# Show that the Lie derivative is equal to the commutator

Let $\Omega \subseteq \mathbb{R}^d$ be open. Let $\epsilon > 0$. Let $(\phi_t)_{t \in (-\epsilon , \epsilon)}$ be a family in $\mathrm{Diff}(\Omega)$ such that $\phi_0 = id_{\Omega}$ and $\phi_s \circ \phi_t = \phi_{s+t}$ for all s,t $\in (-\epsilon , \epsilon)$ such that s+t $\in (-\epsilon,\epsilon)$ and such that $t \to \phi_t(x)$ is continuously differentiable for each fixed x $\in \Omega$. Let $X$ be the vector field defined by $X(f)(x) = \frac{\partial}{\partial t}|_{t=0}f(\phi _t(x))$.

We define the Lie derivative $L_XT$ of a tensor field $T$ of type $\binom{r}{s}$ on $\Omega$ w.r.t. $X$ by $$(L_XT [\psi] ^{i_1...i_r}_{j_1...j_s}(y) = \frac {\partial}{\partial t}|_{t=0} T[ \psi \circ \phi _{-t}] ^{i_1...i_r}_{j_1...j_s}(y)$$ for $y \in \Omega$, $\psi \in \mathrm{Diff}(\Omega)$.

And finally, let g be a metric field on $\Omega$ and let $\nabla$ be the covariant derivative constructed from g.

Show for $\psi \in \mathrm{Diff}(\Omega))$

a) $(L_Xf )[\psi] = X[\psi]^j(\partial_jf)[\psi]$ for f a smooth function on $\Omega$

b) $(L_XY )[\psi]^j = (X^k \partial _k Y^j -Y^k\partial_kX^j)[\psi] = (X^k \nabla _k Y^j -Y^k \nabla_kX^j)[\psi]$

and therefore it is the commutator

and finally

c) $(L_Xg [\psi]_{jk} = (X^l \partial _l g_{jk} - g_{jk} \partial _j X^l + g_{jl} \partial_kX^l)[\psi] = (\nabla_jX_k + \nabla _kX_j)[\psi]$

where as usual $X_k = g_{kl}X^l$.

Well. I just got that by definition the $\nabla _k$ is equal to $\partial_k$

but for the other things I have no Idea..

thanks for help.

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any hints for it ? –  lukas Jan 21 '14 at 20:32
When you said that $\nabla_k=\partial_k$, you were kind of mistaken: actually, for an $(r,s)$ tensor along the basis vector $e_c$, $$(\nabla_c T)^{a_1 \ldots a_r}{}_{b_1 \ldots b_s} = \frac{\partial}{\partial x^c}T^{a_1 \ldots a_r}{}_{b_1 \ldots b_s}+\,\Gamma ^{a_1}{}_{dc} T ^{d \ldots a_r}{}_{b_1 \ldots b_s} + \cdots + \Gamma ^{a_r}{}_{dc} T ^{a_1 \ldots a_{r-1}d}{}_{b_1 \ldots b_s}$$ The $\Gamma^k_{\ i j}$ are the Christoffel symbols. –  user122283 Jan 22 '14 at 1:37
I've fixed some typos and factual errors in the main body of the question, and separated your update by the horizontal line. –  Yuri Vyatkin Jan 22 '14 at 6:23
No, $\nabla_k$ is the covariant derivative, not the partial derivative $\partial_k$, as @Sanath has correctly pointed out. –  Yuri Vyatkin Jan 22 '14 at 6:27

Just get all your definitions handy, and verify a number of things. Here is a to do list.

1) Make sure that you understand why $L_X T$ is a tensor field again.

2) Use the chain rule to show a) and observe that you can replace $\partial_k$ with $\nabla_k$ just because on functions these are the same thing.

3) The first equality in b) is actually shown here.

4) To understand why the partial derivatives can be replaces by covariant derivatives (induced by the metric!) use the formula given by @Sanath in the comments. The main observation is that the Christoffel symbols are symmetric w.r.t. the bottom indices.

5) Put the things together and work out the rest.

Good luck!

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