# Relation between Lie bracket and curvature

Let be $$M$$ a Riemannian mainfold with $$\dim M = 3$$. Let $$p\in M$$ and let $$\phi:U\subset\mathbb{R}^3\rightarrow V\subset M$$ a parameterization around $$p$$. Let $$X, Y, Z$$ the vector fields $$\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$$, where $$\phi^{-1}(q) = (x, y, z)$$ are local coordinates. We know, in $$p$$, the Lie bracket $$[X, Y] = 0$$. Does this imply $$R(X, Y)Z(p) = 0$$? Prove it or show a counterexample.

I'm stuck in this problem because I don't see a way to "jump" from Lie bracket to curvature.

The curvature definition says $$R(X, Y)Z = \nabla_Y\nabla_XZ - \nabla_X\nabla_YZ + \nabla_{[X, Y]}Z$$

and I know from symmetry $$[X, Y]=\nabla_XY-\nabla_YX$$.

In my efforts I tried to use this to develop a formula to double connection funding

$$\nabla_Y\nabla_XZ = \nabla_Y\nabla_ZY + \nabla_{[X, Z]}Y + [Y, [X,Z]]$$ and $$\nabla_X\nabla_YZ = \nabla_X\nabla_ZY + \nabla_{[Y,Z]}X + [X,[Y,Z]]$$

where I conclude that $$R(X,Y)Z = \nabla_Y\nabla_ZY - \nabla_X\nabla_ZY + \nabla_{[X, Z]}Y - \nabla_{[Y,Z]}X$$

but I don't fell like I'm getting anywhere so, any hint will be appreciate.

Very false. $$R$$ is a tensor, so $$(R(X,Y)Z)_p$$ depends on $$X_p$$, $$Y_p$$ and $$Z_p$$ only, so that the way you extend these particular vectors to vector fields near $$p$$ to compute the curvature via the definition (which involves Lie brackets) doesn't matter. You can take any non-flat $$M$$ to produce a counter-example, such as $$\Bbb S^3$$.