Bianchi identity proof : why we can consider $[X,Y]=[Y,Z]=[X,Z]=0$?

Question

I recall that the Riemann curvature tensor is defined by \begin{align*} R:\Gamma(M)\times \Gamma(M)\times \Gamma(M)&\longrightarrow \Gamma(M)\\ (X,Y,Z)&\longmapsto [\nabla _X,\nabla _Y]Z-\nabla _{[X,Y]}Z. \end{align*} Notice that $\nabla $ denote Levi-Civita connexion. We denote $$R_{XY}:\Gamma(M)\longrightarrow \Gamma(M)$$ by $$R_{XY}Z=R(X,Y,Z).$$

We want to prove that $$R_{XY}Z+R_{YZ}X+R_{ZX}Y=0.$$ The proof start by : We can suppose WLOG that $[X,Y]=[Y,Z]=[X,Z]=0$.

Question : I really don't understand why we can suppose that. My teacher told me that we can always use a coordinate system where the bracket vanish, but I don't understand why. Do you have any explanation ?

Answer 1

Your teacher's explanation means that: For any local coordinate system $(x,U)$ of $M^n$ around a point $p$, we can always write

$$X=\sum_{i=1}^{n}X_i\frac{\partial}{\partial x_i} \hspace{1cm}Y=\sum_{j=1}^{n}Y_j\frac{\partial}{\partial x_j}\hspace{1cm}Z=\sum_{k=1}^{n}Z_k\frac{\partial}{\partial x_k} $$ and so on ...

Therefore if you make a simple calculation using the properties of the bracket you will have $$\bigl[\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\bigr]=0$$ for any $i,j=1\ldots n$