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We know that a connection $\nabla$ in a manifold M hashas the purpose of performing the same role as the covariant derivative of vector fields of surfaces in $\mathbb{R}^3$. Such analogies are clearly perceived in the connection definition.

When M has a Riemannian metric,$~~ g~~$, to make things even more similar to the case in $\mathbb{R}^3$ we introduce the concept of Riemannian connection or Levi-Civita connection, which is a connection satisfying:

  1. compatibility with the metric: $Xg(Y,Z)=g(\nabla_X Y,Z)+g(Y,\nabla_X Z)$

  2. torsion-free: $T(X,Y)=\nabla_X Y-\nabla_YX-[X,Y]\equiv 0$

where $ X, Y, Z $ are vector fields on $M.$

What I understand is the need / intuition of the second property, torsion-free,what would work well if not we assume this hypothesis?

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3  
You might find this mathoverflow thread interesting: mathoverflow.net/questions/20493/…. One motivation for requiring $\nabla$ to have 0 torsion is that this (along with 1) singles out a unique connection. People do however study geometries with torsion (and I've heard these have applications to string theory). –  Eric O. Korman Apr 3 '12 at 20:37
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One good motivation for choosing a torsion free connection is the fact that it gives you equality of mixed partial derivatives, i.e. if $X$ and $Y$ are coordinate vector fields then $\nabla_X Y = \nabla_Y X$. –  treble Apr 3 '12 at 23:24

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