Does there exist an affine connection is not a Riemannian connection?

Question

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.

Answer 1

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 $\nabla$.