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In loop quantum gravity, the canonical (Ashtekar) variables are chosen to be the densitized inverse triad $\mathbf{E}$ and some rotation connection field $\mathbf{A}$. To get the ordinary triad from $\mathbf{E}$, we need to take its "square root". But we know as long as the volume factor, i.e. the square root of the determinant of $\mathbf{E}$ is nonzero, we always have two solutions for the triad. They are related by a branch cut around $|\mathbf{E}| = 0$. We also know that if this determinant is negative, the corresponding triad will be imaginary! OK, you might now say there's no problem as long as we stick to solutions where $|\mathbf{E}|$ is always positive everywhere.

But here is where another problem comes in. The state about which loop quantum gravity expansions are made is the state where $\mathbf{E}=0$ everywhere. Excitations over this state by Wilson loops contain wavefunctional components with negative determinants as frequently as we have components with positive determinants. Does that mean loop quantum gravity predicts imaginary distances, negative areas and imaginary volumes? And besides, aren't we expanding about a highly irregular solution?

Another related question is: People sometimes multiply the Hamiltonian constraint by the factor $\sqrt{| \mathbf{E} |}$, but this goes to zero whenever $| \mathbf{E} | = 0$, which actually weakens the constraints by introducing additional solutions which were previously forbidden. Is such a procedure really justified?

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The question could be improved by posting relevant links. :-) –  Sklivvz Jan 9 '11 at 11:50
@Sklivvz: links to what? All that is needed is Ashtekar variables but those are presumably known to anyone who might try to answer this question :-) –  Marek Jan 9 '11 at 11:54
@Marek: thanks for providing the link (so there is a link to be provided, see?). I am fully aware that someone that can answer the post necessarily knows what the post is about - whereas not all people that will read the post in the future might be, and the link will benefit them. –  Sklivvz Jan 9 '11 at 12:53
@Sklivvz: well, sure. But if you start with providing links for everything someone might not know, where will you stop? In this question I can imagine every second word not being understood by a layman. Do you propose we provide links for everything? Or am I correct in assuming that you wanted those links just for your own sake (i.e. links only for things you are not acquainted with)? :-) –  Marek Jan 9 '11 at 12:57
@Marek: when I ask questions I try to always include relevant links. They tend to make the question more precise. The level of this forum is pretty much university level (undergraduate+) so this is the level that should be assumed, in my opinion. Links don't cost anything, and I didn't down vote. It was merely a suggestion... on how to get my up vote :-) –  Sklivvz Jan 9 '11 at 13:00

2 Answers 2

No, LQG does not predict imaginary distances, negative areas and imaginary volumes. It is sufficient to remember the geometrical interpretation of the triad. A triad with a negative determinant is simply a left handed rather than a right handed triad. Therefore it does not describes funny spaces with imaginary volumes, but just the same good old space with where the basic triad is left handed rather than right handed. This simply guides you in giving the proper definitions of area and volume, putting the right sign and the right absolute values where it belongs. There may be papers where those absolute values have been placed wrong, but when things are done properly, no imaginary volumes ! carlo rovelli

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But the triad is $\bf e$, while the densitized inverse triad $\bf E$ is quadratic in $\bf e$. It's $\bf E$ which shows up in the canonical variables. And if $\bf E$ has a negative determinant, $\bf e$ has to be imaginary. –  QGR Jan 27 '11 at 16:30
No. There are sereval ways of seeing this. One is the following (a la Thiemann). The exact relation between E and e is: E=|det e|e^{-1}. It follows that if E change signs, so does e. And the determinant of the two has the same sign. carlo rovelli –  Carlo Rovelli Jan 28 '11 at 14:25
Is it $\left| \text{det} \mathbf{e} \right|\,\mathbf{e}^{-1}$, or $\text{det} \mathbf{e} \;\mathbf{e}^{-1}$? Because if it's the former, the canonical relations break down whenever the determinant goes to zero. –  QGR Jan 28 '11 at 17:12
I am not sure what you mean, but even if what you say is correct, there would be nothing bad. In the classical limit, the quantum theory we seek needs not to agree with GR in the extreme regions where the determinant of the metric is zero. Indeed, Ashtekar theory is known to be different from GR in these regions, since it admits solutions with singular metrics that are not admissible in GR. But this is not bad. Because the part of GR which supported by the observations is that for det e different from zero. Any theory behaving a bit differently for det e=0 is as empirically credible as GR. –  Carlo Rovelli Jan 28 '11 at 21:24
@QGR. Maybe I was not clear enough. I agree with you that if we expanded aroynd det e ≠ 0 there is no problem. By this does imply that there is a problem if we expand around e = 0. Why should there be a problem? You say: because we have a theory different from GR? So what? We have a theory that agrees with GR where it should, namely where GR is tested. –  Carlo Rovelli Feb 6 '11 at 8:41

Yes, absolutely, the "proper distances vanish" is a singular point of the configuration space of general relativity and it is not permissible to "expand around it".

And you are also totally right about the "imaginary roots" problem. The map was chosen to be bilinear exactly because the loop quantum gravity physicists wanted the areas to be quantized in a simple way. So the triad $E$ really measures the proper areas and the proper distances must be taken as the square root of it.

There is nothing physically legitimate about this operation - it is just a trick to pretend that the metric tensor is equivalent to a bulk gauge field. While the counting of components of the SU(2) gauge field could work, the physics doesn't work and the field redefinition is not one-to-one. The appearance of the imaginary roots is just one manifestation of the problem.

There are other manifestations of the illegitimacy of the field redefinition from gravity to the "new variables". In particular, the quantization of the areas themselves proves that the map is only valid locally on the configuration space, but not globally. The holonomies or Wilson loops constructed out of the gauge field are periodic variables; that's essentially why their canonical momenta are quantized - and the canonical momenta are the areas.

However, in the proper gravity, with the continuous metric tensor, there is no quantization of the areas. In fact, such a quantization violates the local Lorentz symmetry because the "spin network" always picks a preferred reference frame, much like the luminiferous aether. This brutally breaks the Lorentz symmetry and destroys inertia: the entropy density is pretty much Planckian and an object moving through this material will instantly dissipate its energy to the "spin network degrees of freedom" and stop.

So the field redefinition is only a game that is valid locally at the configuration space - because of the right counting of the degrees of freedom - but is incompatible globally, because it is not one-to-one, and it incompatible with the required dynamics for the degrees of freedom. In the new variables, one can never get a theory that behaves as general relativity.

That's why the word "gravity" in "loop quantum gravity" is a misnomer. There can't be any gravity.

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This is complete nonsense. The map was not "chosen to be bilinear exactly because the loop quantum gravity physicists wanted the areas to be quantized in a simple way", for the simple reason that the map was introduced in physics 10 years before anybody mentioned quantization of the area! –  Carlo Rovelli Feb 6 '11 at 8:43

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