paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0008047 | We prove the proposition by induction on MATH in the descent order. First, consider the case MATH. Suppose MATH. Then, MATH due to MATH, and MATH due to MATH. Therefore, the claim reduces to the well-known power series expansion MATH which converges for MATH. Next, let us assume REF holds for MATH. Then, for MATH such ... |
math/0008047 | By REF , MATH, where MATH with MATH if MATH, MATH, MATH if MATH, and MATH otherwise. Then, the claim follows from the fact that MATH mod MATH, where MATH is the ideal in REF, and that MATH by the convergence property (Q̃-II). |
math/0008047 | We define series of MATH which depend also on MATH. By REF is the limit MATH of the formula MATH . To prove REF , we use MATH where REF is easily shown by elementary transformations. Let MATH . By REF is equivalent to the following equality: MATH . |
math/0008047 | We recall that MATH, MATH, and MATH. Then, by REF , we have MATH . Therefore, by REF , MATH . |
math/0008047 | It is well-known that MATH and the Right-hand side is characterized by REF the coefficient of MATH is REF; REF it is skew MATH-invariant. Therefore, it is enough to show that MATH satisfies the same properties. REF follows from the fact MATH. Under the assumption that MATH are MATH-invariant, REF is equivalent to the f... |
math/0008048 | Let MATH satisfy MATH and MATH be framed NAME disks pairing all the double points of MATH. The MATH may be assumed to have disjointly embedded boundaries after applying the move of REF . We now describe three modifications of MATH and the collection of NAME disks which can be used to geometrically realize the relations... |
math/0008048 | First note that since the MATH are ordered all the NAME disks are canonically oriented via our convention. Thus the signs associated to the intersections between the NAME disks and the MATH are well-defined. Also, the element of MATH associated to such an intersection point does not depend on the choice of whisker for ... |
math/0008049 | Let us recall the construction of the class MATH. Since MATH is an immersion, MATH is a closed subset in MATH. Since MATH is a generic immersion, MATH is transversal to MATH outside MATH. Therefore MATH is a compact oriented submanifold without boundary MATH. Then the composition MATH is MATH (see details in CITE). By ... |
math/0008049 | To prove REF , it suffices to interpret the right hand side of REF in terms of the differential MATH. Since MATH, we have MATH where MATH is the differential in the exact bordism sequence of pair. Denote by MATH the inclusion MATH. Obviously, the map MATH is a homotopy equivalence. Since MATH is a nowhere zero cross-se... |
math/0008049 | It suffices to substitute REF into REF with MATH. |
math/0008049 | follows immediately from REF . |
math/0008053 | In the paper CITE , NAME proved the equivalence similar to REF for NAME functions. Namely, MATH with a constant independent of MATH and MATH . This implies the equivalence MATH . Next, by the definition of the equivalence in distribution, it is obvious that MATH . Therefore we must prove only that MATH . Fix MATH (we c... |
math/0008053 | Suppose that a system MATH satiesfies REF . Then for MATH and MATH REF is fulfilled. Hence, MATH for any MATH and MATH . Analogously, the systems MATH and MATH satisfy REF , also. Therefore, MATH . Using REF , we obtain: MATH and a constant of this equivalnce depends only on MATH and MATH (see REF ). The opposite asser... |
math/0008053 | Under REF there exists a subsequence MATH such that MATH where a constant MATH does not depend of MATH and MATH ([REF ], detailed proof see in CITE ). Then, if MATH and MATH we have MATH . Hence, by the definition of the MATH -functional, MATH . In the proof of the opposite inequality we shall use REF and an estimate o... |
math/0008053 | First suppose there exists a subsystem MATH satisfying REF - REF. By REF , we can select a subsystem MATH such that equivalence REF holds. Using REF , we obtain: MATH . Conversely, suppose that MATH and MATH . Clearly, the NAME system possess the properties similar to REF - REF. Therefore MATH also, satisfies REF. As w... |
math/0008053 | By REF , MATH contains a subsequence MATH that is equivalent in distribution to the NAME system. It means that MATH where MATH and MATH is a constant independent of MATH . In particular, it yields MATH . Therefore, by definition of NAME functions, MATH . Denote MATH and MATH . Then if MATH where MATH . |
math/0008053 | By REF , it is enough to find a subsystem MATH and a set MATH such that for every segment MATH and for some MATH we have MATH (a constant of this equivalence depends on MATH and MATH). Next, we can use arguments similar to those in the proof of REF . Hence we make only some remarks. First of all, it is clear that REF h... |
math/0008053 | in view of REF we can assume that the consructed subsystem MATH satisfies REF for an arbitrary set MATH of positive NAME measure. Next, we argue as in the proofs of REF . |
math/0008053 | For any positive integer MATH we have MATH . Let us fix an one-to-one mapping from the collection MATH to the set of intervals MATH . Suppose MATH is the element of the collection MATH corresponding to an interval MATH . Denote by MATH the set of all MATH such that MATH . For every MATH we define MATH . Then the image ... |
math/0008053 | In view of REF it is sufficient to prove that there exists a subset MATH such that MATH and MATH with a constant depending only on MATH . As before, here MATH . Using REF , we can select a subset MATH satisfying REF . Since MATH then REF holds also for functions MATH . NAME, by REF , the set MATH can be extended to a m... |
math/0008053 | NAME the converse. Let MATH be a linear operator such that MATH and MATH . Here, as before, MATH . Denote MATH for MATH where MATH and MATH if MATH . Using REF , we obtain: MATH and MATH for some constants MATH and MATH . Hence, in particular, MATH . In addition, from REF it follows that MATH and MATH for any MATH . So... |
math/0008057 | Let MATH be the NAME space on the unit disk, MATH the space of restrictions of functions in MATH to MATH. We view MATH as a NAME space in the inner product such that the restriction mapping from MATH onto MATH is a partial isometry. The reproducing kernel for MATH is the NAME kernel MATH, and the reproducing kernel for... |
math/0008057 | It is sufficient to prove the result when MATH and MATH. For suppose that the result is known in this case, and consider the general situation. Let MATH be the linear fractional mapping of MATH onto itself given by MATH. Thus MATH and MATH. Put MATH, MATH, and MATH . Define MATH on MATH by REF using MATH and MATH in pl... |
math/0008057 | The last statement follows from CITE. It is convenient to assume that MATH and MATH. If the result is known in this case, then as in the proof of REF , define MATH and MATH on MATH, where MATH. As in the same proof, introduce the kernel MATH and isomorphism MATH from MATH onto MATH. Under MATH, the functions in MATH wh... |
math/0008057 | Write MATH and MATH for the NAME spaces with reproducing kernels MATH and MATH. Define a relation MATH . It is easy to see that MATH is isometric. We show that the domain MATH of MATH contains a maximal uniformly negative subspace of MATH. To this end, consider a NAME matrix of the form MATH where MATH are any points i... |
math/0008057 | We repeat the constructions in the proof of REF . The partial isometry MATH is again a contraction in the present situation. In general, the operator MATH is not a contraction, but it is a bounded operator and so MATH is defined for MATH sufficiently small. The argument goes through if we restrict attention to a suitab... |
math/0008057 | The first equality in REF can be shown by induction. The second equality follows from the first. To see this, use the identities MATH, MATH, MATH, and MATH to obtain MATH which is the second equality in REF. We get REF on replacing the entries of the matrices by their complex conjugates. We prove REF . Assume MATH. The... |
math/0008057 | CASE: By the first equality in REF, MATH . By REF, MATH . If MATH, then MATH is a submatrix of MATH obtained by deleting a set of rows and corresponding columns, and therefore MATH, yielding REF . CASE: The first and third equalities hold by REF. Since MATH and hence MATH the second equality also holds. CASE: By REF , ... |
math/0008057 | This follows on expressing all of the quantities in terms of the sequences MATH and MATH. For example, for the negative eigenvalues, if one of the quantities has constant value MATH from some point on, then MATH for all sufficiently large MATH, and all have constant value MATH from some point on. |
math/0008057 | Let MATH be the complex numbers viewed as a NAME space in the Euclidean metric. Define MATH by REF. Then by REF, the matrices REF have MATH negative eigenvalues for all sufficiently large MATH, that is, the sequence MATH belongs to MATH. As in NAME and NAME\ĭn CITE, construct a NAME dilation for MATH; that is, we const... |
math/0008057 | Suppose that REF with the data MATH has a solution in MATH. Let MATH be the NAME expansion of this solution. By the necessary conditions for REF discussed above, MATH has MATH negative eigenvalues for all sufficiently large MATH. Define MATH so that MATH for all MATH. Then REF implies that MATH has MATH negative eigenv... |
math/0008057 | All of the citations below are from CITE. MATH . Corollary on p. CASE: MATH REF , p. CASE: MATH REF , p. CASE: MATH REF , p. CASE: MATH REF , p. REF, and REF , p. CASE: MATH REF , p. REF, and REF , p. CASE: MATH . Corollary on p. REF , p. CASE: MATH . The ``if" part follows from MATH and MATH, the ``only if" part from ... |
math/0008057 | For any extension of the given sequence by numbers MATH, we assume that MATH are defined as in REF. CASE: According to MATH there are infinitely many MATH such that MATH. For such MATH we have MATH and MATH. Hence by MATH there is an extension MATH of MATH such that MATH for all MATH. REF implies that MATH and hence MA... |
math/0008057 | Since a holomorphic Hermitian kernel has the same number of negative squares on subregions CITE, by a change of scale we may assume that MATH. We may also assume without loss of generality that MATH is a NAME space. Let MATH be the NAME class of MATH-valued functions on the unit disk MATH. Assume that MATH. By REF, we ... |
math/0008065 | The NAME metric tensor in MATH extends continuously to the NAME metric tensor in MATH. CITE This implies that MATH. On the other hand, suppose a length-minimizing path in the completion metric joining MATH and MATH enters MATH, and is thus a NAME geodesic MATH there. Without loss of generality we can assume MATH lies i... |
math/0008065 | If not, there exists a sequence MATH and a sequence MATH of maximally pinched frontier points on the frontier of MATH with MATH and MATH; here, MATH of course refers to the curves MATH. Since there are but a finite number of homotopy classes of maximally pinched surfaces, we can find subsequences, again called MATH and... |
math/0008065 | If the lemma were not true, there would be sequences MATH and a sequence MATH of curve families such that MATH, MATH, and MATH, while MATH. From the first two conditions we can find subsequences again denoted MATH and a sequence MATH such that MATH and MATH converge to the same point MATH. Fix MATH a compact subset of ... |
math/0008065 | Since MATH is an open map, we need to show that MATH is proper. Suppose on the contrary that there is a sequence MATH leaving every compact set in MATH such that MATH lies in a compact set MATH of MATH. We first show that no subsequence of MATH can project to a precompact open set MATH in the moduli space MATH. For as ... |
math/0008065 | For each MATH we are interested in its MATH orbit. Let MATH . Since MATH is an isometry and fixes the frontier of MATH by the induction hypothesis, each point of MATH is the same distance from the frontier of MATH as MATH is. Fix a small MATH and let MATH . It is easy to see MATH is open and connected. Let MATH denote ... |
math/0008066 | We are considering the knot MATH. Because MATH is algebraic order REF, we assume MATH is even. Let MATH be the two - fold branched cover of MATH. Let MATH. By CITE, we have MATH . Here MATH is the triangle on the MATH - plane with vertices at MATH, MATH, and MATH. The notation int refers to a count of integral points o... |
math/0008066 | We use the notation that MATH for the connected sum of MATH copies of MATH, or the direct product of MATH copies of MATH, depending on whether MATH is a space or a module. Take a linear combination of twisted doubles of the unknot, MATH, MATH when MATH. The NAME polynomial of this knot is MATH. If MATH is slice, its NA... |
math/0008066 | The NAME surface and NAME form of a MATH - twisted double of a knot are identical to those of the MATH - twisted double of the unknot, thus for MATH prime, a MATH - twisted double of any knot is of algebraic order REF. Let MATH the MATH - fold branched cover of MATH. A trivial consequence of REF , is that for any chara... |
math/0008070 | If MATH is a finite domination with right inverse MATH, let MATH be a homotopy from the identity to MATH. Since MATH, the closure of MATH is compact in MATH. Conversely, if the closure of MATH is compact in MATH, let MATH be a finite subcomplex of MATH containing MATH. Setting MATH equal to the inclusion MATH and MATH ... |
math/0008070 | The key result is the trick of CITE, which shows that if MATH, MATH are maps such that MATH then MATH is homotopy equivalent to the mapping telescope of MATH. This requires the calculus of mapping cylinders, which we now recall. By definition, the mapping cylinder of a map MATH is the identification space MATH . We sha... |
math/0008070 | The mapping torus of a map MATH is defined (as usual) by MATH . For any maps MATH, MATH there is defined a homotopy equivalence MATH . If MATH and MATH is a finite MATH complex we thus have homotopy equivalences MATH with MATH a finite MATH complex. Conversely, if MATH is homotopy equivalent to a finite MATH complex MA... |
math/0008075 | The assertion is equivalent to the statement that for all MATH the following identity holds: MATH where MATH is given by MATH . In order to prove this identity it is sufficient to prove that for each MATH the terms MATH occur as many times in REF as in REF . In fact, MATH and MATH occurs in REF exactly MATH times if MA... |
math/0008075 | MATH and MATH are the MATH sections of the infinite matrices MATH and MATH of the previous proposition. Let MATH be the MATH sections of the infinite matrix MATH. Because of the triangular structure of MATH, it follows that MATH. Noting that the entries on the diagonal of MATH are equal to MATH, we obtain the desired a... |
math/0008075 | The moments of MATH are given by MATH . Here we have made a change of variables MATH and written MATH using the binomial formula. With regard to REF this completes the proof. |
math/0008075 | Note first that MATH is well defined since MATH. Moreover, MATH and MATH. By rearranging rows and columns of MATH in an obvious way, it is easily seen that this matrix is similar to MATH where MATH and MATH. This matrix equals MATH where MATH and MATH are NAME matrices of size MATH and MATH, respectively. Multiplying t... |
math/0008075 | Since MATH it follows from REF that MATH. Now we can apply REF in order to obtain the identity MATH. |
math/0008075 | First of all we multiply the matrix MATH from the left and right with MATH. We obtain the matrix MATH by observing that MATH. Next we claim that MATH with a certain matrix MATH. If we take the determinant of this equation, we obtain the desired determinant identity. In order to proof the above matrix identity it suffic... |
math/0008075 | In REF the numbers MATH are defined in terms of the numbers MATH. By a simple inspection of this formula, it is easy to see that for any given sequence MATH there exists a sequence MATH such that REF and MATH holds for all positive MATH. Now let us define the numbers MATH not by REF but by REF . Then with MATH and MATH... |
math/0008075 | Obviously, MATH. Hence MATH. It is sufficient to verify REF for the NAME coefficients and moments. First of all, MATH . Hence MATH . The expression in the big braces equals (by a change of variables MATH in the second part of the sum) MATH . Hence MATH . Now it is easy to see that MATH are the moments of the function M... |
math/0008075 | From the assumptions MATH it follows that the NAME coefficients MATH are zero. Hence MATH has a checkered pattern, and rearranging rows and columns it is easily seen that MATH is similar to the matrix MATH. The NAME coefficients MATH of MATH are equal to zero. By rearranging the rows and columns of MATH in the same way... |
math/0008075 | Define MATH. Then REF implies that MATH. Since MATH the function is well defined and MATH. Now the formula follows from REF and by taking square roots. |
math/0008077 | This follows from REF , from the remark made above, from REF and from the fact that if the infinite group MATH is normally embeddable into some group MATH and lies in MATH, then without loss of generality MATH is of the same cardinality as MATH. Let us show it. The infinite group MATH has a system of generators of the ... |
math/0008077 | We consider the wreath product MATH. For each MATH we take a MATH (the first element MATH is situated in the place number MATH). Then it is easy to compute that MATH (MATH is in the place number MATH). We should now just note that the set of all strings on the right for all MATH forms the first copy of MATH in MATH. Le... |
math/0008077 | It is clear that if for each group of the family MATH there is an embedding of type stated in REF (into a group MATH, MATH), then for the (direct or cartesian) product MATH there is an embedding of mentioned type as well, namely into the (direct or cartesian) product MATH. And since each finite nilpotent group is a dir... |
math/0008077 | CASE: We have MATH. Thus MATH (see in CITE). Since the set of all finite groups generates the variety MATH of all groups, there is a finite group MATH. Thus MATH because MATH is an extension of MATH by MATH. The finite group MATH satisfies the requirement of the first point. We can now repeat this step for the variety ... |
math/0008077 | Since MATH and since multiplication of varieties is a monotonic operation we have MATH (see REF). The product of two locally finite varieties is locally finite (NAME. CITE). Thus the sets MATH and MATH of finite groups of products MATH and MATH are not equal too. Let MATH. Since MATH, necessarily MATH. |
math/0008079 | Since the torus is a compact set, it suffices to show that all joint moments agree. In other words, we need to show that if MATH is a monic NAME monomial, then MATH . But this follows immediately from REF . |
math/0008079 | Recall CITE that the eigenvalue distribution of MATH is given by the density MATH . Equivalently, if we view MATH as acting on the eigenvalues as MATH we have the density MATH . By the Main Lemma, we must extract the MATH-divisible monomials. Clearly, the action of MATH preserves divisibility, so we may restrict our at... |
math/0008079 | Indeed, by the theorem, MATH . But this is precisely the desired result. |
math/0008079 | Again, referring to CITE, the eigenvalue density for a random orthogonal matrix is (up to an overall constant, and ignoring fixed eigenvalues) given by MATH where MATH is the hyperoctahedral group (signed permutations), MATH is a certain character of MATH, MATH is a vector in MATH, acted on in the obvious way by MATH, ... |
math/0008079 | It suffices to observe that the conclusion is true whenever MATH is equivalent to a union of cosets MATH for MATH and MATH for MATH. By a case-by-case analysis, this holds whenever MATH. |
math/0008079 | We claim that it suffices to prove MATH where MATH is an arbitrary function on the unit circle. Indeed, we may take MATH to have the form MATH at which point comparing coefficients of the MATH tells us that all joint moments of the polynomials MATH and MATH agree, where MATH, MATH, MATH are uniform and independent. But... |
math/0008079 | The key observation is that an element MATH is in MATH if and only if there exists a translation MATH such that MATH . Thus instead of considering the stabilizer in MATH of the translate MATH, we can consider the stabilizer in MATH of the vector MATH; we have a canonical isomorphism between MATH and MATH. Since MATH is... |
math/0008079 | Indeed, MATH is a finite subgroup of MATH generated by reflections, so by definition is parabolic. |
math/0008079 | Certainly, there exists an element MATH such that MATH is in the fundamental chamber of MATH; the point MATH is then independent of the choice of MATH. The remaining freedom in the choice of MATH is simply that we can apply any element of MATH. Since MATH there exists an element MATH of MATH such that MATH is in the fu... |
math/0008079 | The main complication is the fact that MATH might not equal MATH. If not, let MATH be the orthogonal complement of MATH in MATH, and write MATH with MATH and MATH. Then we can write MATH . Thus MATH if and only if MATH. We may therefore assume that MATH, and thus MATH. Furthermore, since the desired result is invariant... |
math/0008079 | Writing MATH as above, we observe that MATH has connected center if and only if for every element MATH, there exists some element of MATH that projects to MATH. Dualizing, this is precisely the requirement that MATH. |
math/0008079 | The key observation is that MATH is not stabilized by any element of MATH. Consequently, the sum MATH is equal to REF precisely when the group MATH is trivial. If the sum is REF, then clearly MATH; conversely, if MATH, then the sum must equal REF. It thus remains to consider the group MATH. Aside from diagram automorph... |
math/0008080 | The first property follows from the fact that each blow-up increases the rank of second homology by MATH. Thus MATH has rank MATH, so MATH must have MATH irreducible components. Notice that this implies easily the well-known result CITE that MATH where MATH is the number of horizontal curves of MATH and MATH is the num... |
math/0008080 | If the curve is an exceptional curve then it has self-intersection MATH. If MATH, then the curve must have valency at least three (since any MATH exceptional curve that could be blown down is a cutting divisor). Any three adjacent curves must include two horizontal curves, which contradicts the fact that the dual graph... |
math/0008080 | Assume we have a horizontal curve MATH of type MATH with MATH. It intersects each of the three MATH curves MATH times (counting with multiplicity) so in order to break cycles - REF - we have to blow up at least MATH times on each MATH horizontal curve, so the proper transforms of the three MATH curves have self-interse... |
math/0008080 | We blow up at a point on MATH precisely when at least two horizontal curves meet in a common point there. In general, if a horizontal curve meets MATH with a high degree of tangency then we blow up repeatedly there. But, since all horizontal curves are MATH and MATH curves, they meet MATH transversally, so a point on M... |
math/0008080 | Since the two branches consist of curves of self-intersection MATH, they cannot be reduced before the other branches are reduced. If the rest of the configuration of curves is blown down first then the valency MATH curve becomes a valency MATH curve with non-negative self-intersection and no more blow-downs can be done... |
math/0008080 | We will assume otherwise and reduce to the situation of REF to give a contradiction. Thus, assume that two MATH curves do not intersect in two points contained in the union of the MATH curves. Then in order to break cycles these curves must be blown up at least four times - once each for at least three of the MATH curv... |
math/0008080 | Assume that there are more than three MATH horizontal curves in MATH and at least two MATH curves, say MATH and MATH. CASE: MATH and MATH meet on MATH. Then they meet each of at least two MATH curves in distinct points, so after blowing up to destroy cycles, these MATH curves have self-intersection number MATH and REF ... |
math/0008080 | The statement is trivial for one MATH horizontal curve so assume there are at least two MATH horizontal curves in MATH. By the previous lemma, there are exactly three MATH horizontal curves. If there are exactly two MATH horizontal curves in MATH then the lemma is clear since the curves cannot be tangent by REF . When ... |
math/0008080 | By assumption and REF there are at least three MATH horizontal curves and some MATH horizontal curves in MATH. If there is exactly one MATH horizontal curve then the proposition is clear. If there is more than one MATH horizontal curve, then by REF there are precisely three MATH horizontal curves and two of the MATH ho... |
math/0008080 | Let MATH denote the number of horizontal curves and MATH denote the number of MATH vertical curves in MATH. Thus MATH consists of MATH irreducible components and by REF , when blowing up to get MATH from MATH we must leave MATH exceptional curves behind as cutting divisors. By REF we must break all cycles. The minimum ... |
math/0008080 | In each of the configurations of REF we can perform a NAME transformation by blowing up MATH and MATH for MATH or MATH and then blowing down MATH and the proper transform of the MATH vertical curve that contains MATH. The exceptional divisors MATH and MATH become MATH curves and the MATH vertical curve that contains MA... |
math/0008080 | Suppose otherwise, that MATH is not a cutting divisor and for MATH nor is MATH a cutting divisor. The exceptional curves MATH introduce an extra intersection and hence an extra cutting divisor is required. There is one such extra intersection in the configuration with MATH and two such extra intersections in the config... |
math/0008080 | The proper transform of each horizontal curve has self-intersection less than or equal to MATH and all curves in MATH beyond horizontal curves have self-intersection strictly less than MATH. If the two horizontal curves that meet MATH, MATH and MATH, have self-intersection strictly less than MATH, then since all curves... |
math/0008080 | Since MATH must be separated from at least one of MATH and MATH then at most one of MATH and MATH can be true. Similarly MATH must be separated from at least one of MATH and MATH so at most one of MATH and MATH can be true. By REF , if MATH then MATH so MATH. Similarly, MATH implies that MATH and MATH. |
math/0008080 | Suppose otherwise. Assume that MATH and MATH. If this is not the case, then by REF we may assume that MATH and MATH and argue similarly. The curves beyond MATH have self-intersection strictly less than MATH. The curve immediately adjacent and beyond MATH is MATH and this has self-intersection strictly less than MATH. T... |
math/0008080 | Since the self-intersection of each of the curves beyond MATH is less than MATH each branch beyond MATH cannot be blown down before MATH. Thus, there are at most two branches. Furthermore, since the self-intersection of each of the curves beyond MATH, MATH is less than MATH, the branch beyond MATH can be blown down bef... |
math/0008080 | If the non-separating blow-up sequence occurs on the branch beyond MATH, MATH then that branch cannot be blown down. By the proof of REF , in order to obtain a linear graph we must blow down MATH of the branches beyond MATH, MATH. Thus, if the non-separating blow-up sequence does occur beyond MATH for some MATH, then t... |
math/0008080 | By REF if the non-separating blow-up sequence occurs beyond MATH then MATH. In particular, MATH. Thus, MATH, MATH. With only four horizontal curves, we can perform a NAME transformation to make MATH the MATH curve and hence we are in the first case of REF . |
math/0008080 | Arguing as previously, if the non-separating blow-up sequence occurs anywhere else, or if it is more complicated, then it introduces a new branch preventing the divisor MATH from blowing down to a linear graph. |
math/0008080 | The calculation involves the relation between plumbing graphs and splice diagrams described in CITE or CITE, with which we assume familiarity. In particular, we use the continued fractions of weighted graphs described in CITE. If one has a chain of vertices with weights MATH, its continued fraction based at the first v... |
math/0008080 | For the following computations we continue to assume the reader is familiar with the relationship between resolution graphs and splice diagrams described in CITE. The arrows signify places at infinity of the generic fibre, one on each horizontal curve. The fact that MATH is next to MATH in the dual graph says that the ... |
math/0008080 | The first part was part of the proof of REF . For the second part, note that if MATH then certainly MATH must fail, so MATH and the nonseparating blow-ups were on the left. The continued fraction on the left was MATH which is integral, showing that the left chain consisted only of the exceptional curve before the non-s... |
math/0008080 | The splice diagram prescribes the number of horizontal curves and the separating blow-up sequences at each point of intersection. The only freedom is in the placement of the horizontal curves in MATH, and in the choice of points, on prescribed curves, on which to perform the string of blow-ups we call the non-separatin... |
math/0008080 | We can argue as in the previous section. The curve over infinity, MATH is not blown up since there are no triple points. If there is more than one vertical curve over a finite value then there are precisely three MATH horizontal curves since otherwise there would be at least two MATH horizontal curves that would be blo... |
math/0008080 | By the classification of ample rational polynomials of simple type, the proposition is true in this case. So, we may assume that there is a horizontal curve of type MATH for MATH. Suppose there is no MATH horizontal curve with the property of the proposition. Then by the proof of REF there are at least two MATH horizon... |
math/0008080 | The statement is true for MATH by REF so will assume MATH. A curve of type MATH will intersect the MATH horizontal curves MATH times, with multiplicity, unless possibly if the MATH curve is singular at these points of intersection. The latter possibility is ruled out by the assumption of the corollary. Hence the MATH h... |
math/0008085 | Since the NAME function is constant on components of flat connections and since MATH classifies the MATH-cover MATH, it follows that MATH is the trivial MATH-cover. Thus, every such MATH is a homeomorphic copy of a component of MATH and is therefore compact. Choose MATH. Then MATH is nonabelian, and we can assume after... |
math/0008085 | We first prove that MATH depends only on the perturbation MATH. Since we have already seen that the spectral flow terms are gauge invariant, we just need to show that MATH is independent of the choices of MATH for MATH . Suppose then that MATH also satisfy REF . Taking lifts MATH compatible with MATH it follows from ad... |
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