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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 that MATH and MATH, the LHS of REF is equal to MATH by the induction hypothesis. Here, we used the fact that, for MATH, MATH. Substituting REF for MATH in REF , we have MATH where MATH . Suppose MATH. Then, MATH holds. It follows that MATH and MATH. Thus, applying REF to the second factor of REF , we obtain REF . It is easy to check that the Right-hand side in REF converges for MATH. |
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 fact that MATH is skew MATH-invariant, This is easily seen by using the following well-known transformation property under the simple reflection MATH: MATH . |
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 FR, INT, and BC so that MATH vanishes in the quotient of MATH by the single relation SC. REF A finger move on MATH guided by an arc representing MATH creates a cancelling pair of double points of MATH which are paired by a clean NAME disk MATH. By performing boundary twists and interior twists on MATH one can create intersections between MATH and MATH so that MATH for any integers MATH and MATH such that MATH modulo REF. CASE: By similarly creating a clean NAME disk MATH and tubing into any immersed sphere representing MATH it can be arranged that MATH. CASE: If a NAME disk MATH has an interior intersection point MATH with MATH that contributes MATH to MATH then MATH may be eliminated by a finger move at the cost of creating a new pair MATH of double points of MATH which admit a NAME disk MATH (with embedded boundary disjoint from existing NAME disks) such that MATH has a single intersection with MATH and MATH (See REF ). By using these three modifications we may assume that our collection of NAME disks satisfies MATH in MATH modulo the SC relation. We can now move pairs of intersection points which have algebraically cancelling contributions to MATH on to the same NAME disk as follows (see CITE for a detailed description of the simply-connected case.) The finger move illustrated in REF exchanges a point MATH that contributes MATH to MATH for a point MATH that contributes MATH to MATH. This finger move also creates two new double points of MATH which admit a NAME disk MATH (with MATH embedded and disjoint from all other NAME disks) such that MATH. By performing this finger move through the negative arc of MATH instead of the positive arc one can similarly exchange a point MATH that contributes MATH to MATH for a point MATH that contributes MATH to MATH. In this way it can be arranged that all double points of MATH are paired by NAME disks MATH such that MATH for all MATH. |
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 the NAME disk because we are modding out by the diagonal right action MATH of MATH. Since any two NAME disks with the same boundary differ by an element MATH, MATH does not depend on the choices of the interiors of the NAME disks because it is measured in the quotient of MATH by MATH. In order to show that MATH does not depend on the boundaries of the NAME disks it is convenient to generalize the definition of MATH to allow weak NAME disks along the lines of REF. For each MATH, where MATH and MATH agrees with the orientation of MATH at MATH, define MATH from the following three group elements: Give MATH and MATH a common whisker at MATH then take the positive and negative group elements of MATH together with the group element of MATH corresponding to the sheet MATH. Define the sign of MATH to be equal to the sign of the permutation MATH. The generalized definition of MATH includes the sum of MATH over such MATH. Note that this generalization does not require any new relations and reduces to the original definition after eliminating the intersections and self-intersections between boundaries of the NAME disks in the usual way REF . The arguments of REF now apply to show that MATH does not depend on the boundaries of the NAME disks: pushing the boundary of a NAME disk across an intersection point REF creates an interior intersection in the collar of the NAME disk and a boundary intersection with cancelling contributions to MATH. Independence of the choice of pairings of the intersection points (see REF ) also follows. Note that in the present setting there are no subtleties concerning the pre-images of intersection points. Thus MATH is well-defined. To see that MATH is invariant under homotopies of the MATH it suffices to check invariance under finger moves and NAME moves on embedded NAME disks. NAME moves only create embedded NAME disks which clearly don't change MATH. NAME move on an embedded NAME disk MATH which eliminates a pair of self-intersections on MATH will create a new pair of intersections between MATH and MATH for each intersection between MATH and MATH, but these new intersections have cancelling contributions to MATH which remains unchanged. The same applies to a NAME move that eliminates a pair of intersections between two different spheres and since the embedded NAME disk can be assumed to have been included in any collection used to compute MATH again the invariant is unchanged. Thus MATH only depends on the homotopy classes of the MATH. It follows directly from the definition that MATH satisfies REF can be checked as follows: MATH where the MATH are parallel copies of MATH. Since the normal bundle of MATH is trivial, each self-intersection of MATH gives nine intersection points among the MATH of which exactly three are self-intersections. Thus each NAME disk MATH for a canceling pair of self-intersections of MATH yields six canceling pairs of intersections among the MATH paired by NAME disks which are essentially parallel copies of MATH. If MATH has an interior intersection with MATH that contributes MATH to MATH (expressed in the notation of this section) then there will be six corresponding terms contributing MATH to MATH. See REF for a schematic illustration where we have circled the points corresponding to one of the contributions, an intersection between MATH and a NAME disk on MATH and MATH. To complete the proof of REF we now show that the MATH can be homotoped to pairwise disjoint maps if MATH (the converse is clear). First use MATH to separate MATH and MATH by pushing MATH off the NAME disks MATH and then doing NAME moves on MATH. Now MATH is given completely in terms of NAME disks pairing MATH and NAME disks pairing MATH. The arguments of REF (tubing the NAME disks into spheres and using the move of REF ) can now be applied to get a second layer of NAME disks which pair all intersections between MATH and the NAME disks for MATH. After pushing any intersections between the secondary NAME disks and MATH down into MATH, the secondary NAME disks can be used to make MATH disjoint from the NAME disks for MATH. After pushing down any intersections with MATH, the NAME disks on MATH may now be used to eliminate all intersections between MATH and MATH. After similarly applying the arguments of REF to get secondary NAME disks pairing the intersections between MATH and the NAME disks for MATH, one can push down intersections and do NAME moves to eliminate all intersections between MATH and MATH so that the MATH are pairwise disjoint. |
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 definition CITE of NAME duality MATH . Since MATH is transversal to MATH outside MATH, we have MATH . |
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-section of MATH, we have MATH where MATH is the differential in the exact cobordism sequence of pair, and MATH is the NAME duality on the total manifold MATH. Denote by MATH the isomorphism MATH. From the explicit formula CITE for the MATH-equivariant isomorphism, that identify a small neighborhood of zero section of MATH with the neighborhood MATH it follows that the following diagram commutes (double arrows here denote isomorphisms). MATH . Therefore, MATH. From the naturality of the MATH-product CITE we have MATH where MATH is the inclusion. Then, by REF MATH . It remains only to note that MATH. |
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 can assume that not all MATH are equal to zero). We shall show that the tail distribution of a polynomial MATH depends only on the norms MATH or, by hypothesis, on the function MATH . Let MATH be a constant of equivalence REF . The MATH -functional MATH is a concave function with respect to MATH [REF,MATH]. Therefore from the NAME inequality [REF,MATH] we get for MATH . Since MATH increases, then MATH for all MATH . On the other hand, by NAME 's inequality and hypothesis, we have MATH for MATH and MATH . In particular, if MATH then MATH and MATH . Following CITE , we define the functionals MATH . Suppose that MATH . By REF , MATH for some MATH . Therefore from REF it follows MATH . Hence, by the definition of MATH we have MATH and MATH where MATH . Next, as MATH is a concave function, the definition of MATH and MATH implies MATH and MATH where MATH . If we combine this with REF , then we get MATH . In addition, from REF it follows MATH and therefore MATH . Thus we have MATH . Next, by REF we can find MATH such that MATH . Then from REF we obtain MATH . The last yields MATH where MATH . In the same way, using REF , we can prove also the formulas similar to REF - REF for the NAME functions. Namely, if MATH is a constant of equivalence REF and MATH then MATH and MATH . Let us denote MATH . Note that MATH and MATH are universal constants. Hence MATH depends only on the constant MATH of equivalence REF . In the case MATH formulas MATH and REF give MATH . If MATH then MATH by REF . Hence MATH yields: MATH and REF holds for all MATH . Conversely, if MATH then from REF , MATH and REF we have MATH . If MATH then REF implies MATH . Taking into account REF , we get MATH and REF is valid again. From REF it follows that MATH . This completes the proof of REF . |
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 assertion is an easy consequence of REF . This completes the proof. |
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 of MATH -norms of modified NAME products from above for MATH (applications of usual NAME products to similar problems see, for example, in [REF,MATH] and [REF,MATH]). First note that from REF concerning to MATH it follows MATH . Therefore, without loss of generality, we can assume that MATH for all MATH . Let MATH be a sequence of real numbers such that MATH . It is readily seen that MATH is a weakly compact sequence in the space MATH . Hence there are a r.v. MATH and a subsequence MATH such that MATH and MATH weakly in MATH . Futhermore, MATH weakly in MATH . Therefore there exists a positive integer MATH such that MATH . Denote MATH . Suppose that positive integers MATH and functions MATH are chosen. Let MATH be a positive integer such that MATH and for the function MATH where MATH and the summation is taken over all sets of indices MATH . Let us show that REF is valid for a such sequence MATH . First let MATH be a positive integer. Suppose that MATH is an arbitrary partition of MATH . For MATH and MATH define sets MATH (it is possible some of them are empty). Introduce block NAME products corresponding to this partition: MATH where MATH such that MATH . Denote MATH . Let us estimate the integral MATH from below. Here MATH where MATH and MATH . Then MATH where the sums MATH are defined as follows. Note that MATH . The sum MATH contains all terms with integrals of the form MATH or MATH where MATH . In the sums MATH and MATH we include all terms with integrals of the form MATH (MATH) and MATH respectively. Combining REF , and REF , we get MATH . Every term of the sum MATH contains the factor MATH . Therefore from REF we have MATH . Futhermore, if MATH then the sums MATH and MATH in REF are absent. Hence, MATH . Now, let MATH . From REF - REF , and REF MATH for every MATH . REF imply MATH . Let MATH such that MATH . For every MATH choose MATH such that REF holds and also MATH . By the last equality and REF , and REF , we obtain MATH . So, for every MATH and a partition MATH of positive integers there exist enough large positive integer MATH and real numbers MATH satisfying REF such that MATH . Now, our aim is to prove an estimate of the integral MATH from above with an expression of the form MATH with a positive constant MATH . By NAME 's inequality, MATH where MATH . Since MATH then it is enough to estimate for this from above the quantity MATH . Let MATH . As before, MATH is a partition of MATH and real numbers MATH satisfy REF . From the obvious inequality MATH it follows that MATH . Denote by MATH the number of elements of a set MATH . Then the inner product in the last expression is equal to the sum: MATH where the summation in the last term is taken over all non-empty subsets MATH of the set MATH . In view of REF the expression in braces is not greater than MATH if MATH . So, combining REF , and REF again, we obtain: MATH . Let us note that for MATH . Since the sets MATH are mutually disjoint, then the integrands in the last expression are the distinct products of pairwise distinct functions MATH . In addition, the number of these functions is equal to MATH for terms of the first,second,.,last sum, respectively. Therefore the maximal index of them in every integrand is not less than MATH respectively. Finally, by REF , there holds MATH where the inner summation is taken over all sets of indices MATH . From this inequality, REF , and REF it follows MATH . So, by REF , we get MATH . Taking into account REF , from the last inequality we have MATH . Therefore, by REF , MATH for all positive integers MATH and MATH . Let us prove an analogous inequality in the cases MATH and MATH . First, the constructed system MATH is a NAME basic sequence. It means that MATH . In fact, by assumption, MATH . Therefore REF implies that MATH for all MATH . Similarly, MATH and REF is proved. REF shows that the system MATH is a MATH -system. Therefore MATH hase the same property, and hence it is a NAME system [REF ]. It means that MATH where a constant MATH depends only on MATH . The inequality MATH implies that MATH . Hence, using the properties of the MATH -functional, REF , and REF we obtain: MATH and MATH . Suppose now that MATH is an arbitrary real number. Choose a positive integer MATH such that MATH . Then REF , and REF yield MATH where MATH depends only on MATH . To conclude the proof, it remains to note that REF holds, in particular, for the system MATH . |
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 we remarked in REF, the NAME system is a MATH -system. Hence MATH have this property, too. The possibility of the selection of subsystem MATH satisfying REF is a consequence of well-known NAME 's theorem [REF ]. This completes the proof. |
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 holds for functions MATH is any segment from MATH . Next, as in the proof of REF from CITE , we can select a subsequence MATH such that the relations similar to REF hold for the integrals taken over an arbitrary segment MATH . Moreover, by well-known NAME 's lemma (see [REF[ or [REF,L. REFEF]), we can assume that there is a some set MATH such that MATH and for any subset MATH there holds: MATH . So, we can repeat next the arguments of the proof of REF replacing only the integrals over MATH with the integrals over a segment MATH such that MATH . |
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 of the function MATH includes the segment MATH . By REF , the functions MATH satisfy the inequality: MATH . Therefore it can be found MATH such that MATH . First suppose MATH . Denote by MATH step functions defined for MATH and having the following properties: MATH (for example, a such functions can be constructed using the NAME functions). Let MATH be an analogous system, but defined on the interval MATH . Let us define the functions MATH as follows. If MATH then MATH and MATH . By REF , we have MATH . At the same time, from the definition of functions MATH and REF it follows MATH for every another set of indices MATH . In the case MATH we define MATH . Then equalities REF hold again. So, the functions MATH are defined on MATH . In addition, let MATH for MATH and MATH for MATH . Since REF - REF yield that MATH is a multiplicative set on the segment MATH and MATH then this completes the proof. |
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 multiplicative set MATH on the segment MATH . It is easy to check that the set of functions MATH is a multiplicative system on the segment MATH and MATH . Therefore, by REF from CITE , MATH where MATH and a positive constant MATH depends only on MATH . Hence, MATH . Using the last inequality and arguing as in the proof of REF (see REF ), we obtain: MATH where MATH depends only on MATH . Let us prove the opposite inequality. Denote MATH . At first, suppose that MATH . For MATH by the definition of the MATH -functional, we have: MATH . Therefore it is sufficient to consider the case when MATH . Let MATH be an arbitrary partition of the set MATH (probably, some of them are empty) and MATH . Introduce the NAME products MATH and the integral MATH . After regrouping its terms we get MATH where MATH . Since MATH then REF and the NAME inequality yield MATH . By REF , MATH and we have MATH . Let MATH satisfy REF and MATH . Then from REF we obtain MATH . Thus, by REF , we have MATH . Now, let us estimate the integral MATH from above. By NAME 's inequality, MATH where MATH . Arguing as in the proof of REF , we can prove that MATH . So, REF , the embedding MATH and the NAME inequality imply MATH . Therefore MATH and from REF it follows MATH . Analogous relations hold also for MATH and MATH . In fact, since MATH is an orthonormal set and MATH then MATH . In addition, as in the proof of REF , using REF from CITE , we obtain MATH where a constant MATH depends only on MATH . Finally, we extend, as usual, REF - REF to all real MATH and get MATH where MATH . Note that this constant depends only on MATH . Therefore in view of REF equivalence REF holds with a conctant depending only on MATH . This completes the proof. |
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 is a basis in the closed linear span MATH generated by this system in the space MATH [REF,Pr. REF]. In other words, MATH is a NAME basic sequence, and therefore MATH as MATH for every MATH [REF,MATH]. For an arbitrary MATH we write: MATH where MATH is the orthogonal complement to MATH . Then, obvious, MATH as MATH . Finally, we have MATH . Relations REF show that the system MATH satisfies REF - REF . Therefore there exists a subsystem MATH that is equivalent in distribution to the NAME system on the segment MATH . NAME and NAME 's inequality (see CITE or [REF,MATH]) imply MATH where a positive constant MATH depends only on MATH and MATH . At the same time, REF yields MATH . Let MATH and MATH . By the last inequalities and [REF ], the application of the real interpolation method MATH (see Introduction) to the NAME couples MATH and MATH gives MATH . Here MATH and a constant of this equivalence depends only on MATH and MATH . Since MATH then the last contradicts with REF , if MATH . This completes the proof. |
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 is its restriction MATH to MATH. The identity MATH shows that MATH CITE. When MATH and MATH, then MATH . The determinant of this matrix is negative, and so the matrix has one negative eigenvalue. Hence MATH. If REF holds, then MATH is contained in the zero set of a nontrivial function in MATH, and we obtain REF because such functions are of bounded type in MATH. Conversely, assume REF and consider the function MATH where MATH is a NAME product having simple zeros at the points of the NAME sequence MATH (so that MATH) and the number MATH is chosen in MATH such that MATH. Then MATH is the product of the inverse of a NAME factor and a classical NAME function which does not vanish at the zero of the NAME factor, and so it belongs to MATH. Evidently, MATH, MATH. |
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 place of MATH and MATH. A short calculation shows that MATH and so MATH; write MATH for the associated reproducing kernel NAME space. The preceding reproducing kernel identity may be used to show that the mapping MATH acts as an isometry from MATH onto MATH. Writing MATH and MATH for the subspaces defined in the theorem for the original functions MATH and MATH and point MATH, and MATH and MATH for the corresponding subspaces relative to MATH and MATH and point MATH, we find that MATH . Since we assume the result when MATH and MATH, we can find a function MATH for some MATH such that MATH is holomorphic at MATH and MATH for MATH and for all but at most MATH points MATH of MATH. Then MATH has the required properties. Thus without loss of generality, we may assume that MATH and MATH. Define a linear relation MATH in MATH as the span of all pairs MATH with MATH and MATH. A direct calculation shows that MATH is isometric. In fact, consider a second pair with MATH replaced by MATH and MATH replaced by MATH. Expand and simplify the inner products of the first members in MATH and second members in MATH. After simplification, in both cases we obtain MATH and this verifies the assertion. The orthogonal complement of the domain of MATH is MATH, and the orthogonal complement of the range of MATH is MATH. Since these are NAME spaces, it follows from CITE that there is a continuous partial isometry MATH such that MATH has initial space MATH and final space MATH and MATH for all MATH and MATH. Calculating as in CITE, we find that MATH for all MATH and MATH. Since MATH is a partial isometry whose kernel is a NAME space, MATH is a contraction. The embedding mappings MATH and MATH from MATH into MATH and MATH are contractions (in fact isometries), and their adjoints act as projections. The adjoints are also contractions because we assume that MATH and MATH are NAME spaces. Therefore MATH is a contraction on the NAME space MATH. By CITE, the part of the spectrum of MATH that lies in MATH consists of normal eigenvalues. By CITE, the span of root manifolds for eigenvalues in MATH is contained in a nonpositive subspace, and hence the number of such eigenvalues is at most MATH. It follows that MATH is invertible for all but at most MATH points in MATH. Since these exceptional points obviously do not include MATH, MATH is invertible for all MATH for some nonzero numbers MATH in MATH; here MATH and possibly MATH when there are no exceptional points. CASE: If MATH, MATH, and MATH, then MATH and MATH. Since this is trivially true if MATH, assume that MATH. Then MATH . Since MATH, we can take MATH in REF to get MATH. Again by REF, MATH for MATH. Trivially the last identity holds for MATH as well, and we obtain REF: Define MATH for all MATH. Then MATH for all MATH. The case MATH is clear. Assume MATH and MATH. Fix MATH. We use Claim MATH with MATH, MATH. Thus MATH . Claim MATH follows. CASE: MATH for some MATH. It is clear from the definition of MATH that it is a holomorphic function on MATH. For all MATH, by the identity CITE, MATH . MATH-Since MATH in the partial ordering of selfadjoint operators, MATH for some operator MATH, where MATH is a NAME space see, for example, CITE; we can choose MATH so that it has zero kernel, but this property is not needed. Therefore MATH where MATH is a holomorphic function with values in MATH. The first summand on the right of REF has MATH negative squares for some MATH by CITE, and the second summand is nonnegative because MATH is a NAME space. Thus by CITE the kernel REF has MATH negative squares, where MATH. Hence MATH, which proves REF . The function MATH has the required properties by REF . The last statement, which gives the condition for MATH, follows from CITE. |
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 which vanish on MATH correspond to the functions in MATH which vanish on MATH, where MATH. Then as before, the special case implies the general result. In what follows, we assume that MATH and MATH. We apply REF in this situation and also with MATH. It is easy to see that the subspace MATH in REF coincides with the the set of elements of MATH which vanish on MATH and is thus a NAME space by hypothesis. We show that the subspace MATH in REF is a NAME space. By the first part of the proof of REF , MATH is the orthogonal complement of the range of the relation MATH in MATH, and therefore it is the same thing to show that the range of MATH contains a strictly negative subspace of dimension MATH. By CITE, it is sufficient to show that some NAME matrix of elements of the range of MATH has MATH negative eigenvalues. In fact, consider two of the second members of the pairs REF that define MATH, say MATH . By REF, since now MATH, the inner product of these elements in MATH is equal to MATH . Here we can choose MATH and MATH arbitrarily, and then choose MATH so that MATH . Since we assume that MATH, it follows that some NAME matrix of elements of the range of MATH has MATH negative eigenvalues, as was to be shown. This completes the proof that MATH is a NAME space. The hypotheses of REF are thus met, and REF yields a function MATH, MATH, such that MATH for MATH and for all but at most MATH points MATH of MATH. |
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 in MATH and MATH are arbitrary vectors in MATH. Thus MATH . Since we assume that MATH has MATH negative squares, MATH has at most MATH negative eigenvalues no matter how MATH and MATH are chosen, and some such NAME matrix has exactly MATH negative eigenvalues. By CITE, MATH contains a MATH-dimensional subspace which is the antispace of a NAME space in the inner product of MATH. Since MATH, this verifies the assertion. It follows that the closure of MATH in MATH is a regular subspace whose orthogonal complement MATH is a NAME space. By CITE, the closure of the range of MATH is likewise a regular subspace MATH of MATH, and we can construct a partial isometry MATH with initial space MATH and final space MATH such that MATH for all MATH and all MATH. Thus for MATH, MATH and MATH . Hence MATH . Since MATH is a NAME space, MATH is a contraction. As in the proof of REF , because we assume that MATH and MATH are NAME spaces, MATH is a contraction, and the part of the spectrum of MATH that lies in MATH consists of at most MATH normal eigenvalues. Let MATH, where MATH are the points MATH of the unit disk at which MATH is not invertible (MATH). For all MATH and all MATH, MATH by REF. Define MATH . Then MATH, MATH, by REF. The proof that MATH for some MATH is the same as in the proof of REF . The last statement follows from CITE. |
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 suitable neighborhood of the origin. At the end, the identity MATH extends to all but at most MATH points of MATH by analytic continuation. |
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. Then MATH . In REF we use the first equality in REF to obtain MATH . Due to the lower triangular form of MATH, we get MATH . With this definition of MATH and REF , we obtain MATH . Here the matrix MATH is invertible. Note also that MATH so that with MATH defined in this way, we have MATH . We now identify MATH as MATH . From the definition of MATH we find that MATH . It follows that MATH . Finally, we obtain MATH proving REF. Evidently, MATH and MATH . Substituting this in REF we obtain REF . |
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. By the proof of REF , if MATH, then MATH and this proves 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 construct a NAME space MATH that contains MATH isometrically as a regular subspace, and a unitary operator MATH such that MATH where MATH is the projection on MATH with range MATH. Since MATH is a regular subspace of MATH, we can write MATH where MATH is a regular subspace of MATH. Let MATH relative to this decomposition. We show that MATH . The cases MATH are immediate. We prove the formula for MATH assuming it is known for MATH. By REF, MATH so it is the same thing to show that MATH . Put MATH . Then MATH . Since MATH for all MATH, MATH . This allows us to bring REF to the form MATH . Dropping the factor MATH on the right in each term, we easily verify REF by induction: the formula is evident for MATH, and the inductive step follows from the identity MATH. This completes the proof of REF. The identity REF implies that MATH for some positive constants MATH and MATH, and therefore the power series MATH converges in a neighborhood of the origin. |
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 eigenvalues for all MATH. Therefore MATH is a solution to REF with the data MATH. Conversely, assume that REF is solvable with the data MATH, that is, the numbers can be extended to a sequence MATH in MATH. Then the matrices REF have MATH negative eigenvalues for all sufficiently large MATH. Reversing the process above, we obtain a sequence MATH that extends MATH such that the matrices MATH have MATH negative eigenvalues for all sufficiently large MATH. By REF , the series MATH converges in some disk MATH where MATH, and by REF\ĭn and NAME in CITE, the function MATH so defined belongs to MATH. Thus REF is solvable with the data MATH. The argument for the classes MATH and MATH is similar. |
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 REF , p. CASE: MATH . Proposition MATH, p. CASE: MATH and MATH . It is enough to prove these statements for MATH in MATH and MATH in MATH. To do this, we use the proof of REF , p. REF, and REF , p. REF, to construct infinitely many extensions with MATH and infinitely many extensions with MATH (treat the subcases MATH and MATH separately using the argument on p. CASE: Then MATH) and MATH follow from MATH. MATH REF , p. CASE: MATH REF , p. REF. |
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 MATH belongs to MATH. CASE: By MATH there are infinitely many numbers MATH such that MATH and MATH. Therefore MATH and by MATH, MATH. After MATH steps, we obtain numbers MATH such that MATH . Using the same argument with MATH instead of MATH, we obtain numbers MATH (each of which can be chosen in infinitely many ways) such that MATH . Now REF follows from REF . CASE: According to MATH there exists a unique extension MATH of MATH such that MATH . It follows from MATH that also MATH and MATH for MATH. Therefore there exists a unique extension of MATH in the class MATH and this extension belongs to MATH (for the uniqueness part, note that by MATH) the equality MATH can only hold in the present situation when MATH. REF . By REF The unique extension described in REF cannot meet any of the conditions in REF ; since for this extension MATH, in REF we need only consider extensions such that MATH for some MATH. In this situation REF holds with MATH replaced by MATH, and therefore we may restrict attention to extensions satisfying MATH . By MATH, REF holds for all but one choice of MATH; in what follows, we assume that MATH is chosen so that REF is satisfied. The question then is if the sequence MATH can be further extended to an infinite sequence MATH as required in REF . CASE: MATH. Since MATH by REF, by MATH we must have MATH. Thus MATH, and hence REF holds vacuously. REF also holds in this case. For by REF, MATH and MATH and since MATH is invertible, REF follows from REF . CASE: MATH. Then with MATH, in view of REF, MATH . Consider any extension of MATH by a number MATH. By REF, MATH. Applying MATH with MATH replaced by MATH, we obtain MATH and by MATH, MATH . If MATH, we can repeat this argument. We continue in this way for MATH and extend MATH with any numbers MATH, MATH; by MATH and MATH, we have MATH, MATH and MATH provided MATH. If equality holds, that is, MATH then MATH is invertible and the process stops. Hence if such an extension of MATH can be continued to a sequence in a class MATH, then necessarily MATH and according to REF each of the classes MATH and MATH contains infinitely many extensions. Thus the first part of REF will follow once we show that MATH . Since MATH, MATH implies that MATH and since MATH, we therefore have MATH, and by MATH, MATH and MATH. This implies that MATH and also that MATH has the desired value. From the first part of REF it follows that there are no extensions in MATH if MATH whatever the value of MATH. By considering the sequence MATH and its extensions MATH and applying the results just proved (together with MATH) we find that there are no extensions in MATH if MATH whatever the value of MATH. REF is part of REF . REF . By MATH, REF holds for any choice of MATH. This allows us to proceed by an argument which is similar to the proof of REF above; in REF there, the exact value of MATH is unimportant in order to obtain the conclusion. |
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 define a bounded selfadjoint operator MATH on MATH such that MATH and MATH where MATH, that is, MATH for all MATH. For another account of the construction of MATH, see CITE. By the spectral theorem, we can write MATH, where MATH and MATH are selfadjoint operators corresponding to the spectral subspaces MATH, MATH, and MATH for the sets MATH, MATH, and MATH. Since MATH, MATH. Let MATH be MATH. Write MATH for the coset determined by an element MATH of MATH. Define a nondegenerate inner product on MATH by MATH . Using CITE, complete MATH to a NAME space MATH having negative index MATH. The cosets determined by the polynomials are dense in MATH by CITE, and therefore MATH is a total set in MATH. By construction, MATH . Hence by CITE, the matrix MATH has at most MATH negative eigenvalues for all MATH and one such matrix has exactly MATH negative eigenvalues. Since the number of negative eigenvalues of MATH is a nondecreasing function of MATH, this number is MATH for all sufficiently large MATH. Conversely, assume that the matrix MATH has at most MATH negative eigenvalues for all MATH and exactly MATH negative eigenvalues for all sufficiently large MATH. By what we showed above, if we can only show that MATH, it will follow that MATH. Let MATH be the set of nonnegative integers, and define a kernel MATH on MATH by MATH . Our hypotheses imply that MATH. By CITE, there is a unique NAME space MATH of functions MATH on MATH with reproducing kernel MATH. This means that for each MATH and MATH, the sequence MATH belongs to MATH, and for any element MATH of MATH, MATH . By CITE, we can represent the kernel MATH in the form MATH where for each MATH, MATH is the evaluation mapping on MATH to MATH: MATH . By the NAME representation, the operators MATH are uniformly bounded, and therefore for MATH and MATH in a suitable neighborhood of the origin, MATH where MATH. The values of MATH lie in the NAME space MATH, which has negative index MATH. The restriction of MATH to a suitable neighborhood of the origin thus has at most MATH negative squares, and since the number of negative squares is independent of the domain (see CITE), MATH. As noted above, this implies that MATH. |
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 inside MATH except for its endpoints. Then MATH tends to zero near its endpoints, but is positive somewhere in its interior. This contradicts NAME 's convexity result CITE which says that the functions MATH are strictly convex along NAME geodesics. Thus MATH, completing the proof. |
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 and a sequence MATH so that MATH, where MATH is some single maximally pinched frontier point on the frontier of MATH. Then let MATH denote the curve system MATH, so that MATH, and set MATH. Then, since MATH induces a NAME isometry, we have MATH, while MATH: this last equality follows from MATH amounting to but a consistent relabelling of curves and hyperbolic surfaces. But then the first limit implies that the sequence MATH converges to MATH. However this implies MATH and we have a contradiction with the second limit statement. |
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 containing MATH for MATH large. Now MATH is a sequence of curve families such that MATH, while MATH. Since MATH is compact, there are only finitely many curves MATH such that MATH for some MATH. Thus MATH is actually a finite set of curves, and by passing to subsequences we can assume the sequence MATH is a fixed set MATH. But now the length functions MATH are continuous on MATH and we have contradicted the fact that both MATH and MATH converge to MATH. |
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 the compact set MATH has finite NAME diameter, so does MATH; since a set of finite diameter can intersect only finitely many disjoint balls of fixed diameter, MATH could intersect but a finite number of preimages of the precompact open set MATH under the projection map MATH. Thus the subsequence MATH would be contained in the closure of the union of those finite number of precompact preimages, and hence would be contained in a compact set in MATH, contrary to hypothesis. We conclude that the entire sequence projects to a sequence that leaves every compact subset of MATH. However along such a sequence MATH the infimum of the scalar curvature of the NAME metric is MATH CITE, while bounded in the compact set MATH. But this is a contradiction to the fact that an isometry preserves scalar curvature. Thus MATH is in fact surjective. |
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 the curves in MATH. Now let MATH . That is, MATH consists of those points such that the lengths of the curves in MATH are bounded on the entire orbit. We begin by claiming that MATH. We prove the claim by showing that MATH is non-empty, open and closed in the connected set MATH. We first show that MATH is nonempty. Using the isometric action of the mapping class group there exists MATH and MATH and MATH so that MATH and MATH. We may choose MATH (and MATH) small enough so that the hypothesis of REF holds. Then since the orbit of MATH remains within MATH of MATH, REF says that MATH. We now show that MATH is open. Let MATH and let MATH. By REF , there exists MATH such that for all MATH and all MATH in the MATH ball about MATH, MATH . Since MATH is an isometry, the MATH orbit of a MATH ball about MATH is the union of the MATH balls about the points in MATH. Since MATH is open, if we take MATH small enough we can insure the MATH ball about MATH remains in MATH and the inequality above says that it is then contained in MATH, proving MATH is open. Finally, we show MATH is closed. Let MATH be a limit of points MATH and assume MATH. Then there exists a sequence MATH such that MATH. Since MATH there is a compact set MATH and a sequence MATH such that MATH. Let MATH be the collection of image curves. We have MATH . Since MATH is an isometry and MATH induces an isometry, we have that for MATH large enough, the points MATH lie in MATH. Fix such an index MATH and consider the sequence MATH. Now we have MATH . However, given a compact set MATH, there exists a constant MATH (depending only on MATH) such that for any two points MATH and any curve MATH we have MATH. This contradicts the previous assertions that the ratio of MATH and MATH have no bound; thus MATH must in fact lie in MATH, and MATH is closed in MATH. This concludes the proof of the claim. We now conclude the proof of the Proposition. Now let MATH be another maximally pinched frontier point of MATH such that the curves in MATH together with those in MATH fill the surface MATH, which means that if we remove the curves in MATH from MATH, the result is a union of simply connected domains. We form the corresponding MATH, and again the claim above shows that this coincides with MATH. Thus every point in MATH lies in MATH. Let MATH be a point at distance MATH from the frontier and MATH from MATH. Applying REF again, there is a ball MATH of radius MATH about MATH such that if we set MATH then there exists MATH such that MATH for all MATH. We may take MATH. Clearly MATH is MATH invariant. Now we adapt an argument of NAME 's CITE in his proof of the NAME Realization Problem that every finite subgroup of the mapping class group has a fixed point to find a fixed point in the current situation. Consider the subset MATH . By CITE, since the set of curves MATH fills MATH, the set MATH is a compact subset of MATH; also the set MATH is a cell, its boundary is MATH, and it contains MATH in its interior. Now define a function MATH on MATH by MATH where MATH is NAME volume element. As MATH is a subset of the compact set MATH, the set MATH has finite total measure. Further, as MATH is bounded on MATH, we see that MATH is a finite integral for each MATH. Since MATH is MATH-invariant, so is MATH. Since MATH for MATH in the boundary of MATH and MATH, the strict convexity of MATH along NAME geodesics says that the vector MATH points out at each point on the boundary of MATH. Thus as an average of such vectors, MATH points out at each such boundary point. By the NAME index theorem, we see that there is some MATH for which MATH. By the negative curvature and the geodesic convexity of the metric, the distance from a point to a geodesic is a strictly convex function of the parameter along the geodesic. Consequently as an average of such functions, MATH is also strictly convex. This implies that MATH is the unique minimum for MATH. It follows that MATH. Now we wish to estimate MATH. Each point of MATH is distance at most MATH from MATH. Let MATH be any point within distance MATH of MATH and thus MATH is at most MATH from any point of MATH. Since MATH is the average of such distances, MATH. Since MATH is the point that minimizes MATH, MATH . Since MATH is the average of distances frm MATH to points in MATH, MATH . Since we have the bound MATH on the distance from any point of MATH to MATH, MATH . We have thus found a point of MATH within uniform distance MATH of every maximally pinched frontier of MATH. |
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 of the triangle as follows: count REF for every integral interior point, MATH for every non - vertex integral boundary point, and MATH for every integral vertex point except MATH. The character MATH is defined as follows. The character MATH is the character of CITE, pg. REF; let MATH be the MATH - power of MATH. The character is of order MATH, thus MATH takes values in the integers modulo MATH. We can write MATH for some MATH. Since MATH is isomorphic to MATH and MATH is prime, we may consider MATH to be the generator of MATH, written as REF. Then we have MATH. By considering triangles, CITE found that MATH we can further find that MATH . We will abbreviate the first of these polynomials as MATH and the second as MATH. Elementary algebra reveals that the minimum for MATH over MATH occurs at MATH; the closest integer point to that is MATH, and MATH which is negative for all MATH. The only positive value of either MATH or MATH (for MATH) occurs at MATH and MATH respectively, and MATH . We now need to find an appropriate metabolizer and character. A basis for the vector space MATH comes from the isomorphism MATH . Let MATH be a metabolizer of MATH; MATH is isomorphic to MATH. Linear algebra gives us a basis MATH for MATH, written in the given basis for MATH. Taking the sum of these basis elements, every metabolizer contains the element MATH . Multiplying the above element by MATH, we see that MATH also contains, for some MATH's, MATH . Let MATH . Note that MATH is element - wise linking with an element in the metabolizer MATH, and thus is trivial on MATH. We have MATH . We apply REF from CITE and the triangle inequality to calculate MATH . Thus we get that MATH . The expression on the right is strictly negative for all MATH. Therefore MATH is non - zero for all MATH, so MATH is of infinite order in MATH. |
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 NAME polynomial must factor as MATH. Thus in order that our knot to be slice, we must have that each of the MATH are even. Let MATH. The two - fold branched cover of the knot MATH is MATH. Let MATH be a character MATH . Since there is no non - trivial map MATH for MATH, the character MATH is zero on MATH. We see by the proof of REF, that if MATH is zero, then MATH. Therefore MATH . Thus by the proof of REF above, there is a MATH such that MATH so MATH and so MATH is not slice. |
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 character MATH, we have MATH . It is shown in CITE that characters MATH are in a one - to - one correspondence with characters MATH, thus we abuse notation and use MATH to denote both these characters. The invariant MATH of a knot is defined in CITE; it measures the algebraic sliceness of a knot. The consequence of NAME 's formula is that if we assume that MATH is algebraically slice, then MATH and MATH are zero, and thus MATH and so REF suffices to prove the result. |
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 equal to MATH shows that MATH is finitely dominated. |
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 shall use three general facts about mapping cylinders: CASE: If MATH and MATH are maps and MATH is homotopic to MATH, the mapping cylinder MATH is homotopy equivalent rel MATH to the concatenation of the mapping cylinders MATH and MATH rel MATH. CASE: If MATH with MATH, then the mapping cylinder of MATH is homotopy equivalent to the mapping cylinder of MATH rel MATH. CASE: Every mapping cylinder is homotopy equivalent to its base rel the base. The mapping telescope of a map MATH is the countable union MATH . For any maps MATH, MATH we have MATH with MATH a deformation retract, so that MATH . To see why this holds, note that MATH is homotopy equivalent to an infinite concatenation of alternating MATH's and MATH's which can also be thought of as an infinite concatenation of MATH's and MATH's. Essentially, we're reassociating an infinite product. Here is a picture of this part of the construction. CASE: If MATH, MATH are such that MATH there is defined a homotopy idempotent of a finite MATH complex MATH with MATH. We have homotopy equivalences MATH . The mapping telescope MATH is a countable MATH complex. CASE: As for REF , but with MATH-dimensional. |
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 MATH then the maps MATH are such that MATH, and MATH is dominated by MATH. |
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 MATH and MATH times if MATH, where MATH . In the expression for MATH we have made a change of variables MATH and in MATH a change of variables MATH. Hence it follows that MATH . Moreover, MATH for MATH since then MATH. On the other hand, MATH and MATH occurs in REF exactly MATH times if MATH and MATH times if MATH, where MATH . Thus we are done as soon as we have shown that MATH for each MATH. We distinguish two cases. If MATH is even, then we substitute MATH, MATH, and MATH, MATH in the above expression for MATH and arrive at MATH . If MATH is odd, then we substitute MATH, MATH, and MATH, MATH in the expression for MATH and obtain MATH which also completes the proof. |
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 assertion. |
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 the last matrix from the left and the right with the diagonal matrix MATH we obtain the matrix MATH. Notice in this connection that MATH since MATH. |
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 suffices to show that the following three equations hold: MATH . Notice that REF can be obtained from REF by passing to the transpose. Moreover, by employing REF reduces to MATH . Let us first prove REF . We introduce the MATH column vectors MATH and MATH. Then MATH whence indeed REF follows. Next we remark that from the definition of the sequences MATH and MATH it follows that MATH for all MATH. Introducing the column vectors MATH, MATH and MATH, it can be readily verified that MATH . Using the above relation MATH, it follows that MATH . Since MATH by the same relation, this implies REF . |
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 defined as above it follows from REF that MATH. It remains to show that REF holds. Indeed, we have that MATH . By REF this is equal to 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 MATH. |
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 as above it becomes apparent that MATH is similar to a matrix MATH . From the identity MATH it follows that MATH and MATH. Hence MATH and MATH. Since MATH, this completes the proof. |
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 same cardinality as MATH. So the image MATH is generated by the appropriate images, say, MATH, MATH. For each element MATH there is a representation MATH where MATH are some words from MATH, elements MATH take values MATH or MATH, and where MATH. Let us MATH be the subgroup of MATH generated by this set MATH of elements MATH. Of course MATH and MATH. Furthermore MATH is of the same cardinality as MATH, and so is MATH. |
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. Let MATH . MATH is a countable group and contains MATH as a subnormal subgroup. Besides an isomorphic copy of MATH lies in MATH. Therefore we can use the point f from REF in order to build a special element MATH in the wreath product MATH such that the intersection MATH contains an isomorphic copy of MATH. Thus REF is proved for the group MATH . |
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 direct product of its MATH-subgroup, it is sufficient to prove REF for finite MATH-groups. Modifying REF from CITE for the case of an arbitrary MATH we obtain for each power MATH such a finite MATH-group MATH that its verbal subgroup MATH is of exponent MATH and is from the center of MATH. Then, as it is shown in the proof of REF, the diagonal MATH of wreath product MATH is contained in the verbal subgroup MATH of MATH. If MATH is a MATH-group MATH is a MATH-group, too. And there is an obvious embedding of MATH into MATH (onto MATH): MATH . The base subgroup MATH of MATH is normal in MATH and as a nilpotent group MATH itself has MATH as a subnormal subgroup. |
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 and obtain a new finite group MATH. The next finite groups MATH are found by induction. CASE: Let MATH be a finite complete group that cannot be subnormally embedded into the derived group of some finite group. We can find such a group according to the result of CITE. And let MATH be a finite group with and MATH. We can find such a group, too, because MATH and the set of finite groups generates MATH. The group MATH does not belong to MATH and does not have the mentioned subnormal embedding for MATH does not have the latter. Taking other values for MATH we will obtain infinitely many examples of group MATH. |
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 attention to MATH . A monomial here is MATH-divisible if and only if MATH . Given such a permutation, and given a congruence class MATH, we can define a new permutation MATH by MATH . Then the permutation MATH is determined by the permutations MATH, and we have MATH . It follows that MATH factors: MATH . Since each factor has the form associated to the density of the unitary group of dimension MATH, the result follows. |
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, and we define MATH . The ingredients MATH and MATH are determined as follows: MATH where MATH is the composition of the sign character of MATH with the natural projection MATH, and MATH is the natural sign character of MATH. Consider a signed permutation MATH with MATH. Consider MATH as a permutation of MATH and define MATH for MATH by MATH, MATH. (Thus, for instance, for MATH, MATH.) Then MATH must preserve the partition of MATH induced by MATH. The action of MATH on an individual piece of the partition is either as MATH, if MATH, or as MATH if MATH. Thus, the MATH-divisible part of the inner sum MATH factors; we obtain one MATH-type factor for each pair MATH with MATH, and one MATH-type factor for each MATH with MATH. It remains only to determine the nature of the factors (in particular, the character of MATH that occurs) we obtain in the eight cases. This is an easy case-by-case analysis which we omit. We thus deduce the following identities. For MATH odd: MATH . For MATH even: MATH . The theorem then follows by simple MATH-manipulation. |
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 this implies the desired result. To prove REF, one can express the integrals as determinants (see, for example, REF) then use the main lemma of CITE. An alternate proof, given as REF, involves expressing the integrals in terms of orthogonal polynomials on the unit circle. |
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 the root lattice of MATH, MATH is an affine NAME group. But then we can apply REF to conclude that MATH is generated by the reflections of MATH that fix MATH. The theorem follows immediately. |
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 fundamental chamber of MATH, and again the resulting point is unique. Taking MATH, we are done. |
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 under conjugation by MATH, we may assume that MATH. Fix a coset of MATH in MATH, and let MATH be the representative of that coset that preserves the fundamental chamber. If MATH, then MATH clearly normalizes MATH, and MATH . On the other hand, if MATH then MATH . The result follows. |
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 automorphisms, MATH is the product of the groups corresponding to the simple factors of MATH; we may thus assume that MATH is simple. In this case, we can explicitly verify that when MATH, MATH is nonempty (it is straightforward for the classical groups, and a short computation for the exceptional groups). For MATH, we have the following lemma (from REF): Let MATH be a simple (finite) NAME group, and let MATH. If MATH is the NAME number and MATH the NAME vector of MATH, then MATH is the centroid of the fundamental chamber of MATH, and is thus invariant under all diagram automorphisms of MATH. Thus when MATH, MATH is strictly in the interior of MATH, so any nontrivial element of MATH must come from a diagram automorphism; since this is impossible when the center of MATH is connected, we obtain the desired result for MATH. When MATH, MATH treats the highest root of MATH differently from the other roots, so any diagram automorphism preserving MATH must preserve the highest root. But this precludes the diagram automorphisms corresponding to the cosets of the root lattice in the weight lattice, giving the desired result. |
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 number of irreducible components of MATH. (Both sides are equal to MATH-number of finite curves at infinity-MATH.) The second property follows from the third property. For the third property and sufficiency see CITE. |
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 of MATH has no cycles. If the curve is not exceptional then it is the proper transform of a vertical curve. But we must have blown up at least three times on the vertical curve to get rid of cycles in the dual graph of MATH so in this case the self-intersection is MATH. |
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-intersection at most MATH and the proper transform of the MATH curve has self-intersection at most MATH. By REF , MATH must reduce to a Morrow configuration by a sequence of blow-downs. Thus MATH must contain a MATH curve MATH that blows down. By REF , the curve MATH must be a proper transform of a horizontal curve. The proper transform of each MATH curve has self-intersection at most MATH. Thus MATH must come from one of the MATH horizontal curves. As mentioned above, the proper transform of a MATH curve has self-intersection MATH so MATH must be the proper transform of a MATH curve, MATH. But MATH would intersect MATH, the MATH curve, MATH times and hence MATH since MATH. This is a contradiction so any horizontal curve of type MATH must be a MATH curve. |
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 MATH will be blown up at most once. If there are two such points to be blown up, then after blowing up there will be (in the dual graph) two non-neighbouring MATH curves with valency MATH. The complement of such a configuration cannot be MATH. This is proven by CITE as REF . Actually the result is stated for two MATH curves of valency REF but it applies to valency MATH. Thus, at most one point on MATH is blown up and MATH or MATH. We must show MATH cannot occur. Since there are at least four horizontal curves, if MATH, then MATH has valency at least MATH and every other curve has negative self-intersection. Furthermore, the only possible MATH curves must be horizontal curves, and these intersect MATH in MATH. As we attempt to blow down MATH to get to a Morrow configuration, the only curves that can be blown down will always be adjacent to MATH. Thus the intersection number of MATH will become positive and all other intersection numbers remain negative, so a Morrow configuration cannot be reached. Hence, MATH. |
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. Since there is no MATH curve, we have not reached a Morrow configuration. |
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 curves and at least another time for the intersection of the two MATH curves. Thus they have self-intersection MATH. CASE: Suppose two MATH curves meet on MATH. Then after blowing up (twice if the MATH curves meet at a tangent), the exceptional curves are retained and the final exceptional curve has self-intersection MATH, valency REF and two branches, which we will call MATH and MATH, consisting of the proper transforms of the two MATH curves and any other curves beyond these proper transforms all of which have self-intersection MATH. Thus we are in the situation of REF and we get a contradiction. CASE: Suppose two MATH curves meet MATH at distinct points. Then at least one of the MATH curves, MATH, must meet MATH at a point away from the MATH curves by REF . Also one of the MATH curves, MATH, must meet MATH away from the MATH curves and contain at least two points where it intersects the MATH curves and thus have self-intersection MATH after blowing up to break cycles. We are once more at the situation of REF where the valency MATH curve is MATH which has self-intersection MATH by REF , and the branches MATH and MATH are the proper transform of MATH and any curves beyond it, respectively the proper transform of MATH and any curves beyond it. Thus we have a contradiction. Notice that both cases apply to two MATH curves that may intersect at a tangent point, and shows that this situation is impossible. |
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 applies. CASE: MATH and MATH meet MATH at distinct points. Then one of them, say MATH, meets MATH at a point not on a MATH curve by REF . At least one MATH curve MATH meets MATH and MATH in distinct points. After breaking cycles, MATH and MATH have self-intersections MATH so REF applies again. |
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 there are more than two MATH curves in MATH, apply REF to two of them. If another MATH horizontal curve in MATH does not intersect these two MATH curves at their common two points of intersection then, by REF , it must meet both these MATH curves at the third MATH horizontal curve of MATH. So the first two MATH curves would meet there, which is a contradiction. |
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 horizontal curves contain the common intersection of the MATH curves. Each MATH curve also contains a distinguished point where the curve meets the third MATH horizontal curve. NAME transformation can bring such a configuration to that in REF by blowing up at the two points of intersection of the MATH curves and blowing down the two vertical lines containing the two points. This sends two of the MATH horizontal curves and each MATH curve to MATH horizontal curves and one of the MATH curves to a MATH curve that intersects each of the other horizontal curves exactly once. Note that since we blow up MATH to get the polynomial map, two configurations of curves MATH in MATH related by a NAME transformation give rise to the same polynomial, so we are done. |
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 number of cutting divisors needed to do this is MATH. This is because each of the MATH vertical curves different from MATH must be separated from all but one of the MATH horizontal curves, so we need MATH cutting divisors. Also, the MATH horizontal curve meets each of the MATH horizontal curves and each of the MATH vertical curves once, so that requires MATH cutting divisors (by REF the MATH curve must meet MATH at a triple point with a MATH horizontal curve, so this intersection does not produce a cycle to be broken). We would thus require MATH cutting divisors except that the MATH curve may pass through intersections of the MATH horizontal curves and the MATH vertical curves, so some of the cutting divisors may coincide. The most such intersections possible is MATH and we have then over-counted required cutting divisors by MATH. Hence we get at least MATH cutting divisors. Since the number MATH of cutting divisors is at least MATH, we have MATH, so MATH . The solutions of REF are MATH. Recall by REF that the MATH curve must meet MATH at a triple point with a MATH horizontal curve. Furthermore, by keeping track of when either inequality in REF is an equality, or one away from an equality, we can see that the MATH curve must meet any other MATH vertical curves at a triple point with a MATH horizontal curve. Thus, the only possible configurations for MATH are given in REF . |
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 MATH becomes an exceptional divisor in a new configuration MATH. When MATH is a cutting divisor this operation essentially removes a MATH vertical curve from MATH. |
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 configuration with MATH. As mentioned in the proof of REF the solution MATH gives equality in REF and so it cannot sustain an extra cutting divisor. Similarly the solution MATH is MATH away from equality in REF and so it cannot sustain two extra cutting divisors. Hence we get a contradiction and the lemma is proven. |
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 beyond the two horizontal curves also have self-intersection strictly less than MATH, and since MATH has self-intersection MATH and valence MATH this gives a contradiction by REF . The same argument applies to MATH and MATH together with MATH. |
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. This is because we must blow up between MATH and MATH to separate cycles, and also between MATH and MATH to break cycles and to maintain MATH and MATH. Thus if we blow down MATH the remaining branch beyond MATH consists of curves with self-intersection strictly less than MATH. Also MATH has self-intersection strictly less than MATH since we have to blow up the intersection between MATH and MATH and the intersection between MATH and MATH in order to break cycles and maintain MATH. After blowing down MATH, MATH has self-intersection MATH and valency MATH with two branches consisting of curves of self-intersection strictly less than MATH. Thus we can use REF to get a contradiction. |
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 before MATH only if MATH has self-intersection MATH and each curve beyond has self-intersection MATH. Thus, at most two branches beyond MATH, MATH do not consist of a MATH curve with a string of MATH curves beyond. If there are two such branches then the blow-ups that create them create corresponding branches beyond MATH (or possibly just decrease the intersection number at MATH). These two branches cannot be fully blown down until everything else connecting to the MATH vertex are blown down, but the vertex MATH and any branches beyond it cannot blow down first. Thus MATH cannot blow down to a Morrow configuration. Thus there is at most one such branch, proving the Lemma. |
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 the MATH branch blows down, so we simply swap the labels MATH and MATH. The non-separating blow-up sequence cannot occur on MATH or MATH because the resulting cutting divisor would not be sent to a finite value. If the non-separating blow-up sequence occurs on the branch beyond MATH then we must be able to blow down the branch beyond MATH, hence the branch must consist of MATH with self-intersection MATH. |
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 vertex is defined to be MATH . The dual graph for the curve configuration of REF has chains starting at the vertex MATH and MATH. We claim these chains have continued fractions evaluating to MATH and MATH respectively, where MATH are arbitrary positive integers with MATH. We describe the main ingredients of this calculation but leave the details to the reader. An easy induction shows that the initial separating blow-up sequence leads to chains at MATH and MATH with continued fractions MATH and MATH with positive coprime MATH and MATH. The non-separating blow-up sequence then changes the fraction MATH or MATH that it operates on as follows. If the non-separating blow-up sequence consists of MATH blow-ups at the end of the left chain then MATH is replaced by MATH with MATH and MATH. If the non-separating blow-up sequence is on the right then MATH is similarly changed instead. Renaming, we can describe this in terms of our chosen names MATH as follows. We either have MATH or MATH. If MATH the initial separating blow-up sequence leads to chains with continued fractions MATH and MATH and the non-separating blow-up sequence then consists of a sequence of MATH blowups extending the left chain (and changing its continued fraction to MATH). If MATH the continued fractions are MATH and MATH after the separating blowup and the non-separating blow-up consists of MATH blow-ups extending the right chain (and changing its continued fraction to MATH). To prove the Lemma we must show that the dual graph of our curve configuration blows down to a Morrow configuration. We can blow down the chains starting at MATH, MATH, to get a chain. To check that this chain is a Morrow configuration we must compute its determinant, which we can do with continued fractions as in CITE. We first replace the two end chains by vertices with the rational weights MATH and MATH determined by their continued fractions to get a chain of four vertices with weights MATH . Then, computing the continued fraction for this chain based at its right vertex gives MATH, showing that the determinant is MATH as desired, and completing the proof. |
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 edge determinant of the intervening edge is MATH. This corresponds to the fact that MATH, which we already know. Similarly, MATH is next to MATH for MATH so the weight MATH is determined by the edge determinant condition MATH. The ``total linking number" at the vertex corresponding to each horizontal curve (before blowing down MATH) is zero (terminology of CITE; this reflects the fact that the link component corresponding to the horizontal curve has zero linking number with the entire link at infinity, since at almost all points on a horizontal curve, the polynomial has no pole). The weight MATH is determined by the zero total linking number of MATH, giving MATH. For any MATH the fact that vertex MATH has zero total linking gives MATH. |
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-separating blow-up. Conversely, if the non-separating blow-ups were adjacent to that exceptional curve then the left chain was a single vertex, hence had integral continued fraction, so MATH. The argument for MATH is the same. |
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-separating blow-up sequence. The MATH horizontal curve is a priori the graph of a linear map MATH but can be positioned as the graph of MATH by by an automorphisms of the image MATH. The point in the configuration space of the Theorem determines the placement of the horizontal curves MATH (after putting the MATH curve at MATH and the MATH curve at MATH). The fibre MATH determines the sequence of points for the non-separating blow-up sequence. This proves the Theorem, except that we need to be careful, since some diagrams occur in the form of REF in two different ways, which might seem to lead to disconnected moduli space. But the only cases that appear twice have four horizontal curves and the configurations MATH are related by NAME transformations. |
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 blown up at least twice and since all curves beyond these horizontal curves (exceptional curves or vertical curves) have self-intersection MATH we would get two branches MATH and MATH made up of the proper transforms of these two MATH horizontal curves and all curves beyond these, meeting at a valency MATH curve, MATH, with self-intersection MATH. This is the impossible situation of REF . There can be at most two vertical curves since if there are MATH vertical curves we need to break MATH cycles but since there are precisely three MATH horizontal curves, we begin with MATH curves so we can break at most MATH cycles by REF . Therefore MATH so MATH. The lemma follows when we get rid of the case of two vertical curves and three MATH horizontal curves. The few cases are easily dismissed by hand. |
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 horizontal curves whose proper transforms have self-intersection MATH and meet MATH. By REF any curves beyond these horizontal curves have self-intersection MATH. A horizontal curve of type MATH must meet MATH at exactly one point, and hence with a tangency of order MATH or at a singularity of the curve. This is because if a horizontal curve were to meet MATH twice then we would not be able to break cycles since when we blow up next to MATH, those exceptional curves are sent to infinity under the polynomial and hence must be retained in the configuration of curves. Thus we must blow up there to get a configuration of curves with normal intersections. The final exceptional curve in such a sequence of blow-ups will have self-intersection MATH and valency MATH. If we can blow down the configuration of curves then eventually at least one curve adjacent to the MATH curve is blown down and hence the MATH curve ends up with non-negative self-intersection. But the final configuration is not a linear graph since the proper transforms of the two MATH horizontal curves and any curves beyond give two branches. Thus the final configuration is not a Morrow configuration which contradicts REF . |
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 horizontal curves will be blown up at least MATH times and their proper transforms will have self-intersection MATH. This contradicts the previous proposition so the result follows. |
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 gauge transformation that MATH. The zeroth cohomology MATH then consists of REF which are constant diagonal matrices of the form MATH . In particular, this implies that MATH and MATH is identified with the kernel of MATH acting on MATH-valued forms. Since the dimension of the kernel of a continuous family of NAME operators is upper semicontinuous, MATH is a bounded function on the compact set MATH . Since we are using the MATH convention, it follows that every MATH is contained in a neighborhood MATH such that MATH for all MATH. Taking one such neighborhood for each MATH gives an open covering of MATH . Using compactness to pass to a finite subcover, we conclude that the function MATH is bounded above and below. |
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 additivity of spectral flow and REF that MATH for all MATH. This shows MATH is independent of the choice of MATH satisfying REF . To show MATH, we claim that MATH is divisible by REF for all MATH . Additivity of the spectral flow gives that REF equals MATH . We claim that each of these three terms is divisible by REF. Divisibility of the first term follows because the MATH action on MATH gives rise to a complex structure with which MATH commutes for each t, where MATH is a path from MATH to MATH. This implies MATH is even. Divisibility of the second and third terms is a consequence of the following claim. Claim. If MATH and MATH are MATH connections, then MATH and MATH are divisible by four. To see this, identify MATH with MATH the unit quaternions, and MATH with MATH, the quaternions. The regular representation of MATH on MATH can then be viewed as left multiplication in MATH, and it follows that right multiplication in MATH endows each eigenspace of MATH with a quaternionic structure. This proves the claim. |
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