4
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ Something seems to have gone wrong with your indices in the first term of your final expression, as you have two $b$'s upstairs. $\endgroup$
    – J. Murray
    Jun 5, 2021 at 10:50
  • $\begingroup$ @J.Murray On yes. I just have to bring the indices downstairs that is all. $\endgroup$ Jun 5, 2021 at 11:15
  • $\begingroup$ I am not entirely clear on your formalism, but shouldn't the curvature terms also transform under the Lorentz transform? I would not expect invariance otherwise. $\endgroup$
    – Void
    Jun 5, 2021 at 12:24
  • 1
    $\begingroup$ OP I don’t have time right now to write an answer, but note that since both $F$ and $e$ transform linearly under Lorentz transformations, your problem is a special case of the tensor contraction theorem (that states that a full contraction of indices renders a scalar). $\endgroup$ Jun 5, 2021 at 19:11
  • $\begingroup$ Would a demonstration using finite, rather than infinitesimal, transformations be sufficient for OP? It's much easier to work with finite transformations in this case. $\endgroup$ Jun 7, 2021 at 19:28

3 Answers 3

4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I very much appreciate the explanation but it would have been nice if I could explicitly see it by looking at infinitesimal variation due to local lorentz transformations I mentioned. $\endgroup$ Jun 6, 2021 at 19:53
  • 2
    $\begingroup$ @user44690 I was hoping you could do that yourself after reading my answer. How can you prove $\mathfrak{gl}_4$-invariance of the contraction $A_a B^a$ by applying an infinitesimal transformation? You have $\delta A_a / \delta \varepsilon = - {\lambda^b}_a A^a$, and $\delta B^a / \delta \varepsilon = {\lambda^a}_b B^b$. Using the chain rule, you obtain $\delta (A_a B^a) = 0$. Exactly the same calculation proves the invariance of your expression. $\endgroup$ Jun 7, 2021 at 7:05
  • $\begingroup$ @user44690 the only information not supplied by this answer is the invariance of the Levi-Civitas tensor with respect to the Lorentz group. And that is straightforward to check. $\endgroup$ Jun 8, 2021 at 6:49
0
$\begingroup$

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.

$\endgroup$
-2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.