Considering an action dependent on vierbeins $e_I^a$ and the Riemann curvature tensor $R_{KL}^{ab} = (d \omega^{ab}+ \omega^{ac} \wedge \omega_c^{b})_{KL}$ given by

$$S = \int_M d^4x \det(e) e_I^ae_J^b \epsilon_{KL}^{IJ}F_{ab}^{KL}.$$

Variation by the vierbein would yield zero and together with the condition $d e^{a}+ \omega^{ab} \wedge e_b:= \nabla e^a = 0$ ($\nabla$ is exterior covariant derivative) I would obtain a boundary term after partial integration that looks like

$$S = \int_M d^4x \det(e) \nabla^K (e_I^ae_J^b \epsilon_{KL}^{IJ} \omega_{ab}^L) = \int_{\partial M} d^3x n^K e_I^ae_J^b \epsilon_{KL}^{IJ} \omega_{ab}^L.$$

Is this calculation correct?

Now I guess that this quantity describes the topological degree of the map $G: \partial M \rightarrow P$ with the Poincare group $P$. Since I insert Maurer-Cartan forms $e_I^a = \partial_Ix^a, \omega_{ab}^L = (g^{-1})_a \partial^L g_b$ for tertrad and spin connection I think will obtain $\deg(G)$. Is my idea right?

And is known which topological invariants this topological term yields?


1 Answer 1


The Holst term equals to zero on-shell because its equations of motion is Bianchi identity not because it is total derivative. You can find details in Rovelli & Vidotto 's book.

  • $\begingroup$ When does this topological Holst term plays a role (Examples?)? $\endgroup$
    – kryomaxim
    Commented Feb 14, 2017 at 13:57
  • 1
    $\begingroup$ @kryomaxim There are some investigations on such question. For example arxiv.org/abs/0903.2270 $\endgroup$ Commented Feb 15, 2017 at 8:53

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