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Poincaré Generators

The generators of the Poincaré 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?

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2  
This question is not really suitable for MO, I'm afraid. I suggest you try one of the other sites mentioned in the FAQ. Having said that, the first two terms do not cancel, but they partially cancel the last two terms in the expression yielding the desired result. You should perhaps read about composition of differential operators, though. It might be helpful to have the differential operators act on an arbitrary differentiable function $f$ and then abstract the function at the very end of the calculation. –  José Figueroa-O'Farrill Jun 26 '11 at 19:23
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A separate comment: for the sake of the children, please drop the factors of $i$ in these expressions. You have a real Lie algebra and it is natural that in a unitary representation they should be represented by skew-hermitian operators. –  José Figueroa-O'Farrill Jun 26 '11 at 19:26
1  
Thanks a lot :) –  Burc Jun 26 '11 at 19:39
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