The Master constraint program in LQG

September 2, 2006

Posted by Aaron

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I’m a little reluctant to post much on the master constraint program because I haven’t read much on it. But I thought I’d post this if others want to comment on the subject.

My initial question is how does the master constraint program work in classical mechanics? In particular, say we are given some symplectic manifold and some set of constraints. The master constraint is $M = K^{IJ} C_I C_J .$ Using this, how does one obtain the constrained phase space?

Or is this the wrong question to ask?

Posted at September 2, 2006 5:20 AM UTC

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15 Comments & 1 Trackback

Re: The Master constraint program in LQG

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To formalize the question a bit, let’s review how things normally work. Consider a phase space $M$ , and a set of (for simplicity, first-class) constraints, $C_i$ . The construction of the reduced phase space, $\tilde{M}$ can be described either algebraically or geometrically as follows

Algebraic:

Let

\mathcal{I}\subset R(M)

be the ideal generated by the

C_i

in the ring of functions on

M

. The ring of functions,

R(\tilde{M})

is constructed by

Restrict to the subring $R_0(M) =\{A\in R(M) | \{C_i,A\}\in\mathcal{I}, \forall i\}$ .
Take the quotient $R(\tilde{M})= R_0(M)/\mathcal{I}$ .

Since the constraints are first-class, the ring

R(\tilde{M})

inherits the Poisson-bracket structure,

\{\cdot,\cdot\}

from

R(M)

Geometric:

Restrict to the subspace $M_0=\{C_i=0\}\subset M$ .
The Hamiltonian vector fields, $\{C_i,\cdot\}$ , generate a foliation of $M_0$ . We define $\tilde{M}$ as the space of leaves of this foliation.

Again,

\tilde{M}

inherits its symplectic structure from the symplectic structure of

M

The geometrical characterization is a little simpler to understand. The algebraic one is closer to what we need to do in quantum mechanics.

But, if you refrain from availing yourself of the $C_i$ and, instead, work only with $\mathcal{M}=K^{ij}C_i C_j$ , is there any analogue of either of the classical constructions above?

If not, why should there be a quantum-mechanical construction, which uses only $\mathcal{M}$ and not the $C_i$ themselves, which gives the “right” answer?

Posted by: Jacques Distler on September 2, 2006 6:56 AM | Permalink | PGP Sig | Reply to this

First Class

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For those not versed in the zoology of constrained Hamiltonian Mechanics, the constraints are said to be first class if $\{C_i,C_j\}\in \mathcal{I},\, \forall i,j$ . We usually also assume that the Hamiltonian satisfies $\{C_i,H\}\in \mathcal{I},\, \forall i$ .

Posted by: Jacques Distler on September 2, 2006 8:42 AM | Permalink | PGP Sig | Reply to this

Re: The Master constraint program in LQG

Hi,
As far as I understand(which isnt very much), one cannot define the reduced phasespace using the master constraint.(they do not generate any gauge transformations on the constraint surface obviously.) However M=0 defines the same constraint surface as that defined by C_{j}(x)(for all j,x). Note that this is all one needs to implement the Dirac quantization program. Also the ring of “observables” can still be defined in this program by those functions on phase-space which satisfy {{O,M},O}=0 weakly.

Posted by: anonymous on September 4, 2006 7:55 AM | Permalink | Reply to this

Ring On

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However $\mathcal{M}=0$ defines the same constraint surface …

Not to an algebraic geometer, it doesn’t. The ring of functions is very different.

Also the ring of “observables” can still be defined in this program by those functions on phase-space which satisfy $\{\{O,\mathcal{M}\},O\}=0$ weakly.

By “satisfy … weakly” I presume you mean $\{\{O,\mathcal{M}\},O\}\in \mathcal{I}$ ?

I’m afraid I don’t understand.

This is not a linear condition (if $O_1$ and $O_2$ are two observables satisfying this condition, $O_1+O_2$ need not satisfy it). So the set of such $O$ s does not form a ring.

Posted by: Jacques Distler on September 4, 2006 8:43 AM | Permalink | PGP Sig | Reply to this

Re: Ring On

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Probably the goal here is just to look at $\bigcap_i \mathrm{ker}C_i$ .

As far as I can tell from Robert’s account #, the idea is that forming $\mathcal{M} = K^{ij}C_i C_j$ (with $K^{ij}$ highly non-unique, as far as I can tell) might suggest a way to deal with the fact that in general $\bigcap_i \mathrm{ker}C_i = \{0\}$ by instead looking for $c\in \mathbb{C}$ such that

(1)

\mathrm{ker}(\mathcal{M}-c)

is nontrivial.

I don’t know if anyone discusses nontrivial examples for this procedure, but I gather that this is what following the “master constraint program” amounts to.

Posted by: urs on September 4, 2006 9:11 PM | Permalink | Reply to this

Re: Ring On

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Before getting to quantum mechanics, it behooves us to understand the classical story first.

$M_0$ is the intersection of the vanishing loci of the $C_i$ . It also happens to be the vanishing locus of $\mathcal{M}$ . However, that is far from sufficient, as Thomas notes below.

Indeed, we are not interested in $M_0$ . $M_0$ , in general, need not be even dimensional, and the original symplectic structure is always degenerate, when restricted to $M_0$ . We are interested in the space $\tilde{M}$ , which is the space of leaves of the foliation of $M_0$ . $\tilde{M}$ is even-dimensional, and (assuming we’ve done things correctly) has a nondegenerate symplectic form.

instead looking for $c\in\mathbb{C}$ such that $\ker(\mathcal{M}-c)$ is nontrivial.

I don’t even want to touch the quantum mechanical case yet. There are so many potential pitfalls of the quantum mechanical case (including the operator-ordering issues which force one to take $c\neq 0$ ).

But if you can’t arrive at the right physics classically, it’s rather pointless to go on to the quantum case (with all its attendant difficulties).

Posted by: Jacques Distler on September 4, 2006 9:36 PM | Permalink | PGP Sig | Reply to this

Re: Ring On

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I’d agree. But it’s also true that what Thomas Thiemann discusses is just what I mentioned?

Or is it? I haven’t read his text, just seen Robert’s comment on it.

But I agree, in this entire discussion it is a little hard to tell (for me at least) what - even before we plunge into all the technical details - the goal of the exercise actually is.

To some degree the impression I got is (but people should correct me if that’s wrong) that the goal is to temper with the rules of usual quantum theory and see what happens.

So I got the impression that Thomas Thiemann happens to want to understand the kernel of $\mathcal{M}-c$ .

As you point out, this is likely to be disconnected from “standard quantization”. But so was his “LQG quantization” of the bosonic string (and of the harmonic oscillator).

So (maybe Robert knows this?), does he consider any example for a solution of a “master constraint”. One which would illustrate why one might be tempted to consider this?

Posted by: urs on September 4, 2006 10:26 PM | Permalink | Reply to this

Re: Ring On

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As you point out, this is likely to be disconnected from “standard quantization”. But so was his “LQG quantization” of the bosonic string (and of the harmonic oscillator).

More than that. In “standard quantization”, there’s a Correspondence Principle, a limit in which classical physics is recovered. My complaint is not merely that one does not obtain the “standard” quantum theory. My complaint is that (it appears) that there is no limit in which one obtains the correct classical theory either.

Thomas would claim that, in the case of gravity, no one knows what the correct quantum theory is, so who the hell are you to say that his approach is “incorrect”?

But we bloody well do know what the correct classical theory of gravity is, and I don’t see how to obtain it from an approach which dispenses with the constraints, $C_i$ , and replaces them with a “Master Constraint,” $\mathcal{M}=K^{i j} C_i C_j$ .

Posted by: Jacques Distler on September 5, 2006 6:41 AM | Permalink | PGP Sig | Reply to this

Re: Ring On

Dittrich and Thiemann have a series of five papers called “testing the master constraint program”. What precisely are they testing there? Did anyone read some of that? Robert?

Posted by: urs on September 5, 2006 10:33 AM | Permalink | Reply to this

Re: Ring On

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Nope I have other things to do. However, yesterday, I had a look at ThThs latest opus where the Master Constraint is equation (4.37).

My interpreation of the paragraph below that equation is as follows. As we know you want to restrict yourself to the submanifold where all constraints vanish or alternatively the Master Constraint vanishes. There, you want to go to the space of orbits generated by $\{C_i,\cdot\}$ . You would think that good functions on the space of orbits are the invariants $F$ , i.e. functions for which $\{C_i,F\}=0$ for all $i$ .

Now, you compute (for simplicity using the unit matrix for $K$ ) $\{F,\{F,M\}\}=\{F, 2\sum_i C_i\{F,C_i\}\}= 2\sum_i(\{F,C_i\}^2+C_i\{F,\{F,C_i\}\}).$

As you can see, the second term vanishes on the constraint surface and thus $\{F,\{F,M\}\}$ is equivalent to the vanishing of all $\{F,C_i\}$ . Therefore this seems to be indeed an alternative formulation. And because of this equivalence Jacques’ worries that the condition is quadratic and thus might not define a ring is also unfoudned as it is equivalent to a number of linear conditions.

However, I remember from my algebraic geometry class that one has to be careful that invariants not always are good coordinates on the orbit space: The example were matrices $M$ where the symmetry is similarity $M\sim A^{-1}MA$ for invertible $A$ . The algebraic invariants are formed from the coefficients of the characteristic polynomial $\det(M-\lambda 1)$ (i.e. the elementary symmetric functions of the eigenvalues of $M$ ). However, the orbits are labeled by bringing $M$ to Jordan normal form but $\left(\begin{smallmatrix}\lambda&0\\ 0&\lambda\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}\lambda&1\\ 0&\lambda\end{smallmatrix}\right)$ are not similar but have the same characteristic polynomial (~~These matrixes are supposed to be 2x2 the first lambda,0\\ 0,lambda and the second lamba,1\\ 0,lambda but I cannot get those to \TeX properly. Jacques, please help!!!~~ Fixed>) . Indeed, you can show that there is no continuous invariant that tells the two orbits apart. However, I have no idea of similar complications are relevant here.

Posted by: Robert on September 6, 2006 2:45 PM | Permalink | Reply to this

Re: Ring On

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First off, to produce one of those matrices, try

$
\left( \begin{smallmatrix}
  \lambda & 1 \\
  0 & \lambda
\end{smallmatrix} \right)
$

for inline, or

$$
\begin{pmatrix}
  \lambda & 1 \\
  0 & \lambda
\end{pmatrix}
$$

for display-size matrices.

As to the physics, I have to concede that the set $\{ A\in R(M) | \{A, \{A,\mathcal{M}\}\}\in \mathcal{I} \}$ is isomorphic to $R_0(M) = \{ A\in R(M) | \{A,C_i\}\in \mathcal{I}, \forall i \}$

However, to get the ring of observables, we still need to quotient by $\mathcal{I}$ , $R(\tilde{M})= R_0(M)/\mathcal{I}$

On the geometric side, the subspace $\{\mathcal{M}=0\}\subset M$ is the same as the subspace $\{C_i=0\}\subset M$ . However, $\tilde{M}$ is the space of leaves of the foliation of this space by the Hamiltonian vector fields $\{C_i,\cdot\}$ .

In both approaches, you’ve successfully replaced the first stage of the construction by replacing the set of constraints, $C_i$ , by the “Master Constraint,” $\mathcal{M}$ .

But, in both case, the second stage of the construction of $\tilde{M}$ still requires the $C_i$ themselves, rather than $\mathcal{M}$ .

Posted by: Jacques Distler on September 10, 2006 7:51 AM | Permalink | PGP Sig | Reply to this

Re: Ring On

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Another thing I am confused about on the algebraic side is the importance that we are dealing with real rather than complex algebras. Only over the reals, $X^2+Y^2=0$ is eqivalent to $X=0$ and $Y=0$ . Over $C$ you could try to write a similar thing as $\bar {X}X+\bar {Y}Y=0$ but of course this is no good as $\bar\cdot$ is not holomorphic. Back in my algebraic geometry classes we always assumed the fundametal theorem of algebra, without it things might become weird and this might just be an example of this.

As for the layout side of things: I first tried to write \backslash XX as I would have done in \TeX. But this produced a bar over both X’s and not just the first one (without curly braces I would have expected \backslash bar to operate on the first token following it).

Posted by: Robert on September 20, 2006 2:27 PM | Permalink | Reply to this

Re: Ring On

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Back in my algebraic geometry classes we always assumed the fundametal theorem of algebra, without it things might become weird and this might just be an example of this.

Yes, real algebraic geometry is a bitch, precisely for this reason.

As for the layout side of things: I first tried to write \backslash XX as I would have done in \TeX. But this produced a bar over both X’s and not just the first one (without curly braces I would have expected \backslash bar to operate on the first token following it).

I assume you mean \bar.

One thing that may be confusing is that the tokenization algorithm in itex2MML (influenced by MathML) is slightly different than TeX. xx is one token, set in an upright font. x x is two tokens, set in an italic font. So let’s compare \bar xx, \bar x x, \bar{x x}, \overline xx, \overline x x, \overline{x x}: $\bar xx, \bar x x, \bar{x x}, \overline xx, \overline x x, \overline{x x}$

Note that, if you are viewing this on a Mac, you will not see the difference between \bar and \overline, unless you compensate for this bug

Posted by: Jacques Distler on September 20, 2006 10:46 PM | Permalink | PGP Sig | Reply to this

Regularity conditions?

In the beginning of Henneaux and Teitelboim, there is a discussion about regularity conditions. I don’t have access to the book right now, but IIRC the story goes something like this. If p = 0 defines a constraint surface, then sqrt(p) = 0 or p^2 = 0 define the same surface, so would seem to work just as well. But they don’t, and to eliminate this possibility one introduces the regularity conditions.

I have no idea if this is relevant, but if the master constraint is quadratic, there might be a problem here, like there is for p^2 = 0.

Posted by: Thomas Larsson on September 4, 2006 1:24 PM | Permalink | Reply to this

Re: Regularity conditions?

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Of course this is completely well-known, textbook-level stuff.

One thing to be emphasized is that the criterion is not a property of the particular choice of the $C_i$ , but rather of the ideal, $\mathcal{I}$ that they generate.

For a “first-class ideal”, $\{\mathcal{I},\mathcal{I}\}\subset \mathcal{I}$ , we form $R_0(M) = \{A\in R(M) | \{\mathcal{I},A\}\subset \mathcal{I}\}$ and then $R(\tilde{M}) = R_0(M)/\mathcal{I}$

We require that the ring, $R(\tilde{M})$ , thus-defined, is the ring of functions of a smooth manifold, $\tilde{M}$ .

Apparently, what Thiemann obtains, via his procedure, is not even a ring, let alone the ring of functions of a smooth manifold.

Posted by: Jacques Distler on September 4, 2006 3:08 PM | Permalink | PGP Sig | Reply to this

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