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math/9904014 | Let MATH be the set of all associated semisimple pairs. Obviously, MATH. It follows from REF that MATH is unipotent and therefore MATH is closed in MATH. Since MATH is finite, MATH. On the other hand, MATH. Therefore MATH. Recall that MATH is the commuting variety. Hence, the assertion is equivalent to that MATH is irreducible and of dimension MATH. Our next analysis relies on the structure of MATH described in REF . We have MATH, where MATH and MATH. In both cases described in REF , one has MATH is Abelian and MATH. Let MATH, where MATH, MATH. The MATH-coordinate of the commutator is equal to MATH. Therefore MATH with MATH. Vanishing of the MATH-component yields the equation MATH . Having fixed MATH, we obtain a system of linear equations for MATH. More precisely, consider the family of linear mappings MATH, MATH. Then MATH. By REF , MATH is of dimension MATH. That is, MATH is onto. It follows that MATH for all but finitely many MATH. Therefore MATH has a unique irreducible component passing through MATH and MATH. Recall that the bi-weights of MATH lie in an open half-space in MATH. Therefore there exists a REF-parameter subgroup in the maximal torus MATH which contracts everything in the affine space MATH to the point MATH. Hence MATH is a cone with vertex MATH. Thus, MATH is irreducible and of dimension MATH. |
math/9904014 | From REF , it follows that MATH is a semi-direct product of the unipotent group MATH and a finite group MATH. CASE: Take an arbitrary MATH. It is a semisimple element of finite order. Since MATH is an associated semisimple pair for MATH, it follows from REF that MATH for some MATH. Hence MATH. By REF , one may assume that MATH is NAME in MATH. Since MATH and MATH is regular nilpotent in MATH (see REF), MATH is in the centre of MATH. Because MATH and MATH generate the parabolic subalgebra MATH and MATH, we get MATH for any MATH. This clearly implies that MATH is in the centre of MATH. Since MATH is adjoint, we obtain MATH. CASE: By REF , MATH contains a semisimple element of order two. |
math/9904014 | CASE: Suppose a MATH-triple MATH commutes with MATH. Then we may choose a MATH-triple containing MATH inside of the reductive algebra MATH. CASE: This readily follows from the NAME theory. CASE: In this case MATH satisfy commutator relations REF . From REF , we then conclude that MATH is MATH-conjugate to MATH. |
math/9904014 | The proof is much the same as for the previous assertion. Take a nilpotent element MATH. It then follows from REF that MATH . If MATH, then we must have MATH and MATH for ``parity" reason. Thus MATH is regular in MATH and hence in MATH. This argument can be reversed. |
math/9904014 | CASE: The space MATH possesses the MATH-filtration and MATH. It follows from the definition of MATH-limit that MATH. Furthermore, MATH. Under our assumption, this means that MATH and the eigenvalues of MATH on MATH are nonnegative integers. Assume that MATH for some MATH. Since MATH is killed by some power of MATH, we have MATH is the eigenvalue of MATH on MATH for some MATH, which is impossible. Thus, all the eigenvalues of MATH must be integral. CASE: Set MATH and consider the linear map MATH. By REF , we have MATH. That is, dimension of the kernel is MATH. Thus, MATH must be onto. |
math/9904014 | Since MATH and MATH, we have MATH and MATH. That is, the property of being a dual pair is equivalent to that MATH. We first prove that MATH commutes with MATH. REF says that MATH commutes with MATH. Therefore the subalgebras generated by MATH, MATH and MATH commute. By REF , the subalgebra generated by MATH and MATH is MATH. Under REF , the same applies to MATH in place of MATH. That is, the subalgebra generated by MATH and MATH is MATH. Consider the set MATH. It is immediate that MATH is MATH- and MATH-stable MATH. Assume that MATH, that is, MATH for some MATH and MATH. By successively applying MATH and MATH, we eventually obtain a nonzero commutator MATH with MATH and MATH such that MATH and MATH. It then follows from REF that MATH and MATH. Thus, MATH, MATH and one must have MATH. This contradiction proves that MATH. |
math/9904014 | CASE: Since MATH is Abelian in both cases, REF is satisfied. By REF , the other hypotheses are satisfied, too. The centre of MATH is equal to MATH. CASE: Clearly, MATH is reductive if and only if MATH is reductive. If MATH is reductive, then it contains a suitable MATH-triple together with MATH. The opposite implication follows from REF. CASE: By symmetry, it suffices to prove the first equality. Since MATH, we have MATH . In the proof of the opposite inclusion we use the relation MATH proved in the first part. Let MATH be an arbitrary element. One has to prove that MATH for all MATH. By REF , MATH is just the centre of MATH. Because MATH lies in the connected group MATH, it commutes with MATH. By the very definition, MATH commutes with MATH. Thus, it commutes with MATH. It then follows from REF that MATH for MATH. Consider MATH. Suppose MATH. Choose an element MATH which is killed by the least possible power, say MATH, of MATH. That is, MATH and MATH. Since MATH, we have MATH. Then MATH and hence MATH. In other words, MATH. It follows that MATH and MATH. Therefore MATH for all MATH. However, we have MATH for some MATH and therefore MATH must be zero. This contradiction proves that MATH. CASE: It follows from REF that MATH is semisimple. Assume that MATH, where MATH is a proper regular subalgebra of MATH. Then there exists a maximal semisimple subalgebra MATH such that MATH. This MATH is a regular subalgebra of MATH, too. According to the description of maximal regular semisimple subalgebras of MATH, MATH is contained in the fixed-point subalgebra of some element MATH of prime order (MATH). Then MATH. However, MATH, since MATH is connected and unipotent CITE. |
math/9904014 | CASE: The argument is close to that in REF. By definition, MATH, where MATH consists of nilpotent elements. As MATH, we have MATH. Thus, MATH. In the rectangular case, MATH is reductive. Hence MATH is reductive, too. This clearly forces that MATH. CASE: Since MATH is semisimple, the previous equality means MATH is a NAME subalgebra in MATH. Because MATH is assumed to be even in MATH, the MATH-triple MATH is principal in MATH. CASE: As in the proof of REF , it is enough to prove that MATH. It follows from REF that MATH is generated by MATH, and MATH as NAME algebra. Since MATH and MATH is the centre of MATH, we see that MATH commutes with the stuff just described. |
math/9904014 | CASE: Since MATH, taking the double centralizer gives MATH. Whence MATH is NAME in MATH. Next, MATH, which means MATH is regular in MATH. The centralizer of MATH in MATH is equal to MATH, the centre of MATH. That is, MATH is distinguished in MATH. Since any distinguished element is even (see for example, CITE), the assertion follows. CASE: As MATH, we obtain MATH. In view of REF , MATH is generated by MATH, and MATH as NAME algebra. By definition, MATH centralizes MATH and MATH; and MATH centralizes MATH, because MATH is the center of MATH and MATH is contained in the connected group MATH. Hence MATH. |
math/9904014 | Note first that MATH is the centre of MATH for any MATH. CASE: By definition, the linear space MATH is generated by all elements of the form MATH REF that lie in MATH. Let MATH and MATH. If MATH and MATH, then MATH. CASE: If MATH is even, then MATH. Since MATH, we conclude that MATH. Hence MATH, by the first claim. |
math/9904014 | CASE: Let MATH be the centre of MATH. Then MATH and therefore MATH. In case MATH is even, MATH is generated by MATH, and MATH as NAME algebra. Hence MATH is in the centre of MATH, that is, MATH. CASE: Since MATH and MATH, the assertion follows from the previous Lemma. |
math/9904014 | CASE: Since MATH, the affine space MATH is transversal to the orbit MATH at MATH. Consider the subspace MATH. Since MATH is a principal MATH-triple in MATH (see REF), MATH is a section of the open sheet in MATH. This is a classical result of CITE. Therefore almost all elements in MATH are semisimple and MATH-conjugate to elements in MATH, the latter being both a NAME subalgebra in MATH and the centre of MATH. It follows that MATH and MATH. Consider REF-parameter group MATH, where MATH. It is easily seen that MATH is MATH-stable and MATH for all MATH. Whence MATH. Because MATH is assumed to be even and hence MATH, all MATH-orbits intersecting MATH have the same dimension. Thus MATH. Our next argument relies on results of CITE. He studied the variety MATH for an arbitrary sheet MATH containing an arbitrary nilpotent element MATH. By CITE, we have MATH is closed in MATH, MATH the MATH-orbits in MATH intersect MATH transversally, MATH for any irreducible component MATH of MATH. Applying this to MATH and the irreducible components MATH, we see that MATH. Since MATH and MATH, we have MATH is an irreducible component of MATH and MATH. It follows from transversality condition that the natural map MATH is smooth and hence MATH is smooth, too. CASE: By CITE, the connected group MATH acts trivially on MATH or, equivalently, MATH is contained in MATH. Therefore MATH . Whence MATH. CASE: By CITE, two points MATH lie in the same MATH-orbit if and only if these lie in the same MATH-orbit. Thus, MATH is a section of MATH if and only if MATH acts trivially on MATH. Let MATH be a generic point in MATH. Then MATH is a regular semisimple element in MATH and hence MATH contains a point MATH. We have MATH and MATH for some MATH. Then MATH. By REF, the subgroups MATH and MATH commute. Hence MATH and we are done. CASE: Let MATH be an arbitrary sheet containing MATH. Arguing as in the proof of REF , we obtain MATH. Therefore MATH. Since MATH by NAME 's result, we must have MATH. |
math/9904014 | By REF , each member of a rectangular pn-pair is excellent. |
math/9904016 | Write MATH where MATH and MATH are just maps of the complex plane. Since MATH corresponds to a complete global analytic function MATH, we need to verify that MATH corresponds to some other complete global analytic function MATH. From our definitions we see that MATH is obtained from MATH by substituting each function element MATH by MATH. It is easily checked that MATH is open and MATH is holomorphic for both of the transformations being considered. Thus MATH remains a valid function element. One also verifies that the direct analytic continuation relationships among function elements are unchanged by these transformations. |
math/9904016 | Without loss of generality let MATH and MATH lie on the real axes of, respectively, MATH and MATH. The functions MATH of the function elements MATH, which represent MATH locally, will then have power series on the real axis (of MATH) with real coefficients. Since a real power series when analytically continued along the real axis continues to be real, we can continue MATH until we encounter either a singularity of MATH or a zero of MATH (that is, a singularity of MATH on the real axis of MATH). |
math/9904016 | Near the point of intersection MATH is represented by a function element MATH where MATH is conformal. The equality of angles, formed by a pair of lines in MATH and their images by MATH in MATH, is simply the geometrical statement that MATH is conformal. |
math/9904016 | Suppose MATH and MATH intersect with angle MATH on a nontrivial NAME surface MATH; for convenience, let MATH be the point of intersection. These curves are fixed by NAME reflections MATH and MATH respectively, and MATH is an isometry of MATH. The neighborhood of the point of intersection is the graph MATH where MATH is a neighborhood of the origin in MATH, and MATH is conformal at MATH. The NAME series for MATH at the origin has the form MATH where MATH. A short calculation shows MATH where MATH is again a neighborhood of the origin, and MATH . Since MATH is an isometry, the NAME series for MATH and MATH must agree, term by term. Now if MATH and MATH is irrational, then MATH can be an integer only for MATH (so that MATH). But this requires MATH for MATH which is impossible since MATH is nontrivial. Thus we may assume MATH where MATH and MATH are relatively prime positive integers, MATH (since MATH), and MATH for some integer MATH. Now let MATH be any real curve that intersects MATH at the origin; then MATH by the argument just given, where MATH and MATH are relatively prime positive integers, MATH. However, since MATH, we must have MATH, or that MATH divides the product MATH. This shows that MATH is a multiple of MATH, for some MATH. |
math/9904016 | Suppose MATH contains a real curve and call its completion MATH. We recall that MATH, and its closure in MATH, MATH, are straight lines and MATH cannot have an endpoint within MATH REF . The possible geometrical relationships between MATH and MATH are diagrammed in REF . Since MATH is nontrivial, at least two vertices are singular and are shown circled in each diagram. Either MATH intersects two edges of MATH, as in cases MATH and MATH, or, it intersects an edge and the opposite vertex which may be singular (case MATH) or possibly regular (case MATH). The vertex labels on the diagram refer to our notation for the vertex angles and edges. For example, MATH and MATH are the angles in MATH and MATH, respectively, of vertex REF; MATH is the edge (real curve) bounded by vertices REF, etc. Case MATH is easily disposed of using REF : MATH . This is impossible because vertex REF is singular (MATH). By using NAME reflection to imply the existence of additional real curves, the remaining cases either reduce to case MATH or imply the existence of a singularity within MATH or one of its edges - neither of which is possible. First consider case MATH. Let MATH intersect MATH at MATH and MATH at MATH, forming angles MATH and MATH (see REF ). Any other complete real curve with projection MATH which intersects MATH makes a finite angle with MATH by REF . Thus we may assume the angles MATH and MATH are the smallest possible (for a MATH that intersects both MATH and MATH). Since one of MATH and MATH must be greater than MATH, we assume without loss of generality it is MATH. If we now reflect MATH in MATH we obtain a real curve with projection MATH such that MATH intersects MATH but not MATH. Thus case MATH always reduces to cases MATH or MATH. In case MATH we consider the sequence of real curves with projections MATH, where MATH, MATH, and MATH is the image of MATH under reflection in MATH. Let MATH be the angle formed at vertex REF in MATH by MATH. Clearly for some MATH we arrive at a MATH such that MATH (see REF ). This leads to three subcases: MATH, where MATH, MATH, where MATH and MATH are perpendicular, and MATH, where MATH. In case MATH, MATH since otherwise MATH would have a nontrivial isometry (reflection in MATH). All three subcases immediately lead to contradictions. In MATH, reflecting MATH in MATH presents us with a MATH satisfying case MATH. In MATH, the image of vertex REF under reflection in MATH implies a singularity within MATH; in MATH the same reflection implies a singularity on MATH. REF we either have MATH, case MATH, or MATH, case MATH. Since MATH has no nontrivial isometry, a reflection in MATH in case MATH would place the image of either vertex REF or REF (both singular) somewhere on MATH. In MATH, a NAME reflection of MATH leads to case MATH. |
math/9904016 | We use the real curves to decompose MATH into a set of tiles REF MATH. REF tells us that MATH, so that MATH, where MATH is the edge group of MATH. On the other hand, if MATH is an isometry of MATH, then MATH, where MATH for some MATH. But since MATH has no nontrivial isometries, the map MATH must be the identity and MATH. |
math/9904016 | If MATH is finite we can view MATH as a finite cell complex. We can relate the number of REF, MATH, and the number of REF, MATH, in this complex to the number of REF, MATH. Since every REF-cell is bounded by three REF-cells, each of which bounds exactly one other REF-cell, MATH. Similarly, the boundary of each REF-cell contains three REF-cells (the vertices MATH), each of which belongs to the boundary of a number of REF equal to the order of the corresponding vertex group, MATH. Thus MATH. Finally, since MATH has no nontrivial isometry, and MATH acts transitively on REF-cells of MATH, MATH. The result REF follows from NAME 's formula, MATH. |
math/9904016 | If MATH is not discrete we can find a sequence of MATH such that both MATH and MATH. Let MATH be a regular point of MATH, then MATH. Near MATH we can represent MATH by the graph MATH where MATH is a neighborhood of MATH in MATH, MATH is conformal in MATH, and MATH. Since MATH is an isometry, MATH. From MATH we see that MATH, since MATH as MATH can be arbitrarily small. Since MATH is discrete, MATH is isolated in MATH and there must be a subsequence MATH such that MATH. If, within the sequence MATH, there is a subsequence MATH with MATH, then MATH and we have a contradiction. Thus there must be a subsequence with MATH. Since MATH is conformal at MATH, MATH . But the second limit, above, is independent of MATH so we are forced to conclude that MATH is constant. This is impossible because MATH is nontrivial. |
math/9904016 | Let MATH be an open ball of radius MATH centered at an arbitrary point MATH. Consider the piece of the NAME surface within this ball, MATH, and the projection MATH. MATH is finitely discrete if there is a uniform upper bound on the number of preimages MATH. MATH is covered by the orbit of closed graphs, MATH, where MATH is the edge group of MATH. Let MATH be the lattice group of MATH, then MATH is the union of cosets, MATH, MATH, where MATH is finite because MATH is crystallographic. Again, because MATH is crystallographic, all but finitely many graphs in MATH have empty intersection with a ball of radius MATH, in particular, MATH. Thus we have a bound (independent of MATH and MATH) on the number of graphs in MATH which intersect MATH. But a graph can have at most one preimage of MATH; hence MATH is uniformly bounded above. |
math/9904016 | MATH is the union of countably many closed graphs MATH given by the orbit of MATH under the action of the edge group. Since each graph has at most three singular points, MATH has countably many branch points. If there were uncountably many self-intersection points then uncountably many must arise from one pair of distinct graphs, say MATH and MATH. Let MATH and MATH be the corresponding conformal maps; then MATH would have uncountably many solutions MATH. Thus either MATH, a contradiction, or the zeroes of MATH would not be isolated, another impossibility. |
math/9904016 | We arrive at MATH by composing bijections MATH such that MATH and MATH are (correspondingly) uniformly bounded. The Lemma then follows by application of the triangle inequality. Since MATH is crystallographic, we can partition MATH into translates of a bounded fundamental region, MATH, of its lattice MATH. Thus for any pair MATH we can write MATH where MATH. Consider the point MATH . Since MATH both MATH and MATH have upper bounds independent of MATH and MATH. Now MATH and MATH . But MATH iff MATH, where MATH. Choosing MATH we obtain MATH . As our first bijection we take the translation MATH, where MATH is uniformly bounded from above. The point of this intermediate step is that for MATH we need consider only pairs of sections with bounded separation MATH. In constructing MATH we avoid branch points and crossing points. Since MATH, REF implies MATH. Let MATH be a smooth rectifiable curve with MATH and MATH. To show that MATH exists we recall that MATH is countable. We can then find MATH in the uncountable family of circular arcs with endpoints MATH and MATH, since each point of MATH can eliminate at most one arc. The curve MATH generates a homotopy of the sections MATH and MATH. At each point MATH, MATH is lifted to a unique curve MATH with endpoint MATH and we define our second bijection by MATH. To finish the proof we need to show that MATH is uniformly bounded. Let MATH be the closed subset of MATH that is a suitably small distance MATH or greater from any of its vertices that are branch points of MATH. Let MATH be the corresponding graph. The orbit under the edge group, MATH, is a NAME surface from which all the points of ramification (of the map MATH) have been ``cut out". The complement, MATH, is the disjoint union of the branched neighborhoods of all the points of ramification. It is possible to find curves MATH, such as the circular arcs considered above, where the branched neighborhoods MATH visited by MATH are visited only once, for MATH. Also, because we can bound the length MATH of MATH and there is a minimum distance between branch points (on the branched covering of MATH), the number of such subintervals MATH is bounded, that is, MATH. The bounds on MATH and MATH are uniform bounds, independent of the points MATH and MATH. Two additional bounds are needed before we can proceed to bound MATH. The first is an upper bound MATH on the diameter of the projection of a branched neighborhood, MATH. This follows from the fact that MATH is isometric with the branched neighborhood MATH of a vertex of MATH, and MATH, where MATH is the corresponding vertex group. Clearly the maximum diameter of MATH is bounded because MATH is bounded. The map MATH (which defines MATH), when restricted to MATH is conformal and MATH has a maximum value, MATH, since MATH is closed. This means that if MATH, then MATH . Because MATH is generated from MATH by the action of MATH, this bounds applies globally, for MATH. We are now ready to complete the proof: MATH . For each piece of the curve in a branched neighborhood we have MATH while in the complement (MATH), MATH . REF thus becomes MATH . |
math/9904016 | The expression MATH indicates there exist constants MATH and MATH (independent of MATH) such that for sufficiently large MATH, MATH. Let MATH be a disk of radius MATH centered at the origin and let MATH . We first obtain a bound on the difference, MATH, when neither MATH nor MATH is a crossing point or a branch point of MATH. By REF there exists a bijection MATH such that if MATH, then MATH, where MATH is a constant independent of MATH, MATH, and MATH. This shows MATH . We can cover the annulus MATH by MATH disks MATH of a fixed radius MATH, where, for sufficiently large MATH, MATH and MATH is a constant independent of MATH. By REF , MATH, where MATH is independent of MATH and MATH. Thus MATH. Combining this bound with the bound obtained by interchanging MATH and MATH, we arrive at the statement MATH . We now introduce a disk MATH and consider the region MATH. Since the set of branch points and crossing points is countable and has zero measure in MATH, and MATH is otherwise smooth, MATH . Because MATH is crystallographic, we can partition MATH into translates of a bounded fundamental region of its lattice, MATH. Let MATH be the standard projection on the quotient. On MATH the map MATH is REF-to-REF and MATH . This defines MATH, which we can make positive by appropriate choice of orientation on MATH. Turning now to the region MATH, there is a maximal subset MATH such that MATH and a smallest subset MATH such that MATH. If MATH, then MATH is a proper subset of a fundamental region and MATH . From this it follows that MATH and, from straightforward estimates of MATH and MATH, we conclude MATH . The projection MATH covers the disk MATH multiple times, the multiplicity at the point MATH being the number MATH defined above. Thus MATH . We can again neglect the countable set of branch points MATH and crossing points MATH to argue, for MATH fixed, MATH where in the last step we used REF. Combining REF, and using MATH, we obtain MATH and thus MATH . To evaluate MATH from REF, we regard MATH as MATH equivalence classes of tiles, all isometric to MATH. The result REF follows because the integral of the form MATH over MATH is just the volume of MATH in MATH. |
math/9904016 | First consider the case MATH; then MATH . Now consider the group MATH, where MATH is normal in MATH. Clearly MATH. Moreover, one easily verifies MATH, where MATH is any of the three generators of MATH. These two facts together show MATH; MATH is clearly the lattice group of MATH. Next consider the case MATH. For the generators of MATH we must now use REF. Consider the group MATH, where MATH is normal in MATH. Clearly MATH. In contrast to the previous case, we can now only verify that MATH, where MATH is any of the three generators in REF. Thus MATH is normal in MATH. MATH has index at most two, since multiplication of MATH by the generators of MATH produces at most two, possibly distinct, cosets: MATH and MATH. But if MATH, then we would have some MATH and some MATH such that MATH, or MATH. Since MATH fixes the origin, MATH must be the trivial translation and MATH. This contradicts our assumption MATH and we conclude that MATH and MATH are distinct. Let MATH be the lattice group of MATH. Clearly MATH. Now suppose MATH but MATH. Since then MATH, we must have MATH, that is, MATH for some MATH and MATH. But this implies MATH, a translation, and we arrive at the contradiction MATH. Thus MATH. |
math/9904016 | By REF , MATH has a discrete lattice, that is, the orbit MATH is discrete in MATH. For right triangles, REF gives us MATH if MATH, MATH otherwise. Thus discreteness of MATH implies discreteness of the star MATH in MATH. Since MATH, MATH is just the orbit of MATH under action of the group MATH generated by MATH. Clearly MATH lies in a REF-torus MATH embedded in MATH. Since MATH is discrete, there is a disjoint union of neighborhoods, each containing just one element of MATH. Moreover, since MATH is an isometry of MATH and acts transitively on MATH, there is a uniform lower bound on the volumes of these neighborhoods. This implies the existence of disjoint neighborhoods in MATH, again with a uniform lower bound on their measure. Since MATH has finite measure, this is only possible if MATH has finite order. |
math/9904016 | Let MATH be the isometry group of MATH, MATH its derived point group, and MATH its lattice group. Since conjugation by MATH leaves MATH invariant, consider the automorphisms MATH given by MATH. If MATH has rank MATH, then the homomorphism MATH, where MATH, induces a representation of MATH by integral MATH matrices of determinant MATH. In fact, MATH is an isomorphism since for any of the generators MATH of MATH we can easily find a MATH which is not fixed by MATH (it suffices to look within the star MATH or, if MATH, MATH). Since MATH has order MATH, there must be an element of order MATH in MATH. By REF we must have MATH. On the other hand, if MATH is discrete, then MATH must be discrete (as a lattice in MATH) which is possible only if MATH. Since MATH, we need only consider the values of MATH for powers of small primes: MATH, MATH, MATH, MATH (all other primes and higher powers yield values greater than MATH). From these facts we obtain just the values of MATH given in the statement of the Lemma. |
math/9904016 | Suppose MATH divides MATH and MATH. We may assume that MATH does not divide MATH since otherwise MATH, MATH and MATH would have MATH as a common divisor. Let MATH; then MATH and MATH is a subgroup of MATH of order MATH. By looking in MATH (or MATH), we can find a translation MATH such that MATH and MATH. Now consider the two products of translations MATH where MATH is any element of the coset MATH, and MATH depends on the particular choice of coset elements. Evaluating the products we find MATH . REF implies MATH, and, since MATH does not divide MATH, we conclude MATH. Thus MATH. Similarly, we find MATH . On the other hand, MATH is changed just by making a different choice for one coset element MATH (again because MATH does not divide MATH). Thus we can always make a choice such that MATH, where MATH. Now MATH is a lattice in MATH isomorphic to the cyclotomic lattice MATH, while MATH is a lattice in MATH isomorphic to MATH. Since MATH, MATH. The statement of the Lemma follows from the well known formula for the rank of a cyclotomic lattice. |
math/9904016 | The properties listed in REF are simple consequences of general results. Since MATH we see that MATH is isomorphic to an abstract group generated by an element of order REF, MATH, and an element MATH, which has order MATH, if MATH is odd (and therefore MATH) or MATH is even and just one of MATH and MATH is odd (since then the element of order two in MATH is either MATH or MATH). Otherwise (MATH even, MATH and MATH both odd), MATH and MATH has order MATH. These generators have the relation MATH and imply MATH, the dihedral group of order MATH, or MATH. To compute the genus we use the orders of the vertex groups, REF, in REF : MATH . The geometry of MATH is completely specified by the NAME matrix formed from its generators MATH, MATH: MATH where MATH is the standard inner product. If we consider each MATH as a vector in MATH and form the MATH matrix MATH whose rows are MATH, then MATH and MATH. Using the LLL algorithm it was found that the generators of MATH and MATH (and hence MATH) could always be written as, respectively MATH . To help identify the lattice geometry it was sometimes necessary to define a new basis MATH, where MATH. The new NAME matrix is then given by MATH. Details of this analysis, for the seven combinations of MATH in REF , are provided in the appendix. The only additional data needed in the density REF is the triangle area MATH. The last column of REF identifies which surfaces have doubly periodic sections. In general, since the kernel of the homomorphism MATH is given by the lattice MATH, MATH . Surfaces with doubly periodic sections have MATH, while MATH corresponds to completely quasiperiodic sections. These are the only cases that occur, since (from REF), MATH, the cyclotomic lattice with rank MATH. |
math/9904036 | We proceed by induction on MATH, assuming in addition that the index of MATH is MATH. We just did it for MATH. Assume the construction is done for MATH. Let MATH be an integer as in the proposition, and set MATH . Since MATH, the integer MATH is positive. Also, MATH because MATH and MATH. It implies MATH hence there exists by induction a NAME variety MATH of dimension MATH, index MATH and NAME number MATH, such that MATH . Write MATH, with MATH ample on MATH, and let MATH with projection MATH, so that MATH. As above, it implies that MATH is a NAME variety of dimension MATH when MATH; note also that MATH when MATH, and MATH. We get again MATH . Note that MATH . If MATH, we obtain MATH if MATH, we get MATH . In all cases, MATH . It follows that MATH which proves the proposition. |
math/9904044 | We first replace MATH by its double-adjoint so that we can assume that MATH is closed (it is easy to check that REF remain valid). The problem is to show that it is self-adjoint. Let MATH be the range of the operator MATH. It is a closed subspace of MATH (as MATH, and MATH is closed). Let MATH be the bounded operator onto MATH which is orthogonal projection onto MATH followed with the inverse of MATH. One checks easily that MATH belongs to MATH, hence commutes with its adjoint MATH which will also belong to MATH. Any vector MATH in the kernel of MATH is then in the kernel of MATH (as MATH). So MATH belongs to the orthogonal complement to the range of MATH, that is MATH as the range of MATH is MATH. So MATH and in the same manner MATH. By the basic criterion for self-adjointness (see CITE), MATH is self-adjoint. Let MATH. It commutes with the resolvant MATH hence leaves stable its range MATH. On MATH one has MATH hence MATH so MATH commutes with MATH. |
math/9904044 | We give the proof for completeness. The commutation with MATH-translations is clear. Then MATH contains (the inverse NAME transform of) MATH and all its translates. Hence if MATH is orthogonal to MATH then the function MATH on MATH belongs to MATH and has a vanishing ``inverse NAME transform", hence MATH (almost everywhere). It is also clear using MATH that if MATH extends MATH, then MATH. Let us assume that the sequence MATH is such that MATH and MATH both exist. Let us pick a pointwise on MATH almost everywhere convergent subsequence MATH. Using NAME 's lemma we deduce that MATH belongs to MATH. Using NAME 's lemma again we get the vanishing of MATH, and this shows that MATH is a closed operator. Finally let MATH be in the domain of the adjoint of MATH. There exists then an element MATH of MATH such that for any MATH the equality MATH holds. This implies that the two following functions of MATH: MATH have the same NAME transform on MATH, hence are equal almost everywhere. So MATH and MATH. |
math/9904044 | Let us first assume that MATH is bounded. We use the (inverse NAME transform of the) function MATH and define MATH to be MATH. Let us consider the domain MATH consisting of all finite linear combinations of translates of MATH. It is dense by the argument using unicity of NAME transform in MATH we have used previously. Then MATH, hence MATH is also an extension of the closure of MATH. As MATH is assumed to be bounded this is MATH. But this means that MATH and that MATH (we then note that necessarily MATH is essentially bounded). The next case is when MATH is assumed to be self-adjoint. Its resolvents MATH and MATH are bounded and commute with MATH. Hence they correspond to multiplicators MATH and MATH. The kernel of MATH is orthogonal to the range of MATH which is all of MATH, so in fact it is reduced to MATH. Hence MATH is almost everywhere non-vanishing. Let MATH and MATH. As MATH we get MATH and defining MATH to be MATH we see that MATH is an extension of MATH. Taking the adjoints we deduce that MATH is an extension of MATH. So all three are equal (and MATH is real-valued). For the general case we use the theorem of polar decomposition (see for example CITE). There exists a non-negative self-adjoint operator MATH with the same domain as MATH and a partial isometry MATH such that MATH. Further conditions are satisfied which make MATH and MATH unique: so they also commute with MATH. It follows from what was proven previously that MATH for an appropriate MATH (the product of the multiplicators associated to the self-adjoint MATH and the bounded MATH). The adjoint MATH also has a dense domain and commutes with MATH, so in the same manner MATH for an appropriate MATH. The inclusion MATH implies MATH and MATH. But the double-adjoint coincides with the closed operator MATH. |
math/9904044 | One just checks that MATH intertwines MATH with MATH, and also MATH with MATH and that the inversion MATH also intertwines MATH with MATH, and MATH with MATH. |
math/9904044 | Let us take MATH and consider the linear operator on MATH: MATH . It commutes with the action of MATH hence stabilizes each MATH and is a multiple MATH of the identity there. On the other hand, if we choose MATH and MATH in MATH and consider MATH we obtain a bounded operator MATH on MATH commuting with dilations and such that MATH where the let-hand bracket is computed in MATH while the next two are in MATH. So MATH depends on MATH only through MATH. We then let MATH be the spectral multiplier associated to MATH for an arbitrary MATH satisfying MATH. |
math/9904044 | One applies REF . |
math/9904044 | If the function MATH on MATH is chosen smooth with compact support (so that MATH is entire) then, for any MATH the function MATH, viewed in the additive picture, is smooth on MATH, has compact support, and vanishes identically in a neighborhood of the origin. So its image under the inversion also belongs to the NAME class in the additive picture on MATH. Hence MATH can be written as MATH for some NAME function MATH of the additive variable MATH. One checks that this then implies that MATH is a NAME function of the variable MATH (we assume that MATH does not identically vanish of course), hence that MATH is a NAME function of MATH. The various allowable MATH's have no common zeros so the conclusion follows. |
math/9904044 | We have to check the identity: MATH for all NAME functions MATH with NAME Transform MATH. Both integrals are analytic in MATH, hence both sides are smooth (bounded) functions of MATH. It will be enough to prove the identity after integrating against MATH with an arbitrary NAME function MATH. With the notations of REF , we have to check MATH . But, by REF , and by REF , MATH is just the NAME Transform in MATH of MATH, so this reduces to the MATH-identity MATH . |
math/9904044 | We have to check an identity: MATH for all NAME functions MATH with NAME Transform MATH. Both integrals are analytic in the strip MATH, their ratio is thus a meromorphic function, which depends neither on MATH nor on MATH as it equals MATH on the critical line. Furthermore for any given MATH we can choose MATH, with MATH having very small support around MATH to see that this ratio is in fact analytic. |
math/9904044 | Let MATH and MATH. One checks that MATH (MATH). We choose as homogeneous function MATH. For these choices the identity of REF becomes MATH . Adding a suitable linear combinations of these identities for MATH gives MATH hence the result after evaluating the integrals in terms of MATH. |
math/9904044 | We have to show that MATH is a multiplier of the NAME class. Let MATH. Using REF and the partial fraction expansion of the logarithmic derivative of MATH (as in CITE for the real and complex NAME Gamma functions), or NAME 's formula, or any other means, one finds MATH, MATH, so that MATH. |
math/9904044 | Indeed, it commutes with MATH by REF and it commutes with MATH by construction. |
math/9904044 | Assuming the validity of the estimates we see that both sides of the identity are analytic functions of MATH, so it is enough to prove the identity on the critical line: MATH . As in the proof of REF , it is enough to prove it after integrating against an arbitrary NAME function MATH. With MATH, and using REF this becomes MATH (on the right-hand-side MATH is computed in the multiplicative picture, on the left-hand-side MATH is evaluated in the additive picture). The self-adjointness of MATH reduces this to MATH, which is a valid identity. For the proof of the estimates we observe that MATH is the NAME transform of a MATH-function hence is continuous, so that we only need to show that it is MATH for MATH. For this we use that MATH is additive convolution of MATH with the distribution MATH. The estimate then follows from the formula for MATH given in the next lemma. |
math/9904044 | We have used the notation MATH for the NAME 's constant (MATH). Let MATH for MATH be the homogeneous distribution MATH on MATH. It is a tempered distribution. The formula MATH defines its analytic continuation to MATH, with a simple pole at MATH. Using MATH for MATH, MATH, and expanding in MATH gives MATH . As MATH the result follows. |
math/9904049 | Denote the space obtained at stage MATH by MATH and organize the projections of the fiber squares of all stages as MATH . Then MATH and MATH is the proper transform of MATH in MATH if MATH, while MATH is the component of the exceptional divisor over MATH. The statement will follow once it has been shown that the stated sequence of blowups can indeed be performed. For this, it suffices to check that the centers of those simultaneous blowups will have indeed become disjoint after the previous stages of the construction. The proof will be done by induction on MATH, for all MATH at the same time; after stage MATH, the induction will stop for MATH, and it will continue on for MATH with MATH. For any pair of distinct partitions MATH and MATH of MATH into two blocks, their meet MATH is the `nonpartition', so MATH, the small diagonal of MATH. By REF, the transforms MATH and MATH will be disjoint, making the second stage possible. Assume that stage MATH has been performed; this means that the varieties MATH have been constructed for MATH, and only those for MATH are still being built. Also assume that the proper transforms MATH for MATH with MATH are disjoint. For each partition MATH with MATH, the projection MATH pulls back the obvious isomorphism MATH to an isomorphism MATH. All these subvarieties are disjoint by the inductive assumption, and can all be blown up at the same time. This defines the variety MATH. To provide the inductive step necessary to continue the construction of MATH for MATH, the intersection MATH must be empty for all pairs of distinct MATH, MATH in MATH with MATH. Such a pair automatically satisfies the noncontainment condition; so MATH is a clean intersection. Since MATH, a repeated use of REF shows that MATH is a clean intersection, and then REF implies that MATH is empty. The proper transforms of MATH and MATH become disjoint after stage MATH, and the proof is complete. |
math/9904049 | This has been obtained while proving the theorem, and is formulated separately only for the ease of future reference. |
math/9904049 | In a blowup MATH of a smooth algebraic variety MATH along a smooth center MATH, if MATH is the proper transform of a smooth variety MATH, then in terms of ideal sheaves MATH. Applied at each step, this equality yields MATH, where MATH denotes the composition of the stated blowups. |
math/9904049 | We concentrate on the normal crossing property, which implies the other claims. By construction, the proper transform of every polydiagonal MATH under MATH is a smooth divisor; it will be denoted by MATH. The proper transforms of MATH and MATH become disjoint when that of their intersection MATH is blown up, unless one of MATH and MATH contains the other, that is, unless MATH is a chain. In order to show that for any saturated (maximal length) chain MATH, the union MATH is a normal crossing divisor in MATH, consider the flag of polydiagonals MATH. The blowups of MATH for MATH are irrelevant for the intersection of the components of MATH because their centers are disjoint from MATH hence, the Flag Blowup Lemma can be applied. The normal crossing property of MATH follows by the lemma, and so does the proposition: since any chain MATH is refined by a saturated chain MATH, the components of MATH form a subset of components of MATH. |
math/9904049 | Straightforwardly from the construction of MATH and REF , MATH where MATH is the fiber of the normal bundle to MATH in MATH, which is MATH by an easy dimension count. REF converts this formula into MATH . Since MATH and MATH, there results a recurrence relation MATH and both claims immediately follow. |
math/9904049 | By induction on MATH, each MATH is the closure of MATH in MATH . The basis is clear: MATH. Then, MATH is the blowup of MATH along MATH or in other terms, along MATH . This ideal sheaf becomes MATH upon multiplying by an invertible ideal sheaf, and blowing up MATH is equivalent to taking the closure of the graph of the rational map from MATH to MATH . This provides the inductive step, and eventually MATH. |
math/9904049 | Start with notation for the products from REF : MATH where MATH is the number of essential blocks in a partition MATH. If MATH is the only essential block of MATH, then MATH, so MATH can indeed be used for MATH. MATH . Now take the left of these two diagrams, where MATH and MATH are rational maps defined on MATH, and notice that the projection MATH maps the closure MATH of the graph of MATH onto the closure MATH of the graph of MATH. |
math/9904049 | The normal space MATH at a point MATH to MATH is independent of MATH (assumed positive): its dimension is equal to the cardinality of the nest MATH . The nest alone determines the iterated blowup of MATH induced from REF , and the preimage of the origin under it is isomorphic to MATH. |
math/9904049 | Look at the equations of the large diagonals containing MATH, that is, MATH for all pairs of MATH and MATH belonging to the same block of MATH. Equations coming from different blocks of MATH are independent of each other, leading to the product decomposition. |
math/9904049 | The complement to MATH in MATH is MATH, the configuration space of MATH distinct labeled points in MATH modulo translations. Since MATH is the complement in MATH to the projectivization of the arrangement MATH, it follows that MATH is the orbit space of the diagonal action of MATH on this product by dilations. Separate actions of MATH on each factor together give that of MATH on MATH. Its total orbit space is isomorphic to the product of those coming from the factors, which is MATH. The orbit space MATH maps into this product, with fiber MATH. |
math/9904049 | CASE: The direct factor MATH is the small diagonal MATH. The essential shape of the bottom partition of MATH is the integer MATH, thus by definition, there is a map MATH. The bundle MATH is the pullback by MATH of the tautologial line bundle over MATH; since MATH is an iterated blowup, REF (formulated below) has to be used at each stage. CASE: Affine transformations identify any nondegenerate configuration in MATH with a degenerate one in which all MATH points collide at MATH, cancelling both the direct factor MATH and the fiber of the line bundle MATH. |
math/9904049 | The normal bundle MATH is the pullback MATH. |
math/9904049 | REF follows directly from the definition. CASE: Nothing to be done when MATH and MATH are disjoint. When MATH, denote by MATH the exceptional divisor of MATH, and by MATH the projection MATH, then MATH and the claim follows. |
math/9904049 | Similarly to REF , the definition of MATH implies REF can be checked by induction on MATH, where the inductive step follows by applying the case MATH. Thus, it is enough to show that each divisor MATH is isomorphic to MATH, where MATH and MATH. The argument is based on REF . Every partition from MATH belongs to one of the following six groups: MATH . The proof will be completed by studying the impact of blowups corresponding to partitions in each group on the stratum MATH of the arrangement MATH. Before the blowups, the arrangements induced in MATH and MATH are isomorphic respectively to MATH and MATH. First group, first stage. The exceptional divisor MATH of the first stage has a straight subvariety MATH with the projectivization of MATH in it, and with the arrangement MATH in each normal space to it REF . REF also apply at the subsequent stages, pulling back arrangements inside normal spaces and preserving the straightness of blowup centers. Group REF blowups are irrelevant for the divisor MATH at all stages, and no blowup corresponds to MATH. Group REF blowups turn MATH into MATH. Then the group REF blowup makes it into a divisor isomorphic to MATH. The second factor inherits the projectivization of MATH, and blowups of the remaining group REF transform this divisor into MATH. |
math/9904049 | REF gives MATH, where MATH. By REF , the arrangements MATH transform isomorphically from the normal spaces to MATH in MATH into the normal spaces to MATH in MATH. At the next stage, MATH is a bundle over MATH with fibers isomorphic to MATH. The relevant blowup centers of the subsequent stages are its subbundles; their fibers form in every fiber of MATH an arrangement isomorphic to the projectivization of MATH in MATH. Thus in the end, the fibers of MATH transform into MATH. If in addition MATH, a repeated application of REF shows that MATH is straight in MATH, so MATH and the result follows. |
math/9904049 | Put together REF . |
math/9904049 | Fix a partition MATH of MATH into two-element blocks and let the nest MATH be the set MATH of blocks of MATH. The map MATH takes the divisor MATH of MATH into the stratum MATH of MATH. The divisor is isomorphic to MATH by REF and the stratum is isomorphic to MATH. Since MATH, it follows that MATH maps to MATH. Tracing MATH stage by stage, first transform the factor MATH into MATH; then at stage MATH blow up the proper transform of MATH, turning the second factor into a MATH-bundle over MATH. Since MATH is a normal crossing divisor, the divisors MATH induce in each fiber MATH the projectivized coordinate hyperplane arrangement. All of its strata are blown up at the subsequent stages, turning MATH into MATH. |
math/9904049 | CASE: It is enough to show this for MATH and MATH. The required sublattice of MATH is generated by the union MATH, where the only essential block of MATH REF is MATH (respectively, MATH). CASE: It is enough to consider the same MATH and MATH as in REF and then write explicitly the equations for the large diagonals. CASE: The two lattices MATH determine the sequences of blowups of MATH creating MATH and MATH. It suffices to show that the blowups making MATH can be rearranged, without changing the outcome (up to an isomorphism), into a different sequence so that an intermediate MATH. This situation is quite similar to the consideration of MATH in REF, and similar is the solution. |
math/9904049 | First, reduce to the case of all MATH points colliding at the same point in MATH. Suppose a collision MATH occurs at MATH. If it could be studied near each MATH independently of the other points, as for MATH, the isotropy subgroup would have been MATH, where MATH is the isotropy subgroup of the collision near MATH. It would have corresponded to MATH independent sequences of colored screens and reduced the proof to the case of one collision point, but this does not suit MATH. Fortunately, interdependencies among the corresponding levels in those MATH sequences only put more restrictions on a permutation aspiring to fix MATH. It means that the isotropy subgroup will be a subgroup of the product above, which still does the trick. Pick a MATH-chain MATH and a coherent sequence of colored screens MATH for MATH. A permutation MATH fixes MATH if and only if it fixes all MATH. A colored screen is fixed by MATH exactly when these two conditions are fulfilled: CASE: it is fixed modulo colors; CASE: any two points of the same color go to two points of the same color, not necessarily the original one. Let MATH be the isotropy subgroup at MATH. A permutation MATH satisfies REF for MATH, therefore it induces the scaling of MATH underlying MATH by a scale factor MATH. The map MATH is a group homomorphism, thus there is a group homomorphism MATH, and to show that it is injective suffices to complete the theorem. Take MATH, then MATH does not move points in any of the colored screens MATH, MATH. By coherence, every color in MATH is a point in MATH, since both are but blocks of the partition MATH. Thus, MATH cannot change colors either, in any MATH for MATH, and colors in MATH stay unchanged because there is only one such. This shows that MATH does not move anything at all, and there is only one such permutation: if MATH and MATH, then MATH must induce nontrivial scaling on MATH, where MATH is the maximal index MATH for which MATH and MATH are in the same block of MATH. |
math/9904050 | That MATH is a NAME pair follows from REF . Since REF hold, the difference MATH can be represented as a sum of a contraction (with norm strictly less than one) and a finite-rank operator. Thus MATH is a NAME pair by REF . Since REF holds one can apply REF to conclude that MATH . By REF one gets MATH . Combining REF, and REF one arrives at REF. |
math/9904050 | By REF one concludes that MATH and hence by REF , the pair MATH is a NAME pair. In particular, REF implies MATH . Hence, REF yields MATH proving REF. |
math/9904050 | It suffices to combine REF . |
math/9904050 | By hypothesis, there exist orthogonal projections MATH in MATH such that MATH and the pairs MATH and MATH, MATH are NAME pairs of projections. Since MATH one infers MATH, MATH and hence MATH, MATH are NAME pairs. Moreover, MATH. Thus, MATH by repeatedly using REF. |
math/9904050 | Since MATH and MATH have a bounded inverse, MATH and MATH are well-defined and (compare CITE) MATH . Using standard estimates for resolvents MATH and MATH of the dissipative operators MATH and MATH entering REF (compare REF) one concludes MATH if MATH. Thus, MATH by REF. |
math/9904050 | Since MATH for some MATH and MATH, one infers that MATH. Thus, there exists a closed interval MATH, MATH, such that MATH for all MATH. Hence, for given MATH, one can find a clockwise oriented bounded contour MATH encircling MATH such that MATH and MATH . The second resolvent identity for MATH and MATH then implies MATH, MATH, proving REF. Since MATH, there exists a closed interval MATH, MATH, such that MATH . Thus, choosing the contour MATH the representation REF is valid for all MATH, for sufficiently small MATH. Using the second resolvent identity again and the standard estimate MATH for every self-adjoint operator MATH in MATH, one obtains MATH where MATH denotes the length of the contour MATH. This proves REF. Next, consider the operator-valued NAME function MATH (that is, MATH for all MATH). By hypothesis there exists a MATH such that MATH. Thus, for sufficiently small values of MATH, MATH, the operator MATH, being a small perturbation of the invertible operator MATH, has a bounded inverse. By REF, MATH is invertible for all MATH, proving REF. Since MATH, MATH and MATH are boundedly invertible (dissipative) operators, REF implies MATH. By REF, MATH implying REF. |
math/9904050 | REF follows from REF . Next one notes that for MATH sufficiently small, MATH by REF , and hence MATH . Since by REF (replacing MATH by MATH), MATH one obtains REF. |
math/9904050 | Since by REF for some MATH, REF implies REF. Due to the fact that MATH (compare REF ) one can represent the spectral projections MATH and MATH by the NAME integrals REF and arguing as in the proof of REF then yields MATH . Thus, by REF , the pair MATH is a NAME pair of orthogonal projections, which together with REF proves REF. |
math/9904050 | Define two orthogonal projections MATH and MATH such that MATH and MATH where MATH (taking into account that MATH). Next, we note MATH where the clockwise oriented contour MATH encircles MATH. On the other hand, MATH using REF. Since MATH, the last term in REF is a trace class operator and, hence, MATH if and only if MATH. |
math/9904050 | By hypothesis, MATH for some MATH. Since MATH and MATH as MATH, the operators MATH are also invertible for MATH, MATH sufficiently large. Since the operator-valued logarithm is a continuous function of its (dissipative) argument in the MATH-topology, MATH and hence by REF, MATH . By REF , MATH and thus MATH and MATH . Thus the difference MATH is a compact operator and hence by REF one gets MATH . The estimate REF MATH with remainder term MATH uniform with respect to MATH, then yields MATH . Hence MATH . Combining REF yields MATH proving REF. |
math/9904050 | Fix a MATH, MATH. Denote by MATH the eigenvalues of MATH different from MATH with corresponding multiplicities MATH. First we prove the following representation, MATH with MATH . Since MATH, there exists a MATH such that MATH . For sufficiently small values of MATH, MATH, the self-adjoint operator MATH is well-defined and the trace of MATH reads MATH where by REF , MATH . For MATH one obtains MATH . By REF, using the dominated convergence theorem, one can perform the limit MATH in REF and upon combining REF one arrives at REF using REF . Since the multiplicities MATH of the eigenvalues MATH can also be computed as MATH where MATH are given by REF, the absolutely convergent series MATH can be represented as the NAME integral MATH where MATH is the following eigenvalue counting function MATH . Making the change of variables MATH (separately on MATH and MATH) MATH one infers by REF that MATH where MATH . Combining REF one concludes that MATH for a.e. MATH, where MATH is a left-continuous function given by REF. |
math/9904050 | First one notes that MATH, MATH is invertible. For MATH this holds by hypothesis and for MATH this holds since MATH. Thus, MATH and MATH are well-defined. In the following let MATH. Since MATH, the representation MATH and estimates for the resolvents MATH and MATH, MATH, analogous to those in the proof of REF yield REF. Therefore, MATH . Based on the estimates in the proof of REF mentioned above and the fact that MATH and MATH whenever MATH is continuous with respect to MATH in MATH-topology and MATH for some MATH, one can interchange the integral and the trace in REF and obtain MATH . Next, using the fact that MATH is differentiable with respect to MATH in trace norm for MATH and MATH in trace norm, one concludes MATH and hence MATH integrating REF from MATH to MATH with respect to MATH. Combining REF one obtains MATH . Using the estimate MATH which holds uniformly with respect to MATH, NAME 's theorem implies MATH . Applying REF again, REF implies MATH . Using MATH and combining REF - REF one finally gets MATH . The trace of MATH, MATH, can easily be computed in terms of the eigenvalues MATH of MATH. By REF , MATH has the eigenvalues MATH with associated multiplicities MATH. By NAME 's theorem (compare CITE) MATH . By REF MATH . Since MATH is self-adjoint, one concludes that MATH and that the series MATH converges. Hence, applying the dominated convergence theorem, one can interchange the sum and the integral in REF, arriving at REF. |
math/9904050 | Pick MATH in REF . Taking the imaginary part of both sides of REF, an explicit computation of the integrals in REF yields MATH . |
math/9904050 | Under REF one can apply REF for MATH. Thus, MATH and therefore REF holds due to REF and the fact that the left-hand side of REF is real. |
math/9904050 | Since by hypothesis, MATH, REF implies that MATH for all MATH except possibly at a finite number of points MATH. Introducing the notation MATH one obtains for MATH sufficiently small, MATH . By REF , MATH and by REF (for MATH sufficiently small), MATH while MATH . Combining REF - REF proves REF. Setting MATH one gets by REF, MATH proving REF. |
math/9904050 | By REF , MATH for all MATH sufficiently small. By REF , MATH . By REF one concludes that MATH and from standard properties of the operator logarithm (compare CITE) one also infers MATH . Combining REF - REF one obtains MATH and by REF one infers MATH where MATH . By REF , MATH and therefore, MATH . Since by REF, MATH, one concludes that MATH is a NAME pair of orthogonal projections. Moreover, MATH implies that the pair MATH has a trindex and hence MATH . Now REF follows from the chain REF for the index of a pair of projections, MATH and from the fact that the measure MATH is a probability measure on MATH. |
math/9904050 | Let MATH. Then MATH and MATH prove REF. Next, let MATH. Then MATH and MATH prove REF. |
math/9904050 | By REF the pairs MATH and MATH have a trindex and hence the pair MATH has a generalized trace and MATH . Moreover, the following representations hold MATH . Under REF , MATH, MATH and hence, by REF , MATH . Combining REF - REF proves REF. |
math/9904050 | Introduce the family of trace class operators MATH . Then, MATH and hence the NAME determinant of MATH is well-defined. By the analytic NAME theorem the set of MATH such that MATH does not have a bounded inverse is discrete and therefore there exists a MATH such that MATH is well-defined. By REF , the operator MATH has the eigenvalue MATH if and only if MATH with associated multiplicity equal to MATH. Let MATH different from the eigenvalue MATH with multiplicities MATH. By REF , MATH has the eigenvalues MATH with multiplicities MATH and, in addition, the eigenvalue MATH of multiplicity MATH (if MATH has a nontrivial kernel). Moreover, MATH has no other eigenvalues different from zero. Therefore, by REF, MATH . Moreover, MATH for some MATH. Combining REF results in REF. |
math/9904050 | Differentiating MATH with respect to MATH (compare , CITE) MATH one obtains by REF MATH since MATH. However, MATH can be computed explicitly, MATH iterating the second resolvent identity. Combining REF - REF one infers MATH . Taking traces in REF one gets MATH and thus, differentiating REF with respect to MATH, MATH . Comparing REF one arrives at the trace formula MATH and hence up to an additive constant, MATH coincides with the spectral shift function MATH associated with the pair MATH for a.e. MATH. Next we determine this constant. Introducing MATH, MATH, one obtains MATH . Moreover, MATH where MATH denotes the number of strictly negative eigenvalues of MATH, counting multiplicity. Thus, MATH coincides with the number of strictly negative eigenvalues of MATH. Combining REF, and using REF results in MATH and hence in MATH . Since MATH as MATH, one concludes MATH and hence that MATH is bounded MATH. In particular, REF imply that MATH is integrable, MATH . Since MATH REF yield the trace formula MATH which together with REF proves REF. |
math/9904050 | First of all, one notes that the boundary values MATH and MATH exist MATH a.e. in the topology MATH for every MATH (but in general not for MATH,) CITE, CITE (see also CITE, CITE, CITE). By REF, the operator MATH has a bounded inverse for a.e. MATH. Moreover (see, for example, CITE), MATH . Thus, there exists a set MATH with MATH (MATH denoting NAME measure on MATH) satisfying the following properties. For any REF The boundary values MATH exist in MATH-topology (compare REF). CASE: The operator MATH has a bounded inverse. CASE: MATH converges to MATH as MATH in MATH-topology. For any MATH, MATH introduce the function MATH . Since MATH commute with MATH and the subspace MATH is invariant for MATH one concludes by REF that (compare REF) MATH . On the other hand, by REF , and REF one infers that the function MATH coincides with the spectral shift function MATH associated with the pair MATH where MATH is given by REF. By a result of REF - REF , and REF, one obtains for MATH, MATH . Applying the approximation REF then yields MATH . Convergence REF of MATH to MATH in trace norm implies the convergence of the corresponding spectral shift functions MATH to the spectral shift function MATH in MATH. This in turn implies the existence of a subsequence MATH converging pointwise a.e. Together with REF this proves REF. |
math/9904050 | Since REF have been proven in CITE, we focus on REF. Let MATH be an increasing family of orthogonal projections of rank MATH, that is, MATH, MATH, with MATH . Combining the norm continuity of the logarithm of bounded dissipative operators as discussed in REF with the exponential NAME representation for MATH (that is, the finite-dimensional analog of REF as in REF - REF), one infers for each MATH the existence of a subset MATH, MATH such that MATH . Thus, MATH . Since MATH one concludes REF. |
math/9904050 | Since MATH and MATH, the spectrum of MATH is a discrete set with only possible accumulation points at MATH. Next, the normal boundary values MATH exist in norm for all MATH. Moreover, MATH which can be seen as follows: suppose that REF is false, then by compactness of MATH there exists a MATH such that MATH . Multiplying REF by MATH one infers that MATH . Since MATH and MATH for any MATH, one concludes that there is a MATH such that MATH. Thus, MATH and hence MATH, which contradicts the fact that MATH. This proves REF. By REF , MATH . Moreover, the pair MATH is a NAME pair. Hence the generalized spectral shift function is well-defined and given by MATH . Since the right-hand side of REF is continuous on MATH by REF , MATH has a continuous representative on MATH. Next, assume MATH. Then NAME 's spectral shift function MATH associated with the pair MATH coincides with the right-hand side of REF for a.e. MATH (see, for example, CITE), proving REF for MATH applying REF. The general case of compact operators MATH can be handled using an appropriate approximation argument. Denoting by MATH the eigenvalues of MATH and by MATH the corresponding spectral projections associated with MATH, and introducing the family of the self-adjoint operators MATH one concludes that MATH, MATH, and MATH . Given MATH, there exists a MATH, such that for all MATH the point MATH, MATH, and therefore, by REF (for MATH), for such MATH we have the representation MATH . Here MATH denotes the continuous representative of (the piecewise constant) NAME 's spectral shift function on MATH. Applying REF once again, one can pass to the limit MATH to obtain MATH . By REF we also have MATH . Taking into account that MATH, MATH REF implies MATH by REF . The right-hand side of REF coincides with the continuous representative of the generalized spectral shift function MATH, MATH, which together with REF proves REF. Finally, REF is a consequence of REF and the left continuity of MATH on MATH. |
math/9904050 | The assertion is a direct consequence of REF . |
math/9904053 | If MATH is a sphere or torus, MATH, so that there is nothing to prove. We may thus assume MATH. Let MATH be the bundle projection, MATH the NAME class of the tangent bundle along the fibers, and MATH the fundamental class dual to the orientation class MATH. Let MATH denote integration along the fiber. Then MATH . Denoting by MATH the NAME sup norm dual to the MATH norm used in the definition of the simplicial volume, compare CITE, we deduce: MATH or MATH . As MATH is assumed to be a hyperbolic NAME surface, we have MATH, see CITE. By definition, MATH. Finally, the unit sphere bundle of MATH is a flat MATH-bundle CITE, and so MATH is bounded. By the NAME inequality, the exact bound is MATH (compare CITE), so that REF follows. |
math/9904053 | This follows directly from REF, the equality MATH for closed hyperbolic NAME surfaces MATH, and the multiplicativity of the NAME characteristic in fiber bundles. |
math/9904053 | If MATH, then MATH or MATH must be a circle, so that the right-hand-side of REF vanishes. Similarly, if MATH and either MATH or MATH is a circle the right-hand-side of REF vanishes and there is nothing to prove. Thus the only interesting case is that of a surface bundle, for which we appeal to REF . |
math/9904053 | As MATH is an NAME manifold, we have NAME 's inequality MATH, with equality only if MATH is flat. Flat manifolds are finitely covered by the torus, and so their simplicial volumes vanish and there is nothing to prove in that case. Thus we can assume MATH. This means that neither MATH nor MATH can be a circle or a REF-torus. Moreover, if MATH is a REF-sphere, then so is MATH, and vice versa. In this case both sides of REF vanish. Thus, we only have to consider the case when both MATH and MATH are hyperbolic NAME surfaces. Then MATH, so that we only have to show that the simplicial volume of MATH is bounded above by a universal multiple of its NAME characteristic. But this follows from the NAME inequality MATH for NAME MATH-manifolds proved in CITE. |
math/9904058 | (outlined by NAME): It is known that MATH is a MATH bundle over MATH with monodromy MATH CITE. By standard MATH-manifold theory MATH can be isotoped to a fiber preserving isotopy. By composing with obvious diffeomorphism that extends, we can assume that the fiber orientation is preserved. Since MATH has to commute with monodromy it fixes the ``vanishing cycle" C, corresponding to MATH. So MATH is a composition of NAME twist along the horizontal torus MATH along MATH direction, and NAME twist along the fiber in MATH direction (NAME twist orthogonal to MATH is ruled out since it does not extend to homotopy equivalence MATH), all these diffeomorphisms extend to MATH. |
math/9904058 | By replacing one of the fishtails in MATH by MATH we obtain a homotopy MATH-sphere, which can be easily checked to be MATH REF . Similarly by replacing one of the cusps in MATH we obtain a homotopy MATH which can be checked to be the standard MATH. Unlike the previous case, this check is surprisingly difficult (it requires the proof of NAME 's conjecture). REF is the handlebody of MATH, it is diffeomorphic to REF (to see this, in REF slide one of the small MATH circles over one of the MATH-framed handles, going through the MATH-handle, then slide MATH framed MATH-handle over it). To identify REF by MATH we need to first recall the handlebody picture of MATH, where MATH is the NAME homology sphere CITE: CASE: From CITE we know that surgering MATH (along the loop trivially linking the slice REF-handle) gives MATH, and surgering once more the obvious MATH gives MATH. Performing these two surgeries corresponds to introducing pair of MATH and MATH framed two handles as indicated in REF . By canceling two MATH and MATH-handle pairs from REF gives REF . Via the handle moves of REF one can check that, introducing the two little MATH-framed handles to REF , to obtain REF , has the affect of changing the boundary from MATH to MATH . |
math/9904067 | For completeness we present a short proof based on the following classical characterization of meager sets in MATH. MATH . Suppose that MATH is a meager set. Without loss of generality we can assume that MATH for some partition MATH and real MATH. Let MATH for every MATH. Define MATH . Suppose that MATH. Define MATH. It is clear that MATH . A small modification of the above argument shows that we can consider more than two translations, countably many or even MATH. |
math/9904067 | We start with the following easy observation: Suppose that MATH is a finite set and MATH are such that CASE: MATH and MATH, CASE: MATH. There exist MATH such that MATH . Let MATH . Check that for every MATH, MATH . Thus MATH for all MATH. By NAME theorem there are MATH such that MATH . In particular, MATH . Fix a partition of MATH into finite sets MATH such that MATH. For each MATH chose MATH such that MATH . Let MATH . The following lemma finishes the proof. For every null set MATH there are MATH such that for every MATH . For a closed set MATH, MATH and MATH let MATH . Let MATH . Without loss of generality we can assume that MATH has positive measure. Suppose that MATH is a null set. Let MATH be a set of positive measure such that MATH where MATH . We will construct two reals MATH such that for every MATH . Define by induction an increasing sequence MATH and MATH for MATH. For MATH we put MATH. Suppose that MATH and MATH are defined. We need to define MATH and MATH for MATH. Use the NAME density theorem to find sequences MATH and MATH such that CASE: MATH, CASE: the set MATH has positive measure. For MATH let MATH . Note that MATH . Since the set on the left-hand side has positive measure there must be infinitely many MATH such that MATH since MATH. Let MATH be first such MATH that is bigger than MATH. Apply the lemma to get sequences MATH such that MATH . Define MATH and MATH. This completes the definition of MATH and MATH. Suppose that MATH is given. Let MATH. Without loss of generality we can assume MATH . Let MATH be the set of MATH satisfying the requirement above. We will show that MATH. For each MATH let MATH be such that MATH if possible, that is, if MATH. Let MATH and MATH . Let MATH. Since MATH for all MATH, it follows that MATH. On the other hand MATH since MATH . |
math/9904079 | By REF MATH . Hence, MATH . Since MATH and since MATH and MATH are irreducible, we have MATH . Theorem is proved. |
math/9904079 | The natural homomorphism MATH is, clearly, a homomorphism of MATH-modules. It is not difficult to verify that MATH . Let MATH be a fixed MATH-tableau, MATH and MATH two MATH-semistandard sequences with elements from MATH and MATH, respectively. Then by REF the vectors MATH form a basis of a subspace MATH which is also a MATH-submodule. By the same theorem, MATH and it remains to show that MATH. For this it suffices to show that there exists a MATH such that MATH. Since MATH, it follows that MATH . Therefore, we may assume that the tableau MATH is consequtively filled in along the rows with the numbers REF, etc. Observe that the sequences MATH and MATH remain MATH-semistandard. Let MATH be the sequence MATH where MATH is the partition corresponding to MATH, MATH, MATH. It is not difficult to verify that MATH. Since MATH is a homomorphism of MATH-modules, its restriction onto MATH is an isomorphism. This implies the statement of Theorem for the family MATH. For the family MATH proof is similar. |
math/9904079 | Let MATH be a supercommutative superalgebra, MATH a MATH-module; let MATH and MATH. The elements of MATH may be considered as functions on MATH with values in MATH. Let MATH. Therefore, MATH determines a homomorphism MATH. Set MATH . Observe that MATH naturally acts on MATH and on the algebra of functions on MATH. |
math/9904079 | By REF MATH and MATH hence, MATH. Since homomorphism REF is a homomorphism of MATH-modules, its kernel coinsides with MATH. Let MATH be a MATH rectangle. The condition MATH means that MATH and by REF it suffices to demonstrate that MATH, where MATH is a fixed standard rectangular tableau of size MATH, belongs to the ideal generated by MATH. Let MATH be the corresponding minimal idempotent, MATH the minimal idempotent for a standard tableau MATH of size MATH. Decomposing MATH with respect to the right cosets relative MATH and decomposing MATH with respect to the left cosets relative MATH we obtain a representation of MATH in the form MATH. This implies that MATH is the sum of polynomials of the form MATH, that is, belongs to the ideal generated by the MATH. |
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