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math/9911216
If MATH, then consider the direct MATH-system MATH, where MATH for each MATH and MATH for each MATH with MATH. If MATH, then consider any subset MATH of MATH such that MATH and MATH. Without loss of generality we may assume that MATH contains the unit of MATH. Each MATH can obviously be identified with a subset (denoted by the same letter MATH) of MATH of cardinality MATH. Let MATH be the smallest MATH-subalgebra of MATH containing MATH. If MATH and MATH, then MATH (as subsets of MATH) and consequently MATH. This inclusion map is denoted by MATH. It is easy to verify that the collection MATH is indeed a direct MATH-system such that MATH.
math/9911216
Clearly MATH is dense in MATH (this fact remains true for arbitrary direct systems of MATH-algebras). Consequently, for any point MATH there exists a sequence MATH, consisting of elements from MATH, such that MATH. For each MATH choose an index MATH such that MATH. By REF , there exists an index MATH such that MATH for each MATH. Since MATH, it follows that MATH . Finally, since MATH is closed in MATH, it follows that MATH .
math/9911216
The implication MATH follows from CITE. In order to prove the implication MATH we first consider a direct MATH-system MATH with properties indicated in REF . Next consider the following relation MATH: MATH . Let us verify conditions of REF . Existence. Let MATH and MATH. First we prove the following assertion. CASE: There exist an index MATH, MATH and an invertible element MATH such that MATH . Proof of MATH. By REF , MATH and consequently there exists an invertible element MATH such that MATH. Since MATH is a direct MATH-system it follows from REF that there exists an index MATH such that MATH. Since MATH is a directed set there exists an index MATH such that MATH. This obviously implies that MATH. This finishes the proof of MATH. For a given element MATH consider indeces MATH, satisfying conditions MATH, MATH. By REF , there exists an element MATH such that MATH for each MATH. Obviously MATH, MATH. Note also that CASE: For any MATH there exists an invertible element MATH such that MATH. Since MATH is a direct MATH-system it follows that MATH. Consequently there exists a subset MATH such that MATH and MATH. Let MATH. For each MATH consider an index MATH satisfying condition MATH. Since MATH is a MATH-complete set, we conclude by REF , there exists an index MATH such that MATH for each MATH. We claim that MATH. Indeed, let MATH and MATH. Since, by the above construction, MATH is dense in MATH, there exists a self-adjoint element MATH such that MATH. By REF there exists an invertible element MATH such that MATH. Since MATH, it follows that MATH. This guarantees that MATH. It only remains to note that MATH. Therefore MATH. NAME. Let MATH, MATH, MATH and MATH. Since MATH, there exists an invertible element MATH such that MATH. Since MATH it follows that MATH which shows that MATH and proves that MATH. MATH-closeness. Suppose that MATH is a chain of indeces in MATH with MATH. Assume also that MATH for each MATH where MATH. Our goal is to show that MATH where MATH. Let MATH and MATH. Since MATH is a direct MATH-system it follows that MATH is the direct limit of the direct system generated by MATH-subalgebras MATH, MATH, and corresponding inclusion homomorphisms. Consequently there exist an index MATH and an element MATH such that MATH. Since MATH there exists an invertible element MATH such that MATH. Clearly MATH. This shows that MATH. We are now in position to apply REF which guarantees that the set MATH is cofinal and MATH-closed in MATH. Note here that MATH precisely when for each MATH and for each element MATH there exists an invertible element MATH such that MATH. This means that the direct MATH-system MATH consists of MATH-subalgebras of MATH of real rank zero. Clearly MATH. Proof is completed.
math/9911216
This is an elementary exercise. Let MATH be a self-adjoint element of the product MATH. This obviously means that MATH for each MATH. Let also MATH. Since MATH, MATH, it follows that there exists a self-adjoint and invertible element MATH such that MATH, MATH. Obviously MATH and MATH as required.
math/9911216
Let MATH denote the set (see CITE) of all unital MATH-homomorphisms defined on MATH such that MATH. Next consider the product MATH. Since MATH for each MATH it follows from REF that MATH. The unital MATH-homomorphisms MATH, MATH, define the unique unital MATH-homomorphism MATH such that MATH for each MATH (here MATH denotes the corresponding canonical projection MATH-homomorphism). By REF , MATH can be represented as the limit of the MATH-system MATH such that MATH is a separable unital MATH-algebra of real rank zero for each MATH. Suppressing injective unital MATH-homomorphisms MATH we for notational simplicity assume that MATH's are unital MATH-subalgebras of MATH. Let MATH be a countable dense subset of MATH. By REF , for each MATH there exists an index MATH such that MATH. By REF , there exists an index MATH such that MATH for each MATH. Then MATH for each MATH. This observation coupled with the continuity of MATH guarantees that MATH . Let MATH and MATH denote the unital MATH-homomorphism MATH considered as the homomorphism of MATH into MATH. Note that MATH, where MATH stands for the inclusion. By construction, MATH. Let us show that MATH is MATH-invertible in the sense of Introduction. In our situation for any unital MATH-homomorphism MATH, where MATH is a separable unital MATH-algebra of real rank zero, we need to establish the existence of a unital MATH-homomorphism MATH such that MATH. Indeed, by definition of the set MATH, we conclude that MATH for some index MATH (in particular, MATH for the same index MATH). Next observe that MATH. This allows us to define the required unital MATH-homomorphism MATH as the composition MATH. This completes the proof of MATH-invertibility of MATH and as a consequence (see Introduction) of the universality of MATH.
math/9911218
The four-dimensional abelian variety MATH satisfies REF 's theorem, and so REF applies.
math/9911218
REF is proved in NAME REFa REF . For REF , recall that theorems of NAME and NAME show that the category of abelian motives over MATH, defined using the numerical equivalence classes of algebraic cycles as correspondences, is NAME CITE. Almost by definition, MATH is the fundamental group of the NAME subcategory MATH of this category generated by MATH and the NAME object. When numerical equivalence coincides with homological equivalence, there is a natural map MATH which the theory of NAME categories shows to be bijective. But MATH. REF is proved in NAME REF (see the proof of REF). Almost by definition of MATH, MATH consists of the classes fixed by the NAME germ MATH, and these are exactly the NAME classes.
math/9911218
Since every character of MATH occurs in a space of the form MATH, we see that the subgroups of MATH are determined by their invariants in these spaces. Thus REF is an immediate consequence of the lemma. That ``NAME can be replaced by ``NAME follows from NAME REF. If MATH, then the lemma shows that the MATH-adic NAME conjecture holds for all powers of MATH and all MATH in the set in REF , but if the MATH-adic NAME conjecture holds for one MATH then it holds for all (NAME REF). Conversely, if the MATH-adic NAME conjecture holds for all powers of MATH and a single MATH, then numerical equivalence coincides with MATH-homological equivalence for that MATH (NAME REF), and the preceding lemma then shows that MATH. As MATH, this implies that MATH. The proof of the remaining statement is similar.
math/9911218
In the proof, we ignore NAME twists, that is, we choose an identification of MATH (or MATH). First consider the characteristic zero case. Choose an étale NAME MATH such that MATH is free of rank MATH as a MATH-module and MATH is stable under the NAME involution defined by some ample divisor MATH. The action of MATH on MATH defines a NAME MATH on MATH. We have MATH. Let MATH be the smallest subfield of MATH containing MATH for every homomorphism MATH. It is a NAME, finite and NAME over MATH. Let MATH, and let MATH be the subspace of MATH on which MATH acts through MATH. Then MATH and MATH is one-dimensional. As MATH, it follows that MATH where MATH and MATH are subsets of MATH and MATH respectively, and MATH is the subspace on which MATH acts as MATH - it is of dimension MATH and of NAME type MATH. For MATH, let MATH denote the projection of MATH on MATH. Because MATH is multiplication by an idempotent MATH of MATH, it sends algebraic classes to algebraic classes. Let MATH be the class in MATH of the divisor MATH. Because MATH is algebraic, its isotypic components in MATH are of type MATH, MATH, and, because MATH defines a nondegenerate form on MATH, each such component is nonzero. For each MATH, choose a nonzero element MATH of MATH. Then MATH is a basis for MATH. We may suppose that the MATH have been chosen so that the MATH component of MATH is MATH. Denote MATH by MATH - it is a basis for MATH. For MATH, MATH where MATH runs over the subsets of MATH with MATH. In particular, MATH is algebraic. Moreover, MATH . Only the subsets MATH disjoint from both MATH and MATH contribute to the sum. We shall need REF (NAME REFAREF): Let MATH be the space of algebraic classes in MATH; then for MATH, the map MATH is an isomorphism. Suppose MATH is algebraic with MATH. Let MATH, so that MATH, MATH, MATH. We shall prove by induction on MATH that MATH is also algebraic. If MATH, there is nothing to prove. If not, MATH, and there exists a subset MATH of MATH with MATH elements disjoint from MATH. Then MATH is nonzero and algebraic. By NAME 's theorem, there exists a MATH such that MATH. If MATH occurs with nonzero coefficient in MATH, then it is algebraic. But if MATH is chosen so that MATH occurs with nonzero coefficient in MATH, then MATH, MATH. Since MATH, the induction hypothesis shows that MATH is algebraic. We now prove the theorem (in the case of characteristic zero). We have to show that, for each MATH, the cup-product pairing MATH is nondegenerate. NAME 's theorem shows that the two spaces have the same dimension, and so it suffices to show that the left kernel is zero. Thus, let MATH be a nonzero element of MATH, MATH, and suppose MATH occurs with nonzero coefficient in MATH. It suffices to show that MATH is algebraic, where MATH and MATH are the complements of MATH and MATH in MATH and MATH respectively. From the last paragraph, we know that MATH with MATH algebraic and MATH, MATH, and MATH disjoint. Because MATH is stable under MATH, MATH is algebraic. But MATH where MATH is the complement of MATH in MATH, which is obviously algebraic. Now consider the case MATH. After possibly replacing MATH with an isogenous variety, we may assume that it lifts to an abelian variety MATH with many endomorphisms in characteristic zero (see A. REF). Let MATH be a NAME for MATH as in the first paragraph of the proof. If MATH is such that MATH is in the decomposition group of some prime MATH of MATH dividing MATH, then the same argument as in characteristic zero case applies once one replaces MATH with MATH and MATH with MATH. The NAME density theorem shows that the set of primes MATH such that MATH is the NAME element at a prime MATH dividing MATH has density MATH. For such a prime MATH, MATH is in the decomposition group of MATH.
math/9911218
Suppose MATH is isogenous to MATH with the MATH simple and nonisogenous in pairs. Assume initially that none of the MATH is a supersingular elliptic curve. Then each MATH is a NAME. For each MATH, let MATH, let MATH, and let MATH be the maximal real subfield of MATH. Fix a MATH. For each MATH, there exists a field MATH cyclic of degree MATH over MATH and such that each real and MATH-adic prime of MATH splits in MATH and the local degree at each MATH-adic prime is MATH (NAME and NAME REF, p. REF ). Let MATH. Then MATH is a NAME that splits MATH and can be realized as a subfield of MATH. Therefore (NAME REF/REF, NAME REF), MATH is isogenous to the reduction of an abelian variety MATH with MATH. After replacing MATH with an isogenous variety, we may suppose that it lifts to the abelian variety MATH. The étale algebra MATH acts on MATH diagonally, and satisfies the conditions in the first paragraph of the proof REF . The field MATH generated by the images of MATH in MATH is MATH. Because of our choice of the MATH, every MATH-adic prime in this field is fixed by MATH. This completes the proof of the proposition in this case. When we add a factor MATH to MATH with MATH a supersingular elliptic curve, MATH is replaced with MATH where MATH can be taken to be any quadratic imaginary field in which MATH does not split. If we choose MATH to be MATH, then MATH, and the preceding argument applies.
math/9911218
Let MATH be the space of MATH-algebraic classes in MATH. The proof of the characteristic zero case of the theorem in A. REF will apply with ``algebraic" replaced by ``MATH-algebraic" once we have shown that NAME 's theorem holds for MATH: for MATH, MATH is an isomorphism. This map is automatically injective, and so we only have to prove surjectivity. Let MATH be a MATH-algebraic class in MATH; by assumption, the image MATH of MATH in MATH equals the class MATH of some MATH. There exists a MATH such that MATH, and NAME 's theorem says that there is a MATH such that the cohomology class of MATH is MATH. The images of MATH and MATH in MATH map to MATH and MATH respectively under the isomorphism MATH. As MATH, this proves that MATH is MATH-algebraic.
math/9911218
Immediate consequence of the theorem and the compatibility of the cup-product pairings.
math/9911223
Using the usual embedding theorems, it may be seen that all of the NAME and NAME spaces embed into MATH for some MATH. Let us recall the definition of the NAME space MATH. Let MATH be some tempered NAME function whose NAME transform is non-negative, whose support is ``mostly" contained in a band about MATH, and such that MATH is uniformly bounded above and below. Here MATH. Then MATH is the space of distributions on MATH for which the norm MATH is finite. Write MATH for the function MATH, and set MATH. Since MATH and MATH are real valued, we see that MATH must be an even function. Hence we quickly see that MATH has MATH norm equal to MATH, and is supported in MATH. We will show by induction that MATH for MATH, where MATH, MATH, and MATH. The case MATH follows since MATH. Suppose that our desired inequality holds for MATH. Then MATH whenever MATH, that is, MATH. Let MATH. Since MATH, MATH . It is clear that this is infinite if MATH.
math/9911223
Suppose for a contradiction that there is a solution MATH. By the methods of CITE, we know that there is a number MATH, depending only upon MATH, such that for every MATH that there is a mild solution MATH to the cheap NAME equation, with MATH, that is obtained by starting with MATH, and iterating a function MATH similar to that defined above. By the uniqueness result of CITE, we have that MATH for MATH. From this it is clear that if MATH, then MATH for MATH. Applying this argument several times, we see that MATH for MATH. Then by REF , MATH is not in any NAME space, and hence in particular is not in MATH.
math/9911223
Suppose we have a non-negative solution MATH to the cheap NAME equation with MATH. It is clear that MATH . Thus MATH which is infinite on the support of MATH for MATH, and hence MATH cannot be in MATH.
math/9911224
For any non-degenerate lattice MATH there is a finite number of triangles MATH with vertices in MATH whose sides are generators of MATH and whose angles do not exceed MATH. There are REF such triangles for a rectangular lattice and REF for any other lattice, as shown in the figure. For each non-degenerate lattice MATH choose one of the triangles MATH and trace REF lines which connect the vertices of the triangle with the centre of its circumscribed circle. These lines are distinct unless MATH is rectangular in which case two of them coincide. Each line determines a point in MATH so we obtain a point MATH in MATH for each MATH. It is clear that MATH depends only on MATH and not on a particular choice of a triangle. Now, for a degenerate lattice MATH define MATH as the point of MATH which corresponds to the line containing MATH. A straightforward check shows that MATH is a homeomorphism.
math/9911225
CASE: MATH: Let MATH be such that MATH. It is enough to define MATH for MATH such that CASE: MATH CASE: MATH is below MATH that is MATH and CASE: MATH: Let MATH it is as required. The definition is by induction on MATH. For MATH let MATH. For MATH, suppose MATH is defined. By REF we have MATH, hence there exists MATH extending MATH such that MATH, and let MATH. MATH: Since MATH is below MATH, it is enough to prove by induction on MATH that for every MATH when MATH we have that MATH. For MATH, since MATH clearly for every MATH we have MATH so MATH is a nice MATH-pair. For limit MATH, by the induction hypothesis for every MATH and every MATH we have MATH, hence by REF , MATH. For MATH, by the induction hypothesis for every MATH we have MATH. Let MATH be given. Since MATH is below MATH and MATH, REF implies that MATH; that is, for every MATH we have MATH. So we are done. CASE: Let MATH and MATH be given. It is enough to prove that if MATH then MATH. Using REF it is enough to find MATH such that MATH is below MATH and MATH. We define by induction on MATH such that MATH is below MATH and MATH. For MATH let MATH. For MATH, for every MATH, as MATH by REF there is MATH extending MATH such that MATH. But the number of possible MATH is MATH hence there are a function MATH and a set MATH of cardinality MATH such that MATH. Now take MATH. CASE: Immediate.
math/9911225
CASE: By induction on MATH prove that MATH (see more details in REF ). CASE: Apply REF twice.
math/9911225
CASE: It is easy to show that the pair is MATH-nice. We show by induction on MATH simultaneously for all MATH and every MATH that Min-MATH implies that MATH. When MATH or MATH is a limit ordinal this is easy. Suppose MATH and that MATH; by the definition of rank for MATH there exists MATH extending MATH such that MATH is a nice pair and MATH. By the induction assumption MATH. Hence MATH is as required in the definition of MATH. CASE: Suppose without loss of generality that MATH, let MATH and let MATH. By REF , MATH, by REF , MATH. So we have MATH . Hence the conclusion follows.
math/9911225
CASE: Let MATH, by REF, this is a supremum on a set of ordinals MATH (as MATH) hence is an ordinal MATH. So MATH is as required. CASE: If not, then choose by induction on MATH a sequence MATH such that MATH. So for each MATH, the sequence MATH is a non-increasing sequence of ordinals hence is eventually constant, say for some MATH we have MATH, so as MATH is finite, MATH, so MATH, a contradiction.
math/9911225
CASE: Trivial. CASE: Clearly. Clause REF : MATH. Clause REF : For MATH and MATH, note that MATH so we can use the assumption. CASE: So let MATH and we should find a MATH-witness for MATH. We can choose MATH such that MATH. As MATH, clearly there is a MATH-witness MATH for it. Now the number of possible MATH is MATH (really) even MATH hence for some MATH and MATH we have: MATH and MATH. By renaming MATH, now MATH is a MATH-witness by REF . CASE: Follows by REF. CASE: By REF by the well foundedness of the ordinals as in the proof of REF . CASE: Because for MATH and REF.
math/9911225
Fix MATH. Assume MATH is a counterexample. So MATH. Let MATH be defined as follows: MATH is MATH if MATH and is MATH if MATH. We shall prove that MATH thus getting a contradiction. So let MATH and we shall find a MATH-witness for MATH. Let MATH be such that MATH. As MATH (see REF) there is a MATH-witness MATH for MATH, as earlier without loss of generality MATH for some MATH. We shall prove that MATH is a MATH-witness for MATH. Let MATH for MATH. Clause REF : MATH because MATH. Clause REF : Let MATH. If MATH then MATH. (Why? By group theory, by REFMATH, by choice of MATH, by choice of MATH, respectively). If MATH, then MATH by REFMATH, and proceed as above.
math/9911225
CASE: As MATH is a homomorphism it is enough to prove MATH, hence it is enough to prove MATH which follows from MATH, which means MATH which holds by clause REF. CASE: Follows by REF and MATH being an inverse limit.
math/9911225
Suppose for the sake of contradiction that for some MATH we have MATH. Let MATH be minimal such that MATH, without loss of generality MATH. For MATH let MATH be MATH. We will reach a contradiction by showing that MATH and MATH. Note CASE: MATH if MATH, then MATH as MATH and REF. For MATH, we show that MATH. Let MATH. By the choice of MATH. In this case MATH by the minimality of MATH and, of course, MATH, hence MATH by clause REF. Note: if MATH, then MATH for MATH hence MATH, but MATH immediate contradiction. So assume MATH hence MATH. Now we proceed inductively. We assume that MATH and show that MATH. Let MATH, and let MATH. Observe: CASE: MATH [why? by REF and MATH above.] So CASE: MATH. Now: CASE: MATH why? by clause REF CASE: MATH why? the equality by the definition of MATH, the first inequality by the induction hypothesis and the second inequality was proved above (for MATH), the last inequality by REF clause REF CASE: MATH why? by clause REF. Hence by REF CASE: MATH. Together we get the induction demand for MATH.
math/9911225
For each MATH we define MATH by: MATH . So MATH, so as MATH necessarily MATH, hence for some MATH there is no MATH-witness for MATH and MATH (check the definition of MATH). Let MATH be MATH. Let MATH be large enough (so that it will be possible to use the finite NAME theorem when MATH and when MATH the NAME theorem we require that MATH where MATH). Let MATH be a MATH-witness MATH and even MATH. For each MATH define the two place function MATH from MATH to MATH for MATH let MATH . Define the two-place function MATH from MATH: For MATH let MATH. Clearly MATH. Hence an application of one of the above partition theorems provides us with a set MATH such that MATH is constant. Without loss of generality MATH. For each MATH, clearly MATH is not a MATH-witness for MATH, but the only thing that may go wrong is the inequality, MATH, so for some MATH we have that MATH holds, hence CASE: MATH and MATH. This means clause REF holds and clause REF by definition of MATH is a MATH-witness for MATH. Clause REF follows. So MATH is as required.
math/9911225
of REF Stipulate MATH: if MATH it is MATH, otherwise is it MATH. Assume MATH and MATH is well defined, MATH. Let MATH be: MATH if MATH is a limit ordinal and MATH otherwise that is, MATH, see REF. Note that to construct the family MATH we will combine REF with a second application of the NAME Theorem. Let MATH and MATH be such that MATH (exists by NAME theorem if MATH and by NAME theorem if MATH). Apply REF to get a family MATH satisfying: CASE: MATH, CASE: for MATH and MATH, we have MATH. For MATH such that MATH define a coloring MATH of MATH by two colors according to the following scheme: for MATH, let MATH . By the NAME theorem (if MATH) or NAME Theorem if MATH there is a set MATH such that each coloring is constant on MATH. Let the value of MATH on MATH be denoted MATH. Observe that MATH is never green as this would produce a descending MATH-sequence of ordinals as if MATH for MATH, then MATH, so MATH is strictly decreasing. Let MATH and MATH and MATH is the MATH-th member of MATH the MATH-th member of MATH and let MATH, by clause REF above MATH so MATH and MATH. We claim that MATH remember MATH provides a set that can play the role of MATH. We note CASE: MATH for MATH in MATH [why? clearly MATH are in MATH hence by the choice of MATH we have MATH]. Now clauses REF holds by clause REF above, clause REF holds by MATH and clause REF holds by the choice of the MATH. We are left with clause REF . Let MATH, as above clearly for MATH in MATH we have MATH. Hence for MATH in MATH we have MATH . So giving also clause REF.
math/9911234
In view of CITE we can localise at MATH and prove the analogous result for the local rings MATH and MATH. Let MATH and let MATH with MATH the image of MATH in MATH. Now, using CITE and CITE, we see MATH . Since MATH is local, NAME 's lemma implies that MATH if and only if MATH.
math/9911234
Let MATH be a coherent sheaf on a variety MATH and suppose that MATH for some MATH. Then it follows immediately from NAME 's lemma that there exists a neighbourhood MATH of MATH such that MATH. Since, by REF , the unramified locus consists precisely of the points of MATH on which the (coherent) sheaf of relative differentials disappears, it is clear that the unramified locus is open. Now let MATH be a maximal ideal of MATH containing the minimal prime MATH. Then MATH is unramified (along MATH) if and only if MATH is unramified along the induced map MATH . Thus the unramified locus of MATH is the union of the unramified loci of the irreducible components of MATH.
math/9911234
Let MATH and MATH be the quotient fields of MATH and MATH respectively. Then, by definition, MATH is separable if and only if MATH is a separable extension of MATH. If the latter hypothesis holds, then by CITE the generic (set theoretic) fibre of MATH contains MATH elements. Since the MATH-dimension of MATH is MATH for a generic maximal ideal MATH of MATH, in the generic case the fibre is isomorphic to a direct product of copies of MATH as required.
math/9911234
Suppose MATH is unramified and let MATH. Then MATH where the second equality holds since MATH is regular, CITE, and the third since MATH is unramified. Thus MATH is regular, as required.
math/9911234
First note that since MATH is NAME MATH is a free MATH-module, CITE. It follows that MATH is a (central) subalgebra of MATH. By REF is a maximal ideal of MATH and we have an algebra isomorphism MATH . As the image of MATH in MATH is nilpotent we see that MATH is a local, finite dimensional algebra with simple right MATH-module MATH having dimension MATH. Let MATH be the MATH-projective cover of MATH. Then, as right MATH-modules, we have MATH . Therefore there are algebra isomorphisms MATH . Since MATH is central in MATH we have a map MATH given by MATH. By REF and the freeness of MATH as a MATH-module, this map is injective. By REF we see that as a MATH-module, and so as a MATH-module, there is an isomorphism MATH. Therefore MATH so REF is an isomorphism.
math/9911234
Let MATH, a multiplicatively closed NAME set. For every MATH-module MATH, there is a natural equivalence MATH . Multiplication by elements of MATH on the MATH-module MATH is bijective, so the above isomorphism yields MATH . Combining this with REF proves that if MATH is on the NAME locus then MATH is a projective MATH-module if and only if MATH is a projective MATH-module. It's clear from REF that this happens if and only if MATH, in other words if and only if MATH, that is MATH is unramified. Now by hypothesis the NAME locus of MATH coincides with the smooth locus of MATH. Thus combining the above paragraph with REF yields the theorem.
math/9911234
The existence of the non-degenerate bilinear form on MATH allows us to identify MATH and MATH by CITE. That MATH is a connected, reductive algebraic group follows from CITE and our REF . The precise description of MATH follows from CITE. This description shows that it is a NAME subgroup and so, by CITE, satisfies the hypotheses.
math/9911234
The map MATH induced by orthogonal projection from MATH to MATH, is an isomorphism. To see this note that by CITE MATH for any MATH and MATH, with NAME decomposition MATH. Since MATH is conjugate to element of MATH it follows that MATH is injective. Replacing MATH by MATH in CITE shows that MATH is indeed an isomorphism. The lemma follows by applying the non-degenerate form provided by REF .
math/9911234
For MATH let MATH and let MATH . As observed above there is an isomorphism between MATH and MATH. By REF for all MATH, where MATH is the natural map. Therefore MATH if and only if MATH. But since MATH is non-degenerate on MATH and MATH this happens if and only if MATH, in other words MATH (and hence MATH) is regular.
math/9911234
CASE: Straightforward from the definitions. CASE: Prove this first for MATH and MATH. Both sides are zero if MATH is not in MATH. If MATH, set MATH and prove MATH. Then apply REF . The general case follows by bilinearity. CASE: By linearity one need only treat the case where MATH and MATH are exterior products of MATH-forms. The identity follows for such MATH and MATH by expanding MATH as a determinant down the first column. CASE: Suppose that MATH. Then MATH by REF . Thus MATH, the second equality using REF . That is, MATH, so MATH. Applying this inclusion to MATH and using REF yields the desired conclusion.
math/9911234
Under the identification of MATH with a polynomial algebra in MATH variables given by REF the map MATH can be written as MATH for all MATH. Since MATH and MATH are smooth varieties MATH is smooth at MATH if and only if the differential of MATH at MATH is surjective, CITE. In the above coordinates the differential at MATH is described by the matrix MATH . The first of the above lemmas and the proposition show that this matrix has rank MATH if and only if MATH is a regular element, proving the first claim. The second part is REF .
math/9911234
CASE: This follows from CITE, since the required hypotheses hold by the final sentence of CITE. REF. We apply CITE, which is valid here in light of REF . We conclude that MATH is generated by MATH and MATH and in fact that MATH is a free MATH-module with basis MATH. This proves REF, whilst taking MATH-invariants proves the final claim of REF. That MATH is a polynomial algebra on MATH generators follows from REF . We have a commutative diagram MATH where we have noted that MATH for all MATH. By REF the bottom map is an isomorphism which shows that the whole diagram consists of isomorphisms since MATH is also an isomorphism. So to prove that MATH is a complete intersection over MATH it is enough to show that MATH is a complete intersection over MATH. To see this note that there is a MATH-invariant MATH-algebra isomorphism MATH from MATH to MATH obtained by sending MATH to MATH. Now it's easy to check that MATH, and therefore MATH where MATH. Noting that MATH, where the degree of MATH, considered as an element of MATH, is strictly less than MATH, a straightforward degree argument shows that MATH as required. REF. and REF. These follow exactly as in CITE in light of REF . and REF. and REF .
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CASE: The key point here is that all MATH occur as the centres of primary components in the NAME locus. This follows from REF applied to a semisimple character MATH for which MATH holds for all MATH. For then MATH is a direct summand of the appropriate MATH, by REF . Since MATH is NAME by CITE, it follows immediately that MATH is self-injective. That it is NAME now follows from CITE, since it is commutative and local.
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CASE: Recall that MATH is the root lattice and that MATH is a free abelian group. Let MATH be a decomposition such that MATH and MATH is a finite extension. By CITE MATH. Thus the map MATH has kernel MATH and so induces an isomorphism MATH . It therefore follows from REF and the isomorphism MATH that MATH if and only if MATH, proving the first part of the theorem, in view of the remarks preceding it. CASE: Since MATH involves composition with the winding automorphism MATH, we must apply MATH to REF , so that its conclusions are expressed in terms of representations in MATH of MATH-orbits in MATH. On doing this, REF. is an immediate consequence of REF.
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Suppose first that MATH. By REF , MATH is unramified if and only if MATH for all simple roots MATH, so the second remark following REF forces MATH if MATH is unramified. So we have that MATH showing that the unramified maximal ideals of MATH lying above MATH are of the required form. Now suppose MATH and MATH are in MATH and give rise to the same maximal ideal of MATH. In other words MATH and so MATH and MATH are in the same MATH-orbit. But the stabiliser of MATH is all of MATH, which implies MATH. Thus the proposition is true in this case. To prove the general case apply REF together with the remarks following it.
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Suppose first that MATH is regular. Then, by CITE (noting that MATH by hypothesis), MATH divides the dimension of all the simple MATH-modules, as required. Conversely, suppose that MATH is not regular. Since the simple modules of maximal dimension are necessarily baby NAME modules we see from the remarks following REF that we may assume that MATH. Therefore MATH is not regular nilpotent. There must exist a simple root of MATH, say MATH, such that MATH (for otherwise MATH is a torus, whence MATH is regular). There exists MATH such that MATH. Now the argument of CITE is valid in this context and shows that MATH is not simple. Indeed let MATH be the minimal parabolic subalgebra of MATH with NAME subalgebra spanned by MATH and the elements of MATH. Then MATH restricted to MATH is zero on MATH and MATH so MATH representation theory tells us that MATH has a simple module, MATH, of dimension less than MATH whose highest weight is MATH, CITE. Thus there is a non-zero homomorphism from MATH to MATH. Since the right hand side has dimension equal to MATH we are done.
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As we have seen MATH and MATH. Therefore MATH . The first isomorphism therefore follows from REF . The assumption that MATH implies that MATH for all MATH a root of MATH and so MATH as required.
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Let MATH be the isomorphism of REF . Giving MATH its natural filtration we can see, by passing to integral forms as in CITE, that MATH is a filtration preserving map, where MATH has its natural graded structure. As we have seen in REF the algebra MATH has algebraically independent homogeneous generators MATH of degree MATH respectively. Hence, if we let MATH for all MATH, we see that MATH also has algebraically independent generators of degrees MATH. By CITE, which we have already noted is valid thanks to the earlier results of this chapter, MATH is freely generated as a MATH-module by MATH. Set MATH to be the image of the centre MATH of MATH in MATH, so that MATH. Then we deduce from the foregoing that the maximum degree of an element of MATH is MATH. As before we have MATH, by CITE. We now exhibit an element of MATH with degree MATH, thus proving that the map from MATH to MATH is not surjective as long as MATH. Recall that MATH acts on MATH and MATH by the adjoint action, and this induces an action of MATH on MATH and on MATH, where the latter notation has the obvious meaning. By CITE there is a filtration preserving isomorphism of MATH-modules MATH induced by the NAME map. Up to scalars there is a unique element MATH of degree MATH in MATH, and the space it spans must therefore be MATH-stable, and, since MATH is semisimple, in fact trivial. The MATH-equivariance of MATH shows that MATH is the central element of MATH of large degree which we wanted to find.
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Since MATH it is clear from the definition in REF that MATH is graded. By CITE for every MATH there is an operator, MATH, on MATH which lowers degree by MATH. By CITE the characteristic of the field, MATH, is not a torsion prime (in the sense of CITE) and by hypothesis it is not REF. Thus there exists an element MATH such that the elements MATH give a basis for MATH over MATH, CITE. Let MATH. Then we claim that MATH is MATH-invariant. To see this we use the fact that MATH unless MATH. By the definition of MATH we see then that MATH for all simple reflections MATH in MATH. Thus, by CITE, we have for all simple reflections MATH as claimed. From the previous paragraphs we see that the elements MATH, for MATH, span a subspace of MATH of dimension MATH. Moreover, by REF , MATH and, as we have noted already, it follows from CITE that MATH has dimension MATH. Thus the elements MATH with MATH yield a basis for MATH and since MATH, the NAME series is as claimed. For the last part, simply notice that if a finite dimensional MATH-graded algebra has a homogeneous component of dimension greater than REF then it cannot be uniserial.
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First note that by REF the groups MATH and MATH lie inside the NAME group of MATH. It follows from REF that we can (and will) assume without loss of generality that MATH. Suppose that MATH is a finite dimensional uniserial algebra (with a unique simple module). Then MATH is NAME equivalent to a truncated polynomial ring MATH for some MATH. To see this let MATH be the projective cover of the simple MATH-module and consider MATH. As MATH is uniserial with all composition factors isomorphic so it is easy to check that MATH where MATH is the NAME length of MATH. Suppose that MATH has finite representation type. By CITE MATH is uniserial and has a unique simple module. The above paragraph shows that the centre of MATH is a truncated polynomial ring and so, by NAME 's positive answer CITE to REF A, MATH is a subalgebra of a truncated polynomial ring MATH. Moreover, MATH is a NAME algebra, by REF . Let MATH be an ideal of MATH. By CITE, MATH for some ideal MATH of MATH, and hence MATH. In particular MATH is a uniserial algebra (and in fact a truncated polynomial algebra). Following REF we see that MATH is uniserial only if the set of minimal coset representatives MATH has at most one element of any given length. In particular for MATH to be uniserial it follows that the rank of MATH must be no less than the rank of MATH minus one. On the other hand if MATH then MATH is semisimple, so we need only consider the case when the rank of MATH is one less than the rank of MATH. Let MATH be the NAME graph associated with MATH and suppose that MATH is the simple reflection not in MATH. There are a few cases to consider. Firstly suppose that there is a subdiagram of MATH of the following form, where we label the nodes from left to right as MATH, MATH and MATH, MATH . Then both MATH and MATH are elements of length two in MATH, implying that MATH is not uniserial. Secondly, if MATH occurs as one of the endpoints in the subdiagram of MATH below, then label the nodes from top to bottom and left to right by MATH and MATH, MATH . Then both MATH and MATH are elements of length three in MATH, implying that MATH is not uniserial (this argument applies to type MATH and type MATH). Finally assume that MATH occurs in the subdiagram of MATH below where we label the nodes from left to right by MATH, MATH and MATH, MATH . Then both MATH and MATH are elements of length three in MATH, showing MATH is not uniserial (the same argument works for MATH). This proves the first part of the corollary. Since we are assuming MATH we see by REF that MATH for all roots MATH. Thus there exists MATH such that MATH for all roots MATH. There is a bijection MATH defined by MATH for MATH. Moreover MATH is MATH-equivariant since MATH for all roots MATH. We can combine REF (in particular the refined version of CITE) together with CITE (which applies since we can replace MATH by MATH as above and therefore assume that MATH is nilpotent) to see that if MATH is of standard NAME type and MATH is not simple then MATH has more than one simple module. Thus, by CITE, MATH is not of finite representation type. This proves the second claim. Finally, assume the conditions of REF hold. Then by REF we have MATH . In particular we see that the representation type of MATH depends only on MATH. Therefore we can replace MATH by a regular nilpotent element of MATH, say MATH, and MATH by MATH (as above) and deduce that MATH has finite representation type if and only if MATH has finite representation type. Now the (short) argument in CITE shows indeed that MATH has finite representation type. This proves the final claim and so completes the corollary.
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For MATH let MATH. Then by definition, CITE, MATH . On the other hand since MATH and the decomposition above is unique (for that ordering of positive roots) we must have MATH for all MATH such that MATH. By REF, proving the lemma.
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The algebras MATH and MATH (respectively MATH and MATH) are isomorphic via the map sending MATH to MATH (respectively MATH to MATH). It is clear that these isomorphisms are MATH-equivariant. Moreover MATH is the restriction of the map on MATH which sends MATH to MATH. Under the above isomorphism this becomes the map sending MATH to MATH so it follows that MATH . The second statement is now clear. For the final statement let MATH and note that MATH. Thus there is an isomorphism from MATH to MATH taking MATH to MATH. Under this isomorphism the map MATH obtained by evaluation at MATH is transformed into the map MATH obtained by evaluation at MATH. This proves the claim.
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For ease of notation let MATH. Let MATH be the quotient field of MATH and suppose MATH is a finite group acting faithfully on MATH by automorphisms. Note that MATH. Indeed MATH. For the opposite inclusion observe that since MATH is a finite module over MATH we have that MATH. Thus if MATH we may assume without loss of generality that MATH. It follows that MATH as required. Since MATH acts faithfully on MATH we have that MATH by CITE. Now consider the cases where MATH and MATH. Here we have MATH . It follows from REF that MATH . So generically, for MATH a maximal ideal of MATH the algebra MATH has dimension MATH. Moreover, since the dimension function is upper semicontinuous and MATH is an integral domain, in the non-generic case the dimension of MATH is at least MATH as required.
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As MATH-modules, MATH is a direct summand of MATH since MATH is the image of the reduced trace map and we are in characteristic zero. Thus MATH, as required.
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By definition we have MATH. By REF MATH where MATH is obtained by restriction of MATH. Since MATH and since MATH it follows that MATH depends only MATH and that this map can be evaluated by considering only the MATH component of elements of MATH written in triangular form. Since MATH is fixed pointwise by the map MATH which takes MATH to MATH, the last part of REF implies that the NAME map MATH, of REF yields an isomorphism of MATH with MATH where MATH is restriction. Consider the commutative diagram MATH induced from the inclusions MATH . Then, by the above, MATH. The closed points are given by MATH so MATH parametrises the components of MATH. Let MATH and let MATH be the component of MATH containing the orbit associated to MATH. Now consider the diagram MATH . Since MATH is the stabiliser of MATH, the map MATH is a covering near MATH so the component of MATH containing MATH is isomorphic to MATH. Consider next MATH. Since MATH is the stabiliser of MATH the map MATH is a covering near MATH so the component of MATH containing MATH is precisely the (reduced) point MATH. Now MATH has a unique closed point since MATH stabilises MATH. The neighbourhood of this point is given by the kernel of the map MATH . The tensor product on the right is isomorphic to MATH. Now MATH is a neighbourhood of MATH and we have shown that MATH is a subalgebra of MATH. Therefore MATH is a subalgebra of MATH. Noting that the number of elements in the MATH-orbit associated to MATH such that MATH is precisely MATH we see from REF that MATH . Since MATH by REF , the theorem is proved.
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CASE: Suppose that MATH is regular. Then by definition (see REF ) MATH is a regular semisimple element of MATH and so MATH. Therefore, for MATH such that MATH, we have MATH, and no two such MATH can be MATH-conjugate, so there are MATH summands in the decomposition of MATH given by REF .
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CASE: Trivial. CASE: Suppose that MATH is projective. Then its restriction to MATH is also projective. Since MATH, MATH is a local ring of dimension MATH. Thus, as MATH-module MATH is free of rank one. Moreover, being scalar local and self-injective, MATH has a simple socle which corresponds to MATH. Therefore any non-zero MATH-submodule of MATH must contain MATH, proving that MATH is irreducible. REF : REF applies in view of REF. CASE: In terms of MATH the central character of MATH is described by the NAME map, MATH, that is MATH acts on MATH by restriction from the action of MATH given by MATH. Call the torus element corresponding to this character MATH. Considering this as an element of MATH we consider MATH. By REF the central character of MATH is unramified if and only if MATH. By REF we can assume without loss of generality that MATH is generated by reflections coming from a proper subset of MATH. Now MATH if and only if MATH for all MATH. This occurs if and only if MATH for all MATH. Similarly for MATH. Therefore we deduce that MATH if and only if MATH implies MATH for all roots MATH. By the definition of MATH this last condition is equivalent to MATH implies MATH for all MATH.
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REF can be deduced from CITE. Since MATH is simply-connected it follows from CITE that MATH is the reductive group generated by MATH and the root subgroups MATH where MATH. If MATH it follows from REF that MATH if and only if MATH divides MATH. As MATH is a coefficient of the highest root of MATH this occurs if and only if MATH or MATH.
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Let MATH have reduced expression MATH and let MATH and MATH for MATH. Then, by CITE, MATH . The result now follows from case-by-case analysis using the explicit elements MATH given in the table in the appendix.
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By the universal property it is enough to check that MATH is a vector of weight MATH which is annihilated by MATH for all MATH, MATH. The weight claim follows from the equality MATH . It remains only to show that MATH annihilates MATH for MATH. If MATH this is clear so we can assume that MATH. We pick a reduced expression for the longest word of MATH starting with the simple reflection MATH. Then, in MATH, there is a commutation relation MATH where MATH for some MATH of weight MATH, CITE. Therefore MATH as required.
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Let MATH lie over MATH, as usual. Note that MATH if and only if MATH if and only if MATH is regular semisimple. Suppose all simple MATH-modules have dimension MATH. Then by REF either MATH is regular in MATH or MATH. In both cases this implies that MATH is regular, as required.
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The first statement is an immediate consequence of REF . For the second statement note first that if we know the value of MATH for all MATH then we know the value of MATH for all MATH (since, by REF , MATH has, up to conjugacy, a basis of maximal rank consisting of simple roots of MATH and its longest root). If MATH is prime to the index of connection of MATH then we can calculate the values of MATH given the values of MATH for MATH which means that MATH is completely determined in this case. Hence there is only one possible value for MATH so there are only MATH preimages. Suppose that MATH for two such preimages. Then, by REF , MATH for some MATH. As MATH we see that MATH belongs to the NAME group of MATH. Since both MATH and MATH lie over MATH as above it follows that they are fixed under the dot action of the NAME group of MATH, implying that MATH as required.
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If we put MATH then MATH is unitarily equivalent to MATH for positive real MATH, so its eigenvalues are independent of MATH. The same applies by analytic continuation for complex MATH, in a sense which is familiar in the theory of resonances CITE. When MATH for MATH then the essential spectrum of MATH depends on MATH, being equal to MATH. Since the eigenvalues can only accumulate in the essential spectrum and the only common point of the essential spectra of all of these operators is MATH, this is the only possible limit point of the eigenvalues. In order to obtain bounds on the position of the eigenvalues, we need to introduce the numerical range MATH where we assume that all MATH lie in the MATH-independent domain of MATH. In the current situation MATH is a convex set which contains the entire spectrum of MATH and is contained in MATH . It follows that if MATH is an eigenvalue of MATH then MATH lies in this set for all MATH.
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The main statement of the theorem depends upon the fact that MATH . We now explain why MATH is convex. If we define MATH for MATH, then as the real part of an analytic function MATH is harmonic. Hence MATH is subharmonic and constant on each ray through the origin. This forces MATH to be convex.
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The only new component of the proof is the observation that if we put MATH then we obtain MATH .
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Let MATH be an eigenvalue of MATH where MATH and let MATH be the corresponding eigenfunction, so that MATH. Then MATH so MATH . Putting MATH, MATH and MATH, we deduce that MATH so MATH . This approach is similar to the NAME - NAME principle for finding eigenvalues of self-adjoint operators (see, for instance, CITE). By evaluating the NAME norm of the above operator, whose kernel is MATH we deduce that MATH . This implies MATH.
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Along with REF consider the equation MATH . Its linearly independent solutions are MATH . We shall seek the solution MATH of REF in the form MATH where the functions MATH satisfy MATH . We take MATH which corresponds to MATH. Then MATH . It follows from REF that MATH . We define the function MATH . As is easily seen, MATH where MATH . We observe that if MATH is an eigenvalue of MATH, then MATH. For MATH in the right-hand half-plane we have MATH and, hence MATH . From this formula we get MATH . This together with REF implies MATH . Therefore MATH for sufficiently large MATH. In particular, if MATH then MATH and MATH for all such MATH. Since MATH is analytic in MATH, the function MATH defined by REF is entire. Therefore the number of its zeros in any bounded domain is finite. The eigenvalues of MATH can only be found among the squares of the zeros of MATH, hence the final result.
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Given MATH, every resonance in the stated sector either has not yet been uncovered, in which case MATH, or it has been uncovered, in which case MATH . The proof is completed by a geometrical argument. The set MATH is the intersection of half-planes, and hence is closed and convex.
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An application of NAME 's theorem shows that MATH and this implies that MATH . The inequality MATH follows from the fact that the function MATH is convex. The main statement of the theorem depends upon eliminating MATH from the formulae MATH .
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On choosing MATH, we see that every resonance satisfies MATH.
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Denote MATH then MATH . Let MATH be two linearly independent solutions of REF defined by MATH being some constants. We look for the function MATH in the form MATH where the functions MATH solve MATH . In other words, MATH . Here the Wronskian MATH is constant. REF guarantee the convergence of the integrals in REF and imply MATH . This, in turn, means that MATH as required. The way we construct this solution also implies its uniqueness.
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To prove this we need to show that for MATH and MATH one has MATH. Thanks to REF we obtain MATH and also MATH. Thus we obtain indeed the correct result.
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It is clear from the previous discussion that the data MATH gives rise to MATH with the required properties. So conversely assume that we are given MATH and the action of MATH on objects. We define MATH and the action of MATH on maps as in the statement of the proposition. We first show that MATH is a functor. Indeed let MATH and assume that there are maps MATH and MATH. Then for all MATH we have MATH, but also MATH. Thus by non-degeneracy we have MATH. It is easy to see that the pairing REF defines an isomorphism MATH which is natural in MATH and MATH. The proof is now complete.
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Let us sketch the construction of MATH. The uniqueness will be clear. The objects in MATH are formally written as MATH with MATH and MATH. A morphism MATH is formally written as MATH with MATH where MATH is such that MATH, MATH. We identify MATH with MATH. The functor MATH is defined by MATH and the functor MATH is defined by MATH. It is clear that these have the required properties.
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If we require exactness of MATH and MATH, then there is only one way to make MATH into a triangulated category. First we must define the shift functor by MATH and then the triangles in MATH must be those diagrams that are isomorphic to MATH where MATH is a triangle in MATH (note that the exactness of MATH is equivalent to that of MATH). To show that this yields indeed a triangulated category one must check the axioms in CITE. These all involve the existence of certain objects/maps/triangles. By applying a sufficiently high power of MATH we can translate such problems into ones involving only objects in MATH. Then we use the triangulated structure of MATH and afterwards we go back to the original problem by applying a negative power of MATH.
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By construction MATH is an automorphism on MATH. To prove that MATH is a NAME functor we have to construct suitable maps MATH. Pick MATH, MATH. Then we have MATH . We define MATH as the composition of these maps. It follows easily from REF that the constructed map is independent of MATH, and it is clear that MATH has the required properties.
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We will show that REF. and REF. are equivalent. The equivalence of REF is similar. CASE: By definition we have MATH. Assume that MATH is a non-automorphism. Then by REF we have MATH. CASE: Let MATH be the NAME - NAME triangle associated to MATH. From REF - NAME triangles it follows that there is a morphism of triangles MATH . The fact that MATH together with the fact that MATH is in the socle of MATH implies that MATH must be an isomorphism. But then by the properties of triangles MATH is also an isomorphism. So in fact the triangles REF are isomorphic, and hence in particular REF is an NAME - NAME triangle.
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It is easy to see that MATH linearly independent elements of the (left or right) socle define different triangles. However NAME - NAME triangles are unique. This is a contradiction.
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CASE: Let MATH be an indecomposable object. By NAME duality there is a natural isomorphism MATH as MATH-bimodules. In particular MATH has a one dimensional socle which corresponds to the map MATH. Define MATH and let MATH be a non-zero element of the socle of MATH. We claim that the associated triangle MATH is an NAME - NAME triangle. Let MATH be indecomposable and let MATH be a non-isomorphism. We have to show that the composition MATH is zero. Using NAME duality this amounts to showing that the composition MATH is zero. Since MATH is a non-isomorphism, this is clear. CASE: This is the interesting direction. As pointed out in REF it is sufficient to construct the NAME functor on the full subcategory of MATH consisting of the indecomposable objects. For MATH an indecomposable object in MATH we let MATH be the object MATH. Let MATH be a non-zero element of MATH representing the NAME - NAME triangle MATH . Let MATH and MATH be indecomposable objects in MATH. Then the following hold. CASE: For any non-zero MATH there exists MATH such that MATH. CASE: For any non-zero MATH there exists MATH such that MATH. CASE: Using REF triangles there is a morphism between the triangle determined by MATH (the NAME) and the triangle determined by MATH. MATH . The morphism labeled MATH in the above diagram has the required properties. CASE: Without loss of generality we may assume that MATH is not an isomorphism. We complete MATH to a triangle MATH . Then MATH and since MATH is non-zero, MATH will not be split. Now we look at the following diagram MATH . Since MATH is not split we have by REF that MATH. Hence by the properties of triangles we have MATH for a map MATH. This proves what we want. Having proved the sublemma we return to the main proof. For any indecomposable object MATH choose a linear map MATH such that MATH. It follows from the sublemma that the pairing MATH is non-degenerate. We can now finish our proof by invoking REF .
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This follows from applying REF together with its dual version for left NAME functors.
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This follows using REF and standard properties of almost split sequences and NAME - NAME triangles. We leave the proof to the reader.
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That MATH takes the indicated values on objects follows from the nature of the NAME - NAME triangles in MATH (given by REF ). The assertion about fully faithfulness of MATH and MATH follows from the corresponding property of MATH. REF . follows by inspecting the triangle REF. Finally that MATH and MATH are equivalences in the case that MATH is a NAME functor follows by considering MATH, which is a left NAME functor.
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To show that MATH is hereditary we have to show that MATH. Now MATH is characterized by the property that it is an effaceable MATH-functor which coincides in degree zero with MATH. Hence we have to show that MATH is effaceable. This is clear in degree MATH since there the functor is zero, and for MATH it is also clear since then MATH CITE. Now standard arguments show that as additive categories MATH and MATH are equivalent to MATH. We don't know if this equivalence yields an exact equivalence between MATH and MATH, but recall that the definition of a (right) NAME functor does not involve the triangulated structure. Hence if MATH has a (right) NAME functor then so does MATH and vice versa.
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REF . is an obvious consequence of REF., so we prove REF. Assume there is an infinite number of paths ending in MATH. By REF there must be an arrow MATH such that there is an infinite number of paths ending in MATH. Repeating this we obtain an infinite path MATH such that there is an infinite number of paths ending in every MATH. The existence of such an infinite path contradicts REF .
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Let MATH. Then we need to construct natural isomorphisms MATH . Since MATH and MATH are finite complexes of projectives we can reduce to the case MATH and MATH. So we need natural isomorphisms MATH . Again both of these vector spaces have a natural basis given by the paths from MATH to MATH. This leads to the required isomorphisms.
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First we show that if MATH exists satisfying properties MATH then it is unique. This proof will in particular tell us how to construct MATH. We then show that this construction always yields a category MATH with the required properties. Since MATH and since the objects in MATH remain projective in MATH by REF ., it follows that the derived functor of MATH is fully faithful. It is also clear that MATH is closed inside MATH under the formation of cones. Let MATH be the extended NAME functor. If MATH then MATH can be obtained by starting with objects in MATH and repeatedly taking cones. It then follows from REF that for MATH we have MATH. Hence the triple MATH satisfies the hypotheses of REF . Thus we obtain MATH. According to REF every indecomposable object in MATH will be of the form MATH with MATH. Implicit in the notation MATH is the assumption that MATH is defined for MATH, that is MATH for MATH. In addition we may assume that MATH is minimal. Thus either MATH or else MATH for MATH. The last case is equivalent to MATH for MATH. If MATH then the same is true for MATH. Hence we have to impose REF only for MATH. We conclude that the indecomposable objects in MATH are of the following form: CASE: The indecomposable objects in MATH. CASE: Objects of the form MATH where MATH and MATH is an indecomposable object in MATH, where MATH. This completes the determination of MATH as an additive subcategory of MATH, and finishes the proof of the uniqueness. Let us now assume that MATH is the additive subcategory of MATH whose indecomposable objects are given by REF . We have to show that MATH is a hereditary abelian category satisfying REF.-REF. Since MATH has a NAME functor, it has NAME - NAME triangles. Below we will need the triangle associated to MATH. Using the criterion given in REF we can compute this triangle in MATH. So according to REF the requested triangle is of the form MATH . We now define a MATH-structure on MATH. Using the fact that MATH is hereditary we easily obtain that as additive categories MATH. We now define MATH . We claim that this is a MATH-structure. The only non-trivial axiom we have to verify is that MATH . So this amounts to showing that MATH for MATH and for MATH. We separate this into four cases. MATH . MATH . MATH . MATH . So MATH is indeed an abelian category. Since it is easily seen that MATH, we obtain by NAME duality that MATH for MATH. Hence it follows by REF that MATH is hereditary. Now we verify REF. CASE: This is clear from the construction. CASE: This follows from REF . CASE: This is clear from the construction. CASE: By REF we already know that MATH has a NAME functor, so we can use its properties. An indecomposable object MATH in MATH is projective precisely when MATH. Using the construction of MATH it is easy to see that this happens if and only if MATH. Using REF functors we find that the injectives in MATH must be given by MATH. This proves what we want.
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CASE: This is clear. CASE: Assume that MATH is an injective object in MATH. By decomposing MATH into a direct sum of indecomposables and invoking REF we see that the socle MATH of MATH is non-zero and of finite length. Since MATH is hereditary, MATH is again injective. Repeating this we find a strictly ascending chain MATH such that MATH has finite length. Since MATH is noetherian, this chain must stop, whence MATH has finite length. CASE: We first show that MATH satisfies REF . In fact we show that for a vertex MATH in MATH there is a bound on the length of a chain of non-isomorphisms MATH . Put MATH. Applying MATH we get non-isomorphisms MATH . Since these are indecomposable injectives and MATH is hereditary, all these maps must be surjective. Clearly the required bound is now given by the length of MATH. Now we prove that MATH satisfies REF . Since every MATH has only a finite number of summands in a direct sum decomposition, it is sufficient to show that there exists only a finite number of vertices MATH such that there is a non-zero map MATH. Applying MATH we find that MATH must be a subquotient of MATH. Since MATH has finite length, there is only a finite number of possibilities for MATH. The fact that there is a bound (depending only on MATH) on chains of the form REF implies that every map MATH is a linear combination of products of the MATH. Thus MATH is full, and from this we easily obtain that MATH yields an equivalence between MATH and MATH. Now let MATH be indecomposable injective in MATH. Since MATH has a NAME functor, it follows from REF that MATH is the injective hull of some simple object MATH lying in MATH. By considering the injective MATH and using induction on the length of MATH we find that MATH. Now MATH is clearly the injective hull of MATH also in MATH. Using the fact that MATH is an equivalence there must exist a vertex MATH such that MATH. Furthermore MATH must correspond under MATH to the injective hull of MATH. Since the latter is MATH, we are done. CASE: This is clear. CASE: This proof is similar to that of MATH . Assume that MATH is an indecomposable injective of infinite length, and let MATH be its socle. There exists an indecomposable summand MATH of MATH which is of infinite length. Continuing this procedure we find an infinite sequence of irreducible maps MATH and applying MATH we find a corresponding infinite sequence of irreducible maps between projectives MATH . However such an infinite sequence cannot exist by REF .
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From the fact that MATH and the fact that MATH is closed under cones in MATH, it follows that MATH is closed under MATH. Thus MATH defines a right NAME functor on MATH, and hence via MATH a right NAME functor on MATH. Since right NAME functors are unique, MATH coincides (up to a natural isomorphism) with the standard right NAME functor obtained from deriving the NAME functor which was introduced in REF . It now follows that MATH satisfies REF .,REF.,REF.,REF . Hence we get MATH.
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If MATH is projective, then we are done. If MATH then the fact that MATH is indecomposable plus the fact that MATH is hereditary implies that MATH is projective. Hence this case is covered also. Assume now that MATH is not projective (and hence MATH). Then by faithfulness of MATH we obtain a non-zero map MATH. Induction on MATH now yields that MATH with MATH for some indecomposable projective MATH. Thus MATH and we are done.
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CASE: follows trivially from REF., so we prove REF. Let MATH be a decomposition of MATH into a direct sum of indecomposable objects. If MATH then it maps non-trivially to MATH, whence it is preprojective by the previous lemma. This proves what we want.
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Let us first assume REF. and let MATH, with MATH and MATH. We have to show that MATH is the quotient of a preprojective object. Take a minimal projective presentation MATH with MATH. For MATH the object MATH cannot contain an injective summand since then the resolution REF wasn't minimal. It follows that the object MATH cannot contain an injective summand since then MATH would be injective, contradicting the fact that it is in the essential image of MATH. Hence MATH and MATH are defined, and by the exactness of MATH (see CITE) we obtain an exact sequence MATH . In particular MATH is covered by a preprojective object. Now assume REF. Let MATH be an indecomposable object in MATH, and assume that there is a surjective map MATH with MATH preprojective. Then by REF we have that MATH is also preprojective. Thus for large MATH we have a triangle MATH with MATH and MATH in the image of MATH. Since MATH is closed under cones in MATH and since MATH is indecomposable, it follows that MATH for some MATH. Thus if MATH is defined for MATH then MATH. If MATH is not defined, then MATH is projective and so lies in MATH. Hence in this case we are done also.
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NAME under quotients is clear. Let us now prove closedness under subobjects. Clearly it is sufficient to show that a subobject of a preprojective object is preprojective. But this is precisely REF . Let us now prove closedness under extensions. So assume that we have an exact sequence MATH where MATH. Since we already know that MATH is closed under quotients, we may prove our result for pullbacks of REF. In particular we may assume that MATH is preprojective. But then according to REF we have MATH where MATH and MATH is preprojective. Hence by what we know already we have MATH. This proves what we want. Assume now that MATH satisfies any of the conditions of REF . We have to show that MATH is closed under MATH and MATH in MATH. Since MATH is closed under the formation of cones, it suffices to show that MATH and MATH are in MATH for a preprojective object MATH. The only case which isn't entirely obvious is that MATH is in MATH when MATH is projective, but this follows from REF .
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Since clearly MATH it suffices to prove the implication that MATH connected implies MATH connected. So assume that MATH is a direct sum MATH with MATH and MATH non-trivial. Let MATH be the category of projectives in MATH for MATH. Then clearly MATH. This yields a corresponding decomposition MATH. Since we had assumed that MATH is connected, it follows that for example MATH. By REF the NAME functor on MATH restricts to one on MATH and one on MATH. It is now easy to see that the conditions for REF descend to MATH and MATH. Thus MATH and hence MATH. This is a contradiction.
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Let MATH, MATH have their usual meaning. Since MATH, we easily obtain MATH, and by REF we have that the objects in MATH have finite length and furthermore MATH. Thus in particular MATH. Iterating we find MATH for any MATH. This proves that the preinjective objects have finite length. We also obtain MATH, which yields that the preinjective objects are quotients of projectives.
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For MATH let MATH and MATH be the corresponding projective and injective object in MATH. It is well known and easy to see that sending MATH to MATH defines an injective morphism of quivers MATH, where MATH is the preprojective component of MATH. It is clear that this is surjective if and only if MATH contains no injectives. So it is sufficient to show that MATH contains no injective objects. Assume to the contrary that MATH has some injective object. The projectives are given by the connected quiver MATH, and the arrows of MATH correspond to irreducible maps between indecomposable projectives, in the opposite direction. Because the categories of injectives and projectives are equivalent, the same holds for the injectives. Thus MATH contains all indecomposable injectives. In particular MATH is also the preinjective component of MATH, and hence by REF all objects in MATH have finite length. Assume that there is some finite connected subquiver MATH of MATH which is not NAME. Choose MATH and let MATH be a finite connected subquiver of MATH such that the simple composition factors of all objects in any path from MATH to MATH correspond to vertices in MATH. In the subcategory MATH of MATH we clearly have that MATH and MATH are still projective and injective objects respectively, and it is not hard to see that there is still a path of irreducible maps from MATH to MATH. But since MATH is not NAME, it is well known that the preprojective component of MATH has no injective objects, and we have a contradiction. If all finite subquivers of MATH are NAME, then it is easy to see directly that MATH must be of type MATH, MATH or MATH. These cases can be taken care of by direct computations (see III. REF).
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The proof amounts to verifying the conditions for a split torsion pair. In order to visualize this we have included some pictures of the various NAME involved. Let the functor MATH be defined as usual. CASE: By the dual version of REF we find MATH. If MATH is a non-injective indecomposable object in MATH, then have that MATH. Hence it follows by NAME duality that MATH. Thus MATH is a split torsion pair in MATH. CASE: Now we need to show MATH. First let MATH be indecomposable in MATH and let MATH be an indecomposable which is not in MATH (and hence in particular MATH). Then looking at REF we see that either MATH or MATH. In the first case it is clear that MATH and in the second case this follows from the fact that MATH is a split torsion pair. Let now MATH and MATH be in MATH. By applying MATH for MATH large enough so that MATH and MATH are in the preprojective component of MATH, it is not hard to see that if MATH then there is a path from MATH to MATH in the NAME of MATH. If MATH for some MATH and MATH, we can assume that there is an arrow from MATH to MATH. Write MATH and MATH where MATH and MATH are in MATH and MATH and MATH. Then we have an arrow MATH. Since MATH is a section in MATH we have MATH or MATH. Since MATH we conclude that MATH or MATH, so MATH which is a contradiction. Since MATH is MATH stable, the fact that MATH follows from NAME duality. CASE: That the projectives in MATH are given by MATH follows from the shape of the NAME of MATH.
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Let MATH be an indecomposable object in MATH. We have to show that it is a quotient of a preprojective object. If MATH is itself preprojective, this is clear, and if MATH is preinjective then we can invoke REF . So assume that MATH is neither in the preinjective nor preprojective component. Looking at REF we see that MATH is untouched by the tilting, so we may consider MATH as an object in MATH. Then by assumption there is an exact sequence in MATH of the form MATH where MATH, and consequently MATH, is preprojective. Since all terms in the exact sequence lie in the tilted category MATH, this is also an exact sequence in MATH. Let the functor MATH be defined as usual. Choose MATH large enough such that MATH and MATH are preprojective in MATH, and consider the exact sequence MATH . Thus MATH is in the subcategory MATH of MATH generated by the preprojectives. Since the NAME functor on MATH restricts to one on MATH by REF , it follows that MATH is also in MATH and hence in MATH.
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If MATH and MATH are NAME, this is well-known, so we assume that MATH, MATH are non-Dynkin. It follows from REF that MATH is generated by preprojectives. By REF the category MATH whose projectives are given by MATH, and which is obtained by a sequence of the two tilts described above, also has the property that it is generated by preprojectives. Then it follows from REF that MATH is equivalent to MATH. Since we already know that MATH and MATH are derived equivalent, this finishes the proof.
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The only thing that is not entirely trivial is the fact that MATH is closed under quotients. So let MATH be a surjective map in MATH with MATH. Let MATH be a non-trivial map with MATH preprojective. Finally let MATH be the corresponding pullback diagram. Since MATH is not the zero map, there must exist an indecomposable summand MATH of MATH such that MATH. Since MATH maps non-trivially to MATH, it follows from REF that it must be preprojective. Now clearly MATH. This yields a contradiction.
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CASE: We have by REF. Furthermore it follows from REF that MATH. Hence it is sufficient to show that every MATH is a direct sum MATH with MATH and MATH. From the fact that MATH is noetherian and MATH is closed under extensions REF it follows that there exists a maximal subobject MATH in MATH which lies in MATH, and so MATH lies in MATH. Hence it now suffices to prove that MATH. Since MATH is hereditary, MATH preserves epis, thus it is sufficient to show that MATH with MATH an indecomposable projective and MATH. In addition we may and we will assume that MATH is indecomposable. Assume MATH. Then clearly MATH is not projective and MATH is not injective. Hence by NAME duality MATH. This contradicts the fact that MATH. Hence MATH, which finishes the proof. CASE: That MATH and MATH satisfy NAME duality follows for example from REF . Any projective or injective object in MATH clearly has the same property in MATH and therefore lies in MATH. Since a non-zero object cannot be both in MATH and MATH, we conclude that MATH does not contain nonzero projectives or injectives. Hence MATH is defined everywhere on MATH and is invertible.
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Assume that MATH does not exist. We will show that then MATH is not noetherian. The non-existence of MATH implies that there exists a proper strictly ascending array of integers MATH as well a corresponding array of vertices MATH such that there is an arrow MATH which is not in MATH. The edges MATH induce maps MATH which are the composition of the maps MATH. We now claim that the image of MATH defines a proper ascending chain of subobjects of MATH. This is clear since the path in MATH corresponding to the map MATH in MATH does not pass through MATH for MATH. Hence MATH is not noetherian, and we have obtained a contradiction.
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Assume that MATH is not split. Consider the almost split sequence MATH . From the theory of almost split sequences it follows that we may take MATH and MATH. Also by the theory of almost split sequences it follows that MATH for some maps MATH. Since MATH is projective we have MATH, and hence MATH. Now consider the map MATH given by MATH . Clearly MATH, whence there exists MATH such that MATH . Applying MATH to the first of these equations yields a factorization of MATH, contradicting the hypotheses.
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If MATH is noetherian, then so is MATH since the latter is a full subcategory closed under subobjects. Therefore it follows from the previous lemma that there is some MATH such that for MATH the only arrow starting in MATH is the one which lies in MATH. By dropping some initial vertices in MATH we may assume MATH. Assume now that MATH (as in the statement of the lemma) does not exist. Since injectives have finite length, it is clear that MATH is not injective. We will show that MATH is not noetherian in MATH. The non-existence of MATH implies that there exists a strictly ascending array of integers MATH as well a corresponding array of vertices MATH such that there is an arrow MATH which is not in MATH. The edges MATH correspond to an irreducible map MATH. Using the theory of almost split sequences there is a corresponding map MATH. Also from the path MATH we obtain a map MATH which gives rise to a corresponding map MATH. We denote by MATH the composition MATH . We now claim that the image of MATH defines an ascending chain of subobjects of MATH. If this were not the case, then for some MATH all MATH and MATH must factor through MATH. It follows that the resulting map MATH must be split since otherwise by REF the map MATH factors through MATH, which is impossible given the construction of this map. Thus it follows that MATH. We conclude that some vertex MATH occurs infinitely often among the MATH, but this contradicts REF . Thus we have shown that MATH is not noetherian, and in this way we have obtained a contradiction to the hypotheses.
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Let MATH be the set of equivalence classes of infinite paths in MATH. For every MATH we choose a representative MATH. By the previous lemma we can without loss of generality assume that the only edges in MATH adjacent to MATH REF are those in MATH. Now let MATH be obtained from MATH by removing for all MATH the vertices MATH for MATH as well as the edges in MATH. It is clear that MATH is obtained from MATH by adjoining the strings MATH. Furthermore MATH itself cannot contain any infinite paths since such an infinite path would have to be equivalent to one of the MATH. In particular it would have to contain vertices outside MATH, which is of course a contradiction. It follows that MATH is strongly locally finite, and so MATH is indeed a star.