Structure constants of a group antisymmetric?

jason12345
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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.
 
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It just follows from the definition. If you interchange i and j in the computation, you get a minus sign.
 
zhentil said:
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
 
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