Subset (4)

proofs · 3.68M rows

Split (1)

train · 3.68M rows

paper stringlengths 9 16	proof stringlengths 0 131k
math/0003117	Let MATH be the set of elements MATH of MATH such that a damage rectangle intersects MATH during MATH. Then MATH is covered by an interval of size MATH. At time MATH in MATH, assume the domain of MATH has exposed edges in MATH. If the colony is not an originating colony we can choose the edge to look towards the originating colony. Indeed, if there is no exposed edge pointing in this direction then we can redirect the path by a horizontal stretch to the originating colony, as in REF. Without loss of generality, assume that our edge is a left edge MATH. The conditions of REF (Bad Gap Inference) are satisfied for MATH as the cell MATH in that lemma. REF of the conclusion of the lemma does not happen since it would satisfy the conditions of the Running Gap Corollary for MATH: hence MATH would follow, contrary to the definition of MATH. REF implies REF of the present lemma and it is also easy to see that then MATH is not a member cell. As in the proof of REF (Small Progress), we can show that REF of the Bad Gap Inference Lemma does not occur. Consider REF of that lemma. As in the proof of the Small Progress Lemma, we can conclude that cases other than REF (Exposing) do not occur. These remaining cases result again, in at most one additional time interval MATH f size MATH to be excluded (not two, since the event in question can occur at only one end of the colony MATH). Thus, let MATH.
math/0003117	We can follow the proof of the NAME Lemma for the special case considered here: that the domain is only over MATH, consisting of member cells or of internal germ cells. It is easy to verify that applicable cases of the Bad Gap Inference Lemma remain that lead to a widening gap.
math/0003117	Without loss of generality, assume that MATH is not to the left of the center of its originating colony. Let us build the path MATH with MATH backward in such a way that vertical, parental and horizontal links are chosen in this order of preference. Horizontal links are chosen only in the case of the application of REF (Ancestor). Whenever we have a choice between father and mother we choose the parent towards the center of the originating colony of MATH or towards the center of the current colony, as the construction requries. Assume that MATH is not an outer cell younger than MATH and one of the following holds: CASE: MATH does not leave MATH during MATH; CASE: MATH is not adjacent to its originating colony; CASE: MATH is damage-free; Then it can be attributed to its own colony. Let us direct the path always towards the center of the current colony. If the path did not leave MATH during MATH one finds, applying the NAME Lemma, some time MATH during this period when the colony is covered by a single domain. At time MATH, we can extend the path along horizontal links to the center of the colony MATH of MATH. Continuing backward from there by the original method, the path will never reach the edges of the colony, since the only way for it to move away from the center is by one horizontal link near a damage rectangle, and this can only happen once. Hence an application of the NAME Lemma finishes the proof. If MATH is not adjacent to its originating colony then the backward path constructed by our method never leaves MATH during MATH. Indeed, there is at most one horizontal link in the path during this time, and the parental links do not move the path away from the center. If neither of the above conditions holds but MATH is damage-free then MATH is an endcell of MATH during this time. It must have a sibling in MATH sometime during this interval since otherwise it would be killed by MATH. We can redirect the path into the sibling and continue from there. If MATH is an outer cell then MATH can be redirected to the originating colony during MATH and then continued there without leaving it again, as above. Without loss of generality, suppose that MATH is a right outer cell. The NAME Lemma implies that for MATH, at all times MATH in MATH with MATH as defined there, the cell MATH is in a domain that has no exposed edges in MATH. At any such time MATH the leftmost cell of MATH belongs to the domain since otherwise the left edge of the domain would be exposed. If it has a sibling in MATH then let us then direct MATH to this sibling. According to REF, MATH can be continued from here without having to leave MATH again. If the left end has no left sibling then it must be already a member cell in a new work period. REF shows that during MATH there are no cells to the right end of the domain that could prevent the increase of age as in REF (Address and Age). Therefore according to REF (Small Progress), the minimum age of cells in the domain keeps decreasing along the backward path. But when it decreases by more than MATH the left edge of the colony is also an outer cell and has a left sibling in MATH to which the path can be redirected. Assume that MATH is a member cell, the interval MATH is not damage-free and MATH leaves MATH during this time. Then MATH can still be attributed to its own colony. It follows from the cases already considered in REF, that we can assume that the path leaves MATH on the left and MATH is the left endcell of MATH, that before leaving MATH the path is vertical and the cells on it do not have any right sibling. The path leaves MATH along a horizontal link MATH. Indeed, suppose it is a parental link. The rule animating MATH could not be the growth rule since MATH is a member cell and due to REF, the path is not long enough to get the age from the expansion period to the end of the work period. The healing rule does not act across colony boundaries. The horizontal link across the colony boundary is created as in the proof of REF (Ancestor), therefore the extended damage rectangle is near the colony boundary, near time MATH. If MATH is an outer cell then REF implies that the path can be redirected within the desired time to MATH. Let us show that MATH is not member cell. It could be one only if MATH crosses into MATH at an age when MATH is a newly created colony covered with outer cells and MATH is in the originating colony. Let MATH be such that MATH is still a member cell but MATH is already an outer cell. Then the transition from MATH to MATH involves a transition from outer to member status at the end of the work period. This cannnot happen, however, under our assumption that MATH has no right sibling.
math/0003117	The proof is a repeated application of REF (Small Progress), with some case distinctions and delays due to damage. Let us build the path MATH backward like in the proof of REF (Attribution), till MATH . This path does not leave MATH. Indeed, it could only leave if there was a number of steps in which the path moves towards the edge of the colony. Each such step but one must be a parental link. If the animation was via healing then the healing would be an internal correction and the parent towards the center would be chosen, so this is excluded in all cases but one. The animation cannot be via growth since due to REF, the age of cells along the path MATH cannot decrease enough for them to become growth cells. Let MATH be the maximal domain containing MATH, and let MATH be the ``local" minimum value of age in MATH (taking into account the possible crossing of work period boundary). Let MATH. According to the NAME Lemma, there is a set MATH coverable by REF intervals of size MATH such that for all MATH in MATH the colony MATH is covered by MATH. If MATH then let us change MATH to a little larger value, so that MATH. Then MATH can be led back to the center at time MATH. It remains to show MATH. Let us represent MATH as a union of at most REF disjoint closed intervals MATH (MATH). In each interval MATH, according to REF (Small Progress), MATH keeps increasing to MATH until an exposed edge (say, a right edge) occurs at some time MATH. By the same lemma, for MATH, this edge moves at least MATH cell widths till time MATH towards MATH unless it disappears earlier by reaching MATH. The damage can stop this process and move right the end of MATH by at most REF cells. Therefore except for a union MATH of two time intervals of a total length of MATH accounting for right exposed edges and another exception set MATH accounting for left ones, REF of the Small Progress Lemma is applicable. The set MATH breaks up into a disjoint union of at most REF intervals MATH. The domain MATH covers MATH and has no exposed edges during these intervals, so MATH increases in them as in the Small Progress Lemma, and the total increase during MATH is at least MATH .
math/0003117	Let us prove REF. The damage can change only a single cell MATH of MATH. If this cell is at an end of MATH then there is nothing to prove. Otherwise, MATH and MATH are also in MATH. Assume first that both MATH and MATH survive until MATH. Then we have a case of internal correction; let us show that this correction succeeds. If an internal correction of MATH is also possible before internal correction starts in MATH (it cannot become possible after the correction of MATH started because either MATH was dead or the correction of MATH starts with killing MATH) then the reasoning for MATH can be applied to MATH. Suppose therefore that internal correction of MATH is not possible. The correction begins by killing MATH (in case it is not mild). Let us show that it succeeds by reanimating it. If MATH is not a left edge any time before MATH then the internal correction of MATH will succeed since the animation, needing a neighbor that is not isolated, can use MATH. Similarly if MATH is not a right edge any time before MATH. Let us show that if each of these cells is an edge pointing away from MATH then one of them dies before MATH contrary to our assumption. Certainly, one of them is an exposed edge, suppose that for example, MATH is a left exposed edge at some time MATH in MATH. REF (Bad Gap Inference) implies that either there is a bad gap on the left of MATH or there is a backward path of length MATH leading from MATH to a cell in the same colony that changes, at age MATH, from a protected to an exposed left edge, or undergoes a planned kill. If there is a bad gap of the indicated size then MATH is an isolated cell during MATH. It is easy to see that it is not an endcell of a near-end correction and therefore it will be purged by the time MATH, contrary to the assumption that it stays alive. Suppose now that a backward path of length MATH leads from MATH to a cell in the same colony that changes, at age MATH, from a protected to an exposed left edge. Then it is easy to see that MATH (which also is a right edge at some time before MATH) is an exposed cell for which the Bad Gap Inference Lemma really infers a bad gap on its right, resulting in its purge. Suppose now that for example, MATH dies before MATH. If there will be no cells after MATH that can be traced back to MATH only bey tracing them through the left or MATH, then we are done. If there are at any time MATH, then they must be connected by parental and horizontal links to MATH, therefore they form a domain from MATH left. This domain can only be killed from the edges. MATH cannot be the last element to be killed since then the whole domain disappears by MATH. Therefore there must be moment MATH when MATH and MATH are alive and MATH will be killed. The only rule accomplishing this is MATH, if MATH is a right exposed edge and the gap in MATH cannot be healed. Since MATH is still there this could only happen if no internal correction is possible in MATH, since otherwise this internal correction would have succeeded already before the killing. The only obstacle to the internal correction is if MATH has no right sibling and is not a right protected endcell. It is easy to see using REF (Bad Gap Inference) that there is no domain on the right of MATH with which the successor of MATH could merge in the given time period. Also healing the gap caused by the death of MATH is not possible since (it was created due to the impossibility of healing to begin with). Therefore the the whole domain containing MATH (and possibly containing MATH and even MATH if the latter arose by growth in the meantime) decays within further MATH hours. Let us prove REF now. Suppose that MATH covers MATH at time MATH and all cells of MATH are member cells. Let MATH be the cell changed by the damage. If it is not an endcell or next to an endcell of MATH then (given the absence of planned kill) the change caused by the damage will clearly be corrected by the rule MATH. Suppose that MATH is a left endcell. Then the only obstacle to healing MATH is if there is a competing internal correction in MATH that would create a member cell in MATH. This cell could only belong to a colony MATH different from MATH since MATH is an endcell in MATH. Since there is a correction in MATH the cell MATH must belong to MATH. It is clearly exposed to the right at the observation time of the correction (since the correction would create a member cell, a protected edge is excluded). It must have been there as an exposed cell for a significant time. Indeed, according to REF (NAME Ancestor), MATH has been full for a long time, the growth could not have created MATH during quite a long interval before MATH, and healing could not have created it while MATH was an endcell of MATH. But it could not have been there long since the Decay rule or the Purge rule would have killed it. Suppose now that MATH is a left endcell. Then it is part of an end-correction. According to the healing rule, one possible obstacle to carrying out the end-correction is that there is an internal correction in MATH. In this case, MATH is the endcell of a near-end correction, therefore it will not be purged, and the near-end correction will be carried out. All other possible obstacles listed in the healing rule imply that MATH has been right exposed cell for such a long time before MATH that it would have died. If the healing rule tries to correct MATH it still must be proved that it succeeds in doing so. If MATH is not latent then first MATH will be made latent, then line REF MATH in Subsection REF applies to cell MATH to make a member cell in MATH by setting MATH and triggering thereby rule MATH of Subsection REF. We must show that the animation of the latent MATH succeeds. The only obstacle to the animation would be an end-healing of member cells of some other colony MATH from the left. But this cannot happen since then the rightmost live cell of MATH must have been exposed before the damage, and REF (Bad Gap Inference) would imply that it has a large gap on its right, excluding right end-healing. If MATH is vacant then first it must be made latent by REF of Subsection REF: for this, REF (Creation) can be used. Again, no obstacle will be posed by any opposite end-healing. Adding the time upper bound in the Creation Lemma to MATH, we get the upper bound MATH for the healing time. Due to to REF, this is still smaller than the time needed for the decay rule to kill any cell made exposed by the damage.
math/0003117	We perform the analysis only for MATH. For any MATH, the ``immediate" effect of the damage can affect at most REF symbols of MATH. The damage rectangle can have immediate effect in the cell where it happens, and its healing can change possibly one more cell next to it. If the affected cells happens to contain the symbols from MATH then this can affect two information symbols and two error-check symbols. Assume that the damage rectangle occurs before point REF. Let us look at the damage-free operations in MATH after REF. The first such operation turns MATH into all REF or all REF. If MATH then the last step writes MATH into MATH, and thus the latter will have no errors. Suppose therefore that if MATH. Then before MATH, we have MATH in all but two cells. Let us look at any cell MATH not affected immediately by damage (the large majority is such). Since the majority step computing MATH is correct in MATH, for all MATH there must have been two values of MATH for which MATH wrote the value REF into MATH. For at least one of these MATH, say MATH, the computing period MATH is damage-free. That period wrote an error-free string into MATH, and also determined that this string does not differ from MATH by more than REF symbols. There is another damage-free period MATH. Its MATH could have been different at its beginning (if damage occurred between the two periods), though according to REF by not more than REF symbols. Therefore, since our code corrects REF errors, we have MATH. Except for the immediate effect of new damage, this is going to be the new value of MATH, therefore there can be at most these REF new errors at the end of MATH. Suppose that at the beginning, each MATH has at most REF errors. By REF, the damage rectangle affects at most REF cells of MATH immediately. Therefore during the whole MATH, the value MATH does not change and MATH has at most REF errors. Assume that the damage rectangle occurs on line REF. Then the damage did not occur before and therefore the track MATH contains either all REF's or all REF's. Suppose the damage has occurred before line REF, in iteration MATH of that rule. Then only MATH will be affected, and possibly two more cells, the cell where the damage occurred and a cell next to it due to healing. In REF other iterations, the track MATH still contains either all REF's or all REF's with these exceptions, and the result of these REF iterations is the same as it was without the damage in the third iteration. The majority vote brings out the same result, except for possibly REF immediately affected cells. If the damage occurs after line REF, in REF, that is, in the last step of the rule, then again only the cells immediately affected will not have the correct result. If the damage occurs after line REF then again, this being a one-step action, only the cells immediately affected by the damage would show up as errors.
math/0003117	The proof is similar to, only simpler than the proof of REF : simpler, since at the starting point of MATH, we can already assume that the conditions asserted by that lemma hold.
math/0003117	The discussion in Subsection REF has shown that MATH iff the damage can be covered by a single damage rectangle in MATH where MATH is defined in REF. The space projection of MATH is at least MATH and the time projection is at least MATH. Thus, our assumption implies that during every time interval of size MATH in MATH, the interval MATH is intersected by at most one damage rectangle. Consider a time MATH such that MATH. If MATH is defined then let MATH, while if MATH is defined then let it be some time very close from below to MATH. According to REF, there is a time MATH in MATH such that the colony MATH is covered with member cells belonging to the same work period at time MATH and is not intersected by the extended damage rectangle at this time. REF (Large Progress) says that, by setting MATH, there is a time MATH, and a path MATH in MATH with MATH within distance MATH of the center of MATH such that MATH is not in the wake of a damage rectangle, is covered by a domain, and MATH. If MATH is a member cell let us apply the Large Progress Lemma (backward) repeatedly until either the path crosses a work period boundary, or the age along the path decreases by MATH . Suppose the cell we arrived at is not a member cell and has MATH. Then we perform one more repetition, but stop along the path as soon as the age decreases below MATH. REF (NAME Ancestor) shows that we can stop at a time when the colony is covered by a domain. Let us now denote by MATH the cell at which we arrive this way. Assume that MATH is an outer cell. Then we have MATH and at time MATH, the whole colony is covered by outer cells. REF (Attribution) shows that MATH can be attributed to its own colony and that this colony could not have been occupied by member cells for at least MATH time units before MATH. Therefore MATH. In this case we have MATH. The outer cells at time MATH encode a latent cell of medium MATH. Therefore eventually, when all outer cells turn into member cells, the new colony will encode a latent cell. Now, a vacant-to-latent transition is legal by REF . The reasoning is analogous in case MATH is a germ cell. Assume now that MATH is a member cell in the same work period as MATH: then MATH. Then we have the case when MATH are defined. The Large Progress Lemma can extend the path repeatedly until we arrive at a cell MATH in the same work period, with MATH . One can find a time MATH in MATH such that denoting MATH we have MATH and there is a complete damage-free execution of MATH during MATH. This execution of MATH finds more than MATH cells in MATH with MATH. Assume that this is not true. Then this execution of MATH sets the common value of MATH to a value that was MATH for one of the cells MATH at the time MATH when it examined MATH. By REF, then MATH. It is easy to see that subsequent executions of MATH do not change this value of MATH (except for the spot of damage, which will be corrected). Therefore the work period will have to end at an age before MATH which is not possible given our definition of MATH, since we came back MATH steps within the work period to MATH and went forward at most MATH steps. Let us continue applying the Large Progress Lemma from MATH and extending the path backward from MATH until a cell MATH with MATH or MATH, whichever comes first. Look at the cells MATH which were found by the above-mentioned run of MATH at some time MATH to have MATH and hence MATH. There are so many of them that some of them will be damage-free during the whole work period. Take one such cell MATH. REF keeps increasing MATH between times MATH and MATH, hence MATH, and MATH . Let us show that MATH is not possible. Indeed, this would mean MATH. Then the big cell MATH would go through a switch of a state in MATH but our backward path has never changed a colony work period. Under the assumptions, the switch in MATH is legal. Let us follow the development of the colony MATH, starting from the time MATH found above. Since the big cell is non-vacant during this, location MATH encodes a string with at most REF errors. The rule MATH, begins, at age MATH, with an application of MATH. As the corresponding lemma shows, at the end of this, location MATH will still encode the same string with possibly fewer errors. Thereafter, the MATH track will not change until the next work period, except for the local changes caused by the damage and its corrections by new applications of MATH. The following computation can be analyzed similarly to the first part of the proof of REF (Refresh). According to the definition of REF, the value put into the part corresponding to MATH of MATH is obtained without using the result of MATH, so it is certainly a value legally obtainable from MATH. Any subsequent change of MATH or MATH caused by a damage rectangle will be corrected by a later MATH, and any new inhomogeneity of MATH will be corrected by MATH. If MATH then the whole colony dies, since the exposed right end is impossible to heal. In the case MATH, the rule MATH installs the changes on the track MATH in one step, leaving again no place for more than REF errors. In both cases, the state at time MATH will be a legal consequence of the state at time MATH. Let us lower-bound now the distance between the two switching times of MATH in MATH. (The proof of the upper bound is similar but simpler.) The usual analysis of MATH shows that this rule changes MATH either in all cells (but possibly two) or in none of them (but two). Suppose first that MATH stays that way during the whole work period. Then the end would come only at MATH. Measuring the time for a cell that has never been affected by damage, this would give a lower bound MATH. Suppose now that MATH becomes smaller than MATH. Consider the first execution of MATH when this happens. Then there must have been some damage-free cell MATH such that by the time MATH this execution inspects MATH it has MATH. Let MATH be the time when MATH at the beginning of the work period: then MATH. From here, the desired lower bound follows easily.
math/0003117	Suppose that MATH is a member cell and MATH was covered by member cells at some time in the set MATH (as defined around REF). Using the Large Progress lemma repeatedly, we can go back to a time before age MATH in MATH, just as in the proof of the Legality Lemma above. Then we can follow the development of the colony forward and see that it forms a continuous domain together with its extension. The MATH track has at most REF errors after the first application of MATH, as seen by REF (Refresh). If the computation results in a nonvacant value for the represented big cell then we have REF of the present lemma. The represented field MATH of the colony will be broadcast into the field MATH of its cells, as shown in REF . The homogeneity of this latter field will be maintained by MATH. Thus, depending on the value of MATH of the big cell, growth will take place and the growth forms a continuous domain with the originating colony. Suppose that the computation results in a vacant value. Then MATH will be REF everywhere but in the healable wake of the damage. Growth cannot start accidentally by a temporary wrong value MATH in an endcell since there is enough time to correct this value during the long waiting time of MATH. Also, all cells become doomed. After MATH, the doomed right end becomes exposed by definition, and the whole colony decays within MATH time units. Before that, the colony is full. After that, we have REF of the present lemma. If MATH is an outer cell then we have REF of the present lemma. The Attribution Lemma attributes MATH to the originating colony which is covered by member cells during MATH. It forms a continuous domain with its extension, until the age MATH. From that age on, they form a multidomain. This could only change if the originating colony goes through a next work period and kills itself; however, there is not enough time for this due to REF. Suppose that MATH is a member cell but during MATH, the colony is never covered by member cells. Then we have REF of the present lemma. Let MATH be a non-member cell in MATH during MATH. Then it is an outer cell that can be attributed to its originating colony and then the reasoning of REF above can be repeated.
math/0003117	The proof is similar to the proof of REF (Refresh).
math/0003117	As in the proof of REF (Legality), consider a time MATH such that MATH. If MATH is defined then MATH, while if MATH is defined then it is some time very close from below to MATH. Let us define MATH as in that proof. As in REF of that proof, the Large Progress Lemma can extend the path repeatedly backward from MATH until we either arrive at either an outer or germ cell or at a cell MATH in the same work period, with MATH. Assume that MATH is a germ cell. Then we have REF of the Computation Property. Repeated application of the Large Progress Lemma gives a time interval long enough during which the colony and the adjacent colonies on both sides have been covered by germ cells. This implies that the state of the new big cell is latent and also that for the big cell MATH, we have MATH for some time MATH in MATH and for MATH. Now assume that MATH is a member cell. From time MATH on, we can follow the development of colony MATH. The computation process can be treated similarly to the proof in the Legality Lemma; the additional problem is the retrieval of the needed information from the neighbor colonies in such a way that all retrieved information can be attributed to a single time instant. (This issue was solved once already in the proof of REF (Reach Extension)). Event REF implies that in the whole space-time area in which this communication takes place only the usual damage rectangles must be considered. Due to REF, the whole computation fits into an interval of length MATH and therefore in each direction, at most REF damage rectangles can affect the retrieval (considering communication with a non-adjacent neighbor colony). The extension arms of the colony will be extended by the rule MATH, defined in Subsection REF. The success of carrying out the extension over possible latent or germ cells or an opposing extension arm in the way is guaranteed by REF (Creation). If the extension cells of MATH are growth cells then they are stronger but the age check will fail in this case, so we can assume that this is not the case. If there are extension arms from both colony MATH and a non-adjacent neighbor colony MATH then the strength ordering gives preference to one side and therefore soon only one of the extension arms remains. REF , of REF sends the message MATH along the control line MATH to the cells of the nearest colony MATH (or its extension) on the left communicating with MATH. The existence and condition of MATH will be discussed below. This message will be maintained by the healing rule as a locally maintained field. Its value arrives on control line MATH to MATH (or its extension). There, its value is propagated (carefully) by MATH. Via the rule MATH, the signal causes the cells of MATH to post the needed information. In due time, REF , of REF sends the message MATH along the same control line to colonies MATH and MATH; this causes the track MATH of MATH to pass its information to the right, to track MATH of MATH, which passes it further to the right. Damage can occur during the passing phase, but after healing its effect is confined to information (not the control fields) in the cells (at most two) which it changed. During the same phase, due to the parallelly running REF of MATH, the MATH fields of each cell of MATH snatch the information addressed to them from the right-moving track MATH. At the end of MATH, REF will be applied. By REF (Age Check), at the end of the application of this rule, all but maybe REF cells of the track MATH contain the same value. We will say that the process passed or failed the age check. The rule MATH is iterated REF times but it will not affect the MATH tracks after it passes the age check. The age check can fail at most REF times. For every neighbor colony from which mail is to be retrieved, at most two iterations of MATH can fail the age check. Normally, there can only be one such failure, when the colony was found in a transition age: this gives at most REF failure on the left and REF failure on the right. Due to the idling between iterations of MATH, the next iteration will find the colony past the transition age. It can also happen that the neighbor colony was found in a transition age just before vanishing and next time it will be found in a transition age when it is just being recreated by a farther colony. However, it is easy to see that this can happen only for one neighbor colony on the left and one on the right, bringing the total number of failures to at most REF. Take an iteration of MATH that passes the age check. Let MATH be the earliest time after the beginning of this iteration when the whole area MATH is free of the healable wake of any damage rectangle, and let MATH be the end of this iteration. Let us call any colony MATH with base MATH in MATH a potential left neighbor colony. Suppose that there is a potential left neighbor colony MATH that stays covered during MATH (a temporary hole due to the damage rectangle and healing within MATH time units does not count). It is easy to see by REF (Present Attribution) that all cells between MATH and MATH (if any) not in the wake of a damage rectangle belong to extensions of either MATH or MATH. The retrieval from MATH proceeds according to the program. The mail will be passed from MATH to MATH through the remaining extension arm (if any). The age check guarantees that all the retrieved information can be attributed to a single instant of time before MATH, since it will not change during the at most MATH time units needed for its reading (we count MATH for each mail-passing step, and add MATH more time units for the possible push-back of extension arm by an opposite one). According to REF (Present Attribution), before retrieval, each of the strings to be retrieved represents some string within REF deviations. During retrieval, damage can change at most REF cells, increasing the number of deviations to at most to REF. Since we use a REF-error-correcting code the computation proceeds from here as expected. Suppose now that there is no potential left neighbor colony that stays covered during MATH. There are the following possibilities, according to the Present Attribution REF : CASE: No cells occur in MATH during MATH; CASE: Growth to the right from some colony MATH where MATH is a potential left neighbor colony; CASE: A potential left neighbor colony MATH is beginning to disappear as in REF of REF (Present Attribution). Then within MATH time units hereafter all cells of MATH disappear. Let us look at the MATH track before the age check, for MATH. If for some cell MATH on it MATH is not the direct result of damage and MATH then this value was obtained from a potential neighbor colony MATH by mail during the last retrieval cycle. In all the cases considered, this value is then MATH. Since the age check passes, REF (Age Check) implies that, with the possible exception of REF intervals of REF cells, the cells of MATH have MATH. Thus at the end of MATH, the string on the MATH track represents a vacant value within REF errors. There is a time interval of length at least MATH before MATH in which all big cells corresponding to potential left neighbor colonies MATH are vacant and in which therefore the information collected by MATH can be attributed to any instant. In REF , the big cell corresponding to colony MATH is indeed empty all the time. In REF , the growth process from colony MATH could not have started long before MATH since so few cells reported unsafe age: therefore the duration of growth provides the necessary long interval. In REF , cells of MATH start vanishing within MATH steps after MATH. Otherwise namely the MATH track would snatch values with MATH and then would not pass the age check. Once the vanishing of colony MATH started, it will be over within MATH time units. Repopulating it by growth will take sufficiently long that it does not create a new cell during this retrieval cycle. Assume that MATH is an outer cell. Then we have REF of the Computation Property. The fact that the new big cell is latent can be similarly concluded. The Attribution Lemma allows us to attribute the outer cell in question to a neighbor colony, say with base cell MATH on the left, from which it has grown. This implies MATH for this colony, which, as seen above, implies MATH in the big cell MATH. Now the reasoning of REF above can be applied to this big cell, seeing that it computes its next state according to the transition rule from information retrieved from its neighbors. In particular, it applies the rule MATH and therefore can only have MATH if the big cell MATH is vacant when observed. Therefore, on the other hand, the creation of the latent big cell MATH can be attributed to the rule MATH, hence the transition rule was also satisfied in big cell MATH.
math/0003117	Let MATH be the colony whose base is MATH. If there is any MATH with MATH, MATH and MATH then we are done; assume therefore that there is no such MATH. If there is any MATH in MATH when both MATH and MATH are vacant then at this time there is no potential creator, and we are done. The case remains when for all MATH in MATH there is a MATH, with MATH and there is no big cell MATH during this time whose body intersects the potential body of the big cell MATH. We will show that this case does not occur. Without loss of generality, suppose MATH . Suppose that there is no MATH in MATH with MATH. Then for all MATH in this interval, we have MATH. Tracing backward and forward the evolution of the colony of the big cell MATH and applying REF (Legality) and REF (Retrieval), we will find a whole work period MATH of colony MATH in MATH. At the end of this work period, growth to the right takes place. If the growth succeeds this would contradict our assumption: suppose that does not. This can only happen, according to REF (Creation), if some non-germ cell MATH is in the way. REF (Present Attribution), shows that MATH is a left extension cell of a live big cell MATH in MATH, and is therefore not stronger than the right growth it is keeping up. The rule MATH kills MATH since it prefers right growth to the left one, and therefore MATH does not really prevent the right growth from succeeding. Since this preference is arbitrary assume that the rule MATH actually prefers growth in the left direction and that MATH is a left growth cell. We can trace backward and forward the evolution of the colony of MATH to see that it carries its growth to conclusion, resulting in a new big cell MATH before time MATH, whose body intersects the body of MATH. This has also been excluded. Suppose that there is also a MATH in MATH with MATH. Tracing backward and forward the evolution of the colony of the big cells MATH and MATH (and applying the Legality and Retrieval lemmas) it is easy to see that they are non-vacant for all MATH in MATH (since according to REF , MATH implies MATH). It is easy to see that there is a MATH and MATH such that MATH is a dwell period of big cell MATH with MATH. Assume first that MATH can be chosen. In the colony MATH of MATH between the colonies MATH and MATH, due to REF (Present Attribution), all non-germ cells belong to the extension of one of these two colonies. The growth rule of MATH will try to create a latent big cell in MATH. If the growth rule of MATH is not active then there are no obstacles to this creation and it succeeds, contrary to the original assumption: so we are done. Moreover, it succeeds even if the growth rule of MATH is active since the strength relations favor growth from the left. Assume now that only MATH can be chosen. The reasoning is similar to the above except when the growth rule of MATH is active at the same time when MATH is trying to grow left. This can only happen when the dwell period MATH of big cell MATH overlaps with a dwell period MATH of big cell MATH with MATH. By our assumption, MATH. In the next reasoning, we will use the fact that growth is confined to a time interval of length at most MATH at the end of any colony work period. Assume MATH. Then we must have MATH. Hence the previous dwell period MATH of MATH is in MATH and must have MATH. Therefore MATH in the previous dwell period MATH of MATH. Due to the confinement of growth to the end of the work period, the growth to the left from MATH at the end of the period MATH will be undisturbed by growth to the right from MATH and the creation succeeds near time MATH, contrary to the original assumption. Assume now MATH. Then due to the confinement of growth just mentioned, we have MATH. Hence, MATH, and the next dwell period of MATH, contained entirely in MATH, must have MATH. Then the next dwell period of MATH is still entirely in MATH and has MATH, and the creation to the left succeeds.
math/0003117	In the proof below, controlling always means controlling MATH. Let MATH. For MATH in MATH, let MATH . Let MATH. Assume that MATH is controlled for all MATH. Then MATH contains a cell. By our assumption, for all MATH, each set MATH contains a cell MATH. If it stays until time MATH then we are done. If it disappears the death must have been caused by MATH, due to a cell MATH with MATH about to create another cell whose body intersects with the body of MATH. Cell MATH had MATH to be able to kill and therefore survives until time MATH. Let MATH. Assume that MATH is controlled for all MATH. Then MATH contains a cell. Bu assumption, for all MATH, each set MATH contains a cell MATH. Without loss of generality, suppose MATH. If this is also true for MATH then we are done. Suppose it does not: then one such cell MATH disappears at time MATH. This must have been caused by the rule MATH, due to some cell MATH about to create an adjacent neighbor whose body intersects with the body of MATH. Without loss of generality, assume MATH. According to the MATH, cell MATH must have had MATH, MATH to be able to erase and therefore survives until time MATH. Since according to REF , only the rule MATH changes MATH, and only when a right neighbor has appeared, cell MATH keeps trying to create a right neighbor. Suppose that MATH succeeds in creating MATH before time MATH. If MATH then it will control; otherwise, MATH will try to create MATH. If it succeeds in MATH time units then the created cell will control; if it does not then then according to the Computation Property, another cell in MATH interferes and it will control. Suppose that the creation does not succeed within MATH time units. Then a cell MATH interferes. If MATH then MATH will control, suppose therefore that MATH. We have MATH since MATH. Therefore MATH must have arisen after MATH disappeared. Due to REF above, MATH could not have arisen by spontaneous birth, hence it must have been created by a right neighbor. If MATH was created after MATH then its creating right neighbor is alive after MATH and it will control. Suppose therefore that MATH was created at time MATH and its right neighbor disappeared by time MATH. Suppose that MATH stays for at least MATH time units; then it tries to create MATH. If it succeeds then MATH will control; if it does not then the cell that interferes in the creation will control. Suppose that MATH disappears before MATH. Suppose that MATH now succeeds in creating MATH before time MATH. If MATH then it will control; otherwise, it will try to create MATH. If it succeeds in MATH time units then the created cell will control. If it does not then the cell that interferes will control. Suppose that MATH still does not succeed in its creation. Then another cell MATH appears before MATH that prevents this. This MATH must have been created by a right neighbor that will control.
math/0003117	Assume that for some sites MATH and times MATH, with MATH is blue and MATH is damage-free. We have to prove that MATH is blue. REF implies that this point is controlled; what remains to show is that the controlling cell in MATH is blue. Due to REF above, in the area considered, there is always some cell within distance MATH left or right of any site, and this prevents any germ cell from being born. If therefore the controlling cell would be non-blue then one could construct from it a path of non-blue cells backwards. This path cannot reach outside MATH during MATH since it needs at least MATH time units to move one cell width away from MATH.
math/0003117	Since MATH is controlled there is a cell MATH in MATH. Without loss of generality, assume that MATH. Suppose that MATH stays a cell until time MATH. Since MATH is controlled there is a cell MATH in MATH. If the cell in MATH exists until time MATH then necessarily MATH, and it is easy to see that there is a time MATH such that the points MATH, MATH, MATH, MATH form a path in MATH. Suppose that the cell in MATH lasts until MATH. Then it coexists with MATH at time MATH, therefore MATH. Since MATH is controlled there is a cell MATH in MATH. If the cell in MATH exists until time MATH then MATH, and it is easy to see that there are times MATH such that the points MATH, MATH, MATH, MATH, MATH, MATH form a path in MATH. If the cell in MATH disappears before time MATH then this has been caused by MATH, via a cell MATH about to create another cell intersecting with the body of MATH. As seen in similar discussions above, cell MATH exists during the whole of MATH. Therefore it is not equal to MATH which is assumed to disappear. But its body does not intersect the body of of either MATH or MATH hence MATH. If cell MATH lasts until MATH then it is easy to see that there are times MATH such that the points MATH, MATH, MATH, MATH, MATH, MATH, MATH form a path in MATH. Suppose that MATH disappears before MATH. Now MATH tries to create MATH which (if this succeeds) tries to create MATH, etc. until reaching MATH. If we reach MATH this defines a path again between MATH and MATH. Suppose that one of these creations, for example, the creation of MATH does not succeed. Then a cell exists in MATH . The space is not sufficient for this cell to be born spontaneously, therefore it can be traced back via creations to one of MATH or to some cell existing during MATH between MATH and MATH. This way again a path can be defined. Suppose that the cell in MATH disappears before MATH. This has been caused by MATH, via a cell MATH about to create another cell intersecting with the body of MATH; as above, cell MATH exists during the whole of MATH. Then either MATH or MATH. In the second case, there are points MATH such that MATH, MATH, MATH, MATH, MATH, MATH form a path in MATH. Suppose therefore MATH. After MATH dies, either MATH succeeds creating MATH before time MATH or some cell MATH interferes with this. In the first case, let us call MATH. In both cases, we can continue as in REF above with MATH in place of MATH and MATH in place of MATH. Suppose that the cell in MATH disappears before time MATH. This has been caused by MATH, via a cell MATH about to create another cell intersecting with the body of MATH; again, MATH exists through all the time considered here. If MATH then we can continue as in REF above with MATH in the role of MATH above and MATH in the role of MATH above. If MATH then we can do the same, after a left-right reflection.
math/0003117	Let MATH, MATH for some MATH, for our amplifier, and let MATH be a trajectory of MATH. Let MATH, MATH, etc. Let MATH . Assume that MATH is blue in MATH and MATH is damage-free in MATH. We must prove that MATH is blue in MATH. As mentioned at the beginning of REF, the simulation MATH consists of two parts: the simulation of MATH by a medium MATH of reach REF, and the simulation of MATH by MATH as shown in REF . Control delegation from MATH to MATH is trivial, since the cells of the two media are essentially the same; hence, we need to consider only control delegation from MATH to MATH. REF implies that there are sites MATH and a path in the graph MATH defined on MATH, connecting MATH with MATH. This path consists of blue big cells: indeed, since the area is controlled in MATH birth is excluded, therefore a non-blue big cell would have to trace back its origin to outside MATH, but the complement of MATH is too far to do this in the available time. The path gives rise in a natural way to a sequence MATH of small cells with MATH, MATH, and for each MATH, either MATH and MATH are in two consecutive dwell periods of the same cell, or have the form MATH with MATH. All these cells are member or extension cells of the colonies of the big cells of the path. Since MATH is a broadcast field and MATH in all our cells, these small cells are all blue. Let us form a polygon MATH connecting these points with straight lines. Let MATH be the set of elements of MATH reachable from MATH along any polygon at all without crossing MATH. It is easy to see that MATH is controlled; let us show that it is also blue. Indeed, according to REF , not birth can take place in a controlled area, therefore from any cell in MATH a path of ancestors passes to the outside of MATH. When it crosses MATH one of its cells must be equal to a blue cell and therefore the whole path must be blue.
math/0003117	Let us construct a path MATH backward from MATH in such a way that we keep the path on the left edge of the germ: thus, whenever possible we move left on a horizontal link, otherwise we move back in time on a vertical link. At time MATH the path might stop at a cell MATH one level lower than MATH. There is a time MATH in MATH when MATH is a leader and the germ has size MATH. Suppose this is not so. If the right end is exposed at time MATH then it remains so and the germ decays away in the given time. If the left end is exposed during all of MATH then the left end decays, becomes unhealable, and the germ will decay. Suppose therefore that both ends are protected by some time MATH. If the right end has level MATH at this time, then it has the desired size, too. Suppose the germ at the time MATH has size MATH and the right end has a lower level allowing this size. If the left end is younger than the computation start then the computation would never allow it to switch to a higher level with this small size, so the size MATH must be reached at some time MATH. If it is older than the computation start then (unless an end becomes exposed and the germs dies on that account) the age advances quickly to the new work period, and the whole germ obtains the new, higher level. If the germ is still small its right becomes exposed and then the whole germ will be wiped out. Let us follow the development of the germ from MATH forward. If the left end survives then the germ survives till time MATH and we are done. If the left end will be destroyed by a cell MATH with higher level then we are done. The last possibility is that the left end will be destroyed by an active cell MATH on the same level. Let us construct a path MATH backward from MATH in a way similar to MATH. By the reasoning of REF above we arrive at a time MATH at which MATH is a leader and the germ has size MATH. Following the development of this (active) germ forward, its left end either survives or will be killed by a cell MATH of level MATH (in which case we are done, having found MATH). If the left end survives then the germ of MATH kills the germ of MATH, growing over at least MATH killed cells. (It is easy to see that new germ cells arising in place of the killed ones do not have sufficient time to reach level MATH and are therefore no obstacle to this growth.) This creates a germ of size MATH whose level will increase after its next computation notices this. All this happens within at most REF work periods of the germ of MATH.
math/0003117	All cells in MATH are blue. NAME germ cells cannot arise in a blue area. If a non-blue non-germ cell would occur it could be attributed to a full non-blue colony. It takes two full consecutive colony-occupations to reach into MATH from outside MATH, and this takes at least MATH time units which is, by REF , more than our total time. REF shows that, in the absence of damage, either the conclusion of the present lemma holds or MATH is part of a germ of size at least MATH, without exposed edges. Let MATH be the left endcell of this germ. The following holds during MATH either a cell with level MATH occurs in MATH or a leader cell MATH with level MATH occurs in MATH. In the latter case, at time MATH, all cells between MATH and MATH belong to the germ of MATH, with no place left for cells between them. If MATH is killed during MATH from the left by a higher-level cell then we are done. Suppose that it will be killed by an active cell MATH of the same level. Then REF (Germ Attribution) implies that MATH is contained in a germ without exposed edges, hence of size MATH. This germ conquers the area of the germ of MATH within the next MATH time units, and creates a cell of level MATH, which is one of the desired outcomes. Suppose that this does not happen. If the growth of the germ of MATH will not be held up then it will create a cell of level MATH within MATH time units. It can be held up only by a cell of higher or equal level. A cell of higher level would give what is required: suppose that this does not happen. If it is held up by a cell of the same level, then unless this cell is a leader it decays within time MATH. Therefore the growth succeeds unless a leader cell MATH of the same level is in the way and all cells between MATH and MATH belong to the germ of MATH, with no place left for cells between them. Let MATH, MATH. Let MATH be the event that MATH, there is a leader cell MATH of level MATH in MATH, and at time MATH all cells between MATH and MATH belong to the germ of MATH, with no place left for another cell. The probability that MATH holds and no cell of level MATH occurs during MATH in MATH is at most MATH . If MATH will be killed from the left during MATH then we will be done by the same reasoning as in REF above. Suppose that this does not happen. Suppose that that MATH is the later one among MATH and MATH to make a work period transition after MATH, at time MATH. Let MATH be its next work period transition. Let MATH be the last work period transition time of MATH before MATH and MATH the next work period transitions of MATH. Let MATH be the event that the coin tosses at time MATH make MATH active and the coin toss at time MATH makes MATH passive. Suppose now that MATH is the later one to make its first transition after MATH, at time MATH. Let MATH be its next work period transition. Let MATH be the last work period transition of MATH before MATH and MATH its next work period transition. Let MATH be the event that the coin tosses at time MATH make MATH passive and the coin toss at time MATH makes MATH active. The randomization part of the Computation Property implies that, in the absence of damage, the probability of MATH in both cases is at most MATH. Let us show that in both cases if MATH holds then MATH. that is, the germ of MATH will overrun the germ of MATH while it is passive. In the first case, suppose that the work period MATH of MATH covers the work period MATH of MATH. Then MATH certainly has sufficient time to overrun. If MATH does not covers MATH then it is divided between MATH and MATH, therefore MATH will have sufficient time in one of these work periods for the overrun. The reasoning is analogous in the other case. Repeated application of the above reasoning gives the probability upper bound MATH. Indeed, we reason about MATH disjoint windows (separated by the times MATH where MATH). If MATH fails then MATH still holds in the window between MATH and MATH, so the reasoning is applicable to this new window, etc. The probability bounds will multiply due to the disjointness of the windows.
math/0003117	Repeated application of REF (Level Increase). Notice that MATH . For level MATH, let MATH . Then MATH for MATH. Let MATH be the event that a cell of level MATH appears in MATH before time MATH. The blueness of MATH implies MATH. Repeated application of REF (Level Increase) shows that the probability of MATH for MATH is at most MATH for an appropriate constant MATH. From here, we obtain the required upper bound on the probability of non-birth.
math/0003117	Let MATH be a trajectory of a medium MATH of our amplifier and MATH. We will use the notation MATH, etc. Let us define some quantities with the absolute constants MATH that will be defined later. MATH . Our goal is to show that for every time MATH, if MATH is MATH-blue at time MATH, then MATH is MATH-blue at time MATH. Then we will be done since REF implies MATH. As mentioned at the beginning of REF, the simulation MATH consists of two parts: the simulation of MATH by a medium MATH of reach REF, and the simulation of MATH by MATH as shown in REF . We will concentrate on the self-organization from MATH to MATH. The self-organization from MATH to MATH is similar but much simpler, so we omit it (it refers to the proof of REF ). We will use the inequalities MATH where REF is the same as REF follows from REF. Let MATH be a time, and let MATH be a space interval of length MATH. Let MATH . Assume that MATH is MATH-blue at time MATH. Let MATH be the event that there is no damage in MATH. The Restoration Property gives the upper bound MATH on the probability of MATH, where we used REF. Let MATH be the event that there is a subinterval MATH of MATH of size MATH such that MATH is blue. The probability of MATH is upper-bounded by MATH . The proof is a little similar to the proof of REF , only simpler. If MATH is blue for all MATH then clearly MATH is blue. For some MATH with MATH, let MATH be the event that this set is not blue, for either MATH or MATH. The blueness of MATH at time MATH implies the upper bound MATH on the probability of MATH. The probability that there is a MATH such that MATH holds is at most MATH. If there is no such pair MATH then it is easy to see that MATH holds. Under condition of MATH, let MATH be the subinterval introduced in the definition of MATH. Let MATH . Let MATH be the event that for all MATH such that MATH the set MATH contains a big blue cell. With MATH, REF (Birth) gives the upper bound MATH for an appropriate constant MATH, on the probability of MATH. Using REF, we can compute a constant MATH for upperbounding the above by MATH . We will show that under condition MATH, the trajectory MATH is blue over MATH . Assume that MATH is disjoint from MATH. MATH contains no non-blue cell. Assume that, on the contrary, it contains such a cell. Let us build a path of ancestors from this cell. This path cannot end in a birth inside MATH since REF (Lasting Control) implies that no birth takes place there. Let us show that the path will not leave this set either. We will show that it does not leave on the right. For this, whenever we have a choice between father and mother, we build the path to the left. This way, the path only moves to the right when the cell is obtained from its mother by growth or end-healing. Once the backward path moved into the originating colony of a growth it will stay there at least until the beginning of the work period before it can move right again. It needs therefore at least MATH time units to move over MATH colonies, and its total displacement will be at most MATH. For this to be smaller than the distance of the complement of MATH from MATH, we need MATH . Rearranging and using MATH from REF show that this is satisfied if MATH . Therefore the path never leaves MATH: this is impossible since the path is not blue but the set it is in is blue at time MATH. Let MATH: we will show that MATH becomes controlled in MATH at some time before MATH. Then REF will imply that it stays controlled until MATH and REF above shows that it is actually blue. Let MATH be such that MATH. Event MATH implies that MATH contains a big blue cell MATH. This cell is within distance MATH from MATH, say, to the left. As soon as it arises it begins to create a right neighbor, which also creates a right neighbor, etc., until the chain either reaches MATH or will be prevented by another blue cell MATH within distance MATH on the right. In the latter case, we continue from MATH, and see that the chain controls MATH by time MATH: hence we are done if MATH . Assume that MATH intersects MATH. There are no non-germ non-blue cells in MATH . The size of MATH is MATH according to our assumption about MATH. Therefore if it has any non-blue non-germ cells at time MATH then these cells are in domains with an exposed edge and will decay within MATH time units. It can be seen just as in REF above that non-blue non-germ edges don't have time to grow in from outside MATH. NAME germ cells in MATH may also begin to grow to the right. But within REF cell widths, they meet an existing blue cell MATH, as seen in REF of the proof of REF (Lasting Control). MATH will not be erased by a non-blue germ cell on the left. It will not be erased by a germ cell on the right either since germ cells erase other germ cells only in the right direction. It might be erased by a non-germ cell MATH on the right, that is about to create a left adjacent neighbor to itself. Cell MATH (or, even the neighbor MATH it creates) will then be the next obstacle and it can only be eliminated in a similar way. The time between these successive eliminations (except possibly between the first two) is at least MATH since for the next occurrence of such an event, we must wait for the colony in question to die and for a new colony to overtake and start growing a left extension. Therefore the total number of cells that can be added to the non-blue germ in this exotic way is at most MATH (using MATH again), which still leaves it a germ if MATH . Let MATH: we will show as in REF above that MATH becomes controlled in MATH at some time before MATH. Let MATH be closest possible to MATH while disjoint from that MATH. Event MATH implies that MATH contains a big blue cell MATH. This cell is within distance MATH from MATH, say, to the left. As soon as it arises it begins to create a right neighbor, which also creates a right neighbor, etc., until the chain either reaches MATH or will be prevented by another blue cell MATH within distance MATH on the right. In the latter case, we continue from MATH, and see that the chain controls MATH by time MATH: hence we are done if MATH . We have shown that the probability that MATH is not blue over MATH is at most MATH. We bounded it, referring only to events within a window with space projection MATH. It follows that for a group of MATH disjoint space translations of this window, the probability bounds that MATH is not blue over any of the corresponding translations of MATH, is bounded by MATH.
math/0003120	Let MATH be the corresponding field, which is given by REF , and let MATH. By REF , MATH.
math/0003120	Let MATH be a group such that MATH, and let MATH be a sequence of embeddings MATH. For each MATH, let MATH; and let MATH be a subset of MATH chosen so that CASE: MATH; CASE: MATH is MATH-invariant; and CASE: MATH for all MATH. After passing to a suitable subsequence if necessary, we can suppose that the following conditions hold. CASE: There exists a fixed subset MATH such that MATH for all MATH. CASE: For each MATH, let MATH be the restriction mapping, MATH; and let MATH be the embedding defined by MATH. Then MATH for all MATH. Fix any pair of ordinals MATH, MATH such that MATH. Let MATH be the restriction mapping, MATH. Then MATH, and so we can define a surjective homomorphism MATH by MATH. Clearly MATH; and it is easily checked that MATH.
math/0003120	After performing a preliminary forcing if necessary, we can also suppose that MATH. Let MATH be the free group on MATH generators. Then it is enough to show that MATH satisfies the MATH-compatibility condition. Let MATH be a (necessarily free) group such that MATH, and let MATH be a sequence of embeddings MATH. For each MATH, let MATH. Then MATH is a free group of cardinality at most MATH. By REF, there exists an embedding of MATH into MATH. So REF yields the existence of ordinals MATH and a surjective homomorphism MATH such that MATH and MATH.
math/0003120	Let MATH be the ground model. For each MATH, MATH, let MATH be an infinite cyclic group. For each MATH, let MATH; and let MATH. Define an action of MATH on MATH by MATH for all MATH, MATH; and let MATH be the corresponding semidirect product. Let MATH. Then the members of the normaliser tower of MATH in MATH are CASE: MATH; CASE: MATH; CASE: MATH. Clearly MATH is embeddable in MATH; and so MATH satisfies the MATH-compatibility condition. Let MATH be the notion of forcing, given by REF , which adjoins a graph MATH of cardinality MATH such that MATH. Let MATH be the corresponding field, which is given by REF . Then MATH is a group such that MATH and MATH .
math/0003120	This follows by an easy induction on MATH.
math/0003120	CASE: It is easily checked that MATH and that MATH and that, in general, for each MATH, MATH . CASE: For example, consider the case when MATH. Then for each MATH, MATH and for each MATH such that MATH, MATH .
math/0003120	For each MATH, let MATH be the subgroup of MATH such that MATH. Since MATH, it follows that MATH. For each ordinal MATH, MATH . We will argue by induction on MATH. The result is clear when MATH, and no difficulties arise when MATH is a limit ordinal. Suppose that MATH and that the result holds for MATH. Let MATH; and for each subgroup MATH such that MATH, let MATH. Then MATH and so we must show that MATH . But this is an immediate consequence of the Correspondence Theorem for subgroups of quotient groups, together with the observation that the normaliser of any subgroup MATH is the largest subgroup MATH such that MATH. It is now easy to complete the proof of REF . Applying REF , we see that if MATH, then the normaliser tower of MATH in MATH terminates in exactly MATH steps; and that if MATH, then the normaliser tower of MATH in MATH terminates in exactly MATH steps. This just leaves the cases when MATH. When MATH, we can take MATH; and when MATH, we can take MATH and MATH.
math/0003120	Let MATH. We will work with the symmetric group MATH rather than with MATH. Let MATH be the notion of forcing consisting of the conditions MATH such that the following hold. CASE: MATH. CASE: MATH is a subgroup of MATH such that MATH. CASE: MATH is a function which assigns a permutation MATH to each pair MATH. We set MATH if and only if CASE: MATH; CASE: MATH; CASE: MATH; and CASE: if MATH, then the restriction to MATH of the function, MATH, is an isomorphic embedding of MATH into MATH. MATH is MATH-closed. This is clear. MATH has the MATH . Suppose that MATH for MATH. Using the MATH-System Lemma and the assumption that MATH, after passing to a suitable subsequence if necessary, we can suppose that the following conditions are satisfied. CASE: There exists a fixed ordinal MATH such that MATH for all MATH. CASE: There exists a fixed subgroup MATH such that MATH for all MATH. CASE: There exists a fixed function MATH such that MATH for all MATH. Now fix any two ordinals MATH. Let MATH be the subgroup generated by MATH; and let MATH be any extension of MATH which satisfies REF . Then MATH is a common lower bound of MATH and MATH. For each MATH, the set MATH is dense in MATH. Let MATH. Then we can suppose that MATH. We can define an isomorphic embedding MATH by setting MATH for all MATH. Since MATH, it follows that there exists an isomorphic embedding MATH for each MATH. Hence there exists a condition MATH such that MATH and MATH. For each MATH, the set MATH is dense in MATH. Let MATH. Let MATH be the subgroup generated by MATH, and let MATH be any extension of MATH to MATH which satisfies REF . Then MATH. Let MATH be a generic filter, and let MATH be the corresponding generic extension. Working within MATH, for each MATH, let MATH . Then MATH. Let MATH; and define the function MATH by MATH. Then it is enough to show that MATH is an isomorphic embedding. If MATH, then MATH. Choose a condition MATH such that MATH. If MATH is any ordinal such that MATH, then MATH. Hence MATH. MATH is a group homomorphism. Let MATH, MATH. Let MATH be a condition such that MATH, MATH. Let MATH be any ordinal such that MATH; and let MATH be a condition such that MATH and MATH. Then MATH . It follows that MATH . Finally it is easily checked that MATH; and it follows that MATH. This completes the proof of REF .
math/0003120	Let MATH be an ordinal such that MATH. Then MATH. Let MATH be the notion of forcing obtained by applying REF to MATH. By REF , MATH satisfies our requirements.
math/0003120	This is clear.
math/0003120	Suppose that MATH for MATH. After passing to a suitable subsequence if necessary, we can suppose that the following conditions hold. CASE: There exists a fixed structure MATH such that MATH for all MATH. CASE: There exists a fixed set of restricted atomic types MATH such that MATH for all MATH. CASE: There is a fixed group MATH such that for each MATH, there exists an isomorphism MATH. CASE: For each MATH, let MATH be the embedding defined by MATH. Then MATH for all MATH. Since MATH satisfies the MATH-compatibility condition, there exist ordinals MATH and a surjective homomorphism MATH such that CASE: MATH; and CASE: MATH. Let MATH be the homomorphism defined by MATH. Clearly MATH. Note that if MATH, then MATH . Thus we also have that MATH. Consequently, we can define a condition MATH, MATH by MATH .
math/0003120	Let MATH be the structure for the language MATH such that CASE: the universe of MATH is the disjoint union MATH; CASE: for each relation MATH, MATH. In particular, if MATH, then MATH only realises the trivial restricted atomic type MATH over MATH. Hence none of the restricted atomic types in MATH is realised in MATH. Let MATH be the embedding such that for each MATH, CASE: MATH for all MATH; and CASE: MATH for all MATH. Then MATH.
math/0003120	Let MATH and MATH. We can suppose that MATH. Let MATH. Let MATH be a set of left coset representatives for MATH in MATH, chosen so that MATH. Let MATH be the structure for the language MATH such that CASE: the universe of MATH is the cartesian product MATH; and CASE: for each relation MATH, MATH . By identifying each MATH with the element MATH, we can regard MATH as a substructure of MATH. Once again, each element MATH only realises the trivial restricted atomic type MATH over MATH; and hence none of the restricted atomic types in MATH is realised in MATH. Define an action of MATH on MATH as follows. If MATH and MATH, then MATH where MATH and MATH are such that MATH. It is easily checked that this action yields a homomorphism MATH; and that MATH.
math/0003120	Left to the reader.
math/0003120	Suppose that MATH. Let MATH, MATH be the canonical MATH-names for MATH and MATH; and let MATH be a MATH-name for MATH. Then there exists a condition MATH such that MATH . Using the fact that MATH is MATH-closed, we can inductively construct a descending sequence of REF for MATH such that the following hold. CASE: MATH. CASE: For all MATH, there exists MATH such that MATH. CASE: For all MATH, there exists MATH such that MATH. Let MATH be the greatest lower bound of MATH in MATH. Then MATH, and there exists MATH such that MATH. Let MATH be the structure defined as follows. CASE: The universe of MATH is the disjoint union MATH. CASE: For each relation MATH, MATH. CASE: For each MATH, let MATH be a new binary relation symbol. Then we set MATH iff MATH, MATH and MATH. Once again, it is clear that none of the restricted atomic types in MATH is realised in MATH. Let MATH be the embedding such that CASE: MATH for all MATH; and CASE: MATH for all MATH. Then it is easily checked that MATH. Finally let MATH be the restricted atomic type defined by MATH and let MATH. MATH is omitted in MATH. If MATH realises MATH, then MATH. And if MATH, then MATH, and so MATH. But this contradicts the fact that MATH. Thus MATH. To simplify notation, suppose that MATH; so that MATH. Then for each MATH, we have that MATH, and hence MATH. But this means that MATH realises MATH, which is the final contradiction.
math/0003120	Let MATH, where MATH and MATH. If MATH, then the conjugacy class MATH is contained in MATH, and so MATH. So suppose that MATH. Let MATH be the canonical projection map. Then MATH and MATH. Hence there exists an element MATH such that MATH. Since MATH, it follows that MATH . Arguing as above, we now obtain that MATH.
math/0003120	Let MATH be the least ordinal such that MATH. First suppose that MATH is a limit ordinal. Then MATH, and MATH for all MATH. Consequently, if MATH is the least ordinal such that MATH, then MATH is a nontrivial ascendant subgroup of MATH. But this contradicts the assumption that MATH is strictly simple. Next suppose that MATH is a successor ordinal. Since MATH is a proper normal subgroup of MATH, it follows that MATH. Now notice that MATH and MATH, This implies that MATH. (For example, see REF.) But then MATH, which is a contradiction. The only remaining possibility is that MATH.
math/0003120	Let MATH; and let MATH be the automorphism tower of of MATH. Let MATH; and let MATH be the automorphism tower of MATH. We will prove by induction on MATH that MATH. First consider the case when MATH. Let MATH be any automorphism. Then MATH is a simple nonabelian normal subgroup of the direct product MATH. By REF , either MATH or MATH. Since MATH, it follows that MATH. As MATH is a normal subgroup of MATH, we must have that MATH. Note that MATH. Hence we must also have that MATH. It follows that MATH . Next suppose that MATH and that MATH. Let MATH be any automorphism. By REF , either MATH or MATH. As MATH is a strictly simple group, REF implies that MATH. Since MATH is a simple normal subgroup of MATH, it follows that MATH. Using the facts that MATH and MATH, we now see that MATH and MATH. Hence MATH . Finally no difficulties arise when MATH is a limit ordinal.
math/0003125	By REF, so there exists MATH with MATH . Let MATH, where MATH. Then: MATH which implies (via REF above) that: MATH . Set MATH, so that MATH. Then MATH and MATH as claimed.
math/0003125	See REF for the new presentation and REF for the old presentation.
math/0003125	Since MATH and MATH the assertion follows.
math/0003125	Observe that MATH implies that MATH for every MATH. Therefore: MATH because MATH .
math/0003125	If not, then MATH for some MATH and MATH which contradicts the minimality of MATH.
math/0003125	Our starting point is: MATH which implies that MATH. Since MATH is left-greedy, REF then implies that MATH and so MATH for some positive word MATH. Now: MATH . Since MATH, we conclude that: MATH . Iterating the construction, we obtain MATH for some positive word MATH, also MATH. Putting all of these together we learn that: MATH for some positive word MATH. Let MATH so that MATH . A straightforward calculation shows that MATH . Since MATH, we have MATH . By the minimality of MATH, we must have MATH. REF is proved.
math/0003125	Our first observation is that MATH (see REF ). Since MATH is the maximal head of MATH, it follows that MATH . Our second observation is that by REF and MATH is left-greedy, so that MATH, which implies that MATH and so: MATH . Our third observation is that by the definition of MATH we must have: MATH . Therefore the only thing that we need to prove is that MATH for MATH. We first prove the assertion for MATH. Assume that MATH. We will show that this leads to a contradiction to our choice of MATH. We are given that: MATH where the decomposition on the left comes from REF and the one on the right is the normal form for MATH. By REF the normal form for MATH is MATH . By REF, so MATH for some MATH. Since MATH, it follows that MATH and so MATH . Let MATH be the infimum of MATH. Then CASE: MATH by the above discussion, CASE: MATH by REF , and CASE: if MATH, then MATH since MATH is left-greedy. If MATH, then MATH which contradicts the minimality of MATH. So MATH. Then MATH for some MATH and so MATH so that MATH. However, by our choice of MATH, we know that MATH. Retracing our steps we conclude that the assumption MATH is impossible, so MATH. It remains to attack the cases MATH. The method is identical to the case MATH. Set MATH and let MATH play the role of MATH.
math/0003126	If MATH, the relations in question admit a solution if and only if MATH . The left-hand side of the above equation is precisely, MATH, the MATH spectral residue. It follows that if MATH, then the value of MATH can be freely chosen, and that the solutions are uniquely determined by this value.
math/0003126	The recursive nature of the MATH ensures that MATH vanishes for all MATH. The proposition now follows by inspection of REF .
math/0003126	By REF , the left hand side of the above identity is MATH times the MATH spectral residue corresponding to the potential MATH . Setting MATH and making a change of gauge MATH transforms REF into MATH . Multiplying through by MATH and setting MATH we recover the usual hypergeometric equation MATH . It follows that MATH and hence the MATH spectral residue is given by MATH or equivalently by MATH . The asserted identity now follows from the fundamental invariance property of spectral residues.
math/0003126	As in the preceding proof, REF shows that the left hand side of the present identity is MATH times the MATH spectral residue corresponding to the potential MATH . Setting MATH and making a change of gauge MATH transforms REF into MATH . Dividing through by MATH and setting MATH we obtain the following scaled variation of the confluent hypergeometric equation: MATH . It follows that MATH and hence that MATH . The asserted identity now follows from the fundamental invariance property of spectral residues.
math/0003126	By REF , the left hand side of the present identity is MATH times the MATH spectral residue corresponding to the potential MATH . The rest of the proof is similar to, but somewhat more involved than the proofs of the preceding two Propositions. Suffice it to say that with the above potential, REF can be integrated by means of a hypergeometric function. This fact, in turn, serves to establish the identity in question. The details of this argument are to be found in CITE.
math/0003126	Let MATH be a composition of MATH. Let us begin by noting that the corresponding MATH is sparse if and only if the complimentary composition consists of MATH's and MATH's only. It therefore follows that the enumerating function for MATH such that MATH is sparse is MATH . On the other hand the number of MATH such that MATH for any given MATH is precisely MATH . Hence the enumerating function for MATH such that MATH is MATH-sparse is MATH . The two enumerating functions are equal by REF .
math/0003126	Let us note that the number of points in MATH such that MATH is given by MATH . Hence, the enumerating function for the first class of subsets is given by MATH . Now there is a natural bijection between the set of compositions of MATH by MATH and the set of compositions of MATH by odd numbers. The bijection works by prepending a MATH to a composition of the former type, and then performing substitutions of the form MATH . Consequently, we can write MATH . Turning to the other class of subsets, the number of points MATH that satisfy MATH is given by MATH . Consequently the enumerating function for subsets MATH such that MATH is MATH-sparse is given by MATH where MATH ranges over all positive integers MATH . Next, using the identity in REF we have MATH . Using REF it now becomes a straightforward matter to compare MATH to MATH. Indeed, the desired conclusion follows from the following evident inequality: MATH the inequality being strict for MATH.
math/0003128	REF is nothing but the statement that for any MATH which can be found in for example, CITE and references therein. See also REF. The second part is a corollary of the first part plus a reconstruction theorem proved in CITE, which states that one can reconstructs MATH-point descendants provided that MATH is generated by divisor classes and one-point descendants are known.
math/0003128	It is clear that the vector bundles in this lemma are the push-forwards of two line bundles on the universal curve MATH considered in REF . When the coefficient MATH is negative, one has the following inclusion MATH . MATH is then obtained by pushing-forward the above inclusion of line bundles to MATH. MATH can be obtained in a similar way. When MATH is concave, MATH is positive so that the arrow of REF is reversed.
math/0003128	It is easy to see that the theorem follows from the above proposition by the formula MATH and the excess intersection theory. Note that MATH.
math/0003128	I. (Convex case) The convex case of the proposition is REF. For the convenience of the reader and future references, we reproduce NAME 's proof. As remarked above, it is clear from REF that the equivariant virtual class has its ``main" contribution from MATH . It is also clear that other contributions to MATH come from the boundary strata MATH (and possibly its substrata). It remains to study the boundary terms. This can be done by NAME 's graph construction of vector bundle morphisms. Let MATH and MATH be vector bundles on MATH. By REF there is a homomorphism MATH of vector bundles of the same rank MATH. Let MATH be the NAME bundle over MATH, with universal bundle MATH of rank MATH. There is a canonical embedding MATH taking MATH to MATH. Let MATH be the closure of the image of MATH and let MATH be a cycle in MATH of dimension equal to MATH, where MATH is the multiplicity of MATH. Let MATH be the map induced by projection, and MATH be the image of MATH. The philosophy of the graph construction is that different components MATH are responsible for different types of degeneration of MATH. Since MATH is generically of full rank, there is one component MATH in MATH with multiplicity one such that MATH embeds in MATH via the fibre of MATH. Other MATH map to proper sub-varieties MATH of MATH. It follows that REF MATH . To calculate the contributions from MATH, further study on the behavior of MATH on boundary strata MATH is needed. Let MATH be a generic stable map in strata MATH. MATH can be described fibrewisely (at a generic point) as: MATH where MATH is the parameterized component of MATH. The first map is simply the restriction and the last map is defined by multiplying a factor MATH, where MATH are the nodal points on the parameterized MATH and MATH the homogeneous coordinates on parameterized MATH. It follows that MATH drops ranks only on the comb strata of types MATH described in the previous section. Any hairy comb substrata obtained from MATH by further degenerating the unparameterized components will not further reduce the rank of MATH. Moreover, it has the following transversality property: MATH has generic corank MATH and MATH along MATH and MATH respectively. Then MATH has generic corank MATH along the intersection of two strata. The above fibrewise description actually holds globally. Namely, on the strata MATH, MATH is the following composition MATH where MATH is the projection (see REF), MATH and MATH on MATH are defined in REF . The last map in REF is the push-forward (along the first MATH) of the following map MATH . One can now apply the above study of degeneration type of MATH to the graph construction. Since the rank of MATH decreases only on the strata MATH, REF becomes MATH . Namely, the only MATH in REF are MATH and MATH. To write down MATH explicitly in terms of characteristic classes in MATH one would need a filtration of the above setting. Let us first deal with the simplest case when MATH, that is, MATH is a divisor. The universal curve on MATH (generically) consists of one parameterized and one unparameterized components. The kernel of MATH can be identified with MATH, where MATH and MATH is the kernel of the evaluation morphism MATH of bundles on MATH where MATH and MATH forgets the second marked point. The kernel MATH can be further filtered by the order of zeros of MATH: MATH where MATH consists of those sections of MATH which vanishes at least to the MATH-th order at the marking. Similarly we can filter MATH by the span of the image of MATH on MATH: MATH such that MATH is a filtration of MATH on MATH. Now for each MATH there are MATH components MATH REF because the infinitesimal property of MATH along MATH. Each of MATH is a birational image of a MATH-bundle over MATH. By a local computation CITE, MATH has multiplicity MATH, and the tautological bundle on MATH can be expressed in terms of the filtration of MATH and MATH: MATH . The contribution MATH from the boundary strata MATH can be therefore written as MATH where MATH . This is exactly what we are looking for. When MATH is not a divisorial stratum we may use the above transversality property (of MATH concerning the intersection of two boundary strata). Since every MATH is the intersection of divisors, and the corank of MATH at intersection is equal to the sum of the coranks generically, we can then obtain MATH for general MATH inductively. Let MATH and MATH be the MATH-bundle over MATH. Now apply graph construction to the following vector bundle morphism on (pulled back to) MATH . (Some obvious pull-backs will be omitted.) Then the components of MATH over MATH obtained from this construction map birationally to the components of MATH over MATH. For example, components of MATH corresponding to boundary strata MATH of MATH map to the components of MATH over the boundary strata MATH on MATH. This implies that MATH for any MATH with a fixed ordering of MATH, MATH are (birationally) towers of MATH-bundles over MATH (by first doing MATH then MATH, etc.). An explicit expression of MATH can therefore be obtained: MATH where MATH and MATH is defined to be the filtration of kernel on the MATH-th unparameterized component, similar to that defined in REF and MATH is the push-forward of the inclusion of line bundles on MATH to MATH: MATH . This completes our proof of the convex case. II. (Concave case) The proof of the concave case can in general be carried out in a similar way. However, some crucial modifications will be necessary. Now MATH and MATH (with MATH negative). REF guarantees that there is a bundle map MATH between MATH and MATH. We may apply NAME duality and find MATH. Carry out the graph construction to MATH. REF (on MATH) should be replaced by MATH . Here MATH are the nodal points on the parameterized MATH (see REF ) and MATH is the dualizing sheaf of MATH. A similar modification to REF should also take place: MATH . For the filtration of the kernel of MATH, we may use the following exact sequence (see REF) MATH where MATH is the residue at MATH. Again we can further filter MATH by the order of zeros of MATH, (as did in REF). Similar filtration can be defined on MATH. Namely MATH . Now a similar computation leads to MATH . The rest is straightforward and is left to the reader.
math/0003128	REF requires only to compute MATH, with MATH described by REF MATH . Using the same argument to the following diagram MATH one obtains that MATH . Dimension counting and REF gives us the proof of REF .
math/0003128	From REF for MATH, we only have to compute MATH. A straightforward modification, by fibre square REF and dimension counting, will lead to the conclusion that only degree zero (constant) terms survives in MATH. The difference is that the cohomological degree of MATH is only MATH, because the rank of MATH in this case is one less than that of MATH in convex case. Therefore MATH. This allows us to set MATH.
math/0003128	Set MATH and apply REF .
math/0003128	Expand REF . The MATH term is MATH . By REF the right hand side of REF is linear with respect to MATH. The linear in MATH term on the Right-hand side is exactly REF.
math/0003130	We note that the term in front of MATH in the definition of MATH has another expression : MATH . This follows by noting that MATH satisfies MATH . Then we have MATH which upon integrating gives MATH . Since MATH from integration by parts, subtracting REF from MATH and taking the limit MATH, we find that MATH has mean MATH, as required.
math/0003130	REF are consequences of REF. The result REF is obtained by applying the NAME method to the NAME REF . The specialty of the scaling MATH is related to the fact that the exponent term MATH of the anti-diagonal entry in the jump matrix of the NAME REF has the critical points at MATH, which are the stationary phase points in the asymptotic analysis (see CITE). Recalling MATH, we see that MATH corresponds to one of these stationary phase points. By analyzing MATH around this stationary phase point, we obtain MATH and MATH as MATH with the above MATH. Similar computation appeared in REF, and we omit the detailed computations here.
math/0003130	By l'Hopital's rule and REF , we have MATH where MATH . Then by REF again, we obtain MATH which implies that MATH . Hence as in the proof of REF , MATH using the asymptotics of MATH.
math/0003135	I construct the proof in stages beginning with the end result and finding successive sufficient conditions for the preceding steps. Observe that I actually prove a slightly stronger result: by allowing the even operator MATH to contain a constant term MATH, see REF , I prove the consistency of an invariant manifold model based upon the mode with eigenvalue MATH. When MATH this forms a centre-unstable manifold model for which a relevance theorem also ensures asymptotic completeness. CASE: To solve MATH to errors MATH, expand the centre manifold model REF to MATH as MATH where MATH are some difference operators, and where MATH and MATH implicitly refer to the MATH-th element. Note that superscripts to MATH denote exponentiation whereas those on MATH and MATH denote the index of coefficients in the asymptotic expansion: I do this because MATH and MATH have an implicit subscript MATH denoting the grid element under consideration. Then substitution into the pde and extracting powers of MATH shows that we require MATH . Similarly, substitution into the amplitude REF requires MATH . Whereas substitution into the ibc REF requires, evaluating the left-hand sides at MATH, MATH . REF form a well-posed system of equations for MATH and MATH. In many applications of centre manifold theory, because MATH is singular, we often solve each level in the hierarchy of equations in two steps: the first is to find MATH by ensuring that the right-hand side of MATH is in the range of MATH, this is the so-called ``solvability condition"; the second step is to find MATH. However, here we proceed to construct straightforwardly the solution of the entire set of equations in general. CASE: We show the hierarchy of differential REF are satisfied by functions MATH and give consistent finite difference operators MATH, if MATH satisfy the recursive difference equation MATH . CASE: By the following induction argument REF implies that MATH . Now, MATH . Then since REF is trivially true for MATH, it follows by induction that REF holds for all MATH provided MATH. Further, since MATH is constant by REF , MATH is constant, so higher order differences REF are all zero. CASE: By an ``even" operator MATH I mean one which only involves even order derivatives in MATH. Hence write MATH formally as an infinite sum of even powers of the central difference operator MATH for some coefficients MATH; for example, for the diffusion operator in REF , from CITE, MATH more generally, MATH could be any symmetric convolution operator for which the infinite sum REF forms a reasonable representation. Then REF ensures REF since MATH provided MATH which are precisely the operators required to make the model MATH of MATH consistent to MATH, when truncated as in REF to errors MATH. CASE: By simple substitution, a sequence of functions MATH satisfying the recurrence REF , amplitude REF and the internal boundary REF are MATH where, as always, MATH is a grid scaled coordinate and provided that for MATH and similarly MATH after defining MATH and MATH. CASE: Analysing the difference tables for MATH and MATH and straightforward induction using REF proves that MATH for integer MATH. Then the following MATH and MATH are the unique polynomials, of degree MATH and MATH respectively, also satisfying MATH and MATH: MATH as plotted in REF . For example, MATH and MATH. These polynomial MATH and MATH also need to satisfy the recurrences in REF pointwise in MATH. This is trivially true for MATH. Now, MATH is from REF a polynomial of degree MATH, from its difference table has the same zeros as MATH, it is MATH at MATH, and so must be MATH for all MATH. Similarly for MATH.
math/0003135	As in REF , we expand the the centre manifold ansatz REF to errors MATH: MATH where MATH and MATH are some difference operators, and MATH and MATH are functions defined about the MATH-th gridpoint, and they all implicitly refer to the MATH-th element (as before the superscript to MATH, MATH, MATH and MATH denotes an index in the series, whereas the superscript to MATH denotes exponentiation). After substitution into the pde, terms in MATH determines MATH and MATH as in the previous section. Upon extracting from the pde the coefficients of terms linear in MATH and of various powers of MATH requires us to solve MATH . Substitution of the expansion REF into the amplitude REF and the ibc's REF and equating coefficients of MATH leads to these internal boundary conditions for the MATH: MATH . Since MATH is an odd operator, we formally write it as the following infinite sum of odd powers of centred difference operators MATH for some coefficients MATH; for example (from CITE) MATH and more generally, MATH could be any antisymmetric convolution operator for which the infinite sum REF is reasonable. I prove that there exists solutions MATH, odd functions of MATH (about MATH), with MATH so that the model REF is consistent with the effect of the odd derivatives in MATH to errors MATH. Since we already know MATH and since MATH appears to vary for different problems, the operators MATH are determined by the solvability condition that all terms except MATH appearing in REF combine to be in the range of the singular MATH. CASE: First, prove that the even part of MATH cancels with the even part of MATH and so is eliminated from REF . As a preliminary step consider, using REF , MATH . Then from REF and since MATH is odd and MATH is even, see REF , MATH is odd and so the even part of MATH is MATH . Whereas on the right-hand side of REF the even part of MATH is MATH which exactly cancels with REF from the even part of MATH on the left-hand side of REF . CASE: Second, since the even components of REF that involve the various MATH cancel, and since MATH and MATH are even, then a particular solution MATH of REF may be found that is odd. REF then force these odd MATH to be the unique solutions in the space of finite degree polynomials. CASE: Third, consider evaluating the hierarchy of REF at the centre grid point of the element, MATH or equivalently MATH, and simplify the various contributions. CASE: Now MATH is an even operator and MATH an odd function, so MATH is an odd function, and thus MATH. CASE: Since MATH is a difference operator in MATH it does not affect the amplitude condition that MATH, and thus MATH. CASE: I have already shown that the even parts of MATH and MATH agree pointwise, so they certainly do at MATH, at which the odd parts must also trivially vanish together. Thus the choice REF is the unique one to satisfy this solvability condition for REF . Since the expansion REF then contains the exact differences up to MATH and MATH, the errors in the finite truncation of the finite difference model will be MATH from the even terms and MATH from the odd.
math/0003139	Conditions in the NAME forcing are of the form MATH, where MATH is a finite set of natural numbers and MATH is an infinite subset of MATH such that MATH. If MATH, and MATH, then MATH . Using the fact that truth values in MATH can be decided by pure extensions, we can find a condition MATH which forces that MATH is ``decided continuously" by MATH, more specifically: CASE: there is a function MATH with domain MATH, MATH, such that For all finite MATH and all MATH with sup-MATH: MATH. For any finite MATH let MATH . So MATH is a subtree of MATH. We have CASE: If MATH, then MATH, MATH hence (recall REF ): MATH. It follows that: CASE: for every finite MATH, non empty for simplicity, we have MATH hence lim-MATH is a finite subset of MATH. We also get: CASE: if MATH and sup-MATH, then MATH. [Proof: We know that already MATH decides MATH, and the conditions MATH and MATH (for MATH) are both stronger than MATH.] In particular, we get: CASE: for all finite MATH the sequence MATH converges to MATH. This means that for every MATH for every large enough MATH we have MATH . CASE: for MATH an infinite subset of MATH we let MATH CASE: for MATH finite subsets of MATH we let MATH the smallest MATH such that CASE: whenever MATH are distinct members of MATH then the length of MATH is MATH CASE: MATH . Note that MATH is well defined, as both MATH and MATH are finite. MATH for MATH let MATH so MATH is well defined being the maximum of a finite set of natural numbers. CASE: Note that MATH. So without loss of generality (as we can replace MATH by any infinite subset): CASE: if MATH then MATH hence CASE: if MATH then MATH . CASE: if MATH are finite subsets of MATH and MATH then MATH . Hence CASE: if MATH are finite subsets of MATH, not disjoint for notational simplicity, and MATH then MATH [ why? we can prove this by induction on max-MATH using REF MATH . Hence, letting MATH a finite subset of MATH (=a countable set), we have CASE: if MATH are infinite subsets of MATH, with intersection finite non empty then MATH is included in MATH so MATH is is a finite subset of MATH. [ why ? just use REF MATH and the definition of MATH . Now fix an uncountable family MATH of almost disjoint subsets of MATH. Let MATH. Then by REF, we have MATH. Hence MATH satisfies REF .,REF.,REF.,REF .
math/0003139	CASE: By REF By CITE .
math/0003139	Let MATH be a nontrivial ccc forcing. So for some MATH we have MATH``MATH does not belongs to V" Hence for some quadruple MATH we have: CASE: MATH are ordinals, MATH is a MATH-name and MATH``MATH is not from V" We can choose such quadruple with the ordinal MATH minimal. Necessarily MATH is a limit ordinal and for MATH, MATH is forced by MATH to belong to V. For MATH, let MATH and let MATH. Clearly MATH is a tree with MATH levels, and MATH forces that MATH is a new MATH-branch of it. Now MATH cannot have MATH pairwise incomparable elements, as if MATH for MATH are like that, we can find MATH such that: MATH and MATH``MATH is an initial segment of MATH"; now if MATH are compatible in MATH then MATH are comparable in MATH (being, both, the initial segment of some possible MATH). So MATH are pairwise incompatible contradiction to MATH satisfies the ccc" Also in MATH, by its choice, every member has above it elements of every higher level and there is no node above which the tree has no two distinct members of the same level as then MATH will be forced to belongs to V by some condition above MATH. Also as MATH satisfies the ccc, every level is countable, and by the minimality of MATH as we are allowed to change MATH clearly MATH is a regular cardinal. Now MATH is impossible by ``MATH satisfies the MATH-cc". As there are no NAME tree also MATH is impossible. So clearly MATH forces that MATH add reals so there is MATH as required.
math/0003139	Let MATH be a forcing notion with the Laver property which adds a real, say MATH. Consider any increasing function MATH. The function MATH has only MATH many possible values, that is, is bounded. So, by the Laver property there is a tree MATH and a condition MATH stronger than MATH with MATH . We have thus defined a family MATH. By REF , there is a family MATH such that MATH is finite Clearly, for MATH the conditions MATH and MATH must be incompatible, since any condition MATH stronger than both would force MATH which implies MATH, a contradiction. (Remark: While MATH and MATH can of course contain branches in MATH which did not exist in MATH, the fact that their intersection is a certain finite set is absolute between MATH and MATH.)
math/0003139	Recall the construction of the almost disjoint family after REF , which was used in REF. Instead of using an almost disjoint family of size MATH in the intermediate model we can use a MATH-name of an almost disjoint family of size continuum: Identify the set MATH there with MATH, then every MATH-name MATH of an element of MATH will induce a set MATH. Clearly, MATH . As before, a density argument ensures that there will be MATH many different functions MATH such that MATH appears in one of the generic NAME filters for some MATH, so we can strengthen the conclusion in REF, MATH to get a MATH-size set MATH rather than just an uncountable one.
math/0003144	This follows as every genus REF NAME surface is isomorphic to one of the form MATH. For such NAME surfaces, we have the homomorphism MATH that takes the coset MATH to the translation MATH.
math/0003144	The first bijection was established in REF . The second follows from REF .
math/0003144	For simplicity, we suppose that MATH has complex dimension REF. The relative holomorphic tangent bundle of MATH is the holomorphic line bundle MATH on MATH consisting of holomorphic tangent vectors to MATH that are tangent to the fibers of MATH. That is, MATH . The sheaf of holomorphic sections of its dual is called the relative dualizing sheaf and is often denoted by MATH. The push forward MATH of this sheaf to MATH is a holomorphic line bundle over MATH and has fiber MATH over MATH. Fix a reference point MATH. Let MATH be a local holomorphic section of MATH defined in a contractible neighbourhood MATH of MATH. Since bundle is locally topologically trivial, MATH is homeomorphic to MATH. We can thus identify MATH with MATH for each MATH. Fix a positive basis MATH of MATH. This can be viewed as a positive basis of MATH for all MATH. It is not difficult to show that MATH vary holomorphically with MATH. It follows that MATH varies holomorphically with MATH. This shows that map MATH is holomorphic in the neighbourhood MATH of MATH. It follows that MATH is holomorphic.
math/0003144	The first step is to show that if MATH, then MATH is NAME dense in MATH. To prove this, it suffices to show that MATH is NAME dense in the set of complex points MATH. From NAME theory (or algebraic group theory), we know that all proper subgroups of MATH are extensions of a finite group by a solvable group. Since MATH is isomorphic to MATH, and since MATH is not solvable (as it contains a free group of rank MATH), MATH cannot be contained in any proper algebraic subgroup of MATH, and is therefore NAME dense. An element MATH of MATH stabilizes MATH if and only if MATH for all MATH - that is, if and only if MATH centralizes MATH. But MATH centralizes MATH if and only if it centralizes the NAME closure of MATH. Since MATH is NAME dense in MATH, and since the center of MATH is trivial, it follows that MATH is trivial.
math/0003144	It can be shown, using deformation theory, that the NAME tangent space of MATH at the representation MATH of MATH is given by the relative cohomology group: MATH where MATH is the local system (that is, locally constant sheaf) over MATH whose fiber over MATH is MATH and whose monodromy representation is the homomorphism MATH . Here MATH denotes the NAME algebra of MATH, which is the set of MATH matrices of trace REF, and MATH is the adjoint representation MATH . A result equivalent to this was first proved by CITE. The long exact sequence of the pair MATH with coefficients in MATH gives a short exact sequence MATH and an isomorphism MATH. Deformation theory also tells us that if MATH vanishes, then MATH is smooth at MATH. The Killing form is the symmetric bilinear form on MATH given by MATH . It is non-degenerate and invariant under MATH. It follows that MATH is isomorphic to its dual MATH as a local system. Thus it follows from NAME duality (with twisted coefficients) that MATH . But if MATH is an element of MATH, it follows from REF that MATH vanishes. Consequently, if MATH, then MATH and MATH . We saw above that MATH has dimension MATH, but MATH is not bigger than the dimension of any of its NAME tangent spaces, which we have just shown is MATH. It follows that MATH and that it is smooth as its dimension equals the rank of its NAME tangent space. Finally, since the MATH-action on MATH is principal, MATH is smooth of dimension MATH.
math/0003144	Since a pair of pants has the homotopy type of a bouquet of REF circles, it has NAME characteristic MATH. Since the NAME characteristic of a disjoint union of circles is REF, we have MATH . Thus MATH. Since each pair of pants has REF boundary components, and since each circle lies on the boundary of two pairs of pants, we see that MATH .
math/0003144	A proof can be found in the NAME REF, partie II by NAME in the book CITE. The basic idea is that a pair of pants can be cut into two isomorphic hyperbolic right hexagons, and that two right hyperbolic hexagons are equivalent if the lengths of every other side of one equal the lengths of the corresponding sides of the other. Also, one can construct hyperbolic right hexagons where these lengths are arbitrary positive real numbers.
math/0003144	The case MATH is left as an exercise. Suppose that MATH. We first show that MATH acts fixed point freely on MATH. The isotropy group of a point in MATH lying over MATH is isomorphic to MATH. This is a finite group, and is a subgroup of MATH. It is standard that the natural representation MATH is injective (Exercise: prove this. Hint: use NAME). It follows that the natural representation MATH is injective and that MATH is trivial. (Here we are realizing MATH as a subgroup of MATH as an isotropy group.) It follows from NAME 's Theorem that if MATH, then MATH acts fixed point freely on MATH. The rest of the proof is standard topology. If MATH has a torsion element, then it contains a subgroup MATH of prime order, MATH say. Since this acts fixed point freely on the contractible space MATH, it follows that the quotient MATH is a model of the classifying space MATH of the cyclic group of order MATH. Since the model is a manifold of real dimension MATH, this implies that MATH vanishes when MATH. But this contradicts the known computation that MATH is non-trivial for all MATH. The result follows.
math/0003144	We shall use the fact that each MATH is a quasi-projective variety. As we have seen, this is smooth when MATH and has fundamental group isomorphic to MATH. But a well known result CITE (see also CITE) implies that every smooth quasi-projective variety has the homotopy type of a finite complex. It follows that MATH is finitely presented when MATH. But since MATH has finite index in MATH, this implies that MATH is also finitely presented.
math/0003144	There are several ways to prove this. One is to use NAME cohomology which gives a NAME theoretic computation of MATH. Details can be found in CITE, for example. A more elementary approach goes as follows. First pick a smooth compactification MATH of MATH. Each line bundle MATH over MATH can be extended to a line bundle over MATH. Any two extensions differ by twists by the divisors in MATH that lie in MATH. After twisting by suitable boundary components, we may assume that the extended line bundle also has trivial MATH in MATH. (Prove this using the NAME sequence.) It therefore gives an element of MATH, which, by the discussion at the beginning of the section, is flat. This implies that the original line bundle MATH over MATH is also flat. But if MATH vanishes, then MATH is torsion and MATH is the corresponding character MATH. But since MATH is trivial, this implies that MATH is the trivial character and that MATH is trivial.
math/0003144	Fix a finite orbifold covering of MATH of MATH where MATH is normal in MATH and acts fixed point freely on MATH. The NAME group of the covering is MATH. Suppose that MATH is an orbifold line bundle over MATH with MATH in MATH. This implies that the first NAME class of the pullback of MATH to MATH is also trivial. Since this pullback has a natural MATH-action, this means that the corresponding point in MATH is MATH-invariant. Since MATH it follows that the MATH-invariant part of MATH is finite. It follows that some power MATH of the pullback of MATH to MATH is trivial. This also has a MATH-action. If MATH is a trivializing section of MATH, then the product MATH is a MATH-invariant section of MATH. It follows that MATH has a flat structure invariant under the MATH-action. But since MATH has trivial MATH, it must have trivial monodromy. It is therefore trivial. It follows that MATH is trivial.
math/0003144	We begin with the observation that if MATH is a compact oriented genus MATH surface, then all NAME twists on non-separating SCCs lie in the same homology class as they are conjugate by REF . Denote their common homology class by MATH. Next, using the relations coming from an imbedded genus REF surface with one boundary component, we see that the homology class of any separating SCC that divides MATH into a genus REF and genus MATH surfaces has homology class MATH. Using the relation that comes from an imbedded genus REF surface with REF boundary components, we see that the homology class of every separating SCC is an integer multiple of MATH. It follows that MATH is cyclic and generated by MATH. Now suppose that MATH. Then we can find an imbedded lantern as in REF Since each of the curves in this lantern is non-separating, the lantern relation tells us that MATH, which implies that MATH. This proves the vanishing of MATH when MATH. When MATH, the relation for a genus REF surface with one boundary component implies that MATH as the twist about the boundary is trivial in MATH. Thus MATH is a quotient of MATH. But, as we shall explain a little later, the fact that MATH is at least as big as MATH implies that MATH. A genus REF surface can be obtained from a genus REF surface with two boundary components by identifying the boundary components. The relation obtained for a genus REF surface with REF boundary components gives MATH, so that MATH. This shows that MATH is a quotient of MATH. But, as in the genus REF case, the theory of NAME modular forms shows that it cannot be any smaller. So we have MATH.
math/0003144	All but the generation by MATH follows from preceding results. In genus REF, the generation by MATH follows from the theory of modular forms. Suppose that MATH. Denote the first NAME class of MATH by MATH. To prove that MATH generates MATH, it suffices to show that MATH generates MATH. The following proof of this I learned from NAME. It assumes some knowledge of characteristic classes. A good reference for this topic is the book CITE by NAME and NAME. We begin by recalling the definition of the signature of a compact oriented REF-manifold. Every symmetric bilinear form on a real vector space can be represented by a symmetric matrix. The signature of a non-degenerate symmetric bilinear form is the number of positive eigenvalues of a representing matrix minus the number of negative eigenvalues. To each compact oriented REF-manifold MATH, we associate the symmetric bilinear form MATH defined by MATH . NAME duality implies that it is non-degenerate. The signature MATH of MATH is defined to be the signature of this bilinear form. It is a cobordism invariant. The NAME Signature Theorem (see CITE, for example) asserts that MATH where MATH is the first NAME class of MATH. When MATH is a complex manifold, MATH . Now suppose that MATH is a smooth algebraic surface and that MATH is a smooth algebraic curve. Suppose that MATH is a family whose fibers are smooth curves of genus MATH. Denote the relative cotangent bundle MATH of MATH by MATH. Then it follows from the exact sequence MATH that MATH . Plugging these into REF we see that MATH. Using integration over the fiber, we have MATH . It is standard to denote MATH by MATH. Thus we have MATH . An easy consequence of the NAME Theorem is that MATH. This is proved in detail in the book of CITE. It follows that for a family MATH of smooth curves MATH . The last step is topological. Suppose that MATH is a compact oriented surface and that MATH is an oriented surface bundle over MATH where the fibers of MATH are compact oriented surfaces of genus MATH. Denote the local system of the first integral homology groups of the fibers by MATH. There is a symmetric bilinear form MATH obtained from the cup product and the intersection form. NAME duality implies that it is non-degenerate. It follows from the NAME spectral sequence of MATH that MATH . The local system MATH over MATH corresponds to a mapping MATH into the classifying space of the symplectic group. CITE shows that there is a cohomology class MATH whose value on MATH is the signature of the pairing REF . It follows from this and the discussion above, that under the mapping MATH induced by the canonical homomorphism MATH, MATH goes to MATH. NAME also shows that the image of MATH is exactly MATH, which implies that MATH generates MATH.
math/0003145	Assume MATH for some MATH. According to REF there exist an element MATH such that MATH. Then we have MATH where MATH indicates the holomorphic form on MATH. Consider the composition MATH . The above composite isomorphism induces the dual map MATH with MATH. Moreover, MATH preserves ample classes. Hence the NAME theorem for KREF surfaces assures that there exists the unique isomorphism MATH such that MATH. It is obvious that MATH is an isomorphism of marked surfaces. The converse is derived by the same argument.
math/0003155	Let MATH be the maximal dimension of an irreducible component in MATH. The only possible choices for MATH and MATH. Now MATH and MATH and let MATH. Note that MATH and we may assume by induction on MATH that the chain MATH such that MATH exists and is unique. Then MATH is the unique chain for MATH, which satisfies the condition in the statement.
math/0003155	Assume the second part holds. Take any MATH and let MATH. Let MATH and assume the lemma is proved for dimensions less then MATH. First of all, since MATH is a finite union of its irreducible components, we may proceed assuming that MATH is irreducible. Let MATH be an open subset of MATH such that MATH for all MATH. If MATH then MATH, which has dimension less than MATH and, therefore, MATH is constructible by the induction assumption. If MATH then MATH is constructible, hence so is MATH. It remains to check the case MATH, in which MATH is a finite set of points and is certainly constructible. Conversely, assume that MATH is constructible for every MATH. Let MATH be a closed irreducible subvariety. Then MATH and, since MATH is a finite set and MATH is irreducible, the closure of MATH for some MATH is equal to MATH. But MATH is constructible, hence it is open in its closure MATH.
math/0003155	We need to make some definitions. For a polynomial MATH let MATH be the initial monomial MATH the initial coefficient such that MATH the initial term of MATH. Also for MATH let MATH and MATH be the initial monomial and the initial coefficient of MATH viewed as a polynomial in MATH with coefficients in MATH with respect to MATH, the restriction of MATH to MATH. One obvious observation is that a specialization MATH preserves the order. Let MATH. Consider any MATH not containing MATH. Take a polynomial MATH in ideal of MATH generated by MATH, then there is MATH such that MATH and MATH. Since MATH is a NAME basis in MATH, we have MATH for some MATH, which means that MATH. Now, MATH, because MATH. Thus MATH, which proves that MATH is a NAME basis.
math/0003155	The proof follows from the above algorithm. For the function MATH, MATH the following is true. For every projective MATH there is an open set MATH such that MATH is a constant function. Therefore we may apply REF .
math/0003155	Follows from the algorithm and REF .
math/0003159	As we have already claimed it is enough to prove MATH. We know that there is a good quotient of MATH so it is enough to prove that any closed orbit has maximal dimension. By REF if MATH is closed in MATH then there exists MATH such that MATH. The thesis follows now form MATH and REF .
math/0003159	We prove only REF. MATH . If MATH the last group is isomorphic to MATH and if MATH is isomorphic to MATH. In any case the thesis follows.
math/0003159	REF follows directly from REF Let MATH and MATH. MATH is a MATH module, MATH is a MATH- module and the maps MATH are equivariant with respect the MATH action on MATH. In particular it is enough to prove that if MATH is a partition of MATH, MATH is the MATH-isotipic component of MATH with respect to the MATH action and MATH the MATH-isotipic component of MATH with respect to the MATH action on MATH then MATH . By point REF we have that MATH . In particular MATH is an irreducible representation of MATH. Observe MATH is a MATH-submodule of MATH and that it is clearly nonzero. So MATH as claimed. The proof of REF is equal to the previous one.
math/0003159	The proof is clear by the previous lemma.
math/0003159	We have to compute MATH for all MATH. For all MATH define MATH . Observe that MATH as a MATH-module. So MATH. Observe now that MATH is a quotient of MATH . By the lemmas in the previous section we have that MATH . So in particular MATH if MATH. Hence MATH. The function MATH are clearly MATH-equivariant so the only thing that we have to prove is that they are linearly independent. To prove it we will prove a generalization of it. If MATH and MATH let MATH be the MATH matrix with a MATH in the MATH position and MATH elsewhere. For each MATH and MATH we consider the following sentence MATH: If MATH then MATH is lineraly independent. In the case MATH we call this proposition MATH since it does not depend on MATH and observe that MATH is equivalent to our thesis. For each MATH we consider also the following sentence MATH: If MATH then MATH is true for all MATH, MATH and MATH. First remark: MATH is true. Second remark: let MATH for all MATH and MATH such that MATH. Observe that MATH is linearly independent. Now we prove MATH by induction on MATH. NAME step: MATH. Suppose that there exists MATH such that MATH . If MATH choose a nonzero element MATH (respectively, MATH) and an hyperplane MATH (respectively, MATH) such that MATH (respectively, MATH) and define: MATH and define MATH by MATH . Then we see that MATH . By induction MATH is true for all MATH so we see that MATH for all MATH such that there exists MATH such that MATH and MATH. Now we conclude by the second remark. Second step: MATH if MATH and MATH. Suppose that MATH. We can construct MATH as in the first step and we see that MATH and by MATH we deduce MATH for all MATH. Third step: MATH. By the previous two step we have only to prove MATH for MATH and MATH or MATH. We will suppose MATH, the other case is completely similar. Suppose that MATH. Set MATH and observe that since MATH then MATH. Now choose a non zero vector MATH and for all MATH choose a non zero vector MATH and an hyperplane MATH of MATH such that MATH. Now fix MATH such that MATH and consider MATH and MATH. For all MATH and for all MATH define: MATH . For any MATH we define also MATH by MATH for all MATH, MATH. MATH is a bijection between MATH and MATH: we call MATH the inverse map. Finally we define MATH as in the previou step and we observe that if MATH then MATH if MATH. Hence MATH and applying MATH we obtain MATH for all MATH if MATH and MATH. In a similar way we prove MATH if MATH and MATH and MATH. Finally we observe that if MATH then MATH, hence MATH follows now from MATH that we already know to be true.
math/0003159	CASE: We have to prove MATH and by REF it is enough to prove MATH. So MATH by REF. The proof of REF is equal to the proof of REF. We prove the implication MATH in REF. By REF it is enough to prove that MATH for MATH. MATH . The proof of the converse is completely analougous.
math/0003159	By induction on the length of MATH we can reduce the proof of this lemma to the following identities that are a consequence of REF : MATH for MATH such that MATH.
math/0003159	We prove only MATH. Let's do first the case MATH. If MATH is MATH semistable, then there exists MATH-good such that MATH. Using the notation in REF we have MATH where MATH is a linear map. In our case we can write MATH as MATH and we obseve that no MATH summunds appear in MATH or MATH. Now we construct a new data MATH such that MATH and MATH a MATH-covariant polynomial. Our strategy will be the following: we substitute each MATH with the space MATH in the space MATH and we add MATH copies of MATH to MATH. Let's do it more precise: first of all the new data will not be MATH good so we have to define MATH and MATH: CASE: MATH and MATH, CASE: MATH and MATH for all MATH, CASE: MATH and MATH. Observe that MATH for all MATH so our data will furnish a MATH equivariant function. Moreover if we define MATH we observe that they have the numbers of MATH, MATH, MATH factors specified by MATH. Now we construct the new data MATH in such a way that with respect to the decompositions above we have: MATH . If we construct a data with this property we observe that MATH is an isomorphism if and only if MATH is an isomorphism. Hence MATH implies MATH and the lemma is proved. To construct the new data we choose a basis MATH of MATH and we define the other elements of the data according to the following rules CASE: MATH, CASE: if MATH we set MATH, CASE: MATH for MATH and MATH is an element constructed according to REF in the previous lemma, CASE: MATH for MATH is an element constructed according to REF in the previous lemma, CASE: MATH for MATH and MATH is an element constructed according to REF in the previous lemma. CASE: MATH if MATH and MATH. In this way we garantee that the projection of MATH onto MATH is equal to MATH. To define the remaining part of the new data we do not give details on the indexes, but we explain how to construct it. It is clear that we can choose MATH for the remaining indeces * in such a way that the projection of MATH on MATH is equal to MATH. Finally we observe that a path MATH from MATH to MATH with MATH has to go through a summand of MATH so there exists a path MATH such that MATH. Now we use the previous lemma to change MATH with a MATH such that MATH. More generally if MATH is an element of the path algebra of type MATH with MATH then there exists an element of the MATH-path algebra MATH such that MATH. In this way we define the elements of the MATH-path algebra connecting summunds of MATH and summunds of MATH. In the case MATH we proceed in a similar way: we choose MATH-good and we have MATH . As in the previous case we can find a new data MATH such that: MATH . Now to conclude that MATH is an isomorphism if MATH is we need to know that MATH is an epimorphism and this is not garantee by MATH. But if MATH then, since MATH, we have that MATH is surjective.
math/0003159	If MATH then the result is clear by MATH. Suppose now that MATH and MATH. Let MATH and let MATH. Define now a one parameter subgroup MATH of MATH in the following way: MATH . Since MATH we have that there exists the limit MATH. Let now MATH and MATH a MATH-covariant function on MATH such that MATH. Then MATH . So we must have MATH. The proof of the third case is completely similar to this one.
math/0003159	NAME 's proof extend to this case without changes. Let's prove for REF . We have to prove: MATH that the action on the fiber is free, MATH that it is transitive. First of all we observe that by the previous lemma if MATH then MATH is epi. In particular there exists MATH such that sequence REF is exact, and clearly MATH is univoquely determined up to the action of MATH, moreover this action is free. So MATH and MATH reduce to the following fact: if MATH and MATH is such that sequence REF is exact, then there exists a unique, MATH such that MATH. Since MATH is mono the unicity is clear. To prove the existence we observe that it is equivalent to MATH. But the last statement is clear since we have: MATH and MATH.
math/0003159	This proposition is a straightforward consequence of the previous lemma and the following general fact (see for example REF ): let MATH be an algebraic groups over MATH and MATH, MATH two irreducible algebraic variety over MATH; if MATH acts on MATH and MATH is such that for all MATH the fiber MATH contains exactly one MATH-orbit then MATH is a categorical quotient. If we apply this lemma to the projection MATH, (respectively, MATH) and to the group MATH (respectively, MATH) we obtain the required result.

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