I would like a very simple answer to this question with as little math as possible.

In quantum loop gravity, we want to quantize the curvature of space. I understand that:

  • First, we discretize our manifold (let's say we work in 2D, then we will discretize with triangles)
  • Second, we quantize our discretisation: the length of the edges of our triangles will be the eigenvalues of spin square operators $J^2$.

Actually from what I understood, the curvature is given by some $SU(2)$ group elements that live on the edges of those triangles. The value of the group element will give the curvature of the space.

My questions are:

  1. Am I right by saying that the curvature will be given by a group element? So we discretised our manifold to give a given curvature at each "step" of the space?

  2. What exactly is the link between the length of the edge of the triangles ? Does it says where my group element live? Like if I find $1/2(1/2+1)$ as length of an edge of my triangle (so $j=1/2$), does it mean that my group element describing my curvature will live in the spin $1/2$ representation space?

  3. Why is the group $SU(2)$ used, in particular?


Am I right by saying that the curvature will be given by a group element?

As any observable in any quantum-mechanical theory, the curvature of space in LQG is an operator acting on the kinematical Hilbert space. But how is this operator defined in LQG?

Firstly, let's pass from the differential quantity (the curvature tensor) to a corresponding integral quantity – the holonomy of the Ashtekar-Barbero connection around a small loop. From classical gauge theory we know that the two are related by

$$ h_{\gamma}(A) = \vec{\exp} \int_{\gamma} A_{a} dx ^{a} = 1 + \frac{1}{2} F_{ab} dx^{a} dx^{b} + \mathcal{O}(dx^4),$$

where $dx^{a}$ are the linear elements describing the loop.

Apparently, the holonomy becomes a well-defined operator on the kinematical Hilbert space of LQG. It acts on cylindrical functions by multiplying them by the group element associated to the loop.

Assume that you have a cylindrical function $\Psi(g_1, \dots, g_n)$ of $n$ group elements corresponding to the edges of your graph, and that the loop under consideration is one of those edges (let it be $g_i$). Then the holonomy operator is defined as

$$ \hat{h}_{\gamma} \Psi = g_i \Psi(g_1, \dots, g_n). $$

If the loop under consideration is not among the edges of your graph, it is always possible to extend the graph by adding one more edge (the loop), and passing to a wavefunction

$$ \Psi(g_1, \dots, g_n, g_{n+1}) = \Psi(g_1, \dots, g_n), $$

i.e. independent of the holonomy of the loop. The definition from above thus still makes sense.

Hereby, the holonomy operator in LQG is well-defined. But what about curvature?

So we discretised our manifold to give a given curvature at each "step" of the space?

Yes and no. In LQG, space is neither discrete or continuous – it is "quantum" in the usual sense, meaning that the spectra of geometrical observables are discrete, but the expectation values can still be continuous. It is a very common situation in quantum mechanics, actually. But it is also very unique, because here it is applied to geometry of space, not to dynamical quantities. Such a situation only arises in background independent quantum theories. Some critics of LQG fail to comprehend this, unfortunately, and criticize LQG for being just another theory on the lattice with all the usual flaws of lattice theories, but it is not so.

For instance, consider the curvature operator from the previous question. What should we do to define it, or its analog? Obviously, it corresponds to the the limit of holonomy for when the loop shrinks to a point.

Let's consider a particular state given by a cylindrical function $$ \Psi(g_1, \dots, g_n) $$ which lives on some graph $\Gamma$ (each group element corresponds to the edge of the graph, remember?). We start by introducing a loop $\gamma$ to the graph, and as we shrink it, the loop passes through other edges of the graph.

But because the graph itself is finite, there will be a moment when the loop will no longer pass through other graph edges any longer. Shrinking the loop after this moment makes no difference whatsoever when considered on the space of diffeomorphism-invariant states.

This point deserves a further explanation. Normally, a spin network state is given by a gauge-invariant function on the graph embedded in the spatial manifold $\Sigma$. If we vary the graph a little bit (by changing positions of nodes and edges), we will get a different, orthogonal state. This space suffers from an uncountable basis problem (aka nonseparability), which is why it isn't feasible to use spin networks in QCD. But a beautiful miracle happens in case of gravity (or any other background independent theory) – once the diffeomorphism constraint is taken into consideration, the remaining space (the kernel of the diffeomorphism constraint) becomes separable. Its basis is given by spin network functions on abstract graphs, where the exact embedding of the graph into $\Sigma$ no longer matters.

So back to the contracting loop argument – on the kernel of the diffeomorphism constraint, after a certain moment contracting the loop further doesn't change anything. We are thus presented with a smooth limit of the holonomy operator, which can be used to define curvature.

Note that the limit is defined with respect to a state. Unlike the ordinary situation, where we define the limit of the operator itself, here we define a new operator by taking the limit of the value of the old operator on the state, for each state. This is how ultraviolet divergences of General Relativity are resolved in LQG. This way of taking the limit can be considered as a renormalization of the theory, because it actually changes the theory – this definition of the limit is not the usual "limit of the operator" definition, which diverges.

What is exactly the link between the length of the edge of the triangles?

When you try to come up with an operator that corresponds to the area of a surface (in your 2d example – a length of the curve), it turns out that that the spin network basis diagonalizes this operator. I.e. the area/length operator associated to a surface/curve acts on spin networks as follows:

$$ \hat{A} \Psi = \gamma l_P^2 \sum_i \sqrt{j_i (j_i + 1)} \Psi, $$

where the sum is over intersections between the surface/curve and the links of the graph of $\Psi$, and $j_i$ are spins associated to those links.

It is important to note that spin networks are just a convenient choice of a basis in the space of gauge-invariant cylindrical functions. Here, I've said it. The spins are a purely mathematical artifact, coming from the Peter-Weyl theorem, which allows us to decompose a function on $SU(2)$ in terms of its unitary representations. But it also proves useful in practice, because it enters directly in the spectrum of the area/length operator.

We thus end up with a picture of the background independent quantum theory: the area/length is not pre-determined by initial confitions (in fact, there's only one area/length operator!), but is determined by the quantum state of the system itself.

Does it mean that my group element describing my curvature will live in the spin 1/2 representation space?

Spin networks are eigenstates of area/length, but not eigenstates of curvature. Your curvature operator does not have a single value in this state, it is "fuzzy" – spread across a range of eigenvalues.

It is, in fact, expected. Because in the classical theory, area/length come from the spatial metric field, which comes from the (densitized) triad field. While curvature comes from holonomy, which comes from the Ashtekar-Barbero connection. But those two are known to be a canonically conjugate pair! Thus we should expect noncommutativity between area/length and curvature in the quantum theory.

To see that directly, consider an action of the curvature operator defined above on the spin network state, say, corresponding to a single loop:

$$ \hat{h}_{\gamma} \Psi(g) = g \Psi(g). $$

First, the holonomy is an $SU(2)$-valued operator, which is expected, because in the classical theory it is an element of $SU(2)$. Basically, you need to choose a co-ordinate patch on $SU(2)$ and you will have a normal operator corresponding to each of the co-ordinates.

Each such operator changes the spin network state nontrivially. It becomes a superposition of different spins.

Btw: actual semiclassical (coherent) states in the 4d theory are always quantum-spread in both curvature and area, and this spread is minimized by those states. Thus in the limit of large scales, the quantum nature of the theory isn't observed – we enter the regime of validity of General Relativity. As an analogy, consider a simple harmonic oscillator, or a quantum field – the ground state is spread both in co-ordinate and momentum, but those spreads are classically unobservable.

Finally: why is SU(2) the group elements used? Why this group?

This question has different answers in 3d and 4d.

The 4d answer is – General Relativity with the Holst action in the frame-connection formalism has a natural phase space parametrized by the Ashtekar-Barbero $SO(3)$ connection, and its canonical momenta – the densitized triad (which directly relates to the spatial metric, thus to the area or surfaces).

When quantizing, the projective action of the $SO(3)$ group becomes a normal action of $SU(2)$, as usual.

Also, a small update that will (hopefully) clarify some of the confusion.

LQG uses combinatorics heavily in its definition and computations. Graphs, spin networks, 2-complexes, spinfoams, unitary irreducibles, etc. In fact, graphs and 2-complexes might give you an impression that they actually model space-time in LQG – that space is a graph, and spacetime is a 2-complex.

But it is just not true, and not only technically – LQG is saying something completely different about space-time.

Graphs do not pre-exist like in lattice theories here. Graphs are a basis in the Hilbert space of the quantum theory. States in LQG are superpositions of graphs.

It is just wrong to consider the actual, physical space to be a graph, because – and this is extremely important – any spin network state is wildly quantum, and doesn't have a well-defined classical limit. It doesn't correspond to a nice discretization of the 3-manifold, as you could imagine. Nor does it describe correctly the discrete properties of space-time on microscopic scales! No, graph edges are not the discreteness of space that LQG people are talking about. This could appear very logical and appealing, but it is just false!

Instead, the actual discreteness comes from the spectrum of area/length. Graphs are invisible, they are mathematical tools which we use to encode the quantum properties of spacetime. But the properties are not in direct correspondence with our intuition.

I'll give you an example. It is possible to modify LQG by considering non-compact groups (which will correspond to some other theory rather than General Relativity in the classical limit). The theory is still defined on graphs, but now the set of unitary representations is no longer discrete. In this theory, the spectrum of area is now continuous! We still have graphs and 2-complexes, but the discreteness is just gone. Thus those two were never related.

It is funny actually, because LQG has many points of tension and a lot to be criticized for, but many critics apparently criticize it for something that they have misunderstood. Indeed, it is an extremely common misconception that LQG is a lattice theory where space is a graph.

For example, LQG is sometimes (erroneously) criticized for being non-Lorentz-invariant. While the reality is – local Lorentz symmetry is a gauge group of LQG (just like it is in General Relativity!), which does not receive anomalous contributions on the quantum level. However, the vacuum state corresponding to the Minkowski solution does not possess global Poincare invariance, meaning that according to LQG, special relativity is only valid on large (comparable to Planck) scales. But all of the above has absolutely nothing to do with lattices.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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