Lie Derivative Identity for Contracted Tensors

Question

$\newcommand{\lie}{\mathcal L_X}$Let $M$ be a smooth manifold and $\sigma$ be a smooth covariant $2$-Tensor field on $M$. Further let $X,Y$ and $Z$ be smooth vector fields on $M$. We want to show the equality $$\lie (\sigma(Y,Z))=(\lie \sigma)(Y,Z)+\sigma(\lie Y,Z) + \sigma( Y,\lie Z)$$

Being a physicist I can write everything in coordinates and play around with indices and show this equality, which is not very pleasing as I want to prove it in a manifestly coordinate invariant way. To that end I wrote $\sigma(Y,Z)$ as $ Z \lrcorner (Y \lrcorner \sigma)$, where is the contraction of $Y$ with $\sigma$ e.g. $ (Y \lrcorner \sigma)Z := \sigma(Y,Z)$. Then I used the following identity to prove the above equation:

$$ \lie(Y\lrcorner \tau) = (\lie Y) \lrcorner \tau + Y \lrcorner (\lie \tau) $$

where $\tau$ is a smooth covariant $k$-Tensor field on $M$.

My question is the following: Is it a well-defined^* operation to write $\sigma(Y,Z) = Z \lrcorner (Y \lrcorner \sigma)$ and if not how can I prove the first equality without using coordinate frames?

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^{* It somehow (intuitively) feels like a not well-defined operation but I can't pinpoint, why that should be the case. Maybe my intuition is just wrong.}

Answer 1

Yes it is perfectly valid: The identity you are using is the expression of the fact that Lie derivative commutes with contraction $C$, then you can observe that $$\sigma(Y,Z)= C(C(\sigma \otimes Y \otimes Z )) $$

See Lemma 9 of the following notes for more details https://www.staff.science.uu.nl/~ban00101/lecnotes/lieder.pdf