# Lie brackets and derivatives

1. Jan 29, 2013

Hi i have two questions:
1) When asked to prove $\mathcal{L}_{u}\mathcal{L}_{v}W - \mathcal{L}_{v}\mathcal{L}_{u}W = \mathcal{L}_{[u,v]}$.

I achieved $[u,v]w = \mathcal{L}_{[u,v]}$. This was found by appliying a scalar field <b> to the LHS and simplifying and expanding using + and scalar linearitys to get $[u,v]w$ but I am not sure if these are equivalent.

2) When asked to calculate the Lie bracket [X,Y] where $X=5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x}$ and $Y= y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y}$ is this equivalent to:
$\left((5x^{2}\frac{\partial}{\partial t} -4t\frac{\partial}{\partial x})\frac{\partial (y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})}{\partial x^{a}}-(y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})\frac{\partial (5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x})}{\partial x}\right)\frac{\partial}{\partial x^{b}}$

and if so can it be expanded any further I am not so sure but i dont fully understand $\frac{\partial}{\partial x^{a}}$ derivatives.

2. Jan 29, 2013

never mind figured it out

3. Jan 29, 2013

### dextercioby

I'm sure you're missing out the vector field W in the RHS of 1).

4. Jan 29, 2013