Tuesday, April 12, 2016

general relativity - Dimensional reduction from even to odd dimensions and Chern-Simons terms



I'll be taking some Lagrangians from this paper to try and keep personal typos out of the discussion






I have been looking at the dimensional reduction of Einstein-Maxwell-Dilaton theories and I am trying to understand the correct procedure of working with/the addition of topological terms.


I am used to seeing in odd dimension the addition of the Chern-Simons form together with the metric dependent content of the theory.


For example: \begin{equation} \mathcal{L}_5 = R \star 1 + \frac{1}{{(h^i)^{2}}} \left( \star dh^i \wedge dh^i + \star F^i \wedge F^i \right) + F^1 \wedge F^2 \wedge A^3 \tag{1} \end{equation} where here the index $i=1,2,3$ and is implicitly summed over. $h^i$ are taken to be three non-independent scalar fields and $F^i$ are 2-form field strengths: $F=dA$


My questions are:



  • How do topological invariants arise when dimensionally reducing theories?

  • Are Chern-simons terms in $2n-1$ dimensional theories related to other topological invariants from $2n$ dimensional theories?



If I try to obtain the Lagrangian (1) from the Kaluza-Klein reduction of a $6D$ theory: \begin{equation} \mathcal{L}_6 = R \star 1 - \frac{1}{2} \star d \phi \wedge d \phi - \frac{1}{2} \star H^{(3)} \wedge H^{(3)} \end{equation} for some 3-form field strength $H^{(3)} = dB^{(2)}$.


I will obtain a Lagrangian: \begin{equation} \mathcal{L}_5 = R \star 1 + \frac{1}{{(h^i)^{2}}} \left( \star dh^i \wedge dh^i + \star F^i \wedge F^i \right) \end{equation}



  • Do I insert the Chern-Simons term by hand after the reduction or does this additional term come from the reduction of some even-dimensional topological term which I am neglecting?

  • If I do have to insert it by hand then how do I reconcile the difference between these 4D theories:


\begin{equation} \mathcal{L}_6 \xrightarrow{S^1} \mathcal{L}_5 + F \wedge F \wedge A \xrightarrow{S^1} \mathcal{L}_4 \end{equation} and: \begin{equation} \mathcal{L}_6 \xrightarrow{T^2} \tilde{\mathcal{L}}_4 \end{equation}




No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...