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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 ?

user330587
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1 Answers1

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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 $i,j=1\ldots n$

A s
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  • Thank you for your answer. How can you choose such coordinate system ? – user330587 Jun 22 '16 at 11:55
  • This is nothing but a chart of the atlas of $M$ – A s Jun 22 '16 at 14:04
  • If you search the definition of an atlas of a manifold you will easy understand the meaning of a chart – A s Jun 22 '16 at 14:05
  • I have an atlas $A={(U_i,\varphi_i)}$, now, how can I get that the braket vanish ? I really don't see. Could you explain more please ? Thank you :) – user330587 Jun 23 '16 at 07:48
  • To see that, you take two vector field $X$ an $Y$ as I have shown above, and you start the calculation. of $[X,Y]$. In particular if you take $X=\frac{\partial}{\partial x_i}$ and $Y=\frac{\partial}{\partial x_j}$ you will get $[X,Y]=0$ – A s Jun 23 '16 at 07:54
  • An important point is that the curvature $R$ is a tensor, i.e. $C^{\infty}(M)$-trilinear, even if the Lie bracket is not $C^{\infty}(M)$-bilinear. – Watson Jun 15 '17 at 12:10
  • @As Sorry to bump this after so long (tell me if I should just open a new question?), but doesn't this imply that for any vector fields $X$ and $Y$, we have $[X,Y]=0$? (well, obviously not, but could you explain why?) – boink Jan 27 '21 at 03:07