paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 origin... |
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... |
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 tow... |
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 succ... |
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 symbo... |
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,... |
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 fo... |
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 unti... |
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/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 ... |
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/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 su... |
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 simulatio... |
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 ... |
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 o... |
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... |
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.... |
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 condi... |
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 eac... |
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/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 f... |
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 M... |
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 g... |
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. L... |
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 M... |
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.... |
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/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 , eith... |
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 t... |
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 onl... |
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 ... |
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 ... |
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 m... |
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 M... |
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 pr... |
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... |
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 obta... |
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... |
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... |
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 M... |
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 ... |
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 mod... |
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 in... |
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 ``de... |
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/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 fa... |
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... |
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 th... |
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 ... |
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 homomorphis... |
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,... |
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 o... |
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: pr... |
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 f... |
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... |
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... |
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... |
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 lea... |
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... |
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 c... |
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/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... |
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/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 thin... |
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 ... |
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 mu... |
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... |
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... |
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