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Let $\nabla$ be an affine connection on a smooth manifold $\mathcal M$. We say $\nabla$ is torsion-free if for any vector fields $X, Y$ on $\mathcal M$, we have \begin{equation} \nabla_X Y - \nabla_Y X = [X, Y], \end{equation} where $[X, Y] = X \circ Y - Y \circ X$ is the Lie bracket of $X$ and $Y$. Is it equivalent to state the torsion-free property as \begin{equation} \nabla_X Y = X \circ Y \end{equation} for all vector fields $X, Y$? Why or why not?

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$\nabla_XY$ is a vector field and hence acts as a derivation on $C^\infty(M)$, i.e. $\nabla_XY(fg)=f\nabla_XY(g)+g\nabla_XY(f)$ for any $f,g\in C^\infty(M)$. The object $X\circ Y$ is a second order differential operator \begin{align}X\circ Y(fg)&=X(fY(g)+gY(f))\\ &=X(f)Y(g)+f(X\circ Y)(g)+X(g)Y(f)+gX\circ Y(g)\\ &\neq f(X\circ Y)(g)+g(X\circ Y)(f) \end{align} Since $X\circ Y$ is not a derivation for all $f,g\in C^\infty(M)$ it does not give a well defined vector field so $\nabla_XY\neq X\circ Y$.

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    $\begingroup$ +1. OP should also note that the computation done here is exactly the reason why the Lie bracket $[X,Y]$ is defined the way it is. Originally, one would like $X\circ Y$ to be a vector field, but the presence of the second-order terms is an obstruction. How to get rid of it? Subtract $Y\circ X$ and note that all the second-order terms go away, since second-order mixed partial derivatives commute. $\endgroup$
    – Ivo Terek
    Feb 28 at 20:24
  • $\begingroup$ Crystal clear. Thanks! $\endgroup$
    – markusas
    Feb 29 at 20:41

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