we have the standard inner product on $\mathbb{R}^3$, the spaces of vectors and co-vectors are isomorphic. Is that what Do Carmo meant by that equation? I'm terribly confused at this point.