Local Lorentz invariance of General Relativity

Question

Consider the gravitational action as an integral over a differential 4-form $$ S = \int_{\mathcal{M}} \star F_{ab}\wedge e^a \wedge e^b$$ where $\star F_{ab} = \epsilon_{abcd} F^{cd}$ and $F$ is the curvature two-form and $e^a$ is the tetrad one-form. I wish to prove that the above action is invariant under local lorentz transformations namely $$\delta e^a = \lambda^a_{~~b}e^b\\\delta\omega^{ab} = \omega^{a}_{~~c}\lambda^{cb}+\omega^{b}_{~~c}\lambda^{ac}+d\lambda^{ab} = D\lambda^{ab}$$

Now, having done that I obtain the following $$\delta S = \int_{\mathcal{M}}[\star D\wedge D\lambda_{ab}\wedge e^a \wedge e^b+2\star F_{ab}\wedge e^a\wedge \lambda^b_{~~c}e^c]$$ after which I am compelled to use $$D\wedge D \lambda^{ab}= F^{b}_{~~c}\wedge\lambda^{ac}+F^{a}_{~~c}\wedge\lambda^{cb}$$ to obtain Now, having done that I obtain the following $$\delta S = 2\int_{\mathcal{M}}[\star (F_b^{~~c}\wedge \lambda_{ca})\wedge e^a \wedge e^b+\star F_{ab}\wedge e^a\wedge \lambda^b_{~~c}e^c]$$ which I do not think is zero. Can anyone help?

Answer 1

I think a point you may be missing here is that the indices $a, b, \dots$ are not mere labels, they actually conveniently label the representations involved and can be used to construct manifest invariants of the local Lorentz group. This point is sometimes assumed without emphasizing or explaining, so I thought I'd give you the explanation.

The local Lorentz algebra $\mathfrak{so}_{3,1}$ is a subgroup of $\mathfrak{gl}_4$, the Lie algebra of $4 \times 4$ matrices with the Lie bracket given by the commutator of the matrices. Therefore, representations of $\mathfrak{gl}_4$ are by extension also representations of $\mathfrak{so}_{3,1}$.

As a special case, if something is $\mathfrak{gl}_4$-invariant (that is, belongs to the trivial representation), it must also be $\mathfrak{so}_{3,1}$-invariant. The converse is not true, however. There are objects such as $\eta_{ab}$ and $\varepsilon_{abcd}$ for which $\mathfrak{so}_{3,1}$-invariance must be carefully demonstrated, as they are not $\mathfrak{gl}_4$-invariant.

This formalism is convenient, because we can use the tensor product algebra of the representations of $\mathfrak{gl}_4$ instead of working with the representations of $\mathfrak{so}_{3,1}$. It is easier to work with, because we can use the index notation that is so familiar to physicists. Whenever you see the up/down latin index such as e.g. in ${F^a}_b$, the notation signals which representation of $\mathfrak{gl}_4$ the object belongs to, and therefore it also fixes its $\mathfrak{so}_{3,1}$ representation.

As for $\mathfrak{so}_{3,1}$-invariant objects that are not also $\mathfrak{gl}_4$-invariant, we can still make use of them under this formalism. We can use them in index contraction, such as e.g. $$ \eta_{ab} e^a_{\mu} e^b_{\nu} $$ the result would be $\mathfrak{gl}_4$-invariant if $\eta_{ab}$ was a (2,0)-tensor under $\mathfrak{gl}_4$, but it isn't, so the contraction is not invariant under $\mathfrak{gl}_4$. However, it is trivial to show that it is invariant under $\mathfrak{so}_{3,1}$. To do that, define an auxiliary quantity $g_{ab}$ which is defined to be equal to $\eta_{ab}$ in some basis, and transforms under $\mathfrak{gl}_4$ as a (2,0)-tensor by definition. Then observe that $$ g_{ab} e^a_{\mu} a^b_{\nu} $$ is now a $\mathfrak{gl}_4$-invariant, and so also a $\mathfrak{so}_{3,1}$-invariant. But all actions of the $\mathfrak{so}_{3,1}$ subalgebra preserve the (non-tensorial) relation $g_{ab} = \eta_{ab}$, therefore under the action of the whole subgroup generated by the Lorentz subalgebra, the quantities $g_{ab} e^a_{\mu} e^b_{\nu}$ and $\eta_{ab} e^a_{\mu} e^b_{\nu}$ must coincide. We conclude that $\eta_{ab} e^a_{\mu} e^b_{\nu}$ is a $\mathfrak{so}_{3,1}$-invariant.

By the same logic, $$ (\star F)_{ab} \wedge e^a \wedge e^b = \frac{1}{2} \varepsilon_{abcd} \eta^{c e} \eta^{d f} F_{ef} \wedge e^a \wedge e^b $$ is a full contraction of $\mathfrak{gl}_4$ tensors and objects that are invariant under $\mathfrak{so}_{3,1}$ that we can replace with $\mathfrak{gl}_4$ tensors and then show that they coincide under the action of the local Lorentz group. Hence, it is manifestly Lorentz-invariant.

Answer 2

For a different flavour of answer, you may want to look into the idea of Lie-algebra valued differential forms. This is because the parameter $\lambda^{a}_{~~b}$ that you use for an infinitesimal Lorentz transformation is actually valued in the Lorentz Lie algebra.

This formalism gives rise to a bracket $[\cdot\wedge\cdot]$ between Lie algebra valued forms, since you can't naively combine Lie algebra valued differential forms using the wedge product exactly if the Lie algebra is non-abelian. For example, your formula for $D\wedge D\lambda^{a}_{~~b}$ can instead be written as, $$ D^{2}\lambda^{ab} = [F\wedge \lambda]^{ab} = F^{a}_{~~c}\lambda^{cb}-F^{b}_{~~c}\lambda^{ca} $$ which will be equal to what you wrote by keeping in mind that as differential forms, $F$ and $\lambda$ are by definition antisymmetric in their indices.

Now, more to your problem, this bracket also satisfies a cycling condition as, $$ \alpha\wedge[\beta\wedge\gamma] = (-1)^{|\alpha|(|\beta|+|\gamma|)}\beta\wedge[\gamma\wedge\alpha] $$ as well as the almost anti-symmetric property, $$ [\alpha\wedge\beta] = (-1)^{|\alpha||\beta|+1}[\beta\wedge\alpha] $$ where $\alpha,\beta,\gamma$ are Lie algebra valued differential forms of degree $|\alpha|,|\beta|,|\gamma|$ respectively.

Using these properties you should be able to see how the terms in your last line will cancel out.

I have found it hard to find a complete summary of these kinds of things but, some places to start could be section IV of this paper, or appendix A of this paper.

Answer 3

Ok, I think I may have figured this out myself. Using the infinitesimal transformation stated in the question we can show that

$$\delta F^{ab} = D\wedge D\lambda^{ab} = F^{ac}\wedge\lambda^{b}_{~~c}+F^{cb}\wedge\lambda^{a}_{~~c}$$

which proves that the curvature two-form is Lorentz covariant. Now, this may now be used to state covariance of

$$\delta\star F_{ab} = -(\star F_{ac}\wedge\lambda^{c}_{~~b}+\star F_{cb}\wedge\lambda^{c}_{~~a})$$ Now, using the above combined with $\delta e^a = \lambda^a_b e^b$ it is now a simple matter to show the invariance of the gravitational action.