I've been reading a little bit about the definition of symmetries on General Relativity, and they are related with the concept of *Killing vector*, i.e., vectors along which the Lie derivative of the metric vanishes $\mathcal{L}\_X g =0$.


However, afaik the most symmetric geometrical object is the Ricci tensor ([see the post](https://mathoverflow.net/a/123187/25356)), and the a vector $X$ satisfying $\mathcal{L}\_X \text{Ric} = 0$ is known as a *collineation* of the Ricci tensor.


I'd like to know whether is possible to define a sort of Lie derivative for a (general) connection, or a way to somehow define the symmetries of a connection.



Is there a way to define a Lie derivative of a connection?
Of course, yes. Lie derivative is defined for any geometric object (= when it is defined what happends when we change a coordinate system): take the flow $\phi\_t$ of the vector field, consider the pullback $\phi\_t^\*\Gamma$ of your geometric object $\Gamma$ and define Lie derivative as the $\tfrac{d}{dt}$-derviative at $t=0$ of $\phi\_t^\*\Gamma$. 


For affine connection its Lie derivative is an (1,2)-tensor field. It is because derivative is more or less the same as difference and difference of two connections is a (1,2)-tensor field. 


Of course if the Lie derivative of a connection is zero then the connection is presered by the flow so you vector field is a symmetry of the connection. 


Of course there exists an algebraic formula for the Lie derivative in terms of the components of an object in the coordinates ($\sim$ Christoffel symbols, for example), components of the vector field, and their first derivatives. I do not remember this formula by hart but Maples DifferentialGeometry package knows it. It is a sum of two terms, the first part is the usual formula of the Lie derivative for a (1,2)-tensorfield and the other part is something like the vector field plugged in the curvature. 


