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, $$c^{k}_{ij}$$ 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 $$\frac{\partial }{\partial x^{u}}$$

The commutator structure of this group can be written as:

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

Where the $$\epsilon^{i}$$ 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:

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

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

$$f^{u}_{i,v}f^{v}_{j} = c^{k}_{ij}f^{u}_{k}-----(3)$$

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

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 $$f^{u}_{i}$$.

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

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

So that the infinitesimal parameters $$\epsilon^{k}_{3}$$ of the commutator are given by:

$$\epsilon^{k}_{3} = c^{k}_{ij}(\epsilon^{i}_{2}\epsilon^{j}_{1}-\epsilon^{i}_{1}\epsilon^{j}_{2})$$

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

$$c^{k}_{ij} = -c^{k}_{ji}$$

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