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

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 $$Z=<\nabla_ZX,Y>+$$

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.

• Yes, there are also non-Riemannian ones, see Thurston's book Three-Dimensional Geometry and Topology. You can also ask "does there exist a connection that only satisfies $(1),(2),(3)$ and but not $(4)$", right? Jan 20, 2020 at 16:31
• @DietrichBurde Why $(4)$ is independent of the other three?
– LSY
Jan 20, 2020 at 16:33
• Why should it be dependent on $(1),(2),(3)$? Jan 20, 2020 at 16:33
• @DietrichBurde But why a connection also satisfies $(3)$ and $(4)$ ,we give him a new name,Riemannian connection?
– LSY
Jan 20, 2020 at 16:37
• See "Riemannian manifold". There are also affine connections on non-Riemannian manifolds. Jan 20, 2020 at 16:38

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 $$\nabla$$.