paper
stringlengths 9
16
| proof
stringlengths 0
131k
|
|---|---|
math/0003032
|
Let us denote the centralizers in MATH of the actions MATH and MATH by MATH and MATH, respectively. The centralizer MATH contains multiplications by all elements of MATH. For, if one takes any basis in MATH, the multiplication by an element MATH takes elements of the basis into elements of MATH, which are linear combinations with integral coefficients of the basis elements; hence the multiplication is given by an integer matrix. On the other hand any element of each centralizer is a multiplication by an integer in MATH REF . Now assume that the multiplication by MATH belongs to MATH. This means that this multiplication preserves MATH; in particular, MATH. Thus MATH consists of multiplication by elements of MATH. An algebraic isomorphism up to a time change has to preserve both the module of polynomials with integer coefficients in the generators of the action and the centralizer of the action in MATH, which is impossible.
|
math/0003032
|
The action MATH corresponding to the ring MATH is cyclic by definition since the ring coincides with the orbit of MATH. By REF , if MATH were cyclic, it would be algebraically conjugate to MATH, which, by REF , implies that MATH.
|
math/0003032
|
By REF any irreducible action MATH of MATH by automorphisms of MATH is algebraically conjugate to an action of the form MATH for a lattice MATH. Let MATH be the centralizer of MATH in the semigroup of linear endomorphisms of MATH. We fix an element MATH with MATH and consider conjugation of the action MATH by multiplication by MATH; this is simply MATH. The centralizer of MATH acts on the element MATH transitively. By REF the centralizer consists of all multiplications by elements of a certain subring MATH which contains MATH. Thus MATH.
|
math/0003032
|
First, let us point out that it is sufficient to prove the proposition for irreducible actions. For, if MATH is not irreducible, it has a nontrivial irreducible algebraic factor of dimension, say, MATH. Since every factor of an ergodic automorphism is ergodic, we thus obtain an action of MATH in MATH by ergodic automorphisms. By considering a restriction of this action to a subgroup of rank MATH which contains an irreducible matrix, we obtain a NAME action on MATH. By REF . for irreducible actions, the centralizer of this NAME action is a finite extension of MATH, and thus cannot contain MATH, a contradiction. Now assuming that MATH is irreducible, take a matrix MATH with irreducible characteristic polynomial MATH. Such a matrix exists by REF . It has distinct eigenvalues, say MATH. Consider the correspondence MATH defined in REF. By REF for every MATH we have MATH, hence the group of units MATH in MATH contains a subgroup isomorphic to MATH. By the NAME Unit Theorem the rank of the group of units in MATH is equal to MATH, where MATH is the number of real embeddings and MATH is the number of pairs of complex conjugate embeddings of MATH into MATH. Since MATH we deduce that MATH, so the field MATH is totally real, that is all eigenvalues of MATH, and hence of any matrix in MATH, are real. The same argument gives REF , since any element of the centralizer of MATH in MATH corresponds to a unit in MATH. NAME of matrices in MATH is proved in the same way as REF .
|
math/0003032
|
Assume that MATH is not hyperbolic. As MATH is simultaneously diagonalizable with MATH and has real eigenvalues, it has an eigenvalue MATH or MATH. The corresponding eigenspace is rational and MATH - invariant. Since MATH is irreducible, this eigenspace has to coincide with the whole space and hence MATH.
|
math/0003032
|
Suppose the matrix MATH corresponds to the ideal class MATH with the MATH - basis MATH. Then MATH . Since MATH, there exists a matrix MATH having the same eigenvalues which corresponds to a different ideal class MATH with the basis MATH, and we have MATH . The eigenvectors MATH and MATH are chosen with all their entries in MATH. Now assume that MATH is conjugate to MATH. Then MATH contains a matrix MATH conjugate to MATH. Since MATH commutes with MATH we have MATH, and since MATH is conjugate to MATH, MATH is one of the roots of MATH. Moreover, since MATH and all entries of MATH are in MATH, MATH. Thus MATH is one of MATH roots of MATH which belongs to MATH. From MATH REF we deduce that MATH. Since MATH and MATH belong to different ideal classes, MATH for any MATH in the quotient field of MATH, and since MATH is a simple eigenvalue for MATH, we deduce that MATH, and thus MATH can take one of the MATH remaining values. Now assume that MATH corresponds to the third ideal class, that is, MATH and MATH commutes with MATH and is conjugate to MATH, and hence to MATH. Then MATH where MATH is a root of MATH belonging to the field MATH. By the previous considerations, MATH and MATH. An induction argument shows that if the class number of MATH is greater than MATH, there exists a matrix MATH such that no matrix in MATH is conjugate to MATH, that is, MATH and MATH are not conjugate in MATH. Since MATH has the same characteristic polynomial as MATH, continuing the same process, we can find not more than MATH matrices representing different ideal classes having centralizers conjugate to MATH, and the required estimate follows.
|
math/0003032
|
We refer to REF ' from (CITE, ``Corrections."). According to this corollary the measure MATH is an extension of a zero entropy measure for an algebraic factor of smaller dimension with NAME conditional measures in the fiber. But since MATH contains a MATH-automorphism it does not have non - trivial zero entropy factors. Hence the factor in question is the action on a single point and MATH itself is a NAME measure on a rational subtorus.
|
math/0003032
|
First of all, REF is invariant under metric isomorphism, hence MATH also satisfies this condition. But ergodicity with respect to NAME measure can also be expressed in terms of the eigenvalues; hence MATH also satisfies (MATH). Now consider the joining measure MATH on MATH. The conditions of REF are satisfied for the invariant measure MATH of the action MATH. Thus MATH is a translate of NAME measure on a rational MATH - invariant subtorus MATH. On the other hand we know that projections of MATH to both MATH and MATH preserve NAME measure and are one - to - one. The partitions of MATH into pre - images of points for each of the projections are measurable partitions and NAME measures on elements are conditional measures. This implies that both projections are onto, both partitions are partitions into points, and hence MATH and MATH, where MATH is an affine automorphism which has to coincide MATH with the measure - preserving isomorphism MATH.
|
math/0003032
|
Consider a hyperbolic matrix MATH with irreducible characteristic polynomial and real eigenvalues such that the origin is the only fixed point of MATH. Consider a subgroup of MATH isomorphic to MATH and containing MATH as one of its generators. This subgroup determines an embedding MATH. Since MATH and by REF , all matrices in MATH are hyperbolic and hence ergodic, REF is satisfied. Hence by REF , the measure - theoretic centralizer of the action MATH coincides with its algebraic centralizer, which, in turn, and obviously, coincides with centralizer of the single automorphism MATH isomorphic to MATH.
|
math/0003032
|
Since MATH is a measurable factor of MATH, every element which is ergodic for MATH is also ergodic for MATH. Hence MATH also satisfies condition MATH. As before consider the product action MATH which now by the same argument also satisfies MATH. Take the MATH invariant measure MATH on MATH. This measure provides a joining of MATH and MATH. Since MATH is isomorphic to MATH the conditions of REF are satisfied and MATH is a translate of NAME measure on an invariant rational subtorus MATH. Since MATH projects to the first coordinate one-to-one we deduce that MATH is an algebraic epimorphism (mod REF) followed by a translation.
|
math/0003032
|
By REF , MATH and MATH are algebraic factors of each other. This implies that MATH acts on the torus of the same dimension MATH and hence both algebraic factor - maps have finite fibres. Now the statement follows from REF .
|
math/0003032
|
Since action MATH satisfies condition MATH REF we can apply REF and conclude that we only need to consider the case when MATH and MATH are isomorphic over MATH up to a time change. But then, by REF , MATH and MATH are not algebraically isomorphic up to a time change and hence, by REF , they are not measurably isomorphic up to a time change.
|
math/0003032
|
First we notice that a matrix in block form MATH with MATH commutes with MATH if an only if MATH commute with MATH and can thus be identified with elements of MATH. In this case MATH can be identified with a matrix in MATH. Since MATH (compare CITE), the norm of the determinant of the MATH matrix corresponding to MATH is equal MATH. Hence this determinant is a unit in MATH, and we obtain the desired isomorphism.
|
math/0003034
|
Take MATH and MATH in the definition of double concordance.
|
math/0003034
|
There is a copy of MATH in MATH intersecting REF-twist-spin of MATH in MATH. Since MATH splits MATH, there is an invertible concordance from MATH to MATH. Hence MATH is split by MATH and the result follows.
|
math/0003034
|
Let MATH be a copy of the non-straight arc of MATH in REF-ball MATH and let MATH be a copy of the non-straight arc of MATH in MATH as shown in REF . The closed curve MATH bounds an obvious punctured torus MATH that is the shaded region in REF . Consider MATH as the plumbing of two MATH. Let MATH be the cores of the two MATH of MATH and let MATH be disjoint proper line segments in MATH intersecting with MATH exactly once, respectively. See REF . To construct an invertible concordance, we will construct two concordances and then paste them together. First, note that pinching MATH along MATH transforms MATH into the tangle MATH with an unlinked unknotted circle inside which is isotopic to the circle MATH. Now capping off this circle we have a concordance MATH. The tangle MATH in REF represents a slice of this concordance before capping off the circle. In the similar way, pinching MATH along MATH and capping off the unknot gives us another concordance MATH. Let MATH denote the concordance MATH with reversed orientation. We can then paste MATH to MATH along MATH to get a concordance MATH, which will be proved to be isotopic to the product concordance MATH. A few cross-sections of concordance MATH are drawn in REF . Let MATH denote the embedding of two disjoint copies of MATH into MATH as in the definition of concordance in REF. It is obvious from REF that there is a REF-manifold MATH (the union of shaded regions) in MATH bounded by MATH and MATH, whose intersection with MATH at each end of the concordance is the arc MATH and whose cross-section in the middle is the punctured torus MATH. This REF-manifold MATH can be considered as the union of three submanifolds: the product MATH and two REF-dimensional REF-handles MATH. One MATH is glued to MATH along a regular neighborhood of MATH, which corresponds to capping off the circle isotopic to MATH as we constructed the concordance MATH. The other MATH is glued along a regular neighborhood of MATH, which corresponds to capping off the circle isotopic to MATH as we constructed the concordance MATH. Since MATH is a REF-dimensional handlebody with REF handles with cores MATH and MATH, MATH is the manifold that results by adding two REF-handles to a genus MATH solid handlebody along the cores of REF-handles, in this case yielding MATH. Moreover, MATH does not intersect the other straight arc of MATH at any stage. Using this REF-ball MATH, we can isotop MATH to MATH in a regular neighborhood of MATH not disturbing the other arc and MATH. This completes the proof.
|
math/0003034
|
Consider MATH as the complement of the unknot MATH in MATH. The knot MATH in REF is isotopic to MATH in REF . It is obvious from REF that MATH is the link in MATH formed by replacing a trivial REF-tangle in NAME link with MATH (dotted circle in REF ). The proposition now follows from REF .
|
math/0003034
|
Let MATH be a knot in MATH. If MATH is trivial it is prime itself. Suppose now that MATH is nontrivial. Let MATH be MATH satellite of MATH where MATH is the knot in MATH in REF . By REF , MATH splits MATH. We now only need to show MATH is prime. Since MATH is the unknot in MATH, MATH is prime by REF and to complete proof it remains to show its wrapping number MATH. Its winding number is REF, hence its wrapping number is at least one. It is easy to see that the only prime knot in MATH with wrapping number REF is the core. So, if MATH had wrapping number REF, then it is isotopic to the core of MATH. The MATH surgery on the meridian curve MATH in MATH should make MATH unchanged, that is, unknotted. However, the knot in REF , the result of MATH after MATH surgery along MATH, is MATH and hence knotted. Therefore the wrapping number is MATH.
|
math/0003042
|
CASE: First of all, note that MATH; thus the group MATH also acts on the spaces MATH and MATH (and hence on the set MATH). If REF-tuple MATH is admissible, then MATH is connected, and hence the quotient MATH must be connected too. This implies that MATH has a unique connected component. Since MATH has exactly MATH connected components, the cyclic group MATH of order MATH defines a simply transitive cyclic action on the cycles of MATH. CASE: Let MATH the two curves MATH and MATH. Then, the two systems of curves MATH and MATH on MATH define a NAME diagram of genus one. The graph MATH corresponding to MATH is the quotient of the graph MATH corresponding to MATH, respect to MATH. Moreover, the gluings on MATH are invariant respect to MATH. Therefore, the gluings on MATH give rise to the NAME diagram above. This show that REF-tuple MATH is admissible and obviously MATH is the quotient of MATH respect to MATH.
|
math/0003042
|
From MATH we obtain MATH. Thus, MATH.
|
math/0003042
|
If MATH is admissible, then it is straightforward that MATH has a unique cycle. Vice versa, if MATH has a unique cycle, then REF holds. Since MATH implies REF , the result is a direct consequence of the above lemma.
|
math/0003042
|
Since MATH is admissible, the arcs of MATH are precisely the arcs of MATH. Let MATH be the sequence of these arcs, following the canonical orientation on MATH, and let MATH, with MATH. We have: MATH,where MATH. Since MATH if MATH is of type I, MATH if MATH is of type II and MATH if MATH is of type III, the result immediately follows.
|
math/0003042
|
Since the two systems of curves MATH and MATH on MATH define a NAME diagram of MATH, there exist two handlebodies MATH and MATH of genus MATH, with MATH, such that MATH. Let now MATH be the cyclic group of order MATH generated by the homeomorphism MATH on MATH. The action of MATH on MATH extends to both the handlebodies MATH and MATH (see CITE), and hence to REF-manifold MATH. Let MATH (respectively, MATH) be a disc properly embedded in MATH (respectively, in MATH) such that MATH (respectively, MATH). Since MATH and MATH (mod MATH), the discs MATH (respectively, MATH), for MATH, form a system of meridian discs for the handlebody MATH (respectively, MATH). By arguments contained in CITE, the quotients MATH and MATH are both handlebody orbifolds topologically homeomorphic to a genus one handlebody with one arc trivially embedded as its singular set with a cyclic isotropy group of order MATH. The intersection of these orbifolds is a REF-orbifold with two singular points of order MATH, which is topologically the torus MATH; the curve MATH (respectively, MATH), which is the image via the quotient map of the curves MATH (respectively, of the curves MATH), is non-homotopically trivial in MATH. These curves, each of which is a fundamental system of curves in MATH, define a NAME diagram of MATH (induced by MATH). The union of the orbifolds MATH and MATH is a REF-orbifold topologically homeomorphic to MATH, having a genus one REF-bridge knot MATH as singular set of order MATH. Thus, MATH is homeomorphic to MATH and hence MATH is the MATH-fold cyclic covering of MATH, branched over MATH. Since the handlebody orbifolds and their gluing only depend on MATH, the same holds for the branching set MATH. The homeomorphism type of MATH follows from REF .
|
math/0003042
|
Obviously MATH. Since MATH is admissible, it satisfies REF . This proves that MATH satisfies REF , for each MATH. Since MATH and MATH, we obtain MATH, for each MATH, which implies REF, or equivalently REF . Moreover, MATH is odd, since MATH. Thus, REF proves that MATH is admissible. The final result is then a direct consequence of REF .
|
math/0003042
|
From MATH it immediately follows that MATH has a unique cycle in MATH. Since MATH is odd, REF proves that MATH is admissible. Since MATH, all assumptions of REF hold; hence MATH is admissible for each MATH and MATH is a MATH-fold cyclic covering of MATH, branched over a knot MATH which is independent on MATH. In order to determine this knot, we can restrict our attention to the case MATH. Note that MATH and hence MATH is always even. Thus, in the case MATH we can suppose MATH. Let us consider now the genus two NAME diagram MATH. The sequence of NAME moves CITE on this diagram, drawn in REF - REF and described in the Appendix of the paper, leads to the canonical genus one NAME diagram of the lens space MATH (see REF ). Since the representation of lens spaces (including MATH) as REF-fold branched coverings of MATH is unique CITE, the result immediately holds.
|
math/0003052
|
The only property we have to check to make sure that the collection MATH defines a MATH-algebra structure, is the identity looking like MATH where MATH is a quadratic (non-commutative) polynomial. Since MATH has zero differential (this means MATH in our notation) the left-hand side vanishes. The right hand side vanishes for MATH since the collection of MATH does define a MATH-action. Since the polynomials MATH are homogeneous, one has MATH . This proves the claim.
|
math/0003061
|
By CITE we can find simple rank one NAME algebras MATH, MATH such that MATH and MATH. The NAME Theorem for tensor products CITE shows that MATH. Since the algebras involved are all p.i.s.u.n. and satisfy the U.C.T., the result follows from the Classification Theorem CITE.
|
math/0003063
|
As in the case of REF, we need to add a clause to the theorem for the purpose of induction, that is, we will prove the following statement by induction on MATH. For a sufficiently general curve MATH of degree MATH in MATH, MATH for any curve MATH that meets MATH properly. In addition, there exists a set MATH of countably many points on MATH such that if MATH for some MATH, MATH. Let MATH be a pencil of degree MATH curves whose central fiber MATH is the union of a curve MATH of degree MATH and a line MATH and let MATH where MATH. Let MATH and MATH be defined as before. Almost nothing in the argument of REF needs changing except in the case that MATH is trivial and MATH is nowhere vanishing. In this case, following our previous argument, we can show that MATH vanishes at no less than MATH points. The difference is that now we have MATH and we have to verify that there are only finitely many MATH that vanishes at exactly MATH distinct points. This is more or less obvious because the map from MATH to MATH given by REF is dominant and the space MATH has dimension MATH.
|
math/0003063
|
Let MATH be the canonical divisor of MATH. We argue by induction on MATH. For MATH, we need to verify that MATH, which is more or less obvious. Suppose that MATH. Let MATH be a pencil of curves in MATH whose central fiber MATH is a union of MATH and MATH and let MATH. Let MATH and MATH be defined as before. Again, the same argument for REF goes through. We need only to check the following facts, all of which are routine exercises. CASE: MATH. CASE: MATH surjects onto MATH and MATH is base point free on MATH for each MATH. Hence the map from MATH to MATH given by REF is dominant. CASE: In the case that MATH is trivial and MATH is nowhere vanishing, we can prove that MATH vanishes at no less than MATH distinct points as before. But actually, we can do better here since the space MATH has dimension MATH due to the fact that MATH is elliptic instead of rational. Therefore, MATH vanishes at no less than MATH distinct points.
|
math/0003064
|
We show first REF MATH and then REF MATH. The isomorphism MATH can be proved analogously to REF and so is omitted. CASE: Observe that in the case where MATH holds with some nonsingular matrix MATH over MATH, the statement of the lemma obviously holds. Hence by considering MATH as MATH, we can assume without loss of generality that MATH is expressed as MATH with nonzero MATH. Observe now that MATH holds. From this relation and the first observation, we now have REF . CASE: It is sufficient to consider the case MATH with MATH and MATH as in REF . In the case MATH, one can consider MATH. First we define a bijective mapping MATH from MATH to MATH. For convenience we decompose MATH into MATH and MATH as follows MATH . Corresponding to MATH and MATH, we define MATH and MATH as MATH . We now define the mapping MATH as MATH . Since MATH and MATH can be expressed by MATH and MATH as MATH, MATH, the inverse mapping MATH can be defined naturally. Hence, the map MATH is bijective. Now let MATH and MATH. By the straightforward calculation with noting that MATH and MATH are diagonal, we obtain the following relations: MATH . Thus MATH for all MATH. It follows that the ideals MATH and MATH are isomorphic to each other.
|
math/0003064
|
Suppose that MATH is a local ring and MATH is a unimodular row. Thus there exist MATH such that MATH . Since MATH is local, the set of all nonunits is an ideal. From REF , there exists a MATH with MATH such that MATH is a unit. We assume without loss of generality that MATH is a unit. If MATH, then MATH is a unit, which can be considered as a unimodular matrix of MATH. In the following we consider the case MATH. Then we can construct a unimodular matrix MATH: MATH . This MATH contains the row MATH as a submatrix and hence every unimodular row can be complemented. Therefore MATH is NAME.
|
math/0003064
|
This proof mainly follows that of REF. If MATH is in MATH, then we can select the zero matrix as MATH. Thus we assume in the following that MATH is in MATH. Since the determinant of REF is in MATH, there exists a full-size minor of MATH in MATH by NAME 's expansion of REF and by REF . Let MATH be such a full-size minor of MATH having as few rows from MATH as possible. We here construct a matrix MATH such that MATH with a MATH. Since MATH, the full-size minor MATH must contain at least one row of MATH from the matrix MATH. Suppose that MATH is obtained by excluding the rows MATH of MATH and including the rows MATH of MATH. Now define MATH by MATH and MATH for all other MATH, MATH. Observe that MATH is expanded in terms of full-size minors of the matrices MATH and MATH from the factorization MATH by the NAME formula. Every minor of MATH containing more than MATH columns of MATH is zero. By the method of choosing the rows from MATH for the full-size minor MATH, every full-size minor of MATH having less than MATH rows of MATH is in MATH. There is only one nonzero minor of MATH containing exactly MATH columns of MATH, which is obtained by excluding the columns MATH of the identity matrix MATH and including the columns MATH of MATH; it is equal to MATH. From the NAME formula the corresponding minor of MATH is MATH. As a result, MATH is given as a sum of MATH and elements in MATH. By REF , the sum is in MATH and so is MATH.
|
math/0003064
|
Since MATH is local, REF are equivalent by REF . Thus we only prove REF and vice versa. CASE: Suppose that MATH is MATH-stabilizable. Then the MATH-module MATH is projective by REF. Further it is free by REF. Let MATH and MATH be matrices over MATH with MATH. Then the MATH-module MATH is free of rank MATH since MATH is nonsingular over MATH. Let MATH be a basis of the module MATH and MATH the matrix of MATH whose rows are MATH. Then, the matrix MATH can be written in the form MATH by uniquely choosing the matrices MATH in MATH and MATH in MATH. Because of MATH, MATH is a nonzerodivisor. It follows that MATH over MATH. In the following we show that the matrices MATH and MATH are right-coprime over MATH. Since MATH belong to MATH, there exist matrices MATH in MATH and MATH in MATH such that MATH. So we have MATH. Since MATH is nonsingular, we obtain MATH over MATH. Thus MATH is a right-coprime factorization over MATH of MATH. CASE: Suppose that there exists a right-coprime factorization over MATH of the plant MATH; that is, there exist the matrices MATH, MATH, MATH, MATH over MATH with MATH and MATH. If MATH is a nonzerodivisor of MATH, it is obvious that MATH is a MATH-stabilizing controller. Thus in the following we suppose that MATH is a zerodivisor of MATH. By the equivalence between REF , there also exists a left-coprime factorization over MATH of MATH; that is, there exist the matrices MATH, MATH, MATH, MATH over MATH with MATH and MATH. Thus we have the following matrix equation: MATH . Observe that the determinant of the right-hand side of the matrix equation above is in MATH, where MATH denotes the localization of the prime ideal MATH at MATH (Note that MATH is also a prime ideal of MATH). Hence the determinant of the first matrix in REF is in MATH again. Applying REF to the first matrix, we have a matrix MATH over MATH such that the determinant of the matrix MATH is in MATH. Now MATH is a MATH-stabilizing controller.
|
math/0003064
|
Since the following implications are obvious: by virtue of REF , we only show that REF implies REF . Suppose that REF holds. Let MATH, MATH, MATH, and MATH be matrices over MATH with MATH such that MATH and MATH are MATH-nonsingular (recall that MATH is causal). By REF , MATH has both right-/left-coprime factorizations over MATH with MATH. As in the proof of REF , for each MATH in MATH, there exist matrices MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, and MATH over MATH such that MATH hold over MATH. For each MATH let MATH be an arbitrary but fixed element of MATH such that the six matrices MATH, MATH, MATH, MATH, MATH, and MATH are over MATH. For a subset MATH of MATH, denote by MATH the set of all maximal ideals MATH of MATH with MATH, that is, MATH. Since MATH, we have MATH. Thus MATH. Recall that MATH is compact (see REF). Hence there are a finite number of MATH of maximal ideals such that MATH. It follows that MATH and, consequently, MATH. Therefore there exist MATH in MATH with MATH. Next we want to consider that at least one of MATH contains MATH. In the case where every MATH in MATH does not contain MATH, we reconstruct MATH, MATH's, and MATH's as follows. We first pick a MATH with MATH. Then we let MATH be MATH for MATH and MATH. We now let MATH. Then we have again MATH and, in this case, MATH. Hence we can assume without loss of generality that at least one of MATH, say MATH, contains MATH. Observe then that the following equality holds: MATH . At least one of MATH and MATH must be in MATH. Thus in the case MATH, we can reassign MATH's as in REF , so that MATH is in MATH. Therefore we can assume without loss of generality that MATH. Consider here the following matrix MATH which is over MATH. For short we partition REF as MATH . In the case where MATH is MATH-nonsingular, letting MATH we can check that MATH is equal to REF , which implies that MATH is stabilized by MATH. Hence in the rest of this proof we show that if MATH is MATH-singular, then MATH can be made MATH-nonsingular by reassigning MATH and MATH for a MATH. First we show the MATH-nonsingularity of the matrices MATH and MATH. Since MATH, we have MATH. From the first matrix equation of REF , we have MATH. Hence MATH is MATH-nonsingular. Analogously, from the second matrix equation of REF , MATH is MATH-nonsingular. Next consider the following matrix equation over MATH: MATH . Since the matrices MATH, MATH, and MATH are MATH-nonsingular, so is the right-hand side of REF . Thus the first matrix of REF is also MATH-nonsingular. By REF and the first matrix of REF , there exists a matrix MATH of MATH such that the following matrix is MATH-nonsingular: MATH . Now let MATH be MATH. Further we let MATH be the matrix MATH and MATH the matrix MATH, which are consistent with REF . Thus we can now consider without loss of generality that the matrix MATH is MATH-nonsingular and so is MATH.
|
math/0003064
|
We show first the ``Only NAME part and then the ``NAME part. (Only If). Suppose that MATH is stabilizable. Then by REF , for every prime ideal MATH in MATH, MATH is MATH-stabilizable. By REF , MATH has both its right-/left-coprime factorizations over MATH. Suppose that MATH holds over MATH with MATH, where the matrices MATH, MATH, MATH, and MATH are over MATH. Then let MATH. By NAME formula we have MATH. Thus by virtue of REF , the ideal MATH is free (recall that MATH denotes the localization of the full-size minor ideal MATH at MATH), which is also finitely generated. This holds for every prime ideal MATH. From REF, the full-size minor ideal MATH is projective. (If). Suppose that the full-size minor ideal MATH is projective. Let MATH be a prime ideal in MATH. Then MATH is free by REF again. Thus there exist MATH, MATH, and MATH in MATH with MATH and MATH for every MATH. Since MATH and MATH is a nonzerodivisor, we have MATH. Recall here that MATH is local. Hence the set of all nonunits in MATH is an ideal. Thus there exists MATH such that MATH is a unit of MATH. This implies that MATH is a unit of MATH and further that every MATH has a factor MATH over MATH (that is, MATH and MATH are associate). Now let MATH and MATH for every MATH. Then MATH and MATH hold. Since MATH, every MATH has a common factor MATH. Suppose that MATH is an integer with MATH and MATH. Suppose further that MATH are elements in MATH with ascending order. Now let MATH. Then MATH is expressed as MATH where MATH is the MATH-entry of the matrix MATH. Since MATH has a factor MATH, MATH has a factor MATH. This fact holds for all MATH between MATH but MATH. As a result, MATH is a common factor of all entries of MATH. Let MATH over MATH. Since MATH is the identity matrix, the matrix MATH itself is a left inverse of MATH. Let MATH and MATH be matrices with MATH. Further we let MATH and MATH be matrices over MATH with MATH. Then we obtain MATH over MATH, which is a right-coprime factorization over MATH of the plant MATH. Therefore by REF , MATH is stabilizable.
|
math/0003064
|
It is obvious that REF implies REF implies REF . Hence we only show that REF implies REF . CASE: Suppose that REF holds. Let MATH be a maximal ideal of MATH. Since MATH is local, the set of all nonunits in MATH is an ideal. Hence there exists a MATH with MATH such that MATH. Thus there exists MATH in MATH such that MATH. Recalling the proof of REF , we have a finite number of MATH in MATH and MATH such that MATH over MATH. For every MATH to MATH, MATH is an element of MATH with MATH. Therefore we have REF .
|
math/0003064
|
Suppose that MATH is a unique factorization domain. Since the ``NAME part is obvious, we prove only the ``Only NAME part. (Only If). Let MATH be in MATH. Suppose that MATH is projective. If all MATH are zero, the proof is obvious. Thus in the following we suppose that at least one of MATH is nonzero. Since MATH is a unique factorization domain, there exists a nonzero greatest common factor of MATH's, denoted by MATH. Thus there exist MATH's in MATH with MATH. Then MATH is projective again. For any prime ideal MATH in MATH, MATH is free of rank MATH. Since there is no nonunit common factor among MATH's over MATH, MATH. By REF , MATH. Hence MATH, which is free.
|
math/0003064
|
By virtue of REF , we suppose without loss of generality that MATH and MATH are matrices over MATH and MATH and MATH in MATH with MATH and MATH. Let MATH and MATH be the following matrices: MATH . Then we can see that there exists a unimodular matrix MATH with MATH and that MATH. Let MATH be the ideal generated by the full-size minors of MATH and MATH be the ideal generated by MATH's for all MATH and all MATH. Then by REF , MATH is isomorphic to MATH as MATH-modules. Also by NAME formula, MATH. Hence we obtain MATH.
|
math/0003064
|
As in the proof of REF , if all MATH are zero, the proof is obvious. Thus in the following we assume that at least one of MATH is nonzero. Then there exist a nonzero MATH in MATH and MATH in MATH for MATH to MATH such that MATH and MATH. Thus there exist MATH for MATH to MATH with MATH. If MATH was a zerodivisor, the principal ideal MATH could not be free. Hence MATH is a nonzerodivisor. Now we have MATH . Since MATH for all MATH, we have MATH. It follows from REF that we now have REF .
|
math/0003064
|
By the same reason as in the proofs of REF , we assume that at least one of MATH is nonzero. (If). Suppose that REF holds. Then there exist MATH for MATH to MATH such that MATH. By appropriate changes of MATH, we assume without loss of generality that all MATH are nonzero with MATH and all MATH are zero subject to MATH. Observe that for each MATH between MATH and MATH, MATH over MATH, where MATH and MATH denote the localizations of MATH and MATH at MATH, respectively. Hence for each MATH between MATH and MATH, MATH is free over MATH. Therefore by REF, MATH is projective as MATH-module. (Only If). Suppose that MATH is projective. Then again by REF, for each MATH in MATH, MATH is free over MATH. By REF , we have MATH for each MATH in MATH. Then REF can be rewritten as follows by REF : MATH . Since this holds for every MATH in MATH, applying REF to REF we obtain REF .
|
math/0003064
|
We first show REF MATH and then REF the opposite inclusion. CASE: For every MATH, MATH holds, which implies that MATH. Hence MATH. CASE: Suppose that MATH is an element of the quotient ideal MATH. Then for every MATH, there exists MATH such that MATH holds and so MATH. Since this equality holds for every MATH, MATH has a factor MATH. Hence MATH.
|
math/0003064
|
Let MATH be fixed. We first show REF MATH and then REF MATH. They are sufficient to prove this theorem. CASE: Let MATH be an arbitrary but fixed element of MATH. Then there exists a matrix MATH over MATH with MATH. Then for every MATH, we have MATH, so that MATH. This implies MATH. Hence we have MATH. CASE: Let MATH be an arbitrary but fixed element of MATH. Then MATH and hence MATH by REF . This implies that MATH and further that every full-size minor of MATH has a factor MATH over MATH. Since MATH is a factor of MATH, it is a nonzerodivisor of MATH. Now let MATH and MATH for every MATH. Then MATH and MATH hold. Since MATH, every MATH has a common factor MATH. Analogously to the proof of REF , we can show that every entry of MATH has a factor MATH. Let MATH over MATH. Then MATH holds over MATH. Hence there exists an integer MATH such that MATH can be considered over MATH and further MATH holds over MATH. Now letting MATH, we have that MATH is an element of MATH and hence MATH.
|
math/0003064
|
Let MATH denote the elementary factor of the matrix MATH with respect to MATH. Also let MATH denote the reduced minor of MATH with respect to MATH. In the case where MATH is a unique factorization domain, the generalized elementary factor of the plant MATH with respect to MATH is equal to the principal ideal MATH. Thus, by REF , MATH. By virtue of REF , we have MATH.
|
math/0003065
|
See CITE or CITE.
|
math/0003065
|
We give the proof of the cofibration case, as the fibration case is strictly dual. Let MATH and MATH. Then a map MATH and its adjoint MATH are related by MATH, where MATH is the pushout of MATH along MATH. If REF holds and if MATH is a cofibration, we can construct MATH so that it is a weak equivalence to a fibrant object, and it then follows from REF that MATH and hence MATH are weak equivalences, giving REF . Conversely, if REF holds, then MATH is a weak equivalence if and only if MATH is, giving REF . The necessary condition follows from considering the case when MATH is the identity map. That MATH being a trivial cofibration is sufficient is clear; that MATH and MATH being cofibrant is sufficient then follows using REF and the fact that NAME equivalences satisfy a REF out of REF property CITE.
|
math/0003065
|
See CITE.
|
math/0003065
|
Let MATH. REF for MATH given above shows that MATH is computed from MATH and MATH using only colimits and finite products, and both of these commute with filtered colimits and reflexive coequalizers. Hence MATH factors through a functor MATH. To show that MATH and MATH are equivalences, it suffices to show that if MATH, then MATH is an isomorphism. In fact, MATH is clearly an isomorphism when evaluated at any finite set, and the result follows from the fact that every set is a filtered colimit of its finite subsets.
|
math/0003065
|
That limits, filtered colimits, and reflexive coequalizers exist and are created in MATH is immediate from REF. That the free algebra functor is left adjoint is a standard property of monads CITE. Existence of colimits follows from CITE; or note that colimits of a diagram MATH can be constructed explicitly as the reflexive coequalizer in MATH of MATH, the top map being induced by the inclusions MATH and the top map being induced by the algebra structure maps MATH.
|
math/0003065
|
It is clear using REF that MATH factors through the subcategory. It remains to show that MATH is an equivalence. Let MATH denote the full subcategory of finitely generated free MATH-algebras; every object in this subcategory is isomorphic to MATH for some MATH. Consider the sequence of functors MATH the right-hand arrow is the one induced by restriction of functors to the subcategory. The result will follow when we show that the composites MATH and MATH are equivalences. To see that MATH is an equivalence, observe that every MATH-algebra is a coequalizer of a reflexive diagram of free algebras, and that every free MATH-algebra is a filtered colimit of finitely generated free algebras. Thus every functor MATH which commutes with filtered colimits and reflexive coequalizers is determined up to unique isomorphism by its restriction to the subcategory of finitely generated free algebras, and natural transformation between such functors are uniquely determined by this restriction. Any functor MATH extends to an element of MATH by a left NAME extension construction, and therefore this construction gives the inverse to MATH. We now show that MATH is an equivalence. Explicitly, MATH sends MATH to the functor MATH; note that if MATH, then MATH. We will construct an inverse MATH. Given MATH, define MATH by MATH; recall that under the equivalence MATH, the object MATH corresponds to a functor MATH also denoted by MATH, and there is a map MATH natural in MATH. Give MATH the structure of a left MATH-module by MATH using the fact that MATH takes values in MATH-algebras. Give MATH the structure of a right MATH-module by MATH using the MATH-algebra structure of MATH. It follows that MATH is a MATH-bimodule, that our construction MATH is a functor from functors to bimodules, and that MATH and MATH as desired.
|
math/0003065
|
Let MATH. Then MATH is the equalizer of MATH, or equivalently of MATH, where the two arrows send MATH to MATH and MATH respectively, where MATH and MATH denote the algebra structure maps.
|
math/0003065
|
Define MATH to be the evident map MATH, and MATH to be the evident map MATH. Then MATH is easily seen to be a theory, and the evident functor MATH an equivalence.
|
math/0003065
|
This is immediate from the existence of the endomorphism theories.
|
math/0003065
|
Apply REF in each simplicial degree.
|
math/0003065
|
For convenience, we write the proof only in the case MATH and MATH are singleton sets; the general case is only notationally more difficult. We first show that MATH is in fact a natural transformation between functors defined on the full subcategory of free objects in MATH. Consider a map MATH between free objects. This amounts to a collection of maps MATH, MATH, and the induced map MATH factors MATH . To show that MATH commutes with these maps reduces to showing that it commutes with MATH, which is clear. Define MATH on the full subcategory of free objects in MATH. Since every object of MATH is a reflexive coequalizer of a pair of free objects, the MATH extend in a unique way to arbitrary objects in MATH.
|
math/0003065
|
Let MATH be as in REF . For this proof, we will write simplicial operators as acting on the right. Say that a MATH in non-degenerate if it is not of the form MATH for some non-identity MATH and some MATH. We claim: if MATH, MATH are non-degenerate elements such that MATH for some MATH, then CASE: MATH and MATH, and CASE: MATH. From this claim it will follow that for each MATH there is a unique non-degenerate MATH and a unique MATH such that MATH; that is, the underlying degeneracy diagram of MATH is free on the non-degenerate elements, proving REF . To prove the claim, observe that there exist MATH such that MATH and MATH. Then MATH and MATH. Any map in MATH must factor uniquely in the form MATH for an injective MATH and surjective MATH; this fact applied to MATH and MATH together with the non-degeneracy of MATH and MATH implies that MATH and hence that MATH, proving REF . To get REF , observe that the same argument shows that MATH and MATH must admit exactly the same elements of MATH as right inverses, and it is easy to derive REF from this. To show REF observe that MATH, being free, is a disjoint union of free degeneracy diagrams on one generator (in various degrees), and that a free degeneracy diagram on one generator has no non-trivial free sub-diagrams.
|
math/0003065
|
We have that MATH; thus, we must show that MATH is a free degeneracy diagram of MATH-graded sets. First, suppose that MATH and MATH are a discrete theory and algebra. The degeneracy diagram MATH extends to a functor MATH, using the fact that MATH is a monad and and MATH and algebra: the ``face" maps are given by MATH and MATH. Since the extension from a MATH-diagram to a MATH-diagram is natural in MATH and MATH, we see that MATH is the ``diagonal" of a simplicial object in MATH-diagrams, and in particular it is a MATH-diagram, and the result follows using REF.
|
math/0003065
|
The right adjoint MATH is the identity on the underlying simplicial sets, and hence preserves weak equivalences and fibrations, and thus the left adjoint preserves cofibrations.
|
math/0003065
|
Let MATH be a natural transformation of functors MATH such that MATH is product preserving, MATH is a fibrant simplicial set and MATH is a weak equivalence for all MATH; we can use REF or REF . This functor MATH extends to MATH by REF. Now consider MATH where the fiber product is defined using MATH and MATH is defined using MATH. The map MATH is a fibration: it can be factored MATH where both maps are fibrations since MATH and MATH are fibrant. By hypothesis, the dotted arrow exists. Furthermore, MATH is a trivial fibration, and hence MATH and MATH are weak equivalences, and we can conclude that MATH is a weak equivalence.
|
math/0003065
|
Consider the diagram MATH . The left-hand side is obtained by applying MATH to the square used in the proof of REF. By REF the functor MATH must be representable by some right MATH-module, and therefore the horizontal maps on the right-hand side are obtained using REF, and the right-hand square commutes. The top and bottom rows of the rectangle are weak equivalences by the same arguments as used in the proof of REF, and hence we conclude that MATH is a weak equivalence.
|
math/0003065
|
If MATH is a trivial cofibration, this is REF. The theorem follows using REF.
|
math/0003065
|
We can first reduce to the case when MATH is a discrete graded simplicial set, using the diagonal principle REF and the fact that MATH (and similarly MATH) can be obtained as the diagonal of the simplicial object in MATH given by MATH, where MATH is the MATH-th simplicial degree of MATH. Next note that it is enough to show that the conclusion holds when MATH is both discrete and finite, since every graded set is a filtered colimit of its finite subsets, and MATH and MATH commute with such colimits. Now we are done, since MATH is a weak equivalence exactly when MATH is one for all MATH.
|
math/0003065
|
The equivalence of REF is immediate, since the MATH-th graded piece of MATH is MATH. Since for any MATH, MATH is a cofibrant MATH-algebra, REF implies REF . To show that REF implies REF , let MATH be a simplicial object in MATH defined by MATH; then MATH is MATH-free by REF, and hence is cofibrant, and MATH is a weak equivalence by the existence of a contracting homotopy. Now consider MATH . The maps marked MATH are weak equivalences by REF, so to show that MATH is a weak equivalence it suffices to show that MATH is. By the diagonal principle REF, it suffices to show that MATH is a weak equivalence for MATH; this is REF.
|
math/0003065
|
The functors MATH are represented by an appropriate bimodules MATH and MATH, as described in REF. We claim that the map MATH induced by MATH is a weak equivalence, which means that we can derive the corollary as a special case of REF. To see that MATH is a weak equivalence, it suffices to show that it induces a weak equivalence when applied to a free ``algebra", by REF. Translated, this means that we must show that MATH is a weak equivalence when MATH is a free right MATH-module. In fact, this is the case whenever MATH for some MATH, by REF, and so is in particular true for free objects.
|
math/0003065
|
First, note that the pair is a NAME equivalence if and only if the adjunction map MATH is a weak equivalence for every cofibrant MATH-algebra MATH. This is because, given MATH, the adjoint map factors MATH and MATH is a weak equivalence if and only if MATH is. The result now follows from REF, since the adjunction map is isomorphic to MATH.
|
math/0003065
|
We have already seen that MATH is always right proper REF, so we need only consider left properness. That REF implies REF follows by observing that if MATH, then the square MATH is a pushout square in MATH-algebras in which the top arrow is a cofibration; properness implies that MATH is a weak equivalence if MATH is. To show REF implies REF , we must show that for any cofibration MATH, CASE: the functor MATH carries weak equivalences to weak equivalences. We proceed by a series of reductions. First, it suffices to show REF when MATH is a MATH-free map, since cofibrations are strong retracts of such. Next, it suffices to show REF for MATH of the form MATH where MATH is an inclusion of MATH-graded simplicial sets. This is because any MATH-free map can be written as a directed colimit of a series of maps, each of which is a pushout along a map of the form MATH, and because weak equivalences are preserved by directed colimits. Define MATH to be the simplicial object in MATH given by MATH . We claim that if MATH, then the evident augmentation MATH is a weak equivalence. In fact, in each internal degree MATH we have that MATH for some MATH-graded set MATH, and thus MATH which augments to MATH. There is an evident contracting homotopy using the inclusion MATH, showing that MATH is a weak equivalence of (graded) simplicial sets, and hence the claim follows using the diagonal principle REF. Next, it suffices to show REF for MATH of the form MATH for MATH; that is, to show that the functor MATH preserves weak equivalences. This follows using the diagonal principle and the above claim, since then for MATH each MATH must preserve weak equivalences. Next, it suffices to show REF for MATH of the form MATH where MATH is a discrete graded simplicial set; this follows by another application of the diagonal principle to MATH, the diagonal of which is MATH. The theorem now follows using the fact that MATH, with MATH discrete, is a filtered colimit over the diagram of all finite subobjects of MATH, and that weak equivalences are preserved by filtered colimits.
|
math/0003065
|
Recall from REF that MATH is equivalent to a full subcategory of the category of endofunctors on MATH. There is an evident explicit isomorphism MATH natural in MATH, as can be seen by applying REF. More explicit computations show that the monoidal structure on MATH restricts to MATH along MATH.
|
math/0003065
|
It is enough to show that MATH is the free monoid with respect to the MATH-product on MATH; that is, maps MATH are in bijective correspondence with maps MATH of monoids. Then from REF it follows formally that MATH is the free monoid with respect to the MATH-product, that is, it is a free theory. To make MATH into a monoid with respect to the MATH structrue, let MATH be the map classifying the trivial trees, and let MATH be the evident map describing grafting of trees. Now note that MATH is precisely the formula for the free MATH-operad on MATH.
|
math/0003065
|
Using REF, this amounts to showing that MATH, which is a direct translation of REF.
|
math/0003065
|
If MATH has the same image under the two maps, then it can have no vertices labelled by MATH, and hence must be a trivial tree. There is exactly one trivial tree for each element of MATH, and MATH contains only these.
|
math/0003065
|
We first show that it suffices to assume that MATH is a MATH-free theory and that MATH is a MATH-free map of theories. In fact, using the model category structure we see that MATH is a retract of a map MATH, where MATH and MATH are MATH-free. Then there are maps MATH of right MATH-modules, and the composite of these maps is the identity, making MATH a retract of MATH as a right MATH-module. If MATH is cofibrant as a right MATH-module, then MATH is cofibrant as a right MATH-module (since the functor MATH is the left adjoint of a NAME pair), and hence MATH is too. Now suppose that MATH and MATH are MATH-free. Thus MATH and MATH, where MATH and MATH are free degeneracy diagrams in MATH. Then by REF we have MATH. Thus, it suffices to show that MATH is a free degeneracy diagram in MATH. Now MATH, and MATH is free by the hypotheses that MATH and MATH be MATH-free. By REF it suffices to show that MATH is closed inside of MATH. That is, if MATH and MATH such that MATH, then MATH. The operator MATH acts on MATH by relabeling the vertices of MATH according to the way MATH acts on MATH and MATH separately, and it does not change the underlying shape of the tree or whether a given vertex is labelled by MATH or MATH; hence, if MATH is MATH-essential, then so is MATH.
|
math/0003065
|
Using the endomorphism theory technology of REF, it is easy to see that MATH. By REF we see that algebras over MATH are the same as MATH-algebras MATH equipped with a map MATH of graded sets, or equivalently, the same as MATH-algebras MATH equipped with a map MATH of MATH-algebras.
|
math/0003065
|
Suppose that MATH. By REF, MATH is a cofibration between cofibrant theories, and thus MATH carries weak equivalences to weak equivalences by REF. Since there is an isomorphism MATH of underlying MATH-algebras, it follows that MATH is proper by REF.
|
math/0003065
|
Given a simplicial theory MATH, one can construct a weak equivalence MATH from a cofibrant theory MATH, since simplicial theories are a model category REF. Then MATH is a proper simplicial model category by REF, and the induced NAME pair MATH is a NAME equivalence by REF.
|
math/0003065
|
Recall that MATH being pointed means that the initial object MATH is isomorphic to the terminal object, denoted MATH. Choose MATH as in the proof of Theorem B, so that MATH is proper and is NAME equivalent to MATH via MATH. The initial object in MATH is MATH, which is not in general the terminal object. But since MATH is a weak equivalence, MATH is weakly equivalent to MATH. Let MATH denote the theory of MATH-algebras under MATH as in REF, so that MATH. We have restriction functors MATH factoring MATH and hence maps MATH of theories factoring the weak equivalence MATH. Since MATH is proper, the NAME pair induced by MATH is a NAME equivalence by REF, and hence a weak equivalence by REF. Hence MATH is a weak equivalence and so induces a NAME equivalence between MATH and MATH. The theorem is now proved, since MATH is a pointed category, and is proper by REF .
|
math/0003065
|
We first show that it suffices to assume that MATH is MATH-free. In general, MATH is a retract of some MATH-free MATH. Let MATH be a cofibration of MATH-algebras. Write MATH and MATH. Then the diagram MATH is a retract of the diagram obtained by applying MATH to it, which is MATH, and the map MATH is a cofibration of MATH-algebras. If we know that MATH is an effective monomorphism, then this diagram is an equalizer, and so is any retract of it, whence MATH is an effective monomorphism. Now assume MATH is MATH-free. We can also assume that MATH is a MATH-free map, since retracts of effective monomorphisms are again effective monomorphisms. To show that MATH is an effective mono, it suffices to check it in each simplicial degree. Thus, we must show that for MATH, MATH, and MATH, the diagram MATH is an equalizer. Using REF this is the same as MATH where MATH is as defined in REF, and the lemma now follows easily using REF.
|
math/0003065
|
A model category MATH is cellular in the sense of CITE if it is a cofibrantly generated model category with sets MATH and MATH of generating cofibrations and trivial cofibrations with the property that CASE: the domains and codomains of the elements of MATH are ``compact", CASE: the domains of the elements of MATH are ``small relative to MATH", and CASE: the cofibrations are effective monomorphisms. REF say that mapping out of the domains and codomains of the generators commutes with certain kinds of directed colimits (for the precise notions, refer to CITE). They certainly hold for categories of algebras over a simplical theory, since in that case the domains and codomains of the generators are ``small" in the sense that mapping out of them commutes with arbitrary filtered colimits. REF holds for a cofibrant theory by REF, giving the result for the hypotheses of Theorem B. If REF holds in a model category, it also holds in all undercategories, and this gives the result for the hypotheses of Theorem C.
|
math/0003066
|
Consider an operator of the general form MATH . Then it is easily seen that MATH satisfies the NAME equation if and only if MATH and MATH . These equations are satisfied when MATH and MATH. Moreover these are essentially the only such solutions CITE.
|
math/0003066
|
The four relations are routine verifications. The fact that MATH then satisfies the NAME equation is a well-known fact about MATH-matrices extended to this slightly more general situation.
|
math/0003066
|
Recall that MATH . For the second term we note that MATH . Combining this with the binomial expansion of the first term yields the assertion.
|
math/0003066
|
From REF , the coefficients MATH are non-zero only when MATH and in this case, MATH . Hence MATH . Thus interpreting MATH as an operator on MATH we get MATH . In matrix form, MATH .
|
math/0003066
|
Define a shift operator MATH by MATH and let MATH act as usual on operators by conjugation. Then, if MATH, MATH . Choose MATH. Then MATH. This shows that MATH is similar to MATH and hence satisfies the MQYBE when MATH.
|
math/0003066
|
We prove that MATH. In matrix form this is equivalent to MATH . Using the fact that MATH when MATH as required.
|
math/0003067
|
We first note that we have inclusions MATH which give us an inclusion monomorphism MATH. Hence by NAME duality we have a dual homomorphism MATH, where the identification between MATH and MATH is given by MATH, and we easily check that for MATH, we have MATH as defined in the statement of the proposition. Now the action MATH of MATH on MATH via automorphisms again via duality gives us a dual action MATH of MATH on MATH, defined by MATH, MATH. By the theory of group representations for semi-direct product groups (for a reference, see CITE), it is known that for every MATH, and for every MATH, the induced representations MATH and MATH are equivalent, and that for fixed MATH, MATH is irreducible if and only if MATH for every MATH. Now in our case, for MATH and MATH, we have MATH . Since MATH, which is a transformation wavelet set for MATH, we know that the sets MATH are pairwise disjoint. Hence MATH for every MATH. It follows that for every MATH, the representation MATH is irreducible. We now show that MATH is equivalent to the representation MATH given in the statement of REF . Here we use the construction of induced representations given in REF. In this construction, for a locally compact group MATH with closed subgroup MATH, one first needs a NAME cross section MATH. Here, MATH, MATH, and the (right) coset space MATH. Furthermore the groups are discrete, and our cross section MATH is actually a group isomorphism defined by MATH for MATH. We now use MATH to define a one-cocycle MATH for the right action of MATH on the coset space MATH in the standard fashion: MATH is defined by MATH for MATH and MATH. Then we recall from REF that given a unitary representation MATH of the group MATH on the NAME space MATH, the induced representation MATH has as its representation space MATH where the measure on MATH is quasi-invariant under translation by MATH. If this measure is invariant, then the formula for MATH is given by MATH for MATH, MATH, MATH. Here all our groups are discrete, and the translation-invariant measure on the coset space MATH is just the counting measure. Hence for MATH and MATH defined as in the statement of REF we have MATH and MATH so that with respect to these identifications, MATH for MATH, MATH and MATH. Now this is not exactly the same as our formula for MATH, but defining the unitary involution MATH by the formula MATH, one easily checks that MATH for MATH and MATH. This completes the proof of the proposition.
|
math/0003067
|
We have constructed an equivalence between the wavelet representation MATH of REF and the representation MATH of REF, so it only remains to show that MATH is a direct integral of irreducible monomial representations. Now in general the theory of direct integrals of measurable fields of NAME spaces and direct integrals of unitary representations is very technical (see, for example, CITE for a reference). But in our situation, the representation space for the representation MATH is MATH which can also be viewed as MATH, and there are no technicalities involved in showing that this last NAME space is exactly the direct integral MATH. With respect to this decomposition of MATH it is clear that the decomposition of MATH over the measurable subset MATH is given exactly by the representations MATH defined in REF. Finally, REF has shown that for each MATH the representation MATH is an irreducible monomial representation of MATH, and the proof of REF is complete.
|
math/0003067
|
The content of REF shows that under the hypotheses of the Corollary, the proof of REF still works. Keeping the notation of REF , we see that the set MATH will be an orthonormal basis for MATH if and only if the set MATH is an orthonormal basis for MATH. Now MATH where MATH is viewed as a subset of MATH, and since MATH it is clear that for any MATH which is supported on MATH, the function MATH will be supported on MATH, MATH. It follows from this and an inspection of the formula for MATH that if MATH is a set of orthonormal functions whose support lies in MATH which form a basis for MATH, then the set MATH will form an orthonormal basis for MATH. Thus in order to obtain the results of the corollary it is enough to show that the stated conditions on MATH are necessary and sufficient to ensure that the set MATH is an orthonormal basis for MATH. Recall that for MATH we have MATH for MATH and it is well known (compare REF , CITE) that as MATH varies over all of MATH, the functions above will give an orthonormal basis for MATH if and only if MATH is translation congruent to MATH, which is equivalent to the conditions on MATH given in the statement of the corollary.
|
math/0003067
|
Let MATH be a transformation wavelet set for MATH. The proof of REF shows that the wavelet representation MATH is unitarily equivalent to a representation MATH which can be expressed as a direct integral MATH where for each MATH, MATH is an irreducible representation which is unitarily equivalent to MATH defined on the NAME space MATH. We also noted in the proof of REF that for MATH and MATH the representations MATH and MATH are equivalent to one another. Hence the representation MATH, and consequently the wavelet representation MATH, is weakly contained in the set of all of the monomial representations MATH . Let us now consider the regular representation MATH of MATH. Recall that MATH itself is an induced representation, MATH where here MATH represents the trivial subgroup of MATH, and the representation of MATH we are inducing is the trivial one-dimensional representation on MATH. By the theory of induction in stages CITE, we can write MATH . Now let MATH denote the regular representation of the discrete group MATH, that is, MATH . By the NAME theory connecting compact and discrete abelian groups, MATH is equivalent to a representation on the space MATH and in fact can be represented as a direct integral of characters as follows. Setting MATH we have MATH where each MATH, being a character of MATH is exactly a one-dimensional representation on the space MATH. We thus obtain that MATH is equivalent to MATH and since the processes of taking direct integrals of representations and inducing from a subgroup commute, we have that MATH the representation on the right-hand side being defined on the NAME space MATH . Since we have already shown in the proof of REF that MATH we have shown directly that the representation MATH and hence the wavelet representation MATH is weakly contained in MATH . Of course, as with the NAME groups studied in CITE, this also follows immediately from the fact that our group MATH, being the semi-direct product of two abelian groups, is amenable. To show that MATH is weakly contained in MATH, we first note that our hypotheses on MATH together with REF show that if we let MATH be the monomorphism constructed in the proof of REF , then the range of MATH is dense in MATH. Since MATH, MATH is dense in MATH in the usual topology, so that MATH is dense in MATH in the relative topology. It thus follows that MATH is dense in MATH. Hence MATH . The argument that now follows is similar to the proof of REF, which was not used in the proof of REF. We now use the continuity of the induction process (compare CITE) to deduce that MATH where the closure is taken in the hull-kernel topology. We have already shown that MATH is unitarily equivalent to the direct integral of the representations MATH and that for fixed MATH and MATH, MATH is unitarily equivalent to the representation MATH. Since MATH is the direct integral of the representations MATH, it follows from REF that MATH is weakly contained in MATH so that MATH and MATH are weakly equivalent. Finally, as noted in CITE, MATH is faithful since MATH is amenable, and it follows from the definition of weak equivalence that MATH is faithful, thus generalizing REF proved in CITE to the wider class of groups MATH .
|
math/0003068
|
Any self-dual REF-form MATH on any oriented REF-manifold satisfies the NAME formula CITE MATH where MATH is the self-dual NAME tensor. It follows that MATH so that MATH and hence MATH . On the other hand, the particular self-dual REF-form MATH satisfies MATH . Setting MATH, we thus have MATH . But REF tells us that MATH so we obtain MATH . By the NAME inequality, we thus have MATH . Since the NAME inequality also tells us that MATH we thus have MATH as claimed.
|
math/0003068
|
Let MATH be a fixed smooth back-ground metric, and notice that, for MATH, the function MATH is given by MATH by virtue of the weighted conformal invariance of MATH. Thus MATH behaves under conformal rescaling just like the scalar curvature, despite the fact that it might well only be a NAME function. In order to find a suitable choice of MATH, we therefore attempt to minimize MATH on the positive sector of the unit sphere in the NAME space MATH. NAME 's ansatz for doing this CITE is to minimize the functionals MATH and then take the limit of the minimizers as MATH. For each MATH, the existence of minimizers follows from the NAME embedding theorem; indeed, following the proof of CITE, but deleting the very last sentence, one obtains a positive function MATH of class MATH which solves MATH where MATH is the infimum of MATH on the positive sector of the unit sphere of MATH. The convergence as MATH then follows from REF. Namely, since MATH for any metric, and since there is nothing to prove if MATH, our hypothesis that there is no metric of positive scalar curvature in MATH allows us to assume that that MATH. Inspection of REF at a maximum then gives us the MATH estimate MATH for all small MATH. This implies CITE that MATH exists in MATH, and is a weak solution of MATH . NAME theory then tells us that MATH is actually a MATH function. In particular, MATH is a MATH metric, with curvature satisfying MATH in the classical sense.
|
math/0003068
|
Let MATH be the conformal class of some MATH-adapted metric MATH. Since the NAME star operator is conformally invariant in the middle dimension, every metric in MATH is also MATH-adapted. Since MATH is a monopole class for MATH, it therefore follows that MATH does not contain any metrics of positive scalar curvature. REF therefore applies, and tells us that the conformal class MATH contains a metric MATH for which that MATH is a non-positive constant. For this metric, one then has MATH so that MATH by REF . Thus we at least have a MATH estimate concerning the conformally related metric MATH. Let us now compare the left-hand side with analogous expression for the given metric MATH. To do so, we express MATH in the form MATH, where MATH is a positive MATH function, and observe that MATH . Applying NAME, we thus have MATH and hence MATH . This shows that MATH for every MATH-adapted metric MATH. Finally, we observe that MATH for any metric, simply because MATH is trace-free. It follows that MATH at every point, and REF therefore implies the desired REF .
|
math/0003068
|
If equality holds in REF , the last inequality in the proof of REF forces MATH to be the harmonic representative of MATH. Moreover, the NAME inequalities used in that proof shows that equality can only hold if MATH and MATH are constant, and moreover equal. Thus the harmonic form MATH has constant norm, so that MATH is an almost-Kähler form compatible with MATH. Moreover, MATH coincides with the anti-canonical line bundle associated with the almost-complex structure MATH. Since MATH is now a (non-positive) multiple of MATH and since the volume MATH of an almost-Kähler MATH-manifold is exactly MATH, the fact that MATH is a non-positive constant tells us that we now have MATH since by assumption equality holds in REF . It thus follows that MATH with equality iff MATH everywhere belongs to the lowest eigenspace of MATH. However, Blair CITE has shown that MATH for any almost-Kähler MATH-manifold. We thus conclude that MATH is everywhere in the lowest eigenspace of MATH. If we instead have equality in REF or REF , we must in particular be at the minimum of MATH among metrics in the given conformal class. This then implies that MATH is constant, and we therefore also have equality in REF , and the claim follows from the previous argument. The converse assertions follow from CITE and REF . Details are left to the interested reader.
|
math/0003068
|
Let us begin by rewriting REF as MATH where MATH denotes the MATH norm with respect to MATH. By the triangle inequality, we therefore have MATH . We now elect to interpret the left-hand side as the dot product MATH in MATH. Applying NAME, we thus have MATH . Thus MATH so that MATH as claimed. If equality holds, moreover, REF is saturated, and REF applies.
|
math/0003068
|
The proof is a direct extension of the computations in CITE, although for MATH we now use the holonomy-constrained NAME invariant of CITE; compare CITE. Now notice that we may assume that MATH, since otherwise there is nothing to prove. This has the pleasant consequence that any NAME invariant of MATH is independent of polarization, even if MATH. Let MATH denote the first NAME class of a spin-MATH structure on MATH for which the NAME invariant is non-zero, and notice that MATH, because the relevant NAME moduli space must have non-negative virtual dimension. Pull MATH back to MATH via the canonical collapsing map, and, by a standard abuse of notation, let MATH also denote this pulled-back class. Thus, with respect to our given polarization, MATH . Now choose generators MATH for the pull-backs to MATH of the MATH relevant copies of MATH so that MATH . Let MATH be closed curves in MATH which generate of the fundamental groups of the MATH relevant copies of MATH. Then CITE there is a spin-MATH structure on MATH with MATH and MATH . Thus MATH is a monopole class of MATH. But one then has MATH exactly as claimed. If equality held, MATH would be almost-Kähler, and MATH would be the anti-canonical class of the associated almost-complex structure on MATH. But, by construction, MATH whereas MATH by NAME. Since MATH for an almost-complex manifold, equality is thus excluded unless MATH. There is yet more information available, however. Indeed, if equality held, the almost-Kähler class MATH would also necessarily be a non-positive multiple of MATH. On the other hand, our computation shows that equality can only hold if MATH, so it would follow that MATH for all MATH. However, the NAME invariant would then also be non-trivial for a spin-MATH structure with MATH, and REF would then force the homology class MATH to be represented by a pseudo-holomorphic MATH-sphere in the symplectic manifold MATH. But the (positive!) area of this sphere with respect to MATH would then be exactly MATH, contradicting the observation that MATH. Thus equality can definitely be excluded unless MATH and MATH both vanish.
|
math/0003068
|
We may assume that MATH, since otherwise the result follows from the NAME inequality. Now MATH for any metric on MATH on MATH. If MATH is an NAME metric, the trace-free part MATH of the NAME curvature vanishes, and we then have MATH by REF , with equality only if MATH and MATH both vanish. If MATH carries an NAME metric, it therefore follows that MATH . The claim now follows by contraposition.
|
math/0003068
|
For MATH any even integer bigger than MATH million, CITE has constructed a simply connected minimal complex surface MATH of general type (in fact, a hyperelliptic fibration) with MATH and MATH. If we now blow up such a surface at MATH points, where MATH, the resulting simply connected MATH-manifold MATH does not admit NAME metrics by REF . Now for this manifold we have MATH and the right-hand side sweeps out MATH as MATH ranges over MATH. This proves the claim for MATH negative. The case of MATH positive then follows by reversing orientation.
|
math/0003068
|
If MATH, REF becomes MATH and the desired inequality is therefore an immediate consequence. Moreover, if equality holds, REF applies, and MATH is an almost-Kähler manifold with the relevant special properties.
|
math/0003068
|
As in the proof of REF , there is a monopole class MATH of MATH with MATH and such that MATH . If MATH carried an anti-self-dual NAME metric, REF would then tell us that MATH so that MATH . Moreover, equality could only hold if MATH and MATH both vanished for precisely the same reasons delineated in the proof of REF . The result now follows by contraposition.
|
math/0003068
|
We again begin with REF , MATH but this time interpret the left-hand side as the dot product MATH in MATH. Applying the NAME inequality, we then obtain MATH . Thus MATH and MATH as claimed. If equality holds, our use of NAME forces MATH which is to say that MATH . In this case, we then have MATH and it then follows CITE that the metric is NAME, and has constant scalar curvature. Conversely, a NAME surface with constant scalar curvature satisfies MATH and MATH, so that equality is attained for any such metric.
|
math/0003068
|
Since REF tells us that MATH it follows that MATH for any metric on MATH on MATH. The result now follows immediately from REF .
|
math/0003068
|
Since MATH REF tells us that it suffices to prove that MATH . Now observe that the proof of REF tells us that that, for every metric MATH on MATH, there is a spin-MATH structure MATH with non-zero NAME invariant such that MATH . Hence MATH for every metric MATH on MATH. It follows that MATH . To finish the proof, it suffices to produce a family of metrics MATH on MATH such that MATH . To do this, let MATH be the pluri-canonical model of MATH, which carries a NAME orbifold metric MATH by an immediate generalization CITE of the NAME/NAME solution CITE of the MATH case of the NAME conjecture. Now MATH because MATH for any NAME surface, whereas a NAME surface has NAME form given by MATH, where MATH is the NAME form. Let MATH be the orbifold points of MATH, if there are any. Each of these singularities is of type A-D-E, meaning that MATH has a neighborhood is modeled on MATH for some discrete sub-group MATH of MATH. The passage from MATH to MATH is accomplished by replacing each MATH by a collection of MATH-curves with intersections determined by the NAME diagram (NAME graph) of type A-D-E corresponding to MATH. One can now construct smooth metrics on the minimal model MATH by modifying the orbifold metric MATH, without introducing substantial amounts of extra volume, NAME curvature, or self-dual NAME curvature. Indeed, for each MATH there are CITE gravitational instanton metrics on the minimal resolution of MATH, which is precisely obtained by replacing the origin with a set of MATH-curves as above. These gravitational instanton metrics are NAME and anti-self-dual, and they closely approximate the Euclidean metric on MATH, being of the form MATH in the asymptotic region, where MATH is the Euclidean radius; moreover, the first and second derivatives of such a metric are of order MATH and MATH, respectively, in these coordinates. Fix such a metric, and consider the family of gravitational instanton metrics MATH obtained by multiplying it by MATH, and then making the homothetic change MATH of asymptotic coordinates. The metric MATH is then uniformly MATH on the complement of the ball of radius MATH, with first and second derivatives MATH and MATH, respectively. Take geodesic spray coordinates about each of the orbifold points MATH, so that MATH takes the form MATH in these coordinates. Delete the MATH ball around each orbifold point, and replace it in the MATH region of the corresponding gravitational instanton; and on the transitional annulus MATH, consider the metric MATH where MATH is a fixed smooth function which is MATH near MATH and MATH near MATH. This gives us a family of metrics on MATH, the geometry of which will be analyzed in a moment. However, what we really need is a metric on MATH rather than on MATH, so we need to modify MATH a bit more. Now MATH and MATH both admit anti-self-dual metrics of positive scalar curvature - namely, the obvious product metric on MATH, and the NAME metric on the reverse-oriented complex projective plane MATH. The complement of any point in MATH or MATH therefore admits an asymptotically flat metric with MATH and MATH, namely the standard metric rescaled by the NAME Greens function of the appropriate point. One can then find asymptotic charts for these manifolds, parameterized by the complement of, say, the unit ball in MATH, in which these metrics differ from the Euclidean metric by MATH and in which first and second partial derivatives are respectively MATH and MATH. (Notice the fall-off rate is slower than in the previous case; these metrics have positive mass!) Again multiplying by MATH, where MATH is to be viewed as an auxiliary parameter, and making a homothety as before, we thus get metrics on the complement of the MATH ball in MATH of the form MATH, with first and second partial derivatives of order MATH and MATH, respectively. Again, let MATH denote the restriction of the resulting metric on the the MATH regions of MATH disjoint punctured copies of MATH and MATH disjoint punctured copies of MATH. Take geodesic spray coordinates around non-singular points MATH of MATH, and delete MATH-balls around each one, and define a metric MATH on MATH as MATH when MATH, as MATH when MATH, and as REF on the transitional annuli MATH. The sectional curvature of the metrics MATH are then uniformly bounded on the annuli MATH as MATH, while the volumes of these annuli are MATH as MATH. Since the metrics we have used to replace the MATH-balls are all scalar-flat and anti-self-dual, we therefore have MATH and it thus follows that MATH . The theorem follows.
|
math/0003068
|
Complex surfaces of NAME dimension MATH, are, by definition, precisely those of general type; for these, the result is just the MATH case of REF . On the other hand, the analogous assertion for NAME dimensions MATH and MATH immediately follows from the fact CITE that any elliptic surface admits sequences of metrics for which MATH, but for which MATH and MATH are uniformly bounded.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.