As far as I understand, one way to define the field-strength tensor is by using the commutator of covariant derivatives as follows: $$-igT^aF^a_{\mu\nu} = [D_\mu, D_\nu]$$ where $T^a$ is a basis for the Lie algebra, $$D_\mu = I\partial_\mu - igT^aA^a_\mu$$ is the covariant derivative, $g$ is a coupling constant, and $A_\mu^a$ are the components of the gauge field.

By comparing coefficients of basis elements, one obtains that the field strength tensor is given by the following relation: $$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f^{abc}A_\mu^b A_\nu^c$$ where $f^{abc}$ are structure constants of the Lie algebra being studied.

I understand that by interpreting the gauge field $A$ as a Lie algebra valued 1-form on a principle bundle that we can define the field-strength tensor as the exterior derivative of $A$, i.e. $F = dA$. However this doesn't seem to be how physicists first discovered the generalized version of the field-strength tensor (generalized to more than electromagnetism).

1. What was the first motivation for looking at the commutator $[D_\mu, D_\nu]$, before physicists were aware of its geometric interpretation as a measure of curvature?
2. The geometric interpretation of the commutator, based on an answer to this physicsSE quesetion, is that it measures the extent to which the of the covariant derivatives $D_\mu$ and $D_\nu$ fail to commute. That is we consider an infinitesimally small square with edges $e_\mu$ and $e_\nu$ and see how parallel transporting along this square in different orders differs. Is this all that is going on? This is reminiscent of how one defines the full Riemann curvature tensor, but here we omit the Lie bracket term, which with an abuse of notation I write as $D_{[\mu, \nu]}$. Why is this term omitted?

As far as I understand, one way to define the field-strength tensor is by using the commutator of covariant derivatives as follows: $$-igT^aF^a_{\mu\nu} = [D_\mu, D_\nu]$$ where $T^a$ is a basis for the Lie algebra, $$D_\mu = I\partial_\mu - igT^aA^a_\mu$$ is the covariant derivative, $g$ is a coupling constant, and $A_\mu^a$ are the components of the gauge field.

By comparing coefficients of basis elements, one obtains that the field strength tensor is given by the following relation: $$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f^{abc}A_\mu^b A_\nu^c$$ where $f^{abc}$ are structure constants of the Lie algebra being studied.

I understand that by interpreting the gauge field $A$ as a Lie algebra-valued 1-form on a principle bundle that we can define the field-strength tensor as the exterior derivative of $A$, i.e. $F = dA$. However this doesn't seem to be how physicists first discovered the generalized version of the field-strength tensor (generalized to more than electromagnetism).

1. What was the first motivation for looking at the commutator $[D_\mu, D_\nu]$, before physicists were aware of its geometric interpretation as a measure of curvature?
2. The geometric interpretation of the commutator, based on an answer to this phys.SE quesetion, is that it measures the extent to which the of the covariant derivatives $D_\mu$ and $D_\nu$ fail to commute. That is we consider an infinitesimally small square with edges $e_\mu$ and $e_\nu$ and see how parallel transporting along this square in different orders differs. Is this all that is going on? This is reminiscent of how one defines the full Riemann curvature tensor, but here we omit the Lie bracket term, which with an abuse of notation I write as $D_{[\mu, \nu]}$. Why is this term omitted?