# Lie Derivative in coordinates

Lie Derivative is defined through its action on vector fields, as:

$$\mathcal{L}_XY=[X,Y]$$

This is linear in its arguments due to the linearity of Lie Bracket.

While, for functions:

$$\mathcal{L}_Xf=X[f]$$

Ok, now I want to develop the first expression in local coordinates, meaning that:

\require{cancel}\begin{align} \mathcal{L}_{X^\alpha\partial_\alpha}(Y^\beta\partial_\beta)&\overset{\begin{align}line&arity\\ &\uparrow \end{align}}{=}X^\alpha\mathcal{L}_{\partial_\alpha}(Y^\beta\partial_\beta)\overset{\begin{align}Leibn&iz\ rule \\ &\uparrow \end{align}}{=}X^\alpha\left[\mathcal{L}_{\partial_\alpha}(Y^\beta)\partial_\beta+Y^\beta\mathcal{L}_{\partial_\alpha}\partial_\beta \right]\\ &= X^\alpha\left[\partial_\alpha(Y^\beta)\partial_\beta+Y^\beta\mathcal{L}_{\partial_\alpha}\partial_\beta \right] = X^\alpha\left[\partial_\alpha(Y^\beta)\partial_\beta+Y^\beta\cancelto{0}{ \left[\partial_\alpha, \partial_\beta \right]} \right]\\ &= X^\alpha\partial_\alpha(Y^\beta)\partial_\beta \end{align}

But clearly this is wrong cause it is not equal to the commutator of fields. What am I misunderstanding in this procedure?

Cheers

PD: For the covariant derivative the preceding way give us the correct expression, why?

\begin{align} \nabla_XY&=\nabla_{X^\alpha\partial_\alpha}(Y^\beta\partial_\beta)=X^\alpha\nabla_{\partial_\alpha}(Y^\beta\partial_\beta)\\ &= X^\alpha \left[\left( \partial_\alpha Y^\beta\right)\partial_\beta+Y^\beta\nabla_{\partial_\alpha}\partial_\beta \right]\\ &=X^\alpha\left[\left( \partial_\alpha Y^\beta\right)\partial_\beta+Y^\beta \Gamma_{\beta\alpha}^\sigma\partial_\sigma \right]\\ &=X^\alpha\left[ \partial_\alpha Y^\sigma+Y^\beta \Gamma_{\beta\alpha}^\sigma \right]\partial_\sigma \end{align}

• The formula you gave for vector fields is not a definition; it's a theorem. The Lie derivative is defined using the flow of the vector field $X$. Nov 18, 2019 at 17:07

The linearity of the Lie bracket is true taking vector fields on $$M$$ as a $$\mathbb{R}$$-vector space, not as a $$C^\infty(M)$$-module. Indeed, there will be some mixed terms:
$$\mathcal{L}_{fX}Y=f\mathcal{L}_XY-(Yf)X.$$
This is not true for the covariant derivative, which is indedd $$C^\infty(M)$$-linear in its first parameter.