paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003099 | 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 leave... |
math/0003100 | 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/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 pseudodiffer... |
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. |
math/0003100 | 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 o... |
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, f... |
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. |
math/0003100 | 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. |
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 analyt... |
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 ... |
math/0003103 | By a change of coordinates, we can assume without loss of generality that MATH. The result then follows immediately from REF . |
math/0003103 | (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 conver... |
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 subva... |
math/0003107 | 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 s... |
math/0003107 | 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, consideri... |
math/0003107 | 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 t... |
math/0003107 | 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 o... |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | 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 ... |
math/0003113 | 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... |
math/0003113 | We have MATH . In the local case, we have an isomorphism MATH . By class field theory MATH so that MATH . |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | We have MATH which by biduality MATH proves our claim. |
math/0003113 | Apply REF to the exact triangles in REF . |
math/0003113 | 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. |
math/0003113 | 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 ... |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | 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 . |
math/0003113 | Apply MATH to the triangle MATH and use the above corollary. |
math/0003113 | 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... |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | By REF we have MATH and MATH. This implies the claim. |
math/0003113 | 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. |
math/0003113 | The inverse functor is given by MATH if the action of MATH factors through MATH. |
math/0003113 | 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. |
math/0003113 | See the proof of REF. |
math/0003113 | 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/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | Clear from the definition. |
math/0003113 | 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 isomorp... |
math/0003113 | The map MATH maps MATH to MATH. This gives the result. |
math/0003113 | This follows immediately from the definition of the maps MATH and MATH. |
math/0003113 | This follows from MATH and MATH, which is a consequence of the NAME resolution as MATH is a sheaf of regular rings. |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | 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/0003113 | 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 M... |
math/0003113 | 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. |
math/0003113 | Clear because the character group of MATH is MATH. |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | From REF we know that MATH because MATH is prime to MATH. This proves the claim. |
math/0003113 | This follows immediately from the above proposition. |
math/0003113 | The recipe to compute the moment map from REF combined with REF gives immediately the result. |
math/0003113 | 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. |
math/0003113 | 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 g... |
math/0003113 | 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. |
math/0003113 | 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 ... |
math/0003113 | As MATH is prime to MATH, this follows from REF and the definition of MATH in REF. |
math/0003113 | 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. |
math/0003113 | 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. |
math/0003113 | 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. ... |
math/0003117 | 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 tran... |
math/0003117 | 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 ... |
math/0003117 | 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 secon... |
math/0003117 | 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 MAT... |
math/0003117 | 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 . |
math/0003117 | 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 MA... |
math/0003117 | The statement follows by a standard argument from the strong NAME property of the NAME process MATH (see CITE). |
math/0003117 | 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. |
math/0003117 | 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 rand... |
math/0003117 | 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 MAT... |
math/0003117 | 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 ... |
math/0003117 | 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. |
math/0003117 | 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. |
math/0003117 | 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 ... |
math/0003117 | REF can be rearranged as MATH . |
math/0003117 | 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/0003117 | 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... |
math/0003117 | 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 subsect... |
math/0003117 | 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 ob... |
math/0003117 | 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 aff... |
math/0003117 | 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 aw... |
math/0003117 | 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 MA... |
math/0003117 | 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. |
math/0003117 | 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 s... |
math/0003117 | 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 ... |
math/0003117 | Direct consequence of the definition of siblings and exposed edges and REF (Address and Age), REF (Killing). |
math/0003117 | 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. |
math/0003117 | 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 pa... |
math/0003117 | 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, ... |
math/0003117 | 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-consi... |
math/0003117 | 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 edg... |
math/0003117 | 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 ... |
math/0003117 | 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 applica... |
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