Interchanging Duals and Metrics for Riemann Curvature Definition If $F_\nabla$ is the curvature associated to a connection $\nabla$ on a 2-dimensional Riemannian manifold $M$ with metric $g$, then we can define ''the'' Riemann curvature by (https://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds) by
$$
R(X,Y,Z,W) = g(F_\nabla (X,Y)Z, W)
$$
for vector fields $X,Y,Z,W$. 
This is related to a function $K:  M\to \mathbb{R}$ by 
$$
K(x)\sigma(X,Y)\sigma(Z,W) = R(X,Y,Z,W)\mid_x
$$
where $\sigma = e_1^{*}\wedge e_2^*$ is a globally-defined non-vanishing 2-form, and $e_1^*, e_2^*$ are dual to an orthonormal frame $e_1, e_2$. (Side note: is this function the same as the sectional curvature?)
What I would like to understand is whether there is a more direct way to express $K$ in terms of $g$ and $F_\nabla$: for example, a dual section $s^*$ is defined by $s^{*} = g(s,\cdot)$ -  so can we say something like 
$$
K(x)\sigma = \langle\omega , e_1\wedge e_2\rangle\mid_x
$$
where $\omega(s_1, s_2) := g(F_\nabla s_1, s_2)$ for vector fields $s_1, s_2$, and $\langle \cdot, \cdot\rangle$ in this instance is meant to indicate the natural pairing of a 2-form with a bivector.
 A: By the definition of the wedge product we have $\sigma(e_1,e_2)=e_1^*(e_1)e_2^*(e_2)-e_2^*(e_1)e_1^*(e_2)=1$; so just plugging this frame in to the equation $K\, \sigma \otimes \sigma = R$ yields the explicit formula $$R(e_1,e_2,e_1,e_2)=K \sigma(e_1,e_2)\sigma(e_1,e_2) = K.$$ Since $e_i$ is an orthonormal frame, this is indeed the sectional curvature; and also $\sigma$ is the Riemannian area form up to sign. (As an aside, unless your surface is genus $1$ there is no global orthonormal frame; so the equation $\sigma = e_1^* \wedge e_2^*$ should be understood locally.)
If you think of $R$ as a symmetric bilinear form on bivectors then this can be written $$K = R(e_1 \wedge e_2, e_1 \wedge e_2) = R(\sigma^*, \sigma^*).$$
Here the the metric dual relating bivectors and two-forms can be simply defined by $v\wedge w \mapsto v^* \wedge w^*,$ or alternatively by $$(v \wedge w)^* = g_2(v \wedge w,\cdot)$$ where  $$g_2(v\wedge w, x \wedge y) = g(v,x)g(w,y) - g(v,y)g(w,x)$$ is the inner product on bivectors induced by $g$. 
Your proposed equation for $K\sigma$ is what we get by only plugging in the frame once instead of twice: $$R(e_1,e_2,X,Y) = K\sigma(e_1,e_2)\sigma(X,Y) = K\sigma(X,Y),$$ which in terms of bivectors becomes $$K \sigma = R(e_1\wedge e_2, \cdot).$$ Your $\omega$ is really the same thing as $R$, and the natural pairing is the same thing as the partial evaluation I have written.
