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


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

While, for functions:


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?


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

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

  • $\begingroup$ 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 $X$. $\endgroup$ Nov 18, 2019 at 17:07

1 Answer 1


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:


This is not true for the covariant derivative, which is indedd $C^\infty(M)$-linear in its first parameter.


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