Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 5 / c

First verify the difference of affine connections is a tensor. Linearity over smooth functions in $X$ follows from the affine connection axioms. In $Y$, the two derivative terms cancel:Thus $A$ is a smooth $(1,2)$-tensor field, and $A_p(u,v)$ depends only on the tangent vectors $u,v$ at $p$. For any smooth curve, its two covariant accelerations satisfyIf $A(X,Y)=-A(Y,X)$, its diagonal values vanish. The two geodesic equations are therefore identical, and the connections have the same parametrized geodesics.

For necessity, the required local geodesic existence theorem is this: for every smooth affine connection, $p\in M$ and $v\in T_{p}M$, a unique geodesic exists on some interval about zero with $\gamma (0)=p$ and $\dot{\gamma }(0)=v$. Indeed the geodesic equation is the smooth first-order ordinary differential equation on position and velocityso local existence and uniqueness apply. If the two connections have the same parametrized geodesics, apply the acceleration identity to this geodesic at zero to obtain $A_p(v,v)=0$ for every $p,v$. Expanding $A_p(u+v,u+v)=0$ gives $A_p(u,v)+A_p(v,u)=0$. ThereforeThis is the criterion that parametrized geodesics determine the symmetric part of an affine connection. It concerns equality with the same affine parameters. Agreement merely of unparametrized images is weaker: adding $\alpha (X)Y+\alpha (Y)X$ to a connection, for a smooth one-form $\alpha$, changes acceleration by a multiple of the velocity, which can be absorbed by reparametrization. Such a difference need not be antisymmetric.

Finally, since ${\nabla }_X^{\prime }Y={\nabla }_{X}Y-A(X,Y)$, expand the derivative of the metric tensor:Under the assumed metric compatibility of $\nabla$, the first term is zero. ConsequentlyEquivalently, for each fixed $X$, the endomorphism $A(X,\cdot )$ is skew-adjoint for the Riemannian metric. This metric criterion does not require the two connections to have the same geodesics.