Variation of scalar curvature upon frame deformation

Question

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?

mike stone · Answer

A useful equation might be
$$
g^{munu} delta R_{munu} =  (g_{munu}nabla^2 -nabla_mu nabla_nu) delta g^{munu},
$$
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 
$$
nabla_mu{bf e}_a = {bf e}_b {omega^b}_{amu}
$$
which one can write in coordinate-frame components (i.e. where ${bf e}_a=e_a^mu partial_mu$) as
$$
nabla_mu e^nu_a equiv (nabla_mu{bf e}_a)^nu= e^nu_b {omega^b}_{amu}.
$$
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
$$
partial_mu e^nu_a + {Gamma^nu}_{lambdamu} e^lambda_a -  e^nu_b {omega^b}_{amu}=0
$$
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 ${omega^b}_{amu}$ that way. Certainly the expression 
$$
partial_mu e^nu_a + {Gamma^nu}_{lambdamu} e^lambda_a -  e^nu_b {omega^b}_{amu}
$$
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
$$
(delta omega_{ijmu}) e^mu_k 
 =-frac 12left{( eta_{ib}( nabla_j [e^{*b}_alpha delta e^alpha_k]-   nabla_k [e^{*b}_alpha delta e^alpha_j])
+eta_{jb}( nabla_k [e^{*b}_alpha delta e^alpha_i]-   nabla_i [e^{*b}_alpha delta e^alpha_k])
-eta_{kb}( nabla_i [e^{*b}_alpha delta e^alpha_j]-   nabla_j [e^{*b}_alpha delta e^alpha_i])right}.
$$
where I think my $eta_{ib}e^{*b}_alpha delta e^alpha_j= {bf e}_icdot delta{bf e}_j$   are the same thing as your $H_{ij}$

Variation of scalar curvature upon frame deformation

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