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math/9911242
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Assume that MATH is a star. Since by construction MATH is generated by preprojectives, it suffices to show that the indecomposable preprojectives are noetherian. So we need to consider objects of the form MATH with MATH a vertex in MATH and MATH. Taking the sum of all irreducible maps (with multiplicities) going into MATH, we obtain a map MATH, where the sum is finite, which is either surjective or has simple cokernel (the latter happens if MATH). In any case it is sufficient to show that the MATH are noetherian. By REF we know what the MATH can be. They are of the form MATH such that either MATH, or else MATH and the map MATH is obtained from an irreducible map MATH, which in turn corresponds to an arrow MATH in the quiver MATH. Induction on MATH now yields that it is sufficient to show that MATH is noetherian, where MATH lies on one of the rays contained in MATH, and furthermore for a given MATH we may assume that MATH lies arbitrarily far from the starting vertex of the ray. Let MATH be the ray on which MATH lies. We will assume that MATH (if MATH happens to be farther away, then we just drop the initial vertices in MATH.) To check that MATH is noetherian we need to understand the additive category MATH whose objects are direct sums of indecomposables which have a non-zero map to MATH. The theory of almost split sequence easily yields that MATH is the path category of the part of the NAME of MATH spanned by the vertices that have a non-trivial path to MATH. Furthermore, also by the theory of almost split sequences, it is easy to work out what this quiver is. The result is as in REF . By convention we will assume that the relations on the quiver in REF are of the form MATH . Note that from these relations, or directly, it easily follows MATH . To show that MATH is noetherian we have to show that there does not exist an infinite sequence of non-zero maps (unique up to a scalar by REF) MATH, such that the image of MATH is not contained in the image of MATH. Assume to the contrary that such a sequence MATH does indeed exist. Let MATH be the set of all MATH such that MATH for some MATH. For every MATH let MATH be the MATH closest to MATH with the property that MATH and let MATH be the corresponding MATH. It is now clear from the quiver given in REF as well as REF that every MATH factors through one of the MATH, contradicting the choice of MATH. This finishes the proof.
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By REF the category MATH has a NAME functor. As usual we denote the NAME functor by MATH, and we let MATH. This now induces an autoequivalence MATH. Let MATH be an object in MATH and let MATH be the component of the NAME of MATH-containing MATH. Since MATH is an equivalence, it preserves length. Hence we have an almost split sequence MATH, where MATH is uniserial of length MATH. Consider the almost split sequence MATH. Since MATH and MATH are uniserial of length MATH and MATH is simple, it is easy to see that MATH is uniserial of length MATH. Continuing the process, we see that MATH contains only uniserial objects. It is a tube if it contains only a finite number of nonisomorphic simple objects, and of the form MATH otherwise. Assume there is some indecomposable object MATH in MATH which is not in MATH. If MATH, then MATH for some simple object MATH in MATH. Using the properties of almost split sequences and that all objects in MATH are uniserial, we can lift the map MATH to get an epimorphism from MATH to an object of arbitrary length in MATH. Hence MATH, and similarly MATH. It follows that all indecomposable objects in MATH are in MATH. It is easy to see from the above that the component of MATH uniquely determines MATH and that MATH is as in the statement of the theorem. The category of nilpotent finite dimensional representations of MATH or of MATH over MATH is a hereditary abelian MATH-category, and it is easy to see that it has almost split sequences, and hence a NAME functor by REF .
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math/9911242
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Since NAME functors are unique we may assume MATH. It is easy to see that MATH exists and is given by the following formula. MATH where MATH runs through the ideals of MATH with the property that MATH is finite dimensional. In fact such a formula would hold for any autoequivalence of MATH. To describe MATH explicitly consider first the case MATH. In that case local duality implies that MATH where MATH. An explicit computation reveals that MATH. In the case of MATH we could extend local duality theory to the pseudo-compact ring MATH to arrive at the same formula. However it is easier to use that MATH with MATH. In this case graded local duality implies that MATH coincides with tensoring with MATH. Translating this to MATH we see that indeed MATH coincides with tensoring with MATH.
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math/9911242
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In CITE it has been shown that it is sufficient to prove that the projective dimension of each pseudocompact simple is equal to one. This is an easy direct verification.
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Without changing the category MATH we may (and we will) replace MATH by MATH. By the NAME - NAME lemma MATH is finitely generated. Let MATH be homogeneous generators for MATH, respectively of degree MATH. Let MATH be homogeneous generators of MATH as a module over MATH, and let MATH be the maximum of the degrees of the MATH's. Let MATH be the product of the MATH's. We claim that there exists MATH such that for MATH we have MATH. We prove instead that for MATH large we have MATH. This is clearly equivalent. A general element of MATH is a MATH-linear combination of elements of the form MATH. It is sufficient to have that for some MATH : MATH. Assume this is false. Then it follows that the degree of MATH (which is MATH) is less than MATH. This proves what we want. Define MATH . One verifies directly that for MATH one has MATH. Since MATH and MATH are NAME equivalent we clearly have MATH. By REF is covered by the affine open sets MATH. Let MATH be the sheaf of graded rings on MATH associated to MATH. Thus on MATH the sections of MATH are given by MATH. By localization theory MATH is equivalent to the category of coherent graded MATH-modules. We claim that MATH is strongly graded REF and that MATH is coherent over MATH. It is clearly sufficient to check this locally. That MATH is strongly graded follows from REF below. The sheaf MATH is nothing but MATH in the notation of CITE. By the results in CITE MATH is coherent (it is easy to see this directly). Since MATH is strongly graded we have that the category of coherent graded MATH-modules is equivalent to the category of coherent MATH-modules. This finishes the proof.
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For MATH large we have MATH A similar argument shows that MATH.
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This follows easily from the structure theory of hereditary noetherian rings. By REF it follows that MATH is equivalent to the category of coherent MATH-modules where MATH is a coherent sheaf of MATH-algebras and MATH. Let MATH be the center of MATH, and let MATH be the MATH-scheme such that MATH. Now for the purposes of the proof we may replace MATH by its connected components. So we assume that MATH is connected. Let MATH be affine open. Then by REF and CITE it follows that MATH is hereditary. So according to CITE we have that MATH will be a direct sum of a finite dimensional hereditary algebra MATH and a hereditary MATH-algebra MATH which is a direct sum of infinite dimensional prime rings. We identify MATH with a sheaf of finite support on MATH. If MATH then MATH corresponds to a central idempotent in MATH which is zero outside the support of MATH. Hence we can extend it to a central idempotent in MATH. Since we had assumed that the center of MATH is connected, this yields MATH. Hence we only have to consider the case where MATH is a direct sum of prime hereditary rings for all affine MATH. In our case the MATH are then direct sums of orders in central simple algebras. Now it follows from CITE that the center of a hereditary order is a NAME ring. In particular MATH is a non-singular curve. Since MATH is also connected, it follows that MATH is irreducible. Thus if MATH is affine, then MATH is a domain, whence MATH is actually prime. So MATH is a classical hereditary order. This finishes the argument.
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REF . for MATH follows directly from the explicit description of the indecomposable objects in MATH. The claim for MATH can be reduced to the result for MATH, by possibly applying high enough powers of MATH. To show that any component MATH of MATH or MATH is standard, one defines as usual a functor from the path category of MATH to the subcategory of indecomposable objects of MATH given by MATH, in such a way that there is induced a functor from the mesh category of MATH. Using REF., this will be an equivalence.
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We claim that the only simple objects in MATH are the objects of length two in the MATH components of MATH. This clearly implies what we have to show. First we note that the other indecomposables cannot be simple. Indeed let MATH be one of those other indecomposables. Assume first that MATH. Thus MATH can be considered as an object of MATH. Assume now in addition that MATH is not a length one projective. Then it is easy to see directly that we can make a non-trivial short exact sequence MATH with MATH. Hence this sequence remains exact in MATH, and thus MATH is not simple. If MATH happens to lie in MATH, or is given by a length one projective in MATH, then we first apply a power of MATH, and then we apply the above reasoning. Now we prove the converse. So let MATH be an indecomposable object of length two in one of the MATH components of MATH. We have to prove that MATH becomes simple in MATH. Assume now that MATH is not simple in MATH. Thus there is an exact sequence in MATH as in REF, but this time the indecomposable summands of MATH must be in the same component as MATH (by REF ). By applying a high power of MATH we may assume that the summands of MATH are in MATH, and hence REF represents an exact sequence in MATH. It is easy to see that this is impossible.
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This follows directly from the fact that the projectives generate MATH.
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This can be shown in many ways. The most direct method is to compute in MATH, but it is most elegant to consider the function MATH. This function is MATH on MATH and zero on the objects which are not on a path starting in MATH (by REF ). Furthermore MATH is additive on almost split sequences not ending in MATH. This is enough to determine MATH completely.
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Assume MATH with MATH indecomposable. Then we can find an indecomposable MATH which has a path to both MATH and MATH. Thus we obtain. MATH which is a contradiction.
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By REF it suffices to prove that the objects in MATH are noetherian. So let MATH be an object in MATH, and assume that MATH is not noetherian. Then there is some infinite chain MATH . By REF each of the MATH is indecomposable, and by REF each of the MATH is in MATH. Thus by REF each of the MATH is on a path from MATH to MATH. Since there are only a finite number of objects on such paths, this is a contradiction.
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math/9911242
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Assume MATH with MATH indecomposable. Then according to REF we can find an indecomposable MATH such that MATH is non-zero when evaluated on MATH and MATH. Thus we obtain MATH which is a contradiction.
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math/9911242
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As in REF it suffices to prove that the objects in MATH are noetherian. So let MATH be an object in MATH, and assume that MATH is not noetherian. Then there is some infinite chain MATH . After possibly replacing MATH by a subsequence we may by REF assume that either each of the MATH is indecomposable, or else that they all have exactly two indecomposable summands. The first case is dealt with exactly as in REF , so we will concentrate on the second case. Thus we now assume that MATH with MATH and MATH indecomposable. Using REF it is easily seen that possibly after interchanging MATH and MATH we may assume that there exist non-zero maps MATH and MATH. Thus as in the proof of REF we must have MATH and MATH for large MATH, and hence also MATH. From the fact that MATH is MATH-finite, it then easily follows that the inclusion MATH must actually be an isomorphism. This finishes the proof.
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By NAME duality we have MATH with MATH. We will show that MATH for some MATH. By the vanishing of MATH this automatically implies that MATH for all MATH. Since MATH if MATH and MATH is finite dimensional, it suffices to show that this inclusion is not equality for small MATH if MATH. Let MATH be a non-zero map. Since MATH, there must exist some MATH such that MATH and hence some MATH such that MATH. Thus MATH and we are done.
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It is sufficient to show that for any MATH there is a subobject MATH with MATH such that MATH. To prove this we verify the hypotheses for REF for the inverse system of cotorsion subobjects of MATH, that is, subobjects MATH of MATH with MATH. So let MATH be a maximal subobject contained in all MATH with MATH. If MATH then let MATH be a maximal subobject of MATH. Since MATH is simple we have MATH. By NAME we obtain the existence of MATH such that MATH and MATH. This contradicts the definition of MATH. Hence MATH and we are done.
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CASE: From the fact that MATH is torsion free we obtain MATH, and since MATH is simple modulo MATH, the latter is a division algebra. So MATH has no zero divisors. Since by our general hypotheses MATH is finite dimensional, we obtain that MATH. CASE: Pick non-trivial maps MATH, MATH. Then MATH and MATH represent non-zero maps in MATH and MATH. Since MATH and MATH are simple, these maps must be isomorphisms. In particular the compositions MATH and MATH are non-zero in MATH and MATH, and hence they are also non-zero in MATH and MATH. It follows from REF that MATH and MATH are scalar multiples of the identity map. Thus MATH is an isomorphism.
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Assume that there is a surjective map MATH where MATH. CASE: We can reduce to two cases. CASE: MATH and MATH for some simple object MATH. Applying MATH to the short exact sequence MATH yields MATH since MATH using NAME duality. If MATH then MATH and hence MATH since MATH. If MATH then MATH, and we have MATH. CASE: MATH and MATH. Applying MATH to the short exact sequence MATH yields MATH . If MATH then we take MATH and we have MATH. If MATH then we take MATH and we have MATH. In both cases we find MATH. CASE: Assume that there is a non-zero map MATH (with MATH). If MATH then it follows as above that MATH maps non-trivially to MATH. By NAME duality we have that MATH. Let MATH be a non-trivial extension of MATH and MATH. Then MATH is torsion free and maps non-trivially to MATH.
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math/9911242
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Clearly REF. and one direction of REF are trivial. We start by proving the non-trivial direction of REF. So assume that MATH and MATH in MATH are simple modulo MATH and are linked to a common MATH. By REF we may assume that there exists MATH together with surjective maps MATH and MATH. Let MATH be the pullback of these maps and let MATH be the kernel of MATH. Thus we have a commutative diagram with exact rows. MATH . If the lower exact sequence splits, then MATH factors through MATH. In particular there is a non-trivial map MATH. Since MATH and MATH are simple modulo MATH, it follows that MATH must be an isomorphism modulo MATH since it is clearly not zero. Assume now that the lower exact sequence in REF does not split. Then it yields a non-trivial element of MATH. Hence by NAME duality MATH. Since MATH modulo MATH, we obtain MATH modulo MATH. This proves what we want. Now REF. follows easily from REF. For let MATH be simple modulo MATH. Then MATH is linked to the same MATH as MATH. Hence by REF we have that MATH or MATH modulo MATH. Thus we find that in any case MATH modulo MATH. This holds for all simples in MATH, and since this category is semisimple, it follows that MATH is the identity on objects.
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math/9911242
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We first claim that every MATH is linked to at least one object in MATH. This follows from the fact that MATH for MATH. So if MATH then MATH splits off as a factor from MATH, contradicting the connectedness of MATH. Let the simple objects in MATH be represented by MATH. Assume that MATH constitute one MATH-orbit modulo MATH (thus MATH). Let MATH be respectively the union of the orbits which are linked to MATH and MATH. By REF we have MATH and by the previous paragraph MATH. Define MATH as the full subcategory of MATH consisting of objects of the form MATH where MATH is isomorphic to MATH modulo MATH and MATH. We define MATH in a similar way but using MATH and MATH. It is easy to see that MATH. This contradicts again the hypothesis that MATH is connected. So we obtain that MATH. Since every MATH is linked to at least one MATH, and since MATH acts transitively, we obtain that every MATH is linked to every MATH.
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This follows from the NAME property together with REF . See the proof of REF .
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This theorem is proved in the same way as CITE. The only thing that needs to be proved slightly differently is REF. So let MATH. To prove that MATH is projective we take a minimal projective resolution MATH . By the definition of a minimal resolution we have MATH. This yields for MATH: MATH. Thus MATH. Let MATH be indecomposable projective over MATH. Then MATH for some MATH in MATH. The number of times that MATH occurs in a direct sum decomposition of MATH is given by the dimension of MATH, which is finite by hypothesis.
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Assume that the lemma is false. So MATH maps to an infinite number of different MATH with MATH. In particular MATH must be infinite. Without loss of generality we may assume that MATH is torsion free and that MATH has simple image in MATH. Choose an element MATH and write MATH where for MATH the corresponding diagonal idempotent in MATH is denoted by MATH. Thus by the explicit structure of MATH given in REF we find. MATH . Since by REF it follows that MATH is projective with finite multiplicities we have MATH for certain MATH. By REF we obtain MATH . Define MATH as MATH and let MATH be the object in MATH with the property that MATH. Thus MATH is the unique object of length MATH with cosocle MATH. We have MATH . Put MATH and MATH. Then combining the canonical surjective maps MATH yields a canonical surjective map MATH and hence by REF a surjective map MATH. Let MATH be the kernel of this map. Using the exactness of completion we find MATH . Applying MATH to the exact sequence MATH we obtain the exact sequence MATH . Furthermore MATH. This goes to infinity by REF and the definition of MATH. It follows that MATH also goes to infinity. On the other hand we now show that in fact MATH. This is clearly a contradiction. We apply MATH to the exact sequence REF. This yields the exact sequence MATH . From the fact that MATH is simple modulo MATH we obtain by REF that MATH. On the other hand, by completing and using REF we find MATH. This proves what we want.
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math/9911242
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It clearly suffices to show that if MATH then MATH is noetherian. Now we know that MATH is a projective object in MATH mapping only to a finite number of different simples (counting multiplicities). Hence MATH is a direct sum of a finite number of indecomposable projectives. Since the indecomposable pseudocompact projectives over MATH are noetherian, it follows that MATH is noetherian.
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math/9911242
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Let MATH be the linear map corresponding to the identity map MATH under NAME duality. As in the proof of REF it follows that MATH can be chosen freely, subject to the condition that it must be non-vanishing on the almost split sequence ending in MATH. Now define MATH as MATH. Then by functoriality we have a commutative diagram: MATH . It follows that the pairings for MATH and for MATH are compatible. The upper pairing is non-degenerate by NAME duality and the lower pairing is non-degenerate by local duality for MATH (or graded local duality for MATH if MATH). Using these dualities REF follows from REF.
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math/9911242
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This is trivial.
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By REF is a skew field. Since MATH is algebraically closed, it now suffices to show that MATH is finite dimensional. We have MATH . By REF the MATH topology on MATH and MATH is separated. Hence MATH . If MATH runs through the subobjects in MATH with the property that MATH then MATH runs through the subobjects MATH of MATH such that MATH. Thus MATH . By REF this is finite dimensional.
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math/9911242
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Since MATH and MATH are torsion free one has that MATH is equal to the union of MATH where MATH. Hence it suffices to show that the dimension of MATH goes to infinity if MATH runs through the subobjects of MATH satisfying MATH. Put MATH. Applying MATH to the exact sequence MATH yields the exact sequence MATH . Hence it suffices to show that the dimension of MATH goes to infinity. By NAME duality and completion we must show that the dimension of MATH goes to infinity. To this end it is sufficient to show that for indecomposable projectives MATH and MATH over MATH the dimension of MATH goes to infinity where MATH runs through the finite dimensional quotients of MATH. This is an easy direct verification using the explicit structure of MATH given in REF.
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math/9911242
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Assume that there are both finite and infinite orbits. Then by REF there exists MATH, simple in MATH, such that MATH is linked to both a finite and an infinite orbit. This is a contradiction according to REF . Assume now that the orbits are finite but that there exist objects MATH which are distinct and simple modulo MATH. Clearly MATH. On the other hand by REF we know that MATH and MATH are both linked to the same finite orbit, whence by REF it follows that MATH is infinite dimensional. This is clearly a contradiction.
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math/9911242
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Let MATH be the category of triples defined in the statement of the lemma. Define MATH by MATH where MATH is the natural isomorphism MATH. We have to show that MATH is an equivalence. To do this we show that MATH is respectively faithful, full and essentially surjective. Faithfulness : Assume that MATH and that MATH is a homomorphism such that MATH. Thus MATH and MATH. The fact that MATH means that MATH factors as MATH where MATH, the map MATH is an epimorphism and MATH is a monomorphism. Then MATH factors likewise. This then implies that MATH and hence MATH. We conclude MATH. Fullness : Let MATH and MATH be as in the previous paragraph. Now assume that we are given maps MATH and MATH such that MATH. We have to produce a map MATH such that MATH and MATH. Let MATH be a representative of MATH in MATH where MATH is a subobject of MATH such that MATH. Below we denote the inclusion map MATH by MATH. We now have MATH. This means that there exist MATH with MATH such that the compositions MATH and MATH are equal. From REF it follows easily that there exists MATH which is a subobject of MATH with cokernel in MATH such that MATH. We now replace MATH by MATH, and MATH by the composition MATH. Hence we obtain the following commutative diagram MATH . So our problem is now to lift MATH to a homomorphism MATH such that MATH. Let MATH be the pushout of MATH and MATH. Then we have a commutative diagram with exact rows MATH with MATH. It is easy to see that factorizations of MATH through MATH are in one-one correspondence with splittings of MATH. Similarly factorizations of MATH through MATH correspond to splittings of MATH. Now we know that MATH factors through MATH and hence this yields a splitting of MATH. By REF this splitting corresponds to a splitting of MATH and thus to a factorization MATH of MATH through MATH. It is now easy to see that this is the MATH we are looking for. Essential surjectivity : Let MATH. We have to show that this is in the essential image of MATH. Choose MATH representing MATH. By REF is now an isomorphism MATH. As usual MATH is represented by a injective map MATH with cokernel in MATH where MATH is a subobject of MATH also with cokernel in MATH. Thus we have the following maps MATH . By working from left to right and employing REF these arrows and objects can be ``uncompleted". That is, there are arrows and objects in MATH such that completion gives REF. Now MATH and MATH. Checking the appropriate commutative diagrams we find that indeed MATH.
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math/9911242
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NAME of REF follows from the fact that there are only a finite number of MATH such that MATH (see REF ). To prove that REF gives the correct result for MATH on MATH one has to check that it gives the correct result on MATH, MATH and MATH. In each of the cases this is clear.
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The MATH-topology on MATH is separated by REF . So by REF there exists a subobject MATH with MATH and with the property MATH for every subobject MATH of MATH. By REF there exists a MATH such that MATH for MATH. Hence we have MATH for MATH. From this we easily obtain that if we can show REF then for all MATH we have MATH for MATH. Thus REF follows. Now let MATH and let MATH be the largest subobject of MATH which can be written as a quotient of a direct sum MATH with MATH. Assume MATH and write MATH. As above there exists MATH such that MATH for MATH. On the other hand we claim that there exists a non-trivial map MATH for some MATH. By the vanishing of MATH this map then lifts to a non-trivial map MATH, contradicting the choice of MATH. It remains to show that there is a non-trivial map MATH for some MATH. Assume first that MATH contains a torsion free subobject MATH. We may assume that MATH is simple modulo MATH. Then the proof of REF together with REF shows that there is a non-trivial map MATH with MATH. As above we have that there is some MATH. Since MATH is torsion the composition MATH cannot be zero. This proves what we want. Assume now that MATH is torsion and let MATH for MATH be a simple subobject of MATH and let MATH be the MATH-orbit of MATH. Then according to REF there exists MATH together with a non-trivial map MATH. We now consider two possibilities: CASE: MATH. Since MATH modulo MATH for all MATH, it follows that there will also be a non-trivial map MATH for all MATH, whence a non-trivial map MATH. Choose MATH in such a way that MATH and apply MATH. This yields a non-trivial map MATH for all MATH. CASE: MATH. Now we have a surjective map MATH. Since the MATH-orbit of MATH is finite, it follows that we can always find MATH such that MATH maps surjectively to MATH. Hence in both cases we obtain a non-trivial map MATH with MATH. By the above discussion we are done.
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That MATH has the desired properties has already been stated, so we prove the converse. We embed MATH into its closure under direct limits MATH (see CITE). Let MATH be the direct sum of the injective hulls of the simples in MATH. If MATH is simple then clearly MATH and hence MATH for MATH in MATH. For use below we note that it is easy to see that REF holds even for MATH. It follows from REF that MATH defines a cohomological functor MATH. Since MATH is saturated this functor is representable by some object MATH, and since the restriction of MATH is zero on MATH for MATH and exact on MATH it follows that MATH is actually a projective object in MATH. Hence we now have MATH as functors on MATH. Applying MATH it follows that this also holds as functors on MATH. Thus we have MATH. Using REF we find that MATH. Thus MATH is an injective cogenerator of MATH and hence a projective generator of MATH. It follows that MATH is equivalent to MATH where MATH. Since MATH has finite homological dimension, so does MATH. Furthermore MATH (using the functor MATH). This finishes the proof.
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We already know that the category in REF. has the required properties. That this is so for the category in REF. follows in the same way as for MATH where MATH is a non-singular proper curve. We now prove the converse. Assume that MATH is not of the form REF. This means by REF that it has one of the following forms CASE: MATH for MATH a star. CASE: The MATH or the MATH category. CASE: Finite dimensional nilpotent representations over MATH or MATH, with all arrows oriented in the same direction. In the categories in REF all objects are of finite length but they are clearly not of the form MATH since they have no projectives. Hence by REF it follows that MATH is not of type REF . By construction the categories in REF are derived equivalent to the finite dimensional representations over an infinite quiver. Hence again by REF the categories in REF are not saturated. So we are left with REF . Since MATH is a star, there is clearly a section MATH in MATH such that all paths in MATH are finite. Then the indecomposable projective and injective representations of MATH have finite length. By REF we know that MATH is derived equivalent to MATH, and all objects in MATH have finite length. Again by REF we know that MATH is saturated if and only if MATH is finite. But then the same holds for MATH. We conclude that MATH is of the form MATH with MATH finite.
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Let us prove the non-obvious implication. Thus assume that MATH preserves epis. Let MATH be an exact sequence. We have to show that it is NAME equivalent to a split one. Let MATH. Then we have a surjective map MATH, as well as an exact sequence MATH, representing an element of MATH. By the fact that MATH preserves epis, this can be lifted to an element of MATH. Thus we obtain a commutative diagram with exact rows: MATH which can be transformed into the following diagram. MATH from which it follows that REF represents a trivial class in MATH.
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This now follows from the previous lemma using the fact that for MATH one has MATH (since MATH is closed under extensions).
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math/9911242
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CASE: We first show MATH. We have that MATH is a thick REF abelian subcategory of MATH. Let MATH be an epimorphism in MATH, with MATH in MATH. We have by REF, where each MATH is in MATH. Since MATH is noetherian, there is some MATH such that the induced map MATH is an epimorphism. By the dual of CITE we have an isomorphism of derived categories MATH, and in particular MATH for MATH and MATH in MATH. It follows that MATH. CASE: Assume now that MATH. We want to show that for MATH and MATH in MATH we have MATH. If MATH and MATH are both in MATH, this follows from the above, since MATH, which is REF by assumption. If MATH is in MATH, and hence by assumption is noetherian, MATH commutes with MATH, and hence the same is true for the derived functors CITE. It then follows that MATH if MATH is in MATH. Let now MATH be in MATH, and consider the exact sequence MATH where the MATH are injective. Then we have MATH for all MATH in MATH. In particular, for any inclusion map MATH with MATH in MATH there is a commutative diagram MATH where MATH. If we have a commutative diagram MATH with MATH and MATH in MATH and MATH a monomorphism, we then have MATH with MATH and MATH with MATH, but we do not necessarily have MATH. To adjust the map MATH observe that MATH. Hence there is a map MATH with MATH. Since MATH because MATH is in MATH, there is a map MATH such that MATH, and hence MATH. Letting MATH we see that MATH as desired. We now use NAME 's lemma to choose a maximal pair MATH. Assume that MATH. Since MATH is the direct limit of objects in MATH, there is some MATH with MATH in MATH and MATH. Consider MATH and the map MATH. Then there is by the first part of the proof some map MATH such that MATH with MATH, and MATH. Hence the map MATH factors through a map MATH. This map has the property that MATH with MATH. This contradicts the maximality of MATH, and hence MATH. This means that MATH is a split epimorphism, and hence MATH, so that MATH for all MATH in MATH. Hence we have MATH.
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math/9911243
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Let MATH, and define the family of functions MATH by: MATH for every MATH, MATH and MATH. From this we see that: if MATH then MATH so MATH where MATH. Assume that MATH. Then there exists a permutation MATH such that MATH, so there exist MATH positions MATH such that the subsequence MATH is order-isomorphic to one of the patterns in MATH, which contradicts the definition of MATH. So MATH for MATH. Besides MATH (from the definition of this set), hence MATH for MATH.
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math/9911243
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Let MATH. From REF we get that MATH if and only if MATH . So MATH. Besides MATH, hence the theorem holds.
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math/9911243
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We define a function MATH by MATH then evidently MATH for all MATH. Let MATH and MATH. So MATH hence by the isomorphism MATH we have that MATH.
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math/9911243
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Let MATH, and let us consider the possible values of MATH: CASE: MATH. Evidently MATH if and only if MATH. CASE: MATH. Then there exist MATH such that MATH is a permutation of the numbers MATH. For any choice of MATH positions out of MATH, the corresponding permutations preceeded by MATH is order-isomorphic to some permutation in MATH. Since MATH avoids MATH, it is, in fact, order-isomorphic to MATH. We thus get at least MATH occurrences of MATH in MATH, a contradiction. CASE: MATH. Then there exist MATH such that MATH is a permutation of the numbers MATH. As above, we immediately get that MATH is order-isomorphic to MATH. We denote by MATH the set of all permutations in MATH such that MATH, and define the family of functions MATH by: MATH for every MATH, MATH and MATH. It is easy to see that for all MATH, MATH hence MATH. Now we define another function MATH by: MATH where MATH, MATH, MATH. Observe that MATH, since otherwise already MATH contains a pattern from MATH, a contradiction. It is easy to see that MATH for all MATH, hence MATH. So finally, MATH and MATH, since MATH. Since the above cases MATH are disjoint, and by REF we obtain MATH hence MATH for all MATH, MATH.
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math/9911243
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Let MATH be a permutation complement to MATH. By the natural bijection between the set MATH and the set MATH for all MATH we have that MATH have the same cardinality as MATH, which is MATH by REF .
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math/9911243
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Let MATH, MATH, and let us consider the possible values of MATH: CASE: Let MATH. Evidently MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. CASE: MATH. By definition we have that MATH, so let MATH. If MATH then MATH contains at least MATH occurrences of a pattern from MATH, and If MATH then MATH conatins at least MATH occurrences of a pattern from MATH, a contradiction. Since the above cases MATH are disjoint and by REF we obtain MATH for all MATH. Besides MATH, hence MATH.
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math/9911252
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MATH .
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math/9911252
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This result is accomplished by sliding all the algebra on the immersed curve that represents the morphism to the bottom of the curve. The element MATH is the algebraic element so concentrated at the top of the curve. The curve itself is regularly homotopic (fixing the ends at top and bottom) to MATH for a unique integer MATH (the NAME degree of that curve). Since regular homotopy is produced by the topological axioms of the category, this shows that MATH as desired.
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math/9911253
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Proceeding as in the proof of REF , consider the handleslides of the toral REF - handle over the grapes in the reduced forms, as shown in REF . After an isotopy, the toral REF - handle appears as in REF , labelled with a MATH; note that its framing is now MATH. With the exception of MATH, the grapes are written in the usual hexagonally packed notation. We should remark that one is guided in discovering the slides in REF by the multiplicities of the components of the singular fibers, which determine the number of times the toral handle must slide over each grape to achieve the simple pattern in REF . However, one must still be very careful in choosing where to perform the slide in order to avoid knotting and linking. The argument is now completed by a sequence of slips. For singular fibers of type I* (REF ,b) a single slip of grape MATH over grape MATH does the job, recovering the handlebodies given in REF *. For types II* - IV*, the sequence of slips shown in REF will give the clusters of grapes given in the table. Note that if MATH links only one of the grapes, then that grape can be slipped over others while carrying MATH along for the ride. This completes the proof of REF *.
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math/9911254
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Let MATH (MATH), MATH REF be a pair of MATH - compressing disks which gives weak MATH - irreducibility. CASE: We may suppose that MATH (MATH respectively,) is non-separating in MATH (MATH respectively,). Proof of REF Suppose that MATH is separating in MATH. Then MATH cuts MATH into two solid tori, say MATH, MATH. By exchanging the suffix, if necessary, we may suppose that MATH. Then take a meridian disk MATH in MATH such that MATH. We may regard MATH as a (non-separating essential) disk in MATH, and we have MATH. By regarding MATH as MATH, we see that we may suppose that MATH is non-separating in MATH. Suppose that MATH is separating in MATH. Since MATH does not intersect MATH in one point, we have MATH. The disk MATH cuts MATH into two solid tori MATH, MATH, where MATH is a core circle of MATH. If MATH, then the above argument works to show that there exists a non-separating meridian disk for MATH giving weak MATH - reducibility together with MATH. If MATH, then we take a meridian disk MATH for MATH such that MATH, and MATH intersects MATH transversely in one point. We may regard MATH a (non-separating essential) MATH - disk in MATH, and we have MATH. By regarding MATH as MATH, we see that we may suppose that MATH, MATH are non-separating in MATH, MATH respectively. Now we have the following two cases. CASE: MATH. Let MATH be the solid torus obtained from MATH by cutting along MATH. Since MATH is non-separating in MATH and MATH does not contain non-separating REF - sphere, we see that MATH is an essential simple closed curve in MATH. Since MATH does not contain non-separating REF - sphere or punctured lens spaces, MATH is a longitude of MATH, and, hence, there is an annulus MATH in MATH such that MATH. Then MATH gives a disk bounding MATH, and this shows that MATH is a trivial knot. CASE: MATH. Let MATH, MATH, MATH, and MATH. Note that MATH is a core with respect to a natural REF - handle structure on MATH. It is easy to see that MATH is a trivial arc in MATH. Let MATH. We regard MATH as an arc properly embedded in MATH. CASE: MATH is a solid torus and MATH is a trivial arc in MATH. Proof of REF Let MATH be the solid torus obtained from MATH by cutting along MATH and MATH. By the arguments in REF , we see that MATH is a longitude of MATH. Hence MATH is a REF - ball and MATH is a trivial arc in MATH. Since MATH is obtained from MATH by identifying two disks in MATH corresponding to the copies of MATH, we see that MATH is a solid torus, and MATH is a trivial arc in MATH. Hence we see that MATH gives a genus one REF - bridge position of MATH. By the construction of MATH, we see that MATH is isotopic to an unknotting tunnel associated to MATH.
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math/9911254
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Let MATH, MATH, MATH be the closures of the components of MATH such that MATH, MATH are disks, and MATH is an annulus. We divide the proof into several cases. CASE: Either MATH or MATH, say MATH, is separating in MATH. We first show: CASE: The annulus MATH is boundary parallel in MATH. Proof Since MATH is separating in MATH, the component of MATH corresponding to MATH is separating in MATH. Hence, by Lemma C-REF, we see that MATH is compressible or boundary parallel in MATH. Suppose that MATH is compressible in MATH. Since MATH does not contain non-separating REF - sphere, we see that MATH is also separating in MATH, and, hence, MATH and MATH are pairwise parallel in MATH. Let MATH be the annulus in MATH such that MATH. By exchanging suffix, if necessary, we may suppose that MATH is properly embedded in MATH. Since each component of MATH is an unknotted arc, we see that MATH is an unknotted annulus in MATH, and this implies that MATH and MATH are parallel in MATH, and, hence, in MATH that is, MATH is boundary parallel. This completes the proof of REF . By REF , we may suppose, by isotopy, that MATH, and MATH, where MATH is an annulus contained in MATH. CASE: Both MATH and MATH are MATH - incompressible in MATH. Proof Assume, without loss of generality, that there is a MATH - compressing disk MATH for MATH. Note that since MATH consists of two points, MATH and MATH are parallel in MATH. Let MATH be the annulus in MATH bounded by MATH. Let MATH be the disk in MATH bounded by MATH. Then we have the following two cases. CASE: MATH is contained in MATH. We consider REF - sphere MATH in MATH. Let MATH be REF - ball in MATH bounded by MATH. Since MATH does not contain a local knot in MATH, we see that MATH is an unknotted arc properly embedded in MATH. Hence there is an ambient isotopy of MATH which moves MATH to an arc in MATH joining MATH, and which does not move MATH. On the other hand, MATH is a component of the strings of the trivial tangle MATH. This shows that MATH is a trivial knot, a contradiction. CASE: MATH is contained in MATH. In this case, we first consider the disk MATH. By a slight deformation of MATH, we obtain a MATH - compressing disk MATH for MATH such that MATH is contained in MATH. Then, by the argument as in REF , we see that MATH is a trivial knot, a contradiction. This completes the proof of REF . Now we have the following two subcases. CASE:REF MATH and MATH are not MATH - parallel in MATH. In this case, by Lemma D-REF, we see that MATH is isotopic to MATH in MATH. This shows that MATH is isotopic to MATH. CASE:REF MATH and MATH are MATH - parallel in MATH. Let MATH, MATH be the closures of the components of MATH such that MATH. Then MATH is a torus with one hole properly embedded in MATH. By Lemma D-REF, we may suppose, by exchanging suffix if necessary, that there is a MATH - compressing disk MATH for MATH such that MATH, and MATH consists of a point. We consider the genus one surface MATH properly embedded in MATH. By Lemma B-REF, we see that MATH is MATH - compressible in MATH. Let MATH be the MATH - compressing disk for MATH. Now we have the following subsubcases. CASE:REF MATH is contained in MATH. By the MATH - incompressibility of MATH REF , we see that MATH that is, MATH consists of a point. Then MATH cuts MATH into a REF - string trivial tangle which is MATH - isotopic to MATH. Hence MATH is isotopic to MATH. CASE:REF MATH is contained in MATH. In this case, we first show: CASE: MATH. Proof Suppose that MATH. Then, by compressing MATH along MATH, we obtain a disk MATH properly embedded in MATH such that MATH, and MATH separates the components of MATH. Let MATH, MATH be the closures of the components of MATH such that MATH, MATH. Then we can isotope MATH rel MATH in MATH to an arc in MATH without moving MATH. Since MATH and MATH are MATH - parallel in MATH, this shows that MATH is a trivial knot, a contradiction. Let MATH be the closure of the component of MATH such that MATH. Note that MATH is a solid torus in MATH with MATH. By regarding MATH as a very thin solid torus, we may suppose that MATH consists of a disk MATH intersecting MATH in one point. Then MATH is an annulus MATH. CASE: MATH is incompressible in MATH. Proof Assume that MATH is compressible in MATH. Then, by compressing MATH, we obtain a disk MATH in MATH such that MATH. Since MATH intersects MATH in one point, MATH is a non-separating disk in MATH. Hence, we see that MATH is a non-separating REF - sphere in MATH, a contradiction. Then, by Lemma C-REF, there is an essential disk MATH in MATH such that MATH, and, hence, MATH. This shows that MATH is weakly MATH - reducible. CASE: Both MATH and MATH are non-separating in MATH. In this case, we first show: CASE: MATH is boundary parallel in MATH. Proof Assume that MATH is not boundary parallel. Since MATH does not contain non-separating REF - sphere, we see that MATH is incompressible in MATH. Hence, by Lemma C-REF, we see that there is an essential disk MATH for MATH such that MATH, and that MATH cuts MATH into two solid tori MATH, MATH, where MATH. Moreover, since MATH does not contain a punctured lens space, we see that each component of MATH represents a generator of the fundamental group of the solid torus MATH. However this contradicts Lemma C-REF. By REF , we may suppose, by isotopy, that MATH, and MATH, where MATH is an annulus contained in MATH. Then we have the following subcases. CASE:REF Both MATH and MATH are MATH - incompressible in MATH. This case is divided into the following two subsubcases. CASE:REF MATH and MATH are not MATH - parallel in MATH. In this case, by Lemma D-REF, we see that the given unknotting tunnel MATH is isotopic to MATH. CASE:REF MATH and MATH are MATH - parallel in MATH. By Lemma D-REF, there is a MATH - boundary compressing disk MATH for MATH or MATH, say MATH, such that MATH. Let MATH be the closure of the component of MATH which is a torus with two holes. Let MATH. Then MATH is a compressing disk for MATH. Let MATH be the disk obtained by compressing MATH along MATH, and MATH a disk obtained by pushing MATH slightly into MATH. We may regard MATH is properly embedded in MATH. Suppose that MATH is MATH - compressible in MATH. Then we can show that MATH is a trivial knot by using the argument as in the proof of REF of REF Hence MATH is MATH - incompressible in MATH. Hence, by Lemma B-REF , either MATH and MATH are MATH - parallel or MATH bounds a REF - string trivial tangle in MATH, which is not a MATH - parallelism between MATH and MATH. In the former case, we immediately see that the given unknotting tunnel MATH is isotopic to MATH. In the latter case, we have: CASE: Suppose that MATH bounds a REF - string trivial tangle in MATH which is not a MATH - parallelism between MATH and MATH. Then MATH is isotopic to MATH. Proof By Lemma B-REF , we see that MATH and MATH bounds a MATH - parallelism in MATH. Hence, by isotopy, we can move MATH to the position such that MATH, and MATH. Then, by applying the argument of REF with regarding MATH, MATH as MATH, MATH respectively, we see that MATH is isotopic to MATH. CASE:REF Either MATH or MATH is MATH - compressible in MATH. Let MATH be a compressing disk for MATH or MATH, say MATH, in MATH. Then MATH and MATH are parallel in MATH, and let MATH be the annulus in MATH bounded by MATH. Let MATH be a disk properly embedded in MATH which is obtained by moving MATH slightly so that MATH. CASE: MATH. Proof Assume that MATH. Then we may regard MATH is a MATH - compressing disk for MATH in MATH. Then, by using the arguments in REF of the proof of REF of REF , we can show that MATH is a trivial knot, a contradiction. By REF , MATH looks as in REF . Assertion Either "MATH is a spine of MATH"or "there is an essential annulus in MATH". Proof of Assertion Let MATH be a sufficiently small regular neighborhood of MATH, and MATH. Note that MATH is a handlebody, because MATH is an unknotting tunnel for MATH. Let MATH be a non-separating essential disk properly embedded MATH. We may suppose that MATH consists of a disk intersecting MATH in one point. We suppose that MATH is minimal among all non-separating essential disks for MATH. CASE: No component of MATH is a simple closed curve, an arc joining points in MATH, or an arc joining points in MATH. Proof This can be proved by using standard innermost disk, outermost arc, and outermost circle arguments. The idea can be seen in the following figures. CASE: MATH. Proof Assume that MATH. Let MATH be the solid torus obtained by cutting MATH along MATH. Note that MATH is a regular neighborhood of MATH. Since MATH is non-separating in MATH, and MATH does not contain a non-separating REF - sphere, MATH is an essential simple closed curve in MATH, and MATH is not contractible in MATH. This shows that MATH bounds a disk which is an extension of MATH. Hence MATH is a trivial knot, a contradiction. Hence MATH consists of a number of arcs joining points in MATH to points in MATH. Here, by using cut and paste arguments, we remove the components of MATH which are inessential in MATH. CASE: The components of MATH are not nested in MATH. Proof Let MATH be a component of MATH which is innermost in MATH, and MATH the disk in MATH bounded by MATH. Subclaim REF MATH is contained in MATH. Proof Assume that MATH is contained in MATH. Since MATH, this implies that MATH is contained in a regular neighborhood of MATH, contradicting the fact that MATH is an unknotting tunnel. Subclaim REF MATH. Proof Assume that MATH. Then we can show that there is a non-separating disk MATH properly embedded in MATH such that MATH by using the argument as in the Proof of REF of the proof of REF . Then by using the argument as in the proof of REF above, we can show that MATH is a trivial knot, a contradiction. Hence there exists a component of MATH connecting MATH and MATH. This means that MATH is not surrounded by another component of MATH, and this gives the conclusion of REF For each component MATH of MATH, MATH consists of more than one component. Proof Assume that MATH consists of a point. Let MATH be the disk in MATH bounded by MATH. Then MATH and MATH intersects in one point, and this shows that MATH is a trivial arc in MATH, a contradiction. Let MATH. We call the boundary component of MATH corresponding to MATH the outer boundary. Other boundary components of MATH (: the components of MATH) are called inner boundary components. Let MATH be the solid torus obtained by cutting MATH along MATH. Let MATH be an inner boundary component which is "outermost" with respect to the intersection MATH, that is: Let MATH be the union of the components of MATH intersecting MATH. Then except for at most one component, each component of MATH does not intersect other inner boundary components. Let MATH be the disk in MATH bounded by MATH. Let MATH be the components of MATH, which are located on MATH in this order, where MATH cobounds a square MATH in MATH. Let MATH. Let MATH be the image of MATH in MATH. Note that MATH is a torus with two holes. Let MATH. Then by the minimality condition, we see that each MATH is an essential arc properly embedded in MATH. CASE: If MATH are mutually parallel in MATH, then there is an essential annulus in MATH. Proof Note that MATH consists of MATH components, that is, MATH above, and another component, say MATH. Subclaim REF MATH is not parallel to MATH in MATH. Proof Assume that MATH are mutually parallel in MATH. Then we can take a simple closed curve MATH in MATH such that MATH intersects MATH transversely in one point, and MATH is ambient isotopic to MATH in MATH. Let MATH be a regular neighborhood of MATH in MATH such that MATH. Note that MATH is a solid torus, and MATH wraps around MATH longitudally MATH times. This show that REF - sphere contains a lens space with fundamental group a cyclic group of order MATH, a contradiction. By Subclaim REF, we see that we can take simple closed curves MATH, MATH in MATH such that MATH, MATH intersects MATH transversely in one point, MATH is ambient isotopic to MATH in MATH, and MATH is ambient isotopic to MATH in MATH. Let MATH be a regular neighborhood of MATH in MATH such that MATH, and MATH. Then MATH is a genus two handlebody, and MATH is an annulus in MATH. Note that MATH is a regular neighborhood of MATH. Then we denote by MATH the closure of the exterior of this regular neighborhood of MATH. Note that MATH is embedded in MATH. Then attach MATH to MATH along MATH. It is directly observed (see REF ) that we obtain a solid torus, say MATH, such that MATH wraps around MATH longitudally MATH - times. Then, let MATH. Note that MATH is an annulus properly embedded in MATH. Assume that MATH is compressible in MATH. Then the compressing disk is not contained in MATH since MATH is incompressible in MATH. Hence MATH together with a regular neighborhood of this compressing disk produces a punctured lens space with fundamental group a cyclic group of order MATH in MATH, a contradiction. Hence MATH is incompressible in MATH. Then assume that MATH is boundary parallel, and let MATH be the corresponding parallelism. Since MATH, MATH is not MATH. Hence MATH, and this shows that MATH is a solid torus, which implies that MATH is a trivial knot, a contradiction. Hence MATH is an essential annulus in MATH, and this completes the proof of REF . Suppose that MATH contains at least two proper isotopy classes in MATH. We suppose that MATH, MATH belong to mutually different isotopy classes. Let MATH, MATH be the components of MATH. Since MATH and MATH intersects transversely, we easily see that we may suppose that MATH, and MATH. Let MATH be the solid torus obtained by cutting MATH along MATH, and MATH (, hence, MATH is homeomorphic to (torus)MATH). Here we may regard that MATH is obtained from MATH by adding a REF - handle MATH corresponding to MATH, where MATH is a core of MATH. Let MATH, MATH be the components of the image of MATH in MATH, where we may regard that MATH is obtained from MATH by adding MATH. CASE: MATH is "vertical" in MATH that is, MATH is ambient isotopic to the union of arcs of the form MATH, where MATH, MATH are points in (torus). Proof By extending MATH (MATH respectively,) to the cores of MATH, we obtain either an annulus which contains MATH or MATH (if MATH (MATH respectively,) is contained in MATH or MATH), or a rectangle two edges of which are MATH and MATH (if MATH (MATH respectively,) joins MATH and MATH) in MATH. Then we have the following three cases. CASE: Both MATH and MATH join MATH and MATH. In this case, we obtain an annulus MATH by taking the union of the rectangles from MATH and MATH. Since MATH and MATH are not ambient isotopic in MATH, MATH is incompressible in MATH. We note that every incompressible annulus in MATH with one boundary component contained in MATH, the other in MATH is "vertical" (for a proof of this, see, for example, CITE). Hence MATH is vertical, and this shows that MATH is vertical. CASE: Either MATH or MATH, say MATH, join MATH and MATH, and MATH is contained in MATH or MATH. In this case, we see that MATH or MATH is vertical by the existence of the annulus from MATH. Then the existence of the rectangle from MATH shows that MATH and MATH are parallel, and this implies that MATH is vertical. CASE: MATH is contained in MATH, and MATH is contained in MATH. In this case we see that MATH is vertical by the existence of the vertical annuli from MATH and MATH. By REF , and REF, we see that MATH is a spine of MATH or there is an essential annulus in MATH, and this completes the proof of Assertion. Assertion shows that MATH is isotopic to MATH or there is an essential annulus in MATH, and this together with the conclusions of REF shows that we have the conclusions of REF for all cases. This completes the proof of REF .
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math/9911254
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Let MATH, and MATH the closures of the components of MATH such that MATH, MATH are disks and that they are located on MATH successively in this order. CASE: Suppose that there is an annulus component MATH of MATH (MATH or MATH) such that MATH is MATH - compressible in MATH. Then the MATH - compressing disk is disjoint from MATH. Proof Let MATH be the MATH - compressing disk for MATH. Assume that MATH that is, MATH consists of a point. Then, by compressing MATH along MATH, we obtain two disks each of which intersects MATH in one point. But this is impossible, since each component of MATH separates MATH into two disks each intersecting MATH in two points. CASE: Suppose that there is an annulus component MATH in MATH, and an annulus component MATH in MATH. Then either MATH or MATH is MATH - incompressible in MATH or MATH. Proof We first suppose that MATH is MATH - compressible in MATH. Then, by REF , the MATH - compressing disk is disjoint from MATH. Hence, by compressing MATH along the disk, we obtain two disks in MATH which are MATH - essential in MATH and disjoint from MATH. Let MATH be one of the disks. Assume, moreover, that MATH is also MATH - compressible. Then, by using the same argument, we obtain a MATH - essential disk MATH in MATH such that MATH. Note that MATH and MATH are parallel in MATH. This implies that MATH is a two-component trivial link, a contradiction. CASE: If MATH, then we have the conclusion of REF . Proof Note that there are at most three mutually non-parallel, disjoint essential simple closed curves on MATH. Hence if MATH, then there are three components, say MATH, MATH, MATH, of MATH which are mutually parallel on MATH. We may suppose that MATH, MATH, MATH are located on MATH successively in this order. Let MATH (MATH respectively,) be the annulus on MATH bounded by MATH (MATH respectively,). Without loss of generality, we may suppose that MATH (MATH respectively,) is properly embedded in MATH (MATH respectively,). Since MATH is connected, we may suppose, by exchanging suffix if necessary, that each component of MATH separates the boundary points of each component of MATH on MATH. Since each component of MATH is an unknotted arc, we see that MATH is an unknotted annulus. Hence there is an annulus MATH in MATH such that MATH and MATH and MATH are pairwise (MATH - )parallel in MATH. Let MATH be the parallelism between MATH and MATH. If MATH, then we can push the components of MATH out of MATH along the parallelism MATH, still with at least two components of intersection MATH. If MATH, then we can push MATH out of MATH along this parallelism to reduce MATH by two. According to REF and its proof, we suppose that MATH or MATH, and no three components of MATH are mutually parallel in MATH. Note that the intersection numbers of any simple closed curves on MATH with MATH are even, because MATH is a separating surface. This shows that MATH consists of two (in case when MATH) or three (in case when MATH) parallel classes in MATH each of which consists of two components. Hence, each component of MATH (MATH or MATH, say REF) is an annulus. If a component of MATH is MATH - incompressible in MATH, then, by the argument in the proof of REF , we have the conclusion of REF . Hence, in the rest of the proof, we suppose that each component of MATH is a MATH - compressible annulus in MATH. Let MATH be the closure of the component of MATH such that MATH. Note that MATH is a REF - ball such that MATH consists of some components of MATH. Then, by the assumptions, we see that MATH consists of either one, two, or three annuli. CASE: If MATH consists of an annulus, then there is a component of MATH which is MATH - boundary parallel in MATH, and, hence, we have the conclusion of REF . Proof Let MATH. Since MATH is compressible, there is a MATH - compressing disk MATH for MATH in MATH. Note that MATH REF . We may regard that MATH is properly embedded in MATH and MATH is parallel to MATH and MATH in MATH. Since MATH is connected, we see that MATH is non-separating in MATH. By cutting MATH along MATH, we obtain a solid torus MATH such that MATH is a core circle of MATH. Recall that MATH are the closures of the components of MATH. Note that MATH is properly embedded in MATH. Since REF - sphere does not contain a non-separating REF - sphere, we see that MATH in incompressible in MATH. Since every incompressible surface in (torus)MATH is either vertical or boundary parallel annulus (see CITE), MATH is boundary parallel in MATH. Let MATH be the parallelism for MATH, and MATH. Since MATH is connected, and MATH intersects MATH and MATH, we see that MATH is disjoint from the images of MATH and MATH in MATH. Hence we see that MATH is disjoint from the images of MATH in MATH. This shows that the parallelism MATH survives in MATH, and, hence, we have the conclusion of REF by the argument as in the proof of REF If MATH consists of two annuli MATH, MATH, then there is a component of MATH which is (MATH - )boundary parallel in MATH, and, hence, we have the conclusion of REF . Proof By exchanging suffix, if necessary, we may suppose that the annulus MATH is incident to MATH. If MATH, then we have MATH. If MATH, then, by reversing the order of MATH, and changing the suffix of MATH if necessary, we may suppose that MATH. Then let MATH be REF - manifold in MATH such that MATH. Note that MATH is embedded in MATH and MATH. Subclaim Either MATH or MATH, say MATH, is non-separating in MATH. Proof Assume that both MATH and MATH are separating in MATH. Then MATH and MATH are parallel in MATH, but this contradicts the fact that MATH and MATH are connected. Since MATH is a non-separating disk in MATH, and MATH does not contain a non-separating REF - sphere, we see that MATH is incompressible in MATH. Then, since MATH does not contain a punctured lens space with non-trivial fundamental group, we see that MATH is boundary parallel in MATH by Lemma C-REF (see the proof of REF of the proof of REF ). Hence MATH is a parallelism between MATH and MATH, and this shows that MATH is MATH - boundary parallel in MATH along this parallelism to give the conclusion of REF MATH does not consist of three components. Proof Assume that MATH consists of three annuli MATH, MATH, and MATH, where MATH, MATH, and MATH. Since MATH, MATH, MATH are MATH - compressible in MATH, there are mutually disjoint MATH - compressing disks MATH, MATH, MATH for MATH, MATH, MATH respectively. We may regard that MATH, MATH, MATH are properly embedded in MATH. Note that MATH, MATH, MATH are not mutually parallel in MATH. Hence we see that MATH cuts MATH into two components MATH, MATH such that one component of MATH is contained in MATH, and the other component is contained in MATH (see REF ). But this contradicts the fact that MATH is connected. REF , and REF complete the proof of REF .
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math/9911255
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Corresponding to the extension REF there is a fiber bundle MATH, with MATH a classifying space for MATH. As MATH realises the smallest possible NAME characteristic among all orientable MATH - manifolds with fundamental group MATH, compare CITE, we obtain MATH, as claimed, where the last equality follows from the multiplicativity of the NAME characteristic in fiber bundles. For the characterisation of the case of equality, notice that if MATH is regularly fibered, then it is aspherical by the homotopy exact sequence of the fibration. Further, if MATH is aspherical, then by the uniqueness of classifying spaces it is homotopy equivalent to MATH and therefore has the same NAME characteristic. Thus the crucial step is to show that the equality of NAME characteristics implies that MATH is induced by a regular holomorphic map. First of all, as MATH is minimal with MATH and with MATH, it must be of general type. In particular, it is NAME. By REF and NAME, see REF and also CITE, there is a surjective holomorphic map with connected fibers MATH, with MATH a compact complex curve, such that the map MATH in REF factors through MATH. This implies that MATH is a (finitely generated) subgroup of MATH. Thus MATH is the fundamental group of an orientable surface MATH which is a covering of MATH. If MATH were noncompact, MATH would be a free group, contradicting the fact that MATH has cohomological dimension MATH. Thus MATH is compact, with MATH . On the other hand, denoting the generic fiber of MATH by MATH, we have that MATH is a quotient of MATH and so MATH. Now by REF - NAME a singular fiber makes a positive contribution to the NAME characteristic, so we have MATH, so that MATH and MATH imply MATH . Combining REF, we conclude that MATH and therefore MATH. Thus MATH gives a complex structure on MATH and MATH is a holomorphic map inducing MATH. As we are in the case of equality for the NAME - NAME inequality, MATH must be everywhere regular.
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math/9911255
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Suppose MATH admits a complex structure. After possibly reversing the orientation, we may assume that the complex structure is compatible with the orientation. The argument in CITE was as follows: MATH is a minimal surface of general type for which the underlying manifold endowed with the other, non-complex, orientation is symplectic and therefore has non-zero NAME - NAME invariants. Thus REF gives MATH . This, together with the NAME - NAME inequality MATH implies MATH. We can now reach the same conclusion in a different way, and we can also show that the inequality must be strict, as claimed. By REF the surface MATH is regularly fibered, so that the non-negativity of the signature follows from NAME 's theorem. Moreover the NAME - NAME inequality is strict for regularly fibered surfaces, as proved by CITE. Suppose that MATH admits an NAME metric. As it is also symplectic, it has non-zero NAME - NAME invariants and by the result of CITE satisfies MATH. The same argument for the manifold with the other orientation gives MATH. It remains to exclude the case of equality. Suppose that MATH admits an NAME metric and that for a suitable choice of orientation MATH. Then CITE showed that the NAME metric must be NAME - NAME, so that MATH must be a complex surface which is a ball quotient. But then by the above argument for the complex case, we have a contradiction.
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math/9911255
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Suppose MATH admits a complex structure, then by REF MATH is regularly fibered. If the signature vanishes, we are in the borderline case of NAME 's theorem, which says that the signature is nonnegative, and is zero only if all the fibers are isomorphic, so the fibration is isotrivial. In this case we can pull back the fibration to a finite cover of MATH to obtain a product. This implies that the kernel of the monodromy representation has finite index in MATH. Conversely, assume that we have a bundle with finite monodromy. Then it must have zero signature. By the positive resolution of the NAME realisation problem CITE we can choose a complex structure on MATH and a lift of the monodromy representation to the diffeomorphism group of MATH so that the monodromy acts by complex analytic diffeomorphisms of MATH. Fixing an arbitrary complex structure on MATH, we obtain a complex structure on MATH by viewing it as MATH, where MATH acts on MATH by deck transformations and acts on MATH through the chosen lift of the monodromy representation to the diffeomorphism group.
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math/9911255
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Every finitely generated subgroup of MATH is the monodromy group of a surface bundle of zero signature over some base MATH, where the genus of MATH can be taken to be the number of generators of the subgroup, see REF. As the order of the finite subgroups of MATH is bounded, we have an a priori bound on MATH and can apply the previous corollary.
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math/9911257
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Since any two maps of MATH into MATH are mutually homotopic, there is a generic one-parameter family of smooth maps MATH connecting MATH and MATH. By dimensional reasoning, the generic births and deaths of self-intersection points have only two types; finger moves and cusps. In fact at a finite number of MATH's the image of MATH has a cusp or a partial tangency of two local sheets. By a finger move we mean the local deformation including the latter non-transverse case; a finger pushes a local sheet until it penetrates another local sheet. We get a pair of self-intersection points after the penetration with one plus and one minus intersection number. Note that the reverse process of the finger move is a NAME trick. We may think that the finger move follows along a curve MATH that connects two points of MATH. If the fundamental group of the complement is a cyclic group generated by the element encircling the surface, the curve MATH is homotopically trivial relative boundary and hence isotopically trivial; so, we get a REF - disk MATH with two corners and MATH. Now it is not difficult to decompose this finger move into two cusp births. The reverse deformation around the cusp birth, that is, the cusp death process can be described by the collapsing of a REF - disk MATH with one corner such that MATH and the corner point on the boundary is the self-intersection point. Moreover we have two types of cusps for the cusp birth; one gives a positive intersection point and another gives a negative one for the appearing immersed surface. If the finger move is isotopically trivial, we can construct a pair of disjoint collapsing REF - disks for the pair of positive and negative self-intersection points and get a pair of positive and negative cusps. This completes the proof of REF . For another proof one may consult the manuscript written by CITE .
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math/9911257
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Let MATH and MATH be embedded REF - spheres in MATH that are cusp equivalent. The one-parameter family MATH can be deformed so that any cusp birth takes place at MATH and any cusp death takes place at MATH. So we may assume that there is an immersed REF - sphere MATH in MATH such that MATH can be obtained from either MATH or MATH by attaching MATH cusps for some MATH. Let MATH be the embedded surface obtained from MATH by smoothing all the double points of MATH. The result follows from the previous lemma.
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math/9911257
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The surface which is obtained by attaching MATH trivial REF - handles is a connected sum of the original REF - knot and an unknotted surface MATH of genus MATH in the sense of CITE . Since the bounding solid tori for the unknotted surfaces MATH are ambient isotopic to each other by REF , we may assume that there is an orientation preserving diffeomorphism MATH of MATH to itself which not only satisfies MATH but also preserves the circles MATH's determined by the trivial REF - handles. Now we consider each trivial REF - handle. Perform surgery along the loop on the embedded surface which intersects the circle MATH at one point for the trivial REF - handle recursively. We have to respect the stable framing when making the surgery. Then we get a REF - knot in the new ambient manifold, the connected sum of MATH with MATH copies of MATH. Let MATH be the core REF - disk of the surgery disk and MATH the original NAME type REF - disk for the trivial REF - handle. REF - disks MATH's depend on each of the original REF - knots. If we do surgery on the new ambient manifold along each REF - sphere MATH, we get the original REF - knot in MATH. So, doing the reversing surgery MATH times for the complement of each of MATH and MATH again, we get REF .
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math/9911258
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Let MATH be a symplectic basis of MATH. There are MATH members in this basis while if MATH, then there are only MATH positions in the tensor product MATH. It is now a simple matter to construct MATH elements MATH in MATH such that MATH is the identity matrix. Hence the elements MATH are linearly independent. By the obvious duality, the MATH - invariant components of tensors MATH are also linearly independent.
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math/9911258
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For MATH the assertion is empty and for MATH we can check the assertion by a direct computation. Using the formula for the number MATH given above, it can be shown that MATH for any MATH. Hence MATH for MATH. On the other hand, we can inductively construct MATH elements in MATH which are linearly independent for MATH.
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math/9911258
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It can be shown that the elements in MATH are linearly independent because if we express MATH as a linear combination of the standard basis of MATH given in REF , then we find MATH. On the other hand, we can show that the projection under MATH of any member of this standard basis can be expressed as a linear combination of MATH.
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math/9911258
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NAME 's theory CITE implies that MATH can be identified with the kernel of the following homomorphism MATH where MATH on MATH and MATH is the natural injection. The differential MATH is given by MATH. It is easy to deduce from this fact that it is injective. Therefore MATH is isomorphic to the kernel of the following mapping MATH which is given by the NAME triple product. Hain proved in CITE that all of the higher NAME products of MATH vanishes for MATH. Hence passsing to the dual, we see that MATH is determined by the following exact sequence MATH . Here the first mapping is given by MATH and the image of which in MATH under the bracket operation is trivial because of the NAME identity. The irreducible decomposition of MATH is given by MATH . Explicit computations, corresponding to each summand in the above decomposition, show that the mapping MATH hits any summand except MATH. Since we already know that MATH, the result follows.
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math/9911258
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It is easy to see that the irreducible decomposition of the module MATH is given by MATH . Then the result follows from rather long explicit computations, in the framework of symplectic representation theory, of the NAME bracket MATH.
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math/9911258
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By REF , there is a chain MATH such that MATH and MATH is a non-zero element of MATH. On the other hand, explicit computation shows that there is a chain MATH such that MATH and MATH. We can arrange the above elements so that the MATH - cycle MATH is in fact a MATH - invariant one. Thus it defines an element of MATH on which the cohomology class in question takes a non-zero value. The primitivity follows from the property of the trace.
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math/9911258
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We consider the cochain MATH described in REF. If MATH, then MATH so that MATH where MATH is NAME 's signature REF - cocycle for the genus REF mapping class group. Since MATH, MATH is uniquely determined by the above equality. Observe that MATH is not a bounded cochain because MATH is a non-trivial homomorphism. We can now conclude that the bounded cohomology class MATH is non-trivial.
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math/9911258
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If we assume the contrary, then it would follow that the value of the cochain MATH is bounded. But as was mentioned above, this is not the case.
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math/9911260
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The independence of MATH from choice (relative to the endpoints) of the path MATH is identical to that given before, because the only term which could possibly change is the first. The independence from the initial point MATH is more elaborate, as suggested by the discussion above. Let MATH be another generic metric, and MATH a MATH - parameter family of metrics with CASE: MATH a generic path from MATH to MATH; CASE: MATH = a generic path from MATH to MATH; CASE: MATH = a generic path from MATH to MATH; CASE: MATH. As before, we get a MATH - parameter moduli space MATH which is a compact oriented MATH - manifold. Treating the MATH - chains MATH as parameters, in the spirit of the preceding remarks, we consider the following collection of MATH - parameter moduli spaces, which again are MATH - dimensional manifolds with boundary. MATH . The boundary of each of the MATH - dimensional moduli spaces REF,, REF has algebraically MATH points. As discussed before, the boundary of each MATH - parameter moduli space can alternatively be described as the sum of the algebraic counts of points in appropriate MATH - parameter moduli spaces. This leads to MATH equations: MATH . Adding these equations together, most of the terms cancel in pairs, leaving the difference between the invariant calculated with the paths MATH and MATH, plus MATH . However, the isomorphism REF, coupled with the orientation hypothesis that MATH, means that the two terms are equal, and so the invariant doesn't depend on the choice of initial metric MATH. The other choices of parameters involved in the definition of MATH are: the specific surface representing MATH, the choice of section defining MATH, the choice of MATH - chain MATH with MATH, and the section defining MATH. As remarked earlier, the verification that, for fixed MATH, the choices of section don't affect the value of MATH is virtually identical to arguments given above, because sections vary in a contractible space. A similar remark applies to the choice of MATH, given a specific MATH - chain MATH. The independence from the choice of MATH's and MATH's differs in that a substitute must be found for one basic mechanism: the existence of the family MATH derives from the fact that the space of metrics is simply connected. The idea is the same for all of the choices; we will illustrate the point in the simplest instance. So suppose that two MATH - chains MATH and MATH are given, both of which have boundary MATH. The only place in REF in which MATH enters is in the term MATH . Because the MATH - chains have the same boundary, it follows that MATH is a MATH - cycle which is a boundary of a MATH - chain MATH, by our hypothesis that MATH. One can use restriction to connections on MATH to define a MATH - dimensional moduli space MATH. Taking the boundary of this moduli space gives MATH by the standard argument.
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math/9911260
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Choose a path of metrics and a collection of MATH - chains MATH with MATH which define MATH. The path can assumed to be constant near the connected sum point, so it extends to give a path of metrics on MATH. Similarly, the MATH - chains can be assumed to miss the connected sum point, so they are MATH - chains in the connected sum in a natural way. Now we use a standard gluing argument: choose a metric on MATH with a long tube along the MATH. For sufficiently long tube length, we can calculate each term in the definition of the invariant. The MATH - chain MATH with MATH may be taken to be degenerate, so that the last two terms (of the form MATH) are MATH for dimensional reasons. The moduli spaces corresponding to the other terms in the definition, may all be described by the NAME model for the MATH - parameter moduli space, as in CITE. The local picture, and hence the calculation of the coefficients, is the same as in the proof of the usual blowup formula.
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math/9911260
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Suppose that MATH is of the form MATH, where MATH, and notice that restriction defines a homomorphism MATH . Let MATH be a diffeomorphism of the form MATH, as discussed before REF . In particular, MATH should be chosen as a composition of reflections in two different MATH - spheres, as in CITE; the intersection number MATH is computed in that paper to be MATH. Suppose finally that MATH has simple type in the sense of CITE, so that its NAME series MATH is determined by a finite set of basic classes MATH. Rewriting REF in terms of the NAME series of MATH, we see that MATH. In particular, MATH has the form described by the structure theorem of CITE, and so is determined by the same set of basic classes MATH. Moreover, the coefficients MATH in the expression of the series as a sum of exponentials of the MATH, are equal to the corresponding coefficients for MATH. Under composition of diffeomorphisms the NAME series add. For diffeomorphisms MATH whose series are determined by basic classes, this implies the following statement. The set of basic classes for MATH is the union of the set of basic classes for MATH and for MATH, leaving out those basic classes which MATH and MATH have in common but whose coefficients cancel. In other words, a basic class MATH is removed from the union if the coefficient MATH. In the paragraphs which follow, we will show that if MATH is any one of the manifolds described in the statement of REF , then it admits a series of diffeomorphisms MATH which are all homotopic to the identity, with the property that MATH has at least MATH different basic classes. We claim that the image under MATH of the subgroup of MATH generated by the MATH is infinitely generated. Suppose that the diffeomorphisms have been indexed so that MATH has at least one basic class which does not occur in the list of basic classes for the MATH for MATH. Note that if MATH are distinct elements in MATH, then the exponentials MATH are linearly independent elements in the power series ring MATH. Thus in any any linear relation MATH the coefficient MATH must be MATH. The claim follows immediately by induction, and so we have that MATH is infinitely generated. Let MATH denote MATH for MATH odd, and MATH for MATH even. The manifold MATH will be simply MATH, where MATH as before. Let MATH be the simply-connected elliptic surface with MATH and no multiple fiber, and let MATH denote the result of a single logarithmic transform on a fiber in MATH. The standard convention is that MATH is the same as MATH. We will make use of the following facts about these manifolds. CASE: For MATH odd, MATH, and for MATH even, MATH. CASE: MATH decomposes completely into a connected sum of MATH's and MATH's. See CITE or CITE for more details. CASE: The diffeomorphism group of MATH acts transitively on elements in MATH of given square, divisibility, and type (that is, characteristic or not) CITE. CASE: The NAME series for MATH is given CITE by MATH where MATH is the multiple fiber (and therefore the regular fiber MATH in homology). CASE: The NAME series for MATH is MATH where MATH is dual to the exceptional class. The argument differs in minor details between the cases when MATH is even or odd; for simplicity we will concentrate on MATH odd. The main point of this is that MATH is not spin when MATH is odd. Let MATH denote the standard (complex) MATH - sphere in MATH, viewed as a submanifold in MATH, and let MATH denote the analogous sphere in MATH. Using the first two items, choose a diffeomorphism of MATH with MATH. Since MATH is not characteristic, any initial choice of diffeomorphism may be varied by a self-diffeomorphism of MATH to ensure that the image of MATH is homologous to MATH. Denote this image, viewed as a sphere in MATH or in MATH, by MATH. Note that the homology of MATH may be identified with the orthogonal complement to MATH, with respect to the intersection pairing, and hence the image of MATH is MATH. As in CITE, the MATH - spheres MATH in MATH give rise to reflections MATH, and we set MATH . Because MATH and MATH are homologous, the action of MATH on homology is trivial, and thus CITE MATH is homotopic to the identity. The image of MATH under MATH is the NAME series of MATH, and so is given by the formula in REF above. Expanding the hyperbolic sines, we see that MATH has MATH basic classes, and so there are the same number of basic classes for MATH. Thus the MATH generate an infinitely generated subgroup of MATH.
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math/9911272
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Suppose that MATH is of NAME type. Then some element in some tensor construction of the variation of NAME structure on MATH is a NAME class on MATH, but not on MATH. The interpretation of the NAME group as stabilizer of lines generated by NAME classes shows that the generic NAME group on MATH is strictly smaller than MATH. Now suppose that the generic NAME group on MATH is strictly smaller than MATH. Then MATH does carry an extra NAME class. The locus where this class is a NAME class is necessarily of dimension one, hence, MATH, being an irreducible component of it, is of NAME type. The second statement follows now from NAME 's theorem above.
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math/9911272
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Let MATH be some integer, and let MATH be an irreducible component of the inverse image of MATH in MATH. Irreducibility of MATH implies that of MATH. Let MATH denote the polarized variation of MATH-Hodge structure on MATH that we considered before, let MATH be in MATH. We choose an isomorphism of MATH-modules from MATH to MATH. Let MATH be the monodromy representation. REF implies that the NAME closure in MATH of MATH is the subgroup MATH. For MATH prime, the correspondence MATH on MATH is given by a diagram: MATH . For MATH to be irreducible, it suffices that MATH be irreducible, with MATH the covering of MATH obtained from MATH. But this covering corresponds to the MATH-set MATH of MATH-submodules of MATH that are free of rank one. Nori's Theorem CITE (REF below) implies that for MATH large enough, the reduction map from MATH to MATH is surjective. Since MATH acts transitively on MATH, irreducibility follows.
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math/9911272
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Let MATH be the endomorphism algebra MATH of MATH. Then MATH is a semi-simple MATH-algebra containing a commutative semi-simple subalgebra of dimension MATH. Suppose that MATH is simple. Then MATH is a division algebra. Since MATH acts faithfully on MATH, it has dimension dividing MATH, hence MATH is a quadratic extension of MATH. Since MATH has a complex structure commuting with the MATH-action, MATH is a totally imaginary. Suppose now that MATH is not simple. Then MATH is isogeneous to the product of two elliptic curves, MATH and MATH, say. These elliptic curves are in fact isogeneous to each other, because otherwise MATH does not admit a morphism to the endomorphism algebra of MATH. So MATH is isogeneous to MATH, with MATH some elliptic curve. Since MATH is of NAME, MATH is an imaginary quadratic field MATH, and MATH. In this case MATH is an order in the totally imaginary extension MATH of MATH.
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math/9911272
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Let MATH be the morphism induced by the closed immersion of the NAME data MATH. Since MATH is finite, and since the NAME correspondences on MATH permute the irreducible components transitively, the statement we want to prove is equivalent to its analog for MATH. So we will show in fact that there are positive MATH and MATH such that for MATH special in MATH corresponding to MATH, we have: MATH . For MATH special in MATH, let MATH be MATH, and let MATH be the NAME closure in MATH of MATH. Since MATH is of degree at most MATH over MATH, the statement we want to prove is equivalent to the existence of positive MATH and MATH such that for all special MATH in MATH: MATH . So let now MATH be special in MATH, corresponding to some MATH. To study the MATH-orbit of MATH, we construct a zero-dimensional subvariety of NAME type over MATH, containing MATH, and we use the theory of NAME on complex multiplication, rephrased in the language of NAME varieties (see CITE). Let MATH be MATH; it is a MATH-module that is locally free of rank two as MATH-module. Let MATH be an isomorphism of MATH-vector spaces. The NAME structure on MATH given by MATH gives an element MATH of MATH. The lattice MATH in MATH corresponds to an element MATH of MATH. By construction, MATH is the image of MATH. Let MATH. Then MATH gives a closed immersion MATH. Since MATH is a one-dimensional MATH-vector space, MATH is its own centralizer in MATH. It follows that MATH factors through MATH. Hence we have a closed immersion of NAME data: MATH. The reflex field of MATH is contained in MATH, hence we have a canonical model MATH over MATH. We put MATH. Then one easily verifies that we have an injective morphism of NAME varieties MATH, which, on MATH-valued points, is given by MATH. By construction, MATH is the stabilizer in MATH of the lattice MATH; it follows that MATH, hence: MATH . Our next objective is to describe in sufficient detail the action of MATH on MATH induced by the above bijections. Class field theory gives a continuous surjection from MATH to MATH, characterized by the following property. In a representation of MATH that is unramified at a finite place MATH of MATH, the arithmetic NAME element is the image of the class of an idèle that is trivial at all places other than MATH, and the inverse of a uniformizer at MATH. Let MATH be the cocharacter obtained by composing MATH, MATH with the inverse of the isomorphism MATH, MATH. Then MATH is defined over MATH, and one defines: MATH to be the morphism MATH composed with the norm map from MATH to MATH. With these definitions, the quotient MATH of MATH acts on MATH via the morphism MATH, where we view MATH as MATH. It follows that: MATH . We will need a more explicit description of MATH, in terms of the CM type associated to MATH. The morphism MATH extends to a morphism of MATH-algebras MATH. Extending scalars from MATH to MATH gives a morphism of MATH-algebras MATH. Via the isomorphisms: MATH and MATH the idempotent MATH of MATH gives an idempotent in MATH, that is, a partition of MATH into two sets MATH and MATH, where MATH is the complex conjugation on MATH. The set MATH is the CM type corresponding to MATH. Since MATH is the NAME closure of MATH in MATH, MATH. With these notations, we have, for any MATH-algebra MATH: MATH where MATH is the norm map of the extension MATH. Finally, let MATH be in MATH, and define MATH. Then we have: MATH for all MATH-algebras MATH and all MATH in MATH. This is the description of MATH that we work with. Since MATH is generated over MATH by the extension MATH and its conjugate, MATH has degree REF or REF over MATH, and its NAME group MATH is isomorphic to MATH, MATH, or MATH, the dihedral group of order REF. We define MATH to be MATH; note that MATH is a subtorus of MATH, equal to the center of MATH. We will see below that MATH induces an endomorphism of MATH whose image, after passing to MATH-valued points, in MATH is big enough for our purposes. Suppose first that MATH is isomorphic to MATH, say with generator MATH. Then MATH, MATH is the complex conjugation and MATH. After changing MATH, if necessary, one has that MATH. The formula above for MATH shows that MATH is simply given by the element MATH of MATH. Since MATH acts as MATH on MATH, we have MATH on MATH. It follows that: MATH . REF below finishes the proof in this case. Suppose now that MATH is isomorphic to MATH. After changing MATH, if necessary, one has MATH, with MATH of order two and MATH. Let MATH be the order MATH of MATH. Since MATH is given by MATH, the induced map MATH factors through MATH induced by the inclusion MATH. The fact that MATH acts as MATH on MATH and as MATH on MATH implies that the kernel of the map MATH is killed by multiplication by MATH. Since MATH acts as multiplication by MATH on MATH, we get: MATH . Since the order MATH of MATH is contained in MATH, and has discriminant MATH, we have: MATH . The proof in this case is finished by REF . Suppose that MATH is isomorphic to MATH. Let MATH and MATH be generators of MATH, with MATH, and with MATH of order MATH. Then MATH is the complex conjugation, and MATH. After changing MATH, if necessary, we have MATH. It follows that MATH is given by the element MATH of MATH. Using that MATH, a simple computation gives: MATH . It follows that MATH acts as MATH on MATH. We conclude that: MATH . REF finishes the proof in this last case.
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math/9911272
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We will use the following lower bound for class numbers: let MATH be a totally real number field; there exists MATH such that for all totally complex quadratic extensions MATH of MATH, one has: MATH . In order to prove this, one distinguishes two cases: MATH and MATH, and one notes that the regulator MATH is at most MATH. In the case MATH one uses the following consequence of NAME 's REF: for MATH a totally real number field, there exists MATH such that for all totally complex quadratic extensions MATH of MATH, one has: MATH . In the case MATH one applies the NAME theorem (see for example CITE): for MATH and MATH, there exists MATH such that for all NAME extensions MATH of MATH of degree at most MATH one has: MATH . Combining these two results, and using that MATH if MATH gives the inequality we want. We could replace the exponent MATH by MATH if we would just use that MATH if MATH. Let now MATH, MATH and MATH be as in the theorem. Then MATH is the inverse image of a subring MATH of some finite quotient MATH of MATH. We have an exact sequence: MATH . The torsion of MATH is bounded in terms of the degree of MATH, and by NAME 's theorem on units the quotient MATH is finite. The long exact cohomology sequence obtained by taking MATH-invariants of the short exact sequence: MATH gives an injection from MATH into MATH, showing that MATH is of order at most two. We conclude that there exists MATH, depending only on the degree of MATH, such that: MATH . On the other hand, we have: MATH . We claim that for every MATH there exists MATH, depending only on the degree of MATH, such that: MATH . To prove this claim, one notes that: MATH . A simple computation then shows: MATH where MATH is assumed to be at least REF. We conclude that there exists MATH, depending only on MATH, such that: MATH . In order to finish the proof of the theorem, it suffices to prove that for every MATH there exists MATH, depending only on MATH, such that MATH. To do this, we proceed in the same way as we did in CITE. As MATH is a finite commutative group, the two MATH-vector spaces MATH and MATH have the same dimension. The cover of MATH by the disjoint union of MATH and MATH gives an exact sequence: MATH . It follows that MATH is bounded by MATH plus a number depending only on the degree of MATH. We put MATH and MATH. The long exact sequence associated to the multiplication by two on the sheaf MATH on the etale site MATH of MATH shows that MATH is at most MATH. Let MATH be the morphism induced by the inclusion of MATH in MATH. Then MATH is the same as MATH, and we have a short exact sequence: MATH where MATH is the maximal open immersion over which MATH is etale. Let MATH be closed immersion giving the complement of MATH, with MATH reduced. Then the long exact sequences of cohomology groups associated to the exact sequence above and to the exact sequence: MATH show that there exists an integer MATH, depending only on MATH, such that: MATH . As MATH, for MATH, we have proved the first statement of the theorem. If one assumes GRH, then the NAME theorem as stated above is true without the condition that the extension MATH be NAME, see CITE.
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math/9911272
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Let us briefly recall how the compactification MATH is defined. Let MATH be a connected component of MATH. Then each connected component of MATH is of the form MATH, with MATH an arithmetic subgroup of MATH (MATH being the quotient of MATH by its center). The compactification MATH is then defined as the disjoint union of the MATH, where MATH is the union of MATH with its so-called rational boundary components, endowed with the NAME topology. It follows that we can write MATH as MATH, with MATH the disjoint union of the MATH. Let MATH be the MATH-conjugacy class of morphisms from MATH to MATH containing the image of MATH. Each connected component of MATH maps isomorphically to one of MATH (see CITE). We first prove the Theorem above for the NAME datum MATH. The group MATH is a product of simple algebraic groups MATH over MATH, and MATH decomposes as a product of MATH's. For compact open subgroups MATH, MATH and MATH that are products of compact open subgroups of the MATH, the corresponding NAME varieties decompose as a product, so that it suffices to treat the MATH separately. If MATH gives compact NAME varieties, NAME 's theorem CITE implies what we want, for compact open subgroups MATH that are sufficiently small; for arbitrary MATH one takes quotients by finite groups. Suppose now that MATH does give NAME varieties that are not compact. If MATH is of dimension REF, then it is isomorphic to MATH, and we are in the case of modular curves, where the Theorem we are proving is well known (the canonical line bundle with log poles at the cusps on the modular curve MATH, MATH, has degree MATH). Suppose now that MATH has dimension MATH. Then the boundary components are of codimension MATH, and the results we want are given in CITE. The case of arbitrary open compact subgroups of MATH follows by considering quotients by finite groups. The theorem for MATH itself follows from the fact that the connected components of the MATH are of the MATH, with MATH an arithmetic subgroup of MATH.
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math/9911272
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We start with two reductions. First of all, writing MATH and MATH as the unions of their irreducible components, one sees that we may suppose that MATH and MATH are irreducible. Secondly, for MATH in MATH, let MATH and MATH be the morphisms from MATH to MATH and MATH, respectively. Then one has: MATH which shows that MATH is finite if and only if MATH is, and that MATH is at most MATH. This also shows that we may replace MATH and MATH by smaller compact open subgroups. Hence we may suppose, by the previous theorem, that we have very ample line bundles MATH and MATH on the NAME compactifications MATH and MATH such that, for each MATH, MATH and MATH are isomorphic to the same line bundle MATH on MATH. We let MATH and MATH be the closures of MATH and MATH in MATH and MATH, respectively. Let MATH denote the degree of MATH with respect to MATH. Let MATH be in MATH, such that MATH is finite. If the intersection is empty, there is nothing to prove, so we suppose that the intersection is not empty. Then the codimension of MATH is at least the dimension MATH of MATH, and we can choose MATH in MATH such that MATH is contained in MATH, and MATH is finite (because of our assumption on the dimensions of MATH and MATH, MATH is finite). It then follows that MATH is at most MATH times the degree of MATH with respect to MATH. But this degree is MATH times the degree of MATH with respect to MATH, hence we have: MATH .
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math/9911272
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The localization MATH of MATH is the same as that of MATH. Hence, if we let MATH denote MATH, then MATH is free of rank one over MATH. It follows that we can choose the isomorphism MATH to preserve the integral structures on both sides at MATH, that is, such that it induces an isomorphism from MATH to MATH. We note that MATH is split in MATH (that is, MATH is a product of copies of MATH), because MATH is the NAME closure of MATH. Consider now an element MATH of MATH that is equal to MATH at one place above MATH and equal to MATH at all other finite places of MATH. Then MATH, viewed as an element of MATH, is equal to MATH at exactly two places of MATH above MATH that are not in the same MATH-orbit, and equal to MATH at all other finite places of MATH. It follows that MATH is conjugated in MATH, by some element in MATH, to the element MATH that induces MATH (use that MATH acts transitively on the set of free rank one MATH-submodules of MATH). We conclude that MATH is in MATH. But since MATH is in MATH, MATH is also in MATH. It follows that MATH is contained in MATH.
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math/9911272
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By REF below, MATH and MATH generate MATH (here we use that MATH is not ramified in MATH). Let now MATH be in MATH, and let MATH be a preimage of it in MATH under the quotient map for the action by MATH. The fact that MATH is then given by right multiplication on MATH by the element MATH at the place MATH shows that the MATH-orbit of MATH is the image in MATH of the MATH-orbit of MATH. Let now MATH be the subgroup of MATH consisting of MATH such that MATH is in MATH. Then the MATH-orbit of MATH is the image in MATH of the subset MATH of MATH. Now one notes that MATH contains a congruence subgroup of MATH. It follows that the intersection of MATH with MATH is dense in MATH (for the archimedean topology) because MATH is generated by additive subgroups. Since MATH acts transitively on MATH, the lemma is proved.
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math/9911272
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In order to minimize notation, let MATH denote MATH, let MATH denote MATH, and let MATH denote the subgroup of MATH generated by MATH and MATH. Let MATH be the set of MATH-lattices in MATH on which the MATH-bilinear form MATH given by MATH is a perfect pairing of MATH-modules, up to a factor in MATH. The map MATH, MATH induces a bijection from MATH to MATH. Hence, in order to prove our claim, it suffices to show that MATH acts transitively on MATH. So let MATH be in MATH. We note that MATH is either a product of two copies of MATH, or the ring of integers MATH in the unramified quadratic extension MATH of MATH; in both cases, MATH is a product of discrete valuation rings with uniformizer MATH. The theory of finitely generated modules over a discrete valuation ring says that there exists MATH in MATH and MATH and MATH in MATH such that MATH is contained in MATH and has a MATH-basis of the form MATH, with MATH the standard basis of MATH. We note that conjugating MATH by suitable elements of MATH shows that MATH and MATH are in MATH, and that, in the split case, MATH is in MATH. Since the element MATH of MATH is the factor by which MATH differs from a perfect pairing on MATH, it is actually in MATH. It follows that MATH is in MATH. This finishes the proof that MATH is MATH.
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math/9911272
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It only remains to prove that MATH is symmetric (antisymmetric) if and only if MATH is so. For MATH as above, let MATH denote its adjoint, that is, MATH; we will use the same notation for elements of MATH. Then one has MATH. Now MATH is symmetric if and only if MATH, and MATH is antisymmetric if and only if MATH. Hence the result.
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math/9911272
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First of all, we may and do suppose that MATH. The idea is now the following: let MATH act on itself by right translation; then MATH acts on MATH, and MATH is the stabilizer of the ideal MATH; then use that MATH is finitely generated, and that MATH is locally finite. Let us first write down what MATH is, as a MATH-module via right translation on MATH. Well, MATH where the last equality comes from the fact that MATH, as MATH-module given by right translation on MATH, is simply MATH, where MATH is of course the dimension of MATH (note that the MATH-action on MATH extends to a MATH-action). Also, note that MATH is in MATH, and that we have MATH. The scalar subgroup MATH of MATH induces a MATH-grading on MATH. We have: MATH and: MATH . This describes MATH as MATH-module. Let MATH be a finite set of generators of the ideal MATH of MATH. Let MATH be a finite dimensional sub-MATH-module containing the MATH. Then MATH is the stabilizer of the subspace MATH of MATH, hence of the line MATH, with MATH. Now note that MATH is a subrepresentation of a representation of the form MATH.
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math/9911272
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For MATH, see for example NAME 's ``Algebraic number theory", REFnd edition, VIII, REF. In fact, NAME gives the proof when MATH is a field, but for MATH in a product of number fields everything decomposes into products. Let us digress a little bit on the regulator. I find that the regulator MATH should be defined as follows: one considers MATH and puts: MATH with the volume measured with respect to the volume form coming from the standard inner product on MATH, and where MATH is taking log of absolute value at every factor of MATH, with MATH being the factor by which the NAME measures change (MATH for a real place, MATH for a complex one). But this does not give the usual definition, as given in NAME. There one omits any one of the infinite places in order to get a square matrix of which one takes absolute value of the determinant. One easily proves that MATH, which actually makes my definition a bit ugly. Anyway, let's proceed. Since we know the theorem for MATH, all we have to do is to compare our MATH to MATH. Let MATH, MATH, and MATH the morphism induced by the inclusion of MATH in MATH. Then we have a short exact sequence of sheaves on MATH: MATH with MATH a skyscraper sheaf given by MATH in case MATH is given by the subring MATH of the finite quotient MATH of MATH. This gives a long exact sequence: MATH . Let MATH be the cokernel of MATH, and MATH the kernel of MATH. Then one gets: MATH . Putting this all together shows that the right hand side of the equality we want to prove changes by the factor MATH when going from MATH to MATH. So all that we have to do now is to show that the left hand side changes by the same factor. But then note: MATH which is clearly the contribution to MATH of those residue fields.
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math/9911272
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We will first prove this for maximal orders in number fields of bounded degree, then for maximal orders in finite separable MATH-algebras of bounded degree, and then for arbitrary orders of bounded degree. In the case of a maximal order of a number field of bounded degree, we just apply two theorems. The first one is the NAME theorem (see for example CITE), that states that: for MATH and MATH, there exists MATH such that: MATH for all NAME extensions MATH of MATH of degree at most MATH. The second theorem is one of NAME CITE: let MATH. There exists MATH such that for all number fields MATH of degree at most MATH over MATH, one has: MATH . Together, these two results show: let MATH. There exists MATH such that for every number field MATH of degree at most MATH over MATH one has: MATH . This settles the case where the MATH-algebra MATH is a field. The case for a maximal order in a finite separable MATH-algebra of degree at most MATH then follows, because everything decomposes into a product of at most MATH factors, for which one has the result already. So let now MATH be a finite separable MATH-algebra of degree at most MATH, and let MATH be an order in it, given by the subring MATH of some finite quotient MATH of MATH. We have already seen that: MATH . We note that the quotient MATH and its inverse are bounded in terms of MATH only. Hence the theorem follows from the following claim: for MATH and MATH there exists MATH such that for MATH an order in a finite separable MATH-algebra MATH of degree at most MATH, one has: MATH where MATH is the inverse image of the subring MATH of the finite quotient MATH of MATH. We now prove this claim. Let MATH denote MATH. We may and do assume that MATH. Localizing at the maximal ideals of MATH, followed by a simple computation, shows that: MATH . Since MATH, this shows our claim, and hence finishes the proof of the theorem.
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cs/9912013
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Choose MATH to be a multiple of three, sufficiently large that the MATH bound of REF is strictly smaller than MATH, and let MATH. By REF , partition MATH into MATH subsets of at most MATH points, such that any hyperplane cuts few subsets. Let MATH be the family consisting of the largest MATH subsets in the partition. If the smallest member of MATH contains MATH points, then the total size of all the members of the partition would have to be at most MATH, but this total size is just MATH, so MATH and each member of MATH contains at least MATH points. If each MATH-tuple of sets in MATH were transversal, we could apply REF and partition MATH into MATH transversal subfamilies, one of which would have to contain at least MATH subsets. But this violates the MATH bound of REF , so MATH must contain a non-transversal MATH-tuple. This tuple fulfills the conditions of the statement of the lemma.
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cs/9912013
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We form a distribution on the plane by centrally projecting the uniform distribution on a sphere. We show that any nontransversal triple for this distribution must have a set with measure at most MATH times the total measure. The same bound then holds in the limit for discrete point sets formed by taking MATH-approximations of this distribution. Let MATH, MATH denote the three nontransversal subsets of the plane maximizing the minimum measure of any MATH. Without loss of generality, each MATH is convex. Consider the three lines tangent to two of the MATH, and separating them from the third set (such a line must exist since the sets are nontransversal). These lines form an arrangement with seven (possibly degenerate) faces: a triangle adjacent on its edges to three three-sided infinite cells, and on its vertices to three two-sided infinite cells. The sets MATH coincide with the three-sided infinite cells: any set properly contained in such a cell could be extended to the whole cell without violating the nontransversality condition, and if they instead coincided with the two-sided cells we could shrink the arrangement's central triangle while increasing the sizes of all three MATH. The two arrangements in REF can both be viewed as such three-line arrangements, degenerate in different ways. Each line in the plane lifts by central projection to a great circle on the sphere. Consider the great circles formed by lifting four lines: the three lines considered above and the line at infinity. Any arrangement of four circles on the sphere cuts the sphere in the pattern of a (possibly degenerate) cuboctahedron REF . The three-sided infinite cells in the plane lift to quadrilateral faces of this cuboctahedron. Note that the area of a spherical quadrilateral is the sum of its internal angles, minus MATH. Form the dual of the arrangement by treating each great circle as an ``equator" and placing a pair of points at the corresponding two poles. The geodesics between these points have the pattern of a cuboid REF such that the length of each geodesic is an angle complementary to one of the cuboctahedron quadrilaterals' internal angles. Thus, the cuboid minimizing the maximum quadrilateral perimeter corresponds to the cuboctahedron maximizing the minimum quadrilateral area. But any spherical cuboid has at least one face covering at least one-sixth of the sphere, and the minimum perimeter for such a quadrilateral is achieved when the quadrilateral is a square. Therefore, the regular cube minimizes the maximum perimeter and the regular cuboctahedron maximizes the minimum area. The ratio of a regular cuboctahedron's square face area to the area of a full hemisphere is the value given, MATH.
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cs/9912013
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Project the point set onto the subspace spanned by the MATH independent directions, in such a way that the inverse image of each point in the projection is a MATH-flat containing MATH. By REF , we can find a family of MATH subsets of the data points, each with MATH points, such that the MATH-dimensional projection of this family has no transversal. We then let MATH be the MATH-flat determined by applying REF to this family of subsets. Then consider any double wedge bounded by a hyperplane containing MATH and a hyperplane containing MATH. The vertical boundary of this double wedge projects to a hyperplane in MATH, so it must miss one of the MATH subsets in the family. Within this missed subset the double wedge appears to be simply a halfspace through MATH. By REF , the double wedge must therefore contain at least MATH points. Thus if let MATH the theorem is satisfied.
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cs/9912013
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The proof of REF can be viewed as projecting the points onto a horizontal line, dividing the line into two rays at the median of the points, and applying the center transversal theorem to the two sets of MATH points contained in each ray. Instead, we project the points onto the horizontal line as before, but partition this line into three pieces: two rays containing MATH points each, and a line segment in the middle containing the remaining MATH points. We then apply the center transversal theorem to two sets MATH and MATH of MATH points each, formed by the points having a projection in the union of the middle segment and one of the two rays. This theorem finds a line such that no halfspace containing it has fewer than MATH points in either of the sets MATH. We claim that this line has regression depth at least MATH. To prove this, consider any double wedge in which one hyperplane boundary contains the regression line, and the other hyperplane boundary is vertical. The vertical hyperplane then intersects the horizontal projection line in a single point. If this intersection point is in one of the two rays, then the vertical hyperplane misses the set MATH formed by the other ray and the middle segment. In this case, the double wedge contains the same subset of MATH as a halfspace bounded by the double wedge's other bounding plane, and so contains at least MATH points of MATH . In the remaining case, the vertical boundary of the double wedge intersects the horizontal projection line in its middle segment. Within each of the two sets to which we applied the center transversal theorem, the double wedge differs from a halfspace (bounded by the same nonvertical plane) only within the middle set. But any hyperplane bounds two different halfspaces, and the halfspace approximating the double wedge in MATH is opposite the halfspace approximating the double wedge in MATH. Therefore, if we let MATH denote the set of points in the halfspace but not in the double wedge, then MATH and MATH are disjoint subsets of the middle MATH points. The number of points in the double wedge within one set MATH must be at least MATH, so the total number of points in the double wedge is at least MATH. Thus in all cases a double wedge bounded by a hyperplane through the regression line and by a vertical hyperplane contains at least MATH points, showing that the line has depth at least MATH.
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cs/9912013
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We have already proven that MATH, so we need only show that MATH. We work in the dual space, and construct an arrangement of MATH planes in MATH, for MATH any multiple of REF, such that any line has depth at most MATH. Our arrangement consists of five groups of nearly parallel and closely spaced planes, which we label MATH, MATH, MATH, MATH, and MATH. Rather than describe the whole arrangement, we describe the line arrangements in the planar cross-sections at MATH and MATH. Recall that the depth of a line in the three-dimensional arrangement is the minimum number of planes crossed by any vertical ray starting on the line. Limiting attention to rays contained in the two cross-sections (and hence to the planar depth of the two points where the given line intersects these cross-sections) gives an upper bound on the depth of the line, and so a lower bound on MATH. In the first cross-section, we place the groups of lines as shown in REF , with the region where MATH and MATH cross contained inside the triangle formed by the other three groups. Moreover, MATH and MATH do not cross side MATH of the triangle, instead crossing group MATH at points outside the triangle. The points where members of MATH intersect each other are positioned on segment MATH - MATH. Similarly, the crossings within MATH are situated on segment MATH - MATH. The crossings within MATH, MATH, and MATH are situated along the corresponding sides of the triangle formed by these three groups. In the cross-section formed as described above, points from most cells in the arrangement can reach infinity while crossing only one group, and so have depth at most MATH. It is only within the segments MATH - MATH and MATH - MATH that a point can have higher depth. The arrangement is qualitatively similar for nearby cross-sections MATH. Therefore, any deep line in MATH must be either nearly parallel to MATH and not near any MATH, or nearly parallel to MATH and not near MATH. In the second cross-section REF , the groups MATH and MATH reverse roles: the point where MATH and MATH cross is contained in the triangle determined by the other three groups, and the other details of the arrangement are situated in a corresponding manner. Therefore, any deep line would have to be either nearly parallel to MATH and not near any MATH, or nearly parallel to MATH and not near MATH. There is no difficulty forming the two cross-sections described above from a single plane arrangement, since (as shown in REF ) the slopes of the lines within each group can remain the same in each cross-section. But the requirements imposed on a deep line by these two cross-sections are contradictory, therefore no line can have depth greater than MATH in this arrangement.
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cs/9912013
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As in the proof of REF , we project the point set onto the subspace spanned by the MATH independent directions, in such a way that the inverse image of each point in the projection is a MATH-flat containing MATH. By REF , we can find a family of MATH subsets MATH, each with MATH points, such that the MATH-dimensional projection of this family has no transversal. We then find a NAME point MATH and a NAME partition of each set MATH into subsets MATH, for MATH. We let MATH be the MATH-flat spanning these MATH . NAME points. We form each set MATH in our NAME partition as the union MATH. Some points of MATH may not belong to any set MATH, in which case they can be assigned arbitrarily to some set MATH in the partition. Then consider any double wedge bounded by a hyperplane containing MATH and a hyperplane containing MATH. The vertical boundary of this double wedge projects to a hyperplane in MATH, so it must miss one of the MATH subsets MATH. Within MATH the double wedge appears to be simply a halfspace through MATH. It therefore contains at least one point of each set MATH and a fortiori at least one point of each set MATH. Thus if let MATH the theorem is satisfied.
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cs/9912013
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Let MATH be the arrangement of hyperplanes dual to the MATH given points. The distance from points in MATH to MATH is constant within each cell of MATH, and all such distances can be found in time MATH by applying a breadth first search procedure to the arrangement. The depth of a MATH-flat MATH is just the minimum depth of any cell of MATH pierced by MATH. Any two flats that pierce the same set of cells of MATH have the same depth. The space of MATH-flats forms a MATH-dimensional algebraic set MATH, in which the flats touching any MATH-dimensional cell of MATH form a subset of codimension one. The arrangement of these MATH subsets partitions MATH into MATH cells, corresponding to collections of flats that all pierce the same set of cells. We can construct this arrangement, and walk from cell to cell maintaining a priority queue of the depths of the cells in MATH pierced by the flats in the current cell, in time MATH.
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cs/9912013
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We first construct a MATH-approximation MATH of the points, for the range space consisting of double wedges with one vertical boundary, where MATH. Then if a flat MATH has depth MATH with respect to MATH, MATH is within an additive MATH term of the true depth of MATH with respect to the original point set. MATH can be found with MATH, in time MATH, using standard geometric sampling algorithms. We then let MATH be the deepest flat for MATH. Suppose the optimal flat MATH for the original point set has depth MATH. Then the depth of MATH in MATH, and therefore also the depth of MATH in MATH, must be at least MATH. Therefore, the depth of MATH in the original point set must be at least MATH. Since MATH, MATH.
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cs/9912014
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Let MATH be a closest pair, where MATH belongs to a subset MATH created more recently than the subset containing MATH. Then when MATH was created, it contained MATH, so it contained at least one of MATH. Then if MATH was the first of two vertices added to the path, it must have chosen as its neighbor either MATH or a vertex MATH at least as close to MATH. If it chose MATH, edge MATH exists in MATH. If it chose some MATH, then MATH can not have been deleted, since that would have caused MATH to move to a newer MATH, so MATH is at least as good as MATH and still exists in MATH. Similarly if MATH were chosen first then it would have formed edge MATH in MATH or MATH for some vertex MATH at least as close to MATH. Again, MATH could not have been deleted because that would cause MATH to move to a subset MATH created more recently than MATH. So in all cases MATH contains a closest pair.
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cs/9912014
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By construction, the graphs together have at most MATH edges (we rebuild the data structure if this bound is reached), so they take linear space to store. The partition is also easily maintained in linear space.
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cs/9912014
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The correctness and space complexity have already been proven above; it remains to prove the time bounds. First, let us analyze the time for a sequence of updates that do not involve rebuilds to the data structure. We use a potential function argument. Define the potential of set MATH to be MATH, and the potential of the whole data structure to be the sum of the potentials of each subset. This potential is at most MATH (the value it would take for a partition consisting of a single set). The amortized time per operation is MATH, where MATH is the actual time used, MATH is the increase in the upper bound MATH on the potential, and MATH is the increase in the potential. Over the course of a sequence of operations, starting from a situation in which the potential equals MATH, the MATH and MATH terms in this formula telescope, so the total amortized time for the sequence is at most the total actual time; therefore this method provides a valid bound on amortized time. Each time we merge two subsets MATH and MATH, the potential increases by MATH . Since MATH and MATH must be within a factor of two of each other, the two logarithmic terms are constant and this simplifies to MATH. Since the path constructed from the merged subsets has size MATH, and each edge in the path can be found in linear time, the total time for the merge is MATH. Therefore any time spent performing merges can be balanced against an increase in the potential function. Each time we perform an insertion, we create a new set MATH with zero potential, and perform MATH work not counting mergers. However, the bound MATH on the total potential increases by MATH, and the amortized time for each insertion must also include this potential increase, so the total amortized time per insertion is MATH. Each time we perform a deletion, we perform MATH work creating a new subset of at most MATH objects. This work is balanced by a decrease of MATH in the upper bound on the total potential. Further, the new set MATH has some positive potential (up to MATH). However, When we move these MATH objects to a new set, the potential of each set MATH decreases by MATH per object, and this potential decrease dominates the amortized time bound for each deletion, which is therefore MATH. To complete this analysis, we estimate the time spent rebuilding the data structure. Define the excess of graph MATH to be MATH. Initially, all points are in MATH with a total excess of MATH. Each time we merge two subsets, the merged graph's excess becomes nonpositive. The only way to create a positive excess is to move a point out of some MATH, by deleting some other point sharing an edge with the moved point. Each deletion moves MATH points and thus increases the total excess by MATH. Therefore, MATH deletions need to be performed before each rebuild and the amortized time per rebuild step is MATH.
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cs/9912014
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We first consider insertion of a new object. This causes MATH to change only when one of MATH or MATH contains the inserted object. Since for any MATH there is exactly one MATH containing the inserted object, the changed values are in one-to-one correspondence with the sets MATH not containing the inserted object. The result follows from REF . The argument for deletions is similar, except that the deletion of one object and renumbering of another causes roughly twice as many changes to MATH.
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cs/9912014
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As shown above, each update causes MATH changes to the values MATH stored by the data structure. Each changed value can be recomputed in constant time, using the formula MATH if we perform the recomputation for smaller values of MATH before larger ones. The closest pair we seek is MATH.
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cs/9912014
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Each step in these clustering algorithms can be performed by finding the closest pair of clusters, deleting these two clusters from the set of objects represented by our closest pair data structure, and inserting a new object representing the new merged cluster.
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cs/9912014
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This can be seen as a type of hierarchical clustering, in which clusters correspond to fragments, and the distance between two clusters is the length of the shortest edge connecting their endpoints. The sequence of edges added by the hierarchical clustering algorithm of REF is then exactly the same as the sequence added in the multiple fragment method. Alternatively, instead of maintaining the closest pair among a set of clusters, maintain the set of fragment endpoints, with distance MATH between endpoints of the same fragment. Each step of the algorithm then consists of selecting the closest pair, deleting one or both of these endpoints (if they belong to nontrivial fragments) and modifying the distance between the endpoints of the combined fragment.
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cs/9912014
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We use our data structures to maintain a set of MATH objects: the MATH edges in the tour after the MATH-th insertion, and the MATH remaining uninserted sites. The distance between an edge and a site is defined to be the increase in length that would be caused by the corresponding insertion; all other distances are MATH. In this way each successive insertion can be found as the closest pair in this set.
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cs/9912014
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We use the data structures defined above to repeatedly find and delete the closest pair.
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cs/9912014
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Correctness follows from the correctness of the conga line data structure. To prove the time bound, we use a potential function MATH . Each insertion changes this potential by MATH and takes time MATH. Each deletion in which MATH points are moved to a new subset takes time MATH, but increases MATH by MATH. For any MATH, the amortized time (actual time minus difference in MATH) is MATH.
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