paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003159 | Suppose that MATH is an element of minimal length such that MATH and let MATH. We prove the lemma by induction on MATH. The case MATH is trivial. Initial step: MATH. If MATH then we have MATH. Hence MATH, MATH and MATH. So MATH by REF . Inductive step: if MATH then MATH. Let MATH with MATH and MATH, MATH, MATH. By indu... |
math/0003159 | Without loss of generality we can assume MATH. First step: MATH. This is clear since otherwise MATH. Second step: MATH and MATH for all MATH such that MATH. Let MATH and observe that MATH and MATH. Then as in first step we have MATH from which the claim follows. Let now MATH. If MATH then there exists MATH and MATH suc... |
math/0003159 | By REF there exists a closed orbit MATH in MATH with trivial stabilizer. Then by REF there exists MATH such that MATH and by REF MATH is surjective. Now the surjectivity follows by homogeneity. The local triviality over MATH follows also from REF . |
math/0003159 | We prove that the map MATH is surjective. By REF this follows by the following two identities: MATH for each MATH-path MATH and for each admissible path MATH. |
math/0003159 | It's enough to prove that MATH is surjective. Let MATH and consider the sequence (see REF for the notation) : MATH . Since MATH and MATH we have that MATH is not surjective or that MATH is not injective. Suppose that MATH is not surjective, then up to the action of MATH we can assume that MATH. Then, for MATH consider ... |
math/0003159 | We prove this proposition by induction on the order MATH on MATH. First step: MATH. If MATH we can take MATH and MATH. Inductive step. If MATH is not dominant then there exists MATH such that MATH. If MATH we observe that MATH (that is MATH and MATH) and that MATH and so we can apply the inductive hypothesis. If MATH w... |
math/0003159 | Observe that MATH so each irreducible component of MATH must have dimension at least MATH. Suppose now that MATH is dominant. By NAME 's theorem REF MATH is not empty. Observe also that by REF MATH is a smooth subset of MATH of dimension MATH. It is well known that MATH. Hence MATH . By the lemma above we have that if ... |
math/0003162 | CASE: The condition MATH reads MATH this identity is proved for example, in CITE. CASE: Let MATH denote the angle function of MATH and MATH, defined by MATH; we then have MATH . Let MATH and MATH be the NAME forms of MATH, MATH; from REF applied to MATH, we infer MATH; similarly, we have MATH; putting together these tw... |
math/0003162 | Every self-dual NAME Hermitian REF-manifold MATH of non-constant curvature is conformally related (via REF ) to a self-dual NAME metric MATH of nowhere vanishing scalar curvature. A self-dual NAME metric is locally-symmetric if and only if its scalar curvature is constant CITE; thus, the one direction in the correspond... |
math/0003162 | MATH. By REF , MATH has two distinct, non-constant eigenvalues at any point and there exists a positive, non-Kähler Hermitian structure MATH whose NAME form MATH generates the eigenspace of MATH corresponding to the simple eigenvalue. It follows that the Hermitian structure is preserved by the action of MATH, and there... |
math/0003162 | We first observe that REF can be equivalently written as MATH this shows that MATH is function of MATH, say MATH; from the above equality, we get REF , where MATH is a (still unknown) smooth function; in order to determine MATH, we differentiate REF by using REF and substitute into REF ; then, cancellations occur and R... |
math/0003162 | By using REF , the right-hand side of MATH is easily computed; we thus obtain: MATH . Since MATH is self-dual and NAME, MATH, see REF . Then, by projecting REF to MATH, we get REF , whereas the projection of REF to MATH gives REF . |
math/0003162 | If we assume that MATH somewhere on MATH and that the anti-self-dual NAME tensor is identically zero, then, after contracting REF with MATH, we obtain MATH, which contradicts MATH. |
math/0003162 | MATH. According to REF , if the spectrum of MATH is everywhere degenerate, then either MATH vanishes identically (and therefore the hyperhermitian structure is closed by REF ) or MATH has two distinct eigenvalues MATH and MATH at any point. MATH. By REF , we know that a normalized generator MATH of the MATH-eigenspace ... |
math/0003162 | We first show that if MATH admits a non-closed hyperhermitian structure compatible with the negative orientation, then the corresponding NAME form MATH must be one of the forms MATH and MATH given in REF . From REF and the fact that MATH is a MATH-eigenform of MATH, we infer MATH . By differentiating REF and by using R... |
math/0003162 | Any such manifold is either a space of constant curvature, hence conformally flat, or a NAME manifold of constant holomorphic sectional curvature (see REF ). In the former case, the claim follows by REF , whereas in the latter case MATH; we then conclude by using REF . |
math/0003162 | Since MATH is Killing, equality REF MATH holds. For a quaternionic NAME manifold the curvature operator MATH acts on MATH by MATH, where MATH is a positive multiple of the scalar curvature, compare for example, CITE. Thus, projecting REF to MATH we get MATH . |
math/0003162 | We denote MATH any complex MATH -vector field of MATH and MATH the corresponding horizontal lift (considered as complex vector in MATH); then, MATH is integrable if and only if the following identity holds: MATH by the very definition of MATH we have MATH; moreover, since MATH belongs to MATH on MATH, the almost comple... |
math/0003169 | The hermitian metric MATH on MATH is defined by the integration along the fiber: for MATH and MATH, MATH . By CITE, the curvature form of MATH is NAME semipositive. Hence the smooth metric MATH on MATH induced from MATH has semipositive curvature form as well. Moreover, it is extended to a singular hermitian metric MAT... |
math/0003169 | Let MATH be the minimal positive number such that MATH. We take a rational function MATH on MATH such that MATH. Let MATH be the normalization of MATH in the field MATH, and let MATH be a desingularization such that the composite morphism MATH is smooth over MATH. We have MATH . The NAME group MATH acts on MATH such th... |
math/0003169 | Let MATH be a log resolution for the pair MATH, and set MATH. The coefficients of MATH are less than MATH and negative coefficients appear only for exceptional divisors of MATH. We set MATH where MATH is an effective integral divisor and MATH is a MATH-divisor whose coefficients belong to the interval MATH. Since the s... |
math/0003169 | There exists an effective MATH-divisor MATH such that MATH is KLT and MATH. |
math/0003169 | Let MATH be the log resolution of the pair MATH. We have MATH for some MATH-divisor MATH. Then our assertion is proved in REF . |
math/0003169 | By the NAME lemma, there exists an effecive MATH-divisor MATH such that MATH is a MATH-divisor, the pair MATH is KLT, and that MATH is ample. Therefore, we may assume that MATH is a MATH-divisor and that MATH is ample. By the Base Point Free Theorem, there exists a proper surjective morphism MATH with connected fibers ... |
math/0003169 | We may assume that MATH by REF . Let MATH be the minimal resolution of singularities. Since MATH is effective, we can write MATH with MATH being KLT. Therefore, we may assume that MATH is smooth. By REF again, we may also assume that MATH is ample and that MATH is big. Assume first that MATH. Then the assertions follow... |
math/0003169 | By a crepant blowings-up, we may assume that MATH has only terminal singularities. Then we have MATH by CITE, and MATH is pseudo-effective by CITE (see also CITE). By the NAME theorem, we calculate MATH . |
math/0003169 | Assume that MATH is not LC. Let MATH be the LC threshold for MATH so that MATH and MATH is properly LC. Let MATH be a minimal center. By REF , for any positive rational number MATH, there exists an effective MATH-divisor MATH on MATH such that MATH and MATH is KLT. By the perturbation technique, we may assume that MATH... |
math/0003169 | REF is proved in REF . We recall the proof for the convenience of the reader. We set MATH, MATH, MATH, and MATH for MATH. Since MATH and MATH, we can write MATH for some numbers MATH. Hence MATH . On the other hand, we have MATH . Therefore MATH . CASE: We may assume that MATH is ample by the base point free theorem. A... |
math/0003169 | We shall prove REF simultaneously. Let MATH be the smallest positive integer such that MATH. We shall derive a contradiction from MATH. We take a general member MATH. Assume first that MATH is PLT and MATH. Then MATH is NAME canonical. We have an exact sequence MATH . We have MATH and MATH. On the other hand, we have M... |
math/0003179 | For REF , see CITE. For REF , suppose that MATH, then MATH as MATH and MATH by REF . Then, by the MATH-adic criterion CITE, MATH would be a MATH-order, a contradiction. For REF , see CITE. |
math/0003179 | MATH . Since the genus of a non - singular plane curve of degree MATH is MATH, REF follows from CITE. MATH . This is well known property of the Hermitian curve; see for example, CITE or CITE. MATH . If MATH, then from MATH and REF , either MATH or MATH. By hypothesis, MATH can only occur, and so, by REF , MATH is MATH-... |
math/0003179 | The statement on the genus follows immediately from the upper bound on MATH. By REF we have that MATH. If MATH, then MATH and from REF MATH is MATH-Frobenius classical. In particular, REF holds true; that is, we have MATH. It is easy to see that MATH for MATH and that MATH. Moreover, MATH and MATH, and the result follo... |
math/0003179 | If REF holds, then MATH is classical by REF . For MATH, the hypothesis on MATH yields MATH and hence MATH. Thus MATH is classical by REF . Note that the following computations will provide another proof of this fact. For the rest of the proof we assume MATH to be non-classical, and we show that no one of REF ,, REF hol... |
math/0003179 | REF follows from REF . To prove REF , we first note that MATH (compare REF ), and that MATH is a power of MATH (see REF ). Now, with the same notation as in the proof of the previous proposition, we get MATH with MATH. So MATH. Furthermore, MATH and so MATH by REF . Hence MATH and REF follows. |
math/0003179 | Let MATH be the ramification divisor associated to MATH and suppose that MATH for each MATH. Then from CITE, MATH which is a contradiction as MATH and MATH. |
math/0003179 | By some computations we obtain that MATH is bigger than MATH in REF . So the curve MATH is classical for MATH. Let MATH be as in REF . Then the MATH-orders are MATH and MATH with MATH (compare CITE). Therefore, the MATH-orders are MATH and MATH with MATH. Since MATH, from REF , MATH, and hence MATH by the MATH-adic cri... |
math/0003179 | Suppose that MATH. By means of some computations, MATH and hence REF holds true. With the same notation as in the proof of that lemma, we can then use the following two facts: MATH, and MATH. Actually, we will improve the latter one. MATH is a power of MATH . Indeed, by MATH and the MATH-adic criterion CITE, a necessar... |
math/0003179 | Let MATH and MATH. Then the image of the morphism MATH is the curve defined by MATH. This proves the lemma. |
math/0003179 | If REF holds, it is clear that MATH is MATH-covered by the Hermitian curve. This property extends to MATH via the previous lemma. For both curves, the MATH-maximality now follows from CITE. |
math/0003179 | Let MATH, MATH, and MATH. Then MATH and MATH so that MATH . This shows that MATH is contained in the NAME semigroup MATH at MATH. In particular, MATH. Since MATH (see CITE), the result follows. |
math/0003179 | If REF is satisfied, the result follows from REF . Now if MATH is MATH-maximal, then MATH is also MATH-maximal by REF and CITE. Then the corollary follows from REF . |
math/0003179 | REF follows from the identity MATH and REF . To show REF , it is enough to check that MATH. Recall that the NAME symbol MATH is defined by: MATH . In our case, since MATH, MATH. By the quadratic reciprocity low MATH from MATH and MATH we get MATH. Now, as MATH, we have that MATH. In other words, MATH viewed as an eleme... |
math/0003179 | The curve MATH is MATH-covered by MATH via the morphism MATH, where MATH and MATH. |
math/0003179 | Let MATH. It is not difficult to see that MATH and MATH. Hence, for MATH, MATH and hence MATH provided that MATH and MATH. Let MATH denote the set introduced in REF . Then MATH, and it is easily checked that MATH is a semigroup. By means of some computations we see that MATH, whence MATH follows. |
math/0003179 | The ``if" part follows from REF and here we do not use the hypothesis that MATH is prime. For the ``only if" part, we first notice that each MATH is MATH-rational. Now the case MATH and MATH in the proof of REF gives MATH. Therefore MATH because MATH CITE. As MATH and MATH is prime, the result follows. |
math/0003179 | Similar to the proof of REF . |
math/0003180 | Since MATH does not divide MATH, the equation has MATH solutions in MATH. If MATH is a solution, then MATH, as MATH, and hence MATH. |
math/0003180 | Suppose that MATH is not complete. Then there exists MATH such that for any MATH-rational line MATH through MATH, MATH . If MATH is a line such that REF holds and that MATH, then MATH . Let MATH. Then, as MATH is MATH-rational, MATH and thus, as MATH, MATH is the tangent line of MATH at MATH (see REF ). Therefore MATH ... |
math/0003185 | We construct a functor MATH from the first category into the second category and investigate its properties. Let MATH be a NAME module over MATH. Suppose, MATH is already MATH-linearly and continuously embedded into a certain standard NAME module MATH for some NAME space MATH. Consider the intersection MATH of MATH wit... |
math/0003185 | Every isometric copy of MATH as a MATH-submodule of another NAME MATH-module MATH is an orthogonal summand of MATH since MATH is self-dual by assumption, compare CITE. In case MATH for some NAME space MATH we have the isometric algebraic embedding MATH, where MATH denotes the algebraic tensor product. Let us fix the st... |
math/0003185 | The equivalence of REF has been shown for countably generated NAME MATH-modules in CITE. The implication MATH can be shown to hold setting MATH for the existing unitary operator MATH and for MATH. The demonstration of the inverse implication requires slightly more work. For the given normalized tight (modular) frames M... |
math/0003185 | The set of algebraic generators MATH of MATH is a frame with respect to any MATH-valued inner product on MATH which turns MATH into a NAME MATH-module. That is the inequality MATH is satisfied for two finite positive real constants MATH and any MATH, see CITE. What is more, for any frame of MATH there exists another MA... |
math/0003185 | Since both the modular NAME bases are sets of modular generators of MATH we obtain two MATH-valued (rectangular, without loss of generality) matrices MATH and MATH with MATH and MATH such that MATH . Combining these two sets of equalities in both the possible ways we obtain MATH for MATH, MATH. Now, since we deal with ... |
math/0003187 | For a definition of MATH-equivalence and NAME forms, see for example CITE. It is well-known that MATH and MATH are MATH-equivalent iff they have isometric NAME forms; for an algebraic-topological proof of that combine NAME and NAME, CITE, or alternatively NAME, CITE. Thus REF is equivalent to REF . If MATH is obtained ... |
math/0003187 | The action will be defined in terms of MATH-moves. Let us recall surgery on a clasper with no trivalent vertices, a so-called MATH-move: MATH . A MATH-move can be thought of as a right-handed finger move, or a right-handed NAME twist, and has an inverse given by a left-handed finger move (indicated by a stroke in the o... |
math/0003187 | If MATH is an oriented edge, let MATH denote the pair of flags such that MATH has the same orientation as MATH and opposite from MATH. This gives rise to a map MATH and MATH, which is an isomorphism. |
math/0003187 | It follows immediately using REF above. |
math/0003187 | We now claim that the proof of REF works without change, and proves REF . Indeed, REF work for MATH-null claspers as stated. Furthermore, if MATH is a MATH-null clasper, then each leaf of MATH lies in the commutator group MATH, where MATH. Since MATH, this implies that each leaf of MATH bounds a surface in MATH whose b... |
math/0003187 | Using REF , we will rather describe a canonical isomorphism MATH. Recall the exact sequence MATH where MATH and MATH are the abelian groups of MATH-valued functions on the vertices and oriented edges of MATH. Let MATH. This gives rise to the element MATH where the product is taken over all edges of MATH. It is easy to ... |
math/0003187 | Our Sliding Lemma implies that an edge of MATH can slide past two arcs of MATH with opposite orientations. It can also slide past another edge. It is easy to see that this implies our result. |
math/0003187 | Since MATH, it suffices to consider the case of even MATH. Recalling REF , suppose that MATH (for MATH) is a collection of claspers in MATH each of degree MATH, and let MATH denote the set of arms of MATH. REF and its following discussion implies that MATH and that the nonzero contribution to the right hand side come f... |
math/0003187 | It suffices to consider a collection MATH of claspers in MATH each of degree MATH. Let MATH denote the set of arms of MATH. The counting argument of the above Proposition shows that the contributions to MATH come from complete contractions of a disjoint union MATH of MATH vortices. A vortex is the diagram MATH, the nex... |
math/0003187 | Since MATH is an unlink and MATH is the disjoint union of an unknot MATH and the link MATH, it suffices to consider the case of a hairy strut MATH. In this case, MATH is a meridian of MATH. The formula for the NAME integral of the NAME Link of CITE (applied to MATH), together with the fact that MATH has linking number ... |
math/0003187 | We have that MATH . In other words, the diagrams that contribute in MATH are those whose components either lie in MATH or are hairy struts of the shape MATH. Because of the restriction of the linking matrix REF, the pair of legs MATH of a hairy strut of the above shape must be glued to a pair of legs labeled by MATH as... |
math/0003187 | It is easy to see that linking numbers satisfy the above axioms. For Uniqueness, assume that MATH is another such function, and let MATH. consider a link MATH, and a surface MATH that bounds MATH. Using the Cutting Property, as in the proof of REF , it follows that MATH is a linear combination of MATH and MATH, and hen... |
math/0003187 | The Symmetry, Cutting and MATH-Sliding Properties follows immediately from REF . The Specialization Property follows from the well-known fact that given a covering space MATH and a cycle MATH in MATH that lifts to MATH in MATH, and a cycle MATH in MATH, then the intersection of MATH with MATH equals to the intersection... |
math/0003187 | It will be more convenient to present the proof in the algebra MATH. Let MATH denote a change of scaling of a string-link MATH, as shown in the following figure in case MATH has three stands: MATH . The locality of the NAME integral implies that MATH for an associator MATH. Write MATH for an element MATH, where MATH is... |
math/0003187 | Consider the special link MATH, (this is an abbreviation for MATH) whose connected sum of the first two components gives MATH. How does the NAME integral of MATH determine that of MATH? The answer, though a bit complicated, is known by CITE. Following that notation, we have MATH where MATH is the operation that glues a... |
math/0003187 | Consider the link MATH. Recall that a slide move is given by: In an artistic way, the next figure shows the result of a slide move (compare also with REF ) that replaces MATH by MATH, where MATH is the orientation reversed knot. MATH . As in the previous lemma, we have that MATH where MATH is the operation that replace... |
math/0003187 | This follows from REF once we show that MATH satisfies the Symmetry, Specialization, MATH-Sliding, Cutting and Initial Condition stated in that lemma. The symmetry follows by REF . Specialization follows from the fact that MATH coincides with the coefficient of a strut in the NAME integral, which equals to MATH. The MA... |
math/0003189 | MATH . Let us recall three different moves on the set MATH of REF knots in MATH: CASE: Changing a crossing, that is, doing a MATH-modification in the language of CITE and CITE. CASE: Doing a MATH-move, that is, doing a MATH-modification in the language of CITE. CASE: Doing a null-move, in the language of CITE. These th... |
math/0003189 | MATH . Notice that MATH and that MATH. Thus, MATH, where MATH and MATH is a constant. In order to figure out MATH, we need a computation. Let MATH be a wheel with two legs attached to an unknot MATH. Consider the MATH-link of degree MATH (also denoted by MATH) in the complement of MATH. Let MATH for MATH and MATH denot... |
math/0003190 | Let MATH, let MATH be homogeneous and let MATH. By NAME commutator formula, MATH . Then the first part follows immediately. Since MATH is a submodule of MATH, we have MATH. It is easy to see that MATH. This completes the proof. |
math/0003190 | We shall prove the first part by induction on MATH. From the definition of MATH, the lemma is true for MATH. Assume it is true for any MATH homogeneous vectors in MATH. Now let MATH be homogeneous and let MATH with MATH . Set MATH . Since MATH, in REF , we may take MATH. Let MATH be any nonnegative integer such that MA... |
math/0003190 | Let MATH be the set defined by REF . We shall prove MATH . Let MATH and let MATH be homogeneous. Then for any MATH, MATH because REF MATH . This proves REF . Since MATH for MATH, REF implies MATH . By changing variable we get MATH . By REF , MATH and by REF MATH . Consequently, MATH . That is, MATH . Conversely, let MA... |
math/0003190 | Recall the conjugation REF of CITE: MATH . Because MATH is locally nilpotent on MATH, we may set MATH, so that we have MATH . Then the first part of the proposition follows immediately. By changing variable MATH we get MATH . This completes the proof. |
math/0003190 | From REF we easily get MATH . Using this and the fact that MATH, we get MATH . This proves REF . The second part follows from REF immediately. |
math/0003190 | First, using REF we get (CITE, REF ): MATH . Because MATH for MATH, MATH . Since for any homogeneous MATH, MATH REF , we have MATH noting that MATH for MATH. Using REF and all the above information we have MATH . Similarly, using the fact MATH we have MATH . This completes the proof. |
math/0003190 | Let MATH. For homogeneous MATH we set MATH . Then extend the definition by linearity. It follows from REF that MATH gives rise to a MATH-structure on MATH . In particular, MATH . The following arguments are classical and routine in nature. Let MATH be homogeneous and let MATH. For any MATH, because MATH in view of REF ... |
math/0003190 | In view of REF we only need to prove MATH for MATH, MATH. Let us assume MATH is homogeneous. First, from the proof of REF we have MATH . Then using the fact MATH we get MATH . Using REF , we get MATH . (This argument is also similar to one in the proof REF .) |
math/0003191 | Write MATH and MATH. Then MATH and MATH. The result follows. |
math/0003191 | The NAME theorem tells us that MATH if and only if there exists a nonzero continuous linear functional on MATH which vanishes on MATH. The result now follows from REF . |
math/0003191 | Let MATH such that MATH and define a nonzero function MATH by MATH. For MATH, set MATH. Then MATH . Conversely suppose there exists MATH such that MATH. This means that MATH for all MATH, for all MATH except on a set MATH of measure zero. Also MATH for all MATH except on a set MATH of measure zero. Since MATH, we may c... |
math/0003191 | Define a NAME space isomorphism MATH by MATH for MATH. We want to show that this isomorphism commutes with the action of MATH. Clearly it will be sufficient to show that MATH commutes with the action of MATH. If MATH, then MATH . Thus the action of MATH commutes with MATH. We deduce that for MATH, there exists MATH suc... |
math/0003191 | We have MATH. By the above remarks, MATH is a sum of elements of the form MATH. Therefore we need only prove that MATH. But MATH if MATH and REF if MATH, and the result follows. |
math/0003191 | Let MATH. It was shown in CITE that MATH. Write MATH. If MATH and MATH, then by NAME 's inequality CITE applied to the function MATH for MATH, MATH consequently MATH . Therefore MATH is a continuous map from MATH into MATH for MATH. It is also continuous for MATH. The lemma follows because the map MATH is continuous; s... |
math/0003191 | Let MATH and write MATH. Then MATH . Applying this in the case MATH, we obtain MATH. Using REF , we deduce that if MATH and MATH, then MATH. Since MATH, the result now follows from REF . |
math/0003191 | If MATH, then MATH and MATH by REF . Conversely suppose there exists MATH such that MATH. Then MATH for some MATH, so replacing MATH with MATH, we may assume that MATH. If MATH, then MATH and MATH. Using REF we see that MATH, and we deduce from REF that MATH for all MATH. It follows that MATH for all MATH, consequently... |
math/0003191 | Since MATH, we see from REF that MATH. Of course MATH. We now prove the stronger statement that MATH for all MATH. We have MATH . Therefore MATH and the result follows. |
math/0003191 | We shall use induction on MATH, so assume that the result is true with MATH in place of MATH. Let MATH denote the NAME graph of MATH with respect to the generators MATH. Thus the vertices of MATH are the elements of MATH, and MATH are joined by an edge if and only if MATH for some MATH. Also MATH acts by left multiplic... |
math/0003191 | This follows immediately from REF : replace MATH with MATH for all MATH. |
math/0003191 | The above elements generate the subgroup generated by MATH . The result follows from REF . |
math/0003191 | Let MATH and let MATH be the free group on MATH. By REF there is a monomorphism MATH determined by the formula MATH . Note that MATH induces a NAME space monomorphism MATH. Set MATH. Since MATH is a MATH-zero divisor by REF , we see that MATH is a MATH-zero divisor, say MATH where MATH. Write MATH where MATH is a right... |
math/0003191 | Let MATH such that MATH and suppose MATH. Since MATH is a positive definite function by CITE, we can apply CITE to deduce that MATH. By REF MATH and the result is proven. |
math/0003192 | Given MATH and MATH, let MATH be the result of varying MATH by multiplying the MATH coordinate by MATH and adjusting the last coordinate so as to remain on MATH (that is, MATH). Differentiating the relation MATH implicitly with respect to MATH at REF yields MATH . By minimality of MATH, we know that the modulus of MATH... |
math/0003192 | For MATH, let MATH be the torus MATH shrunk in the last coordinate by MATH, that is, the set of MATH for which MATH, MATH and MATH. Write NAME 's formula as an iterated integral MATH . Here MATH is the circle of radius MATH. Let MATH be a compact set not containing MATH. For each fixed MATH, the function MATH has radiu... |
math/0003192 | The first statement is immediate. To prove the second, let MATH and see from the definition of MATH that MATH . By definition of MATH, the ratio MATH is some constant MATH independent of MATH, hence MATH . The right hand side of this is the derivative of MATH with respect to MATH at MATH. By definition of MATH this van... |
math/0003192 | Differentiate the equation MATH to get MATH which is the same as REF . Differentiate again to get MATH and use REF to eliminate MATH, giving REF . The formula for MATH follows from the definitions of MATH and of the partial derivative. |
math/0003192 | Changing variables to MATH, the integral becomes MATH the curve MATH is the image of MATH under MATH, so MATH and MATH remains in the right half plane, strictly except at REF. For MATH write MATH as MATH, where MATH is a polynomial of degree MATH and MATH is bounded; this can be done since MATH may be approximated by a... |
math/0003192 | The two-sided integral is the sum of two one-sided integrals on intervals MATH and MATH. The integral over MATH may be written as an integral over MATH of the function MATH. With MATH still denoting the coefficients resulting form the application of REF to the first integral, let MATH denote the coefficients when REF i... |
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