Structure constants of a group antisymmetric?

[jason12345]

First, how do you put in spaces without them being deleted when i post?

How does one show that the structure constants, [tex]c^{k}_{ij}[/tex] of a group are antisymmetric? The context is from page 12 of Anderson's "Principles of Relativity Physics", with the statement right at the bottom:

With u = 1,2,3,4 and ',' denoting partial differentiation so ,u means [tex]\frac{\partial }{\partial x^{u}}[/tex]

The commutator structure of this group can be written as:

[tex]\zeta^{u}(x) = \epsilon^{i}f^{u}_{i}-----i = 1,2,...,N,-----(1)[/tex]

Where the [tex]\epsilon^{i}[/tex] are the group parameters. Our basic requirement is that the commutator of two such infinitesimal mappings must again be of this form. We find for this commutator:

[tex]\zeta^{u}_{3} = (\epsilon^{i}_{2}\epsilon^{j}_{1}-\epsilon^{i}_{1}\epsilon^{j}_{2})f^{u}_{i,v}f^{v}_{j}-----(2)[/tex]

In order that it be of the form (1) the functions [tex]f^{u}_{i}[/tex] must be related to each other by an equation of the form:

[tex]f^{u}_{i,v}f^{v}_{j} = c^{k}_{ij}f^{u}_{k}-----(3)[/tex]

Where the [tex]c^{k}_{ij}[/tex]are constants independent of the [tex]\epsilon^{i}[/tex] and the [tex]x^{u}[/tex] .

they are called the structure constants of the group and serve to characterize it in

a manner that is independent of the particular form taken by the [tex]f^{u}_{i}[/tex].

If we substitute (3) back into (2) we obtain:

[tex]\zeta^{u}_{3} = (\epsilon^{i}_{2}\epsilon^{j}_{1}-\epsilon^{i}_{1}\epsilon^{j}_{2})c^{k}_{ij}f^{u}_{k}[/tex]

So that the infinitesimal parameters [tex]\epsilon^{k}_{3}[/tex] of the commutator are given by:

[tex]\epsilon^{k}_{3} = c^{k}_{ij}(\epsilon^{i}_{2}\epsilon^{j}_{1}-\epsilon^{i}_{1}\epsilon^{j}_{2})[/tex]

From the manner of their construction we see that the structure constants
are antisymmetric in the two lower indices, that is,

[tex]c^{k}_{ij} = -c^{k}_{ji}[/tex]

Thanks for any help in advance.

 

[zhentil]

It just follows from the definition. If you interchange i and j in the computation, you get a minus sign.

 

[jason12345]

It just follows from the definition. If you interchange i and j in the computation, you get a minus sign.

Can you show me then?

Thanks