Does the discreteness of spacetime in canonical approaches imply good bye to STR?
Here is as I see the problem. The best way to understand it is to think what happens with rotations and quantum theory. Suppose that a certain vector quantity $V=(V_1,V_2,V_3) $ is such that its components are always multiples of a fixed quantity $d $. Then one is tempted to say that obviously rotational invariance is broken because if I take the vector $V=(d,0,0) $ and rotate it a bit, I get $V=(\cos(\phi) d, \sin(\phi) d,0) $, and $\cos(\phi) d $ is smaller than $d $. Therefore, either rotational invariance is broken, or the vector components can be smaller. Right? No, wrong. Why? Because of quantum theory. Suppose now that the quantity $V$ is the angular momentum of an atom. Then, since the atom is quantum mechanical, you cannot measure all the 3 components together. If you measure one, you can get say either $0 $, or $\hbar$, or $2\hbar$ ..., that it, precisely multiples of a fixed quantity.
Now supose you have measured that the $x$ component of the angular momentum was hbar. Rotate slowly the atom. Do you measure them something a bit smaller that $\hbar$? No! You measure again either zero, or $\hbar$ ... what changes continuously is not the eigenvalue, namely the quantity that you measure, but rather the probabilities of measuring one or the other of those eigenvalues. Same with the Planck area in LQG. If you measure an area, (and if LQG is correct, which all to be seen, of course!) you get a certain discrete number. If you boost the system, you do not meqsure the Lorentz contracted eigenvalues of the area: you measure one or the other of the same eigenvalues, with continuously changing probabilities. And, by the way, of course areas are observables. For instance any CERN experiment that measures a total scattering amplitude amplitude is measuring an area. Scattering amplitudes are in $\mathrm{cm}^2$, that is are areas.
Yes, sb1, all your statements are totally correct.
A discrete spacetime immediately implies lots of bulk degrees of freedom that remember the detailed arrangement of the discrete blocks - unless it is regular and unique. Consequently, the "vacuum" carries a gigantic entropy density - the Planck entropy density if the concept is applied at the Planck scale.
Such an entropy density $(huge,0,0,0)$ immediately violates the Lorentz symmetry because in different reference frames, it will have nonzero spatial components.
Also, the privileged reference frame will mean that any moving object will instantly - within a Planck time or so - dissipate all of its kinetic energy to the extra degrees of freedom in the discrete structure. So the "vacuum" will actually behave as a superdense liquid that immediately stops any "swimmer": inertia becomes totally impossible. One could continue with other flagrant contradictions.
All these things dramatically violate basic observations of the real world. In fact, the Fermi satellite has verified that even at the Planck scale, the Lorentz symmetry works much better than up to O(100%) errors. Also, one can see that any non-stringy, non-QFT unification attempt for quantum gravity has considered the discreteness in the very sense we discuss, which is why it is instantly excluded.
One can use special dictionary and special proofs for each of them. For example, spin foam models define the proper area of surface $\Sigma$ more or less as the number of its intersections with the spin foam. However, it's clear that this number can't go to zero even if the shape of $\Sigma$ in spacetime is chosen to be light-like or near-light-like.
It's worth making an analogy here. LQG does not impose an fixed lattice structure. The only discreteness existing is that area and volume operators have a discrete spectrum, specifically meaning that there is a "smallest" area and volume possible before zero. This should be compared to the familiar spin algebra --- there is rotational invariance, but the states are not rotationally invariant (individually)! The situation is almost exactly analogous.