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math/0103204 | Pull back to MATH, restrict to the fibers of the two projections and use the NAME Theorem. For the arithmetic genus use the exact sequence MATH and the results of REF. |
math/0103204 | For any points MATH of MATH, the image of MATH in MATH is contained in the divisor MATH. Conversely, if MATH is contained in a translate MATH of MATH, then we have MATH, for all MATH. Equivalently, for all MATH, we have MATH, that is, MATH and MATH is effective. |
math/0103204 | By REF the space MATH can be identified with MATH. So MATH corresponds to a section MATH. The support of the divisor MATH of zeros of MATH is the set of divisors MATH such that MATH. So our lemma is equivalent to saying that the intersection MATH is not empty. This follows from the following computation of the intersection number of MATH and MATH where we use MATH and MATH. MATH . The analogous calculation with MATH instead of MATH proves the second asertion. |
math/0103204 | This follows from the positivity of the intersection number of MATH and MATH: MATH . Tha analogus calculation with MATH instead of MATH proves the second assertion. |
math/0103204 | It is sufficient to prove that the map MATH is dominant. A general point of MATH is a complete MATH on MATH. By the previous lemma, there is a divisor of MATH which contains MATH. So there is MATH with MATH. The pair MATH maps to MATH by MATH. To see that MATH for a general choice of MATH, it is sufficient to prove that MATH for a general choice of MATH. If MATH is base-point-free, then this is automatic. If MATH has base points, then it is sufficient to prove that no divisor MATH is contained in its base divisor. It follows from REF (see CITE p. REF) that a general MATH is base-point-free unless MATH is either trigonal, bielliptic or a smooth plane quintic. In all these cases, the base divisor can be chosen to be general so that it contains no divisor MATH. The assertion about MATH is proved similarly, using the corresponding statements for MATH. |
math/0103204 | By REF , the variety MATH is not empty. To see that the dimensions of MATH and MATH are everywhere MATH, note that MATH is equivalent to MATH. Requiring MATH (respectively, MATH) imposes at most one condition on the pair MATH. Since the dimension of MATH is MATH CITE, the proposition follows. |
math/0103204 | Using the exact sequence MATH we need to understand the sections of MATH which vanish on MATH. Equivalently, translating everything by MATH, we need to understand the sections of MATH which vanish on MATH. For this, we use the embedding of MATH in MATH: MATH . Since MATH and MATH (see REF for this notation), by REF this gives the exact sequence of cohomology MATH . Since MATH, by REF the elements of MATH all vanish on MATH, hence they also vanish on MATH. So if the coboundary map is not injective, then there must be elements of MATH which are not restrictions of elements of MATH. In particular, by the above exact sequence, we must have MATH. |
math/0103204 | Let MATH be general so that in particular we have MATH (see REF). Then, as we noted in REF, the image MATH of MATH in MATH is not zero. Since MATH deforms with MATH, by REF , the image of MATH in MATH is zero. So MATH is in the kernel of the coboundary map MATH which is therefore not injective. It follows, by REF , that MATH. Since the dimension of MATH is at least MATH (see REF ) and MATH is one-dimensional, the dimension of MATH is at least MATH. If the genus is MATH, then since MATH is not empty, the dimension of MATH is at least MATH. If MATH has dimension MATH, then by the above discussion we have MATH for a MATH-dimensional family of MATH (in MATH). So MATH and, by NAME 's Lemma, since MATH is not hyperelliptic, we have MATH or MATH. If MATH, then clearly MATH. Suppose now that every component MATH has dimension MATH. Then MATH by the above and MATH. Here NAME 's Lemma only gives us MATH so we use the following argument. Since MATH generically on MATH, the MATH form a MATH-dimensional family of linear systems and so do the MATH. Writing MATH, the MATH vary in a family of effective divisors of dimension MATH. Therefore the degree of MATH is at least MATH and MATH, that is, MATH. Next MATH has dimension MATH and the general fibers of MATH are one-dimensional, all equal to a union of components of MATH, say MATH. Since we can suppose MATH (see REF), the condition MATH for all MATH means that MATH for all MATH. Therefore the projection of center MATH from the canonical curve MATH to MATH is not birational to its image. So there is a nonconstant map MATH of degree MATH with MATH smooth such that MATH and MATH where MATH is a two-dimensional linear system MATH on MATH and MATH is the base divisor of MATH. Consider now the linear systems MATH. As MATH varies in MATH, the divisors of these linear systems form a MATH-dimensional family of divisors. Therefore we have MATH . Combining this with the equality MATH we obtain MATH . Since MATH has degree at least MATH we first obtain MATH . If MATH, then MATH and MATH is a conic in MATH. Hence MATH is trigonal and MATH. In this case MATH since for any MATH, if we take MATH, then MATH. So MATH is of dimension MATH which is contrary to our hypothesis. Therefore MATH has degree MATH and MATH. In this case, since MATH is not hyperelliptic, MATH is birational to a plane curve of degree MATH or MATH and has genus MATH or MATH. If MATH is elliptic, then any divisor MATH is in MATH and MATH is of dimension MATH which is against our hypothesis. If MATH has genus MATH, then its plane model has degree MATH. If MATH has genus MATH, then it has only one MATH which is then MATH. This implies that MATH has dimension MATH since MATH for MATH in a MATH-dimensional family of effective divisors. Therefore MATH by the NAME Theorem. However, this is impossible as MATH is a complete linear system of dimension MATH for a general MATH. So MATH has genus MATH and its plane model has a double point: MATH for some points MATH and MATH on MATH such that MATH. We obtain MATH and, for MATH general, MATH is a general effective divisor of degree MATH on MATH. In particular, MATH. Furthermore, since MATH is general and MATH for all MATH, we obtain MATH . Now, since the MATH vary in a family of dimension MATH, MATH must vary in a family of dimension MATH, that is, the points MATH and MATH are general in MATH. Since MATH is fixed this gives MATH and MATH has degree MATH. |
math/0103205 | If MATH, denote by MATH the blow-up of MATH along MATH, so that one can consider the following diagram of morphisms: MATH . Thus, MATH . If we tensor the exact sequence defining MATH in MATH with MATH, we get MATH . Observe that MATH-since MATH is isomorphic to the NAME algebra of the NAME group MATH, which is finite by assumption on MATH (see CITE); thus, the cohomology sequence associated to REF reduces to MATH . A sufficient condition for MATH is therefore MATH. By NAME duality on MATH, we have MATH . Since MATH is locally free, then MATH, so MATH becomes MATH . Denote by MATH the MATH-exceptional divisor in MATH such that MATH. From standard computations with blow-ups, we get MATH . Therefore, the right-hand side of MATH becomes MATH. Since we have MATH-from the fact that MATH on MATH, we get MATH. For REF , that is, MATH, one can directly use the exact sequence MATH . |
math/0103205 | By definition, MATH is a smooth projective variety. From the assumptions on MATH and MATH and from the NAME vanishing theorem (see, for example, CITE, page REF), it follows that MATH . Consider the natural projection morphism MATH and recall that MATH(see REF ). From the relative NAME sequence MATH-and from the exact sequence MATH-we get that MATH . Therefore, if we consider MATH in REF , we get MATH . By projection formula, MATH for each MATH. Since the fibres of MATH are isomorphic to MATH and since MATH is relatively ample, all the higher direct image sheaves in REF are zero; thus, by NAME spectral sequence and by REF , we get the statement. |
math/0103205 | First of all, we want to show that REF implies REF . To prove this, we will use REF . Therefore, the first step of our analysis is to apply such vanishing result to the vector bundle MATH-where MATH is a positive integer. The problem reduces to finding which "twists" of MATH are big and nef on MATH. In the sequel we shall write for short MATH instead of MATH. From the exact sequence MATH-it is useful compute for which positive integers MATH the vector bundle MATH is ample or globally generated (see CITE). From the NAME sequence of MATH one deduces the exact sequence MATH(see CITE, page REF, and CITE, page REF); therefore, one trivially has MATH-that is, MATH and, so, MATH are globally generated whereas MATH and MATH are ample, for MATH. Recall now that MATH is the universal line over the NAME MATH of lines in MATH (see, for example, REF and C, or CITE, page REF). By standard properties of projective bundles, MATH, thus we have MATH-with the natural projection MATH on the MATH-th factor, MATH. If MATH denotes the NAME embedding of MATH in MATH, one determines the map MATH . On the other hand, we can consider the complete tautological linear system MATH, which is free since MATH is globally generated. From the NAME spectral sequence, the NAME sequence and the NAME formula (see CITE, page REF), we get that MATH-where here MATH. Therefore, the complete linear system MATH defines a morphism MATH . One easily sees that MATH and MATH coincide, so the global sections of MATH contract the MATH-fibres of MATH in MATH, which are lines in MATH. From the fact that MATH, the restriction of MATH to MATH is generically finite since MATH, being of general type, is not filled by lines. Thus the rank REF vector bundle MATH is globally generated and big and nef. By REF , MATH, for each nef divisor MATH. Since MATH, we have that MATH for each nef divisor MATH. Therefore, if MATH, with MATH nef, then MATH . The vanishing result MATH is a fundamental tool for the following second part of the proof. On MATH we can consider the exact sequence MATH which determines the restriction map MATH: MATH . By REF , MATH is surjective. Next, by tensoring the exact sequence MATH with MATH, we get MATH-Thus, the map MATH is surjective if and only if MATH. From the first part of this proof, REF implies that MATH, so we have MATH . By REF , the surjectivity of MATH implies therefore that MATH and so the statement. The last step is to determine if, with the given hypotheses, the map MATH is surjective. Consider the map MATH . By REF , for each MATH, the sheaf morphism MATH-is surjective; thus, for each MATH there exist two global sections MATH which generate the stalk MATH as a MATH-module, that is, MATH . If MATH is the set of nodes of MATH, then MATH. The surjectivity of MATH implies there exist global sections MATH such that MATH . Therefore, MATH and MATH-where MATH and MATH where MATH for MATH. This means that the map MATH is surjective. Moreover, since the condition for a point MATH to be regular is an open condition in the family, it follows that the component of MATH containing MATH has the expected number of moduli. |
math/0103205 | For the first part of the statement, one can repeat the procedure at the beginning of the proof of REF . From MATH we get the second part of the statement. |
math/0103205 | CASE: The first part of the statement is a straightforward computation. We shall briefly recall the fundamental steps of its proof. If MATH is the blow-up of MATH along MATH, by the hypotheses on MATH and by the pull-back to MATH of the NAME sequence, we get MATH . Since MATH is g.l.n. and non-degenerate in MATH, by NAME duality and by the pull-back on MATH of the NAME sequence we have MATH-where MATH and where-MATH-is the NAME map of MATH. Since MATH, from MATH it follows that MATH iff MATH. In this case, by standard NAME theory (see REF , page REF), there is no first-order deformation of MATH, with MATH fixed, induced by first-order deformations of the linear system MATH; so all such deformations are induced by elements of MATH. To get the second part of REF observe that, by the regularity of MATH and by REF , MATH. Assume, by contradiction, that MATH and let MATH be a non-zero vector. Such MATH corresponds to a tangential direction MATH since, by REF , we have MATH. By the regularity assumption of MATH, all directions in MATH are unobstructed. This means there exist a one-dimensional base scheme MATH, smooth at the central point MATH, and a family MATH such that MATH-where MATH . Since MATH, the family MATH corresponds to a family of maps MATH, for which MATH . By composing MATH with the map MATH, where MATH, we get a family of maps MATH for which MATH . From REF , we know that MATH and, from the above computations, we have MATH. Therefore, the element MATH is induced by first-order projectivities, so the family MATH is determined by a family MATH, where MATH-such that MATH and MATH, for each MATH whereas MATH. Since MATH is of general type, then MATH. Therefore, MATH-then MATH-where MATH is a smooth surface projectively equivalent to MATH, for each MATH, and MATH. We therefore obtain a family of maps MATH such that MATH, for each MATH. By composing such family of maps with MATH, MATH, we thus get a family of maps MATH-where MATH . Since MATH and MATH, the element MATH is also an element of MATH. This leads to a contradiction; indeed, by tensoring the exact sequence defining MATH in MATH with MATH, we get MATH-From the above computations, we know that MATH, which implies MATH. CASE: In this case MATH on MATH and MATH is non-degenerate in MATH, then MATH(with abuse of notation, we denote always by MATH the map MATH). From the hypotheses on MATH, we get MATH . By using the same computations of REF , we get-MATH-where MATH . Thus, as in REF , MATH if and only if MATH. Note that MATH . From the pull-back of the NAME sequence and from the hypotheses on MATH, we get MATH . Observe that MATH: indeed, the first vanishing trivially holds whereas, by NAME 's isomorphism and by NAME duality, we have MATH; from the regularity of MATH, REF and the hypothesis MATH, we get MATH. Since MATH is effective by assumption, MATH also imposes independent conditions to MATH. By standard NAME 's vanishing theorem, we have MATH, so MATH. We therefore obtain MATH where MATH is the MATH-exceptional divisor. Since MATH is a fixed divisor, MATH. Moreover, since MATH, the elements of such a vector space correspond to first-order projectivities fixing pointwise the hyperplane MATH. Turning back to REF , MATH if and only if MATH is an isomorphism. In such a case, all first-order deformations of MATH, with MATH fixed, are induced up to first-order by projectivities not fixing pointwise the curve MATH. For the second statement in REF , one can follow the same procedure in REF . By supposing there exists a non-zero vector MATH, one determines a family MATH, where MATH, such that MATH . As before, one obtains MATH, so the family MATH is contained in the sugroup MATH, whose elements pointwise fix the curve MATH. Therefore, we have MATH, for each MATH, contradicting the existence of the non-trivial, one-dimensional family MATH. |
math/0103205 | We start by considering the vanishing MATH. By contradiction, assume that MATH does not impose independent conditions to MATH. Let MATH be a minimal MATH-dimensional subscheme of MATH for which this property holds and let MATH. This means that MATH and that MATH satisfies the NAME condition (see, for example CITE). Therefore, a non-zero element of MATH gives rise to a non-trivial rank REF vector bundle MATH fitting in the following exact sequence MATH with MATH and MATH. Hence MATH . Since MATH is effective and irreducible with MATH, from MATH it follows that MATH is a big and nef divisor (see REF ). By applying the Index theorem to the divisor pair MATH and by MATH, we get MATH . Therefore, MATH-which means that MATH is NAME (see Definition MATH and Remark MATH), hence MATH. Twisting MATH by MATH, we obtain MATH . We claim that MATH; otherwise, MATH would be an effective divisor, therefore MATH, for each ample divisor MATH. From MATH, it follows that MATH, so, by MATH and MATH, MATH . Thus MATH; next by MATH it follows that MATH, hence MATH. The cohomology sequence associated to MATH ensures there exists a divisor MATH such that MATH and such that the irreducible nodal curve MATH, whose set of nodes is MATH, is not a component of MATH (otherwise, MATH would be an effective divisor, which contradicts the non-effectiveness of MATH). Next, by NAME 's theorem, we get MATH . On the other hand, taking MATH maximal, we may further assume that the general section of MATH vanishes in codimension MATH. Denote by MATH this vanishing-locus, thus, MATH; moreover, MATH, which implies MATH . (Note that MATH since MATH and MATH is effective). By applying the Index theorem to the divisor pair MATH, we get MATH . Note now that, from REF and MATH it follows that MATH, since MATH hence MATH. From MATH and from the positivity of MATH, it follows MATH . We observe that the left side member of REF is non-negative, since MATH, where MATH is effective and, by (MATH), MATH. Squaring both sides of (MATH), together with (MATH), we find MATH . On the other hand, by MATH, we get MATH-that is, MATH . Next, we define MATH . Putting together MATH and MATH, it follows that MATH. We will show that, with our numerical hypotheses, one has MATH, proving the statement. Indeed, the discriminant of the equation MATH is MATH, and it is a positive number, since MATH, by MATH, and MATH. We remark that MATH iff MATH, where MATH and MATH so we have to show that, MATH. From MATH, it follows that MATH. Note that MATH: indeed, if MATH then MATH, which contradicts the Index Theorem, since MATH and MATH nef. Moreover, we have MATH: for simplicity, put MATH; thus MATH iff MATH . If MATH, MATH trivially holds; on the other hand, if MATH, by squaring both sides of MATH we get MATH which means MATH, that is, REF. With analogous computations we get that MATH, which ensures there exists at least a positive integral value for the number of nodes. In conclusion, our numerical hypotheses contradict MATH, therefore the assumption MATH leads to a contradiction. For what concerns the other vanishing, that is, MATH, if we consider the exact sequence MATH-by NAME duality we get MATH since, by MATH, MATH cannot be effective. |
math/0103205 | See REF . |
math/0103205 | By the hypothesis on MATH and by the facts that MATH and MATH is as in REF , we get that MATH is g.l.n. (see REF ). Therefore, as in REF , MATH . By combining the pull-back to MATH of the NAME sequence in MATH and the exact sequence defining MATH in MATH, we get the following diagram: MATH . From the regularity of MATH, we get MATH . From the second row, MATH is surjective if and only if MATH is. Since MATH and since MATH is surjective, it suffices to prove that MATH is surjective. With the given hypotheses, we shall prove that MATH-holds. By NAME duality and by NAME 's isomorphism, MATH . The regularity of MATH at MATH implies that the restriction map MATH-is surjective. By tensoring the exact sequence MATH-with the vector bundle MATH, we get MATH . Since MATH is a non-degenerate c.i. (in particular projectively normal), from standard computations involving the NAME sequence restricted to MATH we find MATH (for details, see CITE). Therefore, the vanishing MATH holds if and only if the map MATH is surjective. By the assumption MATH, with MATH, the vector bundle MATH is globally generated; then one concludes as in REF . In the same way, one concludes also in the case MATH and MATH. Since we have proven that MATH is g.l.n and that the map MATH, by REF we get the statement. |
math/0103205 | Suppose, by contradiction, that MATH; thus, MATH. Since MATH is by assumption a regular point, it corresponds to an unobstructed curve in MATH. Therefore, an element of MATH is induced by an effective algebraic deformation. From what observed in REF , such deformations must be induced by projectivities. Then, one can conclude by using REF . |
math/0103208 | We know from CITE that the space MATH is smooth at MATH. Consequence: we can find an embedded unit disc MATH, centered about MATH such that MATH holds. In this situation we can apply CITE to the family MATH, and the claim is shown. |
math/0103208 | If MATH is any line, then it follows immediately from REF that the kernel of the evaluation map MATH is contained in MATH - for this, note that the the vector MATH is contained in MATH and is perpendicular to any other vector. Recall from CITE that the vector bundle MATH is globally generated. The evaluation map MATH is therefore surjective, and we obtain a vector space morphism MATH . Since MATH is smooth, it is elementary to see that for any tangent vector MATH, the union of the sections MATH gives a section MATH. |
math/0103208 | Since the pull-back MATH intersects MATH-fibers with multiplicity one, and since the section MATH will always vanish on MATH, it is clear that the support MATH must be of the form MATH . The proof is finished if we show that any section MATH with MATH is contained in MATH if and only if the orthogonality holds. That, however, is exactly the statement of REF . |
math/0103208 | By REF , the restriction MATH is injective and the base locus of the linear sub-system MATH is exactly the contracted section MATH. The claim is therefore shown if we note that for every tangent vector MATH, which is not contained in MATH, the section MATH does not vanish on MATH. For this, we refer to REF above. |
math/0103208 | As a first step, we claim that the curve MATH is contained in MATH. CASE: As MATH was chosen generally, we can find a birational morphism MATH with MATH such that the reduced scheme MATH is smooth at MATH and such that tangent map of the restricted universal morphism MATH has maximal rank at MATH. This enables us to find an analytic open subset MATH and a lifting MATH with curves which are tangent to MATH for points MATH. More precisely, we can find an embedded polycylinder MATH such that CASE: The restricted universal morphism MATH induces a morphism MATH. CASE: For every morphism MATH, the curve MATH meets MATH tangentially in MATH. We will now apply REF to sections MATH which come from the deformation family MATH. It is clear from the construction that if MATH is any tangent vector, then MATH . Since MATH is known to be Legendrian, the tangent space MATH will be isotropic with respect to the NAME MATH; the restriction MATH is identically zero: MATH . In particular, the tangent vector MATH is perpendicular to MATH. Applying REF , we see that the tangent space MATH is associated with sections in MATH. This implies that the image MATH must be MATH-integral - see CITE. Since, on the other hand, MATH already contains the Legendrian submanifold MATH, it turns out that MATH is contained in MATH so that the claim MATH follows. CASE: As a next step we consider the strict transform MATH of MATH in the universal family MATH. We will now argue by contradiction, assume that MATH is not a fiber of the map MATH, and derive a contradiction. For this, note that the curve MATH intersects the strict transform MATH of MATH, which is a MATH-fiber, transversally in the preimage of MATH. We claim that this is impossible; a contradiction is thus reached. In order to see this contradiction, observe that MATH intersects the curve MATH with multiplicity one; this is because MATH is general and the morphism MATH is birational. On the other hand, by REF we can always find a vector MATH such that the associated section MATH vanishes on the fiber MATH, but not on the curve MATH. Thus, the curve MATH intersects the divisor MATH tangentially, and the intersection number cannot be one. This ends the proof of REF . |
math/0103208 | Using the notation of above, if MATH is the unique curve containing both MATH and MATH, and if MATH is a curve which contains MATH, then REF shows that MATH and MATH intersect transversally. Since MATH was generically chosen, the claim follows. |
math/0103208 | Since all curves which are associated with points in MATH are smooth, and since MATH is smooth, the preimage MATH is a section over MATH which can be contracted to a point. By definition, MATH is a cone if we can show that CASE: MATH is isomorphic to the image of MATH under the contraction of MATH. CASE: There exists another section MATH which is disjoint from MATH. See the introductory remarks in the paper CITE for more information on this. In order to show REF , recall NAME 's NAME argument CITE which asserts that MATH is finite. Likewise, recall from CITE that MATH is birational. Consequence: the NAME factorization of MATH yields a decomposition as follows: MATH . In order to construct the divisor MATH, choose a tangent vector MATH which is not contained in MATH. By REF , this gives a section MATH. Since MATH intersects the fibers of MATH with mulitplicity one, and since MATH is trivial, it follows that the associated divisor MATH is a section MATH which is disjoint to MATH. |
math/0103212 | REF is proven by CITE; REF, and REF are due to CITE and CITE (see also CITE); REF follows from REF. |
math/0103212 | REF is proven in CITE and CITE. In these papers the base field MATH is assumed to be MATH, but the proofs work for arbitrary algebraically closed field. REF is proven in CITE (it is also a corollary of CITE or CITE). REF follow from REF, and REF. To prove REF one can use the inductive argument in CITE replacing the variety (notation of CITE) MATH (respectively, MATH) by MATH (respectively, a tensor product variety - compare CITE). To prove that the number of MATH-rational points of a tensor product variety depends polynomially on MATH note (compare CITE) that a tensor product variety is a finite union of disjoint subsets each of which is a vector bundle over a fibration with the base equal to a MATH-graded partial flag variety and fibers isomorphic to MATH for some MATH, MATH, MATH (again notation of CITE). Thus it follows from CITE that the number of MATH-rational points of a tensor product variety is given by a computable polynomial in MATH. |
math/0103212 | REF are straightforward. Due to REF the map MATH is well-defined, and due to REF it is surjective. According to CITE there exists a faithful representation of MATH which factors through MATH. Thus MATH is injective. REF follows. |
math/0103218 | Let MATH. We have to show that MATH is finite. First notice that MATH . From MATH we have MATH where we used REF in the last line. |
math/0103218 | We have to show: CASE: MATH CASE: There exists some MATH such that-MATH for all MATH. To see REF , let MATH be given. We know MATH. According to REF we have MATH whenever MATH. To see REF , take MATH. Since MATH equals MATH, we have MATH. We have (from REF): MATH . We can estimate the absolute values of the three summands individually. The first one can be treated analogously to REF, MATH . Very similarly we obtain for the second one MATH and for the third MATH . Since both MATH and MATH are in MATH and MATH, this yields MATH . |
math/0103218 | The elements of MATH with finite MATH-norm form a NAME space with this norm, and MATH is a closed subset of this space. Thus, using REF , the NAME fixed point theorem yields for small enough MATH the existence and uniqueness of an element MATH with MATH. Furthermore, the repeated iteration of MATH with starting point MATH converges to MATH. As long as MATH, the value of MATH has an influence only on the upper bounds for MATH. This proves the proposition. |
math/0103218 | Using REF, both estimates can be easily seen by induction. |
math/0103218 | According to REF it suffices to show that for all MATH . The strategy of the proof is to approximate the discrete distributions MATH by fitting normal densities and use their NAME expansion to obtain the desired bounds. There are several error terms to control. We use REF to obtain an approximation for MATH. For MATH large enough, the lemma yields MATH where MATH is a polynomial of degree four. The coefficients of MATH are rational functions of the moments of MATH up to order four. We now fix MATH as the maximum of MATH and MATH. In particular, together with REF this yields MATH, which we will use later for the error estimates. To simplify the notation, we use the following abbreviations: MATH . We split the left-hand side of REF into several parts, which will be estimated separately: MATH . Before we start to estimate the various sums, we want to state some facts that we will use extensively in this proof. Sums of the form MATH for some real MATH are normally estimated by majorizing them with the appropriate integral MATH. Double sums like MATH are bounded by splitting them in two parts MATH and MATH and treating the halves separately. Another often used inequality yields an upper bound on the discrete folding MATH of two normal densities. We have MATH uniformly for all MATH and MATH. This inequality is proven in REF in the appendix. A last remark worth making is that for positive constants MATH, MATH and MATH with MATH we have MATH for all MATH. This simple fact is obtained by bounding the first factor and the exponential term in MATH separately. Now we come back to, or rather start with, REF. We go from bottom to top and start with bounding REF. A direct consequence of REF is that MATH. So we obtain MATH . In the last step we used the fact, that - in five and more dimensions - the sum MATH is bounded above by MATH, which can easily be seen by splitting the sum. We use REF to see MATH where we used in the last line that MATH (and therefore MATH). Furthermore we have MATH since MATH is a signed measure of bounded steplength. Now we come to REF. Recall the definition of MATH in REF, that is MATH . We analyze the terms in MATH separately, using REF . We write MATH for the polynomial MATH. From REF we obtain MATH . Now we apply first REF and then REF to obtain MATH . We insert REF into REF and obtain MATH with MATH . Using the recursion formula for MATH and the decay rate of MATH (see REF), we obtain MATH . Recall also (see REF ) that MATH implies MATH. More precisely, we have MATH and therefore MATH where we used REF again and the fact that MATH has already been fixed and can thus be bounded by a constant MATH. The error term MATH in REF is given by MATH . We use the local error estimates in REF to estimate the various terms. We have MATH since MATH has bounded steplength. Analogously we can show the same decay for the second term of REF. On the other hand, we bound REF using REF to obtain MATH where we used MATH in the second to last step. Analogously we can show that REF are bounded by MATH and MATH, respectively. We bound REF with REF by MATH where we used MATH to obtain the second to last line. For REF an even better bound without the MATH-term results analogously. Combining the different estimates, we can bound MATH by MATH . Considering this together with REF, we can finally bound REF: MATH . This proves REF . |
math/0103218 | Let MATH with MATH and MATH be given. We want to prove that for small enough MATH we have MATH for some MATH. According to REF, it is sufficient to show that for all natural MATH . Once we know this, we can fix MATH and choose MATH so small that MATH is strictly smaller than one. Since MATH, it suffices to show that MATH . The proof of this estimate is somewhat tedious, but similar to the preceding one. We take the same MATH and we will split the sum in the very same way as before. Again we use some abbreviations to keep the notation as readable as possible: MATH . The left hand side of REF is split in the following parts, which will be estimated separately: MATH . We again start with the last term. First we split MATH into MATH and MATH. We treat the resulting parts of REF separately. For the first part this leads to MATH . For the second part we split the sum and find MATH and finally MATH . Thus MATH is shown. In order to estimate REF, we use REF again, obtaining MATH and therefore MATH . Analogously, we have MATH . The remaining part, REF, will be treated like REF in the proof of the last lemma, using REF . Again we write MATH for the polynomial MATH. From REF we have MATH . For the second sum, we use REF after having replaced MATH by MATH everywhere. We obtain MATH with MATH which results from an argument very similar to that which led to REF. This time the error term MATH in REF is given by MATH . These terms are bounded in exactly the same way as the corresponding ones in REF . So we obtain MATH . Combining these estimates we get MATH as in REF. |
math/0103218 | First recall the space MATH and write MATH for MATH. For the moment, we refer to the sequence MATH as MATH. The following considerations always assume that MATH is small enough. REF yields MATH. From REF we know that MATH is a contraction on MATH. Since MATH is linear, the NAME fixed point theorem now yields the existence of a unique fixed point in MATH. With other words, we have a unique sequence MATH of symmetric measures on MATH with CASE: MATH, CASE: MATH for all MATH, MATH, CASE: MATH. This sequence obviously is the sequence MATH defined in REF. Since as a direct consequence of REF we have MATH the estimate REF now results by choosing MATH small enough. |
math/0103218 | Replace the aperiodic MATH by the periodic MATH everywhere. Instead of using REF, apply a periodic version of REF , namely MATH whenever MATH and MATH have the same parity. The rest of the proof for the aperiodic case carries over word by word (with the constants suitably adapted). |
math/0103218 | The argument uses induction on MATH. As in REF, we define MATH. We freely use the results of appendix MATH. We have MATH and hence MATH too. Now consider MATH: There is only one lace of length two, namely MATH. Therefore we have by REF MATH . On the other hand, we calculate easily (recall REF) MATH . So we have MATH . Since MATH, we obtain MATH . Now we come to the induction step: Fix MATH and assume that MATH for all MATH. Define the truncated sequence MATH by MATH . This sequence satisfies the hypothesis of REF . Thus we obtain a sequence MATH of measures with MATH whenever MATH has the same parity as MATH. As long as MATH is small enough, the positive constants MATH and MATH do not depend on MATH. Defining MATH and using REF as well as the fact that MATH and both are of comparable size, we have MATH where MATH is a positive constant that we fix for the rest of the proof. Since MATH equals MATH for all MATH, we also have MATH for MATH. This can easily be seen by induction. Now we consider MATH. Recall from REF that MATH and therefore MATH . It remains to show that MATH is bounded. We know from the lace expansion formula that MATH where MATH as usual denotes the total mass of MATH. Now we apply REF to the MATH. We obtain MATH and insert this in REF . This leads to MATH . Using this in REF yields MATH . By choosing MATH large enough we see that the lemma follows. |
math/0103218 | Using REF , the second part of the theorem follows directly from REF . In view of REF, the first part is a consequence of REF . |
math/0103218 | The proof of the lemma combines standard large deviation properties with the approximation of MATH obtained by tilting the measure. Let MATH and MATH. Standard large deviation theory (see for example CITE) yields a large deviation principle with entropy function MATH for the laws of MATH. Let MATH denote the convex closure of the set of points with nonzero MATH measure. Then MATH is convex on MATH and even strictly convex on MATH, that is, the interior of MATH. Outside MATH, MATH equals MATH. The function MATH is an analytic diffeomorphism from MATH onto MATH (for a proof see CITE, page REF). Therefore, for any MATH, there exists a unique MATH with MATH. Clearly MATH and MATH. For MATH, we have MATH. Evidently, MATH. Because of symmetry, the odd partial derivatives of MATH vanish at zero. A simple computation yields MATH, and the fourth derivatives at zero depend only on the second and fourth moments of MATH. Now denote by MATH for MATH the tilted measure MATH . Using this, we see that for MATH, we can write MATH . CASE: Since MATH is symmetric and nondegenerate (remember MATH), the boundary of MATH is bounded away from zero. The covariance matrix MATH of MATH is depending analytically on MATH with MATH. It follows that the set MATH of all MATH such that CASE: MATH and CASE: the smallest eigenvalue of MATH is greater or equal MATH, is a compact neighbourhood of zero. Thus for MATH, we have MATH for almost all MATH. It is sufficient to prove the estimate for these MATH, since we can cover the finite number of remaining cases by choosing MATH large enough. So let MATH with MATH. To estimate the first factor in REF, we use NAME expansion for MATH at zero. We obtain MATH where MATH denotes a polynomial containing MATH-th order terms only. The coefficients of the polynomial are rational functions of the moments of MATH up to order four. As a next step we will estimate the second factor in REF, MATH, using a local central limit theorem out of NAME/CITE. REF asserts MATH where MATH are the so called NAME polynomials. They are formal polynomials, consisting of partial derivatives of the normal density in MATH with mean zero and covariance matrix MATH (to keep the notation simple we suppress the MATH-dependance of MATH). For our aims we need the constant implicit in the right hand side of REF to be independent of MATH, which is a priori not guaranteed by CITE. Calculating the constants in the proof of REF , however, shows that they can be choosen such that they only depend on the maximal steplength MATH of MATH and on a lower bound for the smallest eigenvalue of MATH on the other. Therefore the error estimate in REF is uniform on the compact set MATH. Now we come back to the NAME polynomials. In MATH, only derivatives of order MATH, MATH,, MATH appear (see REF ). The coefficients of MATH depend on the moments of MATH up to order MATH. Since MATH, there is only MATH appearing in REF. MATH and MATH vanish at zero, because the odd derivatives of centered normal densities do so. MATH is the centered normal density with covariance matrix MATH itself, so NAME expansion yields MATH. In the NAME expansion of MATH, the odd terms vanish likewise, and we obtain MATH, where the constant MATH and the error term depend only on the moments of MATH up to order four. Therefore REF simplifies to MATH . Inserting REF yields MATH where MATH is a polynomial of degree four with even order terms only. This yields the desired estimate whenever MATH. CASE: For `big' MATH we estimate MATH and MATH separately. Since for fixed natural MATH we have MATH for all MATH, the latter is bounded by MATH. Using MATH in addition, this yields MATH for MATH and for all MATH. Now consider MATH. If MATH, MATH equals zero. If MATH, we bound MATH away from zero with MATH, using NAME expansion: We can do this in a neighborhood of zero with the constant MATH. Since MATH is compact (we even have MATH for all MATH in the boundary of MATH) and MATH convex, we have a nonzero minimum of MATH on MATH and hence we find a constant MATH such that MATH for all MATH. Since MATH is symmetric in each coordinate and rotationally invariant, MATH can be choosen depending only on MATH and the range MATH, but not on the specific law of MATH. If MATH, we use REF to obtain MATH. The arguments leading to REF yield the desired estimate for MATH. If MATH lies in the boundary MATH, use the large deviation principle to obtain MATH for MATH large enough. Using MATH and choosing MATH small enough we obtain MATH. The rest of the argument proceeds as before. Combining the different bounds and using MATH we see that REF holds whenever MATH. |
math/0103218 | Using NAME expansion and the symmetry of MATH we obtain MATH which is REF after inserting MATH. Analogously, first order NAME approximation leads to REF . Now we come to the proof of the error estimates. We write MATH to denote some polynomial of order MATH. The forth partial derivatives of MATH are functions of the form MATH and therefore bounded by MATH. Similarly, the second partial derivatives of MATH are of the form MATH and bounded by MATH. This implies REF. To prove REF, we use the fact that for MATH with MATH we have MATH with MATH depending on MATH, MATH and MATH, but not on MATH. |
math/0103218 | Here we use one dimensional NAME expansion for MATH as a function in MATH to write MATH which implies REF by using MATH. Keeping only the constant term of the NAME approximation leads to REF. To prove the error estimates, we first observe that the second derivative (with respect to MATH) of MATH is of the form MATH and therefore bounded by MATH, while the first derivative of MATH has the form MATH and bounded by MATH. Again, MATH stands for a polynomial of order MATH. Splitting the function MATH and replacing MATH separately by MATH in the first factor and by MATH in the exponential term leads to REF. |
math/0103218 | For MATH we have MATH . The first inequality for general dimension follows immediately from this estimate, because the sum over MATH of the density values equals the MATH-th power of the sum over MATH of the values of the one dimensional normal density with variance MATH. The second inequality follows from the first one by using MATH which comes from the fact that MATH can be bounded uniformly in MATH by MATH. |
math/0103218 | Let MATH, and denote by MATH the shifted cube. Then MATH by NAME 's inequality. Note that for MATH we have MATH, so that MATH because we assumed that MATH. Therefore MATH . |
math/0103218 | It suffices to show that for each path MATH we have (suppressing MATH in the formulas): MATH . Then REF is obtained after insertion of REF into REF followed by factorization of the sum over MATH. To prove REF, we note from REF that the contribution to MATH from all graphs MATH for which MATH is not in an edge is exactly MATH. To resum the contribution from the remaining graphs, we proceed as follows. When MATH does contain an edge ending at MATH, we let MATH denote the largest value of MATH such that the set of edges in MATH with at least one end in the interval MATH forms a connected graph on MATH. We lose nothing by taking MATH, since MATH for all MATH. Then resummation over graphs on MATH gives MATH . With REF this proves REF. |
math/0103218 | Again we abbreviate the notation by writing MATH . We consider the terms MATH separately. For MATH, REF yields MATH since MATH. To keep the notation simple, we set MATH. For MATH we define MATH with the same summation as in REF up to the fact that we allow an additional term for MATH (this corresponds to slashing the line from zero vertically up and is necessary to give the induction below). By REF, we have for all MATH: MATH . Now we show by induction that there is a constant MATH depending only on the dimension MATH such that MATH . For MATH the lace diagram is three-legged: . We have MATH . So together we obtain MATH . Now we come to the induction step. For MATH we will reduce MATH to MATH by merging four subwalks in the lace into two. The following figure illustrates this process: MATH . We use NAME to obtain MATH . Note that for MATH we have MATH and therefore (recall MATH) MATH . Inserting this into REF yields MATH . Using REF with MATH, MATH, MATH, MATH, MATH, MATH and MATH in place of MATH, MATH, MATH, MATH, MATH, MATH and MATH, respectively, gives MATH . Now we choose MATH to be the maximum of the constants appearing in REF. We obtain REF. Now REF together imply MATH . The lemma now follows by inserting REF into REF and choosing MATH small enough. |
math/0103220 | We have: MATH . The second statement follows from MATH and MATH. |
math/0103220 | Suppose MATH is a symmetric MATH-tensor field, which does not vanish at a point in MATH. Using NAME 's theorem, and rescaling MATH and MATH by constants, we may choose a chart MATH, such that MATH and MATH for all MATH, where MATH. Now choose a bump function MATH, such that MATH for MATH and MATH. For MATH we define MATH and MATH. Since the support of MATH is contained in MATH, MATH extends by zero to a compactly supported Hamiltonian vector field on MATH. An easy calculation shows MATH and hence MATH for MATH small enough. |
math/0103220 | As in the proof of REF , we choose a chart MATH, such that MATH and such that MATH on MATH. Take a bump function MATH as above and set MATH . Now define MATH. Then MATH is a compactly supported exact divergence free vector field on MATH and MATH . Again we get MATH and hence MATH for MATH small enough. |
math/0103220 | Recall that the orthogonal complement of MATH in MATH is MATH. The equivalence MATH now follows immediately from REF and the fact that for closed MATH-forms MATH is equivalent to MATH. MATH is obvious from the definition of the curvature MATH and MATH. The integrated NAME equation on MATH-forms, see for example CITE, takes the form MATH and MATH follows. MATH: Suppose MATH. Since MATH it follows from the NAME spectral sequence that the projection MATH induces an isomorphism MATH. So every harmonic MATH-form comes from MATH and hence is parallel. MATH, compare REF and CITE: Suppose MATH is a closed, connected and oriented Riemannian manifold, such that every harmonic MATH-form is parallel. Choose an orthonormal base MATH of harmonic MATH-forms. Since they are parallel they are orthonormal at every point in MATH. Choose a base point MATH, let MATH be the kernel of the NAME MATH and let MATH be the covering of MATH, which has MATH as characteristic subgroup. This is a normal covering, the group of deck transformations is MATH and MATH for smooth functions MATH. Let MATH be a base point in MATH sitting above MATH and assume MATH. Consider the mapping MATH . Obviously this is a proper, surjective submersion and MATH is a compact submanifold. Let MATH. Then the MATH are orthonormal at every point and they are all parallel, especially they commute. Consider MATH . Of course we have MATH, and it follows easily, that MATH is a diffeomorphism. So MATH is closed, connected, oriented and MATH. Moreover MATH is the product metric of the induced metric on MATH and the standard metric on MATH. Every deck transformation of MATH is of the form MATH where MATH and MATH is an orientation preserving isometry of MATH. So MATH is a twisted product, as claimed. To see MATH let MATH be an exact volume preserving vector field, MATH, and let MATH be a harmonic MATH-form. Then MATH where we used MATH to obtain MATH for the second equality. |
math/0103220 | Recall that the orthogonal complement of MATH in MATH is MATH. By REF is equivalent to MATH for all harmonic MATH-forms MATH. On a NAME manifold one has MATH for MATH-forms MATH. Particularly the space of harmonic MATH-forms is MATH-invariant, and so MATH is equivalent to MATH and since harmonic MATH-forms are closed this is equivalent to REF . MATH and MATH are as in the proof of REF . For MATH one needs some extra arguments: One observes, that the span of the MATH constructed in the proof of REF , is MATH-invariant and so is its orthogonal complement. Hence MATH is a complex submanifold and therefore a NAME submanifold. Moreover the complex structure is, locally over MATH, the product structure and so is the symplectic structure as well. MATH follows from the following computation for a function MATH and a closed MATH-form MATH: MATH . For the second equality we used MATH for closed MATH-forms MATH, a relation derived from MATH. |
math/0103220 | Choose a NAME chart centered at MATH, such that MATH and write MATH . Let MATH be a compactly supported function on MATH, such that MATH respectively, MATH locally around MATH. Then the condition MATH shows that MATH, MATH and MATH are all constant locally around MATH. Using MATH one sees, that MATH and MATH for MATH. Using MATH yields MATH. Finally, using MATH shows MATH. So MATH locally around MATH, for some constant MATH. Since MATH is connected this is true globally. |
math/0103220 | Set MATH. Then for every MATH we have MATH . Here MATH has to vanish, since MATH has compact support. So by the previous lemma there exists MATH, such that MATH, that is, MATH. |
math/0103220 | Suppose MATH is bigger than MATH. Then there exists MATH and an open subset MATH, such that MATH, for every all open MATH. For any MATH we choose disjoint subsets MATH. Since MATH . REF yields MATH, such that MATH. But MATH and obviously MATH are linearly independent in MATH. Hence the codimension of MATH in MATH is at least MATH. Since MATH was arbitrary we are done. |
math/0103220 | Suppose conversely such a MATH exists. By REF , MATH has an orthogonal complement in MATH, consisting of Killing vector fields. So this complement has to be finite dimensional, but this contradicts REF . |
math/0103221 | Let MATH be the rectangle given by the definition of NAME boundary. It is easy to check that MATH has a NAME boundary. The conclusion follows now from the Corollary to REF. |
math/0103221 | Let MATH be a NAME sequence in MATH. We can construct an increasing sequence MATH of connected open sets with NAME boundary such that MATH, and MATH. By REF the functions MATH belong to MATH and MATH q.e. on MATH. As MATH for MATH large enough, by the NAME inequality MATH is a NAME sequence in MATH, and therefore it converges strongly in MATH to a function MATH with MATH q.e. on MATH. It is then easy to construct a function MATH such that MATH q.e. on MATH and MATH strongly in MATH for every MATH. As MATH converges strongly in MATH, we conclude that MATH strongly in MATH. Let MATH be a bounded sequence in MATH. As in the previous part of the proof we deduce that MATH is bounded in MATH for every MATH. By a diagonal argument we can prove that there exist a subsequence, still denoted by MATH, and a function MATH such that MATH weakly in MATH for every MATH. Then a sequence of convex combinations of the functions MATH converges to MATH strongly in MATH. This implies MATH q.e. on MATH for every MATH, hence MATH. As MATH is bounded in MATH, we conclude that MATH weakly in MATH. |
math/0103221 | We may assume that MATH and MATH have more than one point, since otherwise the statement is trivial. Let us denote the constant values of MATH on MATH and MATH by MATH and MATH respectively, and let us fix MATH. Since MATH, we may assume that MATH belongs to MATH for some MATH (we use an extension operator if MATH), and that MATH for MATH and MATH. Hence for almost every MATH (the quasi-continuous representative of) MATH takes the values MATH and MATH in two distinct points of MATH. This implies MATH which yields MATH, in contradiction with our assumption. |
math/0103221 | It is clearly enough to prove the statement when MATH. The following concise argument was suggested by NAME. If MATH, MATH, then MATH (the classical proof, see for example, CITE, does not need the hypothesis that MATH is compact). Therefore MATH for every MATH and MATH. This implies that MATH by a standard argument based on the NAME covering lemma (see CITE). |
math/0103221 | Let MATH be the connected components of MATH. As MATH, there exists MATH such that, up to a subsequence, MATH for all MATH. By REF we may also assume that MATH, MATH, MATH in the NAME metric, where MATH are compact and connected. We claim that MATH . Indeed, for every MATH there exists a sequence MATH such that MATH, which implies MATH for some MATH between MATH and MATH. Hence there exists MATH such that MATH for infinitely many indices MATH, and, consequently, MATH. This proves REF , which implies that MATH has at most MATH connected components. By Goa̧b's REF we have MATH for MATH. The conclusion follows now from REF . |
math/0103221 | Given MATH, let MATH. As MATH for MATH large enough, we have MATH. Applying REF with MATH we get MATH . Passing to the limit as MATH we obtain REF . |
math/0103221 | Passing to a subsequence, we may assume, as in the first part of the proof of REF , that there exists a constant MATH such that every MATH has exactly MATH connected components MATH and MATH, MATH, MATH in the NAME metric, where MATH are compact and connected and MATH. As MATH is connected, there exists a finite family of indices MATH, with MATH, such that MATH for MATH. Let us fix a point MATH. By the convergence in the NAME metric there exist MATH and MATH such that MATH and MATH as MATH. Since MATH has a NAME boundary, there exist arcs MATH and MATH in MATH, connecting MATH to MATH and MATH to MATH respectively, such that MATH and MATH as MATH. Let us define MATH . It is clear that MATH in the NAME metric and that MATH. Since MATH we conclude that MATH is connected. |
math/0103221 | It is clear that MATH, since otherwise MATH has exactly one connected component. Since MATH is locally connected (see, for example, CITE), and MATH is open in MATH, the connected components MATH of MATH are open in MATH. Since each MATH is closed in MATH, we have MATH. If MATH, then MATH would contain an open, closed, and non-empty proper subset (recall that MATH), which contradicts the fact that MATH is connected. Therefore MATH. As MATH, we conclude that MATH. Therefore MATH for every connected component MATH of MATH. For MATH let MATH be the union of MATH and of all the connected components MATH of MATH such that MATH. To prove that MATH is open in MATH, we fix a sequence MATH in MATH which converges to a point MATH. If MATH for infinitely many indices MATH, then MATH, hence MATH. If there exists a connected component MATH of MATH such that MATH and MATH for infinitely many indices MATH, then MATH; this implies MATH, since MATH for every connected component MATH of MATH with MATH. In the other cases there exists a sequence MATH of pairwise disjoint connected components of MATH, with MATH, such that, up to a subsequence, MATH. As MATH, by REF MATH, hence MATH, which gives MATH, so that MATH also in this case. Therefore MATH is open in MATH. Since MATH for every connected component MATH of MATH, we have MATH. As MATH is connected, there exists a finite family of indices MATH, with MATH, such that MATH and MATH for MATH. As MATH, there exists a connected component MATH of MATH such that MATH and, consequently, MATH. |
math/0103221 | If MATH, we just define MATH and notice that MATH for MATH large enough. Assume now MATH and let MATH, MATH, be its connected components. If MATH we set MATH. If MATH, by REF there exist a finite family of indices MATH, with MATH, and a family MATH of connected components of MATH, such that MATH for MATH; in this case we set MATH . In both cases we want to construct a sequence MATH in MATH which converges to MATH in the NAME metric and such that MATH and MATH . Let us fix MATH such that the sets MATH, MATH, are pairwise disjoint, and let MATH . It is easy to see that MATH and MATH for MATH large enough, and that MATH converges to MATH in the NAME metric as MATH. By REF there exists a sequence MATH in MATH such that MATH in the NAME metric, MATH, and MATH. If MATH we define MATH. If MATH, for every MATH we fix two points MATH and MATH. By the convergence in the NAME metric there exist MATH and MATH such that MATH and MATH as MATH. Since MATH has a NAME boundary, there exist arcs MATH and MATH in MATH, connecting MATH to MATH and MATH to MATH respectively, such that MATH and MATH as MATH. Let us define MATH . In both cases MATH and MATH it is clear that MATH in the NAME metric and that REF holds, since by REF MATH. As MATH for MATH, and MATH for MATH, we conclude that MATH is connected in both cases. As the connected components MATH of MATH are connected components of MATH, the argument given at the beginning of the proof of REF shows that each MATH is open in MATH and satisfies MATH. Since MATH is separable, the connected components of MATH form a finite or countable sequence MATH. For every MATH we fix a point MATH. As MATH in the NAME metric, there exists MATH such that MATH as MATH. Since MATH has a NAME boundary, for every MATH there exists an arc MATH in MATH, connecting MATH to MATH, such that MATH as MATH. If there are infinitely many connected components MATH, there exists a sequence of integers MATH tending to MATH such that MATH . If there are MATH connected components MATH, REF is true with MATH for every MATH. Let MATH . Then the sets MATH are connected, contain MATH, and converge to MATH in the NAME metric. As MATH by REF , we have MATH which, together with REF , yields MATH . The opposite inequality for the lower limit follows from REF . |
math/0103221 | Let MATH, MATH, be the connected components of MATH. Let us fix MATH such that the sets MATH, MATH, are pairwise disjoint, and let MATH . It is easy to see that MATH and MATH for MATH large enough, and that MATH converges to MATH in the NAME metric as MATH. By REF there exists a sequence MATH in MATH such that MATH in the NAME metric, MATH, and MATH. It suffices now to take MATH. |
math/0103221 | Let MATH be a connected component of MATH and let MATH. Given MATH, let MATH and let MATH be the connected component of MATH containing MATH. For MATH large enough we have MATH. If the boundary of MATH meets MATH, let MATH be the relative interior of MATH in MATH. Since MATH meets MATH, for MATH small enough we have MATH. As MATH q.e. on MATH, we apply REF and deduce that there exists a function MATH, with MATH q.e. on MATH, such that, up to a subsequence, MATH weakly in MATH. Since MATH is arbitrary and MATH, we can construct MATH, with MATH q.e. on MATH, such that MATH weakly in MATH for every open set MATH. If the boundary of MATH does not meet MATH, passing to a subsequence, we can still assume that MATH converges weakly in MATH to some function MATH. Since the space MATH is closed in MATH, we conclude that there exists MATH such that MATH a.e. in MATH, and, as in the previous case, we can construct MATH such that MATH weakly in MATH for every open set MATH. Therefore we have constructed MATH, with MATH q.e. on MATH, such that MATH weakly in MATH for every open set MATH. Assume now that MATH and let MATH. For every MATH there exists MATH such that MATH for MATH. Let MATH be an open set such that MATH. As MATH for MATH large enough, we have also MATH. Then MATH where MATH is an upper bound for MATH and MATH. From the previous part of the lemma MATH and the conclusion follows from the arbitrariness of MATH. |
math/0103221 | If MATH with MATH, we have MATH where the first equality follows from our convention MATH a.e. in MATH, while the second equality follows from REF , since MATH on MATH. Equality REF implies that MATH in MATH, hence MATH in MATH. As MATH is simply connected and has a NAME boundary, there exists MATH such that MATH a.e. in MATH. Since MATH a.e. in MATH, the function MATH is constant q.e. on each connected open subset MATH of MATH, and, by REF , also on MATH. To prove that MATH is constant q.e. on each connected component of MATH we use an approximation argument. We write MATH as the intersection of a decreasing sequence MATH of compact subsets of MATH such that MATH for every MATH, where MATH denotes the interior of MATH in the relative topology of MATH. Note that MATH satisfies MATH since every such function MATH can be extended to a function of MATH by setting MATH in MATH. As MATH, there exists a solution MATH to the problem MATH . Using MATH as test function in REF , we obtain that the norms MATH are uniformly bounded. By REF , there exists MATH, with MATH q.e. on MATH, such that, up to a subsequence, MATH converges to MATH weakly in MATH. Taking MATH as test function in REF , we get MATH . Passing to the limit we obtain that MATH converges to MATH, hence MATH converges to MATH strongly in MATH. Let us prove that MATH . By the uniqueness of the gradients of the solutions, it is enough to prove that MATH is a solution of REF . Let MATH with MATH q.e. on MATH. As MATH and MATH q.e. on MATH, we can use MATH as test function in REF . Then passing to the limit as MATH we obtain REF , and the proof of REF is complete. By the first part of the proof, there exist a function MATH, such that MATH a.e. on MATH. Let MATH be a connected component of MATH. It is easy to see that there exists a connected component MATH of the interior of MATH such that MATH (this is trivial if MATH, and follows from the regularity of MATH if MATH meets MATH). As MATH is constant q.e. on MATH, we obtain that MATH is constant q.e. on MATH. We may assume that MATH for every MATH. Since MATH a.e. on MATH we deduce that MATH converges to MATH strongly in MATH, and by the NAME inequality MATH converges strongly in MATH to a function MATH which satisfies MATH a.e. on MATH. As MATH is constant q.e. on MATH, we conclude that MATH is constant q.e. on MATH. To prove that MATH is constant q.e. on each connected component of MATH, it is enough to show that MATH is constant q.e. on MATH whenever MATH is a rectangle as in the definition of the NAME part of the boundary and MATH. Let MATH be the vector-field defined by MATH a.e. in MATH and MATH a.e. in MATH. As at the beginning of the proof, it is easy to see that MATH in MATH, hence MATH in MATH. Then there exists a function MATH such that MATH a.e in MATH. As MATH a.e. in the connected set MATH, using REF we obtain that MATH is constant q.e. in MATH. As MATH a.e. in the connected set MATH, using REF we obtain that MATH is constant q.e. in MATH. From these facts we deduce that MATH is constant q.e. on MATH. |
math/0103221 | By a standard localization argument, it is enough to prove that for every MATH there exists an open neighbourhood MATH of MATH in MATH such that MATH . For every MATH let MATH be the neighbourhood given in the statement of the theorem. Taking, if necessary, a smaller neighbourhood, we may assume that MATH has a NAME boundary and that MATH is constant q.e. on the closure of each connected component of MATH and of MATH. Let MATH be an arbitrary open neighbourhood of MATH in MATH with MATH. Since MATH is locally connected, the connected components of MATH are open in MATH, so that only a finite number of them meets MATH. Similarly, only a finite number of connected components of MATH meets MATH. Using REF it is easy to prove that there exist a finite family MATH of pairwise disjoint compact sets and a family of distinct constants MATH such that MATH and MATH q.e. on MATH for MATH. We now apply CITE (to a suitable extension of MATH) and construct a sequence of functions MATH, converging to MATH in MATH, such that MATH in a neighbourhood MATH of each MATH. Let MATH, with compact support in MATH, such that MATH q.e. on MATH, let MATH be functions in MATH, with MATH, such that MATH in a neighbourhood of MATH, and let MATH. By REF the function MATH belongs to MATH, and by CITE it belongs to MATH. Since MATH where MATH, we have MATH where the last equality follows from the fact that MATH in MATH and MATH. Passing to the limit as MATH, we obtain MATH showing that MATH is a solution of REF . |
math/0103221 | Let us fix an open ball MATH containing MATH. Using the same extension operator we can construct extensions of MATH and MATH, still denoted by MATH and MATH, such that MATH, MATH weakly in MATH. Given an open set MATH, any function MATH will be extended to a function MATH by setting MATH q.e. in MATH. By CITE we have MATH . Since the complement of MATH has two connected components, from the results of CITE and CITE we deduce that, for every MATH, the solutions MATH of the NAME problems MATH converge strongly in MATH to the solution MATH of the NAME problem MATH . This implies (see, for example, CITE) that, in the space MATH, the subspaces MATH converge to the subspace MATH in the sense of NAME (see CITE). Since MATH by REF , and MATH weakly in MATH, from the convergence in the sense of NAME we deduce that MATH, hence MATH q.e. on MATH by REF . |
math/0103221 | Note that MATH is a minimum point of REF if and only if MATH satisfies REF ; analogously, MATH is a minimum point of REF if and only if MATH satisfies REF with MATH and MATH replaced by MATH and MATH. Taking MATH as test function in the equation satisfied by MATH, we prove that the sequence MATH is bounded in MATH. By REF , there exists a function MATH, with MATH q.e. on MATH, such that, passing to a subsequence, MATH weakly in MATH. We will prove that MATH . As the limit does not depend on the subsequence, this implies that the whole sequence MATH converges to MATH weakly in MATH. Taking again MATH and MATH as test functions in the equations satisfied by MATH and MATH, we obtain MATH . As MATH weakly in MATH and MATH strongly in MATH, from the previous equalities we obtain that MATH converges to MATH, which implies the strong convergence of the gradients in MATH. By the uniqueness of the gradients of the solutions, to prove REF it is enough to show that MATH is a solution of REF . This will be obtained by using REF . First of all we note that MATH by REF , and therefore MATH is locally connected (see, for example, CITE). Let us fix MATH and an open rectangle MATH containing MATH. If MATH, we assume that MATH. If MATH, we assume that MATH is as in the definition of the NAME part of the boundary. Let MATH be an open neighbourhood of MATH in MATH such that MATH. We will prove that there exists a function MATH, with MATH a.e. in MATH, such that MATH is constant q.e. on each connected component of MATH and of MATH. By REF this implies that MATH satisfies REF . Let MATH. Let us prove that there are at most MATH connected components MATH of MATH which meet MATH. Indeed, if MATH meets also MATH, then MATH (since MATH connects a point in MATH with a point in MATH), so that the number of these components can not exceed MATH. On the other hand, it is easy to see that the other connected components of MATH are also connected components of MATH, thus their number can not exceed MATH. Let MATH be the connected components of MATH which meet MATH. As MATH, passing to a subsequence we may assume that MATH for every MATH, and that MATH, MATH, MATH in the NAME metric, where MATH are compact and connected. Arguing as in the proof of REF , we obtain that MATH . Let MATH be the harmonic conjugate of MATH in MATH given by REF . Then MATH a.e. in MATH. We may assume that MATH for every MATH. Since MATH a.e. on MATH, we deduce that MATH converges to MATH weakly in MATH, and by the NAME inequality MATH converges weakly in MATH to a function MATH which satisfies MATH a.e. on MATH. Let us prove that for every MATH there exists a constant MATH such that MATH q.e. on MATH. This is trivial when MATH reduces to one point. If MATH has more than one point, then MATH; since the sets MATH are connected, we obtain also MATH. As MATH q.e. on MATH for suitable constants MATH, using the NAME inequality (see, for example, CITE) it follows that MATH is bounded in MATH, hence the sequence MATH is bounded, and therefore, passing to a subsequence, we may assume that MATH for a suitable constant MATH. Then MATH converges to MATH weakly in MATH, and by REF we conclude that MATH q.e. on MATH. By REF , if MATH, then MATH is constant q.e. on MATH. By REF this implies that MATH is constant q.e. on each connected component of MATH. On the other hand, every MATH is constant q.e. on each connected component of MATH. Since MATH weakly in MATH, a sequence of convex combinations of the functions MATH converges to MATH strongly in MATH, and we conclude that MATH is constant q.e. on each connected component of MATH, hence on each connected component of MATH. Therefore MATH satisfies all hypotheses of REF , which implies that MATH is a solution of REF . |
math/0103221 | For MATH, consider the functions MATH defined by MATH, with the convention that MATH. Then the functions MATH are NAME continuous with constant MATH for every MATH, and the functions MATH are non-increasing for every MATH. Let MATH be a countable dense subset of MATH. For every MATH there exists a countable set MATH such that MATH are continuous at every point of MATH. By REF we have MATH and MATH for every MATH and every MATH, MATH with MATH. This implies that MATH for every MATH and every MATH. Let MATH be the countable set defined by MATH, and let MATH. Then MATH for every MATH, and, by continuity, for every MATH, which yields MATH. This proves that MATH. |
math/0103221 | It is clear that MATH and MATH are increasing and satisfy REF . Therefore MATH is at most countable by REF . Let us fix MATH and a sequence MATH in MATH converging to MATH. By the Compactness REF we may assume that MATH converges in the NAME metric to a set MATH. For every MATH, MATH, with MATH, we have MATH for MATH large enough, hence MATH. As MATH is closed this implies MATH, therefore MATH by REF and by the definition of MATH. |
math/0103221 | Let MATH be a countable dense subset of MATH. Using a diagonal argument, we find a subsequence, still denoted by MATH, and an increasing function MATH, such that MATH in the NAME metric for every MATH. Let MATH and MATH be the increasing functions defined by MATH where MATH denotes the closure. Let MATH be the set of points MATH such that MATH. As MATH and MATH satisfy REF , by REF the set MATH is at most countable. Since MATH for every MATH, we have MATH for every MATH. For every MATH we define MATH. To prove that MATH for a given MATH, by the Compactness REF we may assume that MATH converges in the NAME metric to a set MATH. For every MATH, MATH, with MATH, by monotonicity we have MATH. As MATH is closed, this implies MATH, therefore MATH by the definitions of MATH and MATH. Since MATH is at most countable, by a diagonal argument we find a further subsequence, still denoted by MATH, and a function MATH, such that MATH in the NAME metric for every MATH. Therefore MATH for every MATH, and this implies that MATH is increasing on MATH. |
math/0103221 | Since MATH is a solution of REF which satisfies the boundary REF , by linearity for every MATH we have MATH a.e. in MATH, hence MATH where the last equality is deduced from REF . Dividing by MATH and letting MATH tend to MATH we obtain REF . The continuity of MATH implies that MATH is of class MATH. |
math/0103221 | It is enough to apply REF . |
math/0103221 | For every MATH we have MATH . The conclusion follows dividing by MATH and taking the limit as MATH. |
math/0103221 | Let MATH be defined by MATH. By REF MATH is continuous in MATH for a.e. MATH and every MATH. By REF MATH . The equivalence between REF is now obvious. |
math/0103221 | By REF. Assume by induction that MATH and let MATH be a constant such that MATH. Consider a minimizing sequence MATH of REF . We may assume that MATH for every MATH. By the Compactness REF , passing to a subsequence, we may assume that MATH converges in the NAME metric to some compact set MATH containing MATH. For every MATH let MATH be a solution of the minimum REF which defines MATH. By REF MATH converges strongly in MATH to MATH, where MATH is a solution of the minimum REF which defines MATH. By REF we have MATH and MATH, hence MATH. As MATH, we conclude that MATH. Since MATH is a minimizing sequence, this proves that MATH is a solution of the minimum REF . |
math/0103221 | Let us fix an integer MATH with MATH. From the absolute continuity of MATH we have MATH where the integral is a NAME integral for functions with values in MATH. This implies that MATH where the integral is a NAME integral for functions with values in MATH. As MATH and MATH q.e. on MATH, we have MATH . By the minimality of MATH and by REF we have MATH . From REF , and REF we obtain MATH where MATH by the absolute continuity of the integral. Iterating now this inequality for MATH we get REF with MATH. |
math/0103221 | As MATH is admissible for REF which defines MATH, by the minimality of MATH we have MATH, hence MATH for every MATH. As MATH is absolutely continuous with values in MATH the function MATH is integrable on MATH and there exists a constant MATH such that MATH for every MATH. This implies the former inequality in REF . The latter inequality follows now from REF and from the inequality MATH, which is an obvious consequence of REF for MATH. |
math/0103221 | By REF there exists an increasing function MATH such that, for every MATH, MATH converges to MATH in the NAME metric as MATH along a suitable sequence independent of MATH. By REF we have MATH for every MATH and every MATH. By REF this implies MATH for every MATH. |
math/0103221 | As MATH is a solution of the minimum REF which defines MATH, and MATH strongly in MATH, the conclusion follows from REF . |
math/0103221 | Let us fix MATH and MATH with MATH. Since MATH converges to MATH in the NAME metric as MATH, by REF there exists a sequence MATH in MATH, converging to MATH in the NAME metric, such that MATH and MATH as MATH. By REF this implies that MATH is bounded as MATH. Let MATH and MATH be solutions of the minimum REF which define MATH and MATH, respectively. By REF MATH strongly in MATH. The minimality of MATH expressed by REF gives MATH, which implies MATH. Passing to the limit as MATH and using REF we get MATH. Adding MATH to both sides we obtain REF . A similar proof holds for REF . By REF we have MATH, which implies MATH. Passing to the limit as MATH and using REF we obtain REF . |
math/0103221 | Let us fix MATH with MATH. Given MATH let MATH and MATH be the integers such that MATH and MATH. Let us define MATH and MATH. Applying REF we obtain MATH with MATH converging to zero as MATH. By REF for every MATH we have MATH strongly in MATH as MATH, and by REF we have MATH for every MATH. By REF we get MATH . Passing now to the limit in REF as MATH we obtain REF . |
math/0103221 | Let MATH. From the previous lemma we get MATH . On the other hand, by REF we have MATH, and by REF MATH where MATH is a solution of the minimum REF which defines MATH. Therefore MATH . Since there exists a constant MATH such that MATH and MATH for MATH, from REF we obtain MATH which proves that the function MATH is absolutely continuous. As MATH strongly in MATH when MATH, if we divide REF by MATH, and take the limit as MATH we obtain REF . Equality REF follows from REF . |
math/0103221 | Let us fix MATH, with MATH, and MATH, with MATH. For MATH we have MATH, and from REF we obtain MATH. As the functions MATH and MATH are continuous, passing to the limit as MATH we get REF . |
math/0103221 | Let MATH. Since MATH in the NAME metric as MATH, and MATH by REF , it follows that MATH as MATH by REF . As the function MATH is continuous, we obtain REF . The proof of REF is analogous. |
math/0103221 | Let us fix MATH. For every MATH let MATH be a solution of the minimum REF which defines MATH. As MATH and MATH, from REF we obtain, adding and subtracting MATH, MATH . As MATH is a solution of REF with MATH, and MATH q.e. on MATH, we have MATH. By the monotonicity of MATH, for MATH we have MATH and MATH q.e. on MATH. By the minimum property of MATH we obtain MATH for MATH. Therefore MATH which concludes the proof. |
math/0103221 | Let MATH and MATH be the connected components of MATH. Since MATH and MATH have a NAME boundary, by REF MATH belongs to MATH and MATH. This implies that MATH, and hence MATH. The conclusion follows now from CITE, as shown in CITE. |
math/0103221 | It is enough to adapt the proof of CITE. |
math/0103221 | Let MATH with MATH, let MATH be a solution of the minimum REF which defines MATH, and let MATH be the function defined by MATH on MATH and by MATH on MATH. As MATH q.e. on MATH the function MATH belongs to MATH; using also the fact that MATH q.e. on MATH, we obtain that MATH q.e. on MATH. Therefore MATH . On the other hand, by the minimality of MATH, MATH where the last inequality follows from REF , since MATH has no more than MATH connected components (indeed, MATH has exactly MATH connected components which do not meet MATH, and every connected component of MATH which meets MATH contains a connected component of MATH, so that their number does not exceed MATH). From REF we obtain MATH and the minimality of MATH yields REF . |
math/0103221 | Let MATH be an arbitrary point in MATH and let MATH, MATH, be a family of open balls centred at the points MATH. If the radii are sufficiently small, we have MATH and MATH for MATH. Moreover we may assume that MATH, for suitable constants MATH with MATH, and that the arcs MATH intersect MATH only at the points MATH and MATH, with a transversal intersection. All these properties, together with REF , imply that MATH . In particular this happens for MATH, and for MATH close to MATH if MATH is continuous at MATH. By REF and by REF for every MATH we have that MATH . By REF this implies, taking MATH, MATH which yields MATH . REF follows now from REF applied with MATH. By REF for a.e. in MATH we have MATH. Moreover, for a.e. in MATH the derivative MATH exists for MATH. Let us fix MATH which satisfies all these properties. By REF for MATH close to MATH we have MATH where MATH and MATH. Note that the equality holds in REF for MATH. As the functions MATH and MATH are differentiable at MATH (by REF and by the existence of MATH), we conclude that MATH . By REF we have MATH for MATH, so that the previous equalities yield REF . |
math/0103223 | For MATH, MATH. As MATH is rationally elliptic, the total rank of MATH is finite. By REF , the rank of MATH is also finite, and MATH generates MATH by REF . |
math/0103223 | Let MATH be the morphism of NAME spectral sequences induced by the NAME morphism CITE. By CITE, if MATH, then MATH is generated as an algebra by MATH. But MATH, so MATH. We may identify MATH, so MATH. The result follows by NAME 's Lemma. |
math/0103223 | Let MATH be the submodule of primitive elements. The inclusion extends to a NAME algebra morphism MATH. Recall that MATH is an isomorphism if and only if MATH is an isomorphism for all prime ideals MATH in MATH CITE. Furthermore, each prime ideal MATH in MATH is of the form MATH for some prime MATH, and MATH. Therefore, without loss of generality, we may assume that MATH is a differential NAME algebra over MATH, MATH. To begin, we construct a subalgebra MATH of MATH such that MATH. Recall REF that MATH. Let MATH be a basis of MATH. Since MATH is surjective, each MATH for some MATH. Let MATH. There exists MATH such that MATH whenever MATH. It follows from REF that MATH where MATH has order MATH. Let MATH be a cycle whose homology class is MATH. Let MATH be the free MATH-module on the basis MATH. Define a linear map MATH by MATH. Then MATH extends to a differential MATH-morphism MATH, and MATH. Since MATH is injective and MATH, MATH are MATH-free as modules, MATH is injective. Let MATH be reduction mod MATH. Passing to homology, it is a routine exercise to verify that MATH and MATH. Suppose MATH for some MATH. Then MATH, so MATH for some MATH. Thus MATH, so MATH is a torsion element of MATH. It follows that MATH, whence MATH. So MATH for some MATH, which is again sent to a torsion element of MATH by MATH. Repeat the argument; eventually the process ends since MATH has no infinitely MATH-divisible elements. Therefore MATH and so MATH is injective. Let MATH be the torsion ideal of MATH. Since MATH, the morphism MATH factors to define MATH. Let MATH be the composite MATH. One verifies that MATH. From the definitions, MATH is the identity on MATH. Thus MATH. Since MATH is degree-wise finitely generated, NAME 's Lemma states that MATH. Therefore MATH is surjective and hence an isomorphism of algebras. By REF , MATH is a differential MATH-Hopf algebra. We claim that the diagonal MATH in MATH induces a MATH-Hopf algebra structure on MATH. Indeed, since MATH is a MATH-morphism, MATH. A MATH-morphism MATH is defined by MATH. Since MATH is a MATH-morphism, so too is MATH. Furthermore, MATH inherits coassociativity and the counit from MATH, and so MATH provides MATH with a MATH-Hopf algebra structure. By definition, MATH commutes with MATH ``up to torsion". That is, given MATH, there exist MATH, and MATH such that MATH and MATH. We now define a pairing MATH that exhibits MATH as the NAME algebra dual of the MATH-Hopf algebra MATH. By REF , the proof will then be complete. Use MATH to identify MATH as a differential MATH-subalgebra of MATH to obtain a pairing MATH, where MATH is the differential on MATH. Let MATH. Since the differential in MATH vanishes, MATH. Furthermore, MATH is a domain, hence MATH. Therefore MATH factors to define REF . Observe that REF is non-singular. Indeed, as a chain map, MATH splits, and MATH induces an isomorphism MATH. It follows that MATH, where MATH and MATH. Dually, MATH, and MATH, where MATH is the torsion submodule. Therefore MATH as a module, and one can check that the duality is provided by REF . Since MATH is an algebra morphism, MATH for MATH and MATH. Pass to REF to see that the multiplication in MATH is dual to the diagonal MATH. Suppose that MATH are cycles, and MATH. Using REF , we see that MATH, where MATH for some MATH. Since MATH is a boundary, MATH and MATH are cycles, and MATH is an integral domain, it follows that MATH. Pass to REF to conclude that MATH is dual to the multiplication in MATH. |
math/0103223 | Let MATH be the reduced diagonal on MATH, and let MATH be the NAME morphism. Recall that MATH is an isomorphism. Since MATH, MATH, and so MATH. Observe that MATH and MATH. Furthermore, MATH is the canonical inclusion MATH. As MATH and MATH are torsion-free, the result now follows. |
math/0103223 | By universal coefficients, the natural NAME algebra morphism MATH, defined by MATH for MATH and cycles MATH, is an isomorphism. Let MATH be the quotient morphism. Since MATH is an isomorphism, we get an isomorphism of NAME algebras MATH defined by MATH. By REF and the mapping property of universal enveloping algebras, MATH. It follows that MATH restricts to an isomorphism of NAME algebras MATH. |
math/0103223 | Let MATH be the NAME homomorphism. The inclusion MATH induces the commutative diagram MATH in which the right arrow is a NAME algebra morphism, and the other arrows preserve brackets. The kernels of the vertical arrows are precisely the respective torsion submodules. Therefore the diagram MATH commutes. The left vertical arrow, MATH, and MATH are NAME algebra morphisms. The remaining arrows are NAME algebra morphisms. The vertical arrows are injections, MATH is an isomorphism onto MATH by REF , and the right vertical arrow is of the form MATH. It follows that MATH is an injection and that MATH is torsion. |
math/0103223 | CASE: Set MATH. Then MATH is a MATH-Hopf algebra over MATH. Let MATH be a splitting of the quotient map. By CITE, MATH extends to an isomorphism of MATH-algebras MATH. Let MATH be a basis of MATH indexed over the positive integers. Choose a set of representatives MATH in MATH, and let MATH be the linear span. The inclusion MATH extends to a MATH-morphism MATH. The reduction mod MATH, MATH, is an isomorphism. In particular, MATH is surjective, so by NAME 's Lemma, MATH. It follows that MATH is surjective. Since MATH and MATH are MATH-free and finite type, and MATH is injective, MATH is itself injective and therefore an isomorphism. CASE: To begin, we claim that MATH. From the definitions, MATH. It suffices to show that if MATH, then MATH for all MATH and MATH. Since MATH and MATH, MATH. As MATH is a domain, the claim follows. To show that the canonical algebra morphism MATH is surjective, we use an argument of NAME. As a module, MATH. For MATH, define MATH by MATH and MATH. By definition, each MATH. Let MATH be the set of finite sequences of non-negative integers MATH such that MATH or MATH if MATH is odd, and the last term in MATH is non-zero. Define MATH to be the length of MATH (so MATH). Order MATH by setting MATH if MATH or MATH and MATH. Set MATH and MATH where MATH. Since MATH is an algebra morphism, MATH (exponents with respect to composition). As MATH, MATH is a MATH-derivation; that is, MATH. It follows that MATH, and MATH if MATH. By REF , MATH, while MATH if MATH and MATH. If MATH and MATH, then of course MATH for degree reasons. The set MATH forms an additive basis for MATH. The above formulae for the MATH imply that MATH is a basis for MATH. In particular, the set MATH generates MATH as an algebra, and so MATH is surjective. Let MATH denote the NAME series of the non-negatively graded MATH-space MATH. We extend the definition to a non-negatively graded, free MATH-module MATH: MATH. Note that MATH. Since MATH is surjective, MATH. If MATH, then MATH, so MATH and the natural morphism MATH is an injection. It follows that MATH, so in fact the NAME series are equal. Therefore MATH is an isomorphism. |
math/0103223 | Straightforward. |
math/0103223 | Suppose MATH is a differential MATH-Hopf algebra. As usual, denote by MATH the sub-Lie algebra of primitive elements. The natural NAME algebra morphism MATH is an isomorphism if and only if MATH is an isomorphism for every prime ideal MATH of MATH, and that each prime ideal MATH is of the form MATH for some prime MATH. Furthermore, observe that MATH. Therefore, without loss of generality we may assume that MATH for some odd prime MATH. By REF , MATH as a NAME algebra. Since the differential in MATH is a NAME derivation, it preserves MATH. Thus MATH forms a differential NAME algebra, establishing the `if' assertion. Conversely, suppose MATH. By REF , it suffices to show that MATH is a MATH-algebra. In CITE, NAME constructed a chain isomorphism MATH, where MATH is a MATH-derivation. By construction, MATH factors as MATH for some differential NAME morphism MATH. By the proof of CITE, MATH is a coalgebra morphism. Since MATH is also a coalgebra morphism, MATH is an isomorphism of chain coalgebras. NAME provides MATH with the required MATH-algebra structure. |
math/0103224 | The infimum in the definition of thickness either is achieved for some distinct points MATH, MATH, MATH, or is approached along the diagonal when MATH, say, approaches MATH, giving us MATH. But the first case cannot happen unless the second does as well: consider the sphere of radius MATH with MATH, MATH and MATH on its equator, and relabel the points if necessary so that MATH and MATH are not antipodal. Since this MATH is infimal, MATH must be tangent to the sphere at MATH. Thus MATH, and we see that MATH. This infimum, in turn, is achieved either for some MATH, or in a limit as MATH (when it is the infimal radius of curvature). In the first case, we can check that MATH and MATH must be antipodal points on a sphere of radius MATH, with MATH tangent to the sphere at both points. That means, by definition, that MATH is a doubly critical self-distance for MATH. |
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