Simplicity constraints from $SO(4)$ Plebanski action

Question

The $SO(4)$ Plebanski action yields a first order formulation of Euclidean General Relativity as a constrained (topological) BF-theory. It depends on a $so(4)$ connection 1-form $\omega^{IJ} = \omega_{\mu}^{IJ}dx^{\mu}$ through the curvature $F^{IJ}(\omega) = d_{\omega}\omega^{IJ}$, with $d_{\omega}(\cdot)$ the exterior covariant differential, on a $so(4)$ 2-form $B^{IJ} = B^{IJ}dx^{\mu}\wedge dx^{\nu}$ and a scalar symmetric traceless matrix $\phi_{[IJ][KL]}$ ($\epsilon^{IJKL}\phi_{IJKL} = 0$) with components acting as Lagrange multipliers. Assuming zero cosmological constant ($\Lambda$ = 0), the action reads: \begin{equation} S[\omega; B, \phi] = \int_{M} B^{IJ}\wedge F_{IJ}(\omega) - \frac{1}{2}\phi_{IJKL}B^{IJ}\wedge B^{KL} \end{equation} Varying this action with respect to the multipliers $\phi_{IJKL}$ is supposed to provide the well-known simplicity constraints \begin{equation} B^{IJ}\wedge B^{KL} = e\epsilon^{IJKL}, ~~~ \text{with}~~ e = \frac{1}{4!}\epsilon_{IJKL}B^{IJ}\wedge B^{KL} \end{equation} imposing the on-shell antisymmetric form of the field $B^{IJ}$. However, I don't seem to understand how the variation is to be performed in order to obtain those constraints. I understand it is necessary to take into account the symmetry properties of the fields $\phi_{IJKL}$, but none of the papers I have read regarding this topic detail that calculation.

Does anyone know how this variation should be performed or any papers with further details?

Answer 1

I am currently working on spinfoams, and the same question came to me when I studied this.

The only way I found to obtain the simplicity constraints from the variation of the action is the following;

Following the notation of Baez and Muniain's book, take $(\phi_{s})_{IJKL}:=\phi_{IJKL}+s\delta\phi_{IJKL}$ with $s\in\mathbb{R}$ small. Substituting in the action and calculating $\frac{dS[\omega;B, \phi_s]}{ds}\Bigr|_{s=0}$ we obtain:

$$\frac{dS[\omega;B, \phi_s]}{ds}\Bigr|_{s=0}=-\dfrac{1}{2}\int_{M}B^{IJ}\wedge B^{KL}\delta\phi_{IJKL}.$$

If we take into account that $\epsilon^{IJKL}\phi_{IJKL}=0$, then we also have $\epsilon^{IJKL}\delta\phi_{IJKL}=0$. This last identity means that $e\epsilon^{IJKL}\delta\phi_{IJKL}=0$ for some $e\in\mathbb{R}$. Then, we can write:

$$\frac{dS[\omega;B, \phi_s]}{ds}\Bigr|_{s=0}=-\dfrac{1}{2}\int_{M}B^{IJ}\wedge B^{KL}\delta\phi_{IJKL}-e\epsilon^{IJKL}\delta\phi_{IJKL}.$$

In this case, the condition $\frac{dS[\omega;B, \phi_s]}{ds}\Bigr|_{s=0}=0$ and the arbitrariness of $\delta\phi_{IJKL}$ imply that

$$B^{IJ}\wedge B^{KL}-e\epsilon^{IJKL}=0.$$

In this last expression, if you solve for $e$ you obtain that

$$e=\dfrac{1}{4!}\epsilon_{IJKL}B^{IJ}\wedge B^{KL},$$

as desired.

This is the only way I have found to obtain the simplicity constraints in its correct form.