Lie derivative for tensors

This generalize the Lie derivative of forms and the Lie derivative of vector fields

Approach 1

From @malament2012topics page 53.
For any smooth vector field ${\lambda }^a$, any smooth tensor field ${\alpha }_{b_1\dots b_s}^{a_1\dots a_r}$ and any torsion free affine connection $\mathrm{\nabla }$:

In particular, we have:

• ${\mathcal{L}}_{\lambda }{\alpha }^a={\lambda }^n{\mathrm{\nabla }}_n{\alpha }^a-{\alpha }^n{\mathrm{\nabla }}_n{\lambda }^a$
• ${\mathcal{L}}_{\lambda }{\alpha }_b={\lambda }^n{\mathrm{\nabla }}_n{\alpha }_b+{\alpha }_n{\mathrm{\nabla }}_b{\lambda }^n$
• ${\mathcal{L}}_{\lambda }{g}_{ab}={\lambda }^n{\mathrm{\nabla }}_n{g}_{ab}+g_{nb}{\mathrm{\nabla }}_a{\lambda }^n+g_{an}{\mathrm{\nabla }}_b{\lambda }^n$

Approach 2

From Schuller GR lecture 11
Definition:
The Lie derivative $\mathcal{L}$ on a smooth manifold $(M,\mathcal{O},\mathcal{A})$ sends a pair of a vector field $X$ and a $(p,q)$-tensor field $T$ to a $(p,q)$-tensor field such that:

1. ${\mathcal{L}}_{X}f=Xf$ for any smooth function $f$.
2. ${\mathcal{L}}_{X}Y=[X,Y]$ for any vector field $Y$, where $[X,Y]$ is the Lie bracket of the vector fields $X$ and $Y$.
3. ${\mathcal{L}}_X(T+S)={\mathcal{L}}_{X}T+{\mathcal{L}}_{X}S$, for any two tensor fields $T$ and $S$ of the same type.
4. ${\mathcal{L}}_X(T(\omega ,Y))=({\mathcal{L}}_{X}T)(\omega ,Y)+T({\mathcal{L}}_X\omega ,Y)+T(\omega ,{\mathcal{L}}_{X}Y)$ for any 1-form $\omega$ and vector field $Y$, and similarly for any other valence of $T$.
5. ${\mathcal{L}}_X(\omega \otimes Y)=({\mathcal{L}}_X\omega )\otimes Y+\omega \otimes ({\mathcal{L}}_{X}Y)$ for the tensor product of a 1-form $\omega$ and a vector field $Y$.

There is only one operator satisfying this.