paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0106017 | We have seen in REF that MATH at least twice switches from crossing MATH (perhaps repeatedly) to crossing MATH (and vice versa). To simplify the number of cases we need to consider, note first that we can slit open the part of MATH corresponding to MATH, lengthening MATH while shortening MATH until MATH is so short that it is disjoint from MATH. So, with no loss of generality, we may as well assume that MATH intersects both of the segments MATH but not MATH. Once this is done, the components of MATH consist of three types: subarcs of MATH, subarcs of MATH, and a single component MATH that contains MATH together with all four ends of the two edges MATH . Then MATH intersects each of the segments MATH in an even number of points. Let MATH denote the component of MATH that is exactly half-way along MATH as measured by intersections with MATH. That is, an arc in MATH starting from a point in MATH and ending in MATH will intersect MATH in the same number of points no matter which way along MATH it runs. Denote by MATH the analogous point in MATH. The knot MATH is similarly split up into segments by MATH, some parallel to segments of MATH, some to segments of MATH and some lying on MATH. Any subsegment of MATH that is a union of components of the first (respectively, second) type will be said to be colored MATH (respectively, MATH.) Remembering that MATH is oriented, each segment of MATH can be described as one of three types: CASE: components of MATH that run from an end of MATH to an end of MATH CASE: components of MATH that run from an end of MATH to an end of MATH CASE: components of MATH that run from an end of MATH to the other end of MATH (There are no components of MATH that run from MATH to MATH since MATH.) We will say that these three types of segments of MATH are colored MATH respectively. The notation is meant to suggest that in clockwise rotation, the color changes from MATH to MATH, etc. Since all components of MATH are essential in MATH, MATH cannot be a disk. We will suppose MATH and arrive at a contradiction. There are various ways that the MATH-manifold MATH can lie in MATH. If any component is a closed curve, then the complement of the closed curve is a simple pair of pants (i. e. a REF-punctured sphere) so all arc components of intersection must be parallel to each other. If there are no closed curves, the arcs of MATH fall into (at most) three classes of parallel arcs in MATH. We will assume for the purposes of this argument that MATH consists of three such classes of parallel arcs in MATH; if there are fewer classes of parallel arcs, the same method works, but more easily. Abstractly, the three families of parallel arcs of MATH in MATH give MATH the structure of a hexagon MATH, in which opposite sides are connected via arcs of intersection that are parallel in MATH. See REF . Each end of such an arc of intersection lies in a meridian disk of MATH, corresponding to a point of MATH; if two ends of arcs of intersection lie in the same meridian, we say that the ends have the same label. CASE: Opposite ends of the same arc of MATH cannot have the same label. Proof of REF : Since MATH always crosses each meridian with the same orientation, a normal orientation induced on the intersection arc by a normal orientation of MATH would have to have opposite direction at each end of the intersection arc. CASE: Suppose MATH and MATH are intersection arcs parallel in MATH, connecting opposite sides MATH and MATH of the hexagon MATH. Suppose further that the labels of MATH at MATH and MATH at MATH are the same. Then both are MATH-labels and all labels lying between the ends of the MATH on one of the MATH are MATH-labels. On the other side, between the ends of the MATH, there is exactly one subsegment of MATH-labels. Proof of REF : If there were a counterexample, choose MATH and MATH to be as close as possible (among parallel arcs of intersection in MATH) among all such counterexamples. They cannot be the same intersection arc, by REF . We now show they cannot be adjacent intersection arcs in MATH. For if they were, then the segments of MATH and MATH that lie between them would correspond to parallel segments of MATH on MATH. This is obvious unless the component of MATH on which the segments of MATH lie is MATH, i. e. the segments are colored MATH. But even if the segments are colored MATH, then the fact that this is a counterexample forces both segments to be colored MATH, or both MATH or both MATH, so they are in fact parallel on MATH. Now follow a standard argument that traces its origins to CITE or CITE: Consider the union of the sphere MATH, the single MATH-handle subsection of MATH corresponding to the segments in the MATH between the intersection arcs, and a MATH-handle whose core is the rectangle in MATH lying between the two intersection arcs. (See REF .) This defines a NAME splitting of a punctured Lens space MATH, lying in MATH, clearly an impossibility. Since MATH and MATH are not adjacent, we can consider the intersection arcs MATH and MATH adjacent to MATH and MATH but closer together. Since the intersection arcs MATH and MATH are not a counterexample, either the MATH have different labels at MATH or they have the same labels but the count of segments colored MATH or MATH between them changes. Either outcome is only possible if the segments MATH between MATH and MATH on MATH are colored MATH, for both MATH and at least one, say MATH, is colored MATH or MATH, say MATH. Then MATH is colored either MATH (and the labels of the MATH at MATH are different for MATH) or MATH is also colored MATH. Consider first the former case, MATH is colored MATH. This immediately implies that both of the initial labels were MATH-labels. If another segment colored MATH lies between MATH and MATH on MATH then an intersection arc adjacent to it would be a counterexample closer to MATH than MATH is. So we deduce that all labels on MATH between the MATH are MATH-labels. Because MATH and MATH are a counterexample, another segment colored MATH must lie between MATH and MATH on MATH. Then there is also one colored MATH between MATH and MATH. In order for an intersection arc adjacent to it and the intersection arc MATH not to be a counterexample, there must be a segment on MATH even closer to MATH that is colored MATH. Between it and the segment colored MATH lies every label colored MATH. Opposite to this segment on MATH every label is colored MATH, since no MATH or MATH color appears on MATH between MATH and MATH. So for any of these intersection arcs, whatever the label is on MATH, there's a parallel intersection arc with that label on MATH and between them lies no label MATH or MATH. This creates a closer together pair of intersection arcs that are a counterexample, a contradiction. See REF . If instead MATH, like MATH, is colored MATH, then the labels of the MATH at MATH are the same for MATH and, since these arcs are not a counterexample, they must be MATH-labels, all the labels on one side between the MATH must be MATH-labels, but on the other side there must be a switch to MATH-labels. But this produces exactly the same contradiction as before: all the labels on one side are MATH-labels and an entire sequence of MATH-labels appears on the other side. Hence there can be no counterexamples, completing the proof of REF : Segments corresponding to MATH (or MATH) cannot appear on opposite sides of MATH (including corners). Proof of REF : If MATH appeared on opposite sides, it would immediately contradict REF . Suppose MATH appeared on opposite sides of MATH. Choose those occurrences that are closest together (as measured by arcs between them) and let MATH and MATH denote the arcs adjacent to those occurrences of MATH that are closer together. Let MATH. If the number of arcs between the MATH is less than MATH it would contradict REF , since neither side could then have any MATH-labels. If the number of arcs between them is no less than MATH, then the fact, from REF , that one side consists entirely of MATH-labels would ensure that another label MATH occurs even more closely to the opposite MATH, contradicting the choice of MATH segments that are closest together. Suppose finally that there are between MATH and MATH arcs between them. Then, by REF , on one side between them will be a segment colored MATH or MATH and on the other a segment colored MATH. Moreover, the arcs MATH adjacent to these segments and closer together would have the same MATH-label but would have no MATH-labels between them on either side, again contradicting REF : Two segments, one corresponding to each of MATH cannot occur on the same side of MATH (including corners.) Proof of REF : Let MATH. If both types of MATH occur on one side, say MATH then that side is incident to at least MATH arcs and contains a segment of type MATH or MATH. Also, from REF , the opposite side MATH can be incident to neither type of MATH-interval. Since the MATH are incident to the same number of arcs, this means that MATH contains some segment of the form MATH or MATH, say MATH. In fact, it must be of the form MATH since otherwise its having length MATH would force a MATH segment to appear. For the same reason, the MATH segment in MATH must be opposite a segment in MATH that lies between a MATH label and the MATH label. The arcs MATH adjacent to the MATH and MATH segments and closer together would have the same MATH-label but would have no MATH-labels between them on either side, again contradicting REF . Following REF , we can think of each side of MATH as either a MATH-side, a MATH-side or blank, depending on whether a copy of MATH or no MATH at all appears. Moreover, by REF opposite sides can't both be MATH-sides or both be MATH-sides and (a crucial point) following REF , at least MATH-sides have signs, alternating around MATH as MATH. Combining these facts, the only possible signing of the sides of MATH is, in order, MATH (with some orientation of MATH). Now the fact that the two adjacent blank sides have no sign whereas their adjacent sides have different signs means that the total number of arcs intersecting those two blank sides must be less than MATH. On the other hand, the adjacent sides opposite these blank sides have signs MATH and MATH which guarantees that their combined length is greater than MATH. The contradiction proves the lemma. See REF . |
math/0106017 | Perform a standard NAME move on MATH so that afterwards the new MATH-graph MATH presents MATH as a MATH quasi-cable. Since MATH was disjoint from MATH this has no effect on MATH and, since MATH . REF applies. |
math/0106017 | Suppose, as in the proof of REF , MATH. The case MATH is already settled, so we assume MATH. Set MATH, necessarily even; then MATH and the number of edges of MATH is MATH. Of the intervals in MATH, MATH lie on segments of MATH, MATH contain MATH in their interior, and exactly one contains MATH. Call the last the ``special" component. We now consider how these intervals are distributed around the hexagon MATH described in the proof of REF above. First note that he two vertices corresponding to the corners of the hexagon (when reidentified to give MATH) either lie on the same or opposite sides of MATH, depending on whether MATH is even or odd, and this in turn determines whether the number of intersections of any edge of MATH with MATH is even or odd. The upshot is that every edge of MATH intersects MATH with the same parity and that parity is determined by the parity of MATH. The combinatorial argument gets easier as MATH gets larger, since intervals cooresponding to the same components of MATH appear more often. So for brevity we just do the case MATH (and of course MATH) and merely outline the argument. (In fact the argument for these cases also instantly gives as well all cases in which MATH is a multiple of MATH or MATH.) When MATH there are MATH segments in MATH and it follows that, except perhaps for the special component, centers of opposite edges in MATH represent the same segment of MATH, for there are exactly as many intersection points with MATH (namely MATH) going one way around MATH between them as the other. At most one of these three opposite pairs contains the special component, so the other two display lens spaces MATH in MATH, a contradiction. The only way to avoid this contradiction is if only two (opposite) sides of MATH intersect MATH, so all arcs in MATH are parallel in MATH, and the special component appears in the center of this band of parallel arcs. This only transfers the contradiction: if we let MATH be the annulus which is the complement in MATH of the single band containing all the parallel arcs of MATH then it is easy to see that MATH lies on the boundary MATH of the punctured solid torus obtained by attaching a single component of MATH to MATH. Moreover, the components of MATH are oriented so that they form parallel circles in MATH. This implies that MATH, together with an annulus in MATH, form a NAME bottle in MATH, another (related) contradiction. When MATH then MATH so the number of edges in MATH is MATH. There are two cases: If each edge of MATH intersects MATH the same number of times (as our parity discussion guarantees will happen when MATH), then each triple of corners of the hexagon (corresponding to a vertex of the punctured torus when it's reassembled) represent the same interval of MATH, or perhaps the special segment. At least one of the triples doesn't contain the special segment, and the hexagon in MATH cut off from MATH by the arcs of MATH adjacent to these corners, together with MATH and the segment of MATH the hexagon is incident to describe the spine of a Lens space MATH, a contradiction. If some edges of MATH intersect MATH more often than others (so MATH and so MATH), then consider a longest pair MATH of opposite edges of MATH (length here is shorthand for the number of points of intersection with MATH). Picturing these opposite sides as the top and bottom of the hexagon, consider the distance in MATH between the left ends of the MATH. Our choice of edge guarantees that it is less than MATH the circumference of MATH, that is MATH. Similarly for the right hand ends. It follows readily that there are at least two rectangles in MATH, cut off by a pair of adjacent arcs of MATH running between the MATH that either make up part of the spine of a MATH or contain the special component. At most one can contain the special component, leading to the same contradiction. |
math/0106017 | Here the intervals of MATH that are incident to MATH constitute two adjacent ``special" components. Since we can take MATH (it's necessarily odd, since the hypothesis guarantees that its ends lie on opposite sides of MATH) then MATH. Then the combinatorial arguments of REF applies with little change. |
math/0106017 | We need only consider the case MATH. Let MATH, necessarily odd. Then MATH. In fact, MATH segments lie parallel to segments of MATH, MATH segments (in sequential pairs) are incident to the point MATH and one of these sequential pairs (called the ``special pair") are incident to the point MATH. Much as before, we set MATH as the most difficult but also roughly representative cases. When MATH there are an odd number of arcs in MATH, since MATH is odd and so there are an odd number of arcs incident to each edge. The extra ``special" segments in MATH mean that the intersection arc of MATH that lies in the center of each pair of oppsoite sides of MATH does not have its ends at the same point of intersection of MATH with MATH (an immediate contradiction via its normal orientation) but it does mean that adjacent to each such central arc, at opposite ends, are segments of MATH that are parallel to the same segment of MATH. This would exhibit, as usual, the absurd MATH. This contradiction is only avoided if each side of MATH intersects MATH in exactly one point. But this means that MATH, so MATH as required. When MATH then, since the number of arcs in MATH is MATH, not all edges of MATH intersect MATH the same number of times. We may as well restrict to the case MATH so there are at least MATH arcs in MATH. If there is a single pair of longest edges, each must then intersect MATH at least MATH times. It's easy to see in this case, that wherever the adjacent special edges lie, they cannot disrupt the existence of at least one pair of opposite intervals of MATH (among those lying in these longest sides) that correspond to the same component of MATH and so constitute part of a Lens space MATH. Similarly, if there are two pairs of opposite longest sides, then each must be of length at least MATH and, wherever the adjacent special components lie, they cannot disrupt the existence of a similar Lens space contradiction from at least one pair of opposite longest sides. |
math/0106017 | We have seen that a sphere just below MATH cuts off an upper disk from MATH and that a sphere just above MATH cuts off a lower disk from MATH. The seminal point of thin position (see CITE) is that there cannot simultaneously (even at a critical point of MATH on the interior of MATH) be both an upper and a lower disk, for these disks could be used to push a maximum of MATH below a minimum, thinning MATH. Hence there is a generic height MATH between MATH and MATH for which the level sphere MATH cuts off neither an upper nor a lower disk from MATH. But this means there can be no arcs of MATH which are inessential, for an outermost such inessential arc would cut off either an upper or a lower disk from MATH. |
math/0106017 | Let MATH and MATH be the highest maxima of, respectively, the cycles MATH and MATH. Similarly, let MATH and MATH be the respective lowest minima of these cycles. Let MATH and MATH. Since MATH is in both cycles, we know that any point on MATH lies below MATH and above MATH, so MATH. Since MATH, MATH and MATH are (near) the heights of, respectively, maxima and minima of MATH. Choose a level sphere as in REF between MATH and MATH. By construction, such a level sphere lies at a height between the maximum and minimum of each of the cycles MATH and MATH and so intersects both of them. Now apply REF . |
math/0106017 | The proof is analogous to that of CITE. Suppose that MATH is not in bridge position. Then there is a level sphere MATH that lies between a sequential pair of critical levels for MATH, a maximum just below MATH and a minimum just above MATH. Maximally compress MATH in the complement of MATH. The resulting meridional planar surface MATH is incompressible in the handlebody MATH so each component is parallel to a subsurface of MATH (see CITE). Since the boundary components of MATH are meridians of MATH, each component of MATH can be completed to a sphere in MATH, and the piece of MATH lying in one of the balls bounded by that sphere is an unknotted tree (possibly just an arc) in the ball. For concreteness, choose an innermost such ball MATH and suppose it lies above MATH. Then every arc of MATH lying in MATH has at least one maximum in MATH. If MATH is a single arc MATH, then isotope that arc to lie in MATH. The arc in MATH can be chosen to be disjoint from those disks which are the results of the compressions that created MATH from MATH, so in fact then MATH lies in MATH. After this isotopy, the width of MATH is reduced (since a maximum has been pushed below whatever minimum lay just above MATH) and that of MATH is not increased (since each arc of MATH still has at most one maximum.) This argument shows more: any arc of MATH must have exactly one maximum in MATH, for otherwise a disk of parallelism between that arc and MATH (guaranteed by CITE) could be used to reduce the number of critical points on MATH, contradicting the assumption that MATH is in thin position, hence in minimal bridge position. If the tree MATH contains only maxima (including perhaps MATH-vertices), then pushing one to MATH would push it below the minimum that we know lies (elsewhere) just above MATH, again thinning MATH. So we know that MATH contains at least one minimum and, since it can't be a minimum of MATH, it must be a MATH-vertex, with ends of MATH descending into it. Similarly, if both vertices are in MATH then both must be MATH-vertices. Since MATH is a tree, at most one of MATH lies in MATH so we can assume that, say, MATH intersects MATH. Also MATH intersects MATH since otherwise MATH would contain a minimum in MATH. Let MATH be the arc in MATH consisting of the end of of MATH and the end of MATH at the MATH-vertex. Then the parallelism between MATH and an arc on MATH, guaranteed by NAME 's theorem CITE describes how to pull the end of MATH down to change the vertex into a MATH-vertex. This thins MATH. |
math/0106017 | The proof follows the same line of argument as the proof of the main theorem of CITE. We only need to verify that the argument there does not interfere with the thin presentation of MATH. In fact, the argument here is simpler because some of the more complicated steps in CITE are required only after a step that, in our case, clearly thins MATH or shows that e. CASE: MATH is vertical. So suppose MATH thinly presents MATH but is not in extended bridge position. NAME noted above that MATH is a tunnel number one knot so we know that MATH is in bridge position. So if a maximum of MATH lies below a minimum, either the maximum is a MATH-vertex or the minimum is a MATH-vertex, or both. So there are at most two level spheres with the property that each lies just below a minimum and just above a maximum. Let MATH be the sphere or pair of spheres with this property. Compress MATH as much as possible in the complement of MATH and call the result MATH. A path in MATH between meridians is regular if the corresponding path in MATH is embedded. It is shown in CITE that there is a disk MATH in MATH whose boundary is the union of a path MATH in MATH and an arc MATH in MATH. Moreover the interior of MATH is disjoint from MATH and either MATH is a regular path that is disjoint from some meridian of MATH or MATH has both its ends at the same point MATH of MATH and runs once around either MATH or MATH. Consider each possibility in turn. CASE: The path MATH is a regular path that is disjoint from some meridian of MATH. Say MATH lies below MATH. By general position we can assume that MATH is disjoint from the disks in MATH which are the remains of the compressing disks of MATH, so in fact MATH lies on MATH. If MATH does not pass through a vertex of MATH then just use MATH to isotope the arc of MATH that contains MATH to MATH. During the isotopy, as MATH perhaps passes through MATH (though not through MATH), MATH may get thicker, but once it reaches MATH it will have been thinned, since all that remains of its internal critical points is one minimum, which will have been brought up above the level of the maximum just below MATH. The move similarly cannot thicken MATH. Essentially the same argument applies even when MATH passes through a vertex. MATH is used to slide an end of one of the edges incident to the vertex down to MATH perhaps thereby just extending MATH and not affecting the bridge structure of MATH. In any case MATH is thinned and MATH is not thickened. CASE: MATH has both its ends at the same point MATH of MATH and runs once around either MATH or MATH. Much as in the previous case, MATH can be used to move the cycle MATH or MATH together with the end of MATH between MATH and the cycle to MATH. Unless the cycle was already vertical, this move (once the cycle is tilted again to restore genericity) will thin MATH and will not thicken MATH. So we can assume the cycle is vertical. This means we are done, unless in fact the cycle is MATH and MATH. This case only arises if MATH intersects MATH but not MATH since if it is disjoint from both, we can appeal to CITE directly to get a disk as in REF . (Unless, of course, some component of MATH intersects MATH in exactly one point of MATH. But then MATH would be planar.) Now note that, unless the maximum just below MATH is a MATH-vertex maximum of MATH the move on MATH just described pushes a minimum of MATH (on MATH) past a maximum of MATH, contradicting the thin position of MATH. We deduce that the one and only maximum just below MATH is in fact a MATH-vertex maximum of MATH. If MATH consists of more than one sphere, we could repeat the same argument, just using the component to which we have not just pushed MATH. But that would lead to the contradiction that the MATH-vertex maximum of MATH also lies just below the other plane. We deduce that MATH is a single plane. We have shown then that, aside from the base MATH-vertex of MATH, only minima lie below MATH. Moreover, just above MATH is a minimum and at no other level does a maximum lie below a minimum. It follows that MATH is in extended bridge position. It remains to show that MATH is disjoint from some dividing sphere. Much of the proof mimics CITE. We suppress most of the technical details, except to note that many of the technical problems do not arise in our context. Most importantly, if MATH is a dividing sphere and there are disjoint lower and upper caps, then pushing a vertical cycle which is, say, an extended maximum down past a minimum would immediately thin MATH, even if (as discussed in CITE) passing other maxima might thicken MATH. We deduce that in our context disjoint lower and upper caps cannot arise for elementary reasons. In any case, the upshot of the argument in CITE is that there is a dividing sphere MATH that cuts off from a meridian disk MATH of MATH both an upper disk MATH and a lower disk MATH. Moreover the interior of each is disjoint from MATH. Consider the components MATH and MATH of MATH to which MATH and MATH are incident. If neither MATH nor MATH contain vertices, or if they have no ends in common, or if together they contain at most one vertex and they have a single end in common, then it is easy to use MATH and MATH to push a maximum down past a minimum, contradicting the thinness of MATH or MATH. Note in particular that if the boundary of one of the disks, say MATH, goes once around MATH, although the move described may thicken MATH (compare CITE) it does push a minimum of MATH (namely the minimum of MATH) past a maximum of MATH and so would violate the assumption that MATH thinly presents MATH. We now proceed to dispose of the other cases. Suppose that, say, MATH, contains a vertex and that MATH and MATH have two end meridians in common. We can assume MATH contains only one vertex, else MATH is disjoint from MATH and we are done. Then MATH contains a vertical cycle. If that cycle is MATH or MATH we are done, so we'll assume it's MATH and that MATH. Either MATH and MATH can be used to make MATH (indeed MATH!) thinner or they can be used to isotope MATH into MATH. This last move not only makes MATH non-generic, but it may thicken MATH if MATH winds around MATH. Nonetheless, we persist, inspired by the proof of CITE. That argument shows that, once MATH is level, so the solid torus neighborhood MATH divides MATH into two disks, an innermost disk component of MATH in MATH or a disk cut off by an outermost arc of MATH in MATH can be used to push a maximum (respectively, minimum) of MATH (possibly the maximum near the end of MATH at MATH) down (respectively, up) through the level of MATH. Afterwards, MATH can be tilted slightly to restore genericity and thereby to remove the extra bridges of MATH that may have been introduced when MATH was made perfectly level. Since a maximum has been pushed down (or a minimum up) past MATH, it follows that MATH (indeed MATH, since MATH) has been thinned, the usual contradiction. The possibility remains that MATH and MATH each have a single vertex and they also have a single end in common (they can't have two ends in common since the result would be a cycle in MATH containing both vertices.) Their common end must be a point of MATH, since in MATH that is the only arc that connects the two vertices; in particular, MATH is monotonic. In this case, the disks MATH and MATH either could be used to thin MATH (an immediate contradiction) or they can be used to make MATH level. If neither component MATH or MATH contains all of MATH, then the move simply levels MATH. Once again, this move makes MATH no longer generic and may also thicken MATH, for MATH may wind many times around the edge MATH. But we continue anyway, inspired this time by the proof of CITE. The argument has a number of subcases, but all result in the following conclusion: a maximum (say) of MATH or MATH (possibly contiguous to an end of MATH) can be pushed down to the level of MATH (or below, if it is not contiguous). Once this is achieved, tilt MATH slightly to restore genericity, but leave the pushed down maximum at (or below) the lower end of MATH. The result is a thinning of MATH (indeed MATH). The final possibility is that the move just described levels all of MATH because, say, MATH contains MATH. In this case it seems that the move might pull MATH past other maxima lying below it, thickening MATH so that, after MATH is tilted to restore genericity, MATH actually ends up thicker. Nevertheless, it is argued in CITE that in fact this does not happen, or at least, if it does, the extra thickness (and more) can immediately be removed by a move analogous to that described above when MATH was levelled. This is established by a somewhat complicated combinatorial argument on MATH. We won't repeat the argument here. The upshot is that either MATH ends up thinner, a contradiction, or there were in fact no maxima between MATH and MATH. But in this last case, all the minima (including the base MATH-vertex of MATH) lie below all the maxima, so MATH is in non-extended bridge position. (Only the base vertex of MATH lies between MATH and the level plane MATH for this bridge presentation.) Moreover, all of MATH lies below MATH, verifying the proposition in this case as well. |
math/0106017 | Following REF we only need to consider the case in which MATH is in possibly extended bridge position and MATH is disjoint from a dividing sphere. Suppose first that MATH is in fact in (non-extended) bridge position and, with no loss of generality, suppose MATH lies above the dividing sphere. Then MATH ascends from the lower MATH-vertex and either is monotonic or it has one internal minimum and descends into the other MATH-vertex as well. It's easy to move from one position to the other without affecting the width of either MATH or MATH, so we'll assume for concreteness that MATH is monotonic. By (perhaps) twisting around the other two ends at the lower MATH-vertex we can ensure that the meridian disk for the associated MATH-graph, namely the pre-cable disk in MATH that runs the length of MATH, is disjoint from the descending disk incident to MATH given by the bridge structure. See REF . Then the NAME move has no effect on the bridge structure (hence the width) of MATH, nor the width of MATH: a pair of MATH-vertices with an edge between them is replaced by exactly the same thing. Now assume that MATH has an extended maximum, say. Since the extended maximum contains a vertical cycle, we are done immediately unless the vertical cycle is MATH and MATH. Since MATH is disjoint from the dividing sphere (say it lies above), it runs monotonically from a MATH-vertex to the MATH-vertex base of MATH. Since MATH the knot MATH has a maximum at the MATH-vertex. Now find, somewhere below the MATH-vertex, a level sphere MATH as in REF , so every component of MATH is essential in the NAME surface MATH. Then REF applied to the associated MATH quasi-cable shows that MATH. |
math/0106017 | Following CITE, MATH can be slid to become either an unknotted loop, and we are done, or MATH is a level edge, with its ends incident to the top two maxima (say) of the thinly presented MATH. (In the latter case, regain a generic positioning by slightly perturbing MATH from its level position, changing it to a monotone edge connecting two MATH-vertices.) If MATH is MATH-bridge, it's easy to see that MATH contains an isotopic copy of MATH (compare CITE), and we are done. If MATH is not MATH-bridge and the invariant MATH then defined in CITE is not MATH, it follows from CITE that MATH may be isotoped into MATH and we are done. Consider finally the case MATH for MATH a splitting sphere as in REF . The fact that MATH means that, the pair of meridians MATH cuts off a wave of MATH based at the meridian MATH, say, and that wave intersects MATH in a single arc. In particular (as above) the wave disk can be glued to a subdisk of MATH to get a non-separating meridian MATH of MATH that is disjoint from the wave disk and intersects MATH in a single arc. Appropriately twist the two arcs of MATH that descend from the bottom MATH-vertex of MATH (equivalently, choose an appropriate set of descending disks) as discussed in REF (see REF ) so that the descending disk is disjoint from MATH. Then the standard NAME move on MATH, using MATH not only converts MATH to a graph MATH that thinly presents MATH as a MATH quasi-cable, it does it without thickening MATH, for one pair of MATH-vertices is just replaced with another. (See again REF ). |
math/0106017 | As noted above, according to CITE there is a minimal genus NAME surface MATH for MATH that is disjoint from MATH. The rest follows from REF . |
math/0106017 | Suppose first that the vertices lie on the same side of a dividing sphere MATH, so, with no loss of generality, both are MATH-vertices, say. Then each edge intersects MATH in an even number of points. Since MATH is vertical, together the edges intersect MATH in MATH points. Hence one of these two edges, say MATH, is disjoint from MATH. This means that MATH is the edge that descends from the higher MATH-vertex. MATH can't also descend from this vertex, so MATH intersects MATH. If MATH intersects MATH in two points then MATH is vertical. If MATH intersects MATH in four or more points then there are bridges above and below MATH made up of interior subarcs of MATH, as required. Suppose next that the vertices lie on opposite sides of MATH, so one vertex is a MATH-vertex and the other is a MATH-vertex, and each edge intersects MATH in an odd number of points. Since MATH is vertical, each of these edges intersects MATH in a single point. If MATH also intersects MATH in a single point then every cycle is vertical. If it intersects MATH in three or more points, then there are bridges above and below MATH made up of interior subarcs of MATH, as required. |
math/0106017 | REF notes that MATH. REF shows that MATH is in bridge position. Suppose first that the cycle MATH is vertical. We apply REF , using MATH for MATH. If all cycles are vertical then of course we are done. By hypothesis, neither MATH nor MATH is disjoint from a dividing sphere. So we may assume, following REF , that there are bridges above and below a dividing sphere made up entirely of interior subarcs of MATH. We can of course arrange that the lowest maximum and highest minimum are these bridges. Now choose a level sphere MATH as in REF . Since MATH is vertical and MATH intersects both edges, MATH intersects each edge MATH in a single point. The result then follows from REF . Similarly, suppose the cycle MATH is vertical. Again apply REF this time using MATH for MATH. If all cycles are vertical we are done. By hypothesis, neither MATH nor MATH is disjoint from a dividing sphere so we may arrange that the lowest maximum and highest minimum are from bridges that lie entirely in MATH. Now choose a level sphere MATH as in REF . Since MATH is vertical and MATH intersects both edges, MATH intersects each edge MATH in a single point. The result then follows from REF . |
math/0106017 | The conclusions make sense, since we know from REF that MATH must be in bridge position. If a sphere just above the highest minimum cuts off a lower disk, and a sphere just below the lowest maximum cuts off an upper disk, then some level sphere between them is a critical sphere. This condition is guaranteed unless the highest minimum (or the lowest maximum) is a MATH-vertex (respectively, MATH vertex) with, via the wave condition, an end of MATH descending (respectively, ascending) from the vertex. So suppose the highest minimum, say, is a MATH-vertex with an end of MATH descending. Let MATH be any dividing sphere for MATH. If any component of MATH below MATH is a simple arc, its minimum could be pushed higher than the MATH-vertex, eliminating the problem, so we can assume that all components of MATH below MATH contain vertices. If there is only one such component and it contains a single vertex, then each of the cycles has at most one minimum and so each is vertical. If there is only one such component and it contains both vertices, then the edge between them must be incident to the higher vertex from below, hence that edge is MATH. Moreover, both cycles containing MATH have exactly one minimum, and so both are vertical. The remaining case is when the vertices are in separate components, each lying below MATH, i. e. both of them MATH-vertices. Let MATH denote the higher MATH-vertex. Push the regular minimum on the component containing MATH up to a height just below MATH. Now slide the end of MATH ascending from MATH down to the regular minimum and back up the other side. This has no effect on the width of MATH (since, for example, it doesn't change the number of bridges) but it alters the arrangement of the edges around MATH. In particular, afterwards the end of MATH ascends from MATH so, by the wave condition, we can assume that just above MATH, a level sphere cuts off a lower disk from MATH, and so somewhere between that level and that of the lowest maximum there is a critical sphere as required. See REF . |
math/0106017 | Let MATH be the critical sphere. Let MATH and MATH be the components of MATH to which the lower and upper disks MATH and MATH are incident. Suppose first that one of these components, say MATH, has no vertex (so MATH is a regular minimum). If MATH also has no vertex, then MATH could be thinned, a contradiction. If MATH has two vertices, then it contains an edge disjoint from MATH and we are done. If MATH has one vertex then, since every circuit in MATH has two vertices, MATH and MATH have at most one common end point on MATH. Moreover, by the wave condition, the path MATH has at least one end incident to an end of MATH in MATH. Then MATH and MATH describe how to slide the end of MATH in MATH down to MATH while simultaneously isotoping all of MATH up to MATH. If the slide of the end of MATH is down an end of MATH that would thin MATH; otherwise the effect of the slide is to sew together the ends of MATH extending the end of MATH down to or below the level of the minimum at MATH. (See REF .) This thins MATH (though not necessarily MATH), possibly by changing a MATH-vertex into a MATH-vertex. This contradicts the assumption that MATH is a thinnest such graph. The remaining case is that MATH and MATH both contain a single vertex. As above, MATH and MATH can be used to slide the ends of MATH in MATH and MATH up and, respectively, down, until they lie in MATH. This would thin MATH unless MATH crossed MATH in exactly one point (an end of both MATH and MATH). In the latter case, MATH was monotonic and the slide would level MATH by isotoping it into MATH. If the levelled MATH becomes a loop, then originally it was part of a vertical cycle, and we are done. This is also true if one of the other edges of MATH has a single interior critical point. If neither of these cases occur, then the argument of CITE can be used (as it was in REF ) to move the levelled MATH up (or down) to connect two maxima or two minima. Afterwards MATH is no wider but MATH is thinned, a contradiction completing the proof. |
math/0106017 | Suppose first that some dividing sphere is a critical sphere. Then following REF , if the critical sphere intersects all the edges, then some cycle involving one of MATH is vertical. If it's the cycle MATH we're done. If it's either of the other two cycles, apply REF , observing that either some edge is above a dividing sphere or the dividing sphere will intersect both edges of any vertical cycle. If no dividing sphere is a critical sphere then apply REF . If all the cycles in MATH are vertical then of course we are done. If the wave is based on MATH so MATH then REF says MATH is vertical, as required. Finally, suppose MATH. If MATH then we are done, since MATH is vertical. So suppose MATH and, following REF , MATH is disjoint from a dividing sphere. Find a dividing sphere MATH as in REF and apply REF . |
math/0106017 | Since the lowest maximum is a regular maximum and MATH runs along the edge that contains it, a dividing sphere MATH just below the lowest maximum cuts off an upper disk that is incident to MATH. Similarly, a dividing sphere MATH just above the highest minimum cuts off a lower disk that is incident to MATH. Since some (hence every) dividing sphere intersects MATH more often than it intersects MATH, every dividing sphere cuts off some disk, either upper or lower, incident to MATH. Since at MATH there is a lower one, and at MATH there is an upper one, and at every height between there is one or the other, it follows that at some height there is both an upper and a lower disk incident to MATH. (As usual, this height may be at a saddle tangency of MATH with an interior point of MATH.) |
math/0106017 | We may as well assume MATH lies above MATH, so the vertices are MATH vertices. By thinness we can assume that MATH lies on the MATH-punctured sphere component MATH of MATH lying above MATH. We can think of the components MATH as edges of a graph on MATH, with vertices the four meridian boundary components. That is, if we label the meridian components of MATH as MATH, MATH, MATH, MATH in the obvious way, then the components of MATH consist of a single arc connecting MATH to MATH, a single arc connecting MATH to MATH and MATH arcs connecting MATH to MATH. The MATH arcs are either all parallel or comprise two families of parallel arcs, separating in MATH the points MATH. See REF . Consider the possibilities for MATH: There is only one path in MATH that is disjoint from MATH and has ends at meridians MATH and MATH. And it's parallel in MATH to an arc of MATH. Hence if MATH were that path, MATH could be thinned, using MATH and MATH. Similarly for paths from MATH to MATH. There are (at most) two paths from MATH to MATH disjoint from MATH, each traversing MATH once, and similarly two paths from MATH to MATH. If MATH is any of these paths, then the first conclusion of the lemma follows. Now suppose MATH is the path in MATH (available only if all MATH arcs are parallel) that is disjoint from MATH and runs from MATH to MATH. Suppose, to begin with, that MATH does not also run from MATH to MATH but rather has at least one end at another point of MATH. Consider the MATH eyeglass graph MATH obtained from MATH by a NAME move along MATH, using as the new meridian a neighborhood of MATH (see REF ). Then MATH describes a descent of the new bridge edge MATH for MATH down to MATH. Simultaneously, MATH describes how to move a minimum of (now) MATH above or at least to the level of MATH. In particular, MATH is in bridge position, and a dividing sphere necessarily intersects MATH. Moreover, MATH is thinner than MATH since, in effect, a maximum has been pushed below a minimum. Now put MATH in thin position. According to REF (exploiting the fact here that MATH is a MATH eyeglass) either MATH becomes vertical (which implies that MATH was unknotted) or MATH or the associated MATH-graph, namely MATH can be made as thin as MATH. The last contradicts the hypothesis and the first two possibilities are the conclusions we seek. Now suppose MATH, like MATH, runs from MATH to MATH (so, in particular, MATH is vertical). We could construct MATH as just described; the construction places MATH level in MATH. Unfortunately, when MATH is tilted to restore genericity, MATH is again a level sphere for MATH and no thinning will have occurred. So a different argument will be used, this one exploting the fact that MATH and MATH are disjoint from MATH. Consider the situation once MATH and MATH have been used to put MATH into the plane MATH. Let MATH be a thickened regular neighborhood of MATH and consider the two longitudinal circles MATH. It is easy to see what they are: Before the edge MATH is levelled it intersects MATH in two points (actually the points MATH). The MATH are obtained by banding the corresponding pair of meridians to itself using both MATH and MATH. The important point for our purposes is that MATH. Consider where these points lie. If they both lie on the same longitude, then an arc of MATH they cut off is inessential in the annulus component of MATH in which it lies, and so it can be removed by an isotopy. On the other hand, if one point lies on each longitude, consider the algebraic intersection of MATH with the disk component MATH of MATH bounded by MATH, say. One point is the point MATH. All others come from intersections of MATH, i. e. intersections of the old MATH. Each point MATH contributes MATH points to MATH and they are all of the same sign. So the total algebraic intersection of MATH with MATH is MATH. This contradicts the fact that MATH bounds MATH in MATH. We are left with the conclusion that indeed MATH can be isotoped off of the two longitudes, so MATH only intersects the top of MATH. But in that case, consider MATH. It's easy to see that any component of intersection that is inessential in MATH can be removed (else MATH could be thinned). So every component of MATH is essential. Now simply attach the bottom annulus of MATH to MATH to obtain a sphere intersecting the original MATH only in MATH. Then REF shows MATH. The remaining case is if MATH has one end at each of the meridians MATH and MATH. Exclude any such arc that is parallel to a subarc of MATH, since if MATH were such an arc it would either violate the thinness of MATH or (if MATH has ends at the same meridians) exhibit that MATH is vertical. But the only way that MATH can be disjoint from MATH, connect MATH to MATH, and not be parallel to a subarc of MATH is if all MATH arcs of MATH with ends at these meridians are parallel and MATH is one of the other two paths connecting MATH and MATH. Although these paths are not parallel to a component of MATH in MATH, they are sufficiently parallel in the component of MATH on whose boundary MATH lies to derive the same contradiction. Here is the argument: If the ends of MATH are also at MATH and MATH then together MATH and MATH show that the cycle MATH is vertical. So suppose no end of MATH is at MATH, say. One arc MATH of MATH running from MATH to MATH is visibly isotopic to MATH in the MATH-ball component of MATH on whose boundary MATH lies. (See REF ) The isotopy moves the end of MATH across the meridian MATH and so destroys the property that MATH lies on MATH. Nonetheless, once MATH is moved to MATH then there is no obstruction to pushing MATH below MATH via MATH while simultaneously pushing arcs of MATH parallel to MATH above MATH using MATH. The result is a thinning of MATH, violating the hypothesis. |
math/0106017 | The proof is mostly similar to that of REF . The relevant figure is modified as shown REF , with MATH arcs running between meridians MATH and MATH. There is only one regular path from MATH to MATH (or from MATH to MATH) that is disjoint from MATH. These paths do not cross MATH and so, if MATH is such a path, we could use MATH and MATH to thin MATH (without altering the wave condition), extending MATH down to MATH. The only regular path between MATH and MATH (or MATH and MATH) is parallel to an arc of MATH, so MATH cannot be such a path. If MATH and MATH runs between MATH and MATH we use the same argument as was used for paths from MATH and MATH previously. Suppose finally that MATH runs between MATH and MATH (or, symmetrically, MATH and MATH runs between MATH and MATH.) We would like to use the same trick as was used previously, namely let MATH describe a NAME move that converts MATH into an eyeglass graph. There is a subtle complication, however. Note that the eyeglass graph MATH that is created by this move is in fact a MATH eyeglass, not a MATH eyeglass as before. In particular, the associated MATH-graph MATH to MATH is not MATH, which presented MATH as a MATH quasi-cable. Rather MATH presents MATH as a MATH quasi-cable (or just as MATH if MATH). Nonetheless, we are still in a position to get the same contradiction with REF (for, after all, MATH was chosen to be thinnest among all appropriate MATH-graphs and it was shown that such a graph is no thicker than MATH) as long as we verify that MATH is still appropriate. In other words, we need to verify that MATH still satisfies the wave condition. The pre-cable disk is easy to identify: whereas in the previous argument it was (essentially) a thickened vertical arc in MATH together with a meridian of MATH, here it is obtained from a thickened vertical arc in MATH and a meridian of MATH by adding a half-twist. (See REF .) Notice that MATH intersects this meridian in MATH points, so it is the meridian of MATH (or MATH if MATH) and, whether the wave was based at MATH or at MATH, the extra half-twist guarantees that it is afterwards based at the new meridian, as required. (The language when MATH is: the slope of the wave is still finite.) |
math/0106017 | The upper disk cannot be of type MATH or MATH because a word that is positive in MATH contains neither MATH nor MATH. The result then follows from REF . |
math/0106017 | That MATH follows from REF . We will show that the annulus MATH obtained from MATH-compressing MATH to MATH, using the disk MATH from the splitting sphere, gives rise to an annulus satisfying the conditions of REF . The result then follows from that corollary, possibly by way of REF . Let MATH be the arc described above, which we may as well slide to minimize intersections with the meridians of MATH. In particular, for MATH a level sphere between the lowest maximum and the highest minimimum of MATH, the ends of MATH will lie on the MATH-punctured sphere component MATH of MATH. Let MATH be the word in MATH represented by MATH. Because the wave of MATH is based at MATH CASE: any occurrence of the letter MATH (respectively, MATH) in MATH is preceded and followed by the letter MATH (respectively, MATH). CASE: any occurrence of the letter MATH in MATH is followed by MATH or MATH and preceded by MATH or MATH and CASE: any occurrence of the letter MATH in MATH is followed by MATH or MATH and preceded by MATH or MATH. In particular, by a choice of orientation for MATH, we can assume that MATH is positive (say) in MATH and MATH (e. g. perhaps MATH). Exploiting these facts, together with the symmetries of the diagram, we have three essentially different ways in which the ends of MATH can lie in MATH. These are shown in REF . Now orient MATH from left to right, and read off the words corresponding to MATH (boundaries oriented to be parallel in MATH, not antiparallel). All three cases can be expressed in one of the following two forms, with the details depending on where the ends of MATH are incident to the word given by MATH, namely MATH. Note that, in all cases, the word MATH begins and ends with the letter MATH. The choice of labelling of the components MATH of MATH is made so that the word corresponding to MATH is positive in MATH. Form REF: CASE: MATH CASE: MATH . Here MATH and MATH. Form REF: CASE: MATH CASE: MATH . Here MATH, MATH and MATH. Note that the geometric length of MATH is greater than that of MATH exactly when, for annuli of the first form, MATH and, for annuli of the second form, MATH. In either case, MATH contains an occurrence of the letter MATH, so MATH runs along MATH. If MATH intersects a dividing sphere MATH just twice, then MATH is vertical, and we are done. If MATH, then a regular maximum and minimum (which we can take, respectively, to be the lowest maximum and the highest minimum) lie in the interior of MATH. It follows, then, from REF that we are done if MATH for annuli of the first form, or MATH for annuli in the second form. So we now consider only the alternative possibilities. CASE: MATH . In this case, it follows roughly from the same argument used in REF that there is a dividing sphere MATH so that no arc of MATH cuts off an outermost disk incident to MATH. Indeed, as above, a high dividing sphere cuts off an upper disk incident to MATH and a low dividing sphere cuts off a lower disk incident to MATH. There can't be both an upper and a lower such disk incident to MATH by REF . So at some level there is a dividing sphere MATH that cuts off no outermost disk incident to MATH. But since the geometric lengths of MATH are equal, this implies that MATH intersects MATH only in spanning arcs. We will argue that this is impossible. Suppose (with no loss) that MATH and hence MATH lie above (not below) a dividing sphere. The spanning arcs determine a correspondence between subintervals of MATH and MATH. To be precise, say that a component of MATH and a component of MATH are opposite if there is a (``square") component of MATH incident to both. More generally, a segment of MATH and a segment of MATH are opposite each other if a spanning arc of MATH runs between the beginning of each and another spanning arc runs between the end of each. For example, we first observe that no segment of MATH that is part of a MATH interval (meaning that it comes from an occurrence of a letter MATH, i. e. that it runs along MATH with an orientation opposite to that of MATH) can lie opposite a segment of MATH that is part of a MATH interval. For if this occured then it is easy to see that somewhere on the entire length of the MATH and MATH intervals there would be components of MATH that are opposite to each other on MATH but lie on the same component MATH of MATH. Since they have opposite orientation in MATH and MATH the square component of MATH connecting them can be attached to the punctured solid torus MATH to create a punctured Lens space MATH, which is absurd. Similarly no segment of MATH that is part of a MATH interval can lie opposite a segment of MATH that is part of a MATH interval. This immediately rules out the first form above (again, only under the assumption that MATH) since, following these observations, the only possible segment opposite the transition segment from MATH to MATH in MATH would be exactly a transition segment from MATH to MATH in MATH, and that would lead to the same contradiction. Ruling out the second form (in which no letter MATH appears) is only slightly more complicated. We will focus on the segments of MATH that lie on the middle component of MATH; that is, on the arc component that is equidistant (measuring distance by intersection with MATH) from both ends of MATH. Note that this segment of MATH lies below MATH. Label corresponding segments of MATH by MATH. Note that none of these can be opposite to a segment of MATH lying on MATH (e. g. those segments in MATH that correspond to the transition between different letters) since the segments on MATH lie above MATH. This remark allows us to be a bit casual about length arguments in the next few paragraphs, since it means that inequalities will usually imply strict inequalities. The first observation is that no label MATH occurs opposite to any part of a MATH interval in MATH, for this would allow us to display a Lens space MATH in MATH, as noted above. Let MATH be the segment between the first and last labels MATH in MATH. Then MATH. Since no label MATH lies opposite to MATH it follows that opposite to MATH is part of a segment in MATH corresponding to MATH. Notice that if a label MATH in MATH, say, is opposite to any part of a MATH interval in MATH, than the relation is reciprocal: the label MATH in the MATH interval on MATH is opposite to the MATH interval in MATH containing the original label MATH. (This is not deep, just a reflection that we have taken MATH to lie half way along MATH and so it appears half way along each MATH or MATH interval.) Because MATH there is at least one more label MATH in MATH then there are MATH labels in MATH, not counting the labels MATH in MATH. It follows that some label MATH in MATH is opposite a part of a MATH-interval in MATH, so an entire half of a MATH-interval in MATH is opposite to a subsegment MATH of a single MATH-interval in MATH, for MATH. And, as we've seen, another copy of MATH lies opposite to a subsegment of MATH. This works just as well for construcing a Lens space in MATH as having the half of the MATH-segment itself opposite to MATH. (Two rectangles are glued together along the boundary interval they share on a component of MATH corresponding to part of the MATH intervals.) So we arrive at the same contradiction as previously. CASE: For an annulus of the first form, MATH or, for one of the second form, MATH. We will arrive at the same sort of contradiction, though the argument is a bit more complicated. Again, with no loss, we assume that MATH lies above a dividing sphere and that both the lowest maximum and the highest minimum of MATH are regular critical points on MATH. This implies that just below the lowest maximum (i. e. at a high dividing sphere) the dividing sphere cuts off an upper disk from MATH that is incident to a regular maximum (namely the lowest maximum). Then no outermost disk cut off from MATH by this high dividing sphere can be a lower disk, by thin position. On the other hand, a low dividing sphere does cut off a lower disk from MATH. So there is a height MATH such that a dividing sphere just below MATH cuts off a lower disk from MATH but just above MATH a dividing sphere cuts off no lower disk. But any dividing sphere must cut off some outermost disk, since the geometric lengths of the words represented by the two boundary components of MATH are different. It follows from REF then that either MATH is unknotted or just above MATH all outermost disks cut off of MATH by the level sphere MATH are incident to MATH and, moreover, for each such disk MATH the arc MATH is of the form MATH. That is, there are at most two outermost disks in MATH and they are incident to the subarcs MATH and MATH labelled MATH and MATH of MATH. In this position, the total number of non-spanning arcs in MATH, all of them incident to MATH and each of them cutting off a disk containing either MATH or MATH, is MATH, since each arc has two ends. Since MATH is less than the distance between the ends of MATH (first form) and less than the distance between the ends of MATH (second form), each non-spanning arc cuts off a disk containing exactly one of MATH or MATH and so each arc is parallel to one of the MATH. Let MATH denote the number parallel to MATH, so MATH. Let MATH denote the segment in MATH that is still incident to spanning arcs, and let MATH be its length. See REF . For obvious pictorial reasons, we'll refer to the part of MATH containing the collection of arcs parallel to MATH as the MATH peninsula. CASE: MATH is of the first form. In this case note that MATH, so MATH. CASE: The entire MATH interval of MATH is disjoint from MATH. Say the entire MATH interval lies on the MATH peninsula. Then MATH is entirely made up of powers of MATH and its length is at least MATH. Since MATH is made up of MATH-intervals, at most MATH of the length of the segment opposite to MATH can lie in MATH intervals (half at each end), so in particular, the segment opposite to MATH contains a MATH-segment longer than MATH. In particular, if MATH denotes the terminal segment of MATH at the end of the MATH peninsula, there is also a copy of MATH lying opposite MATH. Since across the ends of the MATH peninsula the orientations of MATH and MATH coincide (that is, the orientations of MATH and MATH are reversed by the folding along MATH) whereas across from MATH they disagree, we obtain the standard Lens space contradiction. See REF i Subcase REFa.ii: MATH is completely contained in the MATH interval. We have MATH and the longest MATH-segment in MATH is of length MATH, since the wave assumption ensures there are no proper powers of MATH in MATH (i. e. no powers greater than one). Hence the total length of MATH-segment REF opposite MATH is at least MATH. In particular, the length of the MATH-segment REF across from MATH must be greater (by at least MATH) than the length of the ends of MATH not contained in MATH. This implies that some subsegment of MATH is opposite a copy of itself, leading to the standard Lens space contradiction. See REF ii. CASE: The MATH interval is completely contained in MATH. The argument of the previous subcase applies as long as MATH, so assume that MATH . To avoid the standard Lens space contradiction, across from the MATH interval in MATH is a segment comprised entirely of MATH-intervals. The largest power of MATH is MATH and MATH is even longer than that, so at least one end of MATH is across from a copy of MATH that lies completely in the segment across from MATH. If the ends of MATH and that copy of MATH coincide, we get a Lens space contradiction via the terminal segment of MATH. If the copy of MATH extends out beyond MATH then we get a Lens space contradiction with the end of MATH adjacent to MATH in MATH. See REF iii. CASE: One end of MATH is contained in MATH, say, and the other end is contained in MATH. Suppose first that the segment opposite the end of MATH at MATH is part of a MATH-interval. Since the longest MATH-intervial comes from MATH and MATH it follows that the MATH-interval ends somewhere in MATH and is followed by a MATH-interval. We get the standard Lens space contradiction with either MATH or MATH, depending on whether the MATH-interval ends across from a point in MATH or a point in MATH. See REF i. Next suppose the segment opposite the end of MATH at MATH is part of a MATH-interval and let MATH denote that part of the single MATH-interval that lies across from MATH. (So MATH is followed either by a MATH-interval or another MATH-interval.) Abusing notation somewhat, let MATH denote that part of the MATH interval that lies on the MATH peninsula. If MATH is longer than MATH we get a Lens space contradiction between MATH and MATH. See REF ii. If MATH is shorter than MATH then, since MATH and MATH is the highest power of MATH, there is more MATH-segment across from MATH than just MATH. If there are some MATH intervals between MATH and the additional MATH-segment, then the far (right-hand in the figure) end of the MATH-segment gives the same Lens space contradiction. So we conclude that MATH is immediately followed by another copy of MATH, which we'll call MATH. If MATH is shorter than MATH, as must happen if most of MATH lies in MATH, we get a Lens space contradiction between MATH and the end of MATH. See REF iii. If MATH is longer than MATH we get a Lens space contradiction, comparing the end of MATH across from a MATH segment with the end of the MATH peninsula. See REF iv. Case REFb MATH is of the second form. Observe then that MATH so MATH. It follows that the distance between the outermost labels MATH (see REF ) is greater than MATH. The rest now follows almost exactly as for the second form in REF . |
math/0106017 | Suppose MATH. Take two parallel copies of MATH and band them together along the part of MATH that does not lie between them. The result is a disk MATH that is disjoint from MATH and separates MATH, leaving one of MATH in the boundary of each of the solid tori components of MATH. Label these solid tori (correspondingly) MATH and denote by MATH the link whose core circles are MATH. Note that MATH is a longitude of MATH and MATH is a MATH cable of MATH, some MATH. MATH is visibly a non-hyperbolic (because of MATH) tunnel number one link (the tunnel is dual to MATH). These have been classified (compare CITE): In particular, MATH is the unknot. But the core of MATH is MATH, as required. If MATH (so MATH) the argument is symmetric, interchanging MATH and MATH. |
math/0106017 | With no loss of generality, assume that MATH lies above the dividing sphere and that MATH is monotonic. The proof now has the same features as the proofs of REF and we use similar notation. Let MATH be the arc as previously, again slid to minimize intersections with the meridians of MATH. Let MATH again be the MATH-punctured sphere lying in MATH, on the neighborhood of the component of MATH that lies above MATH and contains MATH. For the purposes of the argument, we will assume that all MATH arcs of MATH that run between the two copies of MATH in MATH are parallel. If in fact there are two families of parallel arcs, the argument is essentially identical, except for one difference which is noted below. Then MATH consists of three families of arcs. One family of MATH arcs runs between the two copies of meridian MATH in MATH; two arcs each run from a copy of MATH to a copy of MATH. It is natural to parameterize slopes of proper arcs on MATH using these arcs of MATH. Indeed, the discussion will now, in some sense, be parallel to that of CITE. We declare the family of MATH arcs to have slope MATH and the second pair to have slope MATH. An outermost disk of MATH cut off by the pair of meridians MATH defines a wave in MATH; the wave assumption guarantees that such an outermost disk also intersects MATH, so we conclude that the wave has finite slope MATH in the coordinates just defined by the arcs MATH. Moreover MATH is odd since a wave in MATH will be based at each copy of a single meridian (either MATH or MATH) (see CITE for details). An argument will now show that either MATH is unknotted or MATH. The arc MATH is disjoint from the wave. Suppose to begin that MATH intersects both meridians MATH and MATH. Then some arc component MATH of MATH has one end on a copy of each of MATH and MATH in MATH. Then the slope MATH of MATH is odd and can't differ from MATH, the slope of the waves, since if it did its ends would have to run between the base of both waves, i. e. different copies of the same meridian. On the other hand, since MATH is disjoint from MATH, which has one parallel family of arcs of slope MATH and two non-parallel arcs of slope MATH, we have MATH (hence MATH) and MATH. Since we are given that MATH it follows that MATH as claimed. Next suppose that MATH intersects MATH but never MATH. Then any component MATH of MATH that has both ends on copies of MATH in MATH will have slope MATH (since it's disjoint from MATH). The two terminal segments of MATH in MATH will then each have one end on different copies of MATH. But then they can't have their other end (i. e. the end points of MATH) on the same side of MATH. For if they did, then either the arcs cross in the ``square" component of MATH in which they lie or one must be part of a segment of MATH of slope MATH and the other of slope MATH. See REF . (If not all MATH arcs of MATH that run between the two copies of MATH in MATH are parallel, the slopes of both these arcs could be MATH, still sufficient to deduce that this is the slope of the wave.) The only remaining possibility (to avoid the conclusion that MATH) would be that MATH is disjoint from MATH. But that would imply that the boundary of the annulus MATH obtained by MATH-compressing MATH to MATH intersects MATH in a single point. Then by REF MATH is unknotted. So we continue, assuming that the slope of the waves in MATH is MATH. Now apply a NAME move, replacing the meridian of MATH (slope MATH) with the disk whose boundary has slope MATH. This redefines the MATH curve as MATH, presenting MATH as a MATH quasi-cable. Moreover it is no thicker than MATH and it satisfies the wave condition, since the meridian of the new edge MATH has been chosen to be disjoint from the wave. As usual, we can ensure that the NAME disk has no effect on the bridge structure of MATH, so MATH remains in thin position. Recall REF . |
math/0106017 | Without loss we assume MATH lies above a dividing sphere and MATH is monotonic. Suppose MATH. Then MATH has a maximum at the lowest vertex. Find a level sphere as in REF . The result contradicts REF . Having established that MATH, switch the labels of MATH and apply REF . |
math/0106017 | That MATH follows from REF . As previously, let MATH be the annulus obtained from MATH-compressing MATH to MATH using the disk MATH from the splitting sphere. Without loss of generality, assume MATH lies above the dividing sphere. If there are no regular maxima of MATH then MATH is vertical, and we are done. If there is a regular maximum of MATH, we can assume it's the lowest maximum. In that case, a level sphere just below the lowest maximum cuts off an upper disk from MATH and a level sphere just above the highest minimum cuts off a lower disk from MATH. So either some dividing sphere cuts off both an upper disk and a lower disk or some dividing sphere MATH intersects MATH only in essential arcs. In the former REF finishes the proof. The rest of the proof is an extended proof by contradiction. We will show that it is impossible for a dividing sphere to intersect MATH only in spanning arcs. Let MATH be the arc as previously, again slid to minimize intersections with the meridians of MATH. Let MATH again be a level sphere between the highest minimum and the lowest maximum, and MATH again be the MATH-punctured sphere lying in MATH, again supposing (with no loss) that both vertices are MATH vertices, so MATH lies above MATH. Let MATH be the word in MATH represented by MATH. The wave condition now guarantees that (with the right choice of direction for MATH) MATH is positive in MATH and MATH (including as usual the possibility that MATH is the empty word). If the wave is based at MATH then no proper power of MATH occurs in MATH (i. e. no power greater than one); if it's based at MATH then no proper power of MATH occurs in MATH. Exploiting these facts, together with the rotational symmetry of the diagram, we have several essentially different ways in which the ends of MATH can lie in MATH. Representative samples indicating that MATH can begin or end on any letters are shown in REF . We have oriented MATH from left to right. Symmetric figures in which MATH are not shown. The waves themselves are also not shown, but they are described (except for details of how their ends lie near MATH) by the requirement that they are disjoint from MATH, which is shown. Note also, that the number of arcs of MATH connecting the copies of MATH is now MATH. The case MATH is special, since in this case there are no arcs connecting the copies of MATH. Variants that arise in this case are shown in REF . The resulting words corresponding to MATH and MATH can be put in two forms: CASE: MATH CASE: MATH CASE: MATH CASE: MATH . Here MATH . Now apply thin position. It follows immediately that there is a level sphere MATH so that no arc of MATH cuts off an outermost disk incident to MATH, since there can't be both an upper and a lower such disk incident to MATH simultaneously by REF . In particular, MATH consists only of spanning arcs, so MATH . This immediately implies that MATH. Thus it also forces MATH. In the second form above observe then that by regularity of MATH, MATH begins and ends with the letter MATH so by regularity of MATH, MATH. In fact we will show that the second category above does not arise and the first is limited to the case MATH. The letters MATH and MATH do not occur in the words determined by MATH. Consider first the form CASE: MATH CASE: MATH. We will show that MATH so MATH. Suppose, with no loss of generality, that MATH so the length of the segment MATH (the same as the length of the segment MATH) is at least MATH. The easy case is when the wave is based at MATH so there are no repeating MATH's in MATH. Then opposite (in MATH) to the segment MATH is a complete collection of MATH-labels (perhaps separated by a single letter MATH). On the other hand, opposite (in MATH) to MATH must be part of a MATH-segment. Combining the two easily gives a Lens space contradiction. It seems to be harder to establish a Lens space contradiction in the case when the wave is based at MATH, so there are no repeating MATH's in MATH. For the first time we need to use the graph MATH in the dividing sphere MATH whose vertices are the points of intersection of MATH with MATH (we will call these points the MATH and MATH-vertices in MATH) and whose edges are the arcs MATH, viewed both as spanning arcs of MATH and as edges in MATH. Ends of edges at the same vertex in MATH will be said (usually in MATH) to have the same label. The label will be a MATH-label, MATH-label, MATH-label or MATH label depending on whether MATH at that point (as oriented so that MATH is positive in MATH and MATH as above) is passing through a MATH- or MATH-vertex in the direction that the MATH-graph is oriented or in the opposite direction. Our use of the graph MATH in this lemma will be modest, mostly as a book-keeping device. Once the lemma is established we will need to examine MATH much more seriously. Orient all edges in MATH to point from MATH to MATH. We first claim there is an oriented path that begins with an edge incident to a MATH-label and ends with an edge incident to a MATH-label. To see this, remove the MATH-vertices from MATH (but not their incident edges), and let MATH denote those edges, and the MATH-vertices they pass through, that are part of an oriented path beginning with an edge incident to a MATH-label. If there are MATH-vertices in MATH and there are MATH occurrences of MATH in the word MATH then MATH has MATH ends of edges from MATH. Yet only MATH ends of edges in MATH could be in MATH but not at MATH-labels. So some ends of the edges must be at MATH-labels, as claimed. Consider then the shortest oriented paths beginning with a MATH-label and ending with a MATH-label. Among all shortest such paths, pick a path MATH whose ends are closest in MATH as measured by the number of components of MATH that lie between them. We claim that that number is one; i. CASE: MATH begins and ends at the opposite ends of the same interval of MATH. First note that the ends of MATH can't be at the same MATH-vertex, because MATH would then be a loop in MATH whose normal MATH-bundle, as pieced together from neighborhoods of the edges in MATH, would not be oriented. So call the initial MATH-vertex MATH and the terminal MATH-vertex MATH. To be concrete suppose that, in a single MATH-letter of MATH, MATH precedes MATH (we'll say that MATH lies to the left of MATH in the oriented MATH). Unless the labels are precisely adjacent in MATH (which is our claim), we can construct a better path MATH as follows: Start at the MATH-label just to the left of the origin of MATH and construct a path by always using the edge that is one to the left (in MATH) of the edge in MATH. Notice first of all that the collection of edges MATH we have just described is indeed a path in MATH: Suppose MATH and MATH are successive edges of MATH and the edges to their left in MATH are MATH and MATH. We need to show that the end of MATH in MATH is at the same vertex as the end of MATH in MATH. (See REF .) This is obvious unless the end of MATH at MATH (and so the end of MATH in MATH) is the first label of a MATH-segment. But if it were then, since there are no repeating MATH's in MATH, the end of MATH would in fact be a MATH label and we would have found a shorter path. Having established that MATH is in fact a path in MATH, notice that it ends just to the left of the label MATH in MATH hence to the right of the label MATH in MATH. Hence we have found a path of equal length but with ends closer together. Having established that the ends of MATH represent adjacent intersections of MATH with MATH, carry through the above construction of MATH. Consider the sequence of squares that lie between the two paths. When glued together along the components of MATH that their edges in MATH represent, the result is a NAME band whose boundary lies on the level sphere MATH. This is impossible (and represents a different way of viewing a Lens space contradiction). Next consider the form CASE: MATH CASE: MATH . Since MATH and, by the Lens space argument, no part of the MATH interval can lie across from any part of a MATH-interval, it follows that no part of the MATH interval can lie across from any part of the MATH segment. In particular, across from the MATH interval must be a MATH-segment longer than MATH. This means that there are repeated MATH's in MATH, hence no repeated MATH's. What then lies across from the MATH interval? None of it can by part of a MATH-segment, by the Lens space argument, so it must be part of a single MATH-interval. Then MATH. On the other hand, MATH. We conclude that MATH and immediately across from MATH is precisely a single MATH. But if this is the case, then each segment corresponding to a letter in one boundary component will lie exactly opposite a segment corresponding to a single letter in the other boundary component. This is clearly a very special case, and resolving it will begin the process of understanding how to use the graph MATH effectively. NAME already identified (across from MATH) a term in MATH of the form MATH. Across from that same term MATH in MATH must be three letters, at least one of which is also a MATH. There are then two letters MATH exactly aligned opposite each other, exhibiting that each MATH-vertex is the base of a loop in MATH. Ask then what lies in MATH exactly opposite MATH. If any of it is a MATH interval then this fact, together with the established fact that across from the MATH interval is a MATH interval, gives a Lens space contradiction. If any of it is a MATH interval then we will have exhibited that every MATH-vertex in MATH is also the base of a loop. Then, since every vertex is the base of a loop, some such loop will contain no vertices in its interior. The following lemma shows that this is impossible. Any loop in the graph MATH must have vertices in both disks into which the loop divides MATH. A loop without such vertices in its interior would give a problematic MATH-compression of MATH. To see the problem, consider the base of an innermost such loop, that is, the subarc MATH of the meridian to which MATH is MATH-compressed, an arc in MATH. The arc MATH could not be a simple cocore of the band along MATH, otherwise MATH would have been compressible. On the other hand, if MATH were incident to a copy of MATH in MATH from the side in MATH opposite to the band MATH, then the MATH-compression would turn MATH into an essential disk in MATH whose boundary is disjoint from MATH, which is also absurd, for if we attach a neighborhood of the disk to MATH, MATH would lie on the resulting unknotted torus and MATH would be a NAME surface in the solid torus complement, forcing MATH to be trivial. The only remaining possibility is that MATH has both ends incident to the parts of MATH that come from MATH. But MATH crosses each meridian always in the same direction, so two such crossings can't be the ends of a MATH-compression. REF completes the proof of REF . We conclude that the words corresponding to the boundary components of the annulus MATH are exactly CASE: MATH CASE: MATH. In fact more can be said. The fact that there is no occurrence of the letter MATH in MATH or MATH means that neither end of MATH can lie on the segments of MATH connecting the two copies of MATH (when MATH). Then Suppose MATH. If the MATH segments MATH of MATH that connect the two copies of MATH are not all parallel in MATH then MATH begins and ends with the letter MATH. If they are all parallel then MATH begins or ends (perhaps both) with the letter MATH. If some MATH were incident to both ends of MATH then one component of MATH would represent the word MATH and the other one MATH. This is impossible since these words have different lengths. If some MATH contained a single end of MATH then one component of MATH would contain an occurrence of the letter MATH, contradicting REF . So the ends of MATH lie, one each, on the two components of MATH that are not among the MATH. The result follows easily (see REF ). The argument now proceeds by considering every possible type of word MATH. We begin by considering short words, then long words, then words of intermediate length. If MATH (e. CASE: MATH) then MATH is uknotted. If MATH then MATH and MATH is unknotted. If MATH (or is empty) then MATH intersects the meridian MATH in exactly one point, a point in MATH. If MATH then MATH by REF . Then MATH intersects the meridian MATH in exactly one point, a point in MATH. In both cases the result follows from REF . In view of REF we can and will restrict our attention only to words that contain both letters MATH and MATH. To deal with longer words it will be useful to generalize REF . To appreciate how, we examine the local structure of MATH. The key to organizing the information is to orient each edge of MATH, as was done briefly above, so that the edge, when viewed in MATH, points from MATH to MATH. This has the obvious consequence that any MATH-vertex has at least one edge pointing into it, since the word MATH contains the letter MATH, and each MATH-vertex has at least MATH edges pointing out, since the word MATH contains at least MATH occurrences of the letter MATH. Beyond these ends of edges, though, is one more pair at each MATH-vertex (respectively, MATH-vertex) for each occurrence of MATH (respectively, MATH) in MATH. (One of the pair is identified with the occurrence of the letter in MATH and the other with the occurrence in MATH.) Call these ends of edges the MATH-ends. At any MATH-vertex there is a single non-MATH edge pointing in plus a sequence of MATH-ends alternating between pointing in and pointing out. At any MATH-vertex there are MATH non-MATH edge pointing out and a collection of MATH-ends, the latter coming in pairs of adjacent ends, one pointing in and one pointing out. See REF . Between each pair of MATH-ends in MATH or MATH is an arc MATH that is a cocore of the band along MATH. Put another way, banding together the boundary components of MATH along MATH would recover MATH from MATH. Call that side the MATH-side of the MATH-end. Since occurrences of MATH or MATH in MATH occur always with the same sign, MATH crosses MATH always in the same direction, and similarly for MATH. It follows that, at any given vertex of MATH, the MATH-side of any (oriented) MATH-end is always to the right (or always to the left) of the end, as the end is oriented by the orientation of its edge. Moreover, since the ends of MATH are incident to the same side of MATH, that normal direction is well-defined, so if both ends of the same edge in MATH are MATH-ends, then the MATH-side is the same side at both ends. Combining these facts we discover that, throughout any one component MATH of MATH, the MATH-side of any MATH-end at any vertex lies always to the same side (say always to the right as the edge is oriented) of the MATH-end. Under these circumstances, note that a cycle in MATH that has no vertices or edges in its interior, and which moves clockwise around its interior (we call it a clockwise face), must have corners that are always on the MATH-side of MATH-edges. In particular, if such a cycle can be found, then its interior would, before the MATH-compression that changed MATH to MATH, correspond to a compressing disk MATH for MATH. (We know that MATH would be essential in MATH, since it crosses a proper arc in MATH, namely the one MATH-compressed to MATH, always in the same direction.) Such a compression, of course, violates our assumption that MATH is incompressible. Symmetrically, in a component of MATH for which the MATH-side of any MATH-end is to the left of the oriented end, there could be no counterclockwise cycle whose interior is empty (i. e. no counterclockwise face). Although it might not be easy to see if a given component of the graph MATH in MATH is ``right-handed" or ``left-handed" in this sense, it is possible to use the extreme regularity of MATH (guaranteed by the fact that all edges in MATH are parallel in MATH) to identify circumstances in which there are both clockwise and counterclockwise cycles in the same component of MATH with no vertices in the interior of either. That, then, forces one or the other to define a compression of MATH, a contradiction. For example, we have No two faces of MATH can be adjacent and have boundaries that are cycles. Since the cycles are adjacent, one is clockwise and one is counterclockwise. A disk component of MATH will be called a face. If the boundary of the face is a cycle we call it a face cycle. A clockwise (respectively, counterclockwise) face cycle will be called a clockwise (respectively, counterclockwise) face. A face incident only to MATH-vertices (respectively, MATH-vertices) will be called a MATH-face (respectively, MATH-face). Any MATH-face is a face cycle. If MATH any MATH-face is a face cycle. At any MATH-vertex exactly one end of an edge is not a MATH-end and it points into the vertex. Hence there cannot be two adjacent ends of edges pointing out, as there would be in a MATH-face that is not a cycle. No two MATH-faces can be adjacent. If MATH no two MATH-faces can be adjacent. Combine REF It is natural to seek features of the graph which guarantee the existence of cycles. The following lemma suggests a possibility. No distinct MATH-vertices can have edges pointing to (respectively, from) the same vertex. Suppose MATH are two edges in MATH with their heads, say, at the same vertex of MATH. Then the ends of the MATH in MATH are some multiple of MATH apart. (Recall that MATH.) The ends of the MATH in MATH are the same distance apart. So if both ends on MATH are at MATH-labels, they must be at the same MATH-label. It seems from this lemma that bigons may be prevalent. To be precise, define a bigon in MATH to be a pair of edges, each running between the same pair of vertices. A parallel bigon will be a bigon in which both edges of the bigon are oriented in the same direction. An anti-parallel bigon will be one in which the edges are oriented in the opposite direction, forming a cycle in MATH of length two. In the case of a parallel bigon we will denote the vertex from which the edges of the bigon point out as MATH and the vertex into which the edges point as MATH. Suppose there is a parallel bigon in MATH and let MATH be a disk in MATH that it bounds. Suppose in MATH there is an oriented path from MATH to MATH. Then in the interior of MATH there is a cycle that is disjoint from both MATH. With no loss we may assume that MATH contains no other parallel bigon, else we would focus on an innermost one. Since MATH contains both letters MATH and MATH, any vertex is incident to an edge pointing out and an edge pointing in; no vertex is a sink or source. Hence any component of MATH contains a cycle, so we may as well assume that every vertex in the disk belongs to the same component MATH of MATH as the bigon. Suppose, with no loss (as explained above), MATH contains no clockwise face. If there were an oriented path that runs from MATH to MATH inside MATH, then the closed disk would contain both a clockwise and a counterclockwise cycle. Consider an innermost clockwise cycle (perhaps passing more than once through the same vertex, but not crossing at such a vertex) and the disk MATH that it bounds. Suppose MATH contains a vertex. That vertex must be part of an oriented path in the interior of MATH. If that path forms a cycle completely in the interior of MATH we are done. If not (e. g. the ends of the path are at the same vertex of the cycle MATH) the path would cut off a clockwise cycle that would be even further in, a contradiction. (See REF .) So there is no vertex in the interior of MATH. Similarly, if there were an edge in the interior of MATH there would be a further in clockwise cycle. We conclude that MATH would have to be a clockwise face, which is impossible. Suppose there is a parallel bigon in MATH and let MATH be a disk in MATH that it bounds. Then in the closure of MATH (i. e. including the vertices MATH) there is a cycle that includes at most one of MATH. As in the proof of REF we can assume, with no loss, that MATH contains no other parallel bigon, no vertex is a sink or source, that every vertex in the disk belongs to the same component MATH of MATH as the bigon, and that MATH contains no clockwise face. If any end of MATH in MATH points out from MATH, then it is part of an oriented path (since no vertex is a source or sink). If the path ends in MATH we are done by REF . If not, it must contain a cycle not incident to MATH, and we are done. If the only ends of MATH in the bigon that are incident to MATH point into MATH then there can be no edges other than the bigon itself, since even a MATH-vertex can have at most two adjacent ends pointing in. (Recall that MATH-ends alternate between pointing in and pointing out.) So we may as well assume that no end of MATH in the interior of MATH is incident to MATH. In that case, any end lying in the the interior of MATH and incident to MATH must be part of a cycle in the bigon incident only to MATH and we are done. The possibility remains that the interior of the bigon is empty. In that case, at least one end of the edges of the bigon at each vertex is not a MATH-end (since MATH-ends alternate between pointing in and pointing out) so MATH is a MATH-vertex and MATH is a MATH-vertex. If neither end at MATH is a MATH-end then, considering how non-MATH-ends arise, necessarily MATH and the edges of the bigon, when viewed in MATH, are some MATH apart. But then they can't have their other end at the same MATH-vertex, since two ends in MATH with the same MATH-label are at least MATH apart in MATH. Hence we conclude that exactly one end of the bigon at each of MATH and MATH is a MATH-end. Necessarily their MATH-side is the same side and not the side in the bigon. Hence exactly one edge MATH in the bigon (the left one, say) has both of its ends MATH-ends and neither of the ends of the other edge MATH are MATH-ends. In particular, MATH connects, in the annulus MATH, a point in MATH corresponding to a point in the last letter of MATH, to a point in MATH that lies in the final syllable MATH of MATH. Now MATH; suppose the end of MATH in MATH is the MATH-th end in the final syllable MATH of MATH and the end in MATH is the MATH-th end in the final letter MATH of MATH. Suppose for concreteness that MATH. Now consider MATH. Since it's ends are the same as those of MATH, the end of MATH in MATH is the MATH-th end in some letter MATH in MATH. But the length of the terminal segment of MATH is of course the same in MATH as in MATH and so the end of MATH in MATH is the point exactly MATH later than the end is in MATH. That is, it is still part of the same letter MATH of MATH. (See REF .) This is a contradiction, since the other end of MATH is at a MATH-vertex. If MATH we get the same contradiction, using the initial segment of MATH instead of the terminal segment. With a little determination, we have more: The interior of any parallel bigon contains a MATH-vertex. With no loss we can assume that the parallel bigon is innermost. Suppose the disk MATH contained only MATH-vertices in its interior. By REF , MATH contains a cycle that passes through some vertices in its interior. Since the valence of each vertex is greater than two, there are adjacent faces in MATH. Then REF shows that MATH can't both be MATH-vertices. Similarly, all vertices in MATH are in the same component of MATH as the bigon. Let MATH denote the subgraph of MATH, lying in the interior of MATH, obtained by deleting all edges incident to MATH. Claim: MATH contains a cycle. Otherwise consider an oriented path from a source vertex MATH to a sink vertex MATH. All edges (of MATH) pointing into MATH must have their other ends at MATH and not the same vertex, since this would exhibit a further in parallel bigon. There must be at least two such edges, since by REF , MATH contains at least one occurrence of MATH so the word MATH contains at least two. This implies that there are exactly two edges pointing into MATH and one edge comes from each of MATH. Then REF implies that MATH can't both be MATH-vertices. Suppose that MATH were a MATH-vertex. We have already identified three edges pointing out from MATH: the edges of the bigon and an edge to MATH. Then three edges point out from MATH and at least two would have to go to the same vertex in the pair MATH, creating a further in bigon. So MATH is a MATH-vertex and MATH is a MATH-vertex. Now consider MATH. If an edge in MATH pointing out from MATH goes to MATH, then MATH, hence every MATH-vertex, has three edges pointing in. This would force MATH to be part of a further in bigon, a contradiction. So the edge pointing out from MATH goes to MATH and this edge, together with the path from MATH to MATH together with the edge from MATH pointing into MATH give an oriented path from MATH to MATH. Then REF provides a cycle in MATH, as claimed. Having established the claim we continue with the proof of REF . Note that, once we have a cycle in MATH we know that MATH contains a face hence, following REF , a MATH-face cycle. CASE: MATH contains only one occurrence of MATH. In this case each MATH-vertex has valence three - two ends of edges pointing in and one pointing out. (So, for example, MATH must be a MATH-vertex.) We have seen that MATH contains a MATH-face cycle MATH. Now notice that there are only two possible ``corners" of cycles at each MATH-vertex, since the valence is three; one set of corners occurs only in clockwise cycles and one in counterclockwise cycles. As a result, all the corners of MATH come from a single occurrence of the letter MATH in MATH and a single occurrence of the letter MATH in MATH. We claim this is impossible: Of all the vertices of MATH, let MATH be the one that is first encountered when passing along the oriented edge MATH, so, in any occurrence of the letter MATH in MATH, the label corresponding to MATH lies most to the left among all labels coming from vertices of MATH. Similarly, define MATH to be the last vertex of MATH that is encountered along the oriented edge MATH. We repeat: the edges pointing out from MATH and from MATH, when viewed in MATH, leave from the same MATH-interval MATH-in MATH and end in the same MATH-interval MATH in MATH. But, by definition of MATH and MATH, the edge pointing out from MATH in MATH, goes to a label to the left of MATH in MATH whereas the edge pointing out from the label MATH in MATH goes to a label to the right of MATH in MATH. This presents a clear contradiction: there are more ends of edges between the two edges in MATH than there are in MATH. See REF MATH contains three or more occurrences of MATH. In this case we will show that there are two adjacent MATH-faces in MATH, contradicting REF . Let MATH be a component of MATH and let MATH denote the number of vertices in MATH. If MATH are both MATH-vertices then at each MATH-vertex in MATH there can be at most two edges also incident to one of the MATH, one pointing in and one pointing out, since two different MATH-vertices can't have edges pointing toward (or away from) the same MATH-vertex REF and if one MATH-vertex had two edges pointing toward (or away from) the same MATH-vertex it would be a further in bigon. On the other hand, if either vertex (say MATH) were a MATH-vertex then at most two edges could be incident to both MATH and MATH, for otherwise MATH would have adjacent MATH-faces. So clearly MATH (the valence of each vertex is at least MATH) and at most MATH edges connect MATH to MATH in MATH. It follows that MATH contains at least MATH edges. The proof that this would provide two adjacent MATH-faces now follows from this simple observation: Claim: If MATH is a connected graph in the plane with MATH vertices and at least MATH edges then either MATH contains a trivial loop or two faces of MATH (i. e. compact complementary components, necessarily disks) are adjacent. Here's the proof of the claim. Let MATH be the number of edges. If any face is a monogon, we are done. So suppose every face has at least two edges. Either some edge is incident to two faces, and we are done, or the number of faces MATH. Then consider the NAME characteristic of MATH and all its faces: MATH, a contradiction. CASE: MATH contains exactly two occurrences of MATH. Suppose first that MATH are both MATH-vertices. As usual, for each of the vertices in the interior of MATH, there is at most one edge pointing from MATH to the interior vertex by REF and our assumption that MATH is an innermost bigon. Now consider a component MATH of MATH with, say, MATH vertices. We have just shown that there are at least MATH edges in MATH pointing into each MATH-vertex (for at most one edge pointing into the MATH-vertex comes from a MATH-vertex). Hence there are at least MATH edges in MATH. The proof now follows as in the previous case. Suppose finally that one of MATH is a MATH-vertex MATH and one is a MATH-vertex MATH. If, for some component MATH of MATH there are no edges (in MATH) running from MATH to the MATH-vertex, we are done much as before. Similarly, if there is at most one edge MATH in MATH pointing from MATH to MATH we are done: In MATH there are still at least MATH edges pointing into every vertex, except for the single vertex at the end of MATH. Hence there are at least MATH edges in MATH and we can still apply the combinatorial claim above. If there are at least two edges, say MATH and MATH, in MATH pointing from MATH to MATH then, since no adjacent ends at a MATH-vertex point out, there is another end of an edge MATH between the ends of the MATH at MATH. If MATH also goes from MATH to MATH then on either side of it are adjacent MATH-faces, contradicting REF . If instead it goes to another component of MATH then that component is cut off from MATH by MATH so no edge in MATH connects it to MATH, a case we have already established. REF immediately eliminates the possibility that MATH is a long word. Explicitly, we have: The letter MATH occurs at most once in MATH. Suppose MATH and MATH are two edges in MATH that point out from the same MATH-vertex. Then (echoing the argument of REF ) the distance between MATH and MATH as measured along either of MATH is some multiple of MATH. In particular, there are at most MATH candidates for MATH-vertices the other ends of MATH and MATH might be incident to, plus a MATH-vertex. If the letter MATH occurs more than once, then in MATH there are at least MATH occurrences of each MATH-label. And, for each MATH-label, we have just argued that there are at most MATH possible labels in MATH to which they can point, MATH of them MATH-labels and one a MATH label. Hence at least two of the edges point to the same label. This shows that every MATH-vertex is part of a parallel bigon. An innermost parallel bigon then would have to contain only MATH-vertices, contradicting REF . If MATH occurs in MATH then the letter MATH occurs at most once. Following REF we can restrict to the case in which MATH occurs exactly once in MATH. We will assume that MATH occurs MATH times and derive a contradiction. The structure of the proof depends on whether MATH or MATH. If MATH then any two occurrences of the same MATH-label or the same MATH-label in MATH occur a multiple of MATH apart. It follows that at least two edges pointing into any given MATH-vertex in MATH have their other ends at the same vertex. In other words, each MATH-vertex is contained in some parallel bigon. Thus an innermost parallel bigon contains only MATH-vertices. Any MATH-vertex has at least one edge pointing out that goes to a MATH-vertex (since there is only one occurrence of the letter MATH in MATH but two in MATH) and edges that point from MATH-vertices to a given MATH-vertex must all come from the same MATH-vertex by REF . It follows that there are at most two MATH-vertices in the the interior of the bigon. If there were only one, then the two edges coming out from it can't go to the same vertex, for that would form another parallel bigon, so one goes to each of the vertices forming the bigon. This would force MATH to be a MATH-vertex and MATH to be a MATH-vertex. Furthermore, MATH-vertices are of valence MATH, and the edge pointing into the interior MATH-vertex has nowhere to come from but MATH. The two adjacent face cycles contradict REF . See REF i. So there are exactly two MATH-vertices inside the bigon. As noted above, at least one edge pointing out from each of these MATH-vertices must go to a MATH-vertex, and edges can't point from different MATH-vertices to the same MATH-vertex. It follows that both MATH are MATH-vertices. If both of the edges pointing out from a MATH-vertex go to MATH-vertices then, since MATH, they would in fact go to the same MATH-vertex, contradicting our assumption that the parallel bigon is innermost. So each MATH-vertex has an edge pointing from it to the other MATH-vertex. In other words, the two MATH-vertices in the bigon are the vertices of a MATH-cycle, necessarily a MATH-face cycle. (See REF ii.) But this leads to the same contradiction as in the proof of REF . If MATH then we know from REF that MATH either begins or ends in MATH, so the words corresponding to MATH are (up to cyclic rotation) MATH and MATH respectively. Here MATH as usual. Suppose first that no edge in MATH runs from one MATH-vertex to another MATH-vertex. Then the MATH occurrences of any MATH vertex in MATH have their other ends in only the MATH possible MATH-vertices, so there is a parallel bigon at each MATH-vertex. An innermost parallel bigon then contains only MATH-vertices. This contradicts REF . Suppose then that some edge MATH in MATH has both ends at MATH-vertices. Consider the distance in MATH between copies of the same MATH-label, counting distance (i. e. intersection with MATH) along the arcs of MATH that don't intersect MATH. Measured on this side, the distance between any two copies of the same MATH-label in either component of MATH is a multiple of MATH. It follows that the MATH copies of the same label in MATH have their other ends at at most two labels in MATH, one a MATH-label and one a MATH-label. Thus in this case every MATH-vertex is part of a parallel bigon. Then an innermost bigon contains only MATH-vertices. The proof now follows as for the case MATH but is easier, since any MATH-vertex has MATH edges pointing out. At this point there are only two remaining words to consider: MATH and MATH. Eliminating these two requires a bit more detailed argument. MATH or MATH. The cases are symmetric, so without loss of generality suppose MATH. Then MATH is represented by the word MATH and MATH is represented by the word MATH. Let MATH. The type of contradiction depends on how the copies of MATH in MATH are aligned with each other. There are three cases. See REF , where the orientation of MATH is meant to be clockwise around MATH. CASE: The MATH-segment in MATH lies completely opposite a subsegment of the MATH segment in MATH. (REF i) Then symmetrically, the MATH segment in MATH lies completely opposite a subsegment of the MATH segment in MATH. Consider the collection of edges incident to the copy of the MATH-segment in MATH. Among those edges, every MATH-label occurs exactly once in MATH and once in MATH. It follows that these edges, when viewed in MATH, form a collection of MATH-cycles containing every MATH-vertex. Similarly there is a collection of MATH-cycles containing every MATH-vertex. An innermost pure MATH- or pure MATH-cycle can then contain no vertices in its interior and so must be a MATH-face or a MATH-face. Either way, the argument presented in REF presents a contradiction. CASE: The MATH-segment in MATH lies completely opposite a subsegment of the MATH segment in MATH. (REF ii) Then dually the entire MATH-segment in MATH lies completely opposite a subsegment of the MATH segment in MATH. This means that for each MATH-vertex, each of the MATH edges in MATH with that label in MATH can go to at most MATH different MATH-labels in MATH. It follows that every MATH-vertex is part of a parallel bigon. An innermost one can contain only MATH-vertices, contradicting REF The MATH-segment in MATH lies partly opposite an end of the MATH segment in MATH and partly opposite an end of the MATH segment. (REF iii) The argument in this case is a kind of degenerate variant of the argument in REF . Suppose, with no loss and as shown in the figure, that part of the MATH segment in MATH is opposite the beginning end of MATH, overlapping say on MATH edges. (MATH in the figure). Now consider any of the last MATH labels in the first occurrence of MATH in MATH and the corresponding label in the last occurrence of MATH. The distance between them is MATH in MATH. It follows that each of the corresponding MATH-vertices is part of a parallel bigon in MATH, each with a MATH-vertex for MATH. Consider an innermost parallel bigon in MATH and the disk MATH that it bounds. The interior of MATH must contain MATH-vertices, by REF , and it must also contain MATH-vertices, by the argument of REF of the proof of that Lemma. We claim that there is an oriented edge pointing from MATH into MATH and an oriented edge pointing out from MATH into MATH. To see the former, consider the MATH-vertex MATH which, among all MATH-vertices lying in MATH, is the first encountered by MATH. Then the corresponding label in the second copy of MATH in MATH lies across from the label of an earlier vertex in MATH, hence the label is that of MATH, since there is no alternative. So the edge in MATH between them must connect MATH to MATH, pointing toward MATH. A symmetric argument, using the last MATH-vertex MATH in the interior of MATH encountered by MATH, shows that there is an edge pointing from MATH into MATH. Since each MATH-vertex has valence MATH we have now accounted for all edges incident to MATH. In particular one of the two edges of the bigon has a MATH-end at MATH with the MATH side lying within the bigon. What's important here is not that one of the edges of the bigon has a MATH-end at MATH - that fact can be seen simply because one of the ends of the edges of the bigon, viewed in MATH, lies in the second occurrence of MATH in MATH (see REF iii), hence in the MATH section of the word MATH. What is important is that the MATH side of this edge lies in the interior of the bigon. But examining the figure again, we see that the other edge of the bigon lies in the first occurrence of MATH in MATH, hence in the MATH section of the word MATH. That is, the other edge has a MATH end at MATH. So the MATH-side of that edge lies outside of MATH. On the other hand, we've shown that some edge at MATH points into MATH from the interior of MATH and, since any MATH-vertex has only one edge pointing into it, that edge must be adjacent to the (only) MATH-corner at MATH, so that corner must be inside B. This contradiction proves the Lemma, hence the theorem. Having eliminated every possible word for MATH, we deduce that no dividing sphere can intersect MATH only in essential arcs, completing the proof of REF . |
math/0106017 | According to REF if neither of the outcomes above occurs, then there is an appropriate MATH-graph for MATH, thinly presenting it, say, as a MATH quasi-cable. Let MATH be a thinnest appropriate MATH-graph for MATH and, among all such possibilities, choose one with MATH maximal. According to REF MATH = REF and MATH is in bridge position. We claim that if MATH the cycle MATH is unknotted and, if MATH, one of the two cycles MATH is unknotted. This follows immediately from REF unless a dividing sphere is a critical sphere that is disjoint from some edge. Consider the possibilities for such an edge: If the disjoint edge is MATH then the claim is established by REF . If MATH is the disjoint edge then, since MATH has been chosen to have maximal MATH, it follows from REF that MATH and MATH is unknotted, establishing the claim. Similarly, if MATH is the disjoint edge then it follows from REF that the wave is based at MATH. Then REF establishes the claim. So the claim is established in all cases. Now let MATH be the unknotted solid torus neighborhood of MATH in MATH. Since MATH we can apply the ``vacuum cleaner trick": slide the ends of the MATH-handle corresponding to MATH along the arc MATH until MATH has been made disjoint from a meridian of MATH. At that point, MATH has become a tunnel for MATH and remains unknotted. |
math/0106017 | Let MATH be an unknotting tunnel for MATH. If MATH can be slid and isotoped to lie on a genus one NAME surface MATH then MATH is necessarily MATH-bridge (see REF). If not, then according to REF MATH can be slid and isotoped until it is an unknotted loop. The following argument (shown to me by Abby NAME) shows that then MATH is MATH-bridge on an unknotted torus. Let MATH denote the solid torus neighborhood of the loop, containing a short, MATH-parallel arc of MATH. Let MATH denote the arc of MATH that lies outside of MATH. Since MATH is an unknotted handlebody, it follows that the MATH-handle with MATH at its core constitutes a genus two NAME splitting of the solid torus MATH. Any non-trivial splitting of a handlebody (e. g. of MATH) is stabilized CITE, so in fact MATH is also parallel to MATH. This shows that MATH is MATH-bridge with respect to the unknotted torus MATH. CITE has proven the statement for this class of knots. |
math/0106018 | We first need to define the bundle gerbe morphism MATH which maps MATH . Define MATH covering the identity on MATH by sending MATH to the piecewise smooth path MATH given by MATH . Next, we need to define a MATH equivariant map MATH covering MATH and check that it commutes with the bundle gerbe product. So take pairs MATH and MATH where MATH and MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH respectively. Then we put MATH where MATH is the homotopy with endpoints fixed between MATH and MATH given by MATH . We need to check firstly that this map is well defined - that is it respects the equivalence relation MATH - and secondly that MATH commutes with the bundle gerbe products. So suppose MATH and MATH, where MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH and where MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH. We want to show that MATH . Therefore we want to show that for all homotopies MATH with endpoints fixed between MATH and MATH we have MATH . Note that if MATH is a homotopy with endpoints fixed between MATH and MATH and MATH is a homotopy with endpoints fixed between MATH and MATH, then by integrality of MATH we have MATH . Therefore we are reduced to showing that MATH . We have MATH . By the bi-invariantness of MATH, we get MATH, hence MATH which implies the result. Hence MATH is well defined. It is a straightforward matter to verify that MATH respects the bundle gerbe products. It remains to show that there is a transformation of the bundle gerbe morphisms MATH and MATH over MATH which satisfies the compatibility criterion over MATH. This has already been done above for the tautological bundle MATH-gerbe and the proof given there carries over to this case. |
math/0106018 | We will calculate the NAME four class of the bundle MATH-gerbe MATH and show that it is exactly equal to the NAME cocycle obtained by NAME and NAME in CITE and CITE. We then apply REF to conclude that this NAME four class is MATH. We calculate the NAME cocycle MATH as follows. First choose an open cover MATH of MATH relative to which MATH has local sections MATH. Since MATH is a fibration, we can choose sections MATH of the pullback fibration MATH. This is equivalent to choosing maps MATH such that MATH and MATH. Next we choose sections MATH. This amounts to choosing maps MATH such that MATH, MATH, MATH and MATH. Such maps MATH exist because MATH is simply connected. Define a section MATH of the bundle MATH by MATH. In a similar manner construct a section MATH of the bundle MATH. We make MATH into a section of MATH by using the associator section: MATH. Finally we define the cocycle MATH by MATH. We can get an explicit formula for MATH as follows: we choose a homotopy with endpoints fixed MATH such that MATH, MATH, MATH, MATH, MATH and MATH and we set MATH. This is just the integral of MATH over the tetrahedron shown in the following diagram, MATH as described in CITE and CITE. Thus our cocycle agrees with the cocycle defined by NAME and NAME. |
math/0106018 | Suppose the bundle gerbes MATH are given by triples MATH. We first construct the bundle gerbe MATH. Let MATH. Then the fibre product of MATH with itself over MATH is MATH. Suppose the bundle gerbe morphisms MATH are given by MATH. Define a map MATH by sending MATH to MATH. Let MATH. Define a MATH bundle MATH on MATH by setting MATH with projection map MATH induced by the various projections MATH. We want to show that the triple MATH is a bundle gerbe. We first define the product in MATH. This is a MATH bundle isomorphism MATH covering the identity on MATH which satisfies an associativity condition on MATH. Since MATH this amounts to finding a MATH bundle map MATH satisfying an associativity condition over MATH. Let MATH, MATH for MATH. Then MATH and MATH. Apply MATH to MATH. Then MATH. Using the bundle gerbe product in MATH we have that MATH . Let MATH denote the section of the MATH bundle MATH on MATH which descends to MATH. Using the bundle gerbe product in MATH again, we have that MATH . We define a product in MATH by sending MATH to MATH. We have to check that this product is associative. This follows easily from the following equation satisfied by MATH: MATH . This equation is a consequence of the coherency condition satisfied by MATH. Therefore MATH is a bundle gerbe. We now need to define the bundle gerbe morphism MATH. First of all we define a map MATH covering the identity on MATH. If MATH and MATH, then MATH. Since MATH this defines a map MATH. Now suppose MATH and MATH. So MATH. Hence applying MATH to MATH means that MATH. Therefore MATH . This defines a MATH bundle map MATH. It is not hard to check that this map commutes with the bundle gerbe products on MATH and MATH and hence defines a bundle gerbe morphism MATH. Similarly, one can define a transformation of bundle gerbe morphisms MATH which is compatible with MATH. |
math/0106024 | It suffices to show that MATH for all MATH. Let MATH. If MATH is any overlapping partition of MATH with MATH pieces, then (since MATH and MATH have an element in common) MATH will have a repeated entry and therefore the simplex MATH will be degenerate, which means it will represent REF in the normalized chains MATH. |
math/0106024 | It's easy to see that the action of MATH on MATH preserves complexity. The fact that MATH preserves complexity is immediate from REF . Now suppose that MATH and MATH have complexity MATH and choose a diagram MATH of type MATH: MATH . It suffices to show that MATH has complexity MATH, and for this we need to show that the restriction of MATH to MATH has complexity MATH whenever MATH is a two-element subset of MATH. There are two cases: either MATH is contained in some MATH or not. If MATH for some MATH then the complexity of MATH is equal to the complexity of MATH, and this is less than or equal to the complexity of MATH by REF (since MATH is order-preserving on MATH). If MATH is not contained in any MATH then the complexity of MATH is the same as that of MATH. But MATH, and the complexity of MATH is less than or equal to the complexity of MATH by REF (since MATH is order-preserving). |
math/0106024 | NAME has shown CITE that the operad of spaces obtained as the geometric realization of the nerves MATH is weakly equivalent to the operad of little MATH-cubes. Now apply the normalized singular chains functor to get a quasi-isomorphism of chain operads between MATH and MATH. The natural map MATH is a quasi-isomorphism of chain operads and the proof is complete. |
math/0106024 | For MATH let MATH be the subcomplex of MATH generated by the sequences that begin with MATH and have no other occurrences of MATH. In the proof of REF we constructed a chain homotopy MATH that gives a deformation retraction of MATH onto MATH. For MATH let MATH be the transposition MATH and let MATH. Then MATH is a chain homotopy that gives a deformation retraction of MATH onto MATH. If a subcomplex MATH of MATH is invariant under MATH then there is a deformation retraction of MATH onto MATH. In particular, if MATH is the first element of the total order MATH, then MATH is invariant under MATH and its deformation retract is isomorphic to MATH where MATH is the unique order-preserving monomorphism that does not have MATH in its image, MATH is the restriction of MATH to MATH and MATH is the pullback of MATH to a total order of MATH. It then follows by induction that MATH is contractible. |
math/0106024 | This is a general property of homotopy colimits, see CITE Give MATH, the category of chain complexes, the model structure for which the cofibrations are the monomorphisms and the weak equivalences are the quasi-isomorphisms. Then MATH is a Reedy cofibrant diagram and MATH is a quasi-isomorphism. |
math/0106024 | First we simplify REF . We have MATH . Using the fact that MATH and the fact that MATH the last expression simplifies to MATH . From REF it is easy to see that if MATH is of type I, II or III we have MATH . Now suppose MATH is of type I. Define MATH by letting MATH and MATH . Then MATH is of type II, and MATH is the same diagram as MATH, but MATH. Now REF implies that the terms in REF corresponding to MATH and MATH cancel, which completes the proof of REF . For REF , let MATH be a diagram of type III. In this case MATH so we want to show MATH which is immediate from REF For REF , Let MATH be a diagram of type IV. In this case MATH so we need to show MATH . Expanding the definitions and making the obvious cancellations, REF reduces to MATH . Here MATH so REF reduces to MATH and this follows from REF and the fact that MATH . |
math/0106024 | After substituting the definitions of MATH, MATH, and MATH (see REF ) and the definition of MATH (see REF ), REF becomes MATH . Next we substitute the equations MATH and MATH and cancel; this reduces REF to MATH . Now the first and fourth terms on the left side of REF cancel the second term on the right; if we collect the coefficients of MATH in all uncanceled terms, it remains to verify, for each MATH, the following equation (where MATH denotes MATH) MATH and this follows from REF and the equations MATH (this equation follows from the fact that MATH is special) and MATH (this equation follows from the fact that MATH is special). |
math/0106031 | The polyhedron MATH is the feasible region of MATH, the dual program to MATH. The normal fan of MATH is supported on MATH, that is, the union of the normal cones of MATH is MATH, since this is the polar cone of the recession cone MATH of MATH. Suppose MATH is any vector in the interior of a maximal face MATH of MATH. Then by REF , MATH has an optimal solution MATH with support MATH. The optimal solution MATH to the dual of MATH satisfies MATH for all MATH and MATH otherwise, by complementary slackness. Since MATH is a maximal face of MATH, in fact, MATH for all MATH. This shows that MATH is unique, and MATH is contained in the normal cone of MATH at the vertex MATH. If MATH lies in the interior of another maximal face MATH then MATH, the dual optimal solution to MATH satisfies MATH and MATH where MATH. Hence MATH is distinct from MATH and each maximal cone in MATH lies in a distinct maximal cone in the normal fan of MATH. Since MATH and the normal fan of MATH have the same support, they must therefore coincide. |
math/0106031 | Since all data are integral it suffices to prove that the linear relaxation MATH is bounded if and only if MATH. If MATH is a face of MATH then there exists MATH such that MATH and MATH. Using the fact that MATH we see that MATH. This implies that MATH is a positive linear combination of the rows of MATH since MATH. Hence MATH lies in the polar of MATH which is the recession cone of MATH proving that the linear program MATH is bounded. The linear program MATH is feasible since MATH is a feasible solution. If it is bounded as well then MATH is feasible and bounded. Hence the dual of the latter program MATH is feasible. This shows that a superset of MATH is a face of MATH which implies that MATH since MATH is a triangulation. |
math/0106031 | CASE: The lattice point MATH belongs to MATH if and only if MATH is the optimal solution to MATH which is equivalent to MATH being the optimal solution to the reformulation REF of MATH. Since MATH is generic, the last statement is equivalent to MATH. The second statement follows from the fact that REF solves REF if and only if they have the same optimal solution. |
math/0106031 | If MATH, then MATH, and this implies that MATH. If MATH solves MATH, then MATH for all MATH. This implies the result by REF . |
math/0106031 | Suppose MATH. Then for all MATH, MATH. Since MATH can be any vector in MATH, and MATH is bounded by REF , MATH. Hence, by REF , MATH solves MATH. Conversely, if MATH solves MATH, then MATH for all MATH. Since MATH is a relaxation of MATH, MATH for all MATH and hence by REF , MATH. |
math/0106031 | The proof of MATH is the content of REF. The equivalence MATH follows from the definition of a standard pair and REF . |
math/0106031 | The proof follows from REF . |
math/0106031 | The regular triangulation MATH is the normal fan of MATH by REF , and it is unimodular if and only if MATH for every maximal face MATH. This is equivalent to saying that every MATH lies in MATH for every maximal face MATH of MATH. By REF , this happens if and only if MATH has an integral optimum for all MATH. |
math/0106031 | By REF , MATH is a standard pair of MATH for every maximal face MATH of MATH. REF implies that MATH is unimodular (that is, MATH), and therefore MATH for every maximal face MATH of MATH. Hence the semigroups MATH arising from the standard pairs MATH as MATH varies over the maximal faces of MATH cover MATH. Therefore the only standard pairs of MATH are MATH as MATH varies over the maximal faces of MATH. The result then follows from REF . |
math/0106031 | Without loss of generality we can assume that the columns of MATH in MATH form a minimal NAME basis of this cone for any maximal face MATH of MATH. If there were a redundant element, the smaller matrix obtained by removing this column from MATH would still be MATH-normal. For a maximal face MATH, let MATH be the set of indices of all columns of MATH lying in MATH that are different from the columns of MATH. Suppose MATH are the columns of MATH that generate the one dimensional faces of MATH, and MATH a cost vector such that MATH. We modify MATH to obtain a new cost vector MATH such that MATH as follows. For MATH, let MATH. If MATH for some maximal face MATH, then MATH, MATH and we define MATH. Hence, for all MATH, MATH lies in MATH which was a facet of MATH. If MATH is a vector as in REF showing that MATH is a maximal face of MATH then MATH for all MATH and MATH otherwise. Since MATH, we conclude that MATH is a maximal face of MATH. If MATH lies in MATH for a maximal face MATH, then MATH has at least one feasible solution MATH with support in MATH since MATH is MATH-normal. Further, MATH lies in MATH and all feasible solutions of MATH with support in MATH have the same cost value by construction. Suppose MATH is any feasible solution of MATH with support not in MATH. Then MATH since MATH if and only if MATH and MATH is a facet in the lower envelope of MATH. Hence the optimal solutions of MATH are precisely those feasible solutions with support in MATH. The vector MATH can be expressed as MATH where MATH are unique and MATH is also unique. The vector MATH where MATH. Setting MATH for all MATH, MATH for all MATH and MATH otherwise, we obtain all feasible solutions MATH of MATH with support in MATH. If there is more than one such feasible solution, then MATH is not generic. In this case, we can perturb MATH to a generic cost vector MATH by choosing MATH, MATH whenever MATH and MATH otherwise. Suppose MATH are the optimal solutions of the integer programs MATH where MATH. (Note that MATH is the index of MATH in MATH.) The support of each such MATH is contained in MATH. For any MATH, the optimal solution of MATH is hence MATH for some MATH and MATH with support in MATH. This shows that MATH is covered by the affine semigroups MATH where MATH is a maximal face of MATH and MATH as above for each MATH. By construction, the corresponding admissible pairs MATH are all standard for MATH. Since all data is integral, MATH and hence can be scaled to lie in MATH. Renaming MATH as MATH, we conclude that MATH is a NAME family. |
math/0106031 | It is known that if MATH then MATH has a regular unimodular triangulation MATH CITE. The result then follows from REF . |
math/0106031 | The equivalence of REF was established in CITE. REF shows that REF . Hence we just need to show that REF . Suppose that MATH is MATH-normal for every regular triangulation of MATH. In order to show that MATH is supernormal we only need to check submatrices MATH where the dimension of MATH is MATH. Choose a cost vector MATH with MATH if the MATH-th column of MATH does not generate an extreme ray of MATH, and MATH otherwise. This gives a polyhedral subdivision of MATH in which MATH is a maximal face. There are standard procedures that will refine this subdivision to a regular triangulation MATH of MATH. Let MATH be the set of maximal faces MATH of MATH such that MATH lies in MATH. Since MATH is MATH-normal, the columns of MATH that lie in MATH form a NAME basis for MATH for each MATH. However, since their union is the set of columns of MATH that lie in MATH, this union forms a NAME basis for MATH. |
math/0106031 | CASE: If MATH or MATH is graded and MATH, MATH is supernormal and hence by REF , every regular triangulation of MATH supports at least one NAME family. CASE: If MATH and MATH is graded, then we may assume that MATH . In this case, MATH is supernormal and hence every regular triangulation MATH of MATH supports a NAME family by REF . Suppose the maximal cones of MATH, in counter-clockwise order, are MATH. Assume the columns of MATH are labeled such that MATH for MATH, and the columns of MATH in the interior of MATH are labeled in counter-clockwise order as MATH. Hence the MATH columns of MATH from left to right are: MATH . Indexing the columns of MATH by their labels, the maximal faces of MATH are MATH for MATH. Let MATH be the unit vector of MATH indexed by the true column index of MATH in MATH and MATH be the unit vector of MATH indexed by the true column index of MATH in MATH. Since the columns of MATH form a minimal NAME basis of MATH, MATH is the unique solution to MATH for all MATH and MATH is the unique solution to MATH for all MATH. Hence the standard pairs of REF are MATH and MATH for MATH and MATH. Suppose MATH supports a second NAME family MATH. Then every standard pair of MATH is also of the form MATH for MATH, and MATH of them are MATH for MATH. The remaining standard pairs are of the form MATH. To see this, consider the semigroups in MATH arising from the standard pairs of MATH. The total number of standard pairs of MATH and MATH are the same. Since the columns of MATH all lie on MATH, no two MATH-s can be covered by a semigroup coming from the same standard pair and none of them are covered by a semigroup MATH. We show that if MATH is a standard pair of MATH then MATH and thus MATH. If MATH, the standard pairs of MATH are MATH as in REF . If MATH, consider the last cone MATH. If MATH is the second to last column of MATH, then MATH is unimodular and the semigroup from MATH covers MATH. The subcomplex comprised of MATH is a regular triangulation MATH of MATH where MATH is obtained by dropping the last column of MATH. Since MATH is a normal graded matrix with MATH and MATH has less than MATH maximal cones, the standard pairs supported on MATH are as in REF by induction. If MATH is not the second to last column of MATH then MATH, the second to last column of MATH is in the NAME basis of MATH but is not a generator of MATH. So MATH has a standard pair of the form MATH. If MATH, then the lattice point MATH cannot be covered by the semigroup from this or any other standard pair of MATH. Hence MATH. By a similar argument, the remaining standard pairs indexed by MATH are MATH along with MATH. These are precisely the standard pairs of MATH indexed by MATH. Again we are reduced to considering the subcomplex comprised of MATH and by induction, the remaining standard pairs of MATH are as in REF . CASE: The MATH normal matrix MATH of REF has REF distinct NAME families supported on REF out of REF regular triangulations of MATH. Furthermore, the normal matrix MATH has REF distinct NAME families supported on REF out of REF regular triangulations of MATH. |
math/0106038 | Since adjacent entries in the height matrix differ by one, each MATH is either MATH or MATH. There is a well - known REF to MATH correspondence between (halved) alternating sign matrices and perfect matchings of (halved) Aztec diamonds (compare CITE). A MATH . Aztec rectangle is a graph composed of MATH squares (see REF ). A halved Aztec diamond is an Aztec rectangle with the shape of half a square, with some vertices in the two bottom rows missing (see REF ). A perfect matching (REF - factor) of a graph is a set of edges such that every vertex of the graph lies on exactly one of these edges. In the remainder of this paper we will use the term matching instead of perfect matching. We write the entries of the halved alternating sign matrix in the squares of the halved Aztec diamond as shown in REF . The corresponding MATH matchings can be found by demanding that a square surrounding a MATH, REF or REF contains exactly REF or REF edges of the matching, respectively. The edges in the squares containing a REF can be found by joining the two vertices lying in the direction of the next REF's in the same row and column (if there is no REF in the column we take the bottom vertex). There are two choices for the squares containing MATH as shown in REF . This accounts for the weight MATH. Close inspection of the correspondence reveals that the condition on the last row of the height matrix determines which of the vertices of the halved Aztec diamond are missing. The halved Aztec diamond is a MATH . Aztec rectangle with missing last row of vertices and missing vertices in the next row in positions MATH, say. It is easy to see that we have either MATH or MATH corresponding to MATH or MATH, respectively. Therefore, we have to sum over MATH different boundary conditions. Fortunately, we can add pairs of vertices in the last two rows as shown in REF and just count all matchings of the emerging new region (see REF ). The vertices in the bottom row can be matched either to the northeast or to the northwest. This corresponds to the possible choices for the MATH. Now we can apply the following lemma (compare CITE). The number of perfect matchings of a MATH . Aztec rectangle, where all the vertices in the bottom row have been removed except for the MATH-st, the MATH-nd, , and the MATH-th vertex equals MATH . To apply the lemma to our case, we have to set MATH, MATH and MATH. We obtain that our REF - enumeration of halved alternating sign matrices equals MATH as desired. |
math/0106038 | Now we have the weight MATH. We will illustrate all steps of the proof by the example of halved alternating sign matrices of order REF (compare REF ). From there it will be clear what happens in the general case. The first step is another well - known bijection between alternating sign matrices and a family of graphs called fortresses. These are squares arranged in a rectangular shape separated by single edges. On the right, the left and the upper side of the rectangle edges are appended to every other square (see REF for a MATH fortress graph with some extra edges appended to the squares in the bottom row). We have the following replacement rules: CASE: REF's in even places and MATH's in odd places translate to CASE: MATH's in even places and REF's in odd places translate to CASE: REF's translate to The edges of the squares corresponding to MATH determine uniquely which of the four possibilities should be chosen for each REF. An example of the correspondence is shown in REF . The reader should note that there are two choices of edges for MATH's in even places and for REF's in odd places. This accounts for the weight. It is not difficult to see that the restriction on the last row of the height matrix corresponds to a condition on the extra pending edges at the bottom row of the resulting graph. Either both edges in the positions MATH and MATH are contained in the graph or neither. In the following we will repeatedly use a well - known local modification of a graph called urban renewal, which changes the enumeration of perfect matchings only by a global factor (see CITE). The modification is shown in REF . Before we can explain this modification, we have to make a few definitions. Let MATH be a graph with weights assigned to its edges. Then the weight of a matching is the product of the weights of the edges it contains. The weighted enumeration of matchings MATH is now defined as the sum of the weights of all possible matchings of the graph MATH. NAME renewal can now be described as follows. We start with a graph MATH which looks locally like the left - hand - side of REF . Then we contract the four edges of weight REF and change the weights MATH to MATH. We obtain a graph MATH which looks locally like the right - hand - side of REF and like MATH everywhere else. The new edge weights MATH of the resulting MATH are defined by MATH, whereas all other weights stay the same. The weighted enumerations of matchings MATH and MATH of the two graphs are related in the following way: Let MATH be a graph which looks locally like the left - hand - side of REF and let MATH be the graph which looks locally like the right - hand - side and like MATH elsewhere. Then the weighted enumeration of matchings of the new graph MATH equals the weighted enumeration of matchings of the graph MATH multiplied by MATH, that is, MATH . We want to apply urban renewal to all squares in even position in the graph in REF . First, we append two vertical edges to squares in the last row in even position which have no downward - pointing edge appended. In the example in REF , this happens to the fifth square in the bottom row, resulting in the upper graph in REF (at this point, the dotted circles should be ignored). This does not change the enumeration of perfect matchings since there is only one possibility for the new vertices to be paired. We obtain a MATH fortress with some edges appended. Now, we can apply urban renewal to every square in even position (the circled squares in REF ). There are MATH of these squares with MATH, which yields a factor of MATH. The resulting graph is a MATH . Aztec rectangle where every other square has edges of weight MATH (see the dotted lines in the bottom graph in REF ) and some downward - pointing edges appended to the last row of squares. It is easy to see that the original restriction on the last row of the height matrix now translates to the restriction that for each MATH there is exactly one edge in the position MATH or MATH. Similar to the proof of REF , we can interpret the sum of the corresponding MATH terms as the number of matchings of the weighted halved Aztec diamond MATH in REF because every vertex in its last row can be either matched to the left or to the right. Therefore, we have to determine MATH. We will use urban renewal repeatedly to reduce the graph MATH to the graph MATH. The first step consists of replacing every vertex of MATH by a line of three vertices so that the weighted enumeration of matchings remains unchanged. The resulting graph in our example is shown in REF . Now we are in the position to apply urban renewal to all squares in the graph. The factor MATH equals MATH for half of the squares and MATH for the other half. Thus, the factors resulting from urban renewal cancel each other. Square edges of weight REF become edges of weight MATH and vice versa and in our example we obtain the graph shown in REF . The pending edges along the border of the graph have to be in every perfect matching and can be removed together with the two endpoints without changing the enumeration of perfect matchings. For the same reason, we can fill the ``dents" in the bottom row by adding some edges which have to be in every perfect matching. The resulting graph is shown in REF . Now we (almost) repeat the last two steps. We replace each vertex by three vertices to obtain the graph in REF . Note that the squares in the bottom row contain only one edge of weight MATH each. The next step is to apply urban renewal to all squares. The product of the factors MATH is easily seen to be MATH. The new edge weights are MATH, MATH, MATH and MATH (only the forced appended edges have still weight REF). The resulting graph is shown in REF . Now we mark every other vertex in the bottommost row with a dotted circle starting with the second vertex (see REF ). Similarly, we mark all vertices immediately above and to the left of the dotted circles with an unbroken circle. We divide the weight of the edges incident to one of the MATH points marked by an unbroken circle by two and multiply the weight of the edges incident to the MATH points marked by a dotted circle by two. This does not change the weighted enumeration of matchings. Then we strip off all the forced edges (that is, edges that must be contained in every perfect matching) and obtain the graph shown in REF . It is easy to see that every matching contains exactly MATH of the edges with weight MATH or MATH (the double edges and dotted double edges) and exactly MATH edges with weight MATH or MATH. If we now divide the weights of all double edges and dotted double edges by MATH and the weights of all the other edges by MATH, we obtain a graph with edges of weight REF and MATH only. This changes the weighted enumeration by a factor of MATH. The resulting graph is shown in REF . It is clearly the mirror image of MATH (compare REF ). Therefore, we obtain for the weighted enumeration of matchings of MATH: MATH . Since MATH is easily seen to be MATH, we get for our weighted enumeration of halved alternating sign matrices MATH . |
math/0106038 | The proof is analogous to the proof of REF . For example, in the case MATH for all MATH, we use the bijection to fortress graphs and apply urban renewal to all the squares. We obtain a weighted halved Aztec diamond of order MATH (see REF for the case MATH). Imitating the steps in the proof of REF , we can reduce it to the weighted halved Aztec diamond of order MATH. In this way, we again obtain a simple recursion which gives the results stated in REF . |
math/0106042 | For MATH, MATH. Hence MATH is smooth and MATH. |
math/0106042 | Let MATH be an element of MATH. Since MATH is simple and MATH, MATH. Hence MATH. If MATH, then there is a non-zero homomorphism MATH. Then MATH . Therefore MATH and MATH. |
math/0106042 | We set MATH. Assume that MATH. Let MATH be a MATH-stable locally free subsheaf of MATH such that MATH. Then we get a non-zero homomorphism MATH. Since MATH is locally free, MATH must be an isomorphism. Hence MATH. On the other hand, MATH induces an isomorphism MATH. Hence MATH, which is a contradiction. Therefore MATH. We next show that MATH is MATH-semi-stable. Assume that MATH has a torsion submodule MATH. Then MATH is a submodule of MATH containing MATH. By the MATH-semi-stability of MATH, MATH. Hence MATH is of dimension REF. Since MATH is locally free, MATH. Thus MATH is torsion free. Then it is easy to see that MATH is MATH-semi-stable. |
math/0106042 | If MATH, then MATH for MATH. By REF , we get a contradiction. |
math/0106042 | Let MATH be a closed substack of MATH such that MATH belongs to MATH if and only if there is a quotient MATH such that MATH but MATH. Let MATH, MATH be a surjective homomorphism. Then MATH is MATH-semi-stable and does not belong to MATH. Hence MATH is a non-empty open substack of MATH. For pairs of integers MATH and MATH such that MATH, MATH and MATH, let MATH be the substack of MATH consisting of MATH which fits in an exact sequence: MATH where MATH is a MATH-stable sheaf of MATH and MATH is a MATH-semi-stable sheaf of MATH. By CITE or CITE, MATH . By REF , MATH. Hence if MATH or MATH, then we get MATH. If MATH, then by using REF again, we see that MATH. Therefore MATH is a proper substack of MATH, which implies that MATH. By CITE, the locus of non-locally free sheaves is of codimension MATH (use REF). Hence MATH contains a locally free sheaf. |
math/0106042 | For MATH, REF implies that MATH is injective. Hence MATH. Since MATH, MATH. Hence MATH. Then MATH. |
math/0106042 | By REF, we have an exact sequence MATH . Since MATH is isomorphic and MATH, we get that MATH. Therefore we get our claim. |
math/0106042 | By NAME spectral sequence and projection formula, MATH . Since MATH, MATH. |
math/0106042 | By REF, MATH fits in an exact sequence MATH . By REF , MATH is MATH-semi-stable. It is easy to see that MATH. Assume that MATH is not semi-stable and let MATH be a destabilizing subsheaf. Then MATH. By our assumption on MATH, MATH. Hence MATH, which contradicts to REF . |
math/0106042 | We may assume that MATH. Then MATH fits in a universal extension MATH where MATH. Assume that MATH is not MATH-twisted stable. Then there is a MATH-twisted stable subsheaf MATH of MATH such that MATH is MATH-twisted semi-stable. If MATH is contained in MATH, then we get a homomorphism MATH. Since MATH, we get a contradiction. Hence MATH is not contained in MATH. Since MATH is MATH-stable, we get MATH. Hence MATH is injective. Let MATH be a MATH-stable locally free subsheaf of MATH. Then we see that MATH, which implies that MATH is not MATH-twisted stable. Therefore MATH is MATH-twisted stable. |
math/0106042 | If MATH is MATH-stable, then we see that MATH, and hence MATH and MATH, MATH. Assume that there is an exact sequence MATH where MATH is a MATH-stable sheaf of MATH and MATH is a MATH-semi-stable sheaf of MATH. Then we get an exact sequence MATH . Since MATH is MATH-twisted stable, MATH. In particular MATH is MATH-twisted stable. By the stability of MATH, MATH, which implies that MATH. Therefore MATH and MATH. By using REF, we see that MATH is an isomorphism. We note that MATH fits in an exact sequence MATH . By the stability of MATH, REF MATH, or REF MATH, MATH and MATH. Therefore MATH is locally free, or MATH. |
math/0106042 | For a locally free sheaf MATH, CITE implies that MATH . Let MATH be the substack of MATH consisting of MATH which fits in an exact sequence MATH where MATH is a MATH-semi-stable sheaf of MATH, MATH and MATH. By CITE, we see that MATH . Hence by using REF and the assumption MATH, we see that MATH . Therefore we get MATH . |
math/0106042 | We have an exact sequence MATH where MATH. Then MATH. Since MATH, we get our claim. |
math/0106042 | By REF , MATH is normal. Moreover MATH is a dense subset of MATH. Hence MATH. Let MATH be a poly-stable sheaf of MATH, that is, MATH is a direct sum of MATH-twisted stable sheaves. By REF , there are MATH-stable locally free sheaves MATH, MATH of MATH and points MATH, MATH such that MATH. Since MATH and MATH, we see that MATH . Hence MATH belongs to MATH if and only if MATH. Then the last claim follows from this. |
math/0106042 | Since MATH, MATH and MATH are relatively prime. Hence there is a universal family. |
math/0106042 | We have an exact sequence MATH . By REF , MATH is stable and MATH is of MATH-dimensional. Then MATH and MATH is of MATH-dimensional. Hence MATH is stable. We next show that MATH. Since MATH is stable, we get MATH . Combining the fact MATH, we see that MATH. Since MATH, we get MATH. |
math/0106042 | Let MATH be a MATH-twisted coherent system. Since MATH, we have a homomorphism MATH . Then the cokernel is the NAME tangent space of MATH and the obstruction space is MATH. If MATH, then MATH. If MATH, then by using REF and an exact sequence MATH we see that MATH. Hence MATH is smooth. Then we see that MATH . |
math/0106042 | We first assume that MATH. For MATH, REF implies that MATH is injective and MATH is stable. Thus we have a morphism MATH. Conversely for MATH and a MATH-dimensional subspace MATH of MATH, we have an extension MATH whose extension corresponds to the inclusion MATH. Then MATH is stable. Since MATH and there is a universal family, we see that MATH is a (NAME locally trivial) MATH-bundle. Therefore we get our claim. We next treat the second case. For MATH, MATH fits in an exact sequence MATH . Hence MATH defines a point of MATH. Thus we get a morphism MATH. Conversely for MATH, we get a homomorphism MATH. It gives the inverse of MATH (for more details, see CITE). |
math/0106042 | If MATH with MATH, then REF implies that MATH. Hence MATH. By REF , MATH for all MATH. Hence MATH. |
math/0106043 | REF works correctly by the discussion above. REF can be performed in MATH steps by REF shows that we can execute REF in MATH time. Hence, REF in total contribute at most MATH to the running time (since the while-loop is executed once per face). The body of the for-loop has to be executed for each of the MATH arcs in the NAME diagram MATH. REF implies that each execution of the body of the for-loop can be performed in MATH steps. Thus, the claim on the running time follows. Since each node of the face tree corresponds to a face of MATH, the face tree has MATH nodes. Each label on an edge of the face tree needs at most MATH bits, and we can estimate the space requirements of any of the (internal and external) pointers by MATH. Hence, the face tree needs no more than MATH bits. The space required for the storage of MATH is bounded by MATH, if for each pair MATH the set MATH is stored as a bit set, that is, the characteristic vector of MATH is stored bit by bit. |
math/0106045 | By REF , the second inequality of REF is equivalent to MATH and hence MATH, where MATH and the supremum is taken over MATH and over all mappings MATH that satisfy the assumptions. Using the first inequality of REF we conclude, by setting MATH in REF , that MATH . Then, since MATH and MATH, we have MATH . To prove that MATH we take the logarithmic partial derivative with respect to MATH of both sides. Since the left and right hand side of REF are equal when MATH, we only need to prove that the derivative of the left hand side is greater than that of the right hand side for MATH. That is, we need to show that MATH where MATH. The left hand side has an infimum greater than MATH and the right hand side the supremum MATH, so we are done. |
math/0106045 | Follows directly from REF as MATH is MATH-Hölder continuous in MATH with constant MATH by definition. |
math/0106045 | By REF this is clear if MATH or MATH. Let MATH and MATH and MATH be given. We will prove that there exists a point MATH in the intersection of the line through MATH and MATH with MATH such that MATH. Let MATH be the image of MATH in the canonical projection of MATH onto MATH etc. Then, by definition, MATH and so on. Let MATH be the radius of the image of the line containing MATH and MATH under the projection. Of the two possible MATH's, we chose the one on the shorter arc joining MATH and MATH. Then we need to show that MATH . Let us differentiate the left-hand-side with respect to MATH. We the find easily that the left-hand-side has a maximum at MATH. We then have to show that MATH. Now this follows directly from the equation MATH. From REF it follows that MATH . Here we have used the fact that a continuous mapping has the same NAME constant for MATH and MATH. We combine the previous estimates and conclude that MATH . Thus MATH. Note that this estimate is not asymptotically sharp. |
math/0106045 | Let MATH and MATH. Define an auxiliary mapping MATH by: MATH . Since MATH it follows from the definition of MATH that MATH which is equivalent to MATH and thus, by REF holds with the constant MATH indicated. |
math/0106045 | This is proved as REF except that we will use different methods of estimating the upper bounc MATH of MATH. If MATH we use the estimate MATH . In this case we easily see that MATH. For MATH we have, by the NAME continuity of MATH, MATH. Then we proceed exactly as REF to get MATH. It follows that MATH . Recall that in REF the NAME constant in the spherical metric was shown to be less than MATH so we are done. |
math/0106045 | By REF the estimate holds for MATH. Since the inversion MATH is a q-isometry, we conclude that the estimate holds also for MATH. We now deal with the remaining case: MATH. Then MATH . The last inequality holds, since we may assume (otherwise REF is trivial) MATH as MATH. From this it follows immediately that MATH which concludes the proof of REF . For asymptotical sharpness we still need to prove that MATH when MATH for MATH as given in REF , that is, that MATH at the limit. This follows from the following chain of inequalities: MATH . Thus we see that the bound of MATH is asymptotically sharp. |
math/0106045 | The point MATH lies in a region bounded by a line and a circle, as shown in REF . This means that MATH lies in a region bounded by two spheres, as shown in REF . It is immediately clear that MATH is as far from the origin as any other point in the region, so that MATH. We calculate MATH from MATH and MATH . (Note that here MATH refers to an angle as indicated in the picture, not to MATH.) We subtract the second equation from the first to get MATH . Now using this in the first equation gives us the desired formula MATH. |
math/0106045 | Denote the symmetric MATH-qc extension of MATH to a mappin from MATH to MATH by MATH, again (here we use REF ). We assume without loss of generality that MATH and MATH. Since MATH is MATH-qc, MATH. We may assume without loss of generality that MATH. Let MATH be a point, such that the projection of MATH in the real axis equals MATH. Then MATH where the last inequality holds, since MATH is increasing and MATH, since MATH. Therefore we may take MATH in REF to conclude that MATH with MATH the inversion in MATH. Let MATH. Since MATH is a REF-qc, MATH is MATH-qc. Since MATH fixes MATH, MATH fixes MATH. However, REF means that MATH, since MATH. Let MATH be a NAME transformation with MATH and MATH. Then, since MATH is symmetric in MATH, MATH. Now MATH is biLipschitz with constant MATH . (See e. CASE: REF, for these elementary NAME mapping results.) Since MATH is MATH-qc and fixes MATH, REF and MATH, we have MATH, for MATH by REF. It follows from the biLipschitz property of MATH that MATH . When we set MATH and MATH, we get MATH for MATH, which is what we wanted to show. |
math/0106046 | We will assume that the class MATH is indivisible, that is, is not a multiple of another integral class. Our statement clearly follows from this special case. Our purpose is to show that we may find a smooth map MATH with NAME critical points, having the following properties: CASE: The cohomology class of the closed REF-form MATH coincides with MATH, where MATH denotes the standard angular form on the circle MATH. CASE: All fibers MATH are connected, where MATH. CASE: The map MATH has at most one critical value MATH. In other words, all critical points of MATH lie in the same fiber MATH. Having achieved this, we may apply the technique of CITE, pages REF, which allows to collide the critical points of MATH (equivalently, the zeros of closed REF-form MATH) into a single degenerate critical point of a closed REF-form MATH lying in the same cohomology class. Namely, firstly, we may find a piecewise smooth tree MATH containing all the critical points of MATH. Secondly, we may find a continuous map MATH with the following properties: MATH is a single point MATH; MATH is a diffeomorphism onto MATH; MATH is the identity map on the complement of a small neighborhood of MATH; in particular, MATH is homotopic to the identity map MATH. The circle valued map MATH is well-defined and is continuous. Moreover, MATH is smooth on MATH. Applying REF from CITE we see that we may replace the map MATH in a small neighborhood of MATH by a smooth map MATH having a single critical point. MATH is homotopic to MATH and thus closed REF-form MATH lies in the cohomology class MATH and has possibly a single zero. In the rest of the proof we will show that we may find a smooth NAME map MATH with REF above. It is well known that any indivisible class MATH may be realized by a connected codimension one submanifold MATH with oriented normal bundle. Cutting MATH along MATH produces a compact cobordism MATH with MATH, a disjoint union of two copies of MATH. Consider a NAME function MATH having MATH and MATH as regular values and MATH, MATH. We may assume that MATH has no critical points of indices MATH and MATH. Moreover, we may construct MATH, such that all level sets MATH, where MATH, are connected and having the self-indexing property: all critical points of MATH having NAME index MATH lie in MATH, where MATH. The map MATH, where MATH, defines a smooth map MATH in the cohomology class MATH with connected fibers having the following property: for any critical point MATH, MATH, with NAME index MATH, the image MATH equals MATH. In other words, the critical points of MATH with the same NAME index lie in the same fiber, and these critical fibers MATH appear in the order of their NAME indices, while the point MATH moves in the positive direction along the circle MATH. For points MATH on the circle MATH we will write MATH to denote that moving from point MATH in the positive direction along the circle MATH, we first meet MATH, then MATH, until we again meet MATH. Let us now formulate the following approximation to REF : MATH . All critical points of a smooth NAME map MATH with NAME index MATH lie in the same fiber MATH, where MATH, MATH, and MATH . In particular, the critical values MATH coincide. Note that REF is equivalent to REF , which is our purpose. We have found above a smooth map MATH, satisfying REF , and (MATH). On the next step we will show that we may replace MATH by a smooth NAME map MATH with REF and (MATH). Let MATH be the critical values of MATH. Consider a point MATH, lying between MATH and MATH. Cut MATH along the submanifold MATH and consider the obtained cobordism MATH and a NAME function MATH from MATH to an interval, obtained by cutting the circle MATH at point MATH. All level sets of MATH are connected. Moving from the bottom of this cobordism to the top, we meet MATH critical levels; first we meet the level containing the critical points of NAME index MATH, then the levels containing the critical points with NAME indices MATH. We will use the theory of NAME of rearrangement of critical points. Choose a generic gradient like vector field MATH for MATH. Then all integral trajectories of MATH, which go out of the critical points of index MATH, reach MATH without interaction with the other critical points. Therefore we may slide the critical points of index MATH some distance up, putting them on the same level with the critical point of index MATH, see REF . In other words, we may replace MATH by a NAME function MATH, which coincides with MATH near MATH, has the same critical points, but the value at the critical points of index MATH equals the value at the critical points of index MATH. Note that the level sets of MATH are all connected: REF the bottom level MATH is unchanged and so it is connected; REF passing the critical levels with NAME indices MATH may not create nonconnected level sets; REF in principle, nonconnected level sets may appear after passing the top critical value contained the critical points of indices MATH and MATH; however in our situation all higher upper level sets are the same as for the previous function MATH, and so they are all connected. Folding this cobordism back, gives a smooth map MATH having REF , and (MATH). We may proceed similarly to find a smooth map MATH with REF , and (MATH). The critical values of MATH are MATH. We find a point MATH between MATH and MATH and cut MATH along MATH. The NAME function on the obtained cobordism will have MATH critical levels. The lowest will be the level containing all critical points of indices MATH and MATH and then the critical levels of points with the NAME indices MATH. Repeating the above procedure, we may slide the critical points of index MATH and MATH the same distance up putting them on the same level with the critical points of index MATH. Proceeding in this way inductively we arrive at a smooth map MATH having REF , and (MATH) = REF . This completes the proof. MATH . |
math/0106046 | Let MATH be a homotopy inverse of MATH. Choose a closed REF-form MATH on MATH in the cohomology class MATH. Then MATH is a closed REF-form on MATH lying in cohomology class MATH. Fix a homotopy MATH, where MATH, such that MATH and MATH. Compactness of MATH implies that there is a constant MATH such that MATH for any point MATH, where MATH is the track of the point MATH under homotopy MATH, that is, MATH. Suppose that MATH. Given any MATH, there is an open covering MATH, such that MATH are contractible in MATH and there exists a homotopy MATH, where MATH, such that MATH for any MATH, where MATH. Define MATH . These sets form an open cover of MATH. Let us show that the set MATH satisfies REF . Define a homotopy MATH, where MATH, by MATH . Then MATH is the inclusion MATH and for any point MATH holds MATH where MATH is the track of MATH under homotopy MATH. The following diagram MATH is homotopy commutative and the horizontal map below is null-homotopic. This shows that the inclusion MATH is null-homotopic, where MATH. The above argument proves that MATH. The inverse inequality follows similarly. MATH. |
math/0106046 | Denote MATH where MATH. Then MATH is a decreasing sequence of finite-dimensional vector spaces. Hence, there exists an integer MATH, such that MATH coincides with MATH. Any homology class MATH is movable to MATH. Thus, this MATH satisfies REF . By the similar reasons, we may increase MATH, if necessary, such that REF is satisfied as well. MATH . |
math/0106046 | We will show that REF implies REF and that REF implies both REF . The implication MATH follows similarly. Assume that a class MATH is movable to MATH. Realize MATH by a cycle MATH in MATH and specify a compact subset MATH containing MATH and such that MATH. Assume that MATH is large enough, so that it satisfies REF and the subset MATH is disjoint from MATH. Let us show that for any integer MATH the homology class MATH may be realized by a cycle in MATH. Since MATH is movable to MATH, there exists a cycle MATH in MATH representing MATH. Write MATH, where MATH contains MATH, MATH contains MATH and MATH. In the NAME sequence MATH the difference MATH goes to zero. Hence there is a cycle in MATH, which is homologous to MATH in MATH and homologous to MATH in MATH. This proves our claim. Consider MATH . It is a finite dimensional MATH-linear subspace; we observe that MATH is invariant under MATH. Hence, by the NAME Theorem, there exists a polynomial MATH, such that MATH acts trivially on MATH. Since MATH belongs to MATH, we obtain MATH. This shows that REF implies REF . Let us show that REF implies REF . Suppose that MATH is such that MATH for a NAME polynomial MATH . Consider the ring MATH, where MATH is the ideal generated by MATH. The powers MATH, where MATH, form an additive basis of MATH. Since multiplication by MATH is an automorphism of MATH, we obtain that for any integer MATH, the powers MATH form a linear basis of MATH as well. In particular, we may express MATH as a linear combination of MATH in MATH. This means that for any MATH we may find numbers MATH, where MATH, such that MATH . Assume that MATH is a compact subset such that MATH and class MATH can be realized by a cycle MATH in MATH. Then for any integer MATH class MATH may be realized by the cycle MATH . Hence MATH is movable to both ends MATH of MATH. MATH . |
math/0106046 | Consider the following exact sequence of sheaves over MATH . Here MATH denotes the sheaf of locally constant functions, MATH denotes the sheaf of real valued continuous functions, and MATH denotes the sheaf of germs of continuous functions modulo locally constant. More precisely, MATH is the sheaf corresponding to the presheaf MATH. Comparing with our definition of continuous closed REF-form, we find that the space of global sections of MATH, MATH coincides with the space of continuous closed REF-forms on MATH. From REF , using that MATH is a fine sheaf, we obtain an exact sequence MATH . Here MATH is the set of all continuous functions on MATH, and the map MATH acts by assigning to a continuous function MATH the closed REF-form MATH. The group MATH is the MATH-ech cohomology MATH and the map MATH assigns to a closed REF-form MATH its MATH-ech cohomology class MATH. The natural map MATH is an isomorphism assuming that MATH is paracompact, NAME, and homologically locally connected, compare CITE. This implies our statement. MATH . |
math/0106046 | We will assume that MATH, that MATH is compact, and that point MATH is the only critical point of function MATH in MATH. Let MATH be an open neighborhood of MATH with compact closure MATH and such that in MATH there exist local coordinates MATH and MATH is the gradient of MATH with respect to a Riemannian metric MATH. Fix a smooth function MATH, such that MATH and MATH. We want to show that there is a constant MATH such that the derivative of the function MATH along the gradient flow of MATH satisfies in MATH the inequalities: MATH . In the local coordinates MATH and MATH . Both the first and the second terms in this sum can be viewed as symmetric bilinear forms in the partial derivatives of MATH with continuous coefficients, and hence our REF follows. We will define now two smooth functions MATH by MATH . In MATH using REF we have MATH . Hence we conclude that MATH increases and MATH decreases along the gradient flow of MATH in MATH. We set MATH . It is a continuous function with MATH. For any number MATH small enough, the set MATH is an open neighborhood of MATH, contained in MATH. Let MATH be an open neighborhood of MATH with MATH. Let MATH be such that any trajectory MATH of the flow of MATH with MATH, and MATH, where MATH, satisfies MATH . We will show that MATH satisfies the conditions of the Lemma assuming that MATH and MATH is small enough, so that MATH. Since MATH, a trajectory MATH enters the set MATH at MATH if and only if MATH and MATH. Moreover, if MATH and MATH, the trajectory leaves MATH immediately (that is, the trajectory MATH is tangent to the boundary MATH), and if MATH, the trajectory penetrates the interior of MATH. Similarly, if MATH is such that MATH and MATH, the trajectory MATH leaves the set MATH for MATH. We know that, while MATH stays in MATH, the function MATH increases and so the trajectory remains away from MATH. NAME this trajectory come back to MATH for some large time MATH? If this happens, then MATH (because of REF and using MATH) and hence MATH. This proves that under our assumptions on MATH the set MATH coincides with the interval MATH. Now we are left to prove REF. Let MATH denote the set of points MATH, such that MATH belongs to MATH for some MATH. Since MATH, the equation MATH defines a continuous function of MATH. Similarly, the equation MATH defines a continuous function MATH, where MATH. The set MATH equals MATH. This implies our REF . MATH . |
math/0106048 | We have MATH, so MATH therefore MATH, and by the monotonocity of MATH, MATH . |
math/0106048 | One direction is clear : if for some MATH in MATH, a harmonic function MATH as above exists, then the function MATH, where MATH denotes the NAME transform of MATH, is holomorphic and bounded, and MATH, so that MATH is not an essential minorant for MATH. If on the other hand MATH, choose a NAME sequence MATH so that MATH, denote by MATH the NAME product with zeros on the sequence MATH, then MATH will verify MATH. The proof of the converse is essentially the same as that of REF. Suppose that MATH is not an essential minorant. Then there exist MATH in MATH (respectively, MATH) and MATH in MATH such that MATH and MATH for all MATH. Then MATH, where MATH is zero-free and MATH is a NAME product, MATH where MATH, MATH. Let MATH, by the separation condition. First notice that for each MATH, there is at most one point MATH such that MATH, so the sequence MATH is a NAME sequence. Define a sequence MATH by MATH. If MATH, by construction MATH. If MATH, by REF , MATH. In CITE on page REF, lines REF to REF, it is proved that there exists a holomorphic function MATH in the unit disk such that MATH for any MATH such that MATH. Then the function MATH satisfies the conclusion of REF . |
math/0106048 | Proof of sufficiency in REF : Suppose that we have MATH, a function MATH as in REF such that MATH and MATH such that MATH. Then by REF , we may assume that we have in fact MATH, with MATH a harmonic function and MATH in MATH. We set MATH when MATH, MATH otherwise. We now want to relate the sets where MATH is large to those where MATH is large. There exist a constant MATH (depending only on MATH and the constant MATH in the hypotheses above) such that MATH . We prove the contraposite inclusion. Suppose that MATH. Then, for any MATH in MATH, MATH, that is to say, MATH. Then, by REF , we see that MATH is contained in the union of sets MATH for MATH, and therefore, because of REF , MATH . Choosing MATH approximately equal to MATH, we obtain that MATH, therefore, with a slightly modified MATH, MATH. By REF , which applies because of the assumptions made at the start of the proof of REF , the inclusion REF implies that MATH . Then, by the monotonicity of MATH, MATH which contradicts the fact that MATH. |
math/0106048 | Proof of sufficiency in REF : Suppose that we have MATH, a function MATH as in REF and MATH such that MATH. Then by REF , we may assume that we have in fact MATH, with MATH a harmonic function and MATH. REF and its consequence REF then apply to the separated sequence MATH and the decreasing function MATH. To get a contradiction, we assume that for any positive integer MATH and any subset MATH of MATH such that MATH, MATH. Let MATH be a NAME sequence such that MATH for MATH large enough. As in the proof of REF , MATH. Let MATH be the set of indices MATH such that MATH. Since by REF, we do have MATH. For MATH, MATH . Therefore, applying REF , MATH which contradicts our assumption. |
math/0106048 | Proof of REF : This proof is patterned after that in CITE, where instead of Conclusion REF, only the fact that MATH was not NAME was to be obtained. We define points MATH associated to the arcs MATH: MATH (their radial projection is at the center of the corresponding interval). We will define the sequence through the choice of a certain subfamily MATH of the family of all the MATH. We shall define simultaneously a sequence of probability measures MATH, supported on the union of all the MATH for a given index MATH, with the same uniform density on each of the invervals where it is supported. First we define a sequence of integers by MATH . One verifies that MATH, and in fact the sequence we have defined is the largest verifying that property together with MATH, for any nonnegative MATH. The family MATH, where MATH is a subset of MATH which is defined recursively in the following way : if MATH, we want MATH which is ensured by MATH ; if on the contrary MATH, we only select the first half of each interval at the MATH-th level, that is, MATH and MATH . We will denote MATH. By REF , we see that MATH and we thus define the probability measure MATH as having uniform density MATH on the set MATH. Observe that the way we have defined the families MATH implies that MATH and therefore if we denote by MATH a weak limit point of the sequence MATH, MATH. We let MATH be the NAME integral of the measure MATH. This is a positive harmonic function with verifies MATH . In order to ensure that our sequence satisfy Conclusion REF , we need to reduce the sequence MATH. First, define a subset of MATH by MATH, then a subset MATH of MATH by MATH . REF imply that for an appropriate positive constant MATH, MATH, for any MATH in MATH, which yields conclusion REF . Let MATH, where MATH was defined after REF . Let us check conclusion REF . Since for any MATH, MATH, there exists MATH, MATH, such that MATH we see that if MATH (the aperture of the NAME angles) is larger than an absolute constant, then MATH for any MATH in MATH. (To deal with smaller values of MATH, we'd have to "thicken" the sequence by putting MATH equidistant points in each interval MATH, MATH depending on MATH. This has no ill effects on the properties we're interested in. Details are left to the reader). Then we have MATH by the definition of MATH and REF . The sequence MATH is NAME. Let MATH; then MATH . We will show that if MATH and MATH, MATH, then MATH. Accepting this, we then have, using REF , MATH . To prove our first assertion : if MATH, then MATH and REF implies that, for any MATH and MATH, MATH therefore by the definition of MATH as an upper bound, MATH. This implies the desired property. If the sequence MATH is empty, then by the above MATH, for any MATH, so MATH, and MATH in contradiction with our assumption. |
math/0106048 | Proof of necessity in REF : Given a function MATH so that MATH, consider the sequence MATH given by REF . Conclusion REF then implies that MATH, and REF then yields that MATH. Conclusion REF then show that MATH is not an essential minorant for MATH. |
math/0106048 | Proof of necessity in REF : Suppose that MATH is such that there exists a positive integer MATH and a subset MATH of MATH such that MATH and MATH . Without loss of generality, we may assume that MATH for all MATH, and apply REF to the unique function MATH such that MATH is constant on the intervals of the form MATH and MATH . We get sequences MATH and MATH such that MATH is NAME and NAME, and a harmonic function MATH. Let MATH. For any MATH in MATH, MATH, MATH. Define MATH . Thus MATH whenever MATH. We still need to prove that MATH is NAME. For any MATH such that MATH, by REF , MATH so that MATH and MATH is a NAME sequence, therefore MATH is NAME too. Let MATH, MATH, MATH, MATH be the sequences obtained respectively from MATH, MATH, MATH, MATH by adjoining to each point of the sequence a set of MATH separated points located in a hyperbolic neighborhood of fixed size so that there exists MATH greater than MATH such that MATH . It is easy to see that MATH is a NAME sequence and, using NAME 's inequality, that there exists a positive constant MATH such that MATH, for any MATH. Now we want to show that MATH. Take any MATH, and an integer MATH such that MATH. For MATH large enough, MATH, so MATH therefore MATH, and since MATH is NAME, MATH. REF then shows that MATH is not an essential minorant for MATH. |
math/0106048 | Proof of REF : The proof follows exactly the arguments in CITE, so we only sketch it. First, the following "weighted" version of CITE can be proved in exactly the same way. For a real-valued function MATH on the disk, set MATH. We denote by MATH the MATH-dimensional NAME measure. For any MATH as in REF , MATH as above, if MATH then for any (super)harmonic function MATH on MATH, MATH . REF implies, as in CITE, that the given condition is sufficient for MATH to be an essential minorant for MATH (we actually have an if and only if statement, but don't need it right now). Conversely, if we take MATH such that MATH, REF yields a NAME sequence MATH, a separated sequence MATH and a harmonic function MATH so that MATH. We also have, for the sequence MATH, MATH so that MATH and therefore MATH. |
math/0106048 | Proof of REF : Let MATH, MATH separated. We want to prove that MATH, that is to bound the sum in REF . Associate to any point MATH in MATH the arc MATH, and define MATH where MATH stands for the characteristic function of the arc MATH. Since MATH, the sum which we are trying to control is bounded by a constant multiple of MATH. Consider any point MATH where MATH. Then MATH, where MATH, and MATH. As in the proof of REF , we know that in MATH, there can be no more than MATH points of MATH; furthermore the monotonicity of MATH shows that the sum defining MATH is largest when we take the points MATH with the smallest possible moduli. We thus get MATH . Recalling the definition of MATH, we conclude with an integration by parts: MATH . To see that MATH contains no MATH when MATH tends to MATH, consider a sequence of the form MATH . Picking MATH an increasing sequence of integers so that MATH, we see that MATH; on the other hand, it is easy to see that MATH almost everywhere on MATH (for a MATH large enough, and therefore for any MATH). Thus MATH, for any MATH such that MATH. |
math/0106054 | Let MATH be the image of MATH under MATH. Then MATH is finite and invariant under MATH. Let MATH. We can identify the tangent spaces of MATH and MATH, and by the functoriality of exponential functions, the exponential function of MATH is the map MATH which is surjective and has kernel MATH. |
math/0106054 | MATH has no non-zero zero divisors because MATH is assumed to be simple. Now if MATH, then the kernel of MATH is a sub-algebraic group invariant under the action of MATH. Its connected component will be a sub-MATH-module of MATH, and so the kernel of MATH must be finite. Thus REF gives MATH and MATH so that, (as elements of MATH), MATH . Therefore MATH is a division algebra, and its center clearly contains MATH. |
math/0106054 | Replace the word ``isomorphism" in the statement and proof of REF by ``isogeny". |
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