The Fundamental theorem of Riemannian geometry:The assignment $X\rightarrow \nabla X$ on $(M,g)$ is uniquely defined by the following properties
(1)$Y\rightarrow \nabla_Y X$ is a $(1,1)-tensor$ $$\nabla_{fY}X=f\nabla_YX$$
(2)$X\rightarrow\nabla_YX$ is a derivation $$\nabla_Y(fX)=Y(f)X+f\nabla_YX$$
(3)Covariant differentiation is torsion free $$\nabla_XY-\nabla_YX=[X,Y]$$
(4)Covariant differentiation is metric $$Z<X,Y>=<\nabla_ZX,Y>+<X,\nabla_ZY>$$
If one satisfies $(1)$ and $(2)$,it is called an affine connection.Furthermore,if it also satiefies $(3)$ and $(4)$,it is called a Riemannian connection.
We can use $Koszul's formula$ to verify them.
But my consusion is that does there exist a connection that only satisfies $(1)$ and $(2)$ but not $(3)$ and $(4)$?
Because I find the $Koszul's formula$ doesn't tell me how to distinguish whether an affine connection is a Riemannian connection or not.
Best Answer
There are lots of torsion free affine connection which are not compatible with the metric. We can in some sense "enumerate" them in terms of tensor fields.
Let $(M,g)$ be a Riemannian manifold, and $\nabla$ by its Levy-Civita connection.
It is a well known result that any two affine connection differ by a $(2,1)$ tensor. Thus, any affine connection $\tilde{\nabla}$ on $M$ can be written as $$ \tilde{\nabla}_XY=\nabla_XY+A(X,Y) $$ Where $A$ is a $(2,1)$ tensor field. Furthermore, two connections are equivalent iff they correspond to the same tensor field. Thus the set of all affine connections on $M$ is isomorphic to the space of $(2,1)$ tensor fields on $M$. The Levy-Civita connection is given by $A=0$ and no others.
With a computation, one can see that a connection on $M$ is torsion free iff $A$ is symmetric, in the sense that $A(X,Y)=A(Y,X)$. Each such nonzero symmetric tensor field on $M$ corresponds to a torsion free connection distinct from $\nabla$.