Details to Lie derivative of Christoffel symbols

Question

I have a question to an answer given here: Lie derivative of the Christoffel symbol (I would commment there but don't have the necessary reputation yet.)

The question is how to find the expression $$ \mathcal{L}_\xi \Gamma^\mu {}_{\nu\lambda} = \xi^\sigma R^\mu {}_{\lambda\sigma\nu} + \nabla_\nu \nabla_\lambda \xi^\mu.\qquad\text{(1)} $$ in local coordinates for the Lie derivative of the Christoffel symbols.

In the cited answer, the proof is done in a coordinate-free way, however, not all steps are entirely clear to me. There, it is shown that, assuming that the connection is a $(1, 2)$-tensor, $$ (\mathcal{L}_{\xi}\nabla)(X,Y) \stackrel{?}{=} \mathcal{L}_{\xi}(\nabla_X Y) - \nabla_{\mathcal{L}_{\xi} X} Y - \nabla_X (\mathcal{L}_{\xi} Y) = R^{\nabla}(\xi, X) Y + (\nabla^2 \xi)(X,Y). $$ and written in coordinates this will result in Eq. (1).

I don't understand the following:

• From their indices, the Christoffel symbols look like components of a $(1, 2)$-tensor, so assuming that the connection is such a tensor makes sense to me. However, in the equal sign marked with a question mark, this is the formal expression for the Lie derivative of a $(0, 2)$-tensor; also, the third argument, a 1-form, is missing. What am I missing here?

• We also need to turn the coordinate-free expression on the left-hand side into the coordinate expression, i.e. $$ (\mathcal{L}_{\xi}\nabla) \rightarrow \mathcal{L}_\xi \Gamma^\mu {}_{\nu\lambda} $$ How would this work in detail? Since the coordinate-free calculation takes only two vector fields, it's unclear to me how I would the free upstairs index?

I could see that the whole calculation is an equation of vector fields; however, I still don't understand why we use the $(0, 2)$-tensor formula for the Lie derivative when we say we want it to be a $(1, 2)$-tensor.

I have found several posts regarding this topic, however, none of them ultimately tell the formal proof and every explanation I find contains one unexplained detail or two that I cannot figure out myself.

Answer 1

The formula for the Lie derivative of a $\binom12$-tensor is not the definition, but you can easily deduce it from the definition. In the definition, you would view a $\binom12$-tensorfield $A$ as mapping two vector fields and one one-form to a smooth function. But you can also view your tensor field as a map $\hat A$ that takes two vector fields as input and gives one vector field as output. The relation simply looks like $A(X,Y,\phi)= \phi(\hat A(X,Y))$. Now using the definition you get $$ (\mathcal L_\xi A)(X,Y,\phi)=\xi(A(X,Y,\phi))-A(\mathcal L_\xi X, Y,\phi)-A(X,\mathcal L_\xi Y,\phi)-A(X,Y,\mathcal L_\xi \phi). $$ The last term can be written as $-(\mathcal L_\xi \phi)(\hat A(X,Y))=-\xi(\phi(\hat A(X,Y)))+\phi(\mathcal L_\xi (\hat A(X,Y)))$. The second and third term in the right hand side of the above equation can be directly rewritten as $-\phi(\hat A(\mathcal L_\xi X,Y))-\phi(\hat A(X,\mathcal L_\xi Y))$. But the left hand side of the equation should just be interpreted as $\phi((\mathcal L_\xi \hat A)(X,Y))$ so the above equation simply reads as $$ \phi((\mathcal L_\xi \hat A)(X,Y))=\phi(\mathcal L_\xi (\hat A(X,Y)))-\phi(\hat A(\mathcal L_\xi X,Y))-\phi(\hat A(X,\mathcal L_\xi Y)), $$ Since this has to hold for each $\phi$, we get $$ (\mathcal L_\xi \hat A)(X,Y)=\mathcal L_\xi (\hat A(X,Y))-\hat A(\mathcal L_\xi X,Y)-\hat A(X,\mathcal L_\xi Y), $$ which is the formula that is used.

I think you have partly misunderstood the answer you have linked to. The connection is not a $\binom12$-tensor and it is nowhere claimed in that answer that it is. Formula (1) is used there as the definition of the Lie derivative of a connection, the relation to tensor fields is just an analogy. The justification for this definition is explained in the answer, it is via pullbacks of local flows of $\xi$.

The reason I am pointing this out is that in my opinion the formulation "Lie derivatives of the Christoffel symbols" is misleading and this is why you have problems with the coordinate expression. The general version of a Lie derivative can be defined on geometric objects as long as there is a natural action of local diffeomorphisms on them. This is the case for connections, but not for the Christoffel symbols which are just functions defined on the domain of a chart (and taking the Lie derivative in this picture would lead to a different result). What you actually want to get is the analog for $(\mathcal L_\xi\nabla)$ of what the Christoffel symbols are for $\nabla$. This means that you have to look at the value $(\mathcal L_\xi\nabla)(\partial_\nu,\partial_\lambda)$ of $\mathcal L_\xi\nabla$ on two coordinate vector fields and expand the result in coordinate vector fields. Otherwiese put the definition of the symbols you write as $\mathcal L_\xi\Gamma^\mu_{\nu\lambda}$ is that $$ (\mathcal L_\xi\nabla)(\partial_\nu,\partial_\lambda)=\sum_\mu (\mathcal L_\xi\Gamma^\mu_{\nu\lambda})\partial_\mu. $$ Once this is clear, the result is easily obtained from the explanations in the answer you have linked to.