- #1
ledamage
- 36
- 0
Hey there!
I have a simple question concerning infinitesimal generators: In order to get properties of the group, one always linearizes the group element ([itex]g = \exp(\mathrm{i}\theta^a T^a) = 1 + \theta^a T^a + \hdots[/itex]); in this way, one can show, say, that the antisymmetric matrices (where both indices transform in the same representation given by the generators [itex]T^a[/itex]) form an invariant subspace by computing
[tex]\varphi_{ij} \to \varphi'_{ij} = \varphi_{ij} + {(\theta^a T^a)_i}^k {(\theta^a T^a)_j}^\ell \varphi_{k\ell} = \hdots[/tex]
Why is it allowed to drop all orders [itex]\theta^2[/itex] and higher during the computation? Wouldn't that mean that (in the above case) antisymmetry is preserved only to first order in [itex]\theta[/itex]? How do I know that antisymmetry is preserved in higher orders?
I have a simple question concerning infinitesimal generators: In order to get properties of the group, one always linearizes the group element ([itex]g = \exp(\mathrm{i}\theta^a T^a) = 1 + \theta^a T^a + \hdots[/itex]); in this way, one can show, say, that the antisymmetric matrices (where both indices transform in the same representation given by the generators [itex]T^a[/itex]) form an invariant subspace by computing
[tex]\varphi_{ij} \to \varphi'_{ij} = \varphi_{ij} + {(\theta^a T^a)_i}^k {(\theta^a T^a)_j}^\ell \varphi_{k\ell} = \hdots[/tex]
Why is it allowed to drop all orders [itex]\theta^2[/itex] and higher during the computation? Wouldn't that mean that (in the above case) antisymmetry is preserved only to first order in [itex]\theta[/itex]? How do I know that antisymmetry is preserved in higher orders?