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A-and B-resolutions of a crossing.

A-and B-resolutions of a crossing.

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This paper is a brief overview of recent results by the authors relating colored Jones polynomials to geometric topology. The proofs of these results appear in the papers [arXiv:1002.0256] and [arXiv:1108.3370], while this survey focuses on the main ideas and examples.

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... and state graphs. Associated to a diagram D and a crossing x of D are two link diagrams, each with one fewer crossing than D. These are obtained by removing the crossing x, and reconnecting the diagram in one of two ways, called the A-resolution and B-resolution of the crossing, shown in Figure 1. For each crossing of D, we may make a choice of A-resolution or B-resolution, and end up with a crossing-free diagram. ...
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... components of this diagram are called state circles. For each crossing x of D, attach an edge from the state circle on one side of the crossing to the other, as in the dashed lines of Figure 1. Denote the resulting graph by H σ . ...
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... this example, it turns out that the manifold M A = S 3 S A contains a single essential product disk. This disk D is shown in Figure 10. Observe that this lone EPD corresponds to the single 2-edge loop in G A . ...
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... the actual hyperbolic volume of the link in Figure 9 is vol(S 3 K) ≈ 11.3407. Turning to the all-B resolution, Figure 11 shows the graphs H B , G B , and G ′ B , as well as the state surface S B . This time, G ′ B is a tree, thus Theorem 3.2 (applied to a reflected diagram) implies S B is a fiber. ...
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... important point to note is that even though S 3 K is fibered, it does contain surfaces (such as S A ) with quite a lot of guts. Conversely, having |β ′ K | > 0, as we do here, Figure 11. The graphs H B , G B , and G ′ B for the link of Figure 9, and the corresponding surface S B . ...
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... Figure 12, the state surface for this example is shown lying below the projection plane (although soup cans have been smoothed off at their sharp edges in this figure). ...
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... form the disk that we cut along, we isotope the disk by pushing it under the projection plane slightly, keeping its boundary on the state surface S A , so that it meets the link a minimal number of times. Indeed, because S A itself lies on or below the projection plane Figure 12. The surface S A is hanging below the projection plane. ...
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... disk lies slightly below the projection plane everywhere. Figure 13 gives an example. ...
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... a result, each lower polyhedron is combinatorially identical to the checkerboard polyhedron of an alternating sub-diagram of D(K), where the sub-diagram corresponds to region R. This is illustrated in Figure 14, for the knot diagram of Figure 4. ...
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... ideal vertices of the upper polyhedron correspond to strands of the link between undercrossings. In Figure 16, the surface S A is shown in green and gold. There are four ideal edges meeting a crossing, labeled e 1 through e 4 in this figure. ...
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... surface S A runs through the crossing in a twisted rectangle. Looking again at Figure 16, note that the gold portion of S A at the top of that figure is not cut off at the crossing by an ideal edge terminating at v 2 . Instead, it follows e 4 through the crossing and along the underside of the figure, between e 4 and the link. ...
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... visualize these thin portions of shaded face between an edge and the link as tentacles, and sketch them onto H A running from the top-right of a segment to the base of the segment, and then along an edge of H A (Figure 17). Similarly for the bottom-left. ...
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... Figure 17, we have drawn the tentacle by removing a bit of edge of H A . When we do this for each segment, top-right and bottom-left, the remaining connected components of H A correspond exactly to ideal vertices. ...
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... we observed above, edges terminate at undercrossings. Figure 18 shows the portions of ideal edges sketched schematically onto H A . ...
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... that a tentacle will continue past segments of H A on the opposite side of the state circle without terminating, spawning new tentacles, but will terminate at a segment on the same side of the state surface. This is shown in Figure 19. Since H A is a finite graph, the process terminates. ...
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... every non-prime arc indicates that a lower polyhedron violates Definition 5.2. See Figure 21. ...
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... divides the corre- sponding lower polyhedron into two. Figure 21 shows a 3-dimensional view of this cutting process. ...

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... This is the construction of the famous Jones polynomial [93]. Some surveys on the Jones polynomial are [1,71,113]. ...
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