- #1
StarWombat
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- TL;DR
- Trying to prove a result about BF and CS actions, and looking for guidance with some of the working and interpretation
I believe this boils down to lack of familiarity on my part with wedge products of forms, so the answer is probably simple - but it's better to ask a stupid question than to remain ignorant! I've been looking at <https://arxiv.org/abs/hep-th/9505027>, and the idea that the BF [1] action
$$S_{BF}=\int_M tr\left(B\wedge F+\frac{\kappa^2}{3}B∧B∧B\right)$$
can be written as proportional to the difference of Chern-Simons [2] terms, i.e. ##2\kappa S_{BF}=(1/2)(S_{CS}(A+κB)−S_{CS}(A−κB))## where
$$ S_{CS}(A)=\int_M tr(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$$
So, when I expand this out, the factor of ##2\kappa## is easy to obtain - it just comes about because there are two Chern-Simons actions - although it's not quite that simple. The second term in each CS action leads to the expected numerical factor, but the first terms give
$$ (1/2)\left(\int_M tr(A+\kappa B)\wedge dA -\int_M tr(A-\kappa B)\wedge dA\right) = (1/2)\int_M tr(2\kappa B\wedge dA) = \kappa\int_M tr(B\wedge F)$$
This doesn't agree with the factor being ##2\kappa## (as the paper I'm looking at claims, and which I do get for the ##B^3## term), rather than just ##\kappa##. So my first question is why am I off by a factor of two in this term?
There's also an extra term proportional to
$$tr(A\wedge A\wedge B+A\wedge B\wedge A+B\wedge A\wedge A)$$
arising from the second term in each CS action. If this were an ordinary product of matrices, the cyclic nature of the trace would mean that the three terms are equal and I'd get ##3tr AAB##. However, this term must be equal to zero, so my second question is am I right in presuming the wedge product version of the Jacobi identity sends this term to zero? Is there another/better way to understand why this term vanishes?
Thanks in advance.
[1]: https://en.wikipedia.org/wiki/BF_model
[2]: https://en.wikipedia.org/wiki/Chern–Simons_theory
$$S_{BF}=\int_M tr\left(B\wedge F+\frac{\kappa^2}{3}B∧B∧B\right)$$
can be written as proportional to the difference of Chern-Simons [2] terms, i.e. ##2\kappa S_{BF}=(1/2)(S_{CS}(A+κB)−S_{CS}(A−κB))## where
$$ S_{CS}(A)=\int_M tr(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$$
So, when I expand this out, the factor of ##2\kappa## is easy to obtain - it just comes about because there are two Chern-Simons actions - although it's not quite that simple. The second term in each CS action leads to the expected numerical factor, but the first terms give
$$ (1/2)\left(\int_M tr(A+\kappa B)\wedge dA -\int_M tr(A-\kappa B)\wedge dA\right) = (1/2)\int_M tr(2\kappa B\wedge dA) = \kappa\int_M tr(B\wedge F)$$
This doesn't agree with the factor being ##2\kappa## (as the paper I'm looking at claims, and which I do get for the ##B^3## term), rather than just ##\kappa##. So my first question is why am I off by a factor of two in this term?
There's also an extra term proportional to
$$tr(A\wedge A\wedge B+A\wedge B\wedge A+B\wedge A\wedge A)$$
arising from the second term in each CS action. If this were an ordinary product of matrices, the cyclic nature of the trace would mean that the three terms are equal and I'd get ##3tr AAB##. However, this term must be equal to zero, so my second question is am I right in presuming the wedge product version of the Jacobi identity sends this term to zero? Is there another/better way to understand why this term vanishes?
Thanks in advance.
[1]: https://en.wikipedia.org/wiki/BF_model
[2]: https://en.wikipedia.org/wiki/Chern–Simons_theory