paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9903024 | For MATH, we have MATH . |
math/9903024 | First notice that MATH. Indeed, since MATH is connected, the action of MATH on MATH is trivial. Hence for any MATH, MATH is a harmonic form in the same cohomology class as MATH, therefore, MATH. By REF , MATH. Since MATH and MATH has no harmonic part, we have MATH where MATH. Now since MATH, MATH. From MATH we see that... |
math/9903024 | It suffices to show that for any MATH, there exists MATH, such that MATH. Without loss of generality, we can assume that MATH is harmonic. By REF , we can take MATH where MATH. |
math/9903029 | The manifold MATH can be contracted to its MATH-dimensional subcomplex (graph) such that MATH and MATH are its vertices. A connected component in the space of paths joining MATH and MATH in this graph is obviously contractible. |
math/9903029 | Since MATH is a ramified covering, the mapping MATH is a covering, too (nonramified). So it suffices to prove that the set MATH is contractible. The set MATH is discrete. Join its points with a network, a set of smooth nonintersecting arcs such that the complement of these arcs in MATH is a union of discs. Apparently a... |
math/9903029 | Apparently, for any complex structure MATH on MATH there exists exactly one MATH-compatible pair MATH such that MATH. This, together with the Corollary of REF , allows to consider the space MATH as a quotient of the space MATH of complex structures at MATH by action of the group MATH. The point stabilizer of this actio... |
math/9903029 | follows immediately from REF : the NAME space MATH is simply connected, and one has only to observe that the action of the group MATH in MATH is faithful and discrete. |
math/9903029 | Take the ``square root" of the mapping MATH presenting it as MATH where MATH maps the point MATH to MATH, and MATH is defined as MATH. Let MATH be the full preimage of the meridian MATH under MATH. Without loss of generality one can suppose that MATH is the real axis. It divides MATH into two open disks MATH. NAME of t... |
math/9903029 | Recall that the meridians MATH of the slicing cut MATH into slices each one containing exactly one point MATH. Connected components of preimages of these slices are the parts into which MATH divides MATH. Consider, for example, the slice MATH bounded by MATH and MATH, and containing MATH. Split it into a union of curve... |
math/9903029 | follows easily from NAME 's lemma. |
math/9903029 | The mapping MATH in question is an algebraic diffeomorphism of MATH, and thus a linear function: MATH. This function preserves all the MATH critical points (distinct) of the polynomial MATH, and therefore MATH. |
math/9903029 | Prove first that the kernel of the mapping MATH is generated by the involution MATH. Projections MATH and MATH determine an element MATH completely. Thus, it suffices to prove that MATH is the only nontrivial element of the group MATH which projects to the identity mapping of MATH. Apparently any such mapping MATH acts... |
math/9903029 | Consider the quadrangulation MATH together with the biangular faces. Mapping MATH preserves the orientation, and therefore the cyclic ordering of edges in every white vertex of the graph MATH coincides with the ordering of the meridians: MATH. So, if the quadrangulation is not monotone then there exists a pair of faces... |
math/9903029 | To prove that we have defined an action, we should check the defining relations of the group MATH (see CITE for proof; it follows easily from relations REF - REF , too): MATH . The checking is routine. |
math/9903029 | Write MATH where each MATH is either MATH or MATH, and denote MATH. To prove the theorem by induction on MATH suppose that for MATH the theorem is proved, and consider three cases. The symbol MATH denotes the face labelled MATH in the quadrangulation MATH where MATH. Case MATH, and the faces MATH and MATH are not adjac... |
math/9903029 | We prove that already the action of the subgroup MATH generated by MATH is transitive. Take some vertex MATH of the tree as a root, and orient every edge away from the root (downwards). We can speak now about upper and lower end of an edge. Two edges having the same upper end will be called brothers. If the lower end o... |
math/9903029 | Suppose that MATH has a cycle formed by edges MATH. Since MATH do not intersect, they form a polygon MATH. At least one vertex of the original MATH-gon lies inside MATH which is impossible. Thus, MATH is a tree. The second assertion is proved by induction on MATH. Consider the face MATH of MATH marked MATH. The quadran... |
math/9903029 | Indeed, there exist MATH different trees with MATH numbered edges. |
math/9903029 | Let MATH be a covering such that MATH. To prove the statement we must show that the elements MATH and MATH belong to the subgroup MATH. Since the subgroups MATH and MATH have the same index, they coincide, and thus MATH. It is enough to consider the case when MATH is either MATH or some MATH. For brevity we will show i... |
math/9903032 | We give the proof in some detail since this is helpful for the implementations described in the next section. REF preserves the function MATH. We prove that MATH, the symmetric closure of MATH, preserves MATH. Let MATH so that MATH. From the definition of MATH there are two possible situations. For the first case suppo... |
math/9903032 | It is straightforward to verify that irreflexivity, antisymmetry and transitivity of MATH and MATH imply those properties for MATH. The ordering MATH is admissible on MATH because it is made compatible with the right action (defined by composition between arrows on MATH) by the admissibility of MATH on MATH. The orderi... |
math/9903032 | It is immediate from the definition that MATH is length-non-increasing. It is straightforward to verify that MATH is irreflexive, antisymmetric and transitive. It can also be seen that MATH is linear (suppose neither MATH nor MATH then MATH, by the definition, and linearity of MATH, MATH). It is clear from the definiti... |
math/9903032 | This follows immediately from the definition of MATH. |
math/9903032 | Let MATH such that MATH. There are two cases to be considered, by REF . For the first case let MATH, MATH for some MATH, MATH such that MATH. Then MATH. It follows that MATH since MATH is admissible on MATH. For the second case let MATH, MATH for some MATH, MATH such that MATH. Then MATH and so by REF MATH. Hence MATH ... |
math/9903032 | Let MATH be a critical pair. Then there exists a critical term MATH and two rules MATH, MATH such that MATH reduces to MATH with respect to MATH and to MATH with respect to MATH. We first give the two non-overlap cases. Suppose MATH, MATH. Then there exist MATH, MATH such that MATH as shown: MATH . The pair MATH immedi... |
math/9903032 | This is immediate from the Lemma. |
math/9903032 | Let MATH and MATH be a pair of rules. Recall that MATH is the representation of a term MATH as a list. The first type of list overlap occurs when MATH is a sublist of MATH (or vice-versa). This happens in REF . The second type of list overlap occurs when the end of MATH matches the beginning of MATH (or vice-versa). Th... |
math/9903032 | By REF is the result of two different single-step reductions being applied to a critical term MATH. Therefore MATH and MATH. It is immediate that MATH, and so adding MATH to MATH does not add anything to the equivalence relation MATH. |
math/9903032 | The procedure finds all critical pairs resulting from overlaps of rules of MATH. It attempts to resolve them. When they do not resolve it adds them to the system as new rules. Critical pairs of the new system are then examined. When all the critical pairs of a system resolve, then the procedure terminates, the final re... |
math/9903032 | Let MATH present the NAME extension data MATH for the NAME extension MATH. Let MATH be a category finitely presented by MATH and let MATH. Then MATH presents the NAME extension data MATH for the NAME extension MATH. We require to prove that MATH is the NAME extension presented by MATH having data MATH. It is clear that... |
math/9903033 | The quotient morphism MATH, defines a surjection MATH. Then MATH. Define MATH by MATH . To prove that MATH is well-defined we show that it preserves the right congruence MATH. We assume all polynomials of MATH are monic. Let MATH and MATH and suppose that MATH for some MATH, MATH, MATH. By REF. Now either REF MATH and ... |
math/9903033 | Define MATH by MATH. The verification that MATH is a well-defined bijection on the congruence classes is similar to that detailed in the proof of REF . |
math/9903033 | According to the definition, the process of reduction replaces one monomial with monomials which are smaller with respect to MATH (since MATH is a term order on MATH). The existence of an infinite sequence of reductions MATH of polynomials MATH would imply the existence of an infinite sequence MATH of terms MATH. There... |
math/9903033 | The length of a reduction sequence is defined to be the number of one-step reductions of which it is made up. This proof is by induction on the length of the reduction sequence MATH. For the basis of induction suppose the length of the reduction sequence is zero. Then MATH so MATH. For the induction step, assume that i... |
math/9903033 | In REF we proved that MATH is NAME and therefore, by NAME 's Lemma for reduction relations on sets, we need only to prove that MATH is locally confluent. Let MATH such that MATH and MATH. Then MATH and MATH for some MATH, MATH, MATH. Let MATH and MATH. If the reductions do not overlap on MATH, that is, MATH then it is ... |
math/9903033 | The result is proved by showing that, in each of the five cases, an NAME MATH resulting from a match of polynomials MATH can be written in the form MATH and therefore MATH. |
math/9903033 | The result is immediate, since for all MATH then MATH where MATH, MATH. |
math/9903037 | For the first statement, we observe that MATH by assumption contains the four lines MATH. So MATH lies in the ideal MATH. Applying this symmetrically yields the result. The second statement is immediate. |
math/9903037 | The first statement follows because a generic line through MATH meets MATH twice there, and at two other points. The second statement follows by computing the discriminant of MATH, viewed as a quadric in MATH. |
math/9903037 | This is immediate from the equation of the surface. |
math/9903037 | One may check using homological criteria that the linear series MATH has rank REF, and that all of its sections are even functions, so identify points with their negatives. So, the map is MATH, and factors through MATH. The points of MATH are in the base locus, so get blown up. Now, the self-intersection MATH, which di... |
math/9903037 | To begin with, since MATH meets MATH at a smooth point, it must be tangent to the plane MATH, and likewise tangent to the plane MATH. Now, let MATH be the projection away from the point MATH, and let MATH, a conic. Our observations have placed four linear conditions on MATH, so MATH must lie in the pencil MATH . Now ou... |
math/9903037 | We begin by showing that for any MATH satisfying this hypothesis, there are only finitely many conics satisfying the hypothesis of the lemma. But this is easy, since at most two choices of MATH can give acceptable image conics MATH, and each of these will have at most two preimages in MATH. In fact, we observe that sin... |
math/9903037 | Inside MATH, consider REF-plane MATH of cubic forms MATH where as usual MATH. We have been concentrating our attention on the locus in this family where no MATH equals zero, and have been writing MATH. If we pull back our cubic condition MATH in the MATH variables to REF-plane MATH, we see that we have described an ope... |
math/9903037 | This can all be checked by computing the necessary determinants. |
math/9903037 | Given the previous theorem and the proposition, this reduces to a computation. |
math/9903042 | If MATH, MATH then surely MATH. Therefore by REF , one can find a constant MATH such that MATH for all MATH. Let us set MATH where MATH is the constant from REF . The numerical factor REF is somewhat arbitrary. We could chose any factor greater than REF; we take REF for simplicity. Choose MATH and consider the set MATH... |
math/9903042 | Let MATH be the constant given by REF , that is such that MATH for all MATH and all MATH. Let us put MATH where the constant MATH will be determined later. The evolution of the MATH are described by the following ODEs MATH . The analogous equations describe the evolution of the MATH. The methods of the previous section... |
math/9903042 | As in the proof of REF , we estimate separately three sums. MATH . Since in MATH, the norm of MATH, we can write MATH where the constant is defined by the inequality MATH . For this sum to be finite, we need MATH. For MATH we have MATH and hence MATH . Here the constant is the absolute constant defined by MATH . For th... |
math/9903042 | By the energy estimate, we know that MATH for all MATH and MATH. Hence, MATH. Fixing a MATH, set MATH. With this choice, MATH for all MATH, MATH, and MATH with MATH. As before, consider the set MATH . We have to show that if MATH is taken to be sufficiently large, the vector field points inward along MATH. We pick a po... |
math/9903042 | The proof of this theorem is similar to the proof of REF . From the assumptions, we know that MATH for all MATH. We set MATH and MATH for MATH, where MATH is a constant we will set later. Set MATH. Fixing a MATH, choose MATH so that for all MATH, MATH, and MATH with MATH, one has MATH . Notice that by the assumption on... |
math/9903042 | The proof of this theorem begins as the above theorem and then proceeds as the proof of REF . We change variables to MATH and MATH. We use the assumption on MATH to control the lower modes. Then we use the estimates from REF to control the nonlinearity. We omit the details. |
math/9903046 | The NAME bracket on MATH is MATH-equivariant, and so the osculation REF induces an algebraic bracket on the associated graded vector bundle MATH. A neighborhood of the origin in MATH can be identified with the exponential image of MATH in MATH and the NAME bracket in MATH is given by the usual NAME brackets of the left... |
math/9903046 | Let us write MATH, MATH and compute their bracket. By the very definition, we obtain MATH where the first equality follows from the fact that the flows of MATH and MATH through MATH coincide, the next one results from the vanishing of MATH on the image of MATH. Now, repeating the same arguments for MATH, we achieve jus... |
math/9903046 | Recall that the defining equation for the homogeneous components MATH, MATH, MATH, MATH, MATH is MATH while the component of degree zero is MATH . Now, consider vector fields MATH in MATH, MATH in MATH and let us choose elements MATH, MATH such that MATH, MATH with suitable functions MATH, MATH on MATH. Then MATH . The... |
math/9903046 | Let us consider the MATH-modules MATH . By the general theory we know that the complexifications MATH of the dual MATH-modules MATH are the complex cohomologies MATH listed in the table of all complex cohomologies, see REF. Further, let us notice that the two components in MATH have a canonical complex structure. Now, ... |
math/9903046 | We shall discuss only brackets MATH. The other ones are treated analogously. The first part is quite easy. Let us consider MATH. Further, choose MATH, MATH, and MATH such that MATH, MATH. According to REF , there are the projectable vector fields MATH, MATH on MATH such that their projections MATH, MATH satisfy MATH, M... |
math/9903046 | All projections of the NAME brackets MATH are linear over functions and thus algebraic. Therefore, REF implies immediately the first claim. Similarly for the other distribution MATH and the last claim follows by the standard foliation theory. |
math/9903046 | By the defining properties of the regular MATH-structures, the complexified NAME subbundle MATH must be involutive. Thus the obstruction against the integrability of MATH is the NAME tensor MATH. Consequently, the theorem will be proved once we verify the following claim: The NAME tensor MATH, expressed by its frame fo... |
math/9903046 | All considerations are local and so we may suppose that the whole MATH is hyperbolic. If MATH is a product of two REF-dimensional NAME, then we can also consider the product MATH of the corresponding canonical NAME bundles MATH, MATH equipped with the product MATH of the corresponding normal NAME connections. These bun... |
math/9903046 | Exactly as in the hyperbolic case, the complexification of the cohomologies we want to describe is fully described by REF. Because of the complex structure on MATH, each of the real components will produce two copies in the complexification. In order to recognize them, we have to notice that complexifications of comple... |
math/9903046 | Essentially, all technique we need has been developed already. In particular, we may repeat the computation of the NAME tensor from the proof of REF . Since MATH is complex, we can do that with any MATH. The result tells us that the NAME tensor MATH, evaluated on MATH, MATH, is equal to MATH . Now, under the additional... |
math/9903046 | The distributions are integrable if and only if the algebraic NAME brackets of two fields from the same component projected to the other one vanish. This is equivalent to the corresponding condition on the complexified bundles MATH. Now we may use the technique introduced in REF. Thus all the algebraic brackets in ques... |
math/9903046 | In fact, we have nearly proved all necessary facts. Again, the same computation with the NAME tensor reveals, that the antilinear part MATH of the entire curvature obstructs its integrability. Once we assume that all the torsion vanishes, there are no components of the curvature up to homogeneity four. This is not anti... |
math/9903046 | Analogously to the hyperbolic structures, there is the chain distribution in MATH spanned by the horizontal fields MATH with MATH. Again, the straightforward inspection of the possible curvature components reveals that there is no curvature with both arguments in MATH if the torsion vanishes. Thus the chain distributio... |
math/9903051 | Consider the decomposition MATH of MATH into REF-dimensional weight spaces. Our assumption that the weights REF are pairwise linear independent imply that MATH and hence that MATH is two-dimensional. Since the only oriented two-manifold with faithful MATH actions are MATH and MATH, MATH has to be one of them. However, ... |
math/9903051 | The first four assertions are obvious. To prove the last one, let MATH be the identity component of the kernel of the map MATH . Each point of MATH is a MATH-fixed point, so for each MATH one has an isotropy representation of MATH on MATH. The weights of this representation are independent of MATH, so they are the same... |
math/9903051 | Let MATH, be the connected components of MATH and consider a MATH-dimensional wall separating two adjacent MATH's. This wall is defined by an equation of the form MATH for some MATH. Let MATH, MATH, and let's compute the changes in MATH and MATH as MATH passes through this wall: Let MATH and MATH (with MATH and MATH). ... |
math/9903051 | For MATH, the vector field MATH is Hamiltonian and its Hamiltonian function, MATH, is a NAME function whose critical points are the fixed points of MATH. Moreover, the index of a critical point, MATH, is just MATH, so the number of critical points of index MATH is MATH; for more details see CITE. |
math/9903051 | For a vertex MATH, define MATH to be the length of the longest oriented path with terminal vertex at MATH. Then MATH is MATH-compatible and takes only integer values; a small perturbation of MATH produces an injective function, MATH, which is still MATH-compatible. |
math/9903051 | Let MATH be a NAME function as in the theorem of the previous section. Then its restriction to MATH is a MATH-compatible on the vertices, MATH, of MATH. |
math/9903051 | We will prove the ``if" part of this theorem by giving an explicit description of the one-skeleton of MATH. Let MATH be the GKM one-skeleton of MATH. Let MATH be an oriented edge of MATH with vertices, MATH and MATH, for which MATH and let MATH be the embedded two-sphere corresponding to MATH. Then the reduction of MAT... |
math/9903051 | The right and left hand sides of REF are the pull-backs to MATH and MATH of the image of MATH under the map MATH, where MATH and MATH. Since MATH and MATH belong to the same connected component of MATH, the pull-backs coincide. |
math/9903051 | Let MATH be the element of MATH whose restriction to MATH is MATH. Let MATH be the embedded two-sphere in MATH corresponding to MATH and let MATH be the restriction of MATH to MATH. Then the one-point manifold MATH is the reduction of MATH at MATH with respect to MATH. Therefore it suffices to check that MATH is the im... |
math/9903051 | REF is just a restatement of the assumption that MATH and MATH are coplanar; the positivity of MATH is a consequence of the convexity of MATH. If MATH weren't positive, MATH would be in the interior of the intersection of MATH with the plane spanned by MATH and MATH. |
math/9903051 | The fixed point MATH corresponding to MATH is the subspace MATH of MATH spanned by MATH. Therefore the tangent space at MATH is MATH with basis vectors MATH and the weights associated with these basis vectors are REF . |
math/9903051 | From the identification of MATH with the projective space MATH one sees that MATH is an embedded REF-sphere. Moreover, since MATH and MATH satisfy REF , this sphere contains MATH and MATH. To prove the last assertion note that the tangent space to MATH at MATH is MATH . Thus, if MATH and MATH, this tangent space has MA... |
math/9903051 | The index of MATH is the number of edges MATH with MATH; alternatively, it is the number of pairs MATH with MATH, which is the same as MATH. Let MATH be the elements of MATH. The number of elements MATH with MATH is MATH; the number of elements MATH with MATH is MATH and so on; therefore the number of pairs MATH with M... |
math/9903051 | Suppose that there are two distinct oriented edges, MATH and MATH, from MATH to MATH; let MATH. Since MATH and MATH, with MATH, it follows that MATH. Thus the vectors MATH, and MATH are distinct and coplanar, which contradicts the three-independence of MATH at MATH. |
math/9903051 | For MATH let MATH be the fiber over MATH. By definition, MATH is a complete one-skeleton with MATH vertices for which a generating class is the restriction of MATH to MATH. If MATH then MATH, the restriction of MATH to MATH, is an element of MATH and hence, by REF , MATH where MATH if MATH and is REF otherwise. To get ... |
math/9903051 | We will show that every element MATH can be written uniquely as MATH with MATH, MATH, if MATH and REF, if MATH. The restriction, MATH, of MATH to MATH, is an element of MATH, and, therefore, from REF it follows that MATH . But then MATH is constant along fibers of MATH, implying that MATH. Hence MATH can be written as ... |
math/9903051 | Let MATH be a vertex of MATH, corresponding to the edge MATH of MATH and let MATH be a neighbor of MATH, corresponding to the edge MATH and obtained as above by using the edge MATH. Let MATH be the path that connects MATH and MATH, crosses the MATH-level and is contained in REF-dimensional sub-skeleton of MATH generate... |
math/9903051 | The modifications from MATH to MATH are due to the edges that cross one level but not the other one; but these are exactly the edges with one end-point MATH. Let MATH the edges of MATH with initial vertex MATH that point downward and MATH the edges that point upward. Let MATH be the vertex of MATH associated to MATH an... |
math/9903051 | Let MATH be a generic element of MATH, let MATH be MATH-compatible and let MATH be the one-skeleton defined in the previous section. If there is only one vertex MATH with MATH, this one-skeleton is MATH, the complete one-skeleton with MATH vertices and if MATH this one-skeleton is just MATH. Therefore, by studying the ... |
math/9903051 | We first recall a construction used in CITE. For every regular value MATH, let MATH be the subring of those maps MATH that are supported on the set MATH. Now consider regular values MATH such that there is exactly one vertex, MATH, satisfying MATH. Let MATH be the index of MATH and let MATH be the values of the axial f... |
math/9903051 | We will show that every element MATH can be written uniquely as MATH where MATH. Let MATH be the vertices of MATH, ordered so that MATH . Then MATH and, if we let MATH, MATH vanishes at MATH. Now, suppose MATH is supported on MATH. Then MATH . If MATH, then MATH and if MATH let MATH . Then MATH is supported on MATH. Pr... |
math/9903051 | All we need to show is that the image of the map, MATH, is indeed in MATH, that is that MATH satisfies the compatibility REF for the reduced one-skeleton. Let MATH be a basis of MATH such that MATH and MATH is a basis of MATH. Let MATH such that MATH. Then the map MATH given by the identification MATH will send MATH to... |
math/9903051 | If MATH is in the kernel of MATH then MATH for every edge MATH with MATH, where MATH is the vertex that corresponds to this edge of MATH. Since MATH it follows that MATH is divisible by MATH. Similarly MATH is divisible by MATH. Consider now the maps, MATH, of MATH into MATH defined by MATH . It is clear that MATH and ... |
math/9903051 | We will show that MATH is surjective by a dimension count, using induction on the number of vertices MATH that lie under the level MATH. To start, assume there is only one such vertex, MATH. Then MATH is minimum of MATH and the reduced space MATH is a complete one-skeleton with MATH vertices, MATH, one for each edge MA... |
math/9903051 | Our construction of MATH will involve three steps: first we will construct the MATH's corresponding to the edges of MATH; then, for each of these MATH's, we will construct a tubular neighborhood of it in MATH. Then we will construct MATH itself by gluing these tubular neighborhoods together. Let MATH be the one dimensi... |
math/9903051 | Let MATH be a rational deformation. Then MATH since MATH divides both MATH and MATH, it follows that it divides MATH as well, which means that MATH. Conversely, if MATH then REF implies that MATH is a multiple of MATH and, since MATH, for MATH small enough, it is a positive multiple. We can choose MATH small enough for... |
math/9903052 | Since the product MATH is associative, it suffices to check MATH for the case MATH for some MATH, and MATH with MATH. In both of the sub-cases MATH and MATH this is easily verified. |
math/9903052 | The required commutation relations for the operators MATH and MATH follow from those for the elements MATH and MATH. For example MATH shows MATH, and MATH shows MATH. The only non-trivial commutation relation to check is that MATH, or equivalently that MATH is in the center of MATH. In fact MATH is a scalar (compare CI... |
math/9903052 | Using REF we compute MATH . Taking the difference MATH, the terms cubic and linear in MATH's cancel, and the remaining terms yield the Proposition. |
math/9903052 | We calculate: MATH . |
math/9903052 | We have already shown that MATH. Since MATH is MATH-invariant, MATH so that MATH. Furthermore MATH shows MATH. All other commutators are obvious. The formulas for MATH on generators show that the associated graded MATH-differential algebra is just the standard one on MATH. |
math/9903052 | Using REF we compute: MATH . Together with REF this proves REF . |
math/9903052 | This follows by writing MATH and MATH . |
math/9903052 | Let MATH be the multiplication map for MATH. The map MATH is defined by the condition MATH on basic elements MATH. According to REF , under the identification MATH, NAME multiplication is given by composition of the operator MATH followed by wedge product. On basic elements, we can replace MATH by MATH, and therefore M... |
math/9903052 | We compute MATH: MATH where the last term vanishes by the NAME identity for MATH. Similarly MATH . |
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