
The most substantial part of this issue is some remarks by Daniel Ruberman (ruberman@binah.cc.brandeis.edu) on the paper I was talking about last time by Boguslaw Broda. They apparently show that Broda's invariant is not as new as it might appear. But they're rather technical, so I'll put them near the end, and start off with something on the light side, and then note some interesting progress on the Vassiliev invariant scene.
1) Beyond Einstein  is space loopy? by Marcia Bartusiak, Discover, April 1993.
In the airport in Montreal I ran into this article, which was the cover story, with an upsidedown picture of Einstein worked into a bunch of linked keyrings. I bought it  how could I resist?  since it is perhaps the most "pop" exposition of the loop representation of quantum gravity so far. Those interested in the popularization of modern physics might want to compare
Gravity quantized? A radical theory of gravity weaves space from tiny loops, by John Horgan, Scientific American, September 1992.
Given the incredible hype concerning superstring theory, which seems to have faded out by now, I sort of dread the same thing happening to the loop representation of quantum gravity. It is intrinsically less hypeable, since it does not purport to be a theory of everything, and comes right after superstrings were supposed to have solved all the mysteries of the universe. Also, its proponents are (so far) a more cautious breed than the string theorists  note the question marks in both titles! But we will see....
Marcia Bartusiak is a contributing editor of Discover and the author of a book on current topics in astronomy and astrophysics, "Thursday's Universe", which I haven't read. She'll be coming out with a book in June, "Through a Universe Darkly," that's supposed to be about how theories of cosmology have changed down through the ages. She does a decent job of sketching vaguely the outlines of the loop representation to an audience who must be presumed ignorant of quantum theory and general relativity. Or course, there is also a certain amount of humaninterest stuff, with Ashtekar, Rovelli and Smolin (quite rightly) coming off as the heroes of the story. There are, as usual, little boxes with geewhiz remarks like
WITH REAMS OF PAPER SPREAD OUT OVER THE KITCHEN TABLE THEY FOUND SOLUTION AFTER SOLUTION FOR EQUATIONS THOUGHT IMPOSSIBLE TO SOLVE
(which is, after all, true  nobody had previously found solutions to the constraint equations in canonical quantum gravity, and all of a sudden here were lots of 'em!). And there are some amusing discussions of personality: "Affable, creative, and easygoing, Rovelli quickly settled into the role of gobetween, helping mesh the analytic powers of the quiet, contemplative Ashtekar with the creativity of the brash, impetuous Smolin." And discussions of how much messier Smolin's office is than Ashtekar's.
In any event, it's a fun read, and I recommend it. Of course, I'm biased, so don't trust me.
2) Vassiliev invariants contain more information than all knot polynomials, by Sergey Piunikhin, preprint. (Piunikhin is at serguei@math.harvard.edu)
TuraevViro and KauffmanLins invariants for 3manifolds coincide, by Sergey Piunikhin, Journal of Knot Theory and its Ramifications, 1 (1992) 105  135.
Different presentations of 3manifold invariants arising in rational conformal field theory, by Sergey Piunikhin, preprint.
Weights of Feynman diagrams, link polynomials and Vassiliev knot invariants, by Sergey Piunikhin, preprint.
ReshetikhinTuraev and CraneKohnoKontsevich 3manifold invariants coincide, by Sergey Piunikhin, preprint.
I received a packet of papers by Piunikhin a while ago. The most new and interesting thing is the first paper listed above. In "week3" I noted a conjecture of BarNatan that all Vassiliev invariants come from quantum group knot invariants (or in other words, from Lie algebra representations.) Piunikhin claims to refute this by showing that there is a Vassiliev invariant of degree 6 that does not. (However, other people have told me his claim is misleading!) I have been too busy to read this paper yet.
3) Bibliography of publications related to classical and quantum gravity in terms of the Ashtekar variables, by Bernd Bruegmann, 14 pages (LaTeX, 1 figure), available as grqc/9303015.
Let me just quote the abstract; this should be a handy thing:
4) Surgical invariants of fourmanifolds, by Boguslaw Broda, preprint available as hepth/9302092. (Revisited  see "week9")
Let me briefly recall the setup: we describe a compact 4manifold by a handlebody decomposition, and represent this decomposition using a link in S^3. The 2handles are represented by framed knots, while the 1handles are represented by copies of the unknot (which we may think of as having the zero framing). The 1handles and 2handles play quite a different role in constructing the 4manifold  which is why one normally draws the former as copies of the unknot with a dot on them  but Broda's construction does NOT care about this. Broda simply takes the link, forgetting the dots, and cooks up a number from it, using cabling and the Kauffman bracket at an root of unity. Let's call Broda's invariant by b(M)  actually for each primitive rth root of unity, we have b_r(M).
Broda shows that this is a 4manifold invariant by showing it doesn't change under the de Sa moves. One of these consists of adding or deleting a Hopf link 
/\ /\ / \ / \ / \ \ / / \ \ \ \ / / \ \ / \ / \ / \/ \/
in which both components have the zero framing and one represents a 1handle and the other a 2handle. This move depends on the fact that we can "cancel" a 1handle and 2handle pair, a special case of a general result in Morse theory.
But since Broda's invariant doesn't care which circles represent 1handles and which represent 2handles, Broda's invariant is also invariant under adding two 2handles that go this way. This amounts to taking a connected sum with S^2 x S^2. I.e., b(M) = b(M#S^2 x S^2).
Now, Ruberman told me a while back that we must also have b(M) = b(M#CP2#CP2), that is, the invariant doesn't change under taking a connected sum with a copy of CP2 (complex projective 2space) and an orientationreversed copy of CP2. This amounts to adding or deleting a Hopf link in which one component has the zero framing and the other has framing 1. I didn't understand this, so I pestered Ruberman some more, and this is what he says (modulo minor edits). I have not had time to digest it yet:
From this, it follows that b(M) = b(M#S^2 x S^2) = b(M#CP2#CP2). For M is unchanged if you add a cancelling 1,2 pair, independent of the framing on the 2handle. If you change the special circle to an ordinary one, b(M) doesn't change. On the other hand, M has been replaced by its sum with either S2 x S2 or CP2 # CP2, depending on whether the framing on the 2handle is even or odd. (Exercise: why is only the parity relevant?)
Now as I pointed out before, if one replaces all of the 1handles (special circles) of a 4manifold with 2handles, the invariant doesn't change. This operation corresponds to doing surgery on the 4manifold, along the cores of the 1handles. In particular, the manifold has changed by a cobordism. (This is a basic construction; when you do surgery you produce a cobordism, in this case it's M x I U 2handles attached along the circles which you surgered.)
From this, I will now show that Broda's invariant is determined by the signature. (This is in the orientable case. Actually it seems that his invariant is really an invariant of an oriented manifold.) The argument above says that for any M, there is an M', with b(M) = b(M'), where M' has no 1handles, and where M and M' are cobordant. In particular, M' is simply connected. So it suffices to show that b(N) = b(N') if N and N' are simply connected.
So now you can assume you have two simply connected manifolds N,N' which are cobordant via a 5dimensional cobordism W, which you can also assume simply connnected. By highdimensional handlebody theory, you can get rid of the 1handles and 4handles of W, and assume that all the 2handles are added, then all of the 3handles. If you add all the 2handles to N, you get N#k(S^2 x S^2)#l(CP^2#(CP^2)) for some k and l. (Here is where simple connectivity is relevant; the attaching circle of a 2handle is nullhomotopic, and therefore isotopic to an unknotted circle. It's a simple exercise to see what happens when you do surgery on a trivial circle, ie you add on S2 x S2 or CP2 # CP2. On the other hand you get the same manifold as the result of adding 2handles to N'. So
N#k(S^2 x S^2)#l(CP^2#(CP^2)) = N'#k'(S^2 x S^2)#l'(CP^2#(CP^2)),
so by previous remarks b(N) = b(N'), i.e b is a cobordism invariant.
Now: b is also multiplicative under connected sum, because connected sum just takes the union of the link diagrams. The cobordism group is Z, detected by the signature, so b must be a multiple of the signature, modulo some number. (Maybe at this point I realize b should be b_r or some such). If you compute (as a grad student Tianjin Li did for me) b_r(CP^2), you find that b_r lives in the group of rth (or maybe 4rth; I'm at home and don't have my note) roots of unity.
My conclusion: this invariant is a rather complicated way to compute the signature of a 4manifold (modulo r or 4r) from a link diagram of the manifold.
There is an important moral of the story, which is perhaps not obvious to someone outside of 4manifolds. Any invariant which purports to go beyond classical ones (ie invariants of the intersection form) must treat CP^2 and CP^2 very differently. It seems to be the case that many manifolds which are different (ie nondiffeomorphic) become diffeomorphic after you add on CP^2. Thus any useful invariant should get rather obliterated by adding CP^2. On the other hand, nondiffeomorphic manifolds seem to stay nondiffeomorphic, no matter how many CP2's you add on. This phenomenon doesn't seem to be exhibited by any of the quantumgroup type constructions for 3manifolds; as it shouldn't, since (from the 3manifold point of view) an unknot with framing + or 1 doesn't change the 3manifold. So if you're looking for a combinatorial invariant, it seems critical that you try to build in the asymmetry wrt orientation which 4manifolds seem to possess.
Exercise: do the nonorientable case. The answer should be that b is determined by the Euler characteristic, mod 2.
© 1993 John Baez
baez@math.removethis.ucr.andthis.edu
