From nobody Fri Nov 10 21:13:57 2000
Path: newshost.uwo.ca!torn!sunqbc.risq.qc.ca!cpknewshub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 18 Oct 2000 23:51:57 GMT
Organization: UCR
Lines: 81
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8skd9v$pib$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com>
<8rmfr7$qqk$1@icbdnews.scs.agilent.com> <8rtcfh$1vea$1@mortar.ucr.edu>
<87zokcfsuu.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 971913117 11099 129.125.6.17 (18 Oct 2000 23:51:57 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 18 Oct 2000 23:51:57 GMT
Xref: newshost.uwo.ca sci.physics.research:12452
In article <87zokcfsuu.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>As one step in calculating , baez@newmath.UCR.EDU (John Baez)
>writes:
>> Each pair of distinct tetrahedra (i,j) determines a triangle. For each
>> triangle form the quantity
>>
>> K(x_i,x_j) = sin(A(ij) r) / (A(ij) sinh r)
>>
>> where A(ij) is the area labelling that triangle and r is the distance
>> between the points x_i and x_j in the [mass hyperboloid] H.
>>
>> Now: take all 10 of these quantities K(x_i,x_j), multiply them together,
>> and integrate the result over all the variables x_i.
>Is it possible to give any motivation for this definition?
You bet! I keep putting off replying to this question because I've
spent a couple of years thinking about this stuff and it seems like
a hopeless business to explain it nicely in a short article. So
instead of trying to write a massive opus, I'll just mutter a few
sentences and then you can ask some more questions and so on.
>Also, it
>would be great to see a description of the representationtheoretic
>way of defining this, in either the Lorentzian or Riemannian case.
>Depending on the answer, that may be motivation enough for me!
To motivate the above formula (and its Riemannian analogue) we really
need to think about representation theory and its relation to geometry.
Start at the beginning: where do functions on the mass hyperboloid
H come into the game? It's because the space of functions on the mass
hyperboloid is the direct integral of all representations of the Lorentz
group with a certain property. Now, you may not like "direct integrals" 
unless you're an analyst, you probably prefer "direct sums". For this
reason (and to avoid other stuff related to analysis) let's talk about
the Riemannian case, where we use the unit sphere S^3 instead of the mass
hyperboloid. The two cases are *very* similar, so it's good to start
with the easier one.
So: L^2(S^3) is naturally a representation of SO(4) and thus of
its double cover SU(2) x SU(2), so we can ask how it decomposes as
irreps of SU(2) x SU(2). As I'm sure you know [this is the hint for
you to ask a question], it decomposes as the direct sum of all irreps
of the form
j tensor j,
where j ranges over all irreps of SU(2).
So the reason L^2(S^3) gets into the game is that in the BarrettCrane
model, we only work with SO(4) spin networks where the edges are labelled
by irreps of the form j tensor j. These are called "simple" representations
of SO(4).
And then the question is: why do we only use simple irreps of SO(4)
to label spin network edges in the BarrettCrane model?
And the answer is: spin network edges are dual to triangles in a
triangulation of our 4manifold (spacetime). And simple irreps
describe states of a "quantum triangle" in 4 dimensions!
So now you'll ask: "what the heck is a quantum triangle?"
And the answer is (in brief): geometrically quantize the phase space of
"shapes of a triangle in Euclidean R^4", and you'll get a Hilbert space,
which we may call the space of "states of a quantum triangle".
This Hilbert space is a representation of SO(4), for obvious geometrical
reasons. And it happens to be the direct sum of all simple irreps of
SO(4), for slightly less obvious geometrical reasons.
In other words, it's L^2(S^3)!
We're not there yet, but we're getting there... let me know if any of
this makes sense so far, and ask some questions, and then I can continue
if you like.
From nobody Fri Nov 10 21:13:59 2000
Path: newshost.uwo.ca!torn!nntp.cs.ubc.ca!newsspur1.maxwell.syr.edu!news.maxwell.syr.edu!newsfeed.icl.net!diablo.netcom.net.uk!netcom.net.uk!newsfeeds.belnet.be!naxos.belnet.be!news.belnet.be!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 24 Oct 2000 11:58:34 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 73
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <878zrkjg23.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com> <8rmfr7$qqk$1@icbdnews.scs.agilent.com> <8rtcfh$1vea$1@mortar.ucr.edu> <87zokcfsuu.fsf@scratchy.dhis.org> <8skd9v$pib$1@mortar.ucr.edu>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 972388714 2890 129.125.6.17 (24 Oct 2000 11:58:34 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 24 Oct 2000 11:58:34 GMT
Xref: newshost.uwo.ca sci.physics.research:12490
baez@galaxy.ucr.edu writes:
> So: L^2(S^3) is naturally a representation of SO(4) and thus of
> its double cover SU(2) x SU(2), so we can ask how it decomposes as
> irreps of SU(2) x SU(2). As I'm sure you know [this is the hint for
> you to ask a question], it decomposes as the direct sum of all irreps
> of the form
>
> j tensor j,
>
> where j ranges over all irreps of SU(2).
I know that the irreps of a product are exactly the tensor products of
the irreps of the factors, and that the irreps of SU(2) are labelled
by nonnegative halfintegers j and have dimension 2j+1. But I don't
see why L^2(S^3) decomposes into those particular irreps. Could you
say why SU(2) x SU(2) is the double cover of SO(4)? Is it true that
an irrep j tensor k of SU(2) x SU(2) factors through SO(4) iff j + k
is an integer? (And therefore that these are all of the irreps of
SO(4)?) It's probably also related that L^2(G) contains all the
irreps with multiplicity equal to their dimension, but I don't see
how to use this.
> And then the question is: why do we only use [these] simple irreps of SO(4)
> to label spin network edges in the BarrettCrane model?
>
> And the answer is: spin network edges are dual to triangles in a
> triangulation of our 4manifold (spacetime). And simple irreps
> describe states of a "quantum triangle" in 4 dimensions!
I'm a bit confused about spin networks vs. spin foams. A spin network
is a 3d thing, and a spin foam is a 4d thing, right? And in a
triangulation of a 4manifold, an edge should be dual to a 3simplex
(tetrahedron), so I don't really follow what you said. To add to my
confusion, the recipe you described computed a number for a closed
spin foam, going from the empty spin network to itself, so no spin
network edges were labelled at all.
> So now you'll ask: "what the heck is a quantum triangle?"
What the heck is a quantum triangle? :)
> And the answer is (in brief): geometrically quantize the phase space of
> "shapes of a triangle in Euclidean R^4", and you'll get a Hilbert space,
> which we may call the space of "states of a quantum triangle".
I guess the space of shapes of a triangle in R^4 is the same as the
configuration space of three distinct points in R^4, i.e. a subspace
of R^4 x R^4 x R^4, possibly modulo the action of the symmetric group
S_3, depending on whether the vertices are labelled.
So then we look at the Hilbert space of L^2 functions on this space.
(If I understand correctly, geometric quantization starts with the
cotangent bundle viewed as a symplectic manifold, uses the symplectic
form to get a line bundle with connection, and then takes the space of
L^2 sections whose derivatives vanish in the direction of a chosen
polarization. In the special case here, this is supposed to reduce to
L^2 functions on the original manifold, right? I'm guessing that this
is because the line bundle one gets in this way is the top exterior
power of the cotangent bundle, pulled back to the cotangent bundle.
Then sections whose derivative vanish in the vertical direction must
correspond to sections of the notpulledback top exterior power.
And this is trivial for orientable manifolds.)
> This Hilbert space is a representation of SO(4), for obvious geometrical
> reasons. And it happens to be the direct sum of all simple irreps of
> SO(4), for slightly less obvious geometrical reasons.
>
> In other words, it's L^2(S^3)!
Can you explain why and/or go on?
Dan
From nobody Fri Nov 10 21:14:00 2000
Path: newshost.uwo.ca!torn!howland.erols.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 28 Oct 2000 15:55:15 GMT
Organization: UCR
Lines: 286
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8tcot4$17uh$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com> <8rmfr7$qqk$1@icbdnews.scs.agilent.com> <8rtcfh$1vea$1@mortar.ucr.edu> <87zokcfsuu.fsf@scratchy.dhis.org> <8skd9v$pib$1@mortar.ucr.edu> <878zrkjg23.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 972748515 15834 129.125.6.17 (28 Oct 2000 15:55:15 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 28 Oct 2000 15:55:15 GMT
Xref: newshost.uwo.ca sci.physics.research:12601
Dan Christensen writes:
>John Baez wrote:
>> So: L^2(S^3) is naturally a representation of SO(4) and thus of
>> its double cover SU(2) x SU(2), so we can ask how it decomposes as
>> irreps of SU(2) x SU(2). As I'm sure you know [this is the hint for
>> you to ask a question], it decomposes as the direct sum of all irreps
>> of the form
>>
>> j tensor j,
>>
>> where j ranges over all irreps of SU(2).
>I know that the irreps of a product are exactly the tensor products of
>the irreps of the factors, and that the irreps of SU(2) are labelled
>by nonnegative halfintegers j and have dimension 2j+1. But I don't
>see why L^2(S^3) decomposes into those particular irreps.
Okay, I'll explain that.
>Could you say why SU(2) x SU(2) is the double cover of SO(4)?
Yup, I'll also explain that.
>Is it true that an irrep j tensor k of SU(2) x SU(2) factors through
>SO(4) iff j + k is an integer? (And therefore that these are all of the
>irreps of SO(4)?)
Yes, this is true: when you see why SU(2) x SU(2) is a double cover of
SO(4), this will be obvious.
>It's probably also related that L^2(G) contains all the
>irreps with multiplicity equal to their dimension, but I don't see
>how to use this.
Ah, so you know this theorem by Schur! Yes, this is valid for all compact
topological groups, and it's really relevant here.
Okay: I'll start with stuff about representations of a general compact
topological group, and then focus in on SU(2). This is all really good
stuff to know about, and you can't really understand spin foams without
knowing it.
Suppose G is a compact topological group. G acts on L^2(G) by left
translations, but it also acts by right translations! Left and right
translations commute, so L^2(G) becomes a representation not only of G,
but of G x G. The really *good* way to think about Schur's theorem is
this: it tells us how to decompose L^2(G) into irreps of G x G. This is
more informative than the usual textbook way which you mention, where we
just decompose it into irreps of G.
So, how does L^2(G) decompose into irreps of G x G? Pick one unitary
irrep of G from each equivalence class of irreps, make a big list of
all these irreps, and call them R(i). Then
L^2(G) = sum_i R(i) tensor R(i)*
Here I'm using that thing you said: the irreps of a product of groups
are formed as tensor products of irreps of the factors. Thus each guy
R(i) tensor R(i)* is an irrep of G x G.
How do you prove this? The basic idea is simple. Concretely, each
irrep R(i) gives a map from G to n x n matrices where n is the dimension
of R(i). Any matrix entry thus determines a smooth complex function on G.
We thus get a map from the space of matrices to L^2(G). But the space
of matrices is really just R(i) tensor R(i)*. So we get a map from
sum_i R(i) tensor R(i)*
to L^2(G). With a little extra work  the "Schur orthogonality
relations"  you can check that in fact
L^2(G) = sum_i R(i) tensor R(i)*
In other words, they are isomorphic Hilbert spaces. But even better,
they are isomorphic as representations of G x G! This is a little
calculation: just think about what left and right translations do to
the matrix entries of a unitary irrep of G, viewed as functions on G.
As I'm sure you noticed, this marvelous fact
L^2(G) = sum_i R(i) tensor R(i)*
implies the usual claim that each irrep of G shows up in L^2(G) with
multiplicity equal to its dimension. But it's much nicer, because
it explains exactly what's going on: the multiplicity, which was a
mere number, is now accounted for by R(i)*, which is a representation
of the second copy of G, acting by right translations. Only after I
learned this did I truly love Schur's theorem!
Okay, now let's apply this to the group SU(2). The irreps of SU(2)
are listed by spins j = 0, 1/2, 1, ..., with the spinj rep having
dimension 2j+1. So we have
L^2(SU(2)) = sum_j j tensor j*
But all the irreps of SU(2) are isomorphic to their duals! This
is obvious, because the dual of an irrep is an irrep of the same
dimension, and SU(2) has only one irrep of each dimension. So we
have
L^2(SU(2)) = sum_j j tensor j
Next: how does SO(4) and S^3 get into the game? Well, remember that the
quaternions are a 4dimensional normed division algebra, so the unit
quaternions are closed under multiplication. Using the standard trick
for writing quaternions as 2 x 2 complex matrices, we see that the unit
quaternions form the group SU(2). So we have
SU(2) = S^3
Moreover, since the quaternions are a normed division algebra, left
or right multiplication by any unit quaternion acts as a rotation on
the space of quaternions. Left and right multiplications commute
(hmm, I'm experiencing deja vu!) so we get a homomorphism
SU(2) x SU(2) > SO(4)
More concretely, if you give me a pair (g,h) of unit quaternions,
I get an element of SO(4) as follows:
x > gxh^{1}
Now, I claim this map
SU(2) x SU(2) > SO(4)
is a double cover with kernel equal to (1,1) and (1,1). A little
abstract mumbojumbo, or a painful explicit calculation, shows this map
is onto. So let's just see why it's twotoone.
What's the kernel of this map? It's all the pairs (g,h)
where
gxh^{1} = x
for all quaternions x. In other words,
x^{1}gx = h
for all x. Taking x = 1 this means g = h, so we must have
x^{1}gx = g
for all x. So g must be in the center of the group SU(2), i.e.
a diagonal matrix. The only options are g = 1 and g = 1. So the
kernel is (1,1) and (1,1).
In short, what we've got here is a wonderful relation between SU(2),
rotations in 3d space, and rotations in 4d space. As you probably
already know,
SO(3) = SU(2)/{1,1}
Now we're seeing
SO(4) = SU(2) x SU(2)/{(1,1), (1,1)}
The first equation here implies that exactly *half* of the irreps
of SU(2) come from irreps of SO(3), namely the bosonic ones, where
j is an integer. These are the ones where the element 1 in SU(2)
acts as multiplication by 1. The others, the fermionic ones, are
where 1 in SU(2) acts as multiplication by 1.
The second equation here implies that exactly half of the irreps of
SU(2) x SU(2) come from irreps of SO(4), namely those irreps j tensor k
where j+k is an integer. These are the ones where the element (1,1)
in SU(2) x SU(2) acts as multiplication by 1. The others are the ones
where (1,1) acts as multiplication by 1.
Now, we've seen already that
L^2(SU(2)) = sum_j j tensor j
as reps of SU(2) x SU(2), with the two copies of SU(2) acting by
left and right translation. Now we see this in a new light: SU(2)
is the unit sphere in the quaternions, and
SO(4) = SU(2) x SU(2)/{(1,1), (1,1)}
acts on it by rotations! So we have
L^2(S^3) = sum_j j tensor j
as reps of SO(4). Note that j+j is always an integer, so everything
is working as it should. Yay!
And now for some jargon: the irreps
j tensor j
are called "simple" irreps of SO(4).
>> And then the question is: why do we only use simple irreps of SO(4)
>> to label spin network edges in the BarrettCrane model?
>>
>> And the answer is: spin network edges are dual to triangles in a
>> triangulation of our 4manifold (spacetime). And simple irreps
>> describe states of a "quantum triangle" in 4 dimensions!
>I'm a bit confused about spin networks vs. spin foams. A spin network
>is a 3d thing, and a spin foam is a 4d thing, right?
Right. I was being sloppy. Let me try to be less confusing. First let
me restate what I was saying in a way that completely leaves out any
mention of spin networks:
"And then the question is: why do we only use simple irreps of SO(4)
to label spin foam faces in the BarrettCrane model?
And the answer is: spin foam faces are dual to triangles in a
triangulation of our 4manifold (spacetime). And simple irreps
describe states of a "quantum triangle" in 4 dimensions!"
This is what I should have said. But what I said was not exactly wrong!
When we take a slice of a triangulated 4manifold we get a triangulated
3manifold. Spin foam faces in the 4manifold give spin network edges
in the 3manifold. These are now dual to the triangles in the 3manifold!
So my mistake was just that I slipped from talking about 4d spacetime
to 3d space without warning you.
>To add to my
>confusion, the recipe you described computed a number for a closed
>spin foam, going from the empty spin network to itself, so no spin
>network edges were labelled at all.
Right. I was telling you how to compute a partition function for
a 4manifold without boundary. This recipe generalizes to a recipe
for calculating transition amplitudes between spin network states
when we have a 4manifold that's a cobordism from one 3manifold to
another, and spin networks sitting in the 3manifolds. It's a very
simple generalization. But I shouldn't have slipped into talking
about it here!
>> So now you'll ask: "what the heck is a quantum triangle?"
>What the heck is a quantum triangle? :)
See, I'm a mindreader!
>> And the answer is (in brief): geometrically quantize the phase space of
>> "shapes of a triangle in Euclidean R^4", and you'll get a Hilbert space,
>> which we may call the space of "states of a quantum triangle".
>I guess the space of shapes of a triangle in R^4 is the same as the
>configuration space of three distinct points in R^4, i.e. a subspace
>of R^4 x R^4 x R^4, possibly modulo the action of the symmetric group
>S_3, depending on whether the vertices are labelled.
That's very sensible, but you overlooked the fact that I was lying
to you! To get what I called the "Hilbert space of a quantum
triangle", we actually want to take a *quotient space* of the space
of "shapes of a triangle in Euclidean R^4" before we run off and
geometrically quantize it. I didn't tell you about that wrinkle.
Also, you overlooked the fact that I want to treat the space of
shapes of a triangle (or its quotient space) as a *phase* space,
not a configuration space.
So you did something interesting, but not something that gives
L^2(S^3) = sum_j j tensor j
as the answer.
>> This Hilbert space is a representation of SO(4), for obvious geometrical
>> reasons. And it happens to be the direct sum of all simple irreps of
>> SO(4), for slightly less obvious geometrical reasons.
>>
>> In other words, it's L^2(S^3)!
>Can you explain why and/or go on?
Right now I'm pooped, and this post is too long already. So I'll have
to do that later. At least we covered a lot of good stuff about SO(3),
SO(4), and SU(2). This stuff is fundamental to a huge wad of physics,
and it's very beautiful, so it's worth doing. To talk intelligently
about Lorentzian spin foams, we'd need to cover the same ground for
SO(3,1) and SL(2,C) as well. But this is slightly more timeconsuming
since these groups are noncompact and the unitary irreps are infinite
dimensional. One must pay a bit of attention to (shudder) ANALYSIS!
From nobody Fri Nov 10 21:14:01 2000
Path: newshost.uwo.ca!torn!nntp.cs.ubc.ca!cyclone.bc.net!newshog.berkeley.edu!ucberkeley!enews.sgi.com!sdd.hp.com!ihnp4.ucsd.edu!news.ucr.edu!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 28 Oct 2000 12:46:25 0400
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 104
Approved: baez@math.ucr.edu
MessageID: <877l6sgbha.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com> <8rmfr7$qqk$1@icbdnews.scs.agilent.com> <8rtcfh$1vea$1@mortar.ucr.edu> <87zokcfsuu.fsf@scratchy.dhis.org> <8skd9v$pib$1@mortar.ucr.edu> <878zrkjg23.fsf@scratchy.dhis.org> <8tcot4$17uh$1@mortar.ucr.edu>
NNTPPostingHost: mathcln06.ucr.edu
MimeVersion: 1.0
ContentType: text/plain; charset=usascii
XTrace: mortar.ucr.edu 972854352 55482 138.23.203.171 (29 Oct 2000 21:19:12 GMT)
XComplaintsTo: usenet@mortar.ucr.edu
NNTPPostingDate: 29 Oct 2000 21:19:12 GMT
UserAgent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7
ContentLength: 4652
Status: OR
XNewsreader: trn 4.0test69 (20 September 1998)
Originator: baez@mathcln06.math.ucr.edu (John Baez)
Xref: newshost.uwo.ca sci.physics.research:12620
John Baez writes:
> Dan Christensen writes:
> >It's probably also related that L^2(G) contains all the
> >irreps with multiplicity equal to their dimension, but I don't see
> >how to use this.
> Ah, so you know this theorem by Schur! Yes, this is valid for all compact
> topological groups, and it's really relevant here.
I thought it was called the PeterWeyl theorem. In any case, John
goes on to explain a better formulation of this theorem, where L^2(G)
is considered as a representation of G x G using left and right
translation:
> So, how does L^2(G) decompose into irreps of G x G? Pick one unitary
> irrep of G from each equivalence class of irreps, make a big list of
> all these irreps, and call them R(i). Then
>
> L^2(G) = sum_i R(i) tensor R(i)*
Just to be precise: this is a Hilbert space sum, where we allow
infinite convergent sums.
> How do you prove this? The basic idea is simple. Concretely, each
> irrep R(i) gives a map from G to n x n matrices where n is the dimension
> of R(i). Any matrix entry thus determines a smooth complex function on G.
> We thus get a map from the space of matrices to L^2(G).
I don't quite follow these last two sentences. Here's another
definition of a map, and you can tell me whether what I'm saying is
correct. (Well, really, you can tell me whether I can copy things
from Adams correctly...)
We want to construct a map from R(i) tensor R(i)* to L^2(G). Well,
R(i) tensor R(i)* is the same as Hom(R(i),R(i)), the complex linear
maps from R(i) to R(i). Given such a map T, and an element g of G, I
can cook up the complex number Tr(T R(i)(g)), i.e. the trace of the
composite map R(i) > R(i) > R(i), where the first map is the
action of g and the second is the map T. As g varies, this gives
a function in L^2(G).
John goes on to define a map SU(2) x SU(2) > SO(4) using the
quaternions, explains why it's a double cover with kernel (1,1) and
(1,1), and goes on to conclude that
> L^2(S^3) = sum_j j tensor j
The irreps j tensor j are called "simple" irreps of SO(4).
John continues:
> And then the question is: why do we only use simple irreps of SO(4)
> to label spin foam faces in the BarrettCrane model?
>
> And the answer is: spin foam faces are dual to triangles in a
> triangulation of our 4manifold (spacetime). And simple irreps
> describe states of a "quantum triangle" in 4 dimensions!
>
> >What the heck is a quantum triangle? :)
>
> >John wrote:
> >> the answer is (in brief): geometrically quantize the phase space of
> >> "shapes of a triangle in Euclidean R^4", and you'll get a Hilbert space,
> >> which we may call the space of "states of a quantum triangle".
>
> Dan wrote:
> >I guess the space of shapes of a triangle in R^4 is the same as the
> >configuration space of three distinct points in R^4, i.e. a subspace
> >of R^4 x R^4 x R^4, possibly modulo the action of the symmetric group
> >S_3, depending on whether the vertices are labelled.
>
> That's very sensible, but you overlooked the fact that I was lying
> to you! To get what I called the "Hilbert space of a quantum
> triangle", we actually want to take a quotient space of the space
> of "shapes of a triangle in Euclidean R^4" before we run off and
> geometrically quantize it. I didn't tell you about that wrinkle.
>
> Also, you overlooked the fact that I want to treat the space of
> shapes of a triangle (or its quotient space) as a *phase* space, not
> a configuration space.
John's being coy and not giving up the answer, so I've had to resort
to reading some of his papers on this stuff (his survey paper and his
paper with Barrett on the quantum tetrahedron). I see that he doesn't
care about the position of the triangle, so I should quotient by the
action of R^4 by translations. But this doesn't seem to be similar to
what he's written. He considers a triangle to be specified by a
pair of vectors in R^4, modulo the relation that (v,w) = (w,v).
(That is, the pair of vectors (v,w) is thought of as a bivector
v /\ w in the second exterior power of R^4.) This seems kind of
strange to me since it suggests that the triangle whose vertices are
the origin, the tip of v and the tip of w is considered the same as
the triangle whose vertices are the origin, the tip of w and the tip
of (v), but these aren't even similar in general.
Hopefully this confused paragraph will prompt John to clarify things.
(Unless this is a side issue that he thinks isn't worth spending time
on. At some point I want to learn how to efficiently calculate the
expectation values that started this subthread!)
Dan
From nobody Fri Nov 10 21:14:02 2000
Path: newshost.uwo.ca!torn!howland.erols.net!sunqbc.risq.qc.ca!newsfeeds.belnet.be!news.belnet.be!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 30 Oct 2000 13:05:28 GMT
Organization: UCR
Lines: 81
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <200010300359.e9U3xa923006@mathcln06.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com>
<878zrkjg23.fsf@scratchy.dhis.org> <8tcot4$17uh$1@mortar.ucr.edu>
<877l6sgbha.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 972911128 8399 129.125.6.17 (30 Oct 2000 13:05:28 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 30 Oct 2000 13:05:28 GMT
Xref: newshost.uwo.ca sci.physics.research:12637
In article <877l6sgbha.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>John Baez writes:
> Dan Christensen writes:
>> >It's probably also related that L^2(G) contains all the
>> >irreps with multiplicity equal to their dimension, but I don't see
>> >how to use this.
>> Ah, so you know this theorem by Schur!
>I thought it was called the PeterWeyl theorem.
You're right, I goofed.
>In any case, John
>goes on to explain a better formulation of this theorem, where L^2(G)
>is considered as a representation of G x G using left and right
>translation:
>> So, how does L^2(G) decompose into irreps of G x G? Pick one unitary
>> irrep of G from each equivalence class of irreps, make a big list of
>> all these irreps, and call them R(i). Then
>>
>> L^2(G) = sum_i R(i) tensor R(i)*
>Just to be precise: this is a Hilbert space sum, where we allow
>infinite convergent sums.
Right.
>> How do you prove this? The basic idea is simple. Concretely, each
>> irrep R(i) gives a map from G to n x n matrices where n is the dimension
>> of R(i). Any matrix entry thus determines a smooth complex function on G.
>> We thus get a map from the space of matrices to L^2(G).
>I don't quite follow these last two sentences.
You're probably too sophisticated to understand my lowbrow lingo.
'Round these parts, us locals call a linear transformation on an
ndimensional vector space a "matrix". To us, it's just an n x n
box full of numbers. We never learnt no better. And we say a
representation R of the group G is just a doohickey that turns
each element g of G into a matrix R(g), with R(gh) = R(g)R(h).
Now if you look at one entry of this here matrix R(g), why, it's
a number that depends on g. In other words, it's a function on G.
In fact it's a smooth function  smooth as a baby's bottom!  so
it's darn sure in L^2(G) when G is compact. If we do this for all
the entries of our n x n matrix, we get a bunch of functions on G.
We get n^2 of them, in fact!
These span an n^2dimensional subspace of L^2(G), and if you think
about it mighty hard, it turns out we've got ourselves a copy of
the space of n x n matrices sitting right inside L^2(G).
>Here's another
>definition of a map, and you can tell me whether what I'm saying is
>correct.
Okay, will do.
>We want to construct a map from R(i) tensor R(i)* to L^2(G). Well,
>R(i) tensor R(i)* is the same as Hom(R(i),R(i)), the complex linear
>maps from R(i) to R(i). Given such a map T, and an element g of G, I
>can cook up the complex number Tr(T R(i)(g)), i.e. the trace of the
>composite map R(i) > R(i) > R(i), where the first map is the
>action of g and the second is the map T. As g varies, this gives
>a function in L^2(G).
Lord almighty, that's just what I was asayin'! But now it sounds
like citified talk... something a lawyer might say, way up there in
New York City.
:)
(More later.)
From nobody Fri Nov 10 21:14:02 2000
Path: newshost.uwo.ca!torn!howland.erols.net!news.maxwell.syr.edu!upp1.onvoy!msc1.onvoy!onvoy.com!news.state.mn.us!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 31 Oct 2000 17:48:48 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 39
Approved: spr@rosencrantz.stcloudstate.edu (sci.physics.research)
MessageID: <87itq9al7x.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com> <878zrkjg23.fsf@scratchy.dhis.org> <8tcot4$17uh$1@mortar.ucr.edu> <877l6sgbha.fsf@scratchy.dhis.org> <200010300359.e9U3xa923006@mathcln06.ucr.edu>
NNTPPostingHost: rosencrantz.stcloudstate.edu
ContentType: text/plain; charset=usascii
XTrace: news.state.mn.us 973014528 9177 199.17.31.144 (31 Oct 2000 17:48:48 GMT)
XComplaintsTo: usenet@news.state.mn.us
NNTPPostingDate: 31 Oct 2000 17:48:48 GMT
UserAgent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7
Originator: spr@rosencrantz.stcloudstate.edu (sci.physics.research comoderator)
Xref: newshost.uwo.ca sci.physics.research:12680
Mostly for my own sake, let me try to understand the relationship
between John's description of the map occurring in the PeterWeyl
theorem and mine. Suppose R is an ndimensional representation of a
compact Lie group G. Write R* for the dual representation Hom(R,C),
where C is the complex numbers. John described a map
J: R tensor R* > L^2(G)
as follows. First, choosing a basis, each element g of G gives an
n x n matrix R(g). The ij'th entry in this matrix is a smooth function
of g, and so gives an element of L^2(G) that we'll call f_ij. Now
R tensor R* is the same as n x n matrices, and given an n x n matrix M
we can look at the function sum_i,j M_ij f_ij. This is where we send
the matrix M under the map J.
I described a map
D: R tensor R* > L^2(G)
as follows. Given a matrix M and an element g of G, we get a complex
number Tr(M R(g)).
I want to show that these are the same. First let's see what my
definition does when M is the elementary matrix E_ij with a 1 in the
ij spot and zeros everywhere else. Then the diagonal elements of the
product M R(g) are (0, 0, ..., 0, R(g)_ij, 0, ..., 0) with R(g)_ij in
the ith spot. So the trace is R(g)_ij. In other words, for this M my
map does the same thing as John's.
A general matrix M is a sum M_ij E_ij, so D(M)(g) = sum_i,j M_ij R(g)_ij,
just like John's definition (trace is linear). So the top maps are
the same!
And when R is irreducible, this map is a monomorphism.
Ok, so how about that quantum triangle? :)
Dan
From nobody Fri Nov 10 21:14:03 2000
Path: newshost.uwo.ca!torn!utnut!newsflash.concordia.ca!sunqbc.risq.qc.ca!newsfeeds.belnet.be!naxos.belnet.be!news.belnet.be!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 31 Oct 2000 23:10:18 GMT
Organization: UCR
Lines: 131
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8tniqu$2k60$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com>
<877l6sgbha.fsf@scratchy.dhis.org>
<200010300359.e9U3xa923006@mathcln06.ucr.edu>
<87itq9al7x.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973033818 14084 129.125.6.17 (31 Oct 2000 23:10:18 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 31 Oct 2000 23:10:18 GMT
Xref: newshost.uwo.ca sci.physics.research:12699
In article <87itq9al7x.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
[Explanation of PeterWeyl theorem deleted.]
>Ok, so how about that quantum triangle? :)
Right! Sometime I really *should* finish explaining how you
use group representation theory to calculate the crucial quantity
that makes the BarrettCrane model tick: the 10j symbols. But
it'll be a lot more fun if you know a bit about the meaning of
this model. So: quantum triangles.
A triangle with labelled corners in R^n, modulo translations,
is described by a pair of vectors. E.g. if we have a triangle
whose corners are the points A,B,C:
AB
\ /
\ /
\ /
C
we can describe it mod translations by the vectors
u = BA
v = CA
Now, you might think that to quantize the triangle we should take
the space of pairs of vectors, find a nice symplectic structure
on it, and wave the magic wand of geometric quantization over this
space. Or, if not a symplectic structure, at least a Poisson
structure.
But for some reason I don't want to explain now, what really
matters in quantum gravity is not the whole triangle, but
only a certain piece of information about it: the bivector
E = u ^ v
This is an element in the 2nd exterior power of R^n. It keeps track
of the AREA of the triangle, and it keeps track of the UNIT NORMAL
VECTOR to the triangle, but nothing more about the triangle's shape.
Now, suppose we're in Euclidean R^n, Then the space of bivectors
is canonically isomorphic to so(n). Or, for that matter, so(n)*.
And so(n)* has a godgiven Poisson structure on it! To specify it,
we just need to define the Poisson bracket of any two linear functions
on so(n)*, say f and g. But these are elements of so(n)! So we can
define their Poisson bracket by
{f,g} = [f,g]
where [f,g] is the Lie bracket. This powerful idea is due to Kostant
and Kirillov.
If we geometrically quantize so(n)* with this Poisson structure, we
get the Hilbert space of "quantum bivectors in n dimensions". It
is naturally a unitary representation of so(n).
For example, if n = 3, this Hilbert space is a representation of
so(3) = su(2), and thus SU(2). And it works out to be just this:
sum_j j
The Hilbert space direct sum of all the spinj reps! All the irreps
of so(3), in short!
If n = 4, this Hilbert space is a representation of so(4) =
su(2) + su(2), and thus SU(2) x SU(2). And it works out to be just this:
sum_{j,k} j tensor k
The Hilbert space direct sum of all the irreps of so(4)!
This is how it goes in every dimension.
But what about triangles?
In the case n = 3, any bivector can be gotten from a triangle in
the manner described above, so I call
sum_j j
the "Hilbert space of the quantum triangle in 3 dimensions". It
turns out to be fundamental to 3d Riemannian quantum gravity.
In the case n = 4, not any bivector can be gotten from a triangle
in the manner described above. So we really should not geometrically
quantize all of so(4)* if we're interested in quantum triangles in
4 dimensions. We should really quantize only the subspace of so(4)*
that comes from triangles. If we do this, we get the smaller Hilbert
space
sum_j j tensor j
which I call "the Hilbert space of a quantum triangle in 4 dimensions".
It's fundamental to the BarrettCrane model of 4d Riemannian quantum
gravity.
It would be fun to explain where the j = k constraint comes from.
It has a nice geometrical meaning in terms of bivectors and triangles.
But I won't explain it just now. You can either ask me and/or
read my paper with Barrett on the quantum tetrahedron (grqc/9903060).
Anyway, this space
sum_j j tensor j
is none other than the space we were talking about last time!
It's L^2(SU(2))), or in other words, L^2(S^3)! It's the direct
sum of all the simple representations of so(4).
This is ultimately the reason why in the Riemannian 4d BarrettCrane
model, we label triangles in a triangulation of spacetime by simple
representations of so(4).
In the Lorentzian case everything works similarly but with some obvious
modifications: so(4) > so(3,1), L^2(S^3) > L^2(H^3), and so on, where
H^3 = {t^2  x^2  y^2  z^2 = 1}
is hyperbolic 3space. Instead of L^2(H^3) being the direct sum of all
the simple representations of so(3,1), it's a direct integral: there is
one for every nonnegative real number. So instead of sums we get
integrals all over the place. It's not obvious that they all converge.
Analysis! Yuck!
And now I need to go talk about this at the functional analysis
seminar....
From nobody Fri Nov 10 21:14:03 2000
Path: newshost.uwo.ca!torn!howland.erols.net!europa.netcrusader.net!164.67.42.145!awabi.library.ucla.edu!138.23.226.128!news.ucr.edu!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 01 Nov 2000 00:04:22 0500
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 89
Approved: baez@math.ucr.edu
MessageID: <87og00471l.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com> <877l6sgbha.fsf@scratchy.dhis.org> <200010300359.e9U3xa923006@mathcln06.ucr.edu> <87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu>
NNTPPostingHost: mathcln06.ucr.edu
MimeVersion: 1.0
ContentType: text/plain; charset=usascii
XTrace: mortar.ucr.edu 973130408 69093 138.23.203.171 (2 Nov 2000 02:00:08 GMT)
XComplaintsTo: usenet@mortar.ucr.edu
NNTPPostingDate: 2 Nov 2000 02:00:08 GMT
UserAgent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7
ContentLength: 3632
Status: OR
XNewsreader: trn 4.0test69 (20 September 1998)
Originator: baez@mathcln06.math.ucr.edu (John Baez)
Xref: newshost.uwo.ca sci.physics.research:12731
baez@galaxy.ucr.edu (John Baez) writes:
> A triangle with labelled corners in R^n, modulo translations,
> is described by a pair of vectors. E.g. if we have a triangle
> whose corners are the points A,B,C, [...]
> we can describe it mod translations by the vectors
>
> u = BA
> v = CA
>
> [...] for some reason I don't want to explain now, what really
> matters in quantum gravity is not the whole triangle, but
> only a certain piece of information about it: the bivector
>
> E = u ^ v
I'll take this for granted now, although I may bug you about it in
the future!
> This is an element in the 2nd exterior power of R^n. It keeps track
> of the AREA of the triangle, and it keeps track of the UNIT NORMAL
> VECTOR to the triangle, but nothing more about the triangle's shape.
I guess when n is larger than 3, what is really remembered is the area
of the triangle, the plane it lies in, and an orientation of that
plane. (There is no unit normal to a plane in dimensions 4 and up.)
As you say below, not every bivector comes from such a pair of
vectors. The space of planes in R^n has dimension (n1)+(n2)1 =
2n4, I think, and so the space of planes plus an area has dimension
2n4+1 = 2n3. The space of bivectors has dimension n(n1)/2, which
is larger than 2n3 when n > 3.
> Now, suppose we're in Euclidean R^n, Then the space of bivectors
> is canonically isomorphic to so(n). Or, for that matter, so(n)*.
Your paper with Barrett gives an identification: a bivector u ^ v is
sent to the linear map so(n) > R which sends the skewadjoint real
n x n matrix E to the number u' E v, where u' is the transpose of u.
(I'm thinking of u and v as column vectors.) It's easy to see that
the basis element e_i ^ e_j, i < j, is sent to the linear function
which sends E to its ij'th entry E_ij, and these linear functions are
a basis for so(n)*.
I guess this must be measuring the "component" of an infinitesimal
rotation in the plane spanned by u and v (scaled by the area of the
triangle u and v determine).
> And so(n)* has a godgiven Poisson structure on it! To specify it,
> we just need to define the Poisson bracket of any two linear functions
> on so(n)*, say f and g. But these are elements of so(n)! So we can
> define their Poisson bracket by
>
> {f,g} = [f,g]
>
> where [f,g] is the Lie bracket. This powerful idea is due to Kostant
> and Kirillov.
Just to make sure I understand: you specified the Poisson bracket
on linear functions; then you extend it to polynomials using the
Leibnitz rule; then you extend it to all smooth functions by
approximating them by polynomials.
> If we geometrically quantize so(n)* with this Poisson structure, we
> get the Hilbert space of "quantum bivectors in n dimensions". It
> is naturally a unitary representation of so(n).
I vaguely know about quantizing a symplectic manifold, but I don't
know how you quantize a manifold with just a Poisson structure.
However, I don't want to get further away from the goal of this
discussion, so maybe you could just say one or two sentences about
this, or give a reference? Then maybe I can try to figure out
the later parts of your post, such as:
> For example, if n = 3, this Hilbert space is a representation of
> so(3) = su(2), and thus SU(2). And it works out to be just this:
>
> sum_j j
...
> This is ultimately the reason why in the Riemannian 4d BarrettCrane
> model, we label triangles in a triangulation of spacetime by simple
> representations of so(4).
Ok, it's good to have this explained. Now, how can we use this stuff
to get another description of the 10j symbols?
Dan
From nobody Fri Nov 10 21:14:04 2000
Path: newshost.uwo.ca!torn!canopus.cc.umanitoba.ca!newsin.mts.net!cyclone.bc.net!news.maxwell.syr.edu!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 3 Nov 2000 17:03:54 GMT
Organization: UCR
Lines: 158
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8tqk7p$24jd$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu>
<87og00471l.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973271034 22326 129.125.6.17 (3 Nov 2000 17:03:54 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 3 Nov 2000 17:03:54 GMT
Xref: newshost.uwo.ca sci.physics.research:12752
In article <87og00471l.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>baez@galaxy.ucr.edu (John Baez) writes:
>> [...] for some reason I don't want to explain now, what really
>> matters in quantum gravity is not the whole triangle, but
>> only a certain piece of information about it: the bivector
>>
>> E = u ^ v
>>
>> [where u and v are the edge vectors]
>I'll take this for granted now, although I may bug you about it in
>the future!
Heck, I'll say a bit right now: the bivector E here is closely
related to the E field that shows up in the Plebanski
formulation of general relativity. In "week159", I wrote:
........................................................................
Recall that in Plebanski's formalism, we start with:
a) a Lorentz connection A, which can locally be thought of as a
1form taking values in the Lie algebra of the Lorentz group,
and:
b) a field E, which can locally be thought of as a 2form valued
in the Lie algebra of the Lorentz group.
We get a topological field theory by using the Lagrangian
tr(E ^ F)
where F is the curvature of A. The equations of motion say that both
the curvature of A and the exterior covariant derivative of E vanish.
All solutions of these equations are locally gaugeequivalent, so there
are no local degrees of freedom  that's what I mean by saying we get a
topological field theory.
But if we impose the constraint that
E = e ^ e
where e is a "cotetrad"  which locally amounts to a 1form taking
values in R^4  we get the equations of general relativity! We can
impose this constraint by throwing an extra term into the Lagrangian,
involving an extra "Lagrange multiplier" field. The sole purpose of
this extra field is to ensure that when we compute the variation of the
action with respect to it, we get zero iff E = e ^ e.
...........................................................................
This Plebanski stuff also works in the Riemannian case, and the
constraint E = e ^ e is closely related to the constraint that
the bivector associated to any triangle be the wedge product of
two vectors.
Okay, back to our bivector associated to a triangle....
>> This is an element in the 2nd exterior power of R^n. It keeps track
>> of the AREA of the triangle, and it keeps track of the UNIT NORMAL
>> VECTOR to the triangle, but nothing more about the triangle's shape.
>I guess when n is larger than 3, what is really remembered is the area
>of the triangle, the plane it lies in, and an orientation of that
>plane. (There is no unit normal to a plane in dimensions 4 and up.)
Yes, you're exactly right  where by "the plane it lies in" you mean a
plane mod translation, to be painfully precise!
I got lazy and tried to state this stuff more rapidly, and I
screwed up, giving a description that only works in 3d. Your
way works in any dimension.
>As you say below, not every bivector comes from such a pair of
>vectors. The space of planes in R^n has dimension (n1)+(n2)1 =
>2n4, I think, and so the space of planes plus an area has dimension
>2n4+1 = 2n3. The space of bivectors has dimension n(n1)/2, which
>is larger than 2n3 when n > 3.
Right. The bivectors that are wedge products of pairs of vectors
are called "simple" or "decomposable", and there are various nice
characterizations for them, especially in 4 dimensions (the relevant
case here).
>> Now, suppose we're in Euclidean R^n, Then the space of bivectors
>> is canonically isomorphic to so(n). Or, for that matter, so(n)*.
>Your paper with Barrett gives an identification: a bivector u ^ v is
>sent to the linear map so(n) > R which sends the skewadjoint real
>n x n matrix E to the number u' E v, where u' is the transpose of u.
>(I'm thinking of u and v as column vectors.) It's easy to see that
>the basis element e_i ^ e_j, i < j, is sent to the linear function
>which sends E to its ij'th entry E_ij, and these linear functions are
>a basis for so(n)*.
>
>I guess this must be measuring the "component" of an infinitesimal
>rotation in the plane spanned by u and v (scaled by the area of the
>triangle u and v determine).
Right! A crucial part of the geometry here is the map from
triangles to infinitesimal rotations.
>> And so(n)* has a godgiven Poisson structure on it! To specify it,
>> we just need to define the Poisson bracket of any two linear functions
>> on so(n)*, say f and g. But these are elements of so(n)! So we can
>> define their Poisson bracket by
>>
>> {f,g} = [f,g]
>>
>> where [f,g] is the Lie bracket. This powerful idea is due to Kostant
>> and Kirillov.
>Just to make sure I understand: you specified the Poisson bracket
>on linear functions; then you extend it to polynomials using the
>Leibnitz rule; then you extend it to all smooth functions by
>approximating them by polynomials.
Got it.
>> If we geometrically quantize so(n)* with this Poisson structure, we
>> get the Hilbert space of "quantum bivectors in n dimensions". It
>> is naturally a unitary representation of so(n).
>I vaguely know about quantizing a symplectic manifold, but I don't
>know how you quantize a manifold with just a Poisson structure.
Any manifold with a Poisson structure is foliated by symplectic
submanifolds ("leaves"), and you just quantize each leaf and take
the direct sum of all the resulting Hilbert spaces. For example,
if our Poisson manifold is so(3)*, the symplectic leaves are the
concentric spheres centered at the origin. The associated Hilbert
spaces are 0dimensional except for the spheres of integer or
halfinteger radius. The sphere of radius j gives the spinj
rep of SU(2) when we quantize it. So when we quantize so(3)*,
we get:
sum_j j
This is just a fancyass way of talking about the quantum mechanics
of angular momentum, I hope you see. In classical mechanics, angular
momentum in 3d can be any vector in so(3)*  or R^3, to the layman.
In quantum mechanics it gets quantized like this.
>> This is ultimately the reason why in the Riemannian 4d BarrettCrane
>> model, we label triangles in a triangulation of spacetime by simple
>> representations of so(4).
>Ok, it's good to have this explained. Now, how can we use this stuff
>to get another description of the 10j symbols?
Yes, but not right now. Right now I'm worn out from this post!
So read this stuff, don't ask any more questions, and then I'll tell
about the 10j symbols. :)
From nobody Fri Nov 10 21:14:05 2000
Path: newshost.uwo.ca!torn!howland.erols.net!EU.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 3 Nov 2000 17:04:29 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 14
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <87u29rc4xe.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com>
<877l6sgbha.fsf@scratchy.dhis.org>
<200010300359.e9U3xa923006@mathcln06.ucr.edu>
<87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu>
<87og00471l.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973271069 22332 129.125.6.17 (3 Nov 2000 17:04:29 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 3 Nov 2000 17:04:29 GMT
Xref: newshost.uwo.ca sci.physics.research:12744
John said that in quantum gravity, what matters about a triangle is
its area, the plane it lies in, and an orientation of that plane.
Could this be because a tetrahedron is specified uniquely (up to
translation) by the areas of its four faces, the planes they lie in,
and their orientations? And since we only use triangles to build
higher dimensional structures, this is all we need to remember about
triangles?
(Of course, the data listed above *over*specifies the tetrahedron, so
there is a constraint as well.)
Dan
From nobody Fri Nov 10 21:14:05 2000
Path: newshost.uwo.ca!torn!howland.erols.net!EU.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 4 Nov 2000 10:47:15 GMT
Organization: UCR
Lines: 42
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8tuvf8$30q4$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com> <8tniqu$2k60$1@mortar.ucr.edu> <87og00471l.fsf@scratchy.dhis.org> <87u29rc4xe.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973334835 13225 129.125.6.17 (4 Nov 2000 10:47:15 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 4 Nov 2000 10:47:15 GMT
Xref: newshost.uwo.ca sci.physics.research:12766
In article <87u29rc4xe.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>John said that in quantum gravity, what matters about a triangle is
>its area, the plane it lies in, and an orientation of that plane.
>
>Could this be because a tetrahedron is specified uniquely (up to
>translation) by the areas of its four faces, the planes they lie in,
>and their orientations? And since we only use triangles to build
>higher dimensional structures, this is all we need to remember about
>triangles?
Yes! This is a good explanation! I feel kind of silly for not having
given it myself. It's actually no more complicated than this, once one
decides one is going to build space using tetrahedra and spacetime using
4simplexes, and describe the geometry of all this stuff by specifying
information about the triangles.
My longwinded halfexplanation was a feeble attempt to sketch how this
particular decision follows naturally from the Plebanski formulation of
general relativity. That's very important if one is trying to convince
oneself of the merits of the BarrettCrane model. One wants to understand
how quantum gravity can mimic classical general relativity at large length
scales. So one needs to understand as precisely as possible the relation
between:
1) classical Riemannian (or Lorentzian) geometry on a smooth manifold,
2) the geometry of a triangulated manifold with triangles labelled by
areas, and
3) the "quantum geometry" of a triangulated manifold with the triangles
labelled by simple reps of SO(4) (or SO(3,1)).
However, this is a long story and I didn't get very far on it. If
you're happy with a simplicial approach to quantum gravity, we can skip
this stuff for now and focus on how you'd actually do computations in
the BarrettCrane model. In particular, how to compute the 10j symbols!
From nobody Fri Nov 10 21:14:06 2000
Path: newshost.uwo.ca!torn!canopus.cc.umanitoba.ca!newsin.mts.net!cyclone.bc.net!newsfeed.stanford.edu!xfer10.netnews.com!netnews.com!newsfeed.skycache.com!Cidera!cpknewshub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 3 Nov 2000 17:05:12 GMT
Organization: UCR
Lines: 139
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8tsc23$an0$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu>
<87og00471l.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973271112 22447 129.125.6.17 (3 Nov 2000 17:05:12 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 3 Nov 2000 17:05:12 GMT
Xref: newshost.uwo.ca sci.physics.research:12763
In article <87og00471l.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>Now, how can we use this stuff to get another description of
>the 10j symbols?
Okay. Let's talk about the 10j symbols. As usual, I'll do
the simpler Riemannian case, where the relevant group is SO(4).
I've already given an expression for the 10j symbols in terms
of an integral over 5 variables taking values in the 3sphere.
Now I'll give a description in terms of representations of SO(4).
We've already seen how to decompose the space of functions on
the 3sphere into irreps of SO(4). And we've seen that the irreps
which appear are precisely the "simple" ones  those that are of
the form (j tensor j) if we think of them as reps of the double
cover SU(2) x SU(2). More precisely:
L^2(S^3) = sum_j (j tensor j)
Given this, it shouldn't be surprising that we can rewrite integrals
over the 3sphere in terms of sums over these reps. But I won't
give the proof that the integral I wrote down equals the sum I'm
going to write down... and in fact, I'll just sketch the whole
story, and fill in details later as you request them.
1) I assume you know how a graph with edges labelled by (finite
dimensional, unitary) reps of a group and vertices labelled by
intertwining operators determines a number.
2) To evaluate the 10j symbols, we draw a certain graph with edges
labelled by simple reps of SO(4) and vertices labelled by intertwining
operators that I'll explain in part 3). This graph is the complete
graph on 5 vertices  i.e., the 1skeleton of a 4simplex. It
has 10 edges, and labelling these edges with simple reps of SO(4)
is the same as labelling them with spins.
We then use recipe 1) to get a number  this is the 10j symbol!
3) The only missing ingredient is the intertwining operators.
Note that the vertices in the above graph are 4valent, so I need
to describe to you intertwining operators
F: (j tensor j) tensor (k tensor k) > (m tensor m) tensor (n tensor n)
where j,k,m,n are spins and (j tensor j), ..., (n tensor n) are
the corresponding intertwining operators. Here that I've chosen
an orientation for the edges of the 4simplex such that each vertex
has two edges going in and two edges going out.
These interwtining operators are called the "BarrettCrane vertices"
and you can read about them in the papers listed below, as well as
my big fat review article and the original BarrettCrane paper.
But let me briefly summarize them.
a) First you need to learn about these godgiven morphisms
in the category of reps of SU(2):
f: j tensor k > l
where j,k,l label reps of SU(2). These have a nice simple
description which lends itself to computer calculation, and
everything else is based on them.
b) Using these, we get for any spin l a morphism in the category
of reps of SU(2):
f(l): j tensor k > m tensor n
which is simply the composite
f f*
j tensor k > l > m tensor n
(Note that I'm using f to stand for the godgiven intertwiner
from j tensor k to l, regardless of the value of j,k, and l.)
c) Finally, we have
F = sum_l (2l+1) f(l) tensor f(l)
where (2l+1) is of course the dimension of the spinl rep.
To parse this expression, note that
f(l): j tensor k > m tensor n
so
f(l) tensor f(l): j tensor k tensor j tensor k >
m tensor n tensor m tensor n
but we can rearrange the order of the tensor products to think
of it as an operator from
(j tensor j) tensor (k tensor k)
to
(m tensor m) tensor (n tensor n)
which is the sort of thing F is supposed to be.
So, that's it! I should emphasize that unless you really hate
pictures, all of this stuff makes more sense if you use diagrammatic
notation for all the morphisms (=intertwining operators), following
the techniques outlined in my quantum gravity course:
http://math.ucr.edu/home/baez/qg.html
or my big fat spin foam review article:
http://math.ucr.edu/home/baez/foam2.ps
This diagrammatic way of working with intertwining operators lets
you see very clearly how the discrete "quantum geometry" of
spacetime arises naturally from group representation theory. In
particular, the 10j symbol arises from taking the 4simplex with
triangles labelled by areas (spins) and interpreting it as an
intertwining operator from the trivial rep to itself  i.e., a
complex number! What I've discussed above are the basic "building
blocks" of this theory  the trivalent vertex for SU(2) representation
theory, and the 4valent vertex for simple reps of SO(4).
Here is a good book on diagrammatic methods for SU(2) representation
theory:
The classical and quantum 6jsymbols / by J. Scott Carter, Daniel E. Flath,
and Masahico Saito. Princeton, N.J. : Princeton University Press, 1995.
and here are some papers on the BarrettCrane vertex:
David N. Yetter, Generalized BarrettCrane vertices and invariants of
embedded graphs, preprint available as math.QA/9801131.
John W. Barrett, The classical evaluation of relativistic spin networks,
preprint available at math.QA/9803063.
From nobody Fri Nov 10 21:14:07 2000
Path: newshost.uwo.ca!torn!howland.erols.net!newspharm.inet.tele.dk.MISMATCH!news.tele.dk!193.190.198.17!newsfeeds.belnet.be!naxos.belnet.be!news.belnet.be!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 6 Nov 2000 10:42:55 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 38
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <87d7g9r3oj.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu>
<87og00471l.fsf@scratchy.dhis.org> <8tsc23$an0$1@mortar.ucr.edu>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 973507375 5086 129.125.6.17 (6 Nov 2000 10:42:55 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 6 Nov 2000 10:42:55 GMT
Xref: newshost.uwo.ca sci.physics.research:12780
baez@galaxy.ucr.edu writes:
> 1) I assume you know how a graph with edges labelled by (finite
> dimensional, unitary) reps of a group and vertices labelled by
> intertwining operators determines a number.
Yes, I think so. I draw the graph on a piece of paper in a generic
way. Since the tensor product is symmetric monoidal, it doesn't
matter which edge goes under when two edges cross. Then I start at
the top of the page, interpreting caps, cups and crossings as you are
describing in your course [http://math.ucr.edu/home/baez/qg.html (Both
sets of notes are wonderful, by the way. Thanks Toby and Miguel!)]
and using the intertwining operators when I come to vertices.
This defines a linear map from the complex numbers to itself;
in other words, a complex number!
> a) First you need to learn about these godgiven morphisms
> in the category of reps of SU(2):
>
> f: j tensor k > l
>
> where j,k,l label reps of SU(2). These have a nice simple
> description which lends itself to computer calculation, and
> everything else is based on them.
I've heard rumours that j tensor k is a sum of the irreps j+k, j+k1,
..., jk+1, jk, where the bars denote absolute value. In
examples, I kind of see this: if you think of j as the natural action
of SU(2) on x^2j, x^{2j1} y, ..., y^2j, then in j tensor k I can at
least see a subspace that looks like j+k, by taking various weighted
symmetrizations of the variables x, x', y, y'. But I should
understand this better. Can you say more, or give me a reference?
All my books on representation theory seem to spend all their time on
theory, and none on this example!
I'll take a look at the book by Carter, Flath and Saito for starters.
Dan
From nobody Fri Nov 10 21:14:07 2000
Path: newshost.uwo.ca!torn!nntp.cs.ubc.ca!cyclone.bc.net!awabi.library.ucla.edu!138.23.226.128!news.ucr.edu!notformail
From: baez@galaxy.ucr.edu
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 8 Nov 2000 22:16:08 GMT
Organization: UCR
Lines: 122
Approved: baez@math.ucr.edu
MessageID: <8ucjb8$2u29$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com> <87og00471l.fsf@scratchy.dhis.org> <8tsc23$an0$1@mortar.ucr.edu> <87d7g9r3oj.fsf@scratchy.dhis.org>
NNTPPostingHost: mathcln06.ucr.edu
XTrace: mortar.ucr.edu 973721768 96329 138.23.203.171 (8 Nov 2000 22:16:08 GMT)
XComplaintsTo: usenet@mortar.ucr.edu
NNTPPostingDate: 8 Nov 2000 22:16:08 GMT
XNewsreader: trn 4.0test69 (20 September 1998)
Originator: baez@mathcln06.math.ucr.edu (John Baez)
Xref: newshost.uwo.ca sci.physics.research:12848
In article <87d7g9r3oj.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>baez@galaxy.ucr.edu writes:
>> 1) I assume you know how a graph with edges labelled by (finite
>> dimensional, unitary) reps of a group and vertices labelled by
>> intertwining operators determines a number.
>Yes, I think so. I draw the graph on a piece of paper in a generic
>way. Since the tensor product is symmetric monoidal, it doesn't
>matter which edge goes under when two edges cross.
Right! Great! Nothing like categorytheoretic literacy to speed
communication.
>Then I start at
>the top of the page, interpreting caps, cups and crossings as you are
>describing in your course [http://math.ucr.edu/home/baez/qg.html (Both
>sets of notes are wonderful, by the way. Thanks Toby and Miguel!)]
>and using the intertwining operators when I come to vertices.
Yes, thank Toby and Miguel  and hope that they keep producing these
notes! Toby says he has posted notes for the 3rd week, but they seem to
have been swallowed by a black hole. I'm trying to get him to email me
a copy....
>This defines a linear map from the complex numbers to itself;
>in other words, a complex number!
Right!
>> a) First you need to learn about these godgiven morphisms
>> in the category of reps of SU(2):
>>
>> f: j tensor k > l
>>
>> where j,k,l label reps of SU(2). These have a nice simple
>> description which lends itself to computer calculation, and
>> everything else is based on them.
>I've heard rumours that j tensor k is a sum of the irreps j+k, j+k1,
>..., jk+1, jk, where the bars denote absolute value.
The rumours are true. So we get an intertwiner f: j tensor k > l
that's unique up to normalization whenever l is a number of the form
j+k, j+k1, ..., jk+1, jk. Of course, to do serious calculations,
we need to know this intertwiner explicitly, and we need to pick a
normalization.
>In
>examples, I kind of see this: if you think of j as the natural action
>of SU(2) on x^2j, x^{2j1} y, ..., y^2j, then in j tensor k I can at
>least see a subspace that looks like j+k, by taking various weighted
>symmetrizations of the variables x, x', y, y'. But I should
>understand this better. Can you say more, or give me a reference?
I can say more, but now that you've said you're going to look at
this book....
>I'll take a look at the book by Carter, Flath and Saito for starters.
... you've reduced my motivation. This book explains these
intertwiners in painstaking detail.
But looking over it carefully, I see that it's not quite as
lucidly explained as I would like. I am going to cover this
stuff in my quantum gravity course  in fact, I started this
week  so you can either:
1) read Carter, Flath and Saito,
2) wait for the course notes, and/or
3) get me to start telling you stuff now.
In case you pick option 3), I'll start with this. We can think
of i times the standard volume form on V = C^2 as a skewsymmetric
map
a: V tensor V > C
In diagrammatic notation we draw this as a kind of "cup":
\ /
\/
We can then introduce a unique "cap":
/\
/ \
i.e. a map
b: C > V tensor V
such that the following equations hold:
/\
/ \ = 2 "the supervector space for a spin1/2
\ / particle has superdimension equal to 2"
\/
\ /   \ /
\ /   \/
/ =   + "the binor identity"
/ \   /\
/ \   / \
These "cap" and "cup" operators are the building blocks for
the operators f: j tensor k > l. You might wish to work out
their explicit matrix forms, for the purposes of computer calculation.
If you give up, you can cheat and look it up the answer in Carter and
Flath.
Starting with these, we can define the operators f: j tensor k > l
in a cute diagrammatic way...
Puzzle: what the heck is so important about "i times the standard
volume form on C^2"?
Hint: the factor of i is not really important  it's just a convention.
The question is, what's so great about the volume form on C^2?
From nobody Fri Nov 10 21:14:08 2000
Path: newshost.uwo.ca!torn!howland.erols.net!portc.blue.aol.com.MISMATCH!portc01.blue.aol.com!news.maxwell.syr.edu!awabi.library.ucla.edu!138.23.226.128!news.ucr.edu!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 09 Nov 2000 02:48:31 0500
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 105
Approved: baez@math.ucr.edu
MessageID: <874s1hbn74.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com> <87og00471l.fsf@scratchy.dhis.org> <8tsc23$an0$1@mortar.ucr.edu> <87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu>
NNTPPostingHost: mathcln06.ucr.edu
MimeVersion: 1.0
ContentType: text/plain; charset=usascii
XTrace: mortar.ucr.edu 973797774 50207 138.23.203.171 (9 Nov 2000 19:22:54 GMT)
XComplaintsTo: usenet@mortar.ucr.edu
NNTPPostingDate: 9 Nov 2000 19:22:54 GMT
UserAgent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7
ContentLength: 3386
Status: OR
XNewsreader: trn 4.0test69 (20 September 1998)
Originator: baez@mathcln06.math.ucr.edu (John Baez)
Xref: newshost.uwo.ca sci.physics.research:12901
baez@galaxy.ucr.edu writes:
> Dan Christensen wrote:
> >baez@galaxy.ucr.edu writes:
> >> a) First you need to learn about these godgiven morphisms
> >> in the category of reps of SU(2):
> >>
> >> f: j tensor k > l
> >>
> >> where j,k,l label reps of SU(2). These have a nice simple
> >> description which lends itself to computer calculation, and
> >> everything else is based on them.
> >I've heard rumours that j tensor k is a sum of the irreps j+k, j+k1,
> >..., jk+1, jk, where the bars denote absolute value.
> The rumours are true. So we get an intertwiner f: j tensor k > l
> that's unique up to normalization whenever l is a number of the form
> j+k, j+k1, ..., jk+1, jk. Of course, to do serious calculations,
> we need to know this intertwiner explicitly, and we need to pick a
> normalization.
>
> I can say more, but now that you've said you're going to look at
> this book....
> >I'll take a look at the book by Carter, Flath and Saito for starters.
Unfortunately, my library doesn't have this book! It will hopefully
come soon by interlibrary loan. But I'm impatient, and you are
usually pretty patient :), so if it's ok with you I'll keep bugging
you about this.
> 3) get me to start telling you stuff now.
>
> In case you pick option 3), I'll start with this. We can think
> of i times the standard volume form on V = C^2 as a skewsymmetric
> map
>
> a: V tensor V > C
Ok, so this sends (u,v) tensor (w,z) to i(uzvw), where u, v, w and z
are just complex numbers. Let's think about this in coordinates.
Let e1 = (1,0) and e2 = (0,1) be the standard basis of V. Then
a basis of V tensor V is e1 tensor e1, e1 tensor e2, e2 tensor e1
and e2 tensor e2. The map a sends the sum V^ij ei tensor ej to
i V^12  i V^21 = a_ij V^ij, where a_ij is the matrix
0 i
i 0 ,
the negative of one of the Pauli matrices.
> In diagrammatic notation we draw this as a kind of "cup":
>
> \ /
> \/
>
> We can then introduce a unique "cap":
>
> /\
> / \
>
> i.e. a map b: C > V tensor V such that the following equations
> hold:
>
> /\
> / \ = 2 "the supervector space for a spin1/2
> \ / particle has superdimension equal to 2"
> \/
>
> \ /   \ /
> \ /   \/
> / =   + "the binor identity"
> / \   /\
> / \   / \
The cap has to send the complex number 1 to
i e1 tensor e2  i e2 tensor e1. That is, to a^ij ei tensor ej,
where a^ij is the same matrix that showed up above.
> Starting with these, we can define the operators f: j tensor k > l
> in a cute diagrammatic way...
That's good, because I like cute diagrammatic ways.
> Puzzle: what the heck is so important about "i times the standard
> volume form on C^2"?
>
> Hint: the factor of i is not really important  it's just a convention.
> The question is, what's so great about the volume form on C^2?
It's an intertwiner. So its kernel is an invariant subspace, and we
find that 1/2 tensor 1/2 = 1 + 0! (where these numbers denote irreps
of SU(2)). That's pretty great!
(Similarly, b is an intertwiner, and its image is the spin0 rep,
i.e. the trivial rep.)
So how do we handle higher spins?
Dan
From nobody Fri Nov 10 21:14:09 2000
Path: newshost.uwo.ca!torn!howland.erols.net!fuberlin.de!uniberlin.de!rosencrantz.stcloudstate.EDU!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 10 Nov 2000 23:33:49 GMT
Organization: UCR
Lines: 182
Approved: spr@rosencrantz.stcloudstate.edu (sci.physics.research)
MessageID: <8ufdhs$1qf9$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com> <87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu> <874s1hbn74.fsf@scratchy.dhis.org>
NNTPPostingHost: rosencrantz.stcloudstate.edu (199.17.31.144)
XTrace: fuberlin.de 973899229 1925065 199.17.31.144 (16 [28113])
XNewsreader: trn 4.0test72 (19 April 1999)
Originator: ted@rosencrantz.stcloudstate.edu ()
Xref: newshost.uwo.ca sci.physics.research:12935
In article <874s1hbn74.fsf@scratchy.dhis.org>,
Dan Christensen wrote:
>baez@galaxy.ucr.edu writes:
>> We can think of i times the standard volume form on V = C^2
>> as a skewsymmetric map
>>
>> a: V tensor V > C
>>
>> In diagrammatic notation we draw this as a kind of "cup":
>>
>> \ /
>> \/
>Ok, so this sends (u,v) tensor (w,z) to i(uzvw), where u, v, w and z
>are just complex numbers. Let's think about this in coordinates.
>Let e1 = (1,0) and e2 = (0,1) be the standard basis of V. Then
>a basis of V tensor V is e1 tensor e1, e1 tensor e2, e2 tensor e1
>and e2 tensor e2. The map a sends the sum V^ij ei tensor ej to
>i V^12  i V^21 = a_ij V^ij, where a_ij is the matrix
>
> 0 i
>
> i 0 ,
Right.
>[which is] the negative of one of the Pauli matrices.
The negative? Yuck! Luckily this won't affect the final answers:
we could equally well work with i times the standard volume form,
or simply the standard volume form itself. Ultimately, all that
will matter is these identities:
/\
/ \ = 2
\ /
\/
\ /   \ /
\ /   \/
/ =   +
/ \   /\
/ \   / \
and
\ /  
\ /  
/ =   
/ \  
/ \  
\ / \ /
\ / \ /
V V
You may or may not recall Robert Helling griping about physicists
who "raise and lower indices with i times the epsilon tensor". Well,
that's what we're about to do! We'll identify V with its dual, with
the help of the nondegenerate pairing
a: V tensor V > C
Written out as a matrix, this is
0 i
i 0
which is i times the infamous "epsilon tensor"
0 1
1 0
But why are we using i times the epsilon tensor instead of the
epsilon tensor itself? Simple: because that's what Carter, Flath
and Saito did! I don't want to bother with changing all their
formulas to get rid of that damn i  if I did, I would probably
make lots of stupid errors. It's easier to just follow their conventions.
But why did *they* put this i in the formulas? Probably because
that's what Penrose did!
And why did *he* do it? God knows. Probably just to piss off
Robert Helling.
But anyway, this annoying factor of i won't really matter once we
start doing calculations diagrammatically!
Okay... back to our map
a: V tensor V > C
As I said:
>> In diagrammatic notation we draw this as a kind of "cup":
>>
>> \ /
>> \/
>>
>> We can then introduce a unique "cap":
>>
>> /\
>> / \
>>
>> i.e. a map b: C > V tensor V such that the following equations
>> hold:
>>
>> /\
>> / \ = 2 "the supervector space for a spin1/2
>> \ / particle has superdimension equal to 2"
>> \/
>>
>> \ /   \ /
>> \ /   \/
>> / =   + "the binor identity"
>> / \   /\
>> / \   / \
So, the question is, what's this "cap" gadget?
>The cap has to send the complex number 1 to
>i e1 tensor e2  i e2 tensor e1. That is, to a^ij ei tensor ej,
>where a^ij is the same matrix that showed up above.
Righto!
>> Starting with these, we can define the operators f: j tensor k > l
>> in a cute diagrammatic way...
>That's good, because I like cute diagrammatic ways.
Then you're gonna love this stuff!
>> Puzzle: what the heck is so important about "i times the standard
>> volume form on C^2"?
>>
>> Hint: the factor of i is not really important  it's just a convention.
>> The question is, what's so great about the volume form on C^2?
>
>It's an intertwiner. So its kernel is an invariant subspace, and we
>find that 1/2 tensor 1/2 = 1 + 0! (where these numbers denote irreps
>of SU(2)). That's pretty great!
Yes, that's PART of what's so great about it.
But here's another thing: the group of linear transformations of C^2
that preserve the volume form is SL(2,C). In the above constructions,
and everything we're about to do, all we'll ever use about C^2 is
that it's a 2d complex vector space equipped with a volume form.
So all our constructions will automatically be invariant under SL(2,C).
In particular, all the linear maps we cook up  like the cap and cup 
will be intertwining operators between reps of SL(2,C).
So the question is: what's so great about SL(2,C)?
One answer is this: it contains SU(2), and we're interested in
cooking up intertwiners
f: j tensor k > l
where j,k,l are irreps of SU(2)!
But SL(2,C) is actually cool in its own right... why?
>So how do we handle higher spins?
That's the BIG QUESTION. However, I will answer it in a separate
post  this one is getting too long.
Btw, if you wanted to do me a big favor, you could put a bunch of
our discussion here your website, where it'd serve as a kind of
minicourse on how to do calculations in the BarrettCrane model.
From nobody Fri Nov 17 22:45:58 2000
Path: newshost.uwo.ca!torn!howland.erols.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 13 Nov 2000 16:17:24 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 82
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <87g0kzckx9.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu>
<874s1hbn74.fsf@scratchy.dhis.org> <8ufdhs$1qf9$1@mortar.ucr.edu>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 974132244 11248 129.125.6.17 (13 Nov 2000 16:17:24 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 13 Nov 2000 16:17:24 GMT
Xref: newshost.uwo.ca sci.physics.research:12951
John described "cap" and "cup" operators which satisfy:
> /\
> / \ = 2
> \ /
> \/
>
> \ /   \ /
> \ /   \/
> / =   +
> / \   /\
> / \   / \
>
> and
>
> \ /  
> \ /  
> / =   
> / \  
> / \  
> \ / \ /
> \ / \ /
> V V
The last one follows from the previous two, right?
The cup is defined to be i times the standard volume form on C^2, and
the cap is the unique thing which makes the above equations true.
> >> Puzzle: what the heck is so important about "i times the standard
> >> volume form on C^2"?
> >>
> >> Hint: the factor of i is not really important  it's just a convention.
> >> The question is, what's so great about the volume form on C^2?
> >
> >It's an intertwiner. So its kernel is an invariant subspace, and we
> >find that 1/2 tensor 1/2 = 1 + 0! (where these numbers denote irreps
> >of SU(2)). That's pretty great!
>
> Yes, that's PART of what's so great about it.
>
> But here's another thing: the group of linear transformations of C^2
> that preserve the volume form is SL(2,C). In the above constructions,
> and everything we're about to do, all we'll ever use about C^2 is
> that it's a 2d complex vector space equipped with a volume form.
> So all our constructions will automatically be invariant under SL(2,C).
>
> In particular, all the linear maps we cook up  like the cap and cup 
> will be intertwining operators between reps of SL(2,C).
>
> So the question is: what's so great about SL(2,C)?
>
> One answer is this: it contains SU(2), and we're interested in
> cooking up intertwiners
>
> f: j tensor k > l
>
> where j,k,l are irreps of SU(2)!
>
> But SL(2,C) is actually cool in its own right... why?
Well, just like SU(2) is the double cover of SO(3), SL(2,C) is the
double cover of SO(3,1). And SL(2,C) is the complexification of
SU(2). That's pretty cool.
> >So how do we handle higher spins?
>
> That's the BIG QUESTION. However, I will answer it in a separate
> post  this one is getting too long.
Ok, I'll be sitting here at my computer waiting!
> Btw, if you wanted to do me a big favor, you could put a bunch of
> our discussion here your website, where it'd serve as a kind of
> minicourse on how to do calculations in the BarrettCrane model.
Sure thing. For others, the web page is
http://jdc.math.uwo.ca/spinfoams
Dan
From nobody Fri Nov 17 22:45:59 2000
Path: newshost.uwo.ca!torn!nntp.cs.ubc.ca!logbridge.uoregon.edu!usenet01.sei.cmu.edu!nntp.club.cc.cmu.edu!awabi.library.ucla.edu!138.23.226.128!news.ucr.edu!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 13 Nov 2000 00:00:00 GMT
Organization: UCR
Lines: 155
Approved: baez@math.ucr.edu
MessageID: <8uvfe6$ckd$1@mortar.ucr.edu>
References: <8pu2rt$j1u$1@crib.corepower.com> <87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu> <874s1hbn74.fsf@scratchy.dhis.org> <8ufdhs$1qf9$1@mortar.ucr.edu> <87g0kzckx9.fsf@scratchy.dhis.org>
NNTPPostingHost: mathcln06.ucr.edu
XTrace: mortar.ucr.edu 974340358 12941 138.23.203.171 (16 Nov 2000 02:05:58 GMT)
XComplaintsTo: usenet@mortar.ucr.edu
NNTPPostingDate: 16 Nov 2000 02:05:58 GMT
XNewsreader: trn 4.0test69 (20 September 1998)
Originator: baez@mathcln06.math.ucr.edu (John Baez)
Xref: newshost.uwo.ca sci.physics.research:13044
Dan Christensen wrote:
>John described "cap" and "cup" operators which satisfy:
>> /\
>> / \ = 2 [the superdimension of the spin1/2 rep is 2]
>> \ /
>> \/
>>
>> \ /   \ /
>> \ /   \/
>> / =   + [the "binor identity"]
>> / \   /\
>> / \   / \
>>
>> and
>>
>> \ /  
>> \ /  
>> / =    [the cup map is skewsymmetric]
>> / \  
>> / \  
>> \ / \ /
>> \ / \ /
>> V V
>The last one follows from the previous two, right?
Hmm... take the binor identity, stick a cup on the bottom,
use the fact that
/\
/ \ = 2
\ /
\/
and  yup!  you get the third identity.
>The cup is defined to be i times the standard volume form on C^2, and
>the cap is the unique thing which makes the above equations true.
Right.
>> So the question is: what's so great about SL(2,C)?
>Well, just like SU(2) is the double cover of SO(3), SL(2,C) is the
>double cover of [the identity component of] SO(3,1).
>And SL(2,C) is the complexification of SU(2). That's pretty cool.
Right! SL(2,C) is great because it's deeply related to
special relativity, and also general relativity  since
it's the double cover of the of the Lorentz group.
>>>So how do we handle higher spins?
>> That's the BIG QUESTION. However, I will answer it in a separate
>> post  this one is getting too long.
>Ok, I'll be sitting here at my computer waiting!
Okay, here goes. Sorry to have kept you waiting so long.
Take V = C^2, and define cap and cup maps as above. Next,
define a "symmetrization" map from the nth tensor power of
V to itself, which projects down to the space of symmetric
rankn tensors. We draw this map like this:
    

    
where I've shown the example of n = 5. Explicitly, for n = 2
we have
     
    \ /
 = 1/2   + 1/2 /
    / \
     
In general there will be a sum over all n permutations, divided
by n!.
Now, the range of this symmetrization map turns out to be none
other than the spinj rep of SU(2), where j = n/2. If you don't
know this fact, you're free to take this as a definition of the
spinj rep.
Okay, now I'll finally describe the intertwiner
f: j tensor k > l
which I've been promising you for all these years. This will
complete the specification of the BarrettCrane model!  apart
from some piddly little normalization factors we've got to
discuss someday if you ever really do those computer simulations.
I'll show you what it looks like when j = 3/2, k = 1 and l = 3/2:
    
 
\ \ \ / /
\ \ \/ /
  /
  

  
See? It's just built from the cup map and three applications of
the symmetrizer. That's how it always goes. In this case
we've got 3 strands coming in from the upper left since j = 3/2,
2 strands coming in from the upper right since k = 1, and 3 strands
coming out on the bottom since l = 3/2.
Here's another example: j = 3/2, k = 1 and l = 1/2:
    
 
\ \ \ / /
\ \ \/ /
 \ /
 \/



This time we used the cup as many times as possible for this
particular choice of j and k, so we got l = jk.
Here's another example: j = 3/2, k = 1 and l = 5/2:
    
 
\ \ \ / /
    

    
We didn't use the cup at all this time, so we got l = j+k.
See how it works?
Ain't it cool???
Now some people call a state in C^2 a "spinor", but other people
call it a "qubit". And what we're really doing, from the latter
viewpoint, is writing down "quantum logic gates" which manipulate
"qubits" in an SU(2)invariant way  in fact, an SL(2,C)invariant
way! In short, we're seeing how to build little Lorentzinvariant
quantum computers. From this crazy viewpoint, what the BarrettCrane
model does is to build a theory of quantum gravity out of these little
Planckscale quantum computers. Pretty freaky, no? But hopefully
nobody except you will read this whole article, so we won't get a bunch
of untrained loonies running around blathering excitedly about this stuff.
From nobody Fri Nov 17 22:46:00 2000
Path: newshost.uwo.ca!torn!canopus.cc.umanitoba.ca!newsflash.concordia.ca!sunqbc.risq.qc.ca!cpknewshub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 17 Nov 2000 10:54:00 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Lines: 25
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <873dgrqrxj.fsf@scratchy.dhis.org>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu>
<874s1hbn74.fsf@scratchy.dhis.org> <8ufdhs$1qf9$1@mortar.ucr.edu>
<87g0kzckx9.fsf@scratchy.dhis.org> <8uvfe6$ckd$1@mortar.ucr.edu>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 974458440 14996 129.125.6.17 (17 Nov 2000 10:54:00 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 17 Nov 2000 10:54:00 GMT
Xref: newshost.uwo.ca sci.physics.research:13091
John gave the final step in the description of how to calculate the
partition function in the BarrettCrane model, a very cool diagram
involving symmetrizations and the cap operation defined earlier!
Let me see if I remember the context. We were trying to find a map
from the j tensor k representation of SU(2) to the m tensor n
representation. John described maps from j tensor k to l, where
l is in the range jk, jk + 1, ..., j+k1, j+k. So if l is
also in the range mn, ..., m+n, we get a map m tensor n > l,
which dualizes to give a map l > m tensor n. So for l in both
ranges, we have a map j tensor k > l > m tensor n. We then
tensor this with itself, multiply by 2l+1, sum over the l in both
ranges, and do the rest of the recipe.
I think I understand this now, at least at a theoretical level.
Next I'm going to think about how to actually compute all of this
efficiently.
While I'm doing that, maybe John can supply one additional ingredient?
What is an example of an interesting function g that takes a
4manifold that is chopped up into simplices with the triangles
labelled with spins, and produces a number?
Dan
From nobody Sat Apr 7 19:15:26 2001
XFromLine: nobody Sat Dec 2 15:30:26 2000
Path: newshost.uwo.ca!torn!howland.erols.net!surfnet.nl!rug.nl!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 27 Nov 2000 21:39:48 GMT
Organization: UCR
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8vukb4$gf7$1@info.service.rug.nl>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu>
<874s1hbn74.fsf@scratchy.dhis.org> <8ufdhs$1qf9$1@mortar.ucr.edu>
<87g0kzckx9.fsf@scratchy.dhis.org> <8uvfe6$ckd$1@mortar.ucr.edu>
<873dgrqrxj.fsf@scratchy.dhis.org>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 975361188 16871 129.125.6.17 (27 Nov 2000 21:39:48 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 27 Nov 2000 21:39:48 GMT
Xref: scratchy spr:29
Lines: 60
XGnusArticleNumber: 29 Sat Dec 2 15:30:26 2000
MIMEVersion: 1.0
ContentType: text/plain; charset=usascii
Sorry to take so long to reply to this one!
Dan Christensen
>John gave the final step in the description of how to calculate the
>partition function in the BarrettCrane model, a very cool diagram
>involving symmetrizations and the cap operation defined earlier!
>Let me see if I remember the context. We were trying to find a map
>from the j tensor k representation of SU(2) to the m tensor n
>representation. John described maps from j tensor k to l, where
>l is in the range jk, jk + 1, ..., j+k1, j+k. So if l is
>also in the range mn, ..., m+n, we get a map m tensor n > l,
>which dualizes to give a map l > m tensor n. So for l in both
>ranges, we have a map j tensor k > l > m tensor n. We then
>tensor this with itself, multiply by 2l+1, sum over the l in both
>ranges, and do the rest of the recipe.
Right.
>I think I understand this now, at least at a theoretical level.
>Next I'm going to think about how to actually compute all of this
>efficiently.
Great! I'm dying to see some computer simulations of quantum gravity.
I just think I'm not the right person to do them...
>While I'm doing that, maybe John can supply one additional ingredient?
>What is an example of an interesting function g that takes a
>4manifold that is chopped up into simplices with the triangles
>labelled with spins, and produces a number?
Of course these are called "observables" by physicists.
There are a bunch of physically interesting observables, including:
1) traces of holonomies around loops  "the loop variables", or "Wilson
loops" in physics jargon.
2) areas of 2d surfaces.
3) volumes of 3d regions.
4) products of 1)  3), which allow one to study "correlations" between
these various sorts of observables.
I give formulas for some of these things in my big fat review article.
Areas of surfaces are by far the simplest, so let me describe those!
Remember that in the BarrettCrane model we sum over all possible ways
of labelling the triangles in a triangulated 4manifold by spins. For
a given labelling, the "area" of a triangle labelled by the spin j is
just sqrt(j(j+1)). And if we have a surface built from a lot of triangles,
its area is just the sum of the areas of these triangles.
Simple, eh?
I assume you know how to compute the "vacuum expectation value" of
an observable once someone hands you a formula for the observable,
now that you know the rule for computing the partition function of
the BarrettCrane model....
From nobody Sat Apr 7 19:15:29 2001
XFromLine: nobody Sat Dec 2 15:30:26 2000
Path: newshost.uwo.ca!torn!canopus.cc.umanitoba.ca!newsflash.concordia.ca!hammer.uoregon.edu!logbridge.uoregon.edu!howland.erols.net!surfnet.nl!rug.nl!notformail
From: Dan Christensen
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 28 Nov 2000 07:44:24 GMT
Organization: The University of Western Ontario, London, Ont. Canada
Approved: helbig@astro.rug.nl (sci.physics.research)
MessageID: <8vvnoo$qbv$1@info.service.rug.nl>
References: <8pu2rt$j1u$1@crib.corepower.com>
<87d7g9r3oj.fsf@scratchy.dhis.org> <8ucjb8$2u29$1@mortar.ucr.edu>
<874s1hbn74.fsf@scratchy.dhis.org> <8ufdhs$1qf9$1@mortar.ucr.edu>
<87g0kzckx9.fsf@scratchy.dhis.org> <8uvfe6$ckd$1@mortar.ucr.edu>
<873dgrqrxj.fsf@scratchy.dhis.org> <877l5o6etf.fsf@julian.uwo.ca> <8vukb4$gf7$1@info.service.rug.nl>
NNTPPostingHost: gladia.astro.rug.nl
XTrace: info.service.rug.nl 975397464 27007 129.125.6.17 (28 Nov 2000 07:44:24 GMT)
XComplaintsTo: newsmaster@rug.nl
NNTPPostingDate: 28 Nov 2000 07:44:24 GMT
Xref: scratchy spr:28
Lines: 44
XGnusArticleNumber: 28 Sat Dec 2 15:30:26 2000
MIMEVersion: 1.0
ContentType: text/plain; charset=usascii
baez@galaxy.ucr.edu (John Baez) writes:
> >While I'm doing that, maybe John can supply one additional ingredient?
> >What is an example of an interesting function g that takes a
> >4manifold that is chopped up into simplices with the triangles
> >labelled with spins, and produces a number?
>
> Of course these are called "observables" by physicists.
> There are a bunch of physically interesting observables, including:
>
> 1) traces of holonomies around loops  "the loop variables", or "Wilson
> loops" in physics jargon.
>
> 2) areas of 2d surfaces.
>
> 3) volumes of 3d regions.
>
> 4) products of 1)  3), which allow one to study "correlations" between
> these various sorts of observables.
>
> I give formulas for some of these things in my big fat review article.
> Areas of surfaces are by far the simplest, so let me describe those!
[...]
> I assume you know how to compute the "vacuum expectation value" of
> an observable once someone hands you a formula for the observable,
> now that you know the rule for computing the partition function of
> the BarrettCrane model....
I know that one should calculate
sum over all J of g(J) f(J)

sum over all J of f(J)
where J runs over all labellings of the triangles in my triangulated
4manifold with spins, g(J) is the value of the observable for that
labelling, and f(J) is the product of the 10j symbols you get, one
for each 4simplex in the triangulation.
This must be what is called the vacuum expectation value.
Dan
From nobody Sat Apr 7 19:15:31 2001
XFromLine: nobody Sat Dec 2 15:30:26 2000
Path: newshost.uwo.ca!torn!canopus.cc.umanitoba.ca!newsflash.concordia.ca!hammer.uoregon.edu!arclight.uoregon.edu!newsfeed.mathworks.com!nycmny1snh1.gtei.net!news.gtei.net!zeus.visi.com!hermes.visi.com!newsout.visi.com!news.state.mn.us!notformail
From: baez@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: Re: Spin foams and gauge theories
Date: 28 Nov 2000 23:06:11 GMT
Organization: UCR
Approved: spr@rosencrantz.stcloudstate.edu (sci.physics.research)
MessageID: <901dp3$4ap$1@news.state.mn.us>
References: <8pu2rt$j1u$1@crib.corepower.com> <873dgrqrxj.fsf@julian.uwo.ca> <877l5o6etf.fsf@julian.uwo.ca> <8vvnoo$qbv$1@info.service.rug.nl>
NNTPPostingHost: rosencrantz.stcloudstate.edu
XTrace: news.state.mn.us 975452771 4441 199.17.31.144 (28 Nov 2000 23:06:11 GMT)
XComplaintsTo: news@news.state.mn.us
NNTPPostingDate: 28 Nov 2000 23:06:11 GMT
Originator: spr@rosencrantz.stcloudstate.edu (sci.physics.research comoderator)
Xref: scratchy spr:27
Lines: 36
XGnusArticleNumber: 27 Sat Dec 2 15:30:26 2000
In article <8vvnoo$qbv$1@info.service.rug.nl>,
Dan Christensen wrote:
>baez@galaxy.ucr.edu (John Baez) writes:
>> I assume you know how to compute the "vacuum expectation value" of
>> an observable once someone hands you a formula for the observable,
>> now that you know the rule for computing the partition function of
>> the BarrettCrane model....
>I know that one should calculate
>
> sum over all J of g(J) f(J)
> 
> sum over all J of f(J)
>
>where J runs over all labellings of the triangles in my triangulated
>4manifold with spins, g(J) is the value of the observable for that
>labelling, and f(J) is the product of the 10j symbols you get, one
>for each 4simplex in the triangulation.
Got it! I just wanted to check that you were throwing that supercomputer
at the right sort of calculation....
>This must be what is called the vacuum expectation value.
Right. In ordinary quantum field theory on Minkowski spacetime,
this sort of formula (with "f(J)" replaced by the exponential of
the action, and "g(J)" replaced by your favorite observable) is
the recipe for computing the expectation value of an observable
in the vacuum state. Thus we unthinkingly adopt this terminology
in quantum gravity, despite the many problems with the concept of
"vacuum state" in quantum gravity. Don't worry too much though:
it's still a good thing to compute!