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math/0003099
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Let MATH be the part of MATH that lies over MATH. The structure equations on MATH are the same as those on MATH with the difference that, after restriction to MATH the functions MATH and MATH become constant and REF-forms MATH become purely imaginary. Note that the equations MATH define the foliation by the torus leaves. Now going back to the structure equations, just derived above, one sees that, on MATH, the equations MATH hold, where MATH is blocked according to the multiplicities in the descending string of eigenvalues MATH . It follows, of course, that quadratic form MATH is well-defined on the leaf space MATH and that this metric and the symplectically reduced MATH-form MATH define a NAME structure on MATH. Finally, the computation of the curvature forms above shows that this NAME structure is a product of the type described in the proposition.
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math/0003100
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We supposed that our coadjoint orbit admit at least a polarization MATH, satisfying NAME 's condition of irreducibility MATH . The codimension of MATH and therefore the dimension of MATH is MATH. Consider a canonical system of geodesics. The geodesics corresponding to the affine subspace MATH provide linear coordinates MATH. The others are the corresponding MATH. Therefore we can arrange a local system of coordinates, such that exponential map gives linear geodesics on MATH's directions.
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math/0003100
|
Let us denote by MATH the NAME inverse transformation on variables MATH's and by MATH the NAME transformation. Let us denote by MATH the algebra of MATH-invariant pseudodifferential operators on MATH. Locally, the NAME transformation maps symbols (as function on local coordinates of MATH) to MATH-invariant pseudodifferential operators and conversely, the inverse NAME transformation maps the pseudodifferential operators to some specific classes of symbols. For two symbols MATH, their NAME images MATH and MATH are operators and we can define their operator product and then take the NAME inverse transforms, as MATH-product MATH . So this product is again a symbol n the same class MATH. Because of existence of special coordinate systems, linear on MATH's coordinates we can treat for the good strata in the same way as in the cases of exponential groups. And the formal series of MATH-product is convergent.
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math/0003100
|
By the lifting properties of the universal covering, we can easily lift each MATH-product on a NAME manifold onto its universal covering. This correspondence is one-to-one.
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math/0003100
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For local charts, there exist deformation quantization, as said above, MATH by using REF quantization. Also in the intersection of two local charts of coordinates MATH and MATH, there is a symplectomorphism, namely MATH such that MATH . Using the local oscilatting integrals and by compensating the local NAME 's index obstacles, one can exactly construct the unitary operator MATH such that MATH see for example NAME 's book CITE. Because the universal coverings are simply connected, the extensions can be therefore produced because of Monodromy Theorem.
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math/0003100
|
From the description of the canonical coordinates in the previous section, we see that there exists at least a convergent MATH-invariant local MATH-product. This local MATH-product then extended to a MATH-invariant global one on the universal covering, which produces a convergent MATH-product on the coadjoint orbits, following REF .
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math/0003100
|
Let us fix in a basis each of MATH, MATH and MATH. We have therefore a basis of MATH. Our theorem is therefore deduced from the original NAME Theorem CITE.
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math/0003100
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Let us recall CITE, that MATH . From the construction of quantization map MATH as the map arising from the universal property the map MATH . The associated representation of MATH is the solution of the NAME problem for the differential equation MATH . The solution of this problem is uniquely defined.
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math/0003100
|
Our proof is rather long and consists of several steps: CASE: We apply the construction of CITE to the solvable radical MATH of MATH to obtain the NAME open set MATH in MATH and its covering MATH. CASE: The general case is reduced to the semi-simple case MATH, see (CITE, Thm. C). Denote by MATH the corresponding analytic subgroup of MATH. CASE: Take the NAME open set MATH of admissible and well-polarizable strongly regular functionals from MATH and its covering MATH via NAME 's construction CITE. CASE: The desired MATH-invariant NAME open set and its covering are the corresponding Cartesian products MATH REF. There exists a continuous field of polarizations of type MATH, for each MATH. CASE: The NAME derivative commutes with direct integrals, that is, MATH . CASE: The restriction of the NAME representation to coadjoint orbits provides a continuous field of polarizations. In particular, MATH .
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math/0003100
|
The proof also consists of several steps: CASE: From CITE - CITE and CITE it is easy to see obtain a slight unipotent modification( that is, a reduction to the unipotent radical) of the multidimensional quantization procedure. We refer the reader to CITE CITE, and CITE for a detailed exposition. CASE: From the unitary representations MATH in the unipotent context and its differential MATH, it is easy to select the constant term and the vector fields term, (see CITE for the simplest case and notations), MATH . The operator-valued partial (that is, depending on MATH) phase MATH . The phase MATH is the sum of the partial phases, on which the one-parameter group MATH operates by translations for all the factors on the left of MATH in the ordered product MATH . It is easy then to see that our induced representations MATH act as follows MATH . CASE: It suffices now to apply the NAME transforms MATH to the function MATH to obtain MATH where the amplitude MATH is the natural extension of the expression MATH in the correspondence with the fields of polarizations. Hence for each MATH, MATH . Remark that the integral MATH is just a type of NAME path integrals. CASE: If in every repeated induction step, see CITE, CITE, MATH satisfies the compactness criteria, then the operator MATH is trace class and hence MATH . The proof of the theorem is therefore achieved.
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math/0003103
|
By a change of coordinates, we can assume without loss of generality that MATH. The result then follows immediately from REF .
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math/0003103
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(We follow the method of proof of CITE.) The positivity of MATH follows from the positivity of MATH and REF above. To show that MATH extends to the unit disk, we begin by writing MATH where MATH. The function MATH is then given by MATH . Let MATH be the smoothing of MATH by convolution; the MATH are smooth forms converging to MATH in MATH as MATH. On MATH the convergence is in the MATH topology. The functions MATH are smooth in MATH. The condition MATH reads as MATH . So, MATH . Since MATH is holomorphic on a neighborhood of MATH, the current MATH is smooth on MATH and thus MATH converges smoothly to the function MATH given by MATH . But REF defines a smooth function on all of MATH. While MATH in MATH on MATH, so MATH on MATH. This shows that MATH is smooth and smoothly extends onto the disk MATH. Thus MATH is also smooth on MATH.
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math/0003103
|
CASE: We let MATH and we write MATH. We further write MATH. Since the volumes MATH of the MATH are uniformly bounded, by NAME 's theorem (see for example CITE) we can assume, after passing to a subsequence, that MATH converges to a pure REF-dimensional analytic subset MATH of MATH. Note that MATH. CASE: MATH is a subvariety of MATH. In this case MATH is a holomorphic map into an irreducible component MATH of MATH. If MATH denotes a desingularization of MATH, then MATH lifts to a meromorphic map MATH from MATH to MATH. By REF-dimensional version of REF proved in CITE, MATH extends meromorphically onto MATH minus a finite set of points. If this set is nonempty, then MATH would not be homologous to zero in MATH, and hence MATH would not be homologous to zero in MATH. But MATH in the sense of currents. Since MATH, MATH is homologous to zero in MATH and therefore in MATH, a contradiction. So MATH extends onto all of MATH. CASE: MATH is not a subvariety of MATH. Let MATH be the irreducible component of MATH containing MATH. Define the analytic space MATH, where the equivalence relation is defined as follows: The points MATH and MATH are equivalent iff MATH, and necessarily MATH is equivalent to MATH iff MATH. By REF above, this is a proper equivalence relation and hence MATH is a complex space. Let MATH be the projection defined by MATH for MATH and MATH for MATH. Let MATH denote the normalization. By REF , the map MATH is generically one-to-one and thus is a normalization of its image. By the uniqueness of the normalization, MATH lifts to a map MATH, that is, MATH. The map MATH is a biholomorphism onto its image. The boundary MATH, being biholomorphic to MATH, is strictly pseudoconvex after shrinking slightly, so by NAME 's theorem, MATH can be blown down to a normal NAME space. This easily yields an extension of MATH onto MATH. CASE: We denote the extension of MATH onto MATH also by MATH. Let MATH be the maximal compact pure REF-dimensional variety contained in MATH. (In REF above, MATH, whereas in REF .) We consider the pure two-dimensional analytic set MATH in MATH, where MATH is the (finite) set of points of indeterminacy of MATH. CASE: We claim that for all MATH the graph MATH belongs to the MATH-neighborhood of MATH, for MATH. Neighborhoods are taken with respect to the Euclidean metric on MATH and NAME metric on MATH. (In fact, any choice of metric works as well as this one.) This claim follows immediately from REF below. We say that a sequence of meromorphic maps MATH converges to a holomorphic map MATH on a domain MATH if for all compact subsets MATH, MATH for MATH and MATH uniformly on MATH. Let MATH be a sequence of meromorphic maps, where MATH is a compact complex manifold. Suppose that MATH is holomorphic on MATH, where MATH. If there exists a meromorphic map MATH such that MATH, then MATH on MATH. REF is a special case of REF. (REF is stated in terms of ``strong convergence" of meromorphic maps. However, if MATH strongly converges to a holomorphic map, then the sequence converges in the above sense. This is the content of the ``NAME principle" of CITE.) To complete the proof of REF , we consider a point MATH and take any open MATH adapted to MATH, that is, biholomorphic to MATH in such a way that MATH. Then for MATH, we have MATH and thus MATH is a MATH-sheeted analytic covering, where MATH is a natural projection. CASE: The number MATH of sheets is uniformly bounded. Consider the following two cases. CASE: MATH . In this case as MATH we can take the following neighborhood. Let MATH, where MATH and MATH. Take a neighborhood of MATH in MATH of the form MATH such that MATH. Then take some small MATH in MATH and put MATH and MATH. If the number MATH of sheets of the analytic cover MATH is not bounded, it will contradict the fact that MATH has uniformly bounded volume (counted with multiplicities). One should remark now that boundedness of the number of sheets does not depend on the particular choice of the adapted neighborhood of MATH. CASE: MATH. Let MATH be some adapted neighborhood. Find a point MATH such that all its pre-images MATH are smooth points of MATH and MATH is a biholomorphism between neighborhoods MATH on MATH and MATH on MATH. Denote by MATH the projection of MATH into MATH. Take mutually disjoint polydisks MATH with centers MATH. Consider MATH as adapted neighborhoods of MATH in MATH. They are adapted also for MATH . Denote by MATH the corresponding number of sheets. If MATH is not bounded then at least one sequence MATH is also unbounded. Fix MATH with MATH unbounded. If MATH, then everything reduces to REF . So let MATH. Perturbing MATH and thus MATH if necessary, we can suppose that MATH is a point where our map MATH is holomorphic. More precisely MATH for some MATH. Now the contradiction is immediate, because the graphs MATH uniformly approach MATH while MATH converges to MATH in a neighborhood of MATH. CASE: We are exactly under the assumptions of REF, that is, we can apply the ``Continuity Principle." (The condition of boundedness of the cycle geometry is insured by REF from CITE.) This gives us an extension of MATH onto MATH.
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math/0003107
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Assume by induction that MATH and MATH coincide at order MATH. Associativity relation at order MATH implies MATH; thus MATH where MATH is a universal differential operator and MATH is universal, skewsymmetric and of order MATH. Clearly, if MATH corresponds to a graph in MATH, it has MATH arrows and MATH vertices, but since MATH is linear on MATH, at most one arrow can end at any of the first MATH vertices and since MATH is MATH-differential in each argument, exactly one arrow ends at each of the last MATH vertices. Hence MATH so MATH; but if MATH the only graph yields the NAME bracket of functions so this can be cancelled by a change of parameter and if MATH the graph corresponds to a symmetric bidifferential operator, hence does not yield a MATH. So we can assume that MATH vanishes but then the equivalence through MATH builds a star product which is universal and coincide with MATH at order MATH.
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math/0003107
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Indeed if MATH then MATH is differential so MATH is a differential MATH-cocycle with vanishing skewsymmetric part but then, using NAME 's formula, it is the coboundary of a differential MATH-cochain MATH and MATH, being a MATH-cocycle, is a vector field so MATH is differential. One then proceeds by induction, considering MATH and the two differential star products MATH and MATH, where MATH which are equivalent through MATH (that is, MATH).
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math/0003107
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Let us suppose that, modulo some equivalence, the two star products MATH and MATH coincide up to order MATH. Then associativity at order MATH shows that MATH is a NAME MATH-cocycle and so by REF can be written as MATH for a MATH-form MATH. The total skewsymmetrization of the associativity relation at order MATH shows that MATH is a closed MATH-form. Since the second cohomology vanishes, MATH is exact, MATH. Transforming by the equivalence defined by MATH, we can assume that the skewsymmetric part of MATH vanishes. Then MATH where MATH is a differential operator. Using the equivalence defined by MATH we can assume that the star products coincide, modulo an equivalence, up to order MATH and the result follows from induction since two star products always agree in their leading term.
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math/0003107
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On an open set in MATH with the standard symplectic structure MATH, denote the NAME bracket by MATH. Let MATH be a conformal vector field so MATH. The NAME star product MATH is given by MATH and MATH is a derivation of MATH. Now MATH is symplectomorphic to an open set in MATH and any differential star product on this open set is equivalent to MATH so we can pull back MATH and MATH to MATH by a symplectomorphism to give a star product MATH with a derivation of the form MATH. If MATH is an equivalence of MATH with MATH on MATH then MATH is a derivation of the required form.
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math/0003113
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This follows from the theorem and the remark that if we multiply a MATH-module with an element MATH from MATH, then the determinant is multiplied by the norm MATH. But MATH. REF is obvious.
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math/0003113
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This follows from the theorem and the remark that if we multiply a MATH-module with an element MATH from MATH, then the determinant is multiplied by the norm MATH. But MATH.
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math/0003113
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The formula is the combination of two results: REF, which states that MATH where MATH is NAME 's NAME symbol and MATH the horospherical map. Note that what is here called MATH is in CITE MATH. Furthermore, according to REF ., MATH . Note that REF uses an other normalization of the horospherical map and that there is a factor MATH missing because of a wrong normalization of the residue map (the residue of MATH in REF is not MATH but MATH).
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math/0003113
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We have exact sequences MATH where MATH is the residue class field of MATH. By REF. As the order of the residue class field MATH is prime to MATH, we have an exact sequence MATH . As MATH has good reduction at MATH, we now how MATH decomposes in MATH (see REF ). If MATH is split, the ramification degree of MATH in MATH is MATH and the degree of MATH over MATH is MATH. Hence the norm map induces multiplication by MATH on MATH and the inverse limit over these maps is zero. This gives the first claim. In the case where MATH is inert of ramified, MATH is totally ramified in MATH and the norm map from MATH induces the identity on MATH. Putting these sequences together for all MATH and using MATH gives the result.
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math/0003113
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We have MATH . In the local case, we have an isomorphism MATH . By class field theory MATH so that MATH .
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math/0003113
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By local duality we have MATH . On the other hand it is a result of CITE that the cohomology group MATH is zero. As MATH this proves our claim.
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math/0003113
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REF show that we have a canonical map from MATH to MATH because the second cohomology vanishes. The same argument gives the result for MATH.
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math/0003113
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We have MATH which by biduality MATH proves our claim.
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math/0003113
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Apply REF to the exact triangles in REF .
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math/0003113
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The commutativity of the lower square is clear. Let us treat the upper square. The map MATH is the dual of the corestriction MATH . By the local duality theorem the corestriction map is dual to the norm map MATH . This together with the definition of MATH proves our claim.
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math/0003113
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Let MATH be an inert or ramified prime and MATH be the MATH representation on MATH. Then MATH is irreducible (CITE) and because MATH has good reduction the prime above MATH is totally ramified in MATH (CITE) and MATH. Now MATH is two dimensional and MATH is simply a twist of MATH by a power of MATH. Thus MATH acts non trivially on MATH.
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math/0003113
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If MATH is split this follows immediately from REF and if MATH is inert or prime in MATH this follows from the same lemma and the above result because the MATH-eigenspace of MATH is zero.
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math/0003113
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The action of MATH on MATH is via a character MATH. This gives a surjection MATH. As MATH the kernel of this surjection is an ideal with height MATH and hence MATH . This implies MATH. REF then implies the claim.
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math/0003113
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The complexes in the triangle in REF are MATH-modules and we apply MATH. Then MATH by definition of MATH. The same holds for MATH. The result follows with REF .
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math/0003113
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Apply MATH to the triangle MATH and use the above corollary.
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math/0003113
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There is an exact triangle MATH where MATH is the local ring at MATH. As MATH is unramified at the places MATH in MATH, which are in MATH we have by purity MATH . Let us prove that MATH . For this note that MATH are the coinvariants and that MATH. Fix a prime MATH of MATH dividing MATH, then the primes MATH of MATH are permuted by MATH. Fix MATH dividing MATH and let MATH be the stabilizer of MATH. It suffices to prove that MATH acts non trivially on MATH. Let MATH be the inertia group of MATH. This group is non trivial because MATH is ramified above MATH by REF and it acts non trivially on MATH because MATH and by the NAME criterium. This proves our claim.
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math/0003113
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Let MATH and MATH be as in the theorem. By NAME 's ``main conjecture" REF we have MATH . On the other hand, MATH . This together with the above corollaries gives the result.
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math/0003113
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Consider the localization sequence MATH and the purity isomorphism MATH. Moreover MATH for MATH. This together with the above values of MATH gives the desired result.
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math/0003113
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By REF we have MATH and MATH. This implies the claim.
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math/0003113
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The map MATH is finite étale, so that MATH . As MATH is a lisse MATH-sheaf, there is a MATH such that MATH comes from MATH, that is, MATH. Thus MATH which proves our claim.
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math/0003113
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The inverse functor is given by MATH if the action of MATH factors through MATH.
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math/0003113
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We have to compute the transition maps in MATH where now the transition maps MATH are given by multiplication with MATH. This map is zero for MATH.
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math/0003113
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See the proof of REF.
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math/0003113
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Let MATH be a geometric point. Then we choose two topological generators MATH of MATH. It is a standard fact in NAME theory that we have a ring isomorphism MATH, mapping MATH and MATH, with the power series ring in two variables. This implies that the augmentation ideal MATH is mapped to the ideal MATH. The claim that MATH follows immediately. Now MATH corresponds under this isomorphism to the ideal MATH. By induction one sees that this ideal is contained in MATH. As MATH in MATH we have MATH. This gives the map as indicated, which in the limit is clearly an isomorphism.
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math/0003113
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The surjection MATH factors through MATH so that this map is also surjective. As these two sheaves are finite of the same cardinality, this is also an isomorphism.
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math/0003113
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From REF and the universal property of MATH we get the morphism of MATH to MATH. To check that this induces an isomorphism if take pro-sheaves in the index MATH, it is enough to check this after pull-back with MATH. We get the map MATH which is an isomorphism.
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math/0003113
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Clear from the definition.
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math/0003113
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Let us first treat the unipotent case. It suffices to show that MATH is an isomorphism. Using the equivalence of categories REF, this can be tested after pull-back with MATH. The resulting map MATH is on the the MATH-th graded piece MATH induced by the map MATH, which is just the MATH-multiplication and thus an isomorphism. In the geometric case we have MATH and MATH. The map MATH is by definition induced by MATH for MATH. This is obviously an isomorphism if MATH.
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math/0003113
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The map MATH maps MATH to MATH. This gives the result.
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math/0003113
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This follows immediately from the definition of the maps MATH and MATH.
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math/0003113
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This follows from MATH and MATH, which is a consequence of the NAME resolution as MATH is a sheaf of regular rings.
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math/0003113
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We have an isomorphism MATH and a contraction map MATH given on MATH by MATH . This gives the required map and it is straightforward to check that this is indeed a splitting of MATH.
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math/0003113
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Define MATH. Then MATH (see CITE, Expose XVIII REF ), where the higher direct image is taken for the flat topology. The sequence from REF MATH gives an isomorphism MATH.
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math/0003113
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Let MATH be a section and MATH be its preimage in MATH. Let MATH be the embedding and denote by MATH the open immersion of the complement MATH . Denote also by MATH the structure map. We have an exact sequence of étale sheaves MATH . On the other hand let MATH, where MATH is the embedding. Then we get an exact sequence MATH . The long exact cohomology sequence for MATH taken for the flat topology gives MATH because MATH. We have an identification MATH and it is clear that the map MATH is injective. We get an isomorphism of complexes MATH . Using the sequence MATH as in the proof of REF , we see that MATH where MATH. Let us compute MATH. By base change and REF it is a straightforward computation that MATH and by NAME duality we get MATH . If we apply this to the universal section MATH we get the desired result.
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math/0003113
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The sheaf MATH can be described as the pull-back of the sequence MATH by the map MATH where the last map is forgetting the trivialization on MATH. The sheaf MATH is the pull-back by the map MATH of the same sequence. We have to compute the composition of these maps with MATH. This composition factors in both cases as MATH. Thus, the difference is zero. This implies the claim.
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math/0003113
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The composition of MATH with the map MATH to the NAME scheme is zero, because MATH. This gives the first claim. The extension class MATH is given by the pull-back of the sequence REF with the map MATH. Using the explicit description following the proof of REF proves the result.
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math/0003113
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Clear because the character group of MATH is MATH.
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math/0003113
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The section MATH is given by evaluating an isomorphism MATH at MATH. Let MATH be the translation with MATH on MATH, then MATH . The function MATH gives a section of MATH and thus MATH gives a section of MATH . This proves the claim.
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math/0003113
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This is a straightforward computation. The element MATH maps to the mapping, which sends MATH to MATH . The first sum can be rewritten as MATH so that we get MATH. This is the desired result.
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math/0003113
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From REF we know that MATH because MATH is prime to MATH. This proves the claim.
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math/0003113
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This follows immediately from the above proposition.
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math/0003113
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The recipe to compute the moment map from REF combined with REF gives immediately the result.
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math/0003113
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We have by definition and and REF MATH . By definition of MATH we see immediately, that MATH . With the above notation, we have MATH so that REF gives: MATH . This is the desired result.
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math/0003113
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Observe that we identified MATH where MATH has now the conjugate linear MATH-action. In particular, MATH for MATH. We compute MATH where we used the distribution relation for MATH (see CITE II REF) MATH for the last equality. As the NAME group of MATH acts simply transitively on MATH, we get MATH . We have MATH which gives MATH because the norm MATH is the sum over all the NAME translates, which act trivially on MATH. If we finally take the sum over MATH and let MATH get bigger and bigger we get MATH where we used MATH . This is the desired result.
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math/0003113
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If MATH is inert of prime this is just a reformulation of REF . If MATH is split, MATH decomposes into a direct sum for the MATH and the MATH part. Putting them together gives the result.
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math/0003113
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Let MATH be another ideal prime to MATH. Then by REF MATH . It is enough to show that MATH is invertible in MATH, because MATH is a torsion free MATH-module. MATH is a local ring if MATH is inert or prime in MATH and a product of local rings if MATH is split. We have MATH, where MATH is either the maximal ideal or the product of the maximal ideals. The element MATH acts via MATH on MATH and thus MATH is invertible in MATH if MATH. It remains to see that MATH generates MATH. We have MATH which is for MATH a primitive root of unity.
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math/0003113
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As MATH is prime to MATH, this follows from REF and the definition of MATH in REF.
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math/0003113
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By REF we have an isomorphism MATH . This implies that the MATH-module is induced and hence as MATH-module isomorphic to MATH. This implies MATH and the claim of the corollary, because the higher NAME vanish.
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math/0003113
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The complex MATH is isomorphic to MATH because by REF MATH and MATH have the same image and as MATH-modules MATH and MATH. REF implies then the claim.
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math/0003113
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We chow first that MATH finite implies the finiteness of MATH. Note that this is the cokernel of MATH . Computing up to finite groups we get from REF (using REF ) that this cokernel is isomorphic (up to finite groups) to the cokernel of MATH which is contained in MATH. This group is of course finite if MATH is finite. Thus MATH is finite. Using REF . from CITE we see that this implies that MATH is finite. We will now show that this last group controls the kernel of MATH. It suffices to show that the kernel of MATH on MATH is finite because by CITE both MATH and MATH are MATH-modules of rank MATH with MATH a torsion module. So suppose that the image of MATH under MATH has not rank MATH, that is, is finite. Then, because MATH the image of MATH in MATH must be finite as well. The kernel of the map MATH is MATH (group homology). On the other hand, up to finite groups, REF implies that MATH. By REF we get a commutative diagram (up to finite groups) MATH . The kernel of the map MATH is a quotient of MATH which by the above is finite. Thus, we arrive at a contradiction and MATH can not be zero on the free part of MATH. Hence, MATH is non zero on MATH.
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math/0003117
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This theorem is proved somewhat analogously to the theorem on the existence of universal NAME machines. If the universal transition function is MATH then for simulating another transition function MATH, the encoding demarcates colonies of appropriate size with MATH, and writes a string MATH that is the code of the transition table of MATH onto a special track called MATH in each of these colonies. The computation is just a table-look-up: the triple MATH mentioned in the above example must be looked up in the transition table. The transition function governing this activity does not depend on the particular content of the MATH track, and is therefore independent of MATH. For references to the first proofs of universality (in a technically different but similar sense), see CITE .
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math/0003117
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Let us use the amplifier defined in the above lemma. Let MATH be a trajectory of the medium MATH with initial configuration MATH. Let MATH be defined by the recursion MATH. Let MATH be a space-time point in which we want to check MATH. There is a sequence of points MATH such that MATH is a cell of MATH containing MATH in its body with some address MATH. There is a first MATH with MATH. Let MATH be the event that MATH for MATH in MATH. The theorem follows from the bounds on MATH and MATH and from MATH . To prove this inequality, use MATH . By the construction, MATH. Since the duration of MATH is less than MATH we have MATH by the initial stability property, proving MATH. The error-correction property and the broadcast property imply MATH.
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math/0003117
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Let MATH be the configuration of MATH which has state MATH at site MATH and arbitrary states at all other sites MATH, with the following restriction in case of a finite space size MATH. Let MATH only for MATH and MATH only if MATH . Let MATH for MATH. We have MATH where the first equation comes by definition, the second one by fittedness. The encoding conserves this relation, so the partial configuration MATH is an extension of MATH. Therefore the limit MATH exists for each MATH. Since MATH is non-degenerate the limit extends over the whole set of integer sites.
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math/0003117
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The proof is mechanical verification: we reproduce it here only to help the reader check the formalism. Let us construct MATH. For infinite space size, let MATH . For finite space size MATH, define the above only for MATH and MATH in MATH, where MATH on the left-hand side is taken MATH. In all other sites MATH, let MATH. Let MATH be defined again by REF. We define MATH as in the proof of REF . It is sufficient to show REF again to prove that the limits in question exist. By definition, MATH . By the controlling property, its MATH field MATH completely determines the last expression via MATH. By the identification REF and the aggregation REF , MATH . By definition, MATH which proves the statement. Let us show REF. We use induction on MATH, from MATH down to REF. The case MATH says MATH which follows from the definition of MATH. Assume that the statement was proved for numbers MATH: we prove it for MATH. By the definitions of MATH and MATH and by REF we have MATH . By induction, MATH where MATH can also be written in the form MATH for some MATH. Now we have, by the identification REF and the aggregation REF MATH where the third equality is implied by the definition of MATH.
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math/0003117
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For the capacity of the cells in MATH we have MATH where MATH is the number of bits not belonging to MATH or MATH. The state of a cell of MATH must be stored in the fields of the cells of the colony representing it, excluding the error-correcting bits in MATH. Hence MATH .
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math/0003117
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Take MATH satisfying the above conditions and let MATH. Due to the monotonicity of MATH, it is enough to prove that MATH. Take a measure MATH over configurations of MATH, this will give rise to some measure MATH over configurations of period MATH in MATH in a natural way, where MATH is such that for all MATH and all MATH we have MATH. Then, MATH implies MATH and therefore via REF we have MATH .
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math/0003117
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The statement follows by a standard argument from the strong NAME property of the NAME process MATH (see CITE).
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math/0003117
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As in REF, for a small MATH, let MATH with MATH when MATH and MATH otherwise. Then a transition to the limit MATH in REF completes the proof. The transition uses the fact that trajectories of MATH converge in distribution to trajectories of MATH.
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math/0003117
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Let MATH be a trajectory of MATH and MATH a system of disjoint random local conditions in MATH. Let MATH be some time such that MATH for each MATH, and let MATH be an event function measurable in MATH. We must show MATH where MATH is defined as in REF. By the assumption, for each MATH there is a system of disjoint random local conditions MATH for MATH with the properties defined for canonical simulations. Since MATH has a countable range, we obtain measurable functions even if we substitute MATH into MATH here. Let MATH and recall REF. MATH is a rectangle process and MATH is a stopping time. It is easy to see that MATH and that the process MATH is upper semicontinuous almost everywhere. Let us show that for each MATH it is measurable in MATH. We write MATH . Now REF and MATH implies MATH. On the other hand by the definition of random-stopping local conditions, MATH implies MATH. Hence we can write the above expression as MATH . By REF , MATH . By the definition of the rectangle process MATH, also MATH hence indeed MATH. Finally, we have to show MATH . As above, MATH which again can be written as MATH . We have MATH since MATH is a stopping time; combining this with REF completes the proof. By REF we have MATH and MATH where the last sum is over all functions MATH with MATH. Similarly, by REF MATH . Therefore it is sufficient to show MATH . Let us fix a function MATH. Recall that by REF, MATH . For any MATH, the rectangles MATH are all disjoint and belong to MATH. Also the sets MATH are disjoint, therefore all rectangles MATH are disjoint. Recall that also by REF, MATH therefore the system MATH taken for all MATH, is a system of disjoint random local conditions with MATH, and hence the proof is finished by the trajectory property of MATH.
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math/0003117
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Let MATH be a positive integer and MATH be three fields of the states of MATH, where MATH, MATH. The field MATH represents numbers mod MATH. It will be used to keep track of the time of the simulated cells mod MATH, while MATH holds the value of MATH for the previous value of MATH. Let us define MATH. If there is a MATH such that MATH (that is, some neighbor lags behind) then MATH that is, there is no effect. Otherwise, let MATH, and for MATH, let MATH be equal to MATH if MATH, and MATH otherwise. MATH . Thus, we use the MATH and MATH fields of the neighbors according to their meaning and update the three fields according to their meaning. It is easy to check that this transition function simulates MATH in the MATH field if we start it by putting REF into all other fields. Let us check that MATH is commutative. If two neighbors MATH are both allowed to update then neither of them is behind the other modulo MATH, hence they both have the same MATH field. Suppose that MATH updates before MATH. In this case, MATH will use the MATH field of MATH for updating and put its own MATH field into MATH. Next, since now MATH is ``ahead" according to MATH, cell MATH will use the MATH field of MATH for updating: this was the MATH field of MATH before. Therefore the effect of consecutive updating is the same as that of simultaneous updating.
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math/0003117
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Let MATH be the commutative transition function given by REF , with the fields MATH. Let MATH be an arbitrary configuration of MATH and let MATH be a configuration of MATH defined in the statement of the same theorem. Let MATH be a trajectory of MATH, with the starting configuration MATH. An update set MATH similar to REF can be defined now for the trajectory MATH as follows: MATH is in MATH iff MATH is a switching time of MATH. Similarly, MATH can be defined as in REF: MATH . With these, let us define MATH . By the cited theorem, MATH is a trajectory of MATH.
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math/0003117
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This theorem can be proven similarly to REF . The main addition to the construction is that before MATH starts a block simulation of MATH the input will be distributed bit-by-bit to the colonies simulating the cells of MATH. At the end, the output will be collected from these colonies.
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math/0003117
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A detailed proof would be tedious but routine. Essentially, each line of a rule program is some comparison involving some fields: substrings of a state argument MATH. We have MATH squared in the time upper bound since we may have to look up some parameter repeatedly.
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math/0003117
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It is enough to show an expression of the form REF for local conditions of type MATH in MATH. Let MATH be the sizes of the space and time projections of MATH. We define the following lattice of points: MATH . Then the rectangles MATH for MATH form a partition of the space-time. For each MATH in space-time, let MATH be the MATH with MATH. Let MATH be a trajectory of MATH. It is enough to consider the rectangle MATH. If MATH intersects MATH then there are some points MATH in MATH such that MATH are disjoint and even MATH are in MATH. Then it is easy to see that even MATH are disjoint from each other and are contained in MATH. Let MATH be the number of pairs MATH in MATH with this property. We found MATH pairs of rectangles MATH (MATH, MATH) such that MATH, and that if MATH intersects MATH then there is a MATH such that MATH intersects MATH for MATH. We found MATH . To complete the proof, observe that MATH . Since counting shows MATH and the assumption of the lemma says MATH both conditions of a deterministic canonical simulation are satisfied.
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math/0003117
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REF can be rearranged as MATH .
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math/0003117
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We will actually prove the more general variable-period variant of the lemma, with MATH. Let us define a broadcast amplifier frame MATH for example, as in REF . Applying the Amplifier Lemma in the broadcast case, we obtain a broadcast amplifier with media MATH as in REF with a hierarchical code having a broadcast field MATH for both MATH and MATH. Let us denote the whole system of code and the broadcast field (with the additional mappings MATH) by MATH as in REF. For any value MATH of the field MATH, we create an initial configuration MATH as in REF. Then all properties but the initial stability property of an abstract amplifier (defined in REF ) are satisfied by definition. For the initial stability property it is sufficient to note that each medium MATH is a robust medium, with work period lower bound MATH. Therefore, if MATH is a trajectory of MATH and MATH then for each MATH the probability that MATH is less than the probability that damage occurs in MATH. This can be bounded by the Restoration Property.
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math/0003117
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This proof starts analogously to the proof of REF . Let us define an amplifier frame MATH as in REF , with the rider transition function having the property MATH that is, leaving all fields (in particular, the rider and guard fields) unchanged. Applying the Amplifier Lemma, we obtain a uniform amplifier with media MATH as in REF with a hierarchical code having a guarded shared field MATH with guard field MATH as in REF. Let MATH be largest with MATH. For any infinite configuration MATH, we create an initial configuration MATH with a guarded shared field MATH, with active level MATH. As mentioned after REF, if MATH is finite then MATH. Let the trajectories MATH, MATH be defined as before. Then we will have MATH for all MATH, for MATH and MATH. Let MATH for a MATH in MATH (since the code was chosen such that MATH, this is not really a restriction on MATH), and let MATH. Remember the definition of the aggregated input configurations MATH in REF and of MATH in REF. Eventually, we want to estimate the probability of MATH but we will use a generalization corresponding to REF. Let MATH, then REF shows that MATH is a cell of MATH whose body contains MATH with address MATH where MATH was defined before REF. Let MATH be the event that MATH and MATH holds for MATH in MATH. We have MATH. Assume MATH. Then, according to the error-correction property, except for an event of probability MATH, for every MATH in MATH there is a MATH in MATH with MATH . Since MATH is a guarded shared field and MATH for all MATH in MATH, we can replace MATH with MATH in the above equation. MATH implies MATH . The identification property and the aggregation REF implies MATH . Thus, except for an event of probability MATH, we have MATH. Let MATH . Consider the case MATH. By the construction and REF the relation REF holds for MATH, MATH. Since the duration of MATH is less than MATH the probability that MATH undergoes any change during this interval is less than MATH, proving MATH. Consider REF then MATH and the space is finite. According to the definition of MATH in REF, and of MATH in the same proof, the value MATH is defined for MATH in MATH. Thus, at time REF the space is filled as much as possible with adjacent cells of MATH one of which is MATH. The set MATH can be covered by at most MATH copies of the rectangle MATH. Since MATH is a trajectory of the robust medium MATH, the probability of damage on each of these rectangles is at most MATH. Therefore the probability that there is any damage in MATH over MATH is at most MATH. Assume that there is no damage in MATH over MATH. Then the cling-to-life condition and the computation condition imply that all cells of MATH remain nonvacant until time MATH. We defined MATH to leave the rider and guard fields unchanged, and MATH. Hence, by the definition of ``combined" in Subsection REF, for each of these cells MATH we have MATH implying REF for MATH. Hence the total probability upper bound is MATH where we used MATH. With the parameters of REF , we have MATH for small enough MATH, and MATH, hence MATH can be written as MATH with MATH for some constant MATH.
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math/0003117
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We proceed in analogy with the proof of REF using the same notation wherever appropriate and pointing out only what is new or different. We will find out by the end what choice for the functions MATH and MATH works. One can assume that MATH is commutative in the sense of Subsection REF since the methods of that subsection can be used to translate the result for arbitrary transition functions. The transition function MATH on the rider track will be defined to be the aggregated transition function MATH, as in REF , and the MATH will leave the guard field unchanged. With this, an amplifier frame will be obtained as in REF . The Amplifier Lemma gives a uniform amplifier with media MATH as in REF with a hierarchical code having a guarded shared field MATH, MATH as in REF. The parameters of that example are chosen in such a way that MATH and therefore MATH . Let MATH be any trajectory of MATH over MATH with MATH of the form MATH (see REF), satisfying MATH, where MATH. For any MATH and MATH, let a sufficiently large MATH be chosen: we will see by the end of the proof how large it must be. We certainly need MATH which, in view of REF, can be satisfied if MATH. Let MATH. The trajectories MATH, MATH are defined as before. Let MATH for a MATH in MATH. Let MATH be the trajectory of the aggregated cellular automaton MATH with the aggregated initial configuration MATH. Let MATH . Let MATH be the event that MATH and MATH holds for MATH in MATH. We have MATH. Assume MATH. According to the error-correction property, except for an event of probability MATH, for every MATH in MATH there is a MATH in MATH with MATH . Again, we can replace MATH with MATH here. MATH implies MATH . Hence, the identification property and the aggregation REF implies MATH . Thus, except for an event of probability MATH, we have MATH . Consider the case MATH. Then the statement of the theorem holds since the MATH is filled with MATH's. Consider REF then MATH. According to the definition of MATH, at time REF, the space is filled as much as possible with adjacent cells of MATH one of which is MATH. Let us define the intervals MATH . The set MATH can be covered by at most MATH copies of the rectangle MATH. Since MATH is a trajectory of the robust medium MATH, the probability of damage on each of these rectangles is at most MATH. Therefore the probability that there is any damage in MATH over MATH is at most MATH. Assume that there is no damage in MATH over MATH. Then the cling-to-life condition and the computation condition imply that all cells of MATH remain nonvacant until time MATH. We defined MATH to leave MATH unchanged, and to be the aggregated transition function MATH, on MATH. We have MATH. Hence, by the definition of ``combined" in Subsection REF, each cell in each of its transitions during the damage-free computation, applies MATH to the field MATH, making at least MATH steps. By REF, MATH contains the image of a support of MATH, therefore the confinement to this interval does not change the result. Thus MATH and thus MATH holds. This gives a probability upper bound MATH . Using REF (and ignoring integer parts and the additive REF) this can be upper-bounded by MATH . For small enough MATH, by REF, the second term is MATH, and hence MATH will do. This gives MATH for large enough MATH showing that MATH will do.
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math/0003117
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This proof starts analogously to the proof of REF . Let us define an amplifier frame MATH as in REF , with the rider transition function having the property MATH that is, leaving the MATH field unchanged and always setting the guard field to MATH. It will have a broadcast field MATH with MATH bits. Applying REF , we obtain a self-organizing amplifier with media MATH as in REF with a hierarchical code having a guarded shared field MATH, MATH. Also, we will have MATH for all MATH. Let MATH . For a certain value of the field MATH that we call MATH, we create an initial configuration MATH consisting of all blue latent cells covering the whole space, with MATH. Let the trajectories MATH, MATH be defined as before. Let MATH be a space-time point. Eventually, we want to estimate the probability of MATH. Let MATH . If MATH then by the self-organizing property of the amplifier, the evolution MATH is MATH-blue at time MATH for each MATH. Let MATH be the event that MATH is blue in MATH. There is a constant MATH such that MATH . The right-hand side is an upper bound on the probability of damage occurring in MATH in MATH. If damage does not occur there and MATH holds then the control delegation property implies MATH. Let MATH . Consider the case MATH. Then we have MATH. Let MATH . Let MATH be the event that MATH is blue in MATH. Since MATH is MATH-blue at time MATH for each MATH, the probability of MATH is at most MATH for some constant MATH. There is a constant MATH such that MATH . The right-hand side is an upper bound on the probability of damage occurring in MATH in MATH. If damage does not occur there and MATH holds then the lasting control property implies MATH. Thus, the probability that MATH is not blue can be upper-bounded by MATH . Consider REF then the space is finite. Let MATH be the event that the whole space MATH is blue in MATH. Since MATH is MATH-blue at time MATH for each MATH, and MATH, the probability of MATH is at most MATH for some constant MATH. There is a constant MATH such that MATH . The right-hand side is an upper bound on the probability of damage occurring in MATH in MATH. If damage does not occur there and MATH holds then by the lasting control property, MATH also holds. Thus, the probability that MATH is not blue can be upper-bounded by MATH . Thus, the probability that MATH is not blue can be upper-bounded in both REF (ignoring multiplicative constants) by MATH . With the parameters of REF we have MATH and MATH can be written, as in the proof of REF as MATH with MATH for some constant MATH. Also, MATH for some MATH, If MATH then this gives MATH. Since MATH thus MATH . If MATH then MATH. Hence MATH, giving MATH . Both of these bounds are poor if MATH is small. But in this case there is a trivial upper bound MATH on the total probability that MATH is not blue: this bounds the probability that there has been any damage since the beginning that could influence MATH.
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math/0003117
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Some damage occurs near MATH but not in MATH at some time MATH during MATH. When MATH is the end of a dwell period then, since MATH is affected via neighbors, MATH . Then it does not affect MATH directly at all. Clearly, no other damage rectangle affects directly MATH during MATH, and this damage rectangle does not affect MATH before MATH.
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math/0003117
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This is obvious in most cases. One case when it is not is when the exposed edge is a left outer cell that is not an expansion cell. It is imaginable namely that its right neighbor is still in the growth stage. However, our definition of siblings requires the ages of cells to be monotonically nonincreasing as we move away from the originating colony, therefore this is not possible. The situation is similar when a doomed exposed right endcell dies.
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math/0003117
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Let us assume, on the contrary, that MATH is vacant during all of MATH, and we will arrive at a contradiction. The conditions and the rule MATH imply that we have MATH at some MATH and this stays so while MATH is vacant. Now the rule MATH requires that MATH become a cell. However, REF (Computation Property) requires MATH to actually become a cell by time MATH only if the body of no existing cell intersects with the body of MATH: see REF . Suppose therefore that some cell MATH intersects the body of MATH during MATH. Suppose that MATH does not disappear before MATH. Then REF implies that during this time, MATH becomes REF, hence MATH loses the ability to recreate MATH fast. REF of the same rule makes MATH latent within MATH time units. REF then erases MATH within MATH units of time. Any new cell MATH causing a similar obstacle to creating MATH could arise only if MATH creates it. Indeed, the only other way allowed by the Computation Property is REF there; however, this case, reserved for the possible appearence of a latent cell out of ``nothing", (indeed, out of lower-order germs) requires MATH to be vacant for all MATH, that is, that MATH have no (adjacent or non-adjacent) neighbors. But, MATH would be such a neighbor, so MATH will not appear. Thus, MATH will be created within MATH time units after the disappearence of MATH. Suppose that MATH disappears before MATH. If it does not reappear within MATH time units then MATH will be created as above. Suppose therefore that MATH reappears. When it reappears we necessarily have MATH. This turns REF within MATH time units. After it turns REF, the rule MATH erases MATH within MATH units of time and then MATH will be created in MATH time units. Cell MATH will not be recreated to prevent this since MATH for at least MATH time units.
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math/0003117
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The animation requires a sibling for MATH with both MATH and the sibling non-dying. Due to the minimum delay MATH in dying which they did not even begin, these cells remain live till after MATH. Since there is no colony boundary between them, a cut will not break the sibling relation of these cells either.
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math/0003117
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Since MATH is not affected immediately by damage, legality implies that the change is the delayed result of the application of MATH. Let us prove REF first. Suppose first that MATH is not affected by the damage via neighbors at time MATH. Then at the observation time MATH corresponding to the switch MATH, we have the situation described by REF. Suppose now that MATH is affected by the damage via neighbors at time MATH. Then, according to REF , it is not affected via neighbors during MATH. Let MATH be a switching time of MATH in this interval (in a moment, we will see that MATH exists.) Then at the observation time MATH corresponding to the switch MATH, we have the situation described by REF. Indeed, there are at most MATH steps of MATH between MATH and MATH; but the delay parameter of MATH is at least MATH, so since REF implies MATH, the observation time MATH must have occurred during the wait. Let us prove REF. According to REF (Animation), the applied rule was either MATH or MATH. In the healing case, the parents are frozen and non-dying, therefore they will not change their age for a while yet in any way but healing. In the growth case, the age of the child is made equal to the age of the mother at a not much earlier time MATH since once animation has been decided the assignment during animation happens fast (with delay MATH). Therefore the mother has time for at most one increase (with delay MATH) of age in MATH, and this will not break the sibling relation. The mother (and father) does not die since the rule MATH would have announced this at least MATH time units earlier (see REF ) via the field MATH and this would have turned off the animation or healing of MATH (REF implies MATH).
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math/0003117
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If a cell MATH breaks a sibling relation by a rule then one of the cases listed in the statement of the lemma holds. This follows from the definition of siblings, REF (Address and Age), REF (Killing) and the healing rule. We will show that if neither of the possibilities listed in statement of the lemma holds then the cells remain siblings. Suppose that MATH or MATH were animated at some time in MATH; without loss of generality, suppose the latest such time was MATH and the cell was MATH. Then MATH will be without an age change for at least MATH time units which, due to the fact that REF implies MATH, is longer than the whole period under consideration. If MATH also underwent animation during this interval then the same is true for it, hence the two cells remain siblings. Suppose therefore that MATH has been live during MATH. The rule MATH implies that MATH is not a germ unless MATH is one of the parents. If MATH is a parent of MATH then REF implies that the two cells remain siblings for at least MATH time units; after this, both cells have seen each other as siblings and therefore REF (Address and Age) shows that they remain siblings until a cut or a death. Suppose that MATH is not a parent of MATH. If it has changed its age within the last MATH time units then it will not change the age for a long time after, and the two cells remain siblings. If it has not changed its age within this time then for at least MATH time units before the observation time before the animation, it already is a partner of the mother MATH. The rule MATH of Subsection REF implies then that MATH is frozen which keeps MATH and MATH siblings. Suppose now that both cells have been live during MATH. If MATH changes its age within this time then it will not change its age soon and therefore remains a sibling. Suppose therefore that MATH does not change its age during this time. If MATH was a sibling all the time during MATH then MATH sees that MATH is a sibling and will not break the sibling relation. Suppose therefore that MATH changes its age within this interval and becomes a sibling of MATH this way. Then MATH had ample time before this age to observe that MATH is a partner. Therefore MATH is frozen and will not change its age at the next switch.
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math/0003117
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Direct consequence of the definition of siblings and exposed edges and REF (Address and Age), REF (Killing).
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math/0003117
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Parental links can always be replaced with horizontal and vertical links. Horizontal links of size REF jump over a latent cell only. So, paths cannot jump across each other.
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math/0003117
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We call the path to be constructed a desired path. Let us start constructing a steep path MATH with MATH backward from MATH. If we get to MATH then we are done, otherwise, we stop just before we would hit MATH, with MATH being the last element. Then MATH and MATH. Indeed, if this is not so then we could continue the path either by a vertical or by a parental link. The vertical link would be shorter than than MATH, and the parental link would lead to a damage-free cell, so either of them would be allowed. Let us now go back on the path for MATH until the first MATH (counting from MATH) such that either MATH is a parent of MATH or MATH can have a horizontal link at some time during MATH. There will be such a MATH, since otherwise the cell MATH would be isolated throughout MATH with no near-end-correction in a neighbor, and the rule MATH would kill it by the time MATH. Suppose that MATH is not a parent: then let MATH and let MATH be the earliest time in MATH when MATH has a sibling MATH. Let MATH. Suppose that MATH is a mother: then it has a sibling MATH. Let MATH in this case. Suppose that MATH is a father: then let MATH be the corresponding mother with MATH. Let MATH be the part of the original path until MATH, and MATH the new part ending in MATH. Let us build a steep path MATH, MATH, MATH, backwards until either MATH or MATH and let MATH be the part of this path ending with MATH. If MATH then MATH combine to a desired path, suppose therefore that MATH, hence MATH. Then MATH is a parent of MATH, hence by the NAME Lemma, MATH is a sibling of MATH at time MATH. By definition, if MATH is the first time after MATH when MATH has a sibling then MATH; hence MATH. By the NAME Lemma, MATH is also parent of MATH (by the same animation): choosing this as MATH, we are done.
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math/0003117
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It is easy to see that the length of the path is at most MATH. By the definition of the gap path MATH above, path MATH starts on its right. Since age varies by at most REF per link along a path (except for a double link) and since MATH is not a germ cell younger than MATH, no cell on MATH is a germ cell with age MATH, while according to REF , all cells in MATH space-consistent with MATH are germ cells with such age. Therefore MATH never crosses the gap path from right to left. REF gives a lower bound on how fast MATH moves right. Since MATH stays in the colony of MATH, MATH (see REF for the last inequality).
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math/0003117
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Let MATH, and let MATH be the leftmost cell such that the gap MATH has right-age MATH. Let MATH. Assume that for all MATH, a gap path MATH is defined with the desired properties and in such a way that MATH is not a germ cell younger than MATH and is not in the wake of a damage rectangle. Then MATH is a edge space-consistent with MATH that is either exposed or is to become exposed in one step, in one of the cases of REF (Exposing). This last case occurs only if there is a damage rectangle during MATH. Since MATH is not in the wake of a damage rectangle we can build a backward MATH-path MATH according to the Ancestor Lemma. REF and the bound REF gives MATH . The backward path MATH ends in at time MATH. Indeed, if it ended in a birth then MATH would be a germ cell younger than MATH contrary to the assumption. MATH does not cross the gap since otherwise MATH would be a germ cell younger than MATH which was excluded. Therefore defining MATH we have MATH. Without loss of generality, we can suppose MATH. Otherwise, REF (Crossing) implies that path MATH crosses MATH and we can switch from MATH to MATH at the meeting point. Thus, MATH is on the interval MATH, aligned with MATH. Combine MATH and the horizontal path from MATH to MATH into a single chain of links connecting siblings. This chain has at most MATH horizontal links from the left edge MATH to MATH, and then at most MATH links on MATH, giving fewer than MATH links. The assumption that the gap path satisfies the requirements of the Lemma implies MATH, hence MATH is to the right of the left end of the colony of MATH. In the number of steps available, a protected left edge cannot occur on such a path unless MATH is a left outer cell and the path leaves the colony of MATH on the right. Indeed, suppose first that MATH is a member or right outer cell. For a path of ancestors starting from such a cell to turn into a left outer cell this path would first have to walk right to participate in the creation of a colony and then walk back through the growth process from that colony, which is impossible due to REF. Suppose now that MATH is a (exposed) left outer cell. The age of cells on the path is locally monotonically nondecreasing, except when it crosses into another work period. Indeed, since the age of siblings is required to be monotonically nondecreasing towards the originating colony, it is nondecreasing on the horizontal part of the path; it is also nondecreasing on the part MATH constructed by the Ancestor Lemma, except on the horizontal link allowed by that lemma. Therefore the age of MATH cannot be smaller by more than REF. If the age is not smaller at all then MATH is also exposed. If decreasing the age by REF makes it protected then we must have one of the cases in REF . Suppose that MATH is a germ cell. Left germ cells inside a colony, as well as member cells they can turn into on the path MATH are always exposed. We should not rely on this asymmetry in the definition of ``exposed" for germs, however, since we want to apply the lemma also when changing left to right. Let us assume therefore for a moment that left and right is interchanged in the definition of ``exposed" for germs. Then a germ cell that is a left edge is exposed to the left only if its age is MATH. In this case, the same argument works as for left outer cells above. Consider a time interval MATH, assume that the gap MATH with the desired properties was defined up to time MATH, where MATH has size MATH and right-age MATH. Assume that for MATH, no wake of a damage rectangle intersects the area where we define the path further. Then the gap path MATH can be defined further in MATH in such a way that MATH has right-age MATH and size MATH with MATH . Let us define MATH as follows. Suppose that it was defined up to time MATH and let MATH be the next time that is a switching time of either MATH or MATH. We distinguish the following cases. CASE: MATH is a switch of MATH. If the switch is an animation resulting from MATH creating a sibling of MATH then MATH, else MATH. CASE: MATH is a switch of MATH. If MATH dies then MATH is the closest cell to the right of MATH that is not a germ cell younger than MATH or MATH, else MATH. CASE: In all other cases, we leave MATH unchanged. MATH has right-age MATH. Let MATH be a live cell in MATH space-consistent with MATH, and let us build a sequence MATH, MATH, MATH with MATH as follows. Without loss of generality, assume that MATH is a switching time of MATH. If MATH is live at MATH then MATH and MATH is the previous switching time of MATH. Otherwise, either MATH is a newborn germ cell, in which case the sequence ends at MATH, or it has a parent MATH. If MATH then MATH, MATH. Let us show that otherwise, MATH was created by the internal correction part of the healing rule. If MATH then it could not have created MATH by the growth rule since by the definition of MATH, this would have increased MATH, bringing MATH outside the gap. If MATH then it could not have animated MATH by the growth rule since MATH is an exposed edge. The end-correcting part of the rule MATH could not be involved. Indeed, the left end in case could only be the left end of the right neighbor colony of MATH. In this case, MATH would be a member cell and hence MATH would be a left outer cell. For that case, however, we assumed in REF that the path (hence also the gap path which is to its left) stays in the colony of MATH. (This is the part of the Gap Lemma where we use the fact that we upper-bound only the right-age, that is, the age of cells space-consistent with MATH: thus, the end-healing of some other colony inconsistent with MATH can be ignored.) If the internal correction case of healing created MATH, let MATH be the father of MATH, this will be inside the gap. By the parent construction, MATH and MATH. We see that the sequence MATH steps back at least MATH time units in every step and the age of MATH can decrease by at most REF in each such step. Since it can only end in a birth or a germ cell in MATH this and REF proves the age bound. CASE: NAME MATH does not move left during MATH. During the same period, MATH moves right at most once. CASE: For all MATH with MATH, we have MATH. Let us prove REF first. By REF , the gap can contain only latent and germ cells. The size of MATH does not allow MATH to decrease it: indeed, it follows just as as in REF above that end-healing cannot operate. Since MATH is a left gap, the rule MATH can decrease it only on the left. After one such decrease, the size is still MATH. The next application of MATH is away by a waiting period of length MATH. Let us prove REF. Since the conditions of REF are satisfied, MATH is an exposed left edge or is a cell whose age is just one step before the applicability of one of the cases in REF (Exposing). If MATH is exposed then the rule MATH kills it within MATH time units. In the other case, nothing prevents the age of MATH to increase within MATH time units. From now on, the rule MATH applies. To conclude, note that according to REF, every MATH time units, the left edge MATH moves at most one cell width to the right but the right edge MATH moves at least two cell widths. Since the size started from MATH, it will remain MATH. This proves the lower bounds on the size of MATH and on MATH. Let MATH be the wake of the damage rectangle and assume that MATH. Assume also that MATH has size MATH and right-age MATH at time MATH. Then the gap path MATH with the desired properties can be defined for MATH; moreover, MATH has size MATH and right-age MATH. We also have MATH for all MATH. Let MATH. Assume that the interval MATH is closer to MATH than to MATH. Then MATH will be defined just as in the damage-free case. Let MATH. Then we set MATH to be the first site from the left that, at time MATH, is MATH and is aligned with MATH. This defines a gap MATH of size MATH for this time interval. This gap will not decrease during MATH since it is too large for MATH and MATH the damage has insufficient time to trigger MATH. So, the age bound reasoning of point REF above applies and the right-age of MATH will be MATH. Let MATH and MATH. Then MATH, and we set MATH to be the greatest cell MATH and aligned with MATH. Then MATH, hence MATH has size MATH. Again, the reasoning of point REF above applies to show that MATH has age MATH. If MATH then no growth animation could have been pending in MATH from time before MATH and this completes the proof of the claim in REF. Let MATH and MATH. Then we define MATH. At most one cell can be added to MATH by growth during MATH. REF (Ancestor) shows that any live cell in MATH can be traced back to a birth within MATH or to a time in MATH before MATH and therefore the same reasoning as for the other cases gives that MATH has right-age MATH. Assume now that MATH is closer to MATH than to MATH. This case is similar to REF , so we point out only the differences. Now, the gap on the left of MATH may decrease by one during MATH, to size MATH, if a growth step occurs on the left. For MATH, now the cases we distinguish are: Assume MATH. Then MATH. We set MATH to be the smallest cell MATH and aligned with MATH. Then MATH, hence MATH has size MATH, even if a growth step occurred at MATH during MATH. Now assume MATH. Then we define MATH. Since no growth could have occurred on the right-hand side, by applying the earlier reasoning, we will find that MATH has right age MATH. Let us construct MATH for MATH. In the space-time rectangle considered, at most one damage rectangle occurs. Indeed, MATH follows from REF. Let MATH and let MATH be the supremum of the those MATH until which the damage-free construction in the proof of REF is applicable: thus, if there is no damage involved then MATH, else MATH. Applying this construction, we get to MATH with gap size MATH. Applying REF, knowing there is no damage before, we find the age upper bound MATH and the lower bound MATH on MATH. If MATH then we apply the construction of REF to get to MATH. Now, the gap has age upper bound MATH, and lower bound MATH on MATH. Let now MATH take the role of MATH and repeat the damage-free step to MATH. Now, we have age upper bound MATH and lower bound MATH on MATH.
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math/0003117
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Let us show that MATH is a left exposed edge for MATH in MATH. Indeed, MATH can be a weak left exposed edge that is not a left exposed edge only if it has a left sibling that is not strong. Now, if MATH does not have a left sibling and it gets one then it follows from REF (Glue) (and the exclusion of cut, since the edge does not become exposed) that the only way to lose this sibling is if the sibling dies again, which it cannot do before making at least MATH switches, becoming a strong sibling in the meantime. From this, it is easy to see that MATH can have a left sibling only during a time interval adjacent to either MATH or MATH. Since the sibling stays weak these time intervals must be at most MATH long. Due to the above observation, since MATH is a left exposed edge after its first complete work period following MATH, either MATH or MATH will kill it within the following MATH steps. In REF repetitions of this, a gap of width MATH will be created. During this time, by REF, at most one growth step can occur on the left, leaving still a gap of size MATH.
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math/0003117
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Let us construct a backward path MATH made up of horizontal and vertical links such that cell MATH is a weak left exposed edge during every nonempty time interval MATH, and the backward path never passes into the right neighbor colony. Suppose that MATH has already been constructed. CASE: The construction stops in any of the following cases: CASE: the path has reached length MATH; CASE: the path moved REF steps to the left; CASE: MATH belongs to the left neighbor colony; CASE: MATH is not a weak left exposed edge; CASE: If MATH has a switching time MATH immediately before MATH such that MATH is a weak left exposed edge during MATH then let MATH. CASE: Otherwise, let MATH be the lower bound of times MATH such that MATH is a weak left exposed edge. If MATH then let MATH. CASE: Assume now that MATH. Then at time MATH, cell MATH became a weak left exposed edge without taking an action: hence, it must have lost a strong left sibling. (Animation does not produce any exposed edge, see REF (Animation).) According to REF (Glue), this can only happen in one of the ways listed there. It is easy to check that of these, only the killing of the strong left sibling produces an exposed edge in MATH. Each of the rules MATH, MATH and MATH that could have killed MATH presupposes that this cell did not have a left sibling at the observation time MATH of MATH; thus, it did not have a strong left sibling at time MATH. On the other hand, as a strong left sibling of MATH, it has been alive for at least MATH hours. Let MATH. Such a MATH will be called a right jump. By the construction, each jump is surrounded by vertical links. For each jump MATH, cells MATH and MATH are strong siblings at time MATH. Let MATH for MATH be the endpoints of vertical links on the backward path with the property that MATH is not a vertical link; let MATH be the last point of the path. Let us number these points forward in time: MATH, MATH, MATH, etc., creating a left boundary path. If the path has moved REF cells to the left then we are done. If MATH then REF (Bad Gap Opening) is applicable, and it shows that there is a bad gap on the left of MATH. If the path has stopped for some other reason then we have the following cases. CASE: MATH is a weak left exposed edge belonging to a neighbor colony. CASE: MATH is not a weak left exposed edge. In REF , the death of MATH must have created the weak left exposed cell MATH that was a left colony endcell. This is possible only if either MATH is an expanding germ cell closer than MATH to the origin of expansion or it is some other kind of cell in the originating colony of MATH. Continue the construction of the backward path for another MATH steps (it is easy to see that now it will not be stopped earlier) and apply the Bad Gap Opening Lemma, showing that the present lemma is true for this case. In REF we have either one of the cases in REF (Exposing), or MATH dies at time MATH as a left, non-exposed edge.
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math/0003117
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If both endcells of MATH are protected during MATH then the only thing preventing the increase of age in a protected edge, according to REF (Address and Age), is when MATH is frozen. According to REF (Freeze), this MATH can become frozen only when it has a non-dying partner. But REF (Bad Gap Inference) would be applicable to this partner as an edge turned toward MATH and it is easy to verify that in the present case, this lemma implies an impossibly large gap between the partners. Therefore the minimum age of MATH increases every MATH steps. Suppose now, without loss of generality, that the right end of MATH is exposed at some time MATH. REF (Bad Gap Inference) implies the existence of a bad gap on the right of MATH unless one of REF or REF occurs. If none of these cases occurs then REF (Running Gap) will widen the gap as predicted in REF. Suppose that REF of the Bad Gap Inference Lemma occurs. Thus, there is a backward path of length MATH leading from MATH to a protected right colony endcell MATH, just being killed by a right neighbor cell and exposing a left neighbor. There are only two ways this can happen. One: when MATH is the endcell of a growth and is killed by the end-healing of member cells of another colony. This case does not really occur. Indeed, let MATH be the left exposed cell trying to do the end-healing. The Bad Gap Inference lemma can be applied to the MATH and would imply now, when none of the distracting cases applies, the existence of a large gap between MATH and MATH. The other case is when MATH is the endcell of a germ. Then the exposed edge thus created will never become protected, and the Bad Gap Opening lemma implies the creation of a bad and growing gap within MATH hours. Suppose that REF of the Bad Gap Inference Lemma occurs. The cases in REF fall into two categories. In all cases but REF (Bad Gap Opening) applies to the development forward in time from this event, showing that again, the exposed right edge moves left, creating an unhealable gap. In REF , a non-end growth cell at MATH becomes exposed. The edge may later disappear by healing, but only by end-healing. Indeed, for internal healing a close partner would be needed but the Bad Gap Inference Lemma would show (without the distracting other cases, this time) that the partner cannot be close. In order to prevent a bad gap from opening, the end-healing must succeed within MATH hours. Therefore this kind of exposed edge will exist at most for a time interval of length MATH on the right and for a similar interval on the left. The same reasoning applies to REF . We found that in three intervals outside the possibly two exception intervals, the minimum age of MATH increases every MATH steps.
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