# Variation of scalar curvature upon frame deformation

Following the textbook "Ideas and Methods of Supersymmetry and Supergravity" by Ioseph Buchbinder and Sergei Kuzenko (p38 - 40):

If we consider the a frame deformation in our vierbein induced by a symmetric rank 2 Lorentz tensor $H$:

$$e^m_a \rightarrow e_a^m + H_a^b e^m_b$$

One can easily show: $$g_{mn} = e^a_m e^b_n \eta_{ab}$$ $$\implies \delta g_{mn} = -2 e_m^a e_n^b H_{ab}$$ $$\implies H_{ab} = - \frac{1}{2} e^m_a e^n_b \delta g_{mn}$$

The next goal would be to identity how the curvature varies with respect to the variation in the metric so that various invariants could be identified (p40-41), however upon attempting to find the variation in the scalar curvature:

$$\delta R = 2 \nabla^c \nabla_c H^a_a - 2 \nabla^a \nabla^b H_{ab} + 2 H^{ab}R_{ab}$$ $$\implies \delta R = - \nabla^c \nabla_c (\eta^{ab} e_a^{m}e_b^{n} \delta g_{mn}) + \nabla^a \nabla^b (e_a^{m}e_b^{n} \delta g_{mn}) - (e_a^{m}e_b^{n} \delta g_{mn})R_{ab}$$

By using the compatibility of the vielbein with the covariant derivative:

$$\delta R = - \eta^{ab} e_a^{m}e_b^{n} \nabla^c \nabla_c \delta g_{mn} + e_a^{m}e_b^{n} \nabla^a \nabla^b \delta g_{mn} - e_a^{m}e_b^{n} \delta g_{mn}R_{ab}$$

Which does not seem to be of the correct form. Is there anything further I could try or have I approached this incorrectly?

A useful equation might be $$g^{\mu\nu} \delta R_{\mu\nu} = (g_{\mu\nu}\nabla^2 -\nabla_\mu \nabla_\nu) \delta g^{\mu\nu},$$ but may be you are already using this?
I don't think that you have the compatibility correct, though. The connection is defined by specifying the covariant derivative of the basis vectors of $T(M)$. In the case of a vielbein frame ${\bf e}_a$, the covariant derivative of the vector ${\bf e}_a$ is written as $$\nabla_\mu{\bf e}_a = {\bf e}_b {\omega^b}_{a\mu}$$ which one can write in coordinate-frame components (i.e. where ${\bf e}_a=e_a^\mu \partial_\mu$) as $$\nabla_\mu e^\nu_a \equiv (\nabla_\mu{\bf e}_a)^\nu= e^\nu_b {\omega^b}_{a\mu}.$$ The covariant derivative of the vierbein basis is therefore not zero. I have seen people argue for passing vierbeins through covariant derivatives by writing this last equation as $$\partial_\mu e^\nu_a + {\Gamma^\nu}_{\lambda\mu} e^\lambda_a - e^\nu_b {\omega^b}_{a\mu}=0$$ and thinking of this as a kind of "generalized" covariant derivative being zero. In doing this they are making the mistake of imagining that the the "$a$'' in $e_a^\mu$ is an index rather than a label telling us which frame vector ${\bf e}_a$ is. It's perhaps a useful mnemonic, but is also kind of schizophrenic, as they are attempting to work simultaneously with a vielbein frame and a co-ordinate basis for the tangent space $T(M)$. It makes no mathematical sense to interpret the definition of the frame connection ${\omega^b}_{a\mu}$ that way. Certainly the expression $$\partial_\mu e^\nu_a + {\Gamma^\nu}_{\lambda\mu} e^\lambda_a - e^\nu_b {\omega^b}_{a\mu}$$ is not the $\nu$-th component of the covariant derivative $\nabla_\mu {\bf e}_a$! Treating it as if it were will inevitably lead to confusion, and this is what I suspect is happening in your calculation.
Another useful formula for the variation of the tosion-free spin connection under a change of vielbein frame is $$(\delta \omega_{ij\mu}) e^\mu_k =-\frac 12\left\{( \eta_{ib}( \nabla_j [e^{*b}_\alpha \delta e^\alpha_k]- \nabla_k [e^{*b}_\alpha \delta e^\alpha_j]) +\eta_{jb}( \nabla_k [e^{*b}_\alpha \delta e^\alpha_i]- \nabla_i [e^{*b}_\alpha \delta e^\alpha_k]) -\eta_{kb}( \nabla_i [e^{*b}_\alpha \delta e^\alpha_j]- \nabla_j [e^{*b}_\alpha \delta e^\alpha_i])\right\}.$$ where I think my $\eta_{ib}e^{*b}_\alpha \delta e^\alpha_j= {\bf e}_i\cdot \delta{\bf e}_j$ are the same thing as your $H_{ij}$
• @raptakem.Yes eq vanished. we had a power outage as I was ediing it. I don't think torsion-free is relevent. The problem is the multiple roles of the "$a$" in $e_a^\mu$. I'll amend my answer to discuss this. – mike stone Jun 11 '18 at 12:35