paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0004003 | We have already proved most of this. The statement that it defines an equivalence of homotopy theories follows from the fact that it carries suspensions of NAME to suspensions of m-coalgebras and loop-spaces to the cobar construction. |
math/0004003 | Suppose MATH and MATH are semi-simplicial sets such that MATH become equivalent when regarded as objects of MATH. Then there exists a homotopy equivalence MATH which induces a homology equivalence MATH where MATH and MATH are the singular complexes of MATH and MATH, respectively. The conclusion follows from the fact th... |
math/0004003 | In fact MATH will be a strong deformation retract of MATH via a functorial retraction, and the inclusion of MATH will be a homotopy equivalence. We define MATH inductively on simplices via: CASE: If MATH is a MATH-simplex, MATH, where MATH is some point disjoint from MATH and depending on MATH. Note that MATH. CASE: If... |
math/0004003 | The simplicial approximation theorem says that there exist integers MATH and a simplicial map MATH such that MATH is homotopic to MATH - where we have identified MATH with MATH and MATH with MATH, respectively, via the well know homeomorphisms between them. To see that this induced map only depends on the homotopy clas... |
math/0004003 | This follows directly from REF. |
math/0004003 | This is straightforward: the MATH maps shuffle the factors of MATH into the appropriate positions. |
math/0004003 | This follows from the fact that the term MATH is expressible in terms of the composition operations in MATH, so it pulls back to MATH. |
math/0004003 | This follows from: |
math/0004003 | The second statement is the easiest to prove - bearing in mind the remark following REF. The first statement (that MATH, with these compositions, is an operad) requires us to verify the defining identities of an operad in REF. We use the hypothesis and REF to conclude that there exists a coalgebra, MATH, over MATH with... |
math/0004003 | Note that MATH since MATH so that the factor MATH in MATH is closed under compositions. |
math/0004003 | As before, the second statement is the easy one to prove - it follows directly from REF. As in REF, the first statement that MATH is an operad follows from REF - which implies that there exists a coalgebra, MATH, over MATH with an injective structure map MATH . It is not hard to see that the induced map MATH will also ... |
math/0004003 | Simply set all MATH in the formula in REF. |
math/0004009 | Let MATH be a formal metric on MATH, and MATH a nontrivial harmonic MATH-form. Then MATH and MATH are also harmonic and so MATH has constant length. In particular it has no zeros. It follows that MATH and MATH span MATH for all MATH. As MATH is constant for every constant linear combination MATH of MATH and MATH, the N... |
math/0004009 | That the length of any harmonic form is constant follows from REF . The more general statement follows by polarisation. |
math/0004009 | If MATH is harmonic, then MATH is constant by REF . Using that the length of MATH is also constant by REF , we conclude that MATH is constant. The converse is trivial. |
math/0004009 | Fix a formal Riemannian metric on MATH. It follows from the above Lemmas that the number of linearly independent harmonic MATH-forms is at most the rank of the vector bundle MATH. Similarly, when the dimension is MATH, the number of self-dual or anti-self-dual harmonic forms in the middle dimension is bounded by the ra... |
math/0004009 | Fix a formal Riemannian metric MATH on MATH. We consider the NAME or NAME map MATH given by integration of harmonic MATH-forms. As the harmonic MATH-forms have constant lengths and inner products, MATH is a submersion. It induces an isomorphism on MATH, and products of linearly independent harmonic MATH-forms are never... |
math/0004009 | Suppose MATH fibers over MATH with fiber MATH and monodromy diffeomorphism MATH. By NAME 's Lemma, we may assume that MATH preserves a volume form MATH on MATH, so that its pullback to MATH descends to MATH as a closed form which is a volume form along the fibers. We can find a Riemannian metric on MATH for which MATH ... |
math/0004009 | This follows from the obstruction-theory definition of the NAME class, and the fact that every nontrivial harmonic form has no zeros because of REF. |
math/0004009 | Suppose MATH is nontrivial and anti-self-dual. We have MATH which is harmonic if and only if the norm of MATH is constant. If it is constant, it must be a non-zero constant, and then the above equation shows that MATH is a symplectic form inducing the opposite (non-complex) orientation on MATH. In particular, MATH must... |
math/0004009 | It is clear that every metric on every rational homology sphere is formal because there are no nontrivial harmonic forms. Conversely, assume that MATH is a manifold with a non-zero NAME number MATH, for MATH. Let MATH be a Riemannian metric which has positive curvature operator on an open set, say a ball MATH, and assu... |
math/0004010 | REF : notice that every bounded uniformly continuous function extends over the NAME compactification. REF : choose a bounded uniformly continuous pseudometric MATH on MATH subordinated to MATH (that is, MATH) and apply REF to the function from MATH to MATH whose components are distance functions MATH, MATH, with MATH. ... |
math/0004010 | Let MATH be a finite cover of MATH, let MATH, and let MATH be arbitrary. Find an entourage of the diagonal MATH with MATH. At least one element of MATH, denote it by MATH, satisfies the property MATH . By proceeding to a subnet if necessary, we may assume without loss in generality that MATH . In view of the assumed co... |
math/0004010 | The first statement is self-evident. In order to establish the second claim, it is now enough to prove that for every MATH the orbit map MATH is continuous. Let MATH be any. Using the boundedness of the original action, choose a MATH such that for all MATH and MATH, MATH. The set MATH is a neighbourhood of MATH in MATH... |
math/0004010 | Make the set-theoretic disjoint union MATH into a weighted graph, by joining a pair MATH with an edge in any of the following cases: CASE: MATH, with weight MATH; CASE: MATH, with weight MATH, CASE: for some MATH, MATH and MATH, with weight MATH. The weighted graph MATH equipped with the path metric clearly contains MA... |
math/0004010 | We will perform the proof in several simple steps. CASE: By first rescaling the metric MATH on MATH and then replacing it with MATH if necessary, we can assume without loss in generality that the values of MATH are bounded by MATH. CASE: Without loss in generality, we may also assume that MATH supports the structure of... |
math/0004010 | One can assume that MATH, where MATH and MATH. Using REF , choose a finite metric space MATH, elements MATH of MATH, and isometries MATH of MATH such that the naturally indexed finite metric spaces MATH and MATH are MATH-isometric. Using REF , isometrically embed MATH and MATH into a finite metric space MATH in such a ... |
math/0004010 | Let the group MATH act continuously on a compact space MATH. We will show that every finite collection of elements of MATH has a common fixed point in MATH, from which the result follows by an obvious compactness argument. Fix an arbitrary such collection, MATH. Let MATH be an arbitrary element of the unique compatible... |
math/0004010 | Since for every MATH the value of the inner product is uniquely determined by the Euclidean distance between the elements, MATH there is only one way to turn the linear span MATH into a pre-Hilbert space so as to induce the given metric on MATH. The corresponding completion MATH is isometrically isomorphic to the close... |
math/0004010 | The claim consists of two parts: first, that all sets of the form MATH are elements of MATH, and second, that each enourage from MATH contains a set of the above type. CASE: Given MATH, MATH, and MATH as above, choose a bounded uniformly continuous pseudometric MATH on MATH such that MATH, and introduce a bounded unifo... |
math/0004010 | REF : according to REF , the fixed point on compacta property of a topological group MATH is equivalent to the following: for every bounded right uniformly continuous function MATH on MATH taking values in a finite-dimensional Euclidean space, every finite collection of elements MATH, and every MATH, there is a MATH su... |
math/0004010 | REF : assume REF , that is, no finite subspace MATH of MATH containing a copy of MATH is in MATH. Denote by MATH the collection of all finite metric subspaces MATH with MATH. By assumption, MATH. Then for every MATH the set MATH admits a colouring with MATH colours, which we will view as a function MATH, in such a way ... |
math/0004010 | Combine REF . |
math/0004010 | Denote by MATH the two-element metric space with MATH. Partition the set MATH of all isometric embeddings of MATH into MATH into two disjoint subsets MATH in such a way that whenever an injection MATH is in MATH, the `flip' injection MATH is in MATH, and vice versa. Since the space MATH is MATH-discrete, the MATH-neigh... |
math/0004010 | The topological group MATH is isomorphic to the semidirect product MATH of the full orthogonal group MATH equipped with the strong operator topology and the additive group of the NAME space MATH with the usual norm topology, formed with respect to the natural action of MATH on MATH by rotations. (Compare CITE.) Suppose... |
math/0004010 | Without loss in generality, we can assume that MATH is separable (in fact, even countable). Such a MATH can be MATH-embedded into the NAME space MATH (see CITE, or else REF), and therefore we can further assume that MATH. Choose any element MATH and isometries MATH of MATH with the property MATH, MATH. Denote by MATH t... |
math/0004013 | : Let MATH be a simply connected REF-manifold with integral cohomology as in REF and linking form equivalent to a standard form. Pick an orientation on MATH and an isomorphism, MATH so that MATH. Let MATH; then MATH. For any manifold with the above cohomology, MATH, the fourth NAME - NAME class of the tangent bundle. A... |
math/0004015 | By a direct application of REF we have MATH where MATH, MATH. From the assumptions on MATH we see that MATH and the claim follows by some algebra (with MATH). |
math/0004015 | Fix MATH, MATH, MATH, and write MATH for short. We also introduce a dummy tube, setting MATH, MATH. We choose vectors MATH which have the values MATH for MATH, so that MATH and each MATH is a distance MATH from the hyperplane spanned by the remaining MATH vectors MATH. The key observation is that for every MATH, the se... |
math/0004015 | Suppose for contradiction that REF failed. Then we have MATH for some MATH. From REF we thus have MATH . From REF the right-hand side is MATH. From REF we thus have MATH . For every degenerate MATH, let MATH be an affine subspace satisfying REF; one can easily ensure that MATH is a measurable function. Let MATH denote ... |
math/0004015 | When MATH the claim is trivial with MATH empty if MATH is sufficiently small, since MATH while MATH . Now suppose that MATH. We say that two squares MATH are separated if MATH and MATH. We define MATH to be a maximal pairwise-separated set of squares MATH which satisfy MATH . It is easy to see that REF holds. To show R... |
math/0004015 | From REF we have MATH . In particular, we have MATH . Using this, REF, and REF, we find that REF will follow if we can show MATH . By REF , this is equivalent to MATH . On the other hand, from REF we have MATH . Also, from REF, and REF we have MATH . It thus suffices to show that MATH . Consider the right-hand side of ... |
math/0004015 | Let MATH be a maximal MATH-separated set of directions, and for each MATH let MATH be a finitely overlapping cover of MATH by MATH-tubes with direction MATH. We can arrange matters so that every MATH obeys MATH for some MATH and MATH. Call a direction MATH sticky if MATH and define MATH . Clearly REF holds. To prove RE... |
math/0004015 | We can find unit directions MATH and numbers MATH for all MATH such that MATH and MATH where the dot product is taken in MATH. Fix MATH and MATH. Let MATH be a tube in MATH. We can find MATH with MATH such that MATH so in particular we have MATH for MATH. Since MATH is direction-separated, it thus suffices to show that... |
math/0004018 | First, we choose coordinate systems on each space. Let MATH and MATH, so that the discrete quotient map REF is given by MATH, and the continuous quotient map by MATH. Recall that the fiber derivative MATH in these coordinates has the following form (see, for example, CITE) MATH . Then the above diagram is given by: MAT... |
math/0004018 | The proof follows from the results of the previous subsection, in particular, REF relates the DLP dynamics on MATH with the DEP dynamics on MATH which, in turn, is related to the DEL dynamics on MATH via the reconstruction REF . |
math/0004018 | The proof is based on the commutativity of REF and the MATH invariance of the unreduced symplectic forms. Notice that MATH and MATH in REF are NAME manifolds, each being foliated by symplectic leaves, which we denote MATH and MATH for MATH and MATH, respectively. Denote by MATH and MATH the corresponding symplectic for... |
math/0004021 | We will use an induction on the number MATH of normal generators. If MATH the relations imply that all MATH commute in MATH. Since by assumption these elements generate MATH it follows that all commutators vanish in MATH. Now assume the statement holds for groups with MATH normal generators and let MATH be normally gen... |
math/0004021 | The statement for the generators follows from the standard rewriting process in nilpotent groups: If a nilpotent group MATH is normally generated by MATH then it is also generated by these elements. One uses an induction on the nilpotency class of MATH based on the fact that MATH mod MATH implies MATH mod MATH for all ... |
math/0004021 | For MATH ordinary homology, the theorem comes from CITE. To get the general statement, one uses an eventual NAME Theorem as in CITE. It is therefore necessary to reduce to the case of simply-connected spaces. This can be easily done in our context by picking maps MATH, which represent the generators MATH of MATH, and a... |
math/0004021 | The NAME - NAME sequence MATH implies the first statement. The NAME tori represent trivial elements in MATH, since they bound solid tori there. |
math/0004021 | To simplify notation we will assume MATH. The general case can be worked out analogously. Let MATH be a parameterization of a neighborhood of MATH in MATH such that MATH and MATH. We may assume that MATH where MATH . This can in fact be taken as a definition of a finger move. Let MATH be the compactification of MATH. W... |
math/0004021 | Consider MATH. Here MATH is the spin structure on MATH constructed above and MATH is the map that gives MATH on the fundamental group. Since MATH is a spin boundary we have MATH. The factorization of MATH implies that MATH is in the image of the composition MATH for every MATH. Using MATH, REF implies now MATH and henc... |
math/0004021 | Recall that MATH is bounded by MATH by REF . Let MATH be MATH copies of MATH, the manifold from REF . We may write MATH. Now we glue MATH to MATH along parts of their boundary to obtain a manifold MATH with boundary MATH. MATH . Using REF and MATH we can find a map MATH making the triangle MATH commute. By the NAME - N... |
math/0004021 | Let MATH be the spin manifold obtained in REF . Recall that MATH . Now let MATH be the disjoint union of MATH copies of MATH. Then MATH and MATH both contain the disjoint union of MATH copies of MATH which we will denote by MATH. Now glue MATH to MATH along MATH to obtain MATH . Then MATH and MATH bounds disjointly imm... |
math/0004021 | Let MATH (respectively, MATH) be the attaching maps into MATH that guide the construction of MATH (respectively, MATH) upwards (respectively,downwards) from MATH. Now index MATH index MATH implies MATH . By general position we may assume that MATH and MATH are disjoint in the level MATH. But this means that MATH and MA... |
math/0004021 | Using NAME 's jet transversality theorem CITE we can assume that MATH is a generic immersion except for a finite number of cross caps CITE. As in NAME 's original immersion argument, these cross caps can be pushed off the, say, lower boundary MATH. This changes the lower boundary MATH by cusp homotopies which we may as... |
math/0004022 | Let MATH be a local chart of MATH. Then for any MATH and MATH in the domain of this local chart we have: MATH then, using the local coordinates REF , one gets easily all the results of the proposition. |
math/0004022 | Let us first assume REF . Then, using the results of REF and the fact that MATH and MATH are smoothing, one proves easily that MATH is an isomorphism whose inverse is given by: MATH . Now let us prove REF . Following CITE page REF, we recall that the NAME kernel of MATH is the finite sum of a smooth function and of osc... |
math/0004022 | REF are left to the reader. REF is an easy consequence of REF and of REF. |
math/0004022 | For MATH or MATH we set: MATH where MATH denotes the set of smooth functions which vanish of infinite order at MATH. Then we have the following exact sequence: MATH of MATH-algebras. Here MATH denotes the induced formal MATH-deformation of the sphere at infinity. A direct construction of this deformation may be describ... |
math/0004022 | CASE: A standard NAME sequence argument shows that MATH is indeed two dimensional. The fact that MATH defines a basis is left to the reader. CASE: This is an easy consequence from REF and of the properties (see CITE, CITE ) of the trace density map MATH. |
math/0004022 | One obtain this formula by first applying REF and then by letting MATH. As we will see below, the involved characteristic classes of vector bundles on MATH are in fact standard NAME cohomology classes and hence the regularized integral coincides with the orientation class of MATH. |
math/0004023 | As a first step, note that if MATH is a smooth point, MATH an automorphism and MATH a birational morphism, then it is clear that MATH if and only if MATH. To conclude, it suffices to note that - up to composition with automorphisms - there are only finitely many birational morphisms MATH. Recall that for a given bundle... |
math/0004023 | Set MATH where MATH is the ideal sheaf of MATH. It follows immediately that MATH is a coherent sheaf of MATH-algebras. Define MATH. The existence of MATH, MATH and MATH follows by construction. In order to see that fibers of MATH are of the desired type, let MATH be an arbitrary closed point. We are finished if we show... |
math/0004023 | The strategy of this proof is to find a sequence of base changes which modify MATH and MATH so that REF can be applied. First, after finite base change, we may assume that the normalization MATH is a MATH-bundle over MATH. Recall that smooth morphisms are stable under base change. Thus, even after further base changes,... |
math/0004023 | Write MATH, MATH where MATH denotes numerical equivalence and MATH is a general fiber of MATH. Since MATH and MATH are effective, MATH and thus MATH. Therefore MATH. |
math/0004023 | As a first step, we need to find an estimate for the dimension of the subfamily of non-immersed curves. We let MATH be the closed subfamily of non-immersed curves and claim that MATH . Indeed, if MATH, then let MATH be a singular curve which corresponds to a general point of a component of MATH which is of maximal dime... |
math/0004023 | If MATH is a morphism which is birational onto its image, if MATH and the image MATH is a curve which is associated with MATH, then it follows from REF that MATH is smooth in a neighborhood of MATH. Now, to conclude that MATH is a morphism, it suffices to realize that MATH can be written as a composition MATH where MAT... |
math/0004023 | Following an argument of NAME, MATH is generically one-to-one if MATH, and the latter follows from REF . See CITE for a proof of NAME 's result. |
math/0004023 | Let MATH be the normalization of the universal family MATH and consider the diagram MATH . It follows directly from REF that MATH is birational and it follows immediately that MATH is birational as well. But then MATH is a finite birational morphism between normal spaces, and therefore isomorphic. In particular, MATH a... |
math/0004024 | Suppose MATH is a countable set of functions. Let MATH denote the set MATH for every MATH. Every MATH is at most countable because MATH is separable. So choose a MATH in MATH, then we know that for every MATH the set MATH is nowhere dense in MATH. For such a MATH the set MATH is countable. Because the set MATH is also ... |
math/0004024 | The proof of the first inequality is easy. We simply have to observe that the set of all constant functions on the reals is a transitive set of functions. Now for the second inequality. Striving for a contradiction suppose that MATH. Let MATH be a transitive set of functions such that MATH. We define a set map MATH on ... |
math/0004024 | We will show by transfinite induction that MATH is a transitive set in MATH for all MATH. For MATH this is obvious. Suppose the theorem is true for all MATH. Let MATH and MATH be reals in MATH. If MATH is a successor ordinal, MATH, then we use REF in the case that at least one of MATH and MATH is not in MATH to show th... |
math/0004024 | We will construct a fusion sequence MATH such that each MATH will know all the first MATH splitting nodes of every branch of the perfect tree MATH and MATH for every MATH. Because MATH forces that MATH is a new real, there exists an element MATH with maximal length MATH, such that MATH and MATH does not decide MATH. Th... |
math/0004024 | Suppose MATH and MATH are two reals of MATH. We consider two cases. CASE: MATH is a real in MATH. The constant function MATH is a continuous function defined in MATH, thus a member of MATH, and in MATH it maps MATH onto MATH. REF both MATH and MATH are reals not in MATH. Let MATH be a witness of this, so MATH. Accordin... |
math/0004024 | This is an immediate consequence of REF . |
math/0004024 | Define the element MATH as follows for MATH: MATH . In this way we strengthen the tree MATH above MATH keeping the rest of the perfect tree intact (according to MATH anyway). |
math/0004024 | Let MATH denote the set of all MATH extending MATH. Because MATH is a perfect tree there exists a MATH-name MATH such that for every MATH we have MATH . According to REF there exists a MATH such that if MATH is consistent with MATH we have a MATH such that MATH. Put MATH. We have MATH for every MATH. Enumerate MATH as ... |
math/0004024 | By REF we know that there is a MATH and there exist MATH names MATH for a map on the finite sub trees of MATH and MATH for a perfect tree such that MATH. Without loss of generality we assume that MATH. Let us construct a fusion sequence MATH. Let MATH, MATH, MATH and choose MATH in such a way that we are building a fus... |
math/0004024 | By applying REF twice we have a MATH in MATH and MATH names MATH, MATH and MATH, MATH for maps and perfect trees respectively such that MATH. Without loss of generality we can assume that MATH. During the construction of possible finite sub trees MATH for MATH, when constructing the fusion sequence in the proof of REF ... |
math/0004024 | Choose a MATH such that MATH. Let MATH be consistent with MATH and let MATH denote all the MATH extending functions MATH. Because MATH forces that MATH, there is an antichain below MATH of size MATH such that all these elements force different interpretations of MATH in the extension. In other words there exist a seque... |
math/0004024 | For the first part of the theorem suppose that we have MATH such that MATH. We will construct a fusion sequence below MATH and define a continuous function MATH in MATH such that the fusion of the sequence forces that MATH holds in MATH. Let MATH, MATH, MATH, and choose MATH in such a way that we are building a fusion ... |
math/0004024 | Suppose that MATH is a MATH name and MATH an element of MATH such that MATH. So there exists a MATH name MATH and a MATH name MATH such that MATH. Aiming for a contradiction assume MATH is a name for an object not in MATH. There exists a MATH such that MATH does not decide MATH. Now we have MATH does not decide MATH, a... |
math/0004024 | For every MATH we have that there exists a function MATH mapping MATH onto MATH or vice versa. This function MATH is not a member of MATH for the obvious reason that assuming that MATH maps MATH onto MATH we get MATH, which, of course, is false. Using REF and the fact that MATH we see that the size of MATH is at least ... |
math/0004030 | CASE: First we must show that MATH is well defined. Observe first that MATH that is, MATH for every MATH. Let now MATH be arbitrary. Then MATH and MATH for every MATH, hence MATH is well defined. CASE: Now we show that MATH is dense in MATH. First the elements with only finitely many entries not MATH are dense in MATH.... |
math/0004030 | CASE: Let MATH the polar decomposition of MATH, and MATH the spectral resolution of MATH. Then MATH for every unitary MATH and every MATH, that is, MATH for every MATH. Now MATH, and, since MATH, also MATH. CASE: REF shows that MATH, further MATH is dense in MATH. Now MATH is invariant under the unitary group MATH, tha... |
math/0004030 | Set MATH (MATH). Then matrix calculation shows that MATH is a family of orthogonal, equivalent, finite projections, such that MATH and MATH. Now MATH defines a selfadjoint system of matrix units MATH such that MATH for every MATH. Now define operators MATH and MATH . Since MATH is dense in MATH (compare proof of REF) a... |
math/0004030 | REF shows that MATH for every MATH and MATH (compare REF). Now MATH . |
math/0004030 | Let MATH and MATH be bounding sequences for MATH and MATH, respectively, (compare CITE). Then: MATH since MATH and MATH (compare CITE). |
math/0004030 | CASE: Since MATH for all MATH , where MATH is the polar decomposition of MATH, we can write the trace, defined in REF, as follows (MATH is the spectral measure of MATH): MATH . Since the MATH are mutually orthogonal, we have MATH . Further MATH . CASE: Let MATH be cyclic. Then there are MATH with MATH for every MATH, w... |
math/0004030 | Set MATH or MATH. Then MATH fulfills the conditions of REF and is invertible, such that from REF follows that MATH is cyclic and separating. |
math/0004030 | Since MATH is infinite but semifinite there is a countable orthogonal family of finite equivalent projections MATH in MATH, such that MATH. Now there is a selfadjoint system of matrix units MATH with MATH (compare CITE). This shows that MATH is isomorphic to MATH where MATH and MATH is isomorphic to every MATH (MATH). ... |
math/0004030 | The finite case is just REF. In the infinite case the existence and the asserted properties follow from REF infinite case, the uniqueness from REF . |
math/0004030 | Again the finite case is just REF. In the infinite case the necessarity of the trace condition follows from REF and the sufficiency from REF, respectively, |
math/0004030 | CASE: First we observe that MATH for every MATH. For let MATH and MATH, MATH, then MATH . Further MATH such that MATH is an (algebraic) conjugation for MATH. CASE: Let now MATH be bounded ( MATH all the MATH and MATH, respectively, are bounded). Then we show that the NAME operator MATH defined by MATH can be written as... |
math/0004030 | Let MATH now be a type MATH or MATH factor and MATH the operator corresponding to the cyclic and separating vector MATH. Let further MATH the spectral resolution of MATH. Then we can define a positive measure MATH on the MATH-algebra of NAME sets in MATH, such that MATH where MATH for all NAME sets MATH. Now MATH. Assu... |
math/0004030 | We construct the MATH inductively: Since the range of MATH is all of MATH (compare CITE) there is a projection in MATH, such that MATH. Suppose now that for MATH the MATH are pairwise orthogonal with MATH (MATH). Setting MATH the restricted algebra MATH is again a type MATH factor, finite, if MATH is finite, and infini... |
math/0004036 | Put MATH so that MATH. Since MATH is decreasing when MATH or MATH and increasing when MATH. Thus (roughly speaking) MATH attains its maximum at MATH. Since there are MATH terms in the summation formula of MATH, we have MATH . Taking MATH and dividing by MATH we have MATH which turns out to be MATH . Since MATH, we have... |
math/0004047 | (This idea, and in fact the whole proof, is due to CITE.) Assume that the polynomials MATH are pairwise incomparable, where MATH, and MATH abbreviates MATH. For each MATH there is a natural number MATH such that MATH can be written as MATH, where MATH is a MATH-ary term, and MATH (the entries of the vector MATH are the... |
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