Is the Expression Valid in Demonstrating a Poincare Algebra Commutator?

[binbagsss]

Homework Statement

Does $x_p\partial_v\partial_u-x_v\partial_p\partial_u=0$

Homework Equations

I need this to be true to show a poincare algebra commutator.

We have just shown that $[P_u, P_v] =0$, i.e. simply because partial derivatives commute.

Where $P_u=\partial_u$

The Attempt at a Solution

[/B]
I'm guess it's using that partial derivatives commute and renaming the $v$ and $p$ for each other, however I'm a bit confused with this, because, I know this isn't really the context here, but these are free indicies not dummy indices, i..e not an index summing over, so I'm a bit unclear how you could simply rename.

Many thanks

 

[stevendaryl]

Homework Statement

Does $x_p\partial_v\partial_u-x_v\partial_p\partial_u=0$

Homework Equations

I need this to be true to show a poincare algebra commutator.

We have just shown that $[P_u, P_v] =0$, i.e. simply because partial derivatives commute.

Where $P_u=\partial_u$

The Attempt at a Solution

[/B]
I'm guess it's using that partial derivatives commute and renaming the $v$ and $p$ for each other, however I'm a bit confused with this, because, I know this isn't really the context here, but these are free indicies not dummy indices, i..e not an index summing over, so I'm a bit unclear how you could simply rename.

Many thanks

Free indices means that it must be true for every possible choice of $u,p,v$, and for all possible values of $x_p$ and $x_v$. So try the particular case where $u,p,v$ are all different indices, and $x_p = 0$, and $x_v \neq 0$.