# covariant dervative

Let $f:U \to \mathbb{R}^3$ be a surface and Let $\nabla$ be the covariant differential. Given $X,Y$ two tangential vector fields on $f$.

Show that $\nabla_XY-\nabla_YX=[X,Y]$.

Note: $[X,Y]:=(X^iY^j_{,i}-Y^iX^j_{,i})f_i$ is the Lie bracket ( here the Eienstein summation convention is used, thus $i,j$ are sumed over $1,2$. Note also $X^j_{,i}=\frac{\partial X^j}{\partial u^i}$ and $f_i=\frac{\partial f}{\partial u^i}$)

My query: I was introduced by the following definition of $\nabla_XY$ : For each $(u_0,v_0) \in U$ one may define $\nabla X: T_{u_o}f \to T_{u_0}f$ via $$\nabla_XY=\nabla X(Y):=\frac{\nabla X(t_0)}{dt}$$, where $c(t)=f \circ u(t)$ is any curve through $u_0$ with $u_0=u(t_0)$ and $\dot{c}(t_0)=Y$.

Does this definition mean the covariant differential $\nabla X$ is always with respect to a point $(u_0,v_0) \in U$? If so then the equality above is meaningless due to the left hand side is dependent on $t$ whereas the right hand side is dependent on $u^1,u^2$?

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