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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.

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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 $\Bbb S^3$.

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