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math/0105028 | Recall that the matrix part MATH of MATH is an element of MATH and determines the MATH-equivalence class of the MATH-link, as follows from CITE. Recall also that the wrapping relations MATH and MATH depend only on MATH. Thus, it follows that MATH-equivalent MATH-links have equal matrix-part, and that the matrix-free part of MATH makes sense as an element of the graded vector space MATH. CITE implies that MATH is a good boundary link, (that is, MATH) iff its associated matrix MATH is invertible over MATH. In that case, the construction of the MATH invariant implies that MATH. |
math/0105028 | The MATH block follows from the fact that MATH is an unlink. The MATH block follows from the fact that MATH is a meridian of MATH, using the formula for the NAME integral of the NAME Link, together with the fact that the linking number between MATH and MATH is zero. |
math/0105028 | Since MATH, it suffices to consider the case of even MATH. Suppose that MATH (for MATH) is a collection of null claspers in MATH each of degree MATH, and let MATH denote the set of arms of MATH. REF and its following discussion implies that MATH and that the nonzero contribution to the right hand side come from diagrams in MATH that touch all arms. Thus, contributing diagrams have at least MATH-colored legs, to be glued pairwise. Notice that the diagrams in MATH contain no struts. Thus, at most three MATH-colored legs meet at a vertex, and after gluing the MATH-colored legs we obtain trivalent graphs with at least MATH trivalent vertices, in other words of NAME degree at least MATH. Thus, MATH, which implies that MATH is a invariant of acyclic MATH-links of type MATH with values in MATH. |
math/0105028 | It suffices to consider a collection MATH of claspers in MATH each of degree MATH. Let MATH denote the set of arms of MATH. The counting argument of the above Proposition shows that the contributions to MATH come from complete contractions of a disjoint union MATH of MATH vortices. A vortex is the diagram MATH, the next simplest unitrivalent graph after the strut. Furthermore, the MATH legs of MATH should touch all MATH arms of MATH. In other words, there is a REF correspondence between the legs of such MATH and the arms of MATH. Consider a leg MATH of MATH that touches an arm MATH of MATH. If MATH touches MATH, then due to the restriction of the negative inverse linking matrix of MATH (see REF ), it needs to be contracted to another leg of MATH that touches MATH. But this is impossible, since the legs of MATH are in REF correspondence with the arms of MATH. Thus, each leg of MATH touches presicely one edge of MATH. In particular, each leg of MATH is colored by three edges of MATH. Consider a vortex colored by three edges of MATH as part of the MATH invariant. Since the three edges of MATH is an unlink in a ball disjoint from MATH, it follows that the beads on the edges of the vortex are MATH, if the three edges belong to the same MATH, and MATH otherwise. Thus, the diagrams MATH that contribute are a disjoint union of MATH vortices MATH and these votrices are in REF correspondence with the set of claspers MATH, in such a way that the legs of each vortex MATH are colored by the edges of a unique clasper MATH. After we glue the legs of such MATH using the negative inverse linking matrix of MATH, the result follows. |
math/0105028 | Denote the components of MATH and MATH by MATH and MATH respectively. Consider the covariance matrix of MATH with respect to MATH, presented with respect to this basis: MATH where MATH and MATH, where here and below MATH denotes the transpose of MATH, followed by the conjugation MATH. If the canonical decomposition of MATH with respect to MATH is then, MATH (suppressing the summations) then the canonical decomposition of MATH with respect to MATH is MATH . Using the identities of REF , we have MATH . It is easy to see that if MATH and MATH are invertible over MATH, then so is MATH over MATH, see CITE. Thus, the first part of the theorem follows. Continuing, MATH . The identification of the term on the left of the pairing with MATH completes the proof. |
math/0105028 | For the first statement, assume, for convenience, that MATH and MATH, and rename MATH and MATH by MATH and MATH respectively. Applying REF , introducing a labeling set MATH (respectively, MATH), bijective with MATH (respectively, MATH) (and associated sets MATH and MATH), gives the expression in question as: MATH . If the canonical decomposition of MATH with respect to MATH is MATH then the previous expression is equal to MATH . Now, because MATH is a MATH-substantial operator, this gives the required decomposition of MATH with respect to MATH. For the second statement, we compute MATH . To proceed, we must adjust this calculation so that it can distinguish struts, appearing in the left of the pairing, according to which factors, appearing in the right of the pairing, they are glued to. This adjustment is made with REF , and leads to the following (introducing sets MATH, MATH and MATH, bijective copies of MATH): MATH . The calculation proceeds by removing factors appearing in left hand side of the pairing, at the expense of operating with them on the right hand side, follow REF . The first term to be treated in this way will be MATH giving (following an application of REF ): MATH . The next term to be so treated will be MATH which gives: MATH . The result of these operations is MATH . Performing the pairing with respect to the variables MATH, leads to the replacement of the the inner MATH with MATH . Trading the factor MATH gives: MATH . Performing the pairing with respect to MATH sets MATH to zero, and thus we arrive at the following expression. MATH . The proof follows by inspection. |
math/0105028 | Choose generators MATH for the free group MATH and identify MATH with the ring of noncommutative power series MATH. Recall the map MATH given by MATH. The NAME integral is calculated by splitting the quadratic part from the rest of the integrand as follows: MATH . With the following notation for ``NAME and ``NAME struts, MATH it follows that the desired integral equals to the following formal series (which makes sense, since finitely many terms appear in each degree) MATH . Specifically, the answer is recovered if each ``join" is accompanied by a NAME delta. For example: MATH . We focus on a particular summand. It will help in the management of the combinatorics if we label each of the arcs that appear in the struts: MATH . Now we introduce some definitions to assist in the enumeration of the polygons that arise from this pairing. CASE: An NAME of degree MATH is a collection of even-sided polygons whose edges are alternately colored red and green. The total number of edges is MATH. CASE: A labeled NAME of degree MATH is an NAME with an extra bivalent vertex in each edge, such that: CASE: the resulting set of red (respectively, green) edges is equipped with a bijection to MATH (respectively, MATH ), CASE: if an edge is labeled, for example MATH (respectively, MATH), then its neighboring red (respectively, green) edge is labeled MATH (respectively, MATH). CASE: Let the constant MATH, for MATH some NAME, denote the number of labeled NAME which recover MATH when their labels and extra bivalent vertices are forgotten. Observe that: MATH . To calculate this constant describe an NAME of degree MATH, MATH, by a sequence MATH where MATH counts the number of polygons with MATH edges. MATH . Fix a copy of MATH with its set of edges ordered (any old how, so as to distinguish them from each other). There are precisely MATH ways of labeling this so as to respect the conditions of the above definition. But this overcounts by a multiplicative factor of the order of the symmetry group of MATH (that is, with the ordering of the edges forgotten). That symmetry group is precisely the following group (that is, swaps of identical polygons, and rotations and flips of individual polygons): MATH where MATH represents the MATH-th dihedral group. Thus we conclude: MATH . If MATH denotes a connected polygon with MATH edges, it follows that MATH . Substituting the definitions of the ``NAME and ``NAME struts into MATH yields MATH from which the result follows. |
math/0105028 | First, we need to show the existence of such a MATH. A coordinate approach is to define MATH on the ring MATH of REF which is isomorphic to MATH. Given the description of MATH, it is easy to see that MATH exists and is uniquely determined by the above properties. An alternative, coordinate-free definition of MATH is the following. Let MATH denote the NAME localization of the group-ring MATH, followed by a completion with respect to MATH and MATH (where MATH and MATH are generators of MATH). Let MATH and MATH. The map MATH is MATH -inverting, thus it extends to a map of the localization, and further to a map MATH of the completion. MATH is a MATH-graded ring, and we can define MATH by MATH . It is easy to see that this definition of MATH satisfies the properties of the lemma, and that the properties characterize MATH. Finally, REF implies that the restriction of MATH on MATH equals to MATH. |
math/0105028 | In view of REF , it suffices to show that the maps agree on generators MATH of MATH. For MATH, both maps vanish. For MATH, the following identity MATH and REF show that both maps agree. The result follows. |
math/0105028 | Observe the identity MATH for matrices MATH, MATH (not necessarily square) that can be multiplied. It implies that MATH . The lemma follows by induction on MATH and the above identity which solves MATH in terms of quantities known by induction. |
math/0105028 | Consider a matrix presentation of MATH. That is, MATH for MATH and MATH a column vector over MATH. REF implies that MATH . The proposition now follows from REF . |
math/0105028 | The proof follows from a careful combinatorial counting similar to that of the Wheels Identity of REF . Let us precede the proof with an informal examination of what takes place when we repeatedly apply MATH to the empty diagram, MATH. The first application results in a term: MATH . The second application requires the result of applying MATH to MATH. This can be done using the following special case of REF : MATH . So, the second application results in the following terms: MATH . We see that the terms arising from the application of MATH fall into three categories. MATH . Thus, we see that MATH equals to a linear combination of diagrams, each consisting of a number of wheel components, each of which is of the following form (this can be (over-)determined by a sequence MATH of positive integers): MATH . The question, then, is to calculate an expression in terms of such diagrams, counting ``all ways" that they arise from the application of moves REF. We will generate a sequence of such moves with the following object, an edge-list on a shape. The shape determined by a set of positive integers MATH is a set of oriented polygons, one with MATH edges, and so on, such that each has a distinguished bivalent vertex, its basepoint. A labeled shape is a shape with its edges labeled by positive integers. The diagram MATH determined by some labeled shape MATH is the diagram that arises by replacing each polygon by a series of wheels, in the following manner (to distinguish them from beads, the bivalent vertices of the polygons will be drawn by small horizontal segments, and the base-point by an x). MATH . An edge-list on a shape is a finite sequence of the edges of the shape such that each edge is represented at least once. (It is convenient to enumerate the edges along the orientation, starting from the basepoint.) The labeled shape associated to some edge-list on that shape is that shape with each edge labeled by the number of times that edge appears in the edge-list. An edge-list MATH on a shape will determine a sequence of moves REF as follows. CASE: If some MATH is the first appearance in the edge-list of an edge from some polygon with MATH edges, then that term determines a move REF, where for the purposes of discussion the introduced wheel is oriented with MATH beads, with one such distinguished as the ``first", and with the label MATH sitting on bead number MATH, and with all other beads labeled REF. CASE: If some MATH is the first appearance in the edge-list of some edge of a polygon, some other edge of which has already appeared in the edge-list, then that term determines a move REF applied to the factor of MATH that immediately follows the edge MATH on the (oriented) wheel. (That is, the factor of MATH that is introduced``slides" back around the polygon from there to the bead number MATH.) CASE: If some edge MATH has already appeared in the path (say, MATH times) then that step determines the application of a move REF to the factor MATH appearing on the bead number MATH. Observe that the result of these moves will be the diagram determined by the labeled shape associated to the edge-list. Observe also that the number of different edge-lists on a shape which give rise to a given associated labeled shape, for example, MATH . Two edge-lists determine the same sequence of moves if and only if they are related by symmetries of the shape: CASE: rotating individual polygons (that is, shifting the ``base-point"), CASE: permuting a subset of polygons which all have the same number of edges. We now define a function MATH from the set of shapes to the positive integers as follows. If a shape MATH is specified by a sequence MATH, where MATH counts the number of polygons with one edge, MATH counts the number of polygons with MATH edges, and so on, then MATH . We are now ready to prove REF . First consider the case where the given diagram MATH is empty. MATH as needed. In case MATH is not empty, assume for convenience that there are MATH beads on MATH, and explicitly record their labels MATH. Now, MATH and the Proposition follows from the above case of MATH. |
math/0105028 | Consider two pairs MATH and MATH related by a wrapping move MATH as in REF . In other words, for some MATH we have that MATH for MATH where MATH is the identity and MATH. Consider the case that MATH has a diagram with MATH legs labeled by MATH. REF implies that MATH . Let MATH denote the result of gluing all but one of the MATH legs of MATH and renaming the resulting one MATH. In other words, MATH. Let MATH denote the action of MATH for MATH-labeled legs. If MATH has at least one MATH leg, it follows that MATH by the infinitesimal wrapping relations, and it follows that MATH by definition. If MATH has no MATH leg, then MATH. Thus, in all cases, MATH. The converse is obvious; thus the result follows. |
math/0105028 | It is obvious that the MATH relation implies the group-like MATH relation of REF. Since MATH, the result follows. |
math/0105028 | The first part follows from the proof of CITE without essential changes. The second part is standard. |
math/0105028 | Consider the decomposition of MATH given by REF . We begin with a ``MATH-function trick" of REF: MATH . Thus, MATH . Since MATH, the integration by REF implies that MATH . Completing the square, using REF , implies that the above equals to: MATH . Returning to the expression in question: MATH . The second factor equals to MATH, and concludes the first part of the theorem. In order to compute the first factor, we need to separate its MATH-substantial part. With the added assumption on MATH, we can compute the first first factor in a manner analogous to the Wheels Identity of REF MATH . |
math/0105028 | Use the form of the group-like basing relations given in REF (see REF ), and follow the proof of case MATH of REF , using REF . The result follows. |
math/0105030 | By contradiction, suppose that no subset of cardinality MATH of MATH is linearly independent. This means that the matrix whose rows are the rows of MATH indexed by MATH has rank strictly less than MATH. Consequently, we can find rational numbers MATH not all zero, such that MATH, for all MATH. But then at least half of the line MATH is contained in MATH, contradicting that this set is bounded. |
math/0105030 | We use the technique of NAME duality, as introduced in CITE. Consider the vector configuration MATH. Let MATH be its NAME dual configuration. We define MATH as the set of indices MATH such that reversing the MATH-th inequality in MATH produces a bounded set. Now, MATH if an only if the configuration MATH is totally cyclic, that is, if and only if there exists a positive linear dependence among the elements of MATH. Using results from CITE, namely that the dual operation to deletion is contraction, and that dual of a totally cyclic vector configuration is an acyclic vector configuration, we see that MATH if and only if MATH is acyclic, that is, if and only if there is a linear functional MATH such that MATH for all MATH. But this can only happen if MATH is an extreme ray of MATH, which means that there exists MATH such that MATH and MATH for all MATH, MATH. Since a configuration in MATH has at most two extreme rays, we conclude that the cardinality of MATH is at most MATH. The last assertion of this lemma follows from the same arguments used to prove REF . |
math/0105030 | This follows from REF. |
math/0105030 | The first assertion follows from CITE. To see that REF implies REF , notice that, since MATH is a NAME prime ideal , MATH is NAME, and thus free of embedded primes, for all MATH. As a consequence, for each MATH, the ideal MATH, which is the ideal of MATH generated by MATH, is free of embedded primes. The second implication follows from REF. Finally, MATH being NAME implies that MATH is NAME. |
math/0105030 | Suppose that the hyperplanes MATH, for MATH, are pairwise distinct. To check the assertion, first notice that, since MATH is an associated prime of MATH, it is contained in a minimal prime MATH of this initial ideal. Here MATH. Thus, the set obtained from MATH by removing the inequality MATH must be a simplex. Now, by REF , any lattice point MATH in this simplex satisfies MATH, and such lattice points exist. Pick MATH in our simplex such that MATH is minimal. Notice that the lattice point MATH is unique, since the weight vector MATH is generic. This follows from results in CITE. Now let MATH and consider the pair MATH. It is easy to check that this pair satisfies the conditions of REF for the weight vector MATH. Thus, interchanging the first and MATH-th columns of MATH, and replacing MATH by MATH, we obtain a standard pair as we desired. Finally, if the hyperplanes MATH and MATH coincide, for MATH, we use the above argument to change the weight vector MATH by the weight vector MATH, and assume that MATH and MATH are parallel. Of course, the first row of MATH is a multiple of the MATH-th row of MATH. If it were a positive multiple, then for each MATH, MATH does not contain the variable MATH, which implies MATH, a contradiction. Thus the first and MATH-th rows of MATH are negative multiples of each other. To see that MATH, notice that MATH would contradict the last assertion of REF . |
math/0105030 | Suppose that REF does not hold. Then there are MATH, MATH such that MATH, and MATH. Since both MATH and MATH have minimal negative support (and their only integer coordinates are indexed by MATH by the choice of MATH) we conclude that MATH for some MATH, and MATH. Suppose that the standard pair that gives rise to MATH is top-dimensional. Then it is of the form MATH for some MATH. Since MATH, the standard pair corresponding to MATH is of the form MATH. Suppose MATH. Remember that we have MATH such that MATH. Then MATH for MATH, and MATH. Then MATH, a contradiction. Thus MATH. We consider two cases, as in the proof of REF . If two of the hyperplanes MATH, MATH, coincide, then MATH is equal to MATH for some MATH. This implies that MATH, because otherwise the sets MATH and MATH are unbounded. This contradicts the previous paragraph. Now suppose that the hyperplanes MATH, MATH, are pairwise distinct. Since MATH and MATH are simplices, reversing the MATH-th or the MATH-th inequality in our original polytope MATH yields bounded sets. By REF , if we reverse the inequality MATH we also get a bounded set. In view of REF , we derive a contradiction. In conclusion, the standard pair corresponding to MATH cannot be top-dimensional, so it must be of the form MATH, by our choice of MATH. However, by REF , a fake exponent of MATH coming from such a standard pair cannot have a negative integer first coordinate. This contradiction concludes the proof. |
math/0105030 | To see that MATH is a monomial, it is enough to show that MATH. Remember that MATH. Since MATH is a fake exponent with respect to MATH, there is a standard pair MATH of MATH such that MATH for MATH. Pick MATH such that MATH. Then, since MATH, MATH for MATH, and MATH. This means that MATH, so that, by REF , MATH. The rest of the assertions are trivial. |
math/0105030 | Let MATH be a (possibly logarithmic) solution of MATH such that MATH. The function MATH is a linear combination of canonical series with respect to MATH. Write MATH where each MATH is a linear combination of canonical series whose exponents differ by integer vectors, and the exponents in MATH and MATH do not differ by integers if MATH. It is clear that MATH for MATH, so we can reduce to the case when MATH is a linear combination of canonical series solutions whose exponents differ by integer vectors, and we assume this from now on. Write MATH in the form of REF for some MATH, so that MATH where MATH contains only logarithmic terms that less than or incomparable to MATH. Since MATH, MATH is a logarithm-free solution of MATH that is constant with respect to MATH. Thus, MATH is a linear combination of logarithm-free canonical series whose corresponding fake exponents differ by integer vectors, and have first coordinate equal to zero. This means that MATH is a linear combination of functions MATH with MATH, differing pairwise by integer vectors. Now rewrite the function MATH in the form of REF , that is MATH. By the previous reasoning, we can choose MATH such that MATH. We will show that MATH is a constant. Let MATH be a standard pair of MATH such that MATH for MATH. By our choice of MATH, MATH implies MATH, so that MATH. We first consider the case when MATH. Pick MATH. By REF applied to MATH, there exists MATH such that MATH for MATH, MATH and MATH. Now choose a term MATH appearing with nonzero coefficient in MATH such that MATH appears to a maximal power (among the terms in MATH). Since MATH, MATH . Now, MATH contains a summand that is a nonzero multiple of MATH. If MATH, then MATH contains a term that is a nonzero multiple of MATH . By the construction of MATH, this term is nonzero. However, it cannot be cancelled with any other term from MATH, because such a term would have to come from MATH, and MATH was chosen maximal among those terms. This contradiction implies that MATH. We conclude that MATH is constant with respect to the MATH-th variable, and this is true for all MATH. However, by REF MATH where MATH, the vectors MATH belong to the kernel of MATH, and MATH . Since MATH is constant with respect to the MATH-th variable for all MATH, MATH for MATH, MATH, MATH. By REF , MATH of the rows of MATH indexed by MATH are linearly independent. This implies that all the vectors MATH must be zero, and we conclude that MATH is constant. Now we need to show that MATH is constant in the case when the standard pair corresponding to MATH is MATH, and MATH strictly contains MATH. In this case, MATH, for some MATH. Clearly, if MATH, the previous arguments still apply, because no matter what MATH is, the term REF cannot vanish. Thus, we may assume that MATH. The polytope MATH is a simplex, and MATH. This means that REF applies, and so there is a set MATH of cardinality MATH such that, for each MATH, the set obtained from MATH by reversing the inequality MATH is unbounded. Notice that, by REF , we may assume that MATH. If MATH, we can choose MATH such that MATH for MATH, MATH, and MATH. Now we can repeat what we did before, and conclude that MATH is constant with respect to the MATH-th variable, for all MATH . Since the rows of MATH indexed by MATH are linearly independent, using REF , we conclude that MATH is constant. Now, remember that MATH contains a term MATH that is maximal among the logarithmic terms appearing in MATH. Since MATH is constant, MATH, which implies that MATH is logarithm-free. But then it is clear that MATH belongs to MATH. |
math/0105030 | Pick MATH corresponding to a standard pair MATH. Clearly, MATH is the fake exponent of MATH corresponding to this standard pair. Suppose that MATH does not have minimal negative support. Then we can find MATH such that MATH has minimal negative support strictly contained in MATH. In particular, MATH. Notice that MATH, because otherwise MATH. This implies that the functional MATH is minimized in the set MATH. This minimum must be unique since MATH is a generic weight vector. As a consequence, we may assume that MATH is a fake exponent of MATH with respect to the weight vector MATH. Then MATH is a fake exponent of MATH, it has minimal negative support, and MATH. The last two facts follow since MATH has minimal negative support. But now this contradicts REF . |
math/0105030 | Suppose there is a solution MATH of MATH such that MATH lies in MATH. As in the proof of REF , we may assume that MATH is a linear combination of canonical series whose exponents differ by integer vectors. We proceed as in the part of the proof of REF where we show that the functions MATH are logarithm-free. The first step is to write MATH for every MATH as in REF . The function MATH belongs to the kernel of MATH, so we may choose MATH such that MATH appears with a nonzero coefficient in MATH. Now we rewrite MATH as in REF . We want to compute the polynomial MATH. As in REF we consider two cases, according to whether the standard pair MATH corresponding to the fake exponent MATH is embedded or not. In the first case, MATH, and MATH. Choose MATH and a vector MATH such that MATH for MATH, MATH, and MATH. We can do this by REF . Now choose a term MATH appearing with nonzero coefficient in the polynomial MATH such that MATH appears to a maximal power (among the terms in MATH), and consider the term REF from the proof of REF : MATH which appears with a nonzero coefficient in MATH. Since MATH, this function is a further derivative of MATH. As MATH, there are no fake exponents MATH such that MATH. Thus MATH must be a multiple of MATH, which has neither logarithmic terms, nor terms which contain a strictly negative integer power of MATH. We conclude that MATH. The same argument as in REF now implies that MATH is constant. This is a contradiction because MATH can only appear in MATH if MATH has a term MATH. Now we need to consider the case when the standard pair MATH is top-dimensional, that is MATH, for some MATH. If MATH, the same argument as above shows that MATH is constant. If MATH, we need to combine the previous argument with the reasoning from REF to again conclude that MATH is constant. Finally if MATH is a negative integer, we observe that MATH is a linear combination of functions MATH, where the vectors MATH belong to MATH and differ with MATH by an integer vector. But then, for each MATH, a further derivative of MATH has no logarithmic terms, or terms that contain strictly negative integer powers of the variable MATH and MATH. Unless MATH is constant with respect to the MATH-th variable, this contradicts the fact that MATH contains a nonzero multiple of the term REF . As before, we conclude that MATH is constant, a contradiction. |
math/0105030 | Let MATH for MATH. If we can show that the linear functional MATH is minimized uniquely in the set MATH, then this set will contain a fake exponent, which is what we want to prove. Since the weight vector MATH is generic, it is enough to see that MATH is minimized in MATH, or equivalently, that it is maximized in MATH. As in the proof of REF there are two cases. Either the hyperplanes MATH are all distinct, for MATH, or two of those coincide. In the second case, we know that MATH equals MATH for a certain MATH. But then, for each MATH, MATH implies MATH. Since the MATH-th row of MATH is a negative multiple of the first row, the linear functional MATH is bounded above in MATH. Thus it is maximized in MATH, and we obtain logarithm-free canonical solutions MATH of MATH corresponding to MATH for MATH. Now assume that the hyperplanes MATH are pairwise different for MATH, and remember that MATH is an embedded prime of MATH. Then there is a minimal prime of this initial ideal of the form MATH, for some MATH. Since there must be standard pairs corresponding to this minimal prime, we conclude that the set obtained from MATH by deleting the inequality MATH is a simplex. Thus MATH is bounded, so that MATH will be maximized in MATH, and we obtain MATH, the corresponding solution of MATH. Notice that, by REF , the linear functional MATH attains no maximum in MATH for MATH. |
math/0105030 | First notice that if MATH is a logarithm-free solution of MATH, and MATH lies in the span of the functions from REF , then MATH is a linear combination of canonical, logarithm-free solutions of MATH that are either constant with respect to MATH, or whose fake exponents differ by an integer vector with MATH. Since MATH, the only such fake exponent is MATH. Thus MATH. Now let MATH be a logarithmic solution of MATH such that MATH lies in the span of the functions constructed in REF . We may assume that MATH does not contain a summand MATH, MATH. As in the proof of REF , we can write MATH, where MATH contains all the maximal logarithmic terms. Pick MATH, and remember the vector MATH from the paragraph before REF . An argument similar to that in the proof of REF shows that, if MATH contains a power of MATH greater than REF, then MATH cannot be logarithm-free. This and REF imply that MATH, where the vector MATH lies in the kernel of MATH. Thus, MATH where MATH is logarithm-free, and contains no terms MATH, MATH. Notice that, if MATH does not give rise to a solution of MATH as in REF , then MATH. This follows from the same arguments that proved REF . We claim that, once the numbers MATH are fixed, the function MATH itself is fixed. This is because the difference between two such functions would be a logarithm-free solution of MATH whose first derivative lies in the span of the functions from REF . By the first paragraph of this proof, this implies that our difference lies in the kernel of MATH. But since it cannot contain terms in MATH, MATH, we conclude that this difference must be zero. Now we consider two cases, as in the proof of REF . In the case that a hyperplane MATH coincides with MATH, we have MATH linearly independent functions MATH, MATH from REF . But if MATH lies in the span of these functions, we know that MATH where MATH belongs to the kernel of MATH, and MATH. Since the dimension of the kernel of MATH is MATH, we conclude that the dimension of the space of such functions MATH is at most MATH. Thus, the intersection of the image of MATH and the span of the functions MATH, MATH has dimension at most MATH. In the case when all the hyperplanes MATH, MATH are distinct, we have only two functions, MATH and MATH from REF . Thus, if MATH lies in their span, MATH where MATH belongs to the kernel of MATH, and MATH, for MATH. By REF , this implies that MATH is the zero vector, so that MATH vanishes. Thus the intersection of MATH with the image of MATH is MATH, which has dimension MATH. |
math/0105030 | In REF we produced one function in the cokernel of MATH for each function in the kernel of MATH. Furthermore, in REF , we produced at least MATH linearly independent functions in the cokernel of MATH corresponding to MATH. All these functions are clearly linearly independent in the cokernel of MATH. This means that: MATH . Adding MATH to both sides of this inequality we obtain: MATH or equivalently, MATH . Since MATH, it follows that MATH . |
math/0105030 | We will prove our claim by presenting an algorithm to compute MATH for a given homogeneous matrix MATH. This algorithm will rely on NAME basis computations in the NAME algebra. In this proof we will think of MATH as a parameter vector, that is, a vector of indeterminates, instead of an element of MATH. Thus, the MATH-hypergeometric system MATH will no longer be an ideal in the NAME algebra MATH, but an ideal in the parametric NAME algebra MATH, where the parameters MATH commute with the variables MATH and MATH. Given a vector in MATH, the specialization map corresponding to this vector is the map from MATH to MATH that replaces the parameters MATH by the coordinates of our vector. This allows us to introduce the concept of a comprehensive NAME basis of a given left ideal MATH in the parametric NAME algebra. Informally, a set MATH is a comprehensive left NAME basis of MATH if, for every vector in MATH, the specialization of MATH with respect to that vector is a NAME basis of the corresponding specialization of MATH. We are now ready to describe an algorithm to compute MATH. Input: A homogeneous matrix MATH. Output: The exceptional set MATH. CASE: Compute a comprehensive left NAME basis of MATH with respect to the weight vector whose first MATH entries are zeros, and whose last MATH entries are ones. Call this basis MATH. CASE: There are only finitely many initial ideals MATH under specialization of the parameter MATH. These are ideals in the polynomial ring MATH, call them MATH. CASE: For each MATH, the comprehensive NAME basis MATH yields finitely many polynomial conditions of the form MATH or MATH which characterize the (constructible) subset MATH of MATH such that if we specialize MATH to an element of MATH, the initial ideal of the specialized MATH-hypergeometric system is MATH. CASE: For MATH compute MATH . CASE: MATH . By CITE, the number MATH produced in the fourth step of the previous algorithm equals the rank of any MATH-ideal whose initial ideal with respect to MATH is MATH. This justifies the description of MATH from the last step of the algorithm. |
math/0105031 | Let MATH be the reduction of MATH, and MATH the function such that MATH. Write MATH where each MATH has degree at most MATH. Let MATH be the roots of MATH over MATH and MATH the corresponding points on the curve MATH. Then MATH has poles at MATH and possibly at infinity. Let MATH be the completion of the local ring of MATH at MATH, and let MATH the fraction field of MATH; then the maximal ideal of MATH is generated by MATH, and within MATH, MATH can be written as a power series in MATH with integral coefficients. Then the image of MATH in the module MATH of differentials can be written as MATH, and the MATH are integral for MATH (since they coincide with the corresponding coefficients in the expansion of MATH). The map MATH commutes with the passage to the completed local ring, so the image of MATH in MATH is equal to MATH. Now note that MATH has a pole of order at most MATH at each MATH, and its image in MATH has leading term MATH. Consequently, if MATH is an integer such that MATH is integral for MATH and MATH, then MATH is integral. Specifically, we have that MATH is integral for MATH; since the MATH are distinct modulo MATH, that implies that MATH is integral. Applying the same argument to MATH, we deduce that MATH is integral, and so forth. In particular, we may take MATH. Then MATH is integral, as is the reduction of MATH, which yields the desired conclusion. |
math/0105034 | In both cases we are cutting MATH along a two-sided torus or NAME bottle, so MATH-irreducibility is obvious. If MATH, this torus or NAME bottle is incompressible in MATH, and irreducibility of MATH is a general fact CITE. We are left to show that if MATH sharply then MATH and MATH are irreducible. Since they have boundary, it is enough to show that they are prime. Suppose they are not, and consider prime decompositions of MATH and MATH involving summands MATH and MATH. So one summand MATH is assembled to one MATH, and the other MATH's and MATH's survive in MATH. It follows that, up to permutation, MATH is prime, MATH with MATH and MATH prime, MATH and MATH. Sharpness of the original assembling and additivity under MATH now imply that MATH. So REF applies to MATH and MATH. Knowing that MATH it is easy to deduce that MATH and MATH are either MATH or MATH, and that the original assembling was a trivial one. A contradiction. |
math/0105034 | The result is true for MATH, so we proceed by induction on MATH and suppose MATH. By REF we can assume that MATH cannot be split as MATH, because every assembling with MATH is sharp, and we have seen that MATH is a brick. Now if MATH is a brick we are done. Otherwise MATH is either a sharp self-assembling MATH, but in this case MATH and we conclude by induction using REF , or MATH is a sharp assembling MATH. REF states that MATH and MATH are irreducible and MATH-irreducible. If both MATH and MATH have positive complexity we conclude by induction. Otherwise we can assume that MATH and apply REF . Since the assembling is non-trivial, MATH is not of type MATH. It is also not MATH or MATH for MATH, by the property of MATH we are assuming. So MATH is one of MATH, MATH, MATH, MATH. In particular, it is a brick. Now we claim that MATH cannot be split as MATH. Assuming it can, we have two cases. In the first case the assembling of MATH is performed along a free boundary component of MATH, but then we must have MATH, and the assembling is trivial, which is absurd. In the second case MATH is assembled to a free boundary component of MATH, and we have MATH which is again absurd. Our claim is proved. Now we know that MATH again belongs to the finite list of irreducible and MATH-irreducible manifolds which have complexity MATH and cannot be split as an assembling with MATH. However MATH has one more boundary component than MATH, which implies that by repeatedly applying this argument we must eventually end up with a brick. |
math/0105034 | NAME is a property of local nature, and the result is trivial if MATH. For MATH, defining MATH as REF-dimensional portion of MATH and MATH as MATH, the only non-obvious point to show is REF . Of course MATH is either MATH or a union of circles. To check that the only possibilities are those of REF one recalls that MATH is a ball, so MATH is planar, and then MATH is also planar. |
math/0105034 | We first construct a surface with boundary MATH which meets MATH transversely precisely along MATH, as suggested in REF right. Now the portion of MATH which does not lie on MATH consists of a finite number of disjoint circles which can be considered to lie on the boundary of a concentric sub-ball MATH of the ball MATH. These circles bound disjoint discs in MATH, and if we attach these discs to MATH we get the desired MATH. Such a MATH is separating if and only if MATH is because any arc in MATH with ends on MATH can be homotoped to an arc on MATH. This proves the first assertion. The second assertion is obvious. |
math/0105034 | Points REF , in the closed orientable case, are due to CITE. Point REF , which requires a rather careful argument and does not have any closed or even orientable analogue, is new. We first show that if MATH is not standard then either MATH and MATH, or MATH and MATH. Later we will describe standard skeleta without vertices. If MATH reduces to one point of course MATH. Let us first assume that MATH is not purely REF-dimensional, so there is segment MATH contained in REF-dimensional part of MATH. We distinguish two cases depending on whether MATH lies in MATH or on MATH. If MATH, we take a small disc MATH which intersects MATH transversely in one point. As in the proof of REF we attach to MATH a disc contained in the ball MATH, getting a sphere MATH intersecting MATH in one point of MATH. By irreducibility MATH bounds a ball MATH, and MATH is easily seen to be a spine of MATH. NAME now implies that MATH contains vertices, so MATH is a skeleton of MATH with fewer vertices than MATH. A contradiction. If MATH, let MATH be the component of MATH on which MATH lies. Since on MATH there is a circle which meets MATH transversely in one point of MATH, looking at the ball MATH again we see that in MATH there is a properly embedded disc MATH intersecting MATH in a point of MATH. We have now three cases depending on the type of the pair MATH. CASE: If MATH then MATH is a compressing disc for MATH, so by irreducibility MATH is the solid torus. Knowing that MATH meets MATH only in one point it is now easy to show also that MATH and MATH. CASE: If MATH then MATH must be contained in the edge MATH of MATH by REF , and the same reasoning shows that MATH and MATH. CASE: If MATH then MATH must be contained in the edge MATH of MATH by REF . The complement in MATH of MATH is now the union of two NAME strips. If we choose any one of these strips and take its union with MATH, we get an embedded MATH in MATH. Being irreducible and MATH-irreducible, MATH should then be MATH, but MATH: a contradiction. We are left to deal with the case where MATH is purely two-dimensional, so it is quasi-standard, but it is not standard. Let us first suppose that some REF-component MATH of MATH is not a disc. Then either MATH is a sphere, so MATH also reduces to a sphere only, which is clearly impossible because MATH would be MATH, or there exists a loop MATH in MATH such that one of the following holds: CASE: MATH is orientation-reversing on MATH; CASE: MATH separates MATH in two components none of which is a disc. We consider now the closed surface MATH determined by MATH as in REF , and note that MATH is either MATH or MATH. If MATH we deduce that MATH. If MATH irreducibility implies that MATH bounds a ball MATH in MATH. This is clearly impossible in REF , so we are in REF . Now we note that MATH must be a nuclear spine of MATH. Knowing that MATH is not a disc it is easy to deduce that MATH must contain vertices. This contradicts minimality because we could replace the whole of MATH by one disc only, getting another skeleton of MATH with fewer vertices. If MATH is quasi-standard and its REF-components are discs then either MATH is standard or MATH reduces to a single circle. Then it is easy to show that MATH must be the triple hat and MATH. We are left to analyze the case where MATH is standard and MATH, so MATH. Denoting MATH by MATH, REF shows that MATH has MATH faces. We consider first the case MATH. Since MATH has one edge and two faces, it is easy to see that it must be homeomorphic to either MATH or MATH (see REF ) as an abstract polyhedron. This does not quite imply that MATH is MATH or MATH, because in general a skeleton MATH alone is not enough to determine a pair MATH. However MATH certainly does determine MATH, because it is a standard spine of MATH minus a ball, and MATH. We are left to analyze all the polyhedra of the form MATH for MATH, of the form MATH for MATH, and of the form MATH for MATH. Among these polyhedra we must select those which can be thickened to manifolds with two boundary components (a sphere plus either a torus or a NAME bottle). The symmetries of MATH, MATH, and MATH described in REF imply that there are actually not many such polyhedra. More precisely, there is just one MATH, which gives MATH. There are two MATH, one of them is not thickenable (that is, it is not the spine of any manifold), and the other one can be thickened to a manifold with three boundary components (a sphere and two NAME bottles). Finally, there are two MATH, one is not thickenable and the other one gives MATH. This concludes the proof for MATH. Having worked out the case MATH, we turn to MATH, so MATH has MATH edges and MATH faces. If a face of MATH meets MATH in one arc only, then it meets MATH in one edge only, and this edge joins a component of MATH to itself, which easily implies that MATH, against the current assumption. If a face of MATH is an embedded rectangle, with two opposite edges on MATH and two in MATH, then it readily follows that MATH and MATH is either MATH or MATH. As above, to conclude that MATH, we must consider the various polyhedra obtained by attaching MATH, MATH, and MATH to the upper and lower bases of MATH and MATH. Using again REF one sees that there are only six such polyhedra. NAME of them are not thickenable, and the other three give MATH. NAME to the general case with MATH, we note that there is a total of MATH edges on MATH, so there are MATH germs of faces starting from MATH. Knowing that there is a total of MATH faces and none of them uses one germ only, we see that at least one face uses two germs only, so it is a rectangle MATH, possibly an immersed one. If MATH we have three rectangles, one of which must be embedded, and we are led back to a case already discussed. If MATH then MATH must be immersed, so in particular it joins a component MATH of MATH to another MATH component. A regular neighborhood in MATH of MATH is shown in REF . The boundary of this neighborhood is again a graph MATH which determines a separating NAME bottle according to REF . If we cut MATH along MATH we get a disjoint union MATH, which at the level of manifolds gives a splitting MATH. Moreover MATH is a nuclear skeleton of MATH, so MATH, MATH is minimal, and MATH. Now either MATH and MATH or we can proceed, eventually getting that MATH, so MATH for some MATH, and MATH is the corresponding skeleton constructed in REF. The proof is now complete. |
math/0105034 | Assume by contradiction that an edge MATH of MATH is not incident to any vertex of MATH, that is, that both the ends of MATH lie on MATH. If the ends of MATH lie on the same spine MATH then MATH is a connected component of MATH. Standardness of MATH implies that MATH has no vertices, which contradicts the assumption that MATH. So the ends of MATH lie on distinct spines MATH. Let MATH and MATH be the components of MATH on which MATH and MATH lie, and let MATH be a regular neighborhood in MATH of MATH. By construction MATH is a quasi-standard polyhedron with boundary MATH. Here MATH is a trivalent graph with one component homeomorphic to MATH or to MATH, and possibly another component homeomorphic to the circle. Let us first consider the case where MATH has a circle component MATH. This circle lies on MATH and is disjoint from MATH. Standardness of MATH then implies that MATH bounds a disc MATH contained in MATH and disjoint from MATH. In this case we set MATH and MATH. In case MATH is connected we just set MATH and MATH. In both cases we have found a graph MATH homeomorphic to MATH or to MATH which separates MATH. Moreover one component MATH of MATH is standard without vertices and is bounded by MATH. According to REF , the graph MATH determines a separating surface MATH in MATH such that MATH. Since MATH and MATH consists of discs, we have MATH. Of course MATH, for otherwise MATH would be an embedded MATH, but we are assuming that MATH is irreducible and MATH-irreducible and has non-empty boundary. We will now show that if MATH then MATH, and if MATH then MATH splits as MATH. This will imply the conclusion. Assume that MATH, so MATH is a sphere. We denote by MATH the open REF-ball MATH and note that MATH consists of three disjoint open REF-discs, which cut MATH into four open REF-balls. By irreducibility, MATH bounds a closed REF-ball MATH, and MATH is the union of some of the four open REF-balls just described. Viewing MATH abstractly we can now easily construct a new simple polyhedron MATH without vertices such that MATH and MATH consists of three distinct REF-balls, each incident to one of the three open REF-discs which constitute MATH. Let us consider now the simple polyhedron MATH viewed as a subset of MATH. By construction MATH. Moreover MATH is obtained from MATH (which consists of open REF-balls) by attaching each of the three REF-balls of MATH along only one REF-disc (a component of MATH). It follows that MATH still consists of open REF-balls. By puncturing some of REF-components of MATH we can then construct a skeleton of MATH without vertices, so indeed MATH. Assume now that MATH, so MATH is a separating torus or NAME bottle. REF now shows that MATH is obtained by assembling some pair MATH with a pair MATH which has skeleton MATH. By construction MATH is standard without vertices and MATH has three components, and it was shown within the proof of REF that MATH must then be MATH. This completes the proof. |
math/0105034 | A minimal skeleton MATH of MATH is standard by REF , and we have just shown that each edge of MATH joins either MATH to itself or MATH to MATH. Since MATH has MATH quadrivalent vertices, there can be at most MATH edges reaching MATH. Each component of MATH is reached by precisely two edges, so there are at most MATH components. |
math/0105034 | Let MATH be an edge of MATH, and let MATH be the triple of (possibly not distinct) faces of MATH incident to MATH. The number of MATH's that separate MATH from MATH is even; it follows that MATH is a surface away from MATH. Let MATH be a boundary component of MATH, containing MATH. Since MATH is a disc, which is adjacent either to MATH or to MATH (say MATH), then each MATH-component of MATH incident to MATH has MATH on both sides. So MATH is not adjacent to MATH. Finally, since MATH intersects the link of each vertex either nowhere or in a loop, then MATH is a closed surface. It cuts MATH in two components because MATH and MATH lie on opposite sides of MATH. |
math/0105034 | Every region MATH of MATH carries a color MATH given by the number of sheets of the local projection of MATH to MATH. Now we cut MATH open along MATH as explained in CITE, that is, we replace each MATH by its MATH-sheeted cover contained in the normal bundle of MATH in MATH. As a result we get a polyhedron MATH which contains MATH, such that MATH is the disjoint union of an open ball MATH and an open regular neighborhood MATH of MATH in MATH. By removing from each boundary component MATH the open disc MATH we get a polyhedron MATH intersecting MATH in MATH. Now we puncture a MATH-component which separates MATH from MATH and claim that the resulting polyhedron MATH is as desired. Only the inequalities between MATH and MATH are non-obvious. We first prove that all the vertices of MATH which lie on MATH disappear either when we cut MATH along MATH getting MATH or later when we remove MATH from MATH to get MATH. This of course implies the first assertion of the statement. We concentrate on one component MATH of MATH. By REF either both vertices of MATH are vertices of MATH or none of them is. In the latter case there is nothing to show, so we assume that there are three (possibly non-distinct) REF-components of MATH incident to MATH. Let MATH and MATH be the vertices of MATH. Looking first at MATH, we denote by MATH the colors of the six germs at MATH of REF of MATH. Here MATH corresponds to MATH, which is triply incident to MATH. The compatibility equations of normal surfaces now readily imply that that (up to permutation) MATH is even, MATH, and that MATH when MATH. Moreover: CASE: MATH disappears in MATH if MATH; CASE: MATH survives in MATH and remains on MATH, so it disappears in MATH, if MATH and MATH; CASE: MATH survives in MATH and moves to MATH if MATH and MATH. Now if MATH then the same coefficients appear at MATH. The only case where MATH and MATH do not both disappear in MATH is when MATH and MATH. But in this case MATH would contain MATH parallel copies of MATH, which is impossible. The case MATH is easier, because if MATH survives in MATH the situation is as in REF . This is absurd because MATH would contain NAME strips. Now we turn to the second assertion. If MATH the conclusion is obvious, so we proceed assuming MATH. It is now sufficient to show that some face of MATH which separates MATH from MATH contains vertices of MATH, because we can then puncture such a face and collapse the resulting polyhedron until it becomes nuclear, getting fewer vertices. Assume by contradiction that there is no such face. We note that MATH is the union of a quasi-standard polyhedron MATH and some arcs in MATH. REF-components of MATH which separate MATH from MATH are the same as those of MATH, so they give a closed surface MATH by REF . From the fact that MATH we deduce that near a vertex of MATH the transformation of MATH into MATH can be described as in REF , namely MATH can be identified near the vertex with MATH. Of course this does not imply that globally MATH, because the components of MATH playing the role of MATH near vertices may not match across faces. The closed surface MATH cannot be disjoint from MATH, because otherwise MATH would be the obvious sphere MATH. On the other hand we are supposing MATH, so MATH must be a non-empty union of loops. In particular, MATH contains a loop MATH disjoint from MATH. REF now shows that MATH coincides with MATH away from MATH. Using the analysis of the transition from MATH to MATH near MATH already carried out above, we also see that near a component MATH of MATH either MATH coincides with MATH or it is obtained from MATH by adding one edge of MATH, and then slightly pushing the result inside MATH. When MATH the edge added is necessarily MATH. This implies that the loop MATH described above can be viewed as a loop in MATH such that MATH. In addition, if MATH contains a vertex of MATH on a certain component of MATH then it contains also the other vertex in that component. This readily implies that the union of MATH with all the MATH's in MATH touched by MATH is a connected component of MATH. But MATH is standard, so MATH is connected, and we deduce that MATH has no vertices. A contradiction. |
math/0105034 | If MATH the conclusion follows from the classification of standard skeleta without vertices, which was carried out within the proof of REF . So we assume MATH. We proceed by contradiction and assume that there exists an essential sphere, whence a normal one MATH with respect to a standard minimal skeleton MATH. We can now apply the second point of REF to MATH and MATH, getting a polyhedron MATH. By adding an arc to MATH we get a new skeleton of MATH with fewer vertices than MATH: a contradiction. |
math/0105034 | We first assume that MATH cannot be expressed as MATH, we choose a minimal skeleton MATH of MATH, and we apply REF , which easily imply that either MATH with MATH or every face of MATH contains vertices. If we attach MATH and MATH along the map which gives the assembling we get a skeleton MATH of MATH having MATH vertices. Recall now that MATH has a REF-dimensional portion, namely a free segment MATH on MATH. If MATH has vertices we readily deduce that MATH can be collapsed to a subpolyhedron with fewer vertices: a contradiction. So MATH must be of type MATH with MATH. The non-trivial assemblings MATH are easily discussed and the conclusion follows. Assume now that MATH. Noting that MATH has a MATH on its boundary, we deduce that MATH is assembled to MATH. Iterating the splitting of copies of MATH and applying REF we get that MATH, where MATH is irreducible and MATH-irreducible and cannot be split as MATH, and the assembling MATH is sharp. So MATH, but no MATH can be assembled to any of these manifolds. |
math/0105034 | Let MATH be doubly incident to MATH, and let MATH be an arc properly embedded in MATH with endpoints on different edges of MATH. If we cut MATH open along MATH we get a hexagon MATH as in REF , with identifications which allow to reconstruct MATH. The two endpoints of MATH give rise on MATH to four points identified in pairs. Now we choose along MATH a vector field transversal to MATH, and we examine this vector at the four points on MATH. At two of the four points the vector will be directed towards the interior of MATH, and we join these two points by an arc MATH properly embedded in MATH. We also join the other two points by another arc MATH and arrange that MATH and MATH intersect transversely in at most one point. Now MATH is a loop for MATH and, as in the proof of REF , we see that MATH bounds a disc MATH in MATH. This easily implies that MATH actually must be empty, for otherwise MATH and MATH would give rise, in the complement MATH of a regular neighborhood MATH, to two proper discs whose boundaries intersect only once and transversely. Since MATH is empty, MATH is a disc properly embedded in MATH, and the boundary MATH of this disc is essential in MATH, because it intersects MATH in two distinct edges. By irreducibility, MATH is a solid torus or a solid NAME bottle. If it is a solid torus, since MATH meets the meridional disc in two points only, it readily follows that MATH is MATH or MATH, against the hypotheses. If it is a solid NAME bottle, then uniqueness of the embedding of MATH and MATH in MATH implies that MATH is MATH or MATH. |
math/0105034 | A non-trivial loop is isotopic to one which is normal with respect to a triangulation of MATH, that is, it appears as in REF . We must have MATH, MATH, MATH, so MATH, MATH, MATH. If MATH, we further distinguish: if MATH, since we look for a connected curve, we get MATH and MATH, whence the loop MATH; if MATH we do not get any solution; if MATH we get MATH and MATH, whence the loops MATH and MATH. If MATH we must have MATH and MATH, whence the loops MATH and MATH again. If MATH, since the connected curve we look for is also non-trivial, we must have MATH and MATH, whence the loops MATH and MATH. |
math/0105034 | REF shows that the meridian MATH of MATH can be characterized in MATH as the only orientation-preserving loop having connected complement. So every automorphism of MATH maps the meridian to itself and the conclusion follows. |
math/0105034 | It is quite easy to construct commuting order-REF automorphisms MATH and MATH of MATH such that their action on MATH is given by MATH . Given any other automorphism MATH, combining the geometric characterization of MATH with the observation that MATH is isotopic (not only homologous) to itself with opposite orientation, we deduce that (up to isotopy) MATH is the identity on MATH. Up to composing MATH with MATH we can assume that MATH is actually the identity also near MATH, so MATH restricts to an automorphism of the annulus MATH which is the identity on the boundary. The mapping class group relative to the boundary of the annulus is now infinite cyclic generated by the restriction of MATH (but MATH has order REF when viewed on MATH), and the conclusion follows. |
math/0105036 | Suppose MATH is supernormal. Note that MATH contains no other MATH, so MATH and MATH must be a NAME basis for MATH, and thus we have MATH. Conversely, suppose that MATH for all MATH. This means that any lattice point in MATH can be written as a non-negative integer combination of MATH and MATH. Every cone generated by a subset MATH of the MATH can be decomposed as a union of cones of the form MATH, so our assumption implies that every lattice point in MATH can be written as a non-negative integer combination of the vectors in MATH. Therefore MATH is supernormal. |
math/0105036 | Consider the following sequence of vectors in this monoid: MATH where MATH. Explicitly, MATH . At each this iteration, the three vectors MATH generate an index two sublattice of MATH, and MATH is the unique vector which completes the NAME basis for their triangular cone. Suppose there is a finite supernormal generating set MATH for the ambient monoid and consider the smallest index MATH such that MATH is not in MATH. Then the subset MATH violates the defining property of MATH being supernormal. |
math/0105036 | We first prove MATH. Let MATH be supernormal, MATH a triangulation that uses all vectors, and MATH a maximal cell of MATH. If MATH is not a lattice basis of MATH then MATH does not generate the monoid MATH. Supernormality implies that MATH contains at least one other vector MATH, but then this vector MATH cannot be used in the triangulation MATH. This contradicts our hypothesis. The implication MATH is trivial. It remains to show MATH. Suppose REF holds. Let MATH be any subset of MATH. We construct a regular subdivision of MATH which has MATH as one of its faces, and which uses all vectors in MATH as rays. This subdivision can be refined to a regular triangulation MATH of MATH that uses all vectors. By hypothesis, MATH is unimodular, and its restriction to MATH is a unimodular triangulation of MATH. This implies that MATH generates the monoid MATH. We conclude that MATH is supernormal. |
math/0105036 | These statements follow from the fact that MATH is the regular subdivision induced by the vector MATH. |
math/0105036 | We first prove the if direction using REF . Let MATH be a regular triangulation of MATH which uses all vectors. We wish to show that MATH is a unimodular triangulation. By REF there is a simple polyhedron MATH whose normal fan equals MATH. In particular, every vector MATH defines a facet of MATH. Since MATH is a rational polyhedron, there is some MATH such that MATH is integral. The polyhedron MATH has normal fan MATH, and is tight, and so is TDI by assumption. REF implies that every set MATH of MATH vectors in MATH that define a vertex of MATH is a basis of MATH. These cones MATH are the maximal cells of MATH. Hence MATH is unimodular and we conclude that MATH is supernormal. For the only-if direction, suppose that MATH is supernormal and let MATH be such that MATH is tight. Consider any face MATH of MATH, and let MATH be the set of all vectors MATH such MATH holds for all MATH. In view of CITE, it suffices to prove that MATH is a NAME basis. Suppose this is not true. Supernormality implies that MATH contains at least one other vector MATH. Because MATH lies in MATH, we can write MATH where MATH. Since MATH is tight there exists a lattice point MATH with MATH. However since MATH, we know that there is some MATH for which MATH. The first of these two statements implies MATH. The second implies MATH. But these two statements contradict each other, and so we conclude that MATH does not exist, and thus MATH is a NAME basis. It follows that MATH is TDI. |
math/0105036 | According to the definition given in the introduction, two vectors lie in the same cell of the chamber complex if and only if they lie in exactly the same cones spanned by linearly independent MATH-subsets of MATH. This holds if and only if, for every regular subdivision MATH of MATH, they lie in the same cell of MATH. REF completes the proof. |
math/0105036 | This is the second statement of CITE. |
math/0105036 | Let MATH be a convex configuration in MATH and consider any triangulation MATH of MATH that uses all vectors. Now a lattice triangle in the plane which contains no other lattice point has area one half (by Pick's theorem, for example). This implies that the triangulation MATH is unimodular, and so REF implies that MATH is supernormal. |
math/0105036 | The chamber complex of a supernormal configuration MATH is the NAME fan of the associated binomial ideal MATH. Hence MATH is greater or equal to the maximum number of facets of any cone in the chamber complex of a supernormal configuration in MATH. For MATH we can take the chamber complex of a polygon MATH to get a lower bound for MATH. |
math/0105036 | The map MATH is a bijection between the set of lattice points in MATH and the set of monomials in MATH having degree MATH. Hence the condition in the lemma states that every non-zero graded component of MATH contains exactly one monomial which is not in MATH. |
math/0105036 | Let MATH be the vertices of MATH and let MATH be the corresponding monomials in MATH of degree MATH. We first show that every monomial in MATH lies in the monomial ideal MATH. In polyhedral terms, if MATH is any lattice point in MATH, then MATH can be written as MATH where MATH and the MATH are non-negative reals summing to MATH. This means that for the corresponding monomial MATH we have MATH, where MATH, since MATH. There exists an index MATH such that MATH and this implies that MATH, and thus that MATH divides MATH. This shows that MATH lies in MATH. Since our virtual initial ideal MATH must have a standard monomial of degree MATH, it cannot contain the ideal MATH, and we conclude that one of the monomials MATH is not in MATH, as desired. |
math/0105036 | It is clear that the section MATH associated to a triangulation MATH of MATH satisfies the desired conditions, so we need only show that every section MATH satisfying the hypothesis comes from a triangulation. Fix such a MATH. We first observe that MATH for MATH and MATH. If MATH then MATH, and so MATH satisfies MATH. The case that MATH follows from this. We claim that the set of all possible supports of vectors in MATH is a triangulation of MATH. We first show that the subsets of MATH indexed by these supports are linearly independent. Suppose not, so for some MATH there is a vector MATH such that MATH where MATH is a proper subset of MATH. There is some MATH for which MATH, and so MATH. Now MATH, so MATH . This implies that MATH, a contradiction since MATH properly contains MATH. This shows that the cones MATH as MATH ranges over MATH are simplicial and that they cover MATH. We also note that this argument actually shows that for any MATH in the relative interior of MATH we have MATH. Hence the relative interiors of two distinct cones do not intersect. The order ideal hypothesis guarantees that these cones form a simplicial fan. |
math/0105036 | Suppose this is not the case, so there is some MATH with MATH. Since the greatest common divisor of all monomials of the same degree as MATH is MATH, we know that MATH is tight, and so there is some lattice point MATH such that MATH. Because MATH, we also have MATH. This means MATH is a monomial of the same degree as MATH, and is thus divisible by MATH, so MATH. But this implies that MATH, a contradiction. |
math/0105039 | Let's suppose the bundles are complex, in other words MATH. The proof for general MATH is exactly the same. For a MATH-dimensional complex bundle over MATH we have a map MATH and a groupoid structure MATH as in REF . Let MATH be an open set of MATH uniformized by MATH which belongs to its orbifold structure; for MATH and MATH, MATH is an element of MATH ( via the identity on MATH, and the action of MATH in MATH and the conjugation by MATH on MATH thought of as an automorphism of MATH) and we can define MATH by MATH. As MATH, we have MATH, which implies MATH. So MATH with MATH extending the action of MATH in MATH is a uniformizing system for the orbibundle we are constructing, we need to prove now that they define the same germs and then we would get a orbibundle MATH using its bundle orbifold structure. Let MATH be injections of uniformizing systems of MATH, with corresponding bundle uniformizing systems MATH and MATH. For MATH, MATH and MATH, MATH and MATH. As MATH for MATH then MATH; so the bundle uniformizing systems MATH define the same germs, thus they form a bundle orbifold structure over MATH. Conversely, if we have the orbibundle structure for MATH we need to define the function MATH. So, for injections MATH (where MATH extends the MATH's previously defined), MATH will be the element in MATH such that maps MATH, here MATH stands for the projection on the second coordinate; in other words MATH . Because this bundle uniformizing systems define the same germs, MATH satisfies the product formula; the inverse formula is clearly satisfied. |
math/0105039 | We will focus again on complex bundles. To understand what relevant information we have from isomorphic bundles, let's see the following lemmas An isomorphism of bundles over MATH (with maps MATH for MATH) is determined by a map MATH such that MATH satisfying MATH and MATH. It is easy to check that MATH defined in this way is a morphism between the bundles; the equality MATH comes from the diagram of the source map and MATH from the one of the target map, the rest of the diagrams follow from those two. In the same way we could do this procedure for complex orbibundles: An isomorphism of complex orbibundles over MATH (with maps MATH for MATH and MATH orbifold structure of MATH) is determined by the maps MATH such that MATH satisfying MATH. The MATH's form a good map. Because the underlying orbifold structure needs to be mapped to itself, we obtain the MATH's. The equality MATH holds because the good map condition. The proof of the proposition is straight forward from these lemmas. The map MATH that comes from the isomorphism of the complex bundles determines uniquely the MATH's, and vice versa. |
math/0105039 | For MATH a good map between orbifolds, we have a correspondence MATH between open subsets of a compatible cover of MATH and open subsets of MATH, such that MATH , and MATH implies MATH. Moreover, we are provided with local liftings MATH as in REF . Let MATH and MATH be the groupoids constructed from the orbifold structures of MATH and MATH respectively, determined by the compatible cover MATH of MATH and a cover of MATH that uniformizes MATH. Define MATH such that MATH and MATH by MATH, where the MATH's are injections between MATH and MATH and the MATH's are the corresponding injections between MATH and MATH given by the definition of good map; because MATH the function MATH is well defined and together with MATH, satisfy all the conditions for a morphism of groupoids. It is clear that the groupoids just used could differ from the groupoids one obtain after performing the construction defined at the beginning of this chapter, but they are respectively NAME equivalent. On the other hand, if we are given MATH and MATH, we can take a sufficiently small open compatible cover for MATH such that for MATH in its cover there is an open set MATH of MATH with the desired properties. For MATH and MATH uniformizing systems of MATH and MATH respectively, we need to define MATH. The map between MATH and MATH is given by MATH, and the injection between MATH and MATH is given as follows. Let's take MATH and MATH; we have an automorphism of MATH given by the action by MATH on MATH and by conjugation on MATH, call this automorphism MATH; then MATH is an element of MATH, using the properties of MATH and MATH we get that MATH, where MATH; this because every automorphism of MATH comes from the action of an element in MATH (see CITE); moreover, we have that MATH. This will give us an homomorphism MATH sending MATH that together with MATH form the compatible system we required. |
math/0105039 | Noting that the map MATH is a surjection and recalling that MATH we can see that MATH is an étale surjection. Finally because the action of MATH in MATH and MATH is free and MATH, MATH it is immediate to verify that MATH is a fibered square. |
math/0105039 | The proof is the same as in REF . For MATH we get: MATH . |
math/0105039 | We will use the following facts. Let MATH be a NAME. If a class MATH is a torsion element then there exists a principal bundle MATH with structure group MATH so that when seen as an element MATH then MATH. In other words, the image of MATH is exactly the subgroup of torsion elements. Let MATH be a MATH-Hilbert space in which every irreducible representation of MATH appears infinitely many times. Then the equivariant index map MATH is an isomorphism (where MATH acts on MATH by conjugation.) Let MATH be an orbifold where the NAME group MATH acts on MATH. Let MATH define a group extension MATH. Consider the natural homomorphism MATH - it can also be seen as the composition of MATH. Let MATH be the diagonal embedding representation. Then MATH . The orbifold MATH is represented by the groupoid MATH, while the gerbe MATH is represented by the central extension of groupoids REF MATH where MATH. Therefore, using the fact that MATH we get the surjective map MATH using REF and observing that MATH we get the result. Let us consider in the previous lemma the situation where MATH is smooth, MATH is NAME 's principal MATH-bundle associated to MATH, MATH and MATH is the class in MATH labeling the extension MATH . Then using REF we get that MATH . Notice that by REF MATH is defined as the homotopy classes of sections of the bundle MATH. This space of sections can readily be identified with the space MATH and the proposition follows from this. |
math/0105039 | Consider the equations, MATH and take determinants in both equations, we get MATH defining MATH we have MATH this means that the coboundary of MATH is MATH. This concludes the proof. |
math/0105039 | Since all groupoids MATH are NAME equivalent (for any two open covers have a common refinement,) it is enough to consider the groupoid MATH coming from the cover consisting of one open set. The source and target maps of MATH are both identity maps. Then the proposition follows from the fact that the following diagram is a fibered square MATH . |
math/0105039 | Suppose that MATH and MATH are stably isomorphic. Then MATH hence MATH. Therefore from the previous lemma we have MATH and by definition of trivial we get MATH. Conversely if MATH then MATH and then by REF is trivial. Define the trivial bundle gerbe MATH. Then MATH completing the proof. |
math/0105039 | Starting with the data in REF we will construct the category MATH for a small open set MATH. Since MATH is small we can trivialize the gerbe MATH. The objects of MATH are the set of all possible trivializations MATH with the obvious morphisms. Conversely suppose that you are given a gerbe as a sheaf of categories. Then we construct a cocycle MATH as in CITE. We take an object MATH and an automorphisms MATH and define MATH producing a NAME cocycle giving us the necessary data to construct a gerbe as in REF. |
math/0105040 | Note that MATH is a closed NAME subgroup in the group of all conformal diffeomorphisms of MATH. If MATH were noncompact, then by the celebrated result of NAME and NAME (CITE, CITE), MATH would be conformally equivalent with the sphere MATH, MATH. Hence MATH would be simply connected. It is well known that a compact simply connected l.c.K. manifold is conformal to a NAME manifold (compare CITE), which is impossible because the sphere MATH has no NAME structure. |
math/0105040 | Suppose that MATH. Then MATH leaves MATH invariant. As MATH generates MATH, it follows that MATH. In particular, MATH. For any distribution MATH on MATH, denote by MATH the orthogonal complement to MATH with respect to the metric MATH where MATH. Since MATH, if MATH, then MATH, similarly MATH. The equality MATH implies that MATH, MATH for any MATH. Therefore, MATH becomes constant, being a contradiction to REF. |
math/0105040 | Recall that MATH lies in MATH by definition. As MATH is a compact NAME group, its closure MATH in MATH is also compact and so isomorphic to a MATH-torus MATH. Therefore, the lift MATH of MATH to MATH acts properly on MATH. The lift MATH is isomorphic to MATH where MATH. Note that MATH because MATH maps any compact subgroup of MATH to MATH, but the group MATH satisfies MATH. Hence the group MATH has a nontrivial summand in MATH which implies that MATH is closed in MATH. Thus, the group MATH acts properly on MATH. If we note that MATH is isomorphic to MATH, then it acts freely on MATH. |
math/0105040 | As MATH is holomorphic, MATH. Hence, MATH . We note also that MATH . By REF , notice that MATH everywhere on MATH. Since MATH, MATH. For MATH, MATH . This proves that MATH is onto and so MATH is a codimension one smooth regular submanifold of MATH. |
math/0105040 | Let MATH be a component of MATH and MATH be the set MATH. As MATH acts freely and MATH, we have MATH for MATH. Thus MATH is an open subset of MATH. We prove that it is also closed. Let MATH be the closure of MATH in MATH. We choose a limit point MATH. Then MATH. Put MATH, then MATH, so MATH. Since MATH is regular (that is, closed with respect to the relative topology induced from MATH), its component MATH is also closed. Hence MATH. Therefore MATH, proving that MATH is closed in MATH. In conclusion, MATH. Now, if MATH is another component of MATH, the same argument shows MATH. As MATH and MATH, this implies MATH, in other words MATH is connected. |
math/0105040 | First note that MATH . Moreover, from REF, MATH. Hence, MATH on MATH showing that MATH is a contact form. Noting REF and that both MATH and MATH commutes each other, it is easy to see that MATH . Let MATH be the contact subbundle. Since MATH and MATH from REF, if MATH, then MATH. Moreover, MATH, which implies that MATH for all MATH, showing that MATH is the characteristic vector field. |
math/0105040 | Let MATH be the bilinear form defined by MATH. There exist MATH such that MATH, MATH. Then it is easy to see that MATH . Using MATH as above, MATH hence MATH is positive definite. By definition, MATH is strictly pseudoconvex. Let MATH be the canonical splitting of MATH. Then we prove that MATH. Let MATH. Since MATH and MATH is integrable on MATH, MATH. Put MATH for some function MATH and MATH. As MATH from REF, MATH. By definition, MATH. On the other hand, since MATH is MATH-invariant, MATH for MATH. As above, MATH, similarly for MATH, we obtain that MATH . Hence, MATH and so MATH. If we note that MATH is MATH-isomorphic by REF, then MATH is the splitting for MATH, in which we have shown MATH. Therefore MATH is a complex structure on MATH. |
math/0105040 | By REF, MATH (respectively, MATH) preserves MATH (respectively, MATH). Then the equality MATH from REF with REF implies that MATH on MATH. Therefore MATH . |
math/0105040 | First we determine the NAME field MATH. MATH . We start from: MATH because MATH is the characteristic vector field of the contact form MATH. As before, a point MATH can be described uniquely as MATH for some MATH. In particular, using REF , the MATH-coordinate of MATH is MATH. Noting that MATH and MATH, by uniqueness the MATH-coordinate of MATH, MATH. From REF, MATH . The above formula becomes: MATH proving that MATH. Next we observe that the flow MATH acts by isometries with respect to MATH. As MATH is holomorphic, it is enough to prove that each MATH leaves MATH invariant. But MATH . Thus MATH. Now we put MATH in the equality MATH, valid for any MATH-form MATH, take into account MATH and obtain MATH which is equivalent with MATH, so MATH is parallel with respect to MATH as announced. |
math/0105040 | We prove the following two facts: CASE: MATH for every MATH. CASE: MATH where MATH is the homomorphism as before. First note that as MATH centralizes MATH, MATH for MATH. As MATH is holomorphic, MATH. Since MATH acts on MATH as holomorphic homothetic transformations, MATH, MATH preserves MATH. If we recall that MATH is isomorphic, then for MATH, MATH. As MATH is characteristic, it follows MATH. This shows that MATH on MATH. On the other hand, if we note MATH, then MATH where MATH is a positive constant number. Applying MATH to MATH from REF, we obtain MATH. Equivalently, MATH. This shows REF. From REF, MATH . Since MATH, MATH acts through holomorphic isometries of MATH. We have that MATH from REF. Then, MATH . |
math/0105040 | Let MATH be the flow generated by MATH on MATH and MATH its lift to MATH. Denote by MATH the vector field on MATH induced by MATH. Then, MATH. Because MATH is parallel, MATH (respectively, MATH) acts by holomorphic isometries with respect to MATH (respectively, MATH). In particular, MATH preserves MATH. Then, for MATH, we have MATH. As MATH is a homomorphism and MATH is a constant for each MATH (MATH), we can describe as MATH . Recall that MATH is the NAME metric associated to MATH. If MATH acts as holomorphic isometries with respect to MATH, then the above equation implies that MATH, MATH . MATH for every MATH, and so MATH. On the other hand, as MATH, we have: MATH being a contradiction. Thus, MATH with MATH. Hence, MATH is a group of nontrivial homothetic holomorphic transformations isomorphic to MATH. On the other hand, let MATH (respectively, MATH) be the flow generated by -MATH on MATH (respectively, -MATH on MATH). As MATH, MATH-and hence MATH for every MATH. Using the fact that MATH, MATH. This implies that MATH . |
math/0105040 | By definition, any element MATH satisfies MATH. As MATH, choosing MATH, put MATH. Then, MATH. In particular, MATH leaves MATH invariant. Let MATH be the restriction of MATH to MATH. Using REF and MATH, we have that MATH . Hence MATH. If we define MATH, then it is easy to see that MATH is a well defined homomorphism. Let MATH be a point in MATH. As MATH, MATH, so MATH is MATH-equivariant. For MATH, we define a diffeomorphism MATH to be MATH . By definition, MATH and the MATH-coordinate satisfies that MATH. Using REF and MATH, it follows that MATH . To see that MATH is holomorphic, notice that MATH. As MATH, and MATH, MATH . Hence, MATH preserves MATH. Since the complex structure MATH is defined by the commutative REF , MATH for MATH by definition. Then MATH. As a consequence, MATH on MATH. Hence, MATH. It is easy to check that MATH is a homomorphism of MATH into MATH such that MATH. |
math/0105040 | Consider the homomorphism MATH from REF. Then, MATH is a free abelian group of rank MATH. If we note that MATH is an infinite cyclic subgroup of MATH, then we can choose a subgroup MATH of finite index in MATH such that MATH is a direct summand in MATH; MATH. Put MATH and MATH. Then, MATH has finite index in MATH. Obviously MATH maps MATH isomorphically onto MATH which is of finite index in MATH. |
math/0105040 | Let MATH be the inverse isomorphism. For each MATH there exists a unique element MATH such that MATH. As we know that MATH acts properly discontinuously on MATH from the remark below REF, the stabilizer at each point is finite. Suppose that MATH for some point MATH. As MATH, MATH for some MATH. Since MATH is a central element and MATH is a homomorphism, MATH. Thus, MATH, MATH. By definition of the action MATH, MATH. As MATH acts freely on MATH, MATH and so MATH. If we note that MATH, then MATH induces a pseudo-Hermitian structure MATH on MATH. Here we use the same notation MATH to the complex structure on MATH. |
math/0105040 | Let MATH. The equations MATH, MATH show that MATH . Thus MATH applies MATH onto itself. Moreover, if MATH is a dual frame field to MATH (similarly for MATH), then the frame MATH spans MATH. The equation MATH implies that MATH (similary for MATH). Therefore MATH-on MATH. |
math/0105040 | As MATH generates the flow MATH, MATH on MATH by hypothesis and so MATH maps the complex plane field MATH onto MATH. By REF , each MATH preserves MATH. So its lift MATH preserves the MATH-invariant distribution MATH. Since MATH is MATH-isomorphic and each MATH is holomorphic on MATH, MATH is also MATH-isomorphic through the commutative diagram and thus each MATH is holomorphic on MATH; MATH. Therefore, MATH is a closed, noncompact subgroup of MATH-transformations of MATH with respect to MATH. |
math/0105040 | Let MATH be a diffeomorphism defined by MATH. Note that MATH on MATH. As every element of MATH is described as MATH from REF , define a homomorphism MATH by setting MATH . Recall that the action MATH from REF. Then, MATH . Hence, MATH is equivariantly diffeomorphic. Next, since MATH for the MATH-coordinate of MATH and MATH from REF, it follows that: MATH . By definition, MATH. Moreover, when MATH, MATH . Using REF, MATH . Thus MATH. As MATH from REF, MATH maps MATH onto MATH. Consider the commutative diagram: MATH . Here note that MATH on MATH. For MATH, MATH thus, MATH. Hence, MATH is MATH-biholomorphic. Moreover, as MATH and MATH, we obtain that MATH. Therefore, MATH induces a holomorphic isometry from MATH onto MATH. |
math/0105041 | Let MATH, MATH be two local orthonormal triples of Killing fields trivializing MATH on MATH, MATH. Then, on MATH we have MATH. We shall show that MATH are constant. Compute first the bracket MATH . From MATH and MATH (MATH and cyclic permutations), we can derive: MATH . Hence MATH . Thus, for any MATH: MATH. It follows: MATH . Now we use the Killing condition applied to MATH: MATH which yields, on one hand MATH for any MATH and, on the other hand MATH . This and REF imply MATH and the proof is complete. |
math/0105041 | By REF , the bundle MATH is trivial. However, this is not enough to deduce that the trivialization can be realized with Killing fields generating a MATH algebra. For example, the inhomogeneous MATH-dimensional spherical space forms are parallelizable but locally, not globally MATH-Sasakian. To overcome this difficulty, start with the induced locally MATH-Sasakian structure of MATH. Let MATH be the global NAME structure of MATH provided by REF and consider an open set MATH on which MATH is trivialized by a local MATH-Sasakian structure incuding MATH. The manifold MATH is simply connected and NAME, hence analytic (see REF ). By a result of NAME (compare CITE) each local Killing vector field on MATH can be extended uniquely to the whole MATH. We thus extend the above three local Killing fields. Clearly, the extension MATH of MATH coincides with MATH. The extension MATH of MATH is thus orthogonal to MATH and belongs to MATH in every point of MATH. It follows from REF that MATH is a global NAME structure. Now MATH completes the desired global MATH-Sasakian structure. |
math/0105041 | Let MATH be quaternion-Hermitian NAME. Fix a metric MATH and choose an open set MATH on which MATH is trivialized by an admissible basis MATH. Then MATH together with REF of the definition imply MATH, hence MATH . This implies that MATH is a differential ideal and, on the other hand, the derivative of the fundamental four-form is MATH. Differentiating here we get MATH. As MATH is nondegenerate, this means MATH. Consequently, locally, on some open sets MATH, MATH for some differentiable functions defined on MATH. It is now easy to see that for each MATH, the associated MATH-form is closed, hence, taking into account REF , the local metrics MATH are quaternion NAME. Conversely, starting with the local quaternion NAME metrics MATH, define MATH. It can be seen that these local one forms glue together to a global, closed one-form and MATH. Then construct the NAME connection associated to MATH and MATH: MATH . A straightforward computation shows that MATH has the requested properties. The proofs for the global case are completely similar. |
math/0105041 | We have to prove that there exists a unique torsion free connection preserving both MATH and MATH. Indeed, if MATH, MATH are such, let MATH, MATH be the associated NAME forms. Then the fundamental MATH-form MATH satisfies MATH . Using the operator MATH, MATH, REF yields MATH. But MATH is injective, because it is related to its formal adjoint MATH by MATH. Hence MATH. Finally, REF proves that MATH. |
math/0105041 | This result follows directly from REF ), but we prefer to give here a direct proof, adapted to our situation. On each MATH, the relation between the scalar curvatures MATH and MATH of MATH and MATH is (compare CITE, p. CASE: MATH . Hence MATH is constant. If MATH is not identically zero, differentiation of the above identity yields: MATH . As both MATH and MATH are global objects on MATH, it follows that MATH is exact, contradiction. But if MATH on some MATH, then MATH, constant on MATH. This proves that MATH on each MATH. |
math/0105041 | Let MATH be the MATH tensor fields metrically equivalent with the MATH-forms MATH. The identity in the statement can be rewritten as: MATH . Let now MATH be unitary, fixed. In the orthogonal complement of MATH we choose a unitary MATH and let MATH. With these choices, the above identity reads: MATH . Here we use the assumption MATH to obtain: MATH hence MATH have the form MATH which, introduced in REF , gives: MATH . Again using MATH, we may choose MATH and MATH orthogonal to MATH and get: MATH . Now it remains to take MATH to derive the sqew-symmetry of MATH. |
math/0105041 | . Fix MATH and an admissible basis for MATH. Starting from REF and MATH, one can derive the following formula: MATH where MATH. After differentiating REF we get: MATH . The previous Lemma applies and provides: MATH . This yields MATH and MATH with MATH not depending on MATH. Hence, locally MATH and we have MATH an equation similar to REF. The rest and the converse are obvious. |
math/0105041 | On a given submanifold of MATH, locally one can induce the metric MATH and the quaternion NAME one MATH. Correspondingly, there are two second fundamental forms MATH and MATH. As MATH and MATH are conformally related on MATH, the relation between MATH and MATH is MATH where MATH is the part of MATH normal to the submanifold. Now let MATH and let MATH be a quaternion NAME submanifold through MATH as stated. We have MATH. From MATH we then derive MATH. But rank MATH, hence MATH meaning that MATH is normal to MATH: MATH. On the other hand, the same relation MATH shows that MATH is a quaternion NAME submanifold of the quaternion NAME manifold MATH. As quaternion submanifolds of quaternion NAME manifolds are totally geodesic, MATH is totally geodesic in MATH with respect to MATH. It follows MATH on MATH. But MATH is zero from the assumption (MATH is totally geodesic with respect to MATH). This yields MATH on MATH, in particular MATH. Since MATH was arbitrary in MATH, MATH on MATH proving that MATH is quaternion NAME. For the converse, just take MATH. |
math/0105041 | The first statement is a consequence of REF. As for the second one, the bundle MATH is locally generated by the (rescaled to be unitary) local vector fields MATH. Indeed, they are Killing by the last equation of REF; the first condition of REF is given by REF; the transition functions of MATH are in MATH because the transition functions of MATH are so; finally, REF of the definition is implied by REF. |
math/0105041 | Let MATH be a leaf of MATH and let the superscript MATH refer to restrictions of objects from MATH to MATH. A local orthonormal basis of tangent vectors for MATH is provided by MATH. As MATH is totally geodesic, MATH and a direct computation of the curvature tensor of the NAME connection MATH on this basis proves MATH on MATH. Hence MATH is conformally flat and the curvature tensor of the NAME connection is MATH . It follows that the NAME tensor MATH is MATH-parallel and, on the other hand, the sectional curvature is non-negative and strictly positive on any plane of the form MATH. Now recall that the universal Riemannian covering spaces of conformally flat Riemannian manifolds with parallel NAME tensor were classified in CITE. By the above discussion and the reducibility of MATH (due to MATH), the only class fitting from NAME 's classification is that with universal cover MATH equipped with the conformally flat metric written in quaternionic coordinate MATH. We still have to determine the allowed deck groups. Happily, Riemannian manifolds with such universal cover were studied in CITE and, in arbitrary dimension, in CITE. Here it is proved that REF forces the deck group of the covering to contain only conformal transformations of the form (in real coordinates) MATH where MATH and MATH. This leads to the following form of MATH: MATH where MATH is a conformal transformation of maximal module MATH and MATH is one of the finite subgroups of MATH listed in CITE. Finally, as MATH, MATH has an induced integrable quaternionic structure. |
math/0105041 | Only the second statement has to be proved. It is clear that the leaves inherit a hyperhermitian NAME, non hyperkähler (because MATH) structure. The compact hyperhermitian surfaces are classified in CITE and the only class having the stated property is that of NAME surfaces. |
math/0105041 | Note that a submanifold MATH is an integral manifold of MATH if and only if MATH and MATH vanish on MATH. In this case, also MATH vanishes on MATH. Then REF implies that MATH is normal to MATH for any MATH tangent to MATH, that is, MATH is totally real. The statement about the dimension of MATH is now obvious. |
math/0105041 | Let first MATH be a compact quaternion Hermitian NAME manifold as in the statement. The orbits of MATH are closed, hence after rescaling, one may suppose they are circles MATH acting on MATH by isometries because MATH is Killing. The quotient space MATH is an orbifold (a manifold if MATH is regular) and, with respect to the induced metric MATH, the natural projection MATH becomes a Riemannian submersion. Hence, for any leaf MATH of MATH, MATH is a Riemannian covering map. As, according to REF , the leaves of MATH have a locally MATH-Sasakian structure, MATH is locally MATH-Sasakian. Conversely, consider a flat principal MATH-bundle MATH over a compact locally MATH-Sasakian manifold MATH with local Killing field MATH. Choose a closed MATH-form MATH on MATH defining the flat connection of the bundle MATH and define the metric MATH. Also, define an almost quaternionic bundle MATH on MATH by defining its local basis as: MATH where MATH and MATH. It is straightforward to check, as in the complex case (see REF) that MATH is quaternion Hermitian NAME with NAME form MATH. |
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