Tetrad postulate: Implies or results from the metricity of the connection?

Question

Hi,

I see that the tetrad postulate:

$\nabla_{\mu}e_{\nu}^{I}=\partial_{\mu}e_{\nu}^{I}-\Gamma_{\mu\nu}^{\rho}e_{\rho}^{I}+\omega_{\mu J}^{I}e_{\nu}^{J}=0$ Can be merely derived from writing a tensor in two different basis (pure natural-coordinates $\{\partial_\mu\}$ and mixed $\{\partial_\mu\} + \{e_a\}$), my questions are:

1. Does this imply the metricity of the connexion or the inverse?
2. If it is the inverse, how?
3. In writing $g_{\mu\nu}=\eta_{IJ}e_{\mu}^{I}e_{\nu}^{J}$ do we need to impose a metricity on $\eta$?

ps: metricity = metric compatible

Answer 1

Having botched the first attempt at answering this question and not wanting to delete the evidence, let me try again here.

The "tetrad postulate" is independent from metricity and from the condition that the connection be torsion-free. It is simply the equivalence (via the vielbein) of two connections on two different bundles. Here are the details. $M$ is a smooth $n$-dimensional manifold.

First of all we have an affine connection $\nabla$ on $TM$ with connection coefficients $\Gamma^\rho_{\mu\nu}$ relative to a coordinate basis -- that is, $$\nabla_{\partial_\mu} \partial_\nu = \Gamma_{\mu\nu}^\rho \partial_\rho,$$ with $\partial_\mu$ an abbreviation for $\partial/\partial x^\mu$ where $x^\mu$ is a local chart on $M$.

Then we have a connection on an associated vector bundle to the frame bundle $P_{\mathrm{GL}}(M)$. The frame bundle is a principal $\mathrm{GL}(n)$-bundle and given any representation $\rho: \mathrm{GL}(n) \to \mathrm{GL}(V)$ of $\mathrm{GL}(n)$ we can define a vector bundle $$P_{\mathrm{GL}}(M) \times_\rho V.$$ Take $V$ to be the defining $n$-dimensional representation and call the resulting bundle $E$. Relative to a local frame $e_a$ for $E$, a connection $\hat\nabla$ defines connection one-form $\omega$ by $$\hat\nabla_{\partial_\mu} e_a = \omega_{\mu~a}^b e_b.$$

Now the vielbein defines a bundle isomorphism $TM \buildrel\cong\over\longrightarrow E$ and all the "tetrad postulate" says is that the two connections $\nabla$ and $\hat\nabla$ correspond. In fact, the "tetrad postulate" is just the statement that the vielbein is a parallel section of the bundle $T^*M \otimes E$ relative to the tensor product connection.

This works for any affine connection $\nabla$ on any smooth manifold $M$. No metric is involved.

A special case of this construction is when $(M,g)$ is a riemannian manifold and $\nabla$ is the Levi-Civita connection (i.e., the unique torsion-free, metric connection on $TM$). You can without loss of generality restrict to orthonormal frames, which defines a principal $\mathrm{O}(n)$ (or $\mathrm{O}(p,q)$ depending on signature) bundle. The representation $V$ restricts to an irreducible rep of the orthogonal group, possessing an invariant bilinear form $\eta$. This relates $g$ and $\eta$ as in your question.

Answer 2

No it is not the torsionless condition which is:

$T_{\mu\nu}^{I}=D_{[\mu}^{\omega}e_{\nu]}^{I}=\partial_{[\mu}e_{\nu]}^{I}+\omega_{[\mu J}^{I}e_{\nu]}^{J}=0$

This postulate says:

$\nabla_{\mu}e_{\nu}^{I}=\partial_{\mu}e_{\nu}^{I}-\Gamma_{\mu\nu}^{\rho}e_{\rho}^{I}+\omega_{\mu J}^{I}e_{\nu}^{J}=0$

And is merely obtained by expressing a tensor in two different basis then putting the two components equals after returning in the natural basis, it is a sorte of "consistancy" condition.

I see the anticommutativity of the spin-connection not implying the metricity but resulting from it! By imposing $D_{\mu}^{\omega}\eta_{IJ}=0$ (neverthless, after that it will IMPLY it)

So in the same spirit, I wonder if there is a way to obtain the "tetrad postulate" from a same requirement, say, the metricity of $g$ for example (It will neverthless IMPLY it once written down)

I think that the confusion in almost all papers I read is the problem, recently I read a paper which pointed out these confusions (An Ambiguous Statement Called ‘Tetrad Postulate’ and the Correct Field Equations Satisfied by the Tetrad Fields arXiv:math-ph/0411085v12 6 Jan 2008) The author said that this postulate comes from the metricity condition but didn't show how (even if he was very rigourous in the mathematical demonstrations, too rigourous at my tast :-))

Answer 3

Edit: I had originally stated that the so-called "tetrad postulate" is the torsionless condition, but this is not correct, as pointed out by pedro below. The torsionless condition is the statement that the identity map $\mathrm{id} \in \mathrm{End}(TM)$ as a section of $T^*M \otimes TM$ obeys $$d^\nabla \mathrm{id} = 0,$$ where $d^\nabla$ is the exterior covariant derivative: $$d^\nabla : \Omega^1(TM) \to \Omega^2(TM),$$ where $\Omega^p(TM)$ denotes $p$-forms with coefficients in the vector bundle $TM$. And this translates to the vanishing of the antisymmetric (in $\mu\nu$) part of the tetrad postulate.

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Edit: please refer to my second answer!

Let me try to answer the questions as posed:

1. No. The metricity comes from demanding that $$\omega_{\mu~IJ} = - \omega_{\mu~JI}$$ where $$\omega_{\mu~IJ} = \omega_{\mu~I}^K \eta_{KJ}$$

2. I don't understand what you mean by the inverse...

3. You do not impose metricity of $\eta$ because $\eta$ is a fixed inner product on a fixed vector space $E$. The tetrads, which are generically only locally defined on some $U\subset M$, provide smooth isomorphisms between the tangent space and $E$. It is easy to see that provided that the connection coefficients $\omega$ satisfy the above condition, $\nabla g = 0$.

Answer 4

Let me again ask one of my "physics-oriented" questions: (sorry if I'm not mathemacally rigorous)

What Mr. Figueroa-O'Farrill wrote is actually the mathematical background of the so-called "Palatini formalism" of general relativity, this works as follows:

Instead of considering a single field $g$ with which the connection $\Gamma_{0}$ is compatible and torsion-free, ie: $\omega_{\mu J}^{ I}=e_{J}^{\nu}\nabla_{\mu}^{\Gamma_{0}}e_{\nu}^{I}$ where $\nabla_{\mu}^{\Gamma_{0}}$ is the covariant dérivative associated with the unique torsion-free and metric-compatible connection $\Gamma_{0}$ (this can be derived from the 'tetrad postulat'), so $\omega$ is uniquely defined by $e$ and We can take only $e$ as the dynamical field of the theory as it reproduces $g$.

The Palatini formalism then considers the two fields $e$ and $\omega$ as independent fields, (but they remain related by $\omega_{\mu J}^{ I}=e_{J}^{\nu}\nabla_{\mu}^{\Gamma}e_{\nu}^{I}$ where now $\nabla_{\mu}^{\Gamma}$ is an arbitrary connection (not necessarily the Levi-Civita one)), now the equations of motion wrt a variation $\delta \omega$ will precisely (are supposed to!) tell that this connection is the Levi-Civita one (so now $\omega$ and $e$ are no longer independent as they become related by $\omega_{\mu J}^{ I}=e_{J}^{\nu}\nabla_{\mu}^{\Gamma_{0}}e_{\nu}^{I}$), once reinserted in the action, we recover the usual action of Einstein-Hilbert (the other EoM is no more than the Einstein equations)

My problem is that I managed to show that these EoM (wrt $\delta \omega$) are equivalent to a vanishing torsion, but to tell that the connection is the Levi-Civita one, we need metric-compatibility, is this pre-assumed in this theory?

Answer 5

Sorry, can I answer my own question?

Ok I can :-)

In fact what is assumed in the Hilbert-Einstein formulation of GR is only the torsionless condition, which, together with the metricity (which comes from SR) make the connection the Levi-Civita one.

In pulling back things into a Minkowski space (using the tetrad) we want to do some "field theory" which is not well defined on curved spaces, in particular, we want to be able to tackle with spinors.

The tetrad postulate appears therefore as a way to transpose the metricity wrt $\eta$ to that wrt $g$, in particular, $\omega$ is introduced in the Palatini formalism as antisymmetric (ie: metric compatible), this metricity together with the torsionless brought by the EoM wrt $\delta\omega$ makes the connection a Levi-Civita one.

Please comment my reasoning, because I'm not sure about what I'm telling...