paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0008085 | Notice that if MATH for all MATH, the quantity REF equals MATH by CITE. The following argument is a simple generalization of that fact. Suppose MATH is a generic, REF-parameter family of perturbations. Let MATH be the parameterized moduli space. Recall that MATH is a smooth REF-manifold. If all the perturbations MATH i... |
math/0008085 | Since invariance of MATH is proved in CITE, we only need to prove that MATH is independent of all choices made. Now MATH where MATH is a reducible flat MATH connection close to MATH. Of course MATH for some MATH, and making compatible choices for MATH and MATH, we see that MATH . The third step follows by additivity of... |
math/0008085 | We show that MATH for sufficiently small MATH whenever REF holds, that is, whenever MATH for all MATH. This cohomology assumption implies MATH also vanishes for every MATH for any small MATH . (Note that the assumption of smallness of MATH here is stronger than the assumption needed to define MATH . ) Thus MATH and MAT... |
math/0008085 | In CITE, it is proved that MATH. So, the lemma follows once we show that MATH satisfies the same formula. Reversing the orientation of MATH changes the sign of the NAME function but has no effect on the perturbations. Therefore, there is a natural correspondence between the flat moduli spaces MATH and MATH. Obviously, ... |
math/0008085 | Since MATH was shown to satisfy a similar formula in CITE, it suffices to show additivity of MATH under connected sum. We first claim that the numbers MATH and MATH are additive under connected sum. To make this precise, we need to set up the notation. Set MATH . Then every connection MATH on MATH is of the form MATH, ... |
math/0008088 | We begin by considering the vector function MATH defined for MATH by MATH where MATH represents the angle between MATH and the integration variable MATH and MATH where MATH represent some (initial) set of Cartesian coordinates. The vector MATH simply gives the center of mass in MATH of the hypersurface distribution on ... |
math/0008088 | Defining MATH as above REF either MATH vanishes somewhere and we are done or we can pass to MATH. Continuing with the latter case, if MATH has neither a fixed point nor an anti - fixed point (that is, there are no solutions MATH to MATH) then as above we can show that MATH is homotopic to the antipodal map and also to ... |
math/0008088 | In the notation of the preceding paragraph we must show that MATH. The argument proceeds via a simple use of NAME 's theorem as applied to REF rewritten in the forms MATH and MATH . In particular, with MATH in the first of these we have MATH and with MATH in the second we have MATH . By NAME 's theorem, between any two... |
math/0008088 | MATH satisfies MATH in MATH (since MATH follows from the variational characterization of the eigenvalues of MATH via the NAME quotient MATH), which implies that MATH is decreasing in MATH. Hence MATH, which proves the lemma. |
math/0008088 | Using the product representation of MATH, that is, MATH, one has MATH (convergence of the series for MATH here is understood in the sense of symmetric partial sums). From REF we obtain the following representation for MATH . It follows from REF that MATH for MATH. Also from REF we have MATH which is positive for MATH. ... |
math/0008088 | Since MATH (see REF above), we just need to prove that MATH . This inequality follows from REF . Note that it holds for all MATH. |
math/0008088 | Since MATH and MATH, to prove REF and therefore MATH reduces to showing (since MATH can be grouped as MATH) MATH where MATH. From REF this is equivalent to proving MATH which is obviously true for MATH since MATH is positive. For MATH it suffices to show MATH . The fact that MATH is an increasing function of MATH on MA... |
math/0008088 | The proof is similar to that of REF. We assume that MATH through most of the proof, returning to the case MATH (which can be treated explicitly in terms of elementary functions) only at the end. We shall use a suitable trial function (based on MATH) in the NAME quotient for MATH: MATH . It suffices that the trial funct... |
math/0008088 | The proof is similar to that of REF. We use MATH in the NAME quotient for MATH (that is, in REF above). Since MATH we get from REF MATH (again, MATH does not satisfy the equation that MATH does, so REF is strict). Now the lemma follows by the monotonicity of MATH in MATH for MATH. |
math/0008088 | The analog of this result in the Euclidean case was proved in REF. That MATH and MATH as MATH follow from the boundary behavior of MATH. Using the raising and lowering identities REF with MATH and MATH respectively and with MATH fixed at MATH, that is, MATH one obtains MATH and in similar fashion MATH . The function MA... |
math/0008088 | Consider the function MATH. Using REF , and REF one can show that MATH satisfies the equation MATH . Since the function MATH is odd and analytic in MATH for MATH near MATH, we can expand it in odd powers of MATH. Inserting a series expansion in the NAME REF we find MATH for MATH near MATH. Substituting this into the de... |
math/0008088 | Let MATH, respectively MATH, be the lowest eigenfunction, respectively eigenvalue, of the NAME problem on MATH, that is, MATH . Define MATH and MATH. Let MATH, and MATH, where MATH denotes MATH - dimensional measure on MATH. Then we have (see, for example, CITE, p. REF ) MATH for almost every MATH. Applying NAME 's the... |
math/0008089 | See CITE REF and also CITE REF . |
math/0008089 | CASE: is classical, REF. has been proved by CITE, and REF. was given by CITE. |
math/0008089 | Define MATH, MATH, and MATH, then the above functional equation is reduced to the equation from REF, for which it is well-known (compare for example, CITE) that there is only the differentiable function MATH (up to a multiplicative constant) which satisfies the latter equation. |
math/0008089 | It is a straightforward matter to check that the above elements lie in the kernel of MATH. Nevertheless, we give some interrelationships between the various equations. CASE: The symmetry of REF is equivalent to REF : We have to write the following relation MATH as a sum of MATH-term relations. On the left hand side of ... |
math/0008089 | REF-term equation and REF will follow directly from the equation in the next proposition. The equivalence of the two NAME analogues becomes evident after applying the change of variables MATH, MATH, and multiplying the result by MATH. |
math/0008089 | Add REF-term equation to its variant where MATH is replaced by MATH. Four of the terms cancel and the remaining two give the inversion relation. |
math/0008089 | According to REF , the function MATH is a well-defined function on MATH, and as it is not identically zero on MATH, the dimension of MATH is non-zero. By CITE(NAME REF, p. REF), we know that MATH. But as the relations in MATH are given by REF-term equation (that is, the Fundamental Equation of Information Theory) and t... |
math/0008089 | This is a consequence of the following theorem, together with REF . |
math/0008089 | Set MATH, and suppose that MATH verifies MATH and the following identity in MATH . Differentiating the previous equation with respect to MATH gives, MATH with MATH and thus MATH. Setting MATH in the previous identity gives MATH . But as MATH, the previous equality implies MATH . In other words, since MATH we have MATH,... |
math/0008089 | It is a consequence of the following lemma Suppose that MATH is a sequence of integers with MATH (MATH an odd prime fixed), which fulfills the following rules MATH and MATH for all MATH. Then MATH for all MATH. The proof goes by descending induction starting from MATH. First we notice that by the third rule, we have MA... |
math/0008089 | The fact that MATH fulfills REF is a direct computation. For MATH, the family is free for degree reasons, since MATH. Furthermore MATH does not belong to this family for valuation reasons. |
math/0008089 | Let MATH be a polynomial of degree MATH, and suppose that MATH fulfills the following two equations MATH . Then observing that we can deduce the inversion formula as a consequence of REF-term, and taking the derivative with respect to these equations shows that MATH fulfills the inversion formula and the duplication fo... |
math/0008089 | We only need to show that if MATH, assumed to be of degree less than or equal to MATH, setting MATH, MATH fulfills REF-term equation. In order to do that let MATH denote the formal NAME analogue, then taking the derivative with respect to MATH, and rewriting the equation with MATH and finally specializing to MATH, we c... |
math/0008089 | First we observe that the map MATH is well defined. Indeed this is a consequence of the MATH property of the map MATH defined on the units and of the fact that MATH which implies that MATH. Then, the commutativity of the diagram is a direct check. |
math/0008089 | It is a direct consequence of REF and of the construction of MATH. |
math/0008089 | MATH is defined on MATH by assumption. For MATH, we have seen that the function is essentially unique: MATH and the resulting infinitesimal dilogarithm MATH vanishes on MATH (due to REF , it is enough to check that it vanishes on the four term relation, which is straightforward). Now suppose the claim is true for MATH.... |
math/0008089 | Again, the case MATH gives the unique choice for MATH (up to multiplicative constant). Inductively, starting from MATH and MATH, one can form an arbitrary linear combination of them using MATH and MATH which gives a candidate for MATH, with coefficients MATH, say; a subsequent ``integration" (putting MATH and successiv... |
math/0008089 | The infinitesimal polylogarithm MATH vanishes on MATH by REF , and reducing mod MATH obviously conserves this vanishing property. NAME 's result now says that the reduction of MATH is equal to MATH. |
math/0008089 | CASE: Follows from CITE and CITE, respectively. CASE: This follows via the ``derivation map" (see ยง REF). CASE: MATH. |
math/0008089 | CASE: The inversion relation can be checked via a straightforward algebraic manipulation. CASE: In order to prove the distribution relation, let us fix a primitive MATH-th root of unity MATH. Dividing both sides by MATH and developing the fraction into a (finite) series leaves us to prove: MATH and this is true due to ... |
math/0008089 | CASE: We will apply the recipe above. We have MATH and as the degree of either polynomials is less than MATH, we conclude that MATH where MATH is a constant. This, in turn, implies that MATH (specialize MATH and MATH, respectively), and therefore we get as a by-product MATH (in characteristic MATH). CASE: The following... |
math/0008089 | CASE: Set MATH . We want to prove that MATH is REF in MATH. Computing MATH we get MATH . But by REF , MATH and MATH. Moreover by the inversion formula (see REF ) we have MATH. Hence, MATH . As MATH and MATH, we know that MATH and therefore MATH, but using the inversion relation one sees that MATH, which implies MATH. |
math/0008089 | The strategy of proof could be summarized as follows: CASE: Prove that MATH in MATH. CASE: Prove that MATH in MATH. CASE: Prove that the coefficient of MATH in MATH is REF. For the proof of this functional equation we will need several preliminary formulas. First we will use these two relations, in MATH, coming from RE... |
math/0008089 | we can compute explicitly the coefficients of the polynomial MATH. First of all, its zeroth coefficient is MATH. For MATH between MATH and MATH the MATH-th coefficient of MATH is equal to MATH . We use here the standard fact that MATH for any polynomial MATH of degree at most MATH. |
math/0008093 | By the definition of the MATH-th supersymmetric algebra we have MATH where MATH is the diagonal subgroup of MATH. By REF we have therefore MATH where MATH in the previous line is summed over all NAME diagrams satisfying the conditions MATH and MATH. The second to last equality follows from the well-known fact that MATH... |
math/0008093 | By the definition of the MATH-th skew-symmetric algebra we have MATH where MATH is the diagonal subgroup of MATH and MATH is the subspace of MATH that transforms according to the sign character of MATH. By REF we have therefore MATH where MATH in the previous line is summed over all NAME diagrams satisfying the conditi... |
math/0008093 | Since MATH is a highest weight vector for MATH with weight MATH, it belongs to the subspace MATH of MATH which consists of vectors with weight MATH annihilated by the standard NAME in MATH. By REF , MATH is isomorphic to MATH as a MATH-module. There exists a unique vector (up to scalar multiple) in MATH which has weigh... |
math/0008093 | Our argument follows closely the one given in the proof of REF with REF replacing the classical NAME duality. Let MATH denote the the subgroup of MATH, which preserves the partition MATH of MATH. Note that MATH is isomorphic to a semidirect product of MATH and MATH, where MATH acts by interchanging MATH with MATH and M... |
math/0008093 | Observe that MATH has the same MATH-weight as the first NAME diagram of REF . Clearly MATH is non-zero. It is straightforward to verify that MATH is annihilated by the operators in REF . It follows from REF that MATH is also a highest weight vector for MATH. |
math/0008093 | Consider MATH and MATH where MATH. By REF the product MATH is a highest weight vector and thus is annihilated by all operators in REF . In particular applying the operator MATH to MATH and dividing by MATH we obtain the desired identity. |
math/0008093 | We first show that MATH indeed has the correct weight. First note that REF corresponds to the MATH-weight MATH. Let MATH, MATH, denote the MATH disjoint subsets of MATH defined by the condition that MATH if and only if MATH contains a marked box at its MATH-th column and MATH-th row. Put MATH and MATH. The weight of MA... |
math/0008093 | Denoting by MATH and MATH the right-hand side and the left-hand side of REF , respectively, we may regard MATH and MATH as functions of MATH. Since the group MATH acts on MATH, the space of MATH matrices, by left multiplication, it acts on functions of MATH. To be more precise if MATH, then MATH and MATH. We want to st... |
math/0008093 | REF is an obvious consequence of REF. For REF let MATH denote the complementary subset of MATH in MATH put in increasing order. We apply successively the differential operators MATH, MATH, MATH, MATH to REF and find that MATH . Dividing by MATH we obtain REF . By REF we have MATH . Thus if MATH is the permutation arran... |
math/0008093 | Given diagram MATH with MATH rows and MATH columns, for MATH, we want to know how to simplify the expression MATH . We move all MATH to the left and get MATH . Now each MATH is a product of MATH. We arrange all MATH together so that MATH appears to the left of MATH if and only if MATH and call the resulting expression ... |
math/0008093 | Since for a subset MATH of MATH, MATH is a scalar multiple of MATH by REF , it follows that the expression is divisible by MATH and independent of MATH, for MATH, after division. It is clear that it is annihilated by REF . |
math/0008093 | Let MATH, MATH denote the number of marked boxes in MATH that appear in the MATH-th row of some diagram MATH. Then MATH has weight MATH while the expression MATH has weight MATH . So the combined weight is MATH which of course is the weight of the NAME REF . |
math/0008093 | The only thing that remains to prove is that the vector in REF is indeed killed by REF . But this is because of the presence of MATH in the formula of MATH and so is an immediate consequence of REF . |
math/0008093 | It is enough to restrict ourselves to real symmetric MATH matrices MATH. Let MATH be an orthogonal MATH matrix such that MATH, where MATH is a diagonal matrix. We compute MATH where MATH and MATH. But the determinant of the matrix on the right-hand side of REF is zero. |
math/0008093 | First consider the action of the operator MATH, for MATH, on MATH given as in REF . Certainly MATH annihilates the first summand of REF , and furthermore it takes the summand MATH for MATH to MATH which is zero. MATH takes MATH to MATH . The first summand is zero, while the second summand remains. Now we verify similar... |
math/0008093 | The fact that MATH is annihilated by MATH for MATH is a consequence of REF . Now the proof of REF shows that MATH, for every MATH. Thus MATH annihilates MATH as well. So it remains to show that MATH kills MATH. Given a summand in MATH of the form MATH there exists a summand of the form MATH, which is identical to it ex... |
math/0008094 | It suffices to check that MATH is invariant under a transposition of two adjacent MATH's. By the decomposition REF , this reduces to showing that MATH is symmetric in MATH and MATH, identically in MATH for any MATH. Let us single out one of the MATH's and write MATH where MATH, MATH, and the rest of the MATH's are cons... |
math/0008094 | Suppose MATH is not row decreasing. Then there exist two squares MATH and MATH in MATH such that MATH is just left of MATH and MATH . In this case, MATH has the factor of MATH and hence MATH. |
math/0008094 | Recall that MATH has two poles MATH and two zeros MATH. The pole MATH is not an issue in limit REF and the zero MATH has been already accounted for in REF . The pole MATH means that MATH where MATH is the square right on top of the square MATH. This factor is present in MATH if MATH . Similarly, the zero MATH means tha... |
math/0008094 | Let us split the sum in REF in two MATH where MATH and MATH contains summands with nonnegative and negative number MATH, respectively. Using the identity MATH to perform the summation along the rows, we compute: MATH . Similarly, we compute MATH using REF instead of REF and we find that MATH . For any diagram MATH defi... |
math/0008096 | For MATH the statement is obvious. Suppose it is established for some MATH and let MATH be a diagram on MATH chords. Recall that we think of knots and diagrams as being ``long" so the ends of chords of a chord diagram are naturally ordered. Consequently, the chords themselves can also be ordered according to the order ... |
math/0008096 | Say that a diagram with MATH chords on MATH strands is arranged if it has exactly one end of a chord on each of the first MATH strands. We describe an algorithm of connecting any diagram with MATH chords on MATH strands to an arranged one by relations of strand exchange; the lemma follows from the existence of such an ... |
math/0008096 | A REFT relation can be thought of as a resolution of a ``singular" chord diagram whose two chords have a common end. So in order to show that every REFT relation for diagrams of knots up to isotopy comes from a REFT relation for NAME diagrams one can just check that every singular chord diagram on MATH chords can be pr... |
math/0008099 | Using a NAME solid link, we see that for the left side of REF , MATH and for the right, MATH. (This is also seen by REF .) Thus REF are equivalent. Suppose REF . Attach MATH-handles to MATH and MATH. Then MATH and MATH change to MATH-spheres which split by isotopy, and hence MATH is null-bordant. Thus REF . It is obvio... |
math/0008099 | Consider a tri-punctured sphere MATH embedded in MATH whose boundary is MATH with MATH and MATH. There exists an identification of a regular neighborhood MATH in MATH with MATH such that the NAME MATH-links MATH and MATH in MATH correspond to MATH, where MATH is an oriented NAME disk pair. The desired MATH is obtained ... |
math/0008099 | It is a consequence of REF . |
math/0008099 | CASE: Let MATH be a NAME hypersurface for MATH which intersects MATH transversely. For each MATH (MATH), let MATH be the intersection MATH, which is the union of oriented simple loops in MATH (or empty). Let MATH be a regular neighborhood of MATH in MATH. For a component MATH of MATH, let MATH be a solid torus MATH in ... |
math/0008100 | The proof that the NAME relations are preserved under the correspondence MATH and that MATH is sent to MATH is a simple modification of the proof of the quantum analogue of NAME 's theorem presented in REF . The classical analogue of MATH, obtained by specializing MATH to MATH, is easily seen to be injective. This take... |
math/0008100 | In CITE it is shown that MATH and MATH are weakly separated precisely when, after interchanging MATH and MATH if neccessary, either: CASE: MATH and there do not exist three indices MATH such that MATH and MATH or REF MATH and there do not exist four indices MATH such that MATH and MATH REF above indicates that two MATH... |
math/0008100 | Given MATH, the number of chambers in the first MATH strips of the corresponding double wiring arrangement is equal to the number of red and black crossings in the first MATH strips plus MATH - corresponding to the MATH far right chambers . The number of black (respectively red) crossings in the first MATH strips in tu... |
math/0008100 | Take any quantum minors MATH and MATH in MATH. REF proves that if MATH and MATH are chamber sets of a single wiring arrangement then either MATH or MATH. This, taken together with the fact that the single wiring arrangements for the MATH and MATH components of MATH are oppositely labeled, proves the second part of the ... |
math/0008100 | This theorem follows from the well known fact that the any two triangulations are connected by a series of chord exchanges where the diagonal chord of an inscribed quadralateral is "flipped" to its crossing pair. The diagonal "flips" correspond to MATH-moves. |
math/0008100 | Immediate corollary of REF . |
math/0008100 | Since the MATH-action preserves MATH-moves it is enough to verify this assertion in the case where MATH. Proceed by induction on MATH. For MATH the statement is evident. Assume MATH. It is enough to verify the claim for the group elements MATH and MATH, which generate MATH, given by MATH . This follows from the observa... |
math/0008100 | For a MATH-subset MATH of MATH define the diameter of MATH to be the minimal cardinality of a boundary MATH-subset of MATH that contains MATH. Thus the boundary MATH-subsets are precisely those of diameter MATH. Let us call MATH-subsets of diameter MATH almost boundary subsets. It suffices to prove that every maximal c... |
math/0008100 | We proceed by induction on the height. If MATH then we are already done. Assume inductively that the assertion is true for collections of height MATH and let MATH be a collection of height MATH. We need the following: Let MATH and suppose that MATH. Then there exists a unique index MATH such that both MATH and MATH are... |
math/0008100 | Momentary consideration reveals that MATH consists of pairwise weakly separated MATH-subsets of MATH. In virtue of REF we know that MATH will be maximal if and only if MATH. Since MATH it follows that if MATH and MATH then either MATH or MATH. Consequently MATH. For MATH if MATH then either MATH or else there exists MA... |
math/0008100 | Let MATH and suppose that all MATH are real and positive for MATH. We need to show that all other NAME coordinates MATH are real and positive. Take any MATH. Take any maximal collection MATH containing MATH. Since MATH is either MATH or MATH we know that Conjecture REF holds and thus MATH and MATH are connected by a se... |
math/0008101 | For MATH and MATH, we denote by MATH the vector in MATH whose MATH-coordinate is MATH if MATH and zero otherwise. Let MATH, MATH, and MATH be as above. Set MATH and define the sets: MATH . First, we concentrate our attention on the set MATH. We define the sets: MATH . Finally, set MATH . Note that MATH, MATH, and MATH ... |
math/0008102 | Most of the details are contained in the paper CITE, and others will be in a later more detailed paper. But we sketch here the argument for orthogonality in the NAME module. Let the functions MATH be as stated in the lemma. Orthogonality refers to MATH whenever MATH. Hence we must calculate, for MATH: MATH using REF , ... |
math/0008102 | Let MATH be given as in REF , and let MATH. Set MATH . Then MATH satisfies the same orthogonality relations. For MATH proving the assertion. We used the fact that MATH, see REF . Conversely, if MATH and MATH are given representations as described in REF , then MATH . For it follows from the NAME relations REF that the ... |
math/0008107 | Assume first that MATH is a computable function as described in the proposition. Suppose that MATH is a triangulation of MATH with MATH REF-simplices. Let MATH be a triangulation of some closed MATH-manifold MATH containing MATH REF-simplices. Do all possible sequences of NAME moves on the triangulation MATH of length ... |
math/0008107 | Normal triangles and squares chop up any tetrahedron in MATH into several pieces. But at most six of these regions are not of the form MATH or MATH (see REF ). Let MATH be the maximal number of disjoint non-parallel normal REF-spheres in MATH. Then the complement of this family has precisely MATH components. Each of th... |
math/0008107 | First note that after the natural isotopy, all the non-normal arcs we get will give rise to elementary isotopies in the direction of MATH. After each isotopy both MATH and MATH will change, but we'll still denote both resulting spaces by MATH and MATH respectively. After the natural isotopy, MATH and MATH satisfy the f... |
math/0008107 | REF-ball from the lemma can be view as a union of the following two PL REF-balls: the star of the edge MATH and the cone on the disc in the bounding MATH, which is the complement of the star of the vertex MATH on REF-sphere. The triangulation we are aiming for is equal to the triangulation of the latter REF-ball. We th... |
math/0008107 | Since the triangulation of our disc is shellable, we can index all the simplices in it by numbers from MATH to MATH, so that the increasing integers specify a way of reducing our triangulation down to a single triangle. REF-simplex that's left has index MATH. Let's make a MATH move on it. REF-simplex corresponding to M... |
math/0008107 | We will divide the process into three steps. First, we glue a cone on the cone on the boundary of MATH onto the bottom part of the boundary of MATH REF . This is a reversed process to destroying an edge which connects the two cone points of the bit that we glued on. It can therefore, by REF , be accomplished by less th... |
math/0008108 | As in CITE. See CITE as well. |
math/0008108 | As in CITE. |
math/0008108 | Let MATH, say MATH where MATH. Let MATH be the singularity type of MATH at MATH. Let MATH be the chain of rational, smooth curves on MATH that are contracted to MATH in MATH. Assume that MATH intersects MATH and MATH intersects MATH. Put MATH and MATH. For each MATH let MATH be the point of intersection of MATH and MAT... |
math/0008108 | The existence of MATH and MATH satisfying REF is equivalent to the existence of MATH and MATH satisfying the following four conditions. MATH . In fact, suppose that MATH and MATH satisfy REF . Define MATH and MATH by letting for each MATH, MATH . Then MATH and MATH satisfy REF . Conversely, suppose that MATH and MATH s... |
math/0008108 | Note first that MATH where the second isomorphism holds because of REF . Now, for each MATH and each MATH, let MATH denote the dualizing sheaf of MATH. Then MATH . Since MATH is smooth and rational, MATH. So, MATH REF follow immediately. Since MATH by REF , it follows from REF that MATH . In addition, since MATH by REF... |
math/0008108 | Let's prove REF . Since REF holds, REF applies. Let MATH. Let's check first that MATH is injective. In fact, let MATH such that MATH. Then MATH for every MATH. So MATH for every MATH by REF , and hence MATH. By REF , MATH as well. So MATH, and thus MATH for every MATH by REF . Hence MATH. So MATH is injective. Let's pr... |
math/0008108 | Let MATH denote the reducible nodes of MATH. Let MATH be the versal formal deformation of MATH. Then MATH, where MATH for a certain integer MATH. In addition, we may assume that there is an isomorphism of MATH - algebras, MATH for each MATH, where MATH and MATH are the local equations of the branches of MATH meeting at... |
math/0008108 | The ``only if" assertion is clear; see REF Let's prove now the ``if" assertion. Since MATH for every MATH, we may assume that MATH is an invertible MATH - submodule of MATH satisfying MATH. To finish the proof we apply REF with MATH, with MATH and MATH. So we check that the conditions for REF hold. First, since MATH, w... |
math/0008108 | Let MATH be the set of irreducible components of MATH. Let MATH such that MATH . Let's prove REF . Consider first the ``only if" assertion. Let MATH be a regular smoothing of MATH and MATH, where MATH is the dualizing sheaf of MATH. By REF , MATH is the canonical sheaf with focus on MATH. If MATH, then REF holds by REF... |
math/0008108 | Let MATH be the map contracting the subcurve MATH for every MATH. For each MATH let MATH be the blow - up map, MATH and let MATH be the map contracting the subcurve MATH for every MATH. Let's prove REF . First, let MATH be a regular smoothing of MATH such that MATH is the limit canonical aspect of MATH with focus on MA... |
math/0008108 | Let's prove REF . Since MATH, also MATH, and MATH for every MATH by REF . By NAME - NAME, MATH and MATH is surjective. Now, MATH is injective by REF . So MATH by REF . Thus, taking inverse images by MATH gives us a closed embedding MATH. Since MATH and REF holds, the description of MATH given in REF applies. Comparing ... |
math/0008108 | We shall use the set - up of REF. Let's prove REF . Assume first that MATH. Then REF says that MATH is the limit canonical aspect with focus on MATH of any regular smoothing of MATH. Therefore MATH. Assume now that MATH. Then MATH and MATH. By REF , all the NAME coordinates of MATH in MATH are non - zero. In addition, ... |
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