## PoincarĂ© Generators

The generators of the PoincarĂ© algebra (the Lie algebra of the isometry-group of flat minkowski space) obeys these commutator relations:

$ \quad \ \ [P_a,P_b] = 0 $

$-i[M_{ab},P_c] = \eta_{ac} P_b - \eta_{bc} P_a $

$-i[M_{ab},M_{cd}] = \eta_{ac} M_{bd} - \eta_{ad} M_{bc} - \eta_{bc} M_{ad} + \eta_{bd} M_{ac} $

They can also be represented as differential operators on functions on Minkowski space

$ P_a \rightarrow i \frac{\partial}{\partial x^a} $

$ M_{ab} \rightarrow i \left ( x_a \frac \partial {\partial x^b} - x_b \frac \partial {\partial x^a} \right ) - i \mathbb M _{ab} $

Where $ \mathbb M _{ab} $ is a suitable matrix representation.

## Calculations

When I apply the differential forms in the commutator relations I don't quite get the math to work out, and I suspect I am missing something quite fundamental.

The first relation is quite trivial, but the second has got me quite confounded.

I get as far as

$-[M_{ab},P_c] = x_a \dfrac {\partial^2} {x_b x_c} - x_b \dfrac {\partial^2} {x_a x_c} + \dfrac{\partial}{\partial x^c} x_a \dfrac{\partial}{\partial x^b} - \dfrac{\partial}{\partial x^c} x_b \dfrac{\partial}{\partial x^a} $

Now this works out nicely if the first two terms vanish, but I don't see why they do.

Does $ x_a \dfrac {\partial^2} {x_b x_c} - x_b \dfrac {\partial^2} {x_a x_c} = 0 $?

Why?

skew-hermitianoperators. – José Figueroa-O'Farrill Jun 26 '11 at 19:26