Let $\mathcal{X}$ denote the vector space of smooth vector fields on a smooth Riemannian manifold $(M,g)$ . Note that $\mathcal{X}$ is actually a $\mathcal{C}^\infty(M)$ module because we can multiply a vector field by a function to obtain another vector field. The Riemann curvature tensor is the tri-linear $\mathcal{C}^\infty$ mapping $$R:{\mathcal{X}}\times{\mathcal{X}}\times{\mathcal{X}}\to{\mathcal{X}},$$ which is defined by $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$$ where $X,Y,Z\in\mathcal{X}$ are vector fields, where $\nabla$ is the Levi-Civita connection attached to the metric tensor $g$ , and where the square brackets denote the Lie bracket of two vector fields. The tri-linearity means that for every smooth $f\colon M\to\mathbb{R}$ we have $$fR(X,Y)Z=R(fX,Y)Z=R(X,fY)Z=R(X,Y)fZ.$$

In components this tensor is classically denoted by a set of four-indexed components ${R^i}_{jkl}$ . This means that given a basis of linearly independent vector fields $X_i$ we have $$R(X_j,X_k)X_l=\sum_s {R^s}_{jkl}X_s.$$

In a two dimensional manifold it is known that the Gaussian curvature it is given by $$K_g=\frac{R_{1212}}{g_{11}g_{22}-{g_{12}}^2}$$