What is the factor in the definition of [A,A] in differential forms?

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SUMMARY

The discussion centers on the calculation of the field strength F in differential forms, specifically using the connection A = f(r)σ₁ dx + g(r)σ₂ dy, where σ's represent half the Pauli matrices. The user computes dA as f'σ₁ dr ∧ dx + g'σ₂ dr ∧ dy and the commutator [A,A] as 2fgσ₃ dx ∧ dy. A key point of contention is whether a hidden factor of one-half exists in the definition of [A,A], which is clarified by DOX, stating that there are not two contributions, as they arise from the commutation relations of the Pauli matrices.

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Dox
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Hi everyone, I've been studying a paper in which there is a connection given by,

[tex]A = f(r)\sigma_1 dx+g(r)\sigma_2 dy,[/tex]​

where [tex]\sigma[/tex]'s are half the Pauli matrices. I need to calculate the field strength,

[tex]F = dA +[A,A][/tex].​

I have computed it, but a factor is given me problems. I would say,

[tex]dA = f' \sigma_1 dr\wedge dx + g'\sigma_2 dr\wedge dy[/tex]​

and

[tex][A,A] = 2 f g \sigma_3 dx\wedge dy,[/tex]​

with a factor 2 coming from the fact that there are two contributions... like a binomial.

Is it OK or there is a half factor hidden in the definition of [tex][A,A][/tex]?

Thank you so much.

DOX​
 
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There aren't two contributions. They are in the commutator of the sigmas.
 

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