paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005176 | Let MATH denote the first NAME class of the given complex structure MATH, and, by a standard abuse of notation, let MATH also denote the pull-back class of this class to MATH. If MATH are the NAME duals of the exceptional divisors in MATH introduced by blowing up, the complex structure MATH has NAME class MATH . By a r... |
math/0005176 | We may assume that MATH, since otherwise the result follows from the NAME inequality. Now MATH for any metric on MATH on MATH. If MATH is an NAME metric, the trace-free part MATH of the NAME curvature vanishes, and we then have MATH by REF . If MATH carries an NAME metric, it therefore follows that MATH . The claim thu... |
math/0005179 | As MATH, MATH we get that MATH. Hence MATH. |
math/0005179 | We carry the proof inside MATH. The first point to observe is that MATH. So if MATH and MATH then MATH. Let MATH be a club of MATH. Hence MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH . |
math/0005179 | We carry the proof inside MATH. The first point to observe is that MATH. So if MATH, MATH then MATH. Let MATH be a club of MATH. Then MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH . |
math/0005179 | We carry the proof inside MATH. The point to observe is that if MATH, MATH then MATH. Let MATH be a club of MATH. Then MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH . |
math/0005179 | We carry the proof inside MATH. The first point to observe is that MATH. The same technique as for the proof of REF is to be used. In MATH steps we can show that MATH, MATH, MATH. |
math/0005179 | Let MATH. As MATH we can assume without loss of generality that there is MATH such that MATH. Hence we can ignore these lower parts and assume that the set of conditions we start with is MATH. Let MATH. As MATH, MATH we can invoke the MATH-lemma. Hence, without loss of generality, there is MATH such that MATH. As MATH ... |
math/0005179 | Let MATH. Set MATH . We set MATH. It is clear that MATH, MATH, MATH. Let MATH. We define the condition MATH to be MATH with MATH substituted for MATH. Let MATH such that MATH, MATH. Of course the part below MATH poses no problem. So we are left to show that there is MATH such that MATH. By the definition of MATH there ... |
math/0005179 | Fix MATH. Then we can invoke REF on MATH, MATH, MATH to get MATH, MATH such that MATH is pre-dense below MATH. We do the above for all MATH. We let MATH. Of course, MATH is MATH with MATH substituted by MATH. We point out that MATH. We invoke MATH with MATH, MATH, MATH to get MATH, MATH such that MATH is pre-dense belo... |
math/0005179 | We give the proof for MATH. It is essentially the same for all MATH. Let MATH be large enough so that MATH catch `what interests us' and let MATH . Choose MATH such that MATH and set MATH where MATH. Let MATH be a well ordering of MATH such that MATH . We shrink MATH a bit so that the following is satisfied MATH . We s... |
math/0005179 | Construct, by repeat invocation of REF for each MATH, a MATH-decreasing sequence MATH. Let MATH for all MATH. Choose MATH such that MATH. There is MATH such that MATH. By this we eliminated clause REF for MATH. |
math/0005179 | Of course this is completely trivial as MATH. In this case MATH and MATH are the same. |
math/0005179 | Let MATH. As MATH is MATH-closed we get immediately that MATH. We have much more than that. Namely, MATH are not collapsed. For this we remind the reader that the MATH we work with is a generic extension of MATH for a reverse NAME forcing. Let MATH be the reverse NAME forcing up to MATH and MATH be its generic. Let MAT... |
math/0005179 | In order to avoid excess of indices we give the proof of the case MATH. Let MATH be MATH-generic with MATH and let MATH be dense open in MATH. Then MATH is a dense open subset of MATH. By REF for MATH there are MATH and MATH, a MATH-fat tree, such that MATH and MATH . Hence there is MATH which forces the above. That is... |
math/0005179 | Let MATH. Then MATH is dense open in MATH below MATH where MATH. By REF there are MATH, MATH, MATH such that MATH . By the definition of MATH we see that there is a function MATH with domain MATH such that MATH . As MATH there is MATH such that except on a measure MATH set we have MATH. By removing this measure MATH se... |
math/0005179 | Our first observation is that MATH is MATH-c.c. (As opposed to the usual MATH-c.c. we have when MATH). We construct, by induction, the sequence MATH where MATH. Together with it we construct an auxiliary MATH-decreasing sequence MATH. CASE: MATH: By REF and openness of MATH there are MATH, MATH, MATH such that MATH and... |
math/0005179 | In order to avoid too many indices we prove the lemma for the case MATH. Choose MATH large enough so that MATH contains everything we are interested in. Let MATH be such that MATH, MATH, MATH, MATH. Let MATH. Choose MATH such that MATH for all MATH. Let MATH. Note that MATH. Let MATH be a well ordering of MATH such tha... |
math/0005179 | The proof is done by induction on MATH. The case MATH is done in REF for MATH and REF for MATH. The case MATH is done in REF for MATH and REF for MATH. |
math/0005179 | Let MATH . Then MATH . Let MATH be the MATH, MATH such that MATH . We choose MATH such that for all MATH, for all MATH where MATH according to the selection of MATH. By REF, for each MATH there are MATH, MATH such that below MATH is pre-dense. Hence MATH . Let MATH. Let MATH be the condition MATH with its measure MATH ... |
math/0005179 | As usual we give the proof for the case MATH. Let MATH. MATH is a dense open subset in MATH. By REF there is MATH such that MATH . We use the above formula to fix MATH. Then for each MATH we fix MATH, MATH, MATH, MATH. In the same way we fix MATH, MATH, MATH for each MATH. By REF for each MATH there is MATH such that M... |
math/0005179 | Let MATH be large enough, MATH, MATH, MATH, MATH, MATH, MATH. We find MATH such that MATH is MATH-generic. Let MATH be enumeration of all dense open subsets of MATH appearing in MATH. Note that for MATH we have MATH. Let MATH. Choose MATH such that MATH for all MATH. Let MATH. We shrink MATH a bit so that the following... |
math/0005179 | If MATH then the claim is trivial. Hence we assume that MATH. Let MATH. Choose MATH large enough so that MATH contains everything we are interested in. By REF there are MATH, MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH is MATH-generic. Let us set MATH. Note that MATH. Let MATH be MATH-gene... |
math/0005179 | MATH is not collapsed by REF. No cardinals MATH are collapsed as MATH satisfies MATH-c.c. |
math/0005179 | Let MATH be MATH-generic. For each MATH define MATH. It is routine to check that for MATH we have MATH. Hence MATH. For the other direction let MATH. By REF there are MATH large enough, MATH, MATH, MATH and MATH such that MATH is MATH-generic over N where MATH. Hence, MATH. That is MATH. As MATH when MATH and MATH the ... |
math/0005179 | This is immediate due to MATH and REF. |
math/0005179 | Let MATH be the natural factoring. By REF the fate of the cardinals in question is decided by MATH. We note that MATH where MATH. So we factor MATH. As it stands MATH is MATH-closed. So in order to prove the claim we make some finer analysis. We remind the reader that the MATH we work with is a generic extension of MAT... |
math/0005179 | MATH is limit ordinal and by REF, there are unbounded number of cardinals below MATH which are preserved. Hence MATH is preserved. |
math/0005179 | Let MATH, MATH. Let MATH be large enough. By REF we have MATH, MATH, MATH, MATH, MATH, MATH such that MATH for each MATH. Choose MATH such that MATH, MATH. This is possible because MATH, MATH. Choose MATH such that MATH. Let MATH be MATH-generic with MATH. Of course, MATH also. Hence, MATH is MATH-generic over MATH. As... |
math/0005179 | Let MATH such that MATH is inaccessible and MATH be MATH-generic with MATH. (Forcing below an element of MATH eliminates a finite number of exceptions which we might otherwise have. That is if MATH and MATH then the interval MATH is untouched by the forcing). We set MATH . Note that MATH is a NAME generic sequence for ... |
math/0005180 | Let MATH be valleyless, then either MATH or MATH. Add one to every entry of MATH and then append REF to the beginning or end. The new permutation is valleyless of length MATH. The process is reversible so there are twice as many valleyless permutations of length MATH as there are of length MATH. Since MATH we have MATH... |
math/0005180 | Suppose that a permutation MATH with inversion table MATH is valleyless. If MATH for some MATH, MATH, then there is at least one symbol in MATH to the left of MATH that is greater than MATH. If there is any symbol in MATH greater than MATH to the right of MATH, then MATH would be a valley contradicting our assumption. ... |
math/0005181 | Pick MATH, MATH and MATH such that MATH is a MATH quasi-isometry, MATH, and MATH are uniformity data for MATH. Consider MATH such that MATH. We have MATH, and so MATH, from which it follows that MATH. Since MATH we obtain a bound MATH depending only on MATH. The usual ``rubber band" argument, using geodesics in MATH di... |
math/0005181 | We start with the case when MATH is a REF-parameter NAME subgroup, and the proposition follows by examining each NAME block REF. The second case we consider is when MATH has all positive real eigenvalues. By REF we have MATH, and REF follows immediately from the first case applied to MATH, together with the fact that M... |
math/0005181 | We proceed in cases. CASE: Assume that MATH is the unique REF-parameter NAME subgroup such that MATH is conjugate to the absolute NAME form of MATH. Applying REF we have MATH where MATH and REF-parameter subgroup MATH is bounded. Choose MATH and MATH. We must show that the two numbers MATH and MATH have ratio bounded a... |
math/0005181 | We have MATH. The reverse inequality is similar, and so MATH is a quasi-isometric embedding. Since MATH is coarsely onto, an easy argument shows MATH is coarsely onto. |
math/0005181 | By construction, the foliations MATH and MATH are coordinate foliations in MATH; this shows that the flow MATH has a ``global product structure" in the language of hyperbolic dynamical systems. The lemma now follows the proof of the Shadowing Lemma in REF , page REF. A direct proof is also easy to work out, and is left... |
math/0005181 | Let MATH be some center leaf of MATH, of dimension MATH. From REF it follows that MATH is NAME close to some center leaf MATH of MATH, also of dimension MATH. By composition with nearest point projection (which moves points a uniformly bounded amount) we get an induced map MATH. By REF this map is a quasi-isometry. By ... |
math/0005181 | We begin with: For each vertical flow line MATH in MATH, there exists a center leaf MATH in MATH such that MATH is contained in the MATH-neighborhood of MATH, where the constant MATH does not depend on MATH. Before proving the claim, we apply it to prove the proposition as follows. Consider any two vertical flow lines ... |
math/0005181 | This proof will define a sequence of constants which will depend on MATH and on the matrices MATH and MATH. We will indicate the dependence on MATH by writing, for example, MATH, but we will suppress the dependence on MATH. Although each constant in the sequence will depend on previous constants in the sequence, by ind... |
math/0005181 | With what we know, the proof is mostly a matter of chasing through definitions. The quasi-isometry MATH is a bounded distance from a quasi-isometry MATH which takes the horizontal leaf MATH to the horizontal leaf MATH, and which simultaneously takes center leaves of MATH to center leaves of MATH. Now restrict the cente... |
math/0005181 | Consider MATH in the same leaf of MATH but not in the same leaf of MATH, and so MATH are in the same leaf of MATH but not in the same leaf of MATH. Define displacement vectors MATH, MATH, and so MATH and MATH. We know that MATH . We also know that REF is true for MATH sufficiently close to MATH, and so for MATH suffici... |
math/0005181 | To prove REF , consider MATH, let MATH, and let MATH be a geodesic connecting MATH and MATH. Let MATH be the MATH-neighborhood of MATH in MATH, so MATH. Applying REF iteratively, projecting inward starting from the edges of MATH furthest from MATH, it follows that vertical projection MATH distorts any distance MATH by ... |
math/0005181 | Uniqueness of MATH follows obviously from the fact that distinct hyperplanes in MATH have infinite NAME distance. For existence of MATH we follow closely the proof of REF, concentrating on details needed to explicate the difference between the ``quasi-isometric" setting of CITE and the present ``uniformly proper" setti... |
math/0005181 | To prove REF , by REF the inclusion map MATH is uniformly proper and MATH is uniformly contractible, and clearly MATH is a contractible MATH-manifold. Composing with MATH we obtain a uniformly proper map MATH. Now apply REF . The idea of the proof of REF is that bushiness of the tree allows one to gain quasi-isometric ... |
math/0005181 | The idea of the proof is to compare the growth types of the filling area functions for ``quasivertical bigons" in MATH and in MATH. In MATH this growth type will be quadratic, while in MATH it will be exponential. Let MATH, MATH, MATH, or MATH. There is a quotient map MATH whose point pre-images give the horizontal fol... |
math/0005183 | By induction on the length of the proof. It will suffice to show that the axioms are valid, and that the quantifier rules and (tt) preserve validity. The soundness of the quantifier rules is established by observing that corresponding quantifier shifting rules are intuitionistically valid. For instance, since MATH are ... |
math/0005183 | Observe that a hypersequent MATH and its canonical translation MATH are interderivable using the cut rule and the following derivable hypersequents MATH . Thus it suffices to show that the characteristic axioms of IF are derivable; a simple induction on the length of proofs shows that proofs in intuitionistic predicate... |
math/0005183 | Let MATH and MATH be the cut-free proofs of MATH and MATH, respectively. We may assume, renaming variables if necessary, that the eigenvariables in MATH and MATH are distinct. The proof follows NAME 's original NAME. Define the following measures on the pair MATH: the rank MATH, the degree MATH, and the order MATH is t... |
math/0005183 | This is proved exactly as for the classical and intuitionistic case (see CITE). First, observe that all axioms are cut-free derivable from atomic axioms. The cut-elimination theorem thus provides us with a cut-free proof MATH of MATH from atomic axioms. Next, observe that REF rule can be simulated without using cuts by... |
math/0005183 | Easy induction on MATH. Every occurrence of MATH must arise from a weakening, simply delete all these weakenings. |
math/0005183 | By induction on the length of MATH. We distinguish cases according to the last inference MATH in MATH. For simplicity, we will write MATH in what follows below instead of MATH or MATH with the understanding that it denotes an arbitrary multiset of MATH's. CASE: The conclusion of of MATH is so that MATH only occurs on t... |
math/0005192 | Let MATH be a MATH - colored Y - link with MATH components. We call MATH the defect of the coloring. If MATH, the coloring is simple. Suppose MATH; then there are at least two components of the same color MATH. Therefore, the MATH - colored Y - sublink of MATH can be split into two disjoint non-empty parts MATH and MAT... |
math/0005192 | It suffices to prove, that there exist two Y - links MATH and MATH in MATH, each colored in at least MATH colors, such that MATH. Since MATH is a MATH-HS, it may be obtained from MATH by Y - surgery on a Y - link MATH in MATH. Let MATH be a MATH - colored Y - link in MATH, such that its image under Y - surgery on MATH ... |
math/0005192 | Follows from the construction of MATH by an application of MATH. |
math/0005192 | Follows from the construction of MATH by an application of MATH: MATH . |
math/0005192 | The equality MATH follows from the construction of MATH by an application of MATH. Topologically, this corresponds to cutting the corresponding solid handlebody introducing a pair of complimentary handles as shown below: MATH . Let MATH be the component of MATH containing this edge, and MATH, MATH be the components of ... |
math/0005192 | It suffices to prove the surgery equivalence of Y - links obtained by the moves MATH - MATH in the standard handlebody. Instead of drawing the handlebodies we will draw thick lines passing through the handles (encoding a set of surgery and clover components), similarly to REF . By REF , MATH is surgery equivalence. To ... |
math/0005192 | Let us first verify the statement for a Y - graph. By an isotopy and a subsequent application of MATH and MATH, we get MATH . The general case now follows by REF . Finally, MATH by MATH. |
math/0005192 | Let MATH and MATH be the Y - graphs depicted in REF b. Note that MATH looks exactly like MATH, except for the way its lower leaf links the thick line. Turning this leaf to the same position changes the framing of the adjacent edge by a half twist: MATH . By REF , MATH; also, MATH and MATH by REF . Thus MATH and the lem... |
math/0005192 | Rotating the special leaf we can change the framing of the adjacent edge while preserving the isotopy class of MATH. Thus by REF MATH and the lemma follows. |
math/0005192 | Cutting the neighboring internal edge we obtain two new leaves MATH and MATH. Without a loss of generality suppose that the framing of MATH is MATH. Then slide MATH along MATH by MATH as shown below: MATH . Notice that MATH changes the framing of a leaf by MATH, so the new leaf MATH is REF - framed. Splitting MATH as s... |
math/0005192 | We use MATH to pass from MATH to MATH and MATH with MATH as in the proof of REF . Then we split the leaf MATH of MATH by REF introducing new Y - graphs MATH and MATH with MATH: MATH . But MATH by REF , and MATH by REF . Hence MATH. On the other hand, MATH . The comparison of two above expressions for MATH proves the th... |
math/0005192 | It suffices to prove the statement for MATH; the general case then follows by REF . Consider a standard Y - graph MATH in a handlebody MATH of genus REF and attach to MATH an additional handle MATH. We are to slide all three edges of MATH along MATH in a genus REF handlebody MATH as indicated in REF . Each time we will... |
math/0005192 | The map MATH factors through AS and IHX relations by REF . The independence on the choice of MATH follows from REF . The surjectivity follows from the results of REF . The torsion result follows from REF . |
math/0005192 | Any Y - link in MATH, in particular MATH, can be made into a trivial Y - link by framing twists and crossing changes. Moreover, it suffices to use crossing changes which involve only the leaves of Y - graphs. Indeed, instead of a crossing change which involves an edge of a Y - graph, one can do two subsequent crossing ... |
math/0005192 | By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH and that any sublink MATH of MATH also laces MATH. |
math/0005192 | Let MATH be a trivial Y - link in MATH of degree at least MATH and MATH be an arbitrary link lacing MATH. In view of REF , it suffices to prove that MATH belongs to MATH. Suppose that some leaf of MATH is not linked with MATH, that is, bounds a disc which does not intersect MATH; then MATH by REF . Otherwise, all (that... |
math/0005192 | By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH and that any Y - sublink MATH of MATH also laces MATH. |
math/0005192 | Proceeding similarly to the proof of REF , we rotate MATH together with the trivial leaf linked with it. This adds a half twist to the framing of the adjacent edge of MATH, while preserving the isotopy class of MATH and MATH. Thus MATH by REF . |
math/0005192 | Let MATH be a lacing pair in MATH with MATH having at least MATH components. In view of REF , it suffices to prove that MATH. We proceed by downward induction on the degree MATH of MATH. If MATH, then obviously MATH and the theorem follows. Suppose now that the inclusion holds for all Y - links of degree higher than MA... |
math/0005192 | Cut out a tubular neighborhood MATH of MATH; this is a disjoint union of genus REF handlebodies. Components of MATH bound non-intersecting discs in MATH, which intersect the boundary MATH by either one or two circles, which are meridians of MATH. Perform surgery on MATH by cutting, twisting by an appropriate element of... |
math/0005192 | By REF , the image of B under surgery on MATH bounds in the complement of MATH. Thus it will bound also in MATH, for any MATH. |
math/0005192 | Suppose that there is a leaf of MATH, which has at least MATH neighbors, apart from the other leaves of the same Y - graph. Then the corresponding components of MATH comprise MATH. It remains to consider the case when each leaf of MATH has less than MATH neighbors belonging to other Y - graphs. Pick an arbitrary compon... |
math/0005192 | Let MATH be a trivial MATH - component Y - link in MATH and MATH be an arbitrary link lacing MATH. In view of REF , it suffices to prove that MATH. If some leaf of MATH is not linked with MATH, then MATH. Otherwise, all leaves of MATH are linked with MATH and we can use REF to find a good sublink MATH of MATH with at l... |
math/0005192 | Let MATH be a clover of degree MATH in MATH. We proceed by the downward induction on MATH. If MATH, then MATH by REF , and the theorem follows. Suppose that the statement holds for any clover of degree higher than MATH, and let us prove it for a clover MATH of degree MATH. Note that by the induction assumption we can u... |
math/0005192 | By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH, and that any Y - sublink MATH of MATH also MATH - laces MATH. |
math/0005192 | Let MATH be a MATH - lacing pair in MATH such that MATH has at least MATH components. We will show by downward induction on the degree MATH of MATH that MATH. If MATH, then obviously MATH. Inductively suppose that the statement holds for any MATH - lacing pair MATH in MATH with MATH of degree greater than MATH. If for ... |
math/0005193 | MATH makes up the core of the cylinder that is left over if the n-holed sphere is compressed along disks parallel to all of the uninvolved vertices. If the core of the cylinder were knotted (see REF ), then by standard satellite knot theory so is any arc running through the cylinder, but the edge between these two vert... |
math/0005193 | If MATH and MATH were linked, then MATH and MATH (edges of the the original unlinked n-graph running from vertex MATH to vertex MATH and MATH to MATH respectively) would have to be linked. |
math/0005193 | It is well known that if MATH, then MATH produces a MATH REF-bridge link, which is trivial if and only if q=REF. Similarly, of course, MATH is an unknot if and only if MATH so in that case MATH (the picture is just rotated ninety degrees). REF have expositions on the these facts. The classification was first done in RE... |
math/0005193 | By REF the star core is not knotted, so we may assume it consists of one straight edge MATH running from MATH the north pole to MATH at the south pole and another edge MATH meeting this edge at MATH at the origin and winding around in some manner before ending up at a MATH somewhere on the Southern hemi-sphere (as in R... |
math/0005193 | Assume there is a counterexample for MATH and examine what it must look like. MATH minus the interior of the non-standard ball, bounded by the non-standard enveloping REF-holed sphere is homeomorphic to MATH minus an open neighborhood of four arcs MATH which run from the inner sphere to the outer sphere. Since the enve... |
math/0005193 | REF shows a MATH curve in a ball. Such a graph is well known not to be standard, but every subgraph is standard. Let a MATH curve be the star core of an enveloping n-holed sphere. Although it is not standard, it supports a pairwise unlinked n-graph MATH. If we think of the enveloping sphere as bounding a central (round... |
math/0005193 | Assume MATH is chosen with a minimal number of intersections with MATH. Examine an innermost curve MATH on MATH. If MATH is not essential on MATH, then there is an obvious isotopy through which this intersection could have been eliminated, so we may assume it is essential on MATH. Therefore the innermost loop gives us ... |
math/0005193 | Choose a point on MATH to be the origin of MATH. Let MATH be the sphere of radius MATH centered at the origin in MATH, and let MATH be the ball that it bounds. We may alter MATH slightly if necessary so that we may assume that it intersects each MATH transversally. Fix MATH and examine MATH. The pieces of MATH in MATH ... |
math/0005195 | Because MATH, any self-diffeomorphism of MATH preserves orientation. Now MATH is a minimal complex surface of general type, and hence, for the standard `complex' orientation of MATH, the only NAME - NAME basic classes CITE are MATH. Thus any self-diffeomorphism MATH of MATH satisfies MATH . Letting MATH be the NAME dua... |
math/0005196 | To specify a connected reductive group, it is enough to specify a compact torus together with the collection of characters of that torus which will serve as the weights for the semisimple part of the group. The torus in turn can be described as MATH for some lattice MATH, which is the form used in the statement of the ... |
math/0005196 | Following the algorithm provided in CITE one can in fact show that MATH . This statement is also buried in the proof of CITE; even though the author claims it only for MATH. |
math/0005196 | It follows from CITE. |
math/0005196 | It can be verified by inspection and explicit computations. |
math/0005196 | We compute the NAME characteristic of MATH via the structure of elliptic fibration (the NAME characteristic of the general fiber is zero) and NAME - NAME 's sequence. |
math/0005196 | From the previous corollary we have: MATH . Note that MATH, which gives: MATH . Substituting this in the above equation we obtain the statement of the proposition. |
math/0005196 | MATH is the number of cusps away from MATH; our assumptions in REF imply that the cusps are determined by the common zeroes of the polynomials MATH away from MATH (these are ordinary vanishing, see REF ). MATH and MATH might also vanish along MATH, of orders MATH and MATH; MATH and MATH might have a common zero along M... |
math/0005196 | CASE: MATH is regular along MATH (simply laced groups in NAME 's exceptional series.) In this case MATH, which (except for the trivial case) are precisely the simply laced groups in NAME 's exceptional series. Here the singular fibers are of types MATH, and MATH. By assumptions, MATH and MATH becomes REF : MATH . Now w... |
math/0005196 | Case by case checking. |
math/0005196 | As we have already seen in REF, the intersection numbers of the various parts of the discriminant in MATH determine the geometry of MATH and the choice of the group MATH and vice versa. Following REF, we write all the terms in MATH in REF , as coefficients of MATH, the genus of the curve of singularities, the number of... |
math/0005196 | The statement follows from NAME 's formula. |
math/0005196 | CASE: When MATH is finite and there is no monodromy, that is, cases MATH, MATH, MATH, MATH, MATH, MATH, MATH corresponding to the simply laced groups MATH, MATH, MATH, MATH, MATH, MATH, MATH, in NAME 's exceptional series, then the local geometry is given by the following equations: MATH which can be solved since MATH:... |
math/0005197 | It is sufficient to define a composition for elements MATH of the special type MATH, where MATH, MATH, MATH, MATH is a bijection, and MATH. Composition of elements of this type is defined using the composition in the modular operad MATH. |
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