TITLE: Is there a way to define a Lie derivative of a connection?
QUESTION [5 upvotes]: 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), 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.

REPLY [2 votes]: I was expecting the Lie derivative of a connection to be another connection but Vladimir's answer about the difference of two connections being a tensor makes a lot of sense. I would say a coordinate free algebraic formula for the Lie derivative of a connection is:
$\left(\mathcal{L}_{X}\nabla \right)_{Y}Z:=\mathcal{L}_{X}\left( \nabla_{Y}Z\right)-\nabla_{\mathcal{L}_{X}Y}Z-\nabla_{Y}\mathcal{L}_{X}Z$.
This is just the formula for the Lie derivative of a tensor and in this case it does give a $C^{\infty}(M)$-linear object in the variables $Y$ and $Z$. Since this is a tensor field probably a better notation would be
$\left(\mathcal{L}_{X}\nabla\right)\left( Y,Z \right)$.