From nobody Fri Nov 10 21:13:57 2000 Path: newshost.uwo.ca!torn!sunqbc.risq.qc.ca!cpk-news-hub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 971913117 11099 129.125.6.17 (18 Oct 2000 23:51:57 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 representation-theoretic >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 Barrett-Crane 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 Barrett-Crane model? And the answer is: spin network edges are dual to triangles in a triangulation of our 4-manifold (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!news-spur1.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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 972388714 2890 129.125.6.17 (24 Oct 2000 11:58:34 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 non-negative half-integers 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 Barrett-Crane model? > > And the answer is: spin network edges are dual to triangles in a > triangulation of our 4-manifold (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 4-manifold, an edge should be dual to a 3-simplex (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 not-pulled-back 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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 972748515 15834 129.125.6.17 (28 Oct 2000 15:55:15 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 non-negative half-integers 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 spin-j 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 4-dimensional 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 mumbo-jumbo, or a painful explicit calculation, shows this map is onto. So let's just see why it's two-to-one. 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 Barrett-Crane model? >> >> And the answer is: spin network edges are dual to triangles in a >> triangulation of our 4-manifold (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 Barrett-Crane model? And the answer is: spin foam faces are dual to triangles in a triangulation of our 4-manifold (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 4-manifold we get a triangulated 3-manifold. Spin foam faces in the 4-manifold give spin network edges in the 3-manifold. These are now dual to the triangles in the 3-manifold! 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 4-manifold without boundary. This recipe generalizes to a recipe for calculating transition amplitudes between spin network states when we have a 4-manifold that's a cobordism from one 3-manifold to another, and spin networks sitting in the 3-manifolds. 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 mind-reader! >> 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 time-consuming 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!news-hog.berkeley.edu!ucberkeley!enews.sgi.com!sdd.hp.com!ihnp4.ucsd.edu!news.ucr.edu!not-for-mail 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 Message-ID: <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> NNTP-Posting-Host: math-cl-n06.ucr.edu Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: mortar.ucr.edu 972854352 55482 138.23.203.171 (29 Oct 2000 21:19:12 GMT) X-Complaints-To: usenet@mortar.ucr.edu NNTP-Posting-Date: 29 Oct 2000 21:19:12 GMT User-Agent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7 Content-Length: 4652 Status: OR X-Newsreader: trn 4.0-test69 (20 September 1998) Originator: baez@math-cl-n06.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 Peter-Weyl 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 Barrett-Crane model? > > And the answer is: spin foam faces are dual to triangles in a > triangulation of our 4-manifold (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!not-for-mail 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) Message-ID: <200010300359.e9U3xa923006@math-cl-n06.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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 972911128 8399 129.125.6.17 (30 Oct 2000 13:05:28 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 Peter-Weyl 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 n-dimensional 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^2-dimensional 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 a-sayin'! 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!not-for-mail 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) Message-ID: <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@math-cl-n06.ucr.edu> NNTP-Posting-Host: rosencrantz.stcloudstate.edu Content-Type: text/plain; charset=us-ascii X-Trace: news.state.mn.us 973014528 9177 199.17.31.144 (31 Oct 2000 17:48:48 GMT) X-Complaints-To: usenet@news.state.mn.us NNTP-Posting-Date: 31 Oct 2000 17:48:48 GMT User-Agent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7 Originator: spr@rosencrantz.stcloudstate.edu (sci.physics.research co-moderator) 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 Peter-Weyl theorem and mine. Suppose R is an n-dimensional 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!not-for-mail 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) Message-ID: <8tniqu$2k60$1@mortar.ucr.edu> References: <8pu2rt$j1u$1@crib.corepower.com> <877l6sgbha.fsf@scratchy.dhis.org> <200010300359.e9U3xa923006@math-cl-n06.ucr.edu> <87itq9al7x.fsf@scratchy.dhis.org> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973033818 14084 129.125.6.17 (31 Oct 2000 23:10:18 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 Peter-Weyl 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 Barrett-Crane 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: A------B \ / \ / \ / C we can describe it mod translations by the vectors u = B-A v = C-A 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 god-given 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 spin-j 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 Barrett-Crane 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 (gr-qc/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 Barrett-Crane 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 3-space. 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!not-for-mail 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 Message-ID: <87og00471l.fsf@scratchy.dhis.org> References: <8pu2rt$j1u$1@crib.corepower.com> <877l6sgbha.fsf@scratchy.dhis.org> <200010300359.e9U3xa923006@math-cl-n06.ucr.edu> <87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu> NNTP-Posting-Host: math-cl-n06.ucr.edu Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: mortar.ucr.edu 973130408 69093 138.23.203.171 (2 Nov 2000 02:00:08 GMT) X-Complaints-To: usenet@mortar.ucr.edu NNTP-Posting-Date: 2 Nov 2000 02:00:08 GMT User-Agent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7 Content-Length: 3632 Status: OR X-Newsreader: trn 4.0-test69 (20 September 1998) Originator: baez@math-cl-n06.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 = B-A > v = C-A > > [...] 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 (n-1)+(n-2)-1 = 2n-4, I think, and so the space of planes plus an area has dimension 2n-4+1 = 2n-3. The space of bivectors has dimension n(n-1)/2, which is larger than 2n-3 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 skew-adjoint 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 god-given 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 Barrett-Crane > 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!news-in.mts.net!cyclone.bc.net!news.maxwell.syr.edu!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973271034 22326 129.125.6.17 (3 Nov 2000 17:03:54 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 1-form taking values in the Lie algebra of the Lorentz group, and: b) a field E, which can locally be thought of as a 2-form 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 gauge-equivalent, 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 1-form 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 (n-1)+(n-2)-1 = >2n-4, I think, and so the space of planes plus an area has dimension >2n-4+1 = 2n-3. The space of bivectors has dimension n(n-1)/2, which >is larger than 2n-3 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 skew-adjoint 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 god-given 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 0-dimensional except for the spheres of integer or half-integer radius. The sphere of radius j gives the spin-j rep of SU(2) when we quantize it. So when we quantize so(3)*, we get: sum_j j This is just a fancy-ass 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 Barrett-Crane >> 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!not-for-mail 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) Message-ID: <87u29rc4xe.fsf@scratchy.dhis.org> References: <8pu2rt$j1u$1@crib.corepower.com> <877l6sgbha.fsf@scratchy.dhis.org> <200010300359.e9U3xa923006@math-cl-n06.ucr.edu> <87itq9al7x.fsf@scratchy.dhis.org> <8tniqu$2k60$1@mortar.ucr.edu> <87og00471l.fsf@scratchy.dhis.org> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973271069 22332 129.125.6.17 (3 Nov 2000 17:04:29 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973334835 13225 129.125.6.17 (4 Nov 2000 10:47:15 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 4-simplexes, and describe the geometry of all this stuff by specifying information about the triangles. My long-winded half-explanation 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 Barrett-Crane 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 Barrett-Crane 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!news-in.mts.net!cyclone.bc.net!newsfeed.stanford.edu!xfer10.netnews.com!netnews.com!newsfeed.skycache.com!Cidera!cpk-news-hub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973271112 22447 129.125.6.17 (3 Nov 2000 17:05:12 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 3-sphere. 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 3-sphere 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 3-sphere 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 1-skeleton of a 4-simplex. 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 4-valent, 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 4-simplex such that each vertex has two edges going in and two edges going out. These interwtining operators are called the "Barrett-Crane vertices" and you can read about them in the papers listed below, as well as my big fat review article and the original Barrett-Crane paper. But let me briefly summarize them. a) First you need to learn about these god-given 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 god-given 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 spin-l 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 4-simplex 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 4-valent vertex for simple reps of SO(4). Here is a good book on diagrammatic methods for SU(2) representation theory: The classical and quantum 6j-symbols / by J. Scott Carter, Daniel E. Flath, and Masahico Saito. Princeton, N.J. : Princeton University Press, 1995. and here are some papers on the Barrett-Crane vertex: David N. Yetter, Generalized Barrett-Crane 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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 973507375 5086 129.125.6.17 (6 Nov 2000 10:42:55 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 god-given 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+k-1, ..., |j-k|+1, |j-k|, 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^{2j-1} 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!not-for-mail 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 Message-ID: <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> NNTP-Posting-Host: math-cl-n06.ucr.edu X-Trace: mortar.ucr.edu 973721768 96329 138.23.203.171 (8 Nov 2000 22:16:08 GMT) X-Complaints-To: usenet@mortar.ucr.edu NNTP-Posting-Date: 8 Nov 2000 22:16:08 GMT X-Newsreader: trn 4.0-test69 (20 September 1998) Originator: baez@math-cl-n06.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 category-theoretic 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 god-given 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+k-1, >..., |j-k|+1, |j-k|, 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+k-1, ..., |j-k|+1, |j-k|. 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^{2j-1} 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 skew-symmetric 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 spin-1/2 \ / 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!not-for-mail 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 Message-ID: <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> NNTP-Posting-Host: math-cl-n06.ucr.edu Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: mortar.ucr.edu 973797774 50207 138.23.203.171 (9 Nov 2000 19:22:54 GMT) X-Complaints-To: usenet@mortar.ucr.edu NNTP-Posting-Date: 9 Nov 2000 19:22:54 GMT User-Agent: Gnus/5.0808 (Gnus v5.8.8) Emacs/20.7 Content-Length: 3386 Status: OR X-Newsreader: trn 4.0-test69 (20 September 1998) Originator: baez@math-cl-n06.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 god-given 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+k-1, > >..., |j-k|+1, |j-k|, 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+k-1, ..., |j-k|+1, |j-k|. 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 skew-symmetric > map > > a: V tensor V -> C Ok, so this sends (u,v) tensor (w,z) to i(uz-vw), 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 spin-1/2 > \ / 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 spin-0 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!fu-berlin.de!uni-berlin.de!rosencrantz.stcloudstate.EDU!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: rosencrantz.stcloudstate.edu (199.17.31.144) X-Trace: fu-berlin.de 973899229 1925065 199.17.31.144 (16 [28113]) X-Newsreader: trn 4.0-test72 (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 skew-symmetric 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(uz-vw), 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 spin-1/2 >> \ / 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 Barrett-Crane 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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 974132244 11248 129.125.6.17 (13 Nov 2000 16:17:24 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 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 Barrett-Crane model. Sure thing. For others, the web page is http://jdc.math.uwo.ca/spin-foams 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!not-for-mail 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 Message-ID: <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> NNTP-Posting-Host: math-cl-n06.ucr.edu X-Trace: mortar.ucr.edu 974340358 12941 138.23.203.171 (16 Nov 2000 02:05:58 GMT) X-Complaints-To: usenet@mortar.ucr.edu NNTP-Posting-Date: 16 Nov 2000 02:05:58 GMT X-Newsreader: trn 4.0-test69 (20 September 1998) Originator: baez@math-cl-n06.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 spin-1/2 rep is -2] >> \ / >> \/ >> >> \ / | | \ / >> \ / | | \/ >> / = | | + [the "binor identity"] >> / \ | | /\ >> / \ | | / \ >> >> and >> >> \ / | | >> \ / | | >> / = - | | [the cup map is skew-symmetric] >> / \ | | >> / \ | | >> \ / \ / >> \ / \ / >> 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 rank-n 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 spin-j 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 spin-j 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 Barrett-Crane 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 = |j-k|. 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 Lorentz-invariant quantum computers. From this crazy viewpoint, what the Barrett-Crane model does is to build a theory of quantum gravity out of these little Planck-scale 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!cpk-news-hub1.bbnplanet.com!news.gtei.net!npeer.kpnqwest.net!news2.kpn.net!news.kpn.net!surfnet.nl!rug.nl!not-for-mail 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) Message-ID: <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$1@mortar.ucr.edu> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 974458440 14996 129.125.6.17 (17 Nov 2000 10:54:00 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 17 Nov 2000 10:54:00 GMT Xref: newshost.uwo.ca sci.physics.research:13091 John gave the final step in the description of how to calculate the partition function in the Barrett-Crane model, a very cool diagram involving symmetrizations and the cap operation defined earlier! Let me see if I remember the context. We were trying to find a map from the j tensor k representation of SU(2) to the m tensor n representation. John described maps from j tensor k to l, where l is in the range |j-k|, |j-k| + 1, ..., j+k-1, j+k. So if l is also in the range |m-n|, ..., m+n, we get a map m tensor n --> l, which dualizes to give a map l --> m tensor n. So for l in both ranges, we have a map j tensor k --> l --> m tensor n. We then tensor this with itself, multiply by 2l+1, sum over the l in both ranges, and do the rest of the recipe. I think I understand this now, at least at a theoretical level. Next I'm going to think about how to actually compute all of this efficiently. While I'm doing that, maybe John can supply one additional ingredient? What is an example of an interesting function g that takes a 4-manifold that is chopped up into simplices with the triangles labelled with spins, and produces a number? Dan From nobody Sat Apr 7 19:15:26 2001 X-From-Line: nobody Sat Dec 2 15:30:26 2000 Path: newshost.uwo.ca!torn!howland.erols.net!surfnet.nl!rug.nl!not-for-mail From: baez@galaxy.ucr.edu (John Baez) Newsgroups: sci.physics.research Subject: Re: Spin foams and gauge theories Date: 27 Nov 2000 21:39:48 GMT Organization: UCR Approved: helbig@astro.rug.nl (sci.physics.research) Message-ID: <8vukb4$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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 975361188 16871 129.125.6.17 (27 Nov 2000 21:39:48 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 27 Nov 2000 21:39:48 GMT Xref: scratchy spr:29 Lines: 60 X-Gnus-Article-Number: 29 Sat Dec 2 15:30:26 2000 MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii 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 Barrett-Crane 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 |j-k|, |j-k| + 1, ..., j+k-1, j+k. So if l is >also in the range |m-n|, ..., 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 >4-manifold 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 Barrett-Crane model we sum over all possible ways of labelling the triangles in a triangulated 4-manifold 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 Barrett-Crane model.... From nobody Sat Apr 7 19:15:29 2001 X-From-Line: 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!not-for-mail 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) Message-ID: <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> NNTP-Posting-Host: gladia.astro.rug.nl X-Trace: info.service.rug.nl 975397464 27007 129.125.6.17 (28 Nov 2000 07:44:24 GMT) X-Complaints-To: newsmaster@rug.nl NNTP-Posting-Date: 28 Nov 2000 07:44:24 GMT Xref: scratchy spr:28 Lines: 44 X-Gnus-Article-Number: 28 Sat Dec 2 15:30:26 2000 MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii 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 > >4-manifold 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 Barrett-Crane 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 4-manifold 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 4-simplex in the triangulation. This must be what is called the vacuum expectation value. Dan From nobody Sat Apr 7 19:15:31 2001 X-From-Line: 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!nycmny1-snh1.gtei.net!news.gtei.net!zeus.visi.com!hermes.visi.com!news-out.visi.com!news.state.mn.us!not-for-mail 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) Message-ID: <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$1@info.service.rug.nl> NNTP-Posting-Host: rosencrantz.stcloudstate.edu X-Trace: news.state.mn.us 975452771 4441 199.17.31.144 (28 Nov 2000 23:06:11 GMT) X-Complaints-To: news@news.state.mn.us NNTP-Posting-Date: 28 Nov 2000 23:06:11 GMT Originator: spr@rosencrantz.stcloudstate.edu (sci.physics.research co-moderator) Xref: scratchy spr:27 Lines: 36 X-Gnus-Article-Number: 27 Sat Dec 2 15:30:26 2000 In article <8vvnoo$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 Barrett-Crane 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 >4-manifold 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 4-simplex 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!