# “Lorentz-invariant” Morse theory possible?

Hi,

I hope filing this under "Morse theory" is correct.

For the "abstract tensor" approach to knot theory you need: a) a 4D tensor to emulate particle interaction, b) a 2D tensor to emulate pair creation/annihilation. (I formulate it in Feynman diagram terms.) You don't actually need c) a particle propagator since this is the delta function.

Now you can rotate a knot by 90 degrees and you can rotate a Feynman diagram by 90 degrees, but you can't "rotate Morse theory by 90 degrees"! And I find this extremely annoying. A "nice" abstract tensor Sabcd emulating a crossing should have exactly the same symmetry properties as a crossing, i.e. invariance under all 180 degree rotations. And for the pair tensor Pab likewise Pab=Pba (1), and the delta function should be Pab too, instead.

One can introduce "gauges" and use an equivalent tensor formulation, using Pab for left and right turns, but (1) gets you into problems in no time and I never was able to find a "Lorentz-invariant" tensor set. The interesting solutions are even CPT-invariant (for intuitive interpretations of C, P, and T for abstract tensors, I even have charge and baryon number conservation automagically), but T(S)=S^-1 should hold (where I only can get T(S)=S for my solutions).

Am I doing something wrong or is Morse theory (or more precise, Kauffman's abstract tensor approach) symmetry-breaking by nature?

Hauke

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I am not really sure I understand what the question is. Can you say something more precise than "rotate Morse theory by 90 degrees"? – Qiaochu Yuan Feb 24 '11 at 17:46
@Qiaochu: Maybe this is what Hauke is trying to say. One way to search for knot invariants is the following. Any knot (nice map from S^1 to R^3=(x,y,z)) can be perturbed just a little bit so that (1) its projection to R^2=(x,y) has nothing worse than transverse double points (i.e. so that it projects to a "knot diagram"), and (2) its projection to R=(x) has nothing worse than nondegenerate cusps (i.e. its "height" function x is Morse). Then I can chop up the diagram into pieces that look like: cups, caps, crossings. (continued) – Theo Johnson-Freyd Feb 24 '11 at 17:50
(continuation) Two diagrams represent the same knot if they are related by: (1) Reidemeister moves, (2) cap-cup creation/annihilation, (3) switching the heights of distant caps and cups. So what Hauke wants to do is find "Feynman rules", which interpret a cup as a (0,2)-tensor for some vector space V, a cap as some (2,0)-tensor, and a crossing as a (2,2)-tensor, and roughly-vertical lines as the identity map (a (1,1)-tensor). These tensors have to satisfy some quadratic and cubic relations, which look like they overdetermine the tensors. But solutions exist, coming from quantum groups. (cont) – Theo Johnson-Freyd Feb 24 '11 at 17:53
(continuation) So, then, I think that Hauke is just bemoaning the fact that the invariance under Morse and Reidemeister moves must be added in by hand/fiat --- the decomposition of a knot into pieces that can be interpreted as tensors is not manifestly invariant. Of course, that's because it isn't invariant --- Morse theory is a calculational tool, and requires a choice of height function and understanding of how different choices are related, just like the Reidemeister moves are. When Hauke says "Lorentz invariance" he's talking about the SO(2) that acts on knot diagrams. – Theo Johnson-Freyd Feb 24 '11 at 17:56
THX, Theo, you summed up the setup nicely. But it's not the decomposition as such which is the problem: OK, if you rotate a knot by 90 degrees, the pieces will be different. But if I decompose nonstandardly, i.e. assigning a tensor to ALL turns < > ^ v and crossings °/. .\°, instead of letting < and > be the delta function (forgive my ASCII), this is gauge-equivalent to the standard approach. (If < = Pab, > = Qab, °/. = Rabcd, then the standard crossing tensor is sum(PaiRibcjQjd).) So < > swaps with ^ v and °/. with .\°, and the tensor itself should reflect that symmetry. (cont) – Hauke Reddmann Feb 25 '11 at 15:27