paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/9901012 | By direct computation MATH . The second two terms cancel and the first term vanishes immediately. Now MATH by virtue of which and the NAME structural equations the remaining two terms cancel. |
math-ph/9901012 | This follows by virtue of the fact that MATH and MATH commute. Now, using the right graded NAME identity for MATH in the third line one has: MATH . |
math-ph/9901012 | Ker-MATH=the inverse image of the trivial endomorphisms of the right MATH -module MATH so that MATH furthermore, Im-MATH= REF. The zeroth cohomology is then given by MATH . |
math-ph/9901013 | Let MATH, then in terms of local coordinates they may be written as MATH and MATH. So that MATH . |
math-ph/9901013 | Let MATH denote an n-form on MATH and let MATH denote the corresponding section. They are related by a canonical n-form MATH on MATH according to MATH. The point MATH in MATH with coordinates MATH is the n-form MATH at MATH of MATH. Then MATH is uniquely determined to be MATH by the tautology MATH. |
math-ph/9901013 | It the local adapted coordinates MATH takes the form: MATH . Then MATH from which follow NAME 's equations of the NAME formalism. |
math-ph/9901013 | The statement follows from the tautology MATH where MATH is the section corresponding to the NAME n-form MATH at MATH of MATH which corresponds to point MATH of MATH. In local coordinates MATH takes the form MATH. |
math-ph/9901013 | The adjoint operator MATH is nilpotent iff MATH. This may be seen seen from the following identity MATH . Then by direct computation MATH . The generalised multisymplectic structural equation then implies that, MATH then, MATH . The first and second terms vanish by virtue of the NAME identity for the NAME algebra of MA... |
math-ph/9901013 | CASE: Let us firstly consider the case for NAME currents for which the corresponding bundle morphism is the identity on MATH. In this case we have shown that the graded vector spaces MATH and MATH are graded isomorphic. We wish to show that the complexes MATH and MATH are isomorphic. We also learn from REF . that MATH ... |
math-ph/9901013 | Let MATH. In local coordinates they take the form MATH and MATH MATH. |
math-ph/9901013 | The statement follows from the tautology MATH where MATH is the section corresponding to the NAME n-form MATH at MATH of MATH which corresponds to point MATH of MATH. In local coordinates MATH takes the form MATH . |
math/9901004 | Let MATH be a MATH-complement of MATH. Consider a good resolution MATH and put MATH. Write MATH where MATH is the proper transform of MATH and MATH is a subboundary on MATH such that MATH. Then MATH is plt CITE. By Inversion of Adjunction MATH is klt and since MATH is non-singular, MATH. By REF MATH is a MATH-complemen... |
math/9901004 | First we assume that MATH is MATH-ample and so is MATH. By Adjunction REF MATH is klt and MATH has only standard coefficients CITE, CITE. By REF MATH is MATH-complementary for some MATH. Moreover, we can take MATH if MATH is non-compact, see CITE for MATH and CITE for MATH (see also CITE and CITE). Since MATH is MATH-a... |
math/9901004 | Apply REF to MATH. We get a plt blow-up MATH with log canonical MATH. Since MATH, the cone MATH is generated by two extremal rays. Denote them by MATH and MATH. One of them, say MATH, determines the contraction MATH and therefore it is trivial with respect to MATH, so MATH must be positive on MATH. Since MATH, where MA... |
math/9901004 | For sufficiently small MATH the divisor MATH is also plt (see CITE, CITE). Apply MATH-MMP to MATH. After a number of flips we get one of the following: CASE: A log pair MATH, where MATH is nef over MATH. Since MATH, so are both MATH and MATH. This is REF. CASE: A log pair MATH with a non-flipping extremal contraction M... |
math/9901004 | Consider REF . It is easy to see that MATH is also a curve. Denote this curve by MATH. We have a contraction MATH. It is clear that MATH is nef over MATH. By CITE there exists a regular complement of MATH near MATH. By REF this complement can be extended to a regular complement MATH on MATH. The sequence of maps MATH i... |
math/9901004 | Since MATH, the divisor MATH is nef. If MATH is ample over MATH, then obviously, so is MATH. Thus we assume that MATH is not ample over MATH. This is possible only in REF. By of REF MATH is nef and big. Recall that MATH, hence MATH is generated by two extremal rays. If MATH is not ample, then it is trivial on some extr... |
math/9901004 | First we consider REF . By NAME 's REF we have a MATH-complement MATH of MATH, where MATH. By REF MATH can be extended to a MATH-complement MATH on MATH. As in the proof of REF its proper transform on MATH is a MATH-complement of MATH. In REF we have a contraction MATH and a boundary MATH such as in REF . Therefore MAT... |
math/9901004 | Assume that MATH is not klt. As in the proof of REF it is sufficient to show that MATH has a regular non-klt complement. The following is an easy consequence of CITE (see CITE). Let MATH be a projective normal surface and let MATH be boundaries on MATH such that CASE: coefficients of MATH are standard; CASE: MATH is kl... |
math/9901004 | Notation as in the proof of REF . We claim that the proper transform MATH of MATH on MATH is not plt. Indeed, in REF MATH is a sequence of flops with respect to MATH and the fact follows by CITE. Similarly, in REF MATH is not plt. The contraction MATH is log crepant with respect to MATH, so MATH is not plt by CITE. By ... |
math/9901004 | If MATH is birational, then the lemma follows by REF . We assume that MATH is one or two-dimensional. Consider MATH as a sufficiently small neighborhood of MATH. Replace MATH with its MATH-factorial dlt model (see CITE or CITE). Put MATH and MATH. Since MATH is not plt, MATH has at least two irreducible components. Ass... |
math/9901004 | By REF we may assume that MATH is compact and MATH is MATH-ample. Thus MATH. Denote MATH, MATH. If MATH has a regular complement, then by REF it can be extended on MATH, so we assume the opposite. By REF MATH has a MATH-complement MATH for MATH. Moreover, we may assume that MATH is klt (see REF ). Therefore both MATH a... |
math/9901004 | First note that since MATH is a germ along MATH, MATH is connected. Below we will use notation of REF. Taking into account REF we may assume that MATH is not plt. Let MATH be a plt blow-up from REF . By REF we may assume that MATH is a point and By REF there exists a MATH-complement MATH, where MATH. Denote MATH and MA... |
math/9901004 | Let MATH be a contraction from Theorem A. So, MATH is a log conic bundle or a log NAME fibration. Assume that MATH has no regular non-klt complements. Then MATH is exceptional by REF and it is sufficient to prove that singularities MATH (respectively, multiplicities of MATH) are bounded in terms of MATH. First, conside... |
math/9901004 | Let us take any complement MATH which is not klt. Then there is a (not necessarily exceptional) divisor MATH of the field MATH with discrepancy MATH. Let us take another non-klt complement MATH. We claim that MATH. Indeed, we can construct a continuous family of boundaries MATH, MATH such that MATH, MATH, MATH is lc, n... |
math/9901015 | The existence of a MATH-invariant prolongation map is shown by using a MATH-invariant tubular neighbourhood MATH of MATH in MATH: consider the conormal bundle MATH (where MATH denotes the cotangent bundle projection) on which MATH acts in a canonical manner such that MATH is MATH-equivariant; a MATH-invariant tubular n... |
math/9901015 | The first point is a direct consequence of the definitions and of REF . To prove that any function in MATH lies in MATH we can use an explicit description of MATH generated by MATH, MATH (a result which will be proved in REF ). The rest of this point is a consequence of REF . To prove the third point observe that MATH ... |
math/9901015 | Again we shall only be treating the MATH-invariant case and use the notation of the proof of REF . As in CITE we shall first construct the chain homotopies separately on a MATH-invariant open neighbourhood MATH of MATH and on an open MATH-invariant set MATH not meeting MATH such that MATH. The overlap region MATH has t... |
math/9901015 | Since the classical BRST operator is a super-derivation it immediately follows that its kernel is a sub-algebra of the super NAME algebra MATH. Moreover MATH entails that the image of MATH is a super NAME ideal in the kernel. The grading is inherited by the induced ghost number derivation. This proves the lemma. |
math/9901015 | For the first assertion, note that the chain homotopy equation MATH of the `pure' augmented NAME complex (see REF ) still holds on the augmented BRST complex. Hence MATH. Since MATH is obviously nilpotent of order at most MATH the right hand side is invertible. Thanks to MATH the map MATH commutes with MATH. Thus MATH ... |
math/9901015 | The first part is a simple computation using the easily verified fact that MATH. For the second part, thanks to REF it suffices to show MATH. We compute all these terms: Firstly notice that MATH. Since MATH we only have to compute the orders MATH, MATH, and MATH of MATH. The zeroth order is trivially zero since MATH is... |
math/9901015 | It remains to compute the classical limit which is straightforward. |
math/9901015 | This is a straightforward computation. |
math/9901015 | Clearly REF follows from REF and MATH since MATH respectively, MATH are homogeneous of bidegree MATH respectively, MATH. Thus we have to show REF which is a straightforward computation using MATH. In terms of the basis we obtain by a simple computation MATH and similar MATH. Together with REF , and REF applied for MATH... |
math/9901015 | Using the expressions in terms of a basis for these various operators the above commutation relations follow from a tedious but straightforward computation. |
math/9901015 | This follows from the very definitions and the last lemma. |
math/9901015 | First note that the operator MATH vanishes on MATH, hence every element of MATH is equal to a sum of elements of the form MATH where MATH and MATH which immediately shows that MATH is indeed a left ideal of the algebra MATH. Furthermore, MATH is stable under the above representation thanks to quantum covariance and the... |
math/9901015 | Note that the equation MATH means in lowest order that for MATH there is MATH such that MATH. Evaluating on MATH this means that MATH is contained in MATH. Since MATH obviously commutes with every MATH and using associativity we see that MATH has to be in MATH. Moreover in case MATH this means that MATH where MATH when... |
math/9901015 | Since MATH is a deformation of MATH the formal series of differential operators MATH is well-defined. Moreover, we have MATH whence MATH . Because MATH REF we immediately see that MATH (since MATH) which implies that MATH is a projection. This fact together with REF entails that the product MATH vanishes, and since MAT... |
math/9901015 | Firstly, note that the fact that MATH is a deformation of MATH implies that MATH is a well-defined deformation of the classical augmented chain homotopy MATH. Since MATH it follows that MATH commutes with MATH which immediately implies REF . Moreover, since the restriction of MATH to MATH is equal to MATH we see that t... |
math/9901015 | It suffices to show that for any differential operator MATH of order MATH there is a differential operator MATH of order MATH such that the following holds in the tubular neighbourhood of REF : MATH . But this is clear: the tubular neighbourhood is diffeomorphic to an open subset of MATH, hence using the classical mome... |
math/9901015 | Since the map MATH is a projection (see REF ) whose kernel (after restriction to ghostnumber zero) is equal to the quantum vanishing ideal it follows from REF that for all MATH . Hence for all MATH whence the representation identity follows since MATH is injective and MATH is surjective. The rest of the lemma is clear ... |
math/9901015 | The proof is entirely analogous to the proof of the classical REF since only purely cohomological statements are needed. For the last two parts note that MATH is in MATH if MATH and use the proof of REF . |
math/9901015 | The first part is clear: one has equality of quantum and classical representations since MATH is strongly invariant, since MATH intertwines the classical MATH-actions on MATH and on MATH in REF , and since MATH is the identity on MATH by REF . Moreover MATH anticommutes with MATH because the quantized NAME operator, th... |
math/9901015 | Write MATH where MATH are linear endomorphisms of MATH and MATH. We consider the equation MATH where MATH is in MATH. Its solvability is a standard argument for deformed cohomology operators (as for instance in the proof of the existence of the NAME construction): Choose a vector subspace MATH such that MATH. We constr... |
math/9901015 | We know that MATH is a principal fibre bundle over the reduced phase space MATH with structure group MATH. Observe that MATH using right invariant differential forms as a global basis of all differential forms on MATH and letting MATH act on MATH by NAME derivatives of right invariant vector fields. Moreover MATH clear... |
math/9901023 | Let MATH be an arbitrary point of MATH and MATH. Expand the functions MATH in the power series at MATH, which converges absolutely in some disc MATH. Since the first and higher order derivatives of the functions MATH, MATH, MATH over MATH can be also expanded at MATH in the power series which absolutely converge in MAT... |
math/9901023 | In accordance with the generalisation of the NAME theorem CITE, there exists a holomorphic function MATH having the same zeros and with the same orders as MATH. It is easy to show that the function MATH satisfies the requirement of the Proposition. |
math/9901023 | We prove the Proposition by induction over MATH. The case MATH coincides with REF . Suppose that the Proposition is valid for some fixed MATH. Consider a set MATH of holomorphic MATH-valued functions on MATH having rank MATH. It is clear that the rank of the set MATH equals MATH. Hence, one can find a set MATH of holom... |
math/9901023 | Let MATH be some point of MATH. Denote MATH, MATH, MATH. The vectors MATH, MATH, MATH are linearly independent, and one can find vectors MATH, MATH, MATH such that MATH is a basis of MATH. Consider the set MATH of holomorphic MATH-valued functions on MATH. It is clear that the rank of this set equals MATH. Applying arg... |
math/9901023 | Let MATH be a basis of MATH, and MATH be the dual basis of MATH. One has MATH where MATH, MATH, are holomorphic functions on MATH. Since for any MATH the vectors MATH are linearly independent, the matrix MATH is nondegenerate. Hence, there are holomorphic functions MATH, MATH, on MATH such that for any MATH one has MAT... |
math/9901023 | For any MATH the vectors MATH, MATH, MATH are linearly independent, while the vectors MATH, MATH, MATH, MATH are linearly dependent, therefore one has MATH where MATH, MATH, are complex functions on MATH. Due to REF one can construct holomorphic MATH-valued functions MATH, MATH, MATH such that the set MATH is of consta... |
math/9901023 | Without loss of generality we can assume that the rank of the set MATH equals MATH. Thus, in accordance with REF one can find a set MATH of holomorphic MATH-valued functions on MATH such that the set MATH is of constant rank MATH and REF is valid for MATH. It is not difficult to understand that for any MATH such that M... |
math/9901023 | The validity of the Proposition follows from the proof of REF , from the discussion given after that proof, and from the proof of REF . |
math/9901023 | The independence of MATH of MATH can be proved along the lines of the proof of REF . So we have just one subspace of MATH which we denote by MATH. It is quite clear that MATH. Suppose that MATH is a proper subspace of MATH. In this case there exists an element MATH such that MATH for each MATH, but MATH for some elemen... |
math/9901023 | Assume that that we proved the validity of REF for all MATH. Using these equations for MATH, one has MATH . Now differentiating the definition of MATH over MATH and MATH, we can easily show that REF are valid for MATH. |
math/9901023 | The validity of the assertion of the Proposition follows immediately from the NAME formula MATH after taking into account REF and relations REF . |
math/9901023 | From REF it follows that MATH . Relation REF gives MATH . The statement of the Proposition follows now from REF . |
math/9901023 | The vector field MATH is a vector field on MATH. Let MATH be an arbitrary extension of MATH. Consider the vector field MATH . This vector field is, by definition, horizontal and invariant. Let us show that it is an extension of MATH. Indeed, from REF one gets MATH . Since, MATH and MATH are horizontal vector fields, th... |
math/9901023 | Since MATH is horizontal and invariant, there is a unique vector field MATH on MATH such that MATH . It can be easily shown that MATH is an extension of MATH. From REF it follows that MATH . Further, the vector field MATH is horizontal and invariant. Therefore, there is a unique vector field MATH such that MATH . It is... |
math/9901023 | A vector field MATH on MATH is orthogonal to the fibers if and only if MATH . The NAME symbols of the metric MATH are MATH . Using the above relations one sees that the assertion of the Proposition is true. |
math/9901023 | The statement of the Theorem follows from REF . |
math/9901024 | Let MATH be a variety of all trivial representations over field MATH, that are representations in which MATH acts as zero algebra. For MATH the assertion of the Proposition is evident. Therefore, let us assume that MATH. By the condition MATH is free NAME algebra with the system of free generators MATH, and MATH, where... |
math/9901024 | As is well known, MATH can be regarded as MATH-module if the action of MATH in MATH is defined by the formula MATH, where MATH is an arbitrary element of the coset MATH. Proceeding from a decomposition MATH, where MATH is a compliment of the subspace MATH with respect to MATH, we can define a system of representatives ... |
math/9901024 | If MATH, then MATH, and MATH. On the other hand, MATH. By the splitting condition MATH. Hence MATH is the homomorphism MATH into MATH. Furthermore, MATH implies MATH. Consequently, MATH and MATH. Thus the mapping MATH is the injection. The surjective property of MATH is no less obvious: for an arbitrary MATH, MATH is t... |
math/9901024 | MATH is absolutely free algebra with a system of free generators MATH, where MATH is a basis for MATH over MATH, and MATH is well-ordered set CITE. MATH is semidirect product MATH. The mapping MATH - where MATH is the image of MATH in MATH - is extended to homomorphism MATH, and MATH maps MATH into MATH. Since MATH is ... |
math/9901024 | The mapping MATH is extended to the homomorphism MATH, with MATH mapping MATH into MATH. The homomorphism MATH is extended to the homomorphism MATH, with MATH mapping MATH into MATH. The composition of MATH and the canonical homomorphism of MATH onto MATH yields a homomorphism MATH. The homomorphism MATH maps MATH into... |
math/9901024 | In the Theorem take MATH and use MATH. Here MATH. |
math/9901024 | MATH belongs to MATH. Since - by REF - MATH contains a free pair MATH, then MATH generates MATH. |
math/9901034 | The first order derivatives of MATH are conformal if and only if left hand sides of MATH are constant. Subtracting a linear vector field from MATH, one can force them to vanish and thus MATH to be conformal. |
math/9901034 | Let MATH be a polynomial vector field not in MATH. We may suppose that MATH. Indeed, if it is not the case, repeatedly applying REF , we replace MATH by some MATH as long as possible and then subtract from the last built field its homogeneous parts of degree MATH. Besides, as a module of MATH, MATH is split into three ... |
math/9901037 | The partition MATH is calculated by REF as the shape obtained by sliding the skew tableau MATH to partition shape, first into the cell of MATH containing the letter MATH (vacating the cell MATH say) and then into the cell of MATH containing the letter MATH (vacating MATH say). Moreover MATH and MATH. However MATH can b... |
math/9901037 | REF follows by REF , the definition of MATH, and the fact that restriction to subintervals preserves NAME equivalence. For REF, let MATH. If there is a tableau MATH such that MATH, then by REF MATH is obtained from MATH by sliding MATH to the southeast into the cell MATH, and placing a MATH in the northwest corner. If ... |
math/9901037 | By REF , and the definitions, MATH . Using this and REF , we have MATH . |
math/9901037 | Let MATH. It is enough to show that MATH is a MATH-configuration that has the same vacancy numbers as MATH. First we verify that the partitions in MATH have the correct size. To this end set MATH and MATH. Then MATH . Next we check that the vacancy numbers remain the same. Note that MATH valid for MATH. Then for MATH, ... |
math/9901037 | To prove that MATH is well-defined it needs to be shown that MATH is an admissible rigged MATH-configuration. Let us first show that MATH obtained from MATH by removing the corner cell in column of index MATH is indeed a partition. Assume the contrary. This means that MATH where MATH. By CITE MATH for large MATH, so th... |
math/9901037 | The proof proceeds by the induction on MATH given by REF . Suppose first that MATH is the empty sequence. Then both MATH and MATH are the empty set unless MATH is the empty partition, in which case MATH is the singleton consisting of the empty tableau, MATH is the singleton consisting of the empty rigged configuration,... |
math/9901037 | If MATH consists of more than one rectangle or MATH is a single rectangle with more than two columns, the commutativity of MATH and MATH is obvious. If MATH is a single column then both MATH and MATH are the identity and obviously commute. So it may be assumed that MATH is a single rectangle with exactly two columns. T... |
math/9901037 | Let MATH be the lengths of strings selected by MATH acting on MATH. To prove the lemma it suffices to show that MATH for all MATH. By REF , MATH for all MATH. Since MATH is obtained from MATH by adding the string MATH for MATH, it is clear that MATH for all MATH. Let MATH be minimal such that MATH. Then the string MATH... |
math/9901037 | The commuting faces yield the equality MATH. Since MATH is injective this implies MATH as desired. |
math/9901037 | The proof proceeds by the induction that defines the bijection MATH. Suppose the last rectangle of MATH has more than one column; this subsumes the base case MATH. Clearly MATH. Consider the diagram MATH . We wish to show the front face commutes. By REF and the injectivity of MATH, it is enough to show that all the oth... |
math/9901037 | Suppose that the last rectangle of MATH is also a single column; this case subsumes the base case that MATH is a single column. Clearly MATH. MATH . In this diagram there is a map MATH . This is to be understood in the most obvious way, namely, that given a chain of partitions MATH, and MATH, then MATH, that is, the co... |
math/9901037 | If MATH is empty the result holds trivially. Suppose that the last rectangle of MATH has more than one column. Obviously MATH and MATH. Consider the diagram: MATH is injective, the back face commutes by induction, the top and bottom faces are given by the commutative REF , the left face commutes by the definition of MA... |
math/9901037 | Suppose first that MATH and that MATH has more than one column. Note that MATH. MATH . The top face commutes since MATH and MATH are both relabelings that replace different subalphabets. The bottom face commutes since MATH adds singular strings and MATH preserves colabels and hence preserves singularity of strings. The... |
math/9901037 | Consider the diagram MATH . The back face commutes by REF for MATH, the top and bottom faces commute by REF, the left and right faces commute by the Evacuation REF for MATH and MATH, and MATH is injective. Therefore the front face commutes by REF . |
math/9901037 | Consider the diagram MATH . This diagram may be viewed as a prism whose top and bottom are triangles and whose front is REF which must be proven. The back left and back right faces commute by the definition of MATH. The top triangle obviously commutes. It suffices to show that the bottom triangle commutes. This is done... |
math/9901037 | Let MATH and set MATH. Recall that MATH is obtained from MATH by adding a singular string of length MATH to each of the first MATH rigged partitions in MATH. Let MATH be the matrix associated to MATH. Then MATH . Using furthermore that MATH one finds by REF that MATH. Since MATH is an inclusion it follows that the conf... |
math/9901037 | If MATH is empty the result holds trivially. If MATH has more than one column consider the diagram: MATH . Since MATH is an involution the top face commutes by REF. The bottom face commutes by REF and the back face by induction. The left face is the commutative REF and the right face that of REF . Since MATH is injecti... |
math/9901037 | One may reduce to the case MATH using REF, then to the case MATH using REF, and then to the case MATH and MATH using REF again. In this case the sets of tableaux are all empty or all singletons by CITE and the result holds trivially. |
math/9901037 | For MATH the proof of this lemma is the same as the proof of REF with MATH (respectively, MATH) replaced by MATH (respectively, MATH). Note that in the remaining case MATH both MATH and MATH have at most two rectangles so that MATH and MATH are singletons, and the proof follows again by arguments similar to those of th... |
math/9901037 | Recall the intervals of integers MATH REF in the definition of the set MATH. It follows from the definition of MATH that for all MATH, MATH for MATH. Using this fact and the definition of MATH, one may reduce to the case MATH. So it may be assumed that MATH. Obviously MATH. Consider the diagram MATH . The commutation o... |
math/9901037 | MATH by REF . Using the well-known fact MATH it is enough to show that the following diagram commutes: MATH . By CITE MATH for MATH where MATH is the reverse of the complement of the word MATH in the alphabet MATH. Let MATH (respectively, MATH) be the set of standard words MATH such that the standard tableau MATH is in... |
math/9901037 | Again the result follows from the special REF . Consider REF . By the definition of the relabeling map MATH and the fact that in this case both MATH and MATH only change the subtableaux corresponding to the first two rectangles, one may reduce to the case that MATH. By CITE all sets are either singletons or empty. The ... |
math/9901037 | Recall the embedding MATH of REF. By CITE, for all MATH we have MATH. On the other hand by REF , MATH. By replacing MATH by MATH it may be assumed that MATH consists of single rows. The transpose maps are applied to reduce to the case of single columns. Recall the definition of the number MATH CITE. For all MATH we hav... |
math/9901037 | By REF it suffices to prove the result in the case that MATH is any sequence of positive integers, MATH, and MATH is obtained from MATH by removing the rightmost copy of the letter MATH. Set MATH, MATH, MATH and MATH. By CITE and CITE MATH. Hence MATH . Notice that MATH where MATH is a sequence of single columns. Simil... |
math/9901037 | The proof proceeds by induction on MATH. There is nothing to prove unless at least one of MATH and MATH is finite. If one is finite and the other infinite then obviously MATH and MATH choose different strings and MATH and MATH. So it is assumed that both MATH and MATH are finite. For the base case MATH, observe that th... |
math/9901037 | MATH by REF , whose entire proof will be used repeatedly without additional mention. CASE: Consider a string in MATH that is either selected by MATH or MATH, or is such that its image under MATH (respectively, MATH) is selected by MATH (respectively, MATH). It is shown that the image of any such string under both MATH ... |
math/9901037 | By CITE, we have MATH whenever MATH, MATH, and MATH. But in the proof of CITE it is shown that MATH for all MATH, and in particular for MATH. Thus for MATH, REF holds. It suffices to show that the sum over MATH is zero. Suppose not. Then there is a rectangle MATH in MATH whose width is MATH and its height MATH satisfie... |
math/9901037 | Define the cell MATH for MATH; MATH by definition. Also MATH since MATH is assumed to be nonempty which by the NAME rule implies that MATH contains MATH. By induction suppose that the cell MATH has been defined for MATH. Let MATH be the cell just east of MATH if MATH and the cell just south of MATH if MATH. This is wel... |
math/9901037 | Follows directly from REF . |
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