paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0008015 | Consider REF for MATH and MATH in REF respectively. Then the roots of the indicial equations of REF are MATH and MATH at both MATH and MATH. By REF, the log-term coefficients at MATH and at MATH both vanish if and only if MATH. By REF , the corresponding MATH-parameter family of MATH immersions consists of immersions t... |
math/0008015 | For MATH and MATH in REF becomes MATH . By REF , it is enough to show that there exists data MATH such that the difference of the roots of the indicial equation of REF at MATH is an integer and the log-term vanishes. The coefficients of REF expand as MATH for MATH sufficiently close to MATH. Assume the roots MATH, MATH... |
math/0008015 | It is enough to show that MATH is a solution of REF. In fact, by REF, MATH because MATH. |
math/0008015 | By REF, MATH because MATH. |
math/0008015 | Since MATH, MATH for MATH and then MATH REF is well-defined. Hence, by REF, MATH . This completes the proof. |
math/0008016 | Write MATH REF with respect to the decomposition REF. Since MATH, MATH is contained in the NAME algebra MATH of MATH. Hence MATH (MATH). This implies that MATH . The second assertion MATH follows from the well-known formula MATH. The last assertion follows from the fact MATH for MATH. |
math/0008016 | Let MATH be a local coordinate on MATH. Then conformality is equivalent to MATH . Since MATH acts isometrically on MATH as in REF is rewritten as MATH . By REF, it is equivalent to the condition MATH. On the other hand, by differentiating MATH, we have MATH . Hence REF is written as MATH which proves REF. Next we prove... |
math/0008016 | By the conformality, the induced metric is MATH. The proof is similar to that of REF . |
math/0008016 | By the assumption of integrability, there exists a solution MATH of the differential equation MATH for an initial value MATH. Since MATH and MATH, the mapping MATH has the first fundamental form MATH, by REF , and thus it is a conformal immersion. Let MATH be a solution of the differential equation MATH under another i... |
math/0008016 | We set MATH, so it satisfies the integrability REF . Thus by REF , there exists a conformal immersion MATH such that MATH, where MATH is a solution of the ordinary differential equation of MATH. Since MATH is holomorphic, MATH is also holomorphic. Now we have MATH which implies the NAME map MATH of MATH is holomorphic.... |
math/0008016 | Suppose that MATH is a compact NAME surface and MATH a conformal immersion with holomorphic right NAME map. Take a holomorphic lift MATH of MATH. We may assume that MATH is a subgroup of MATH, and MATH is a subset in MATH, by NAME embedding. Then the trace of MATH satisfies MATH where MATH is a complex coordinate of MA... |
math/0008016 | In fact, we have MATH which proves the first assertion. Let MATH be a covering transformation of the universal covering MATH of MATH. Since MATH also induces MATH, there exists MATH such that MATH (see REF ). Hence MATH . This shows that MATH is single-valued on MATH. |
math/0008016 | By REF, we have MATH . Since MATH is single-valued, this implies MATH is single-valued. |
math/0008016 | Suppose that MATH is of finite total curvature (respectively, finite dual total curvature). Since the metric induced by MATH (respectively, MATH) is locally isometric to the metric of a minimal surface in Euclidean space, it has non-positive Gaussian curvature on MATH. So MATH admits a complete metric of non-positive c... |
math/0008016 | Since MATH, it suffices to show the one direction. Let MATH be a path tending to MATH in MATH. Under the assumption that the length MATH of MATH with respect to MATH is finite, it is sufficient to prove MATH. Denoting by MATH one of the lifts of MATH to the universal cover MATH of MATH, we can see from the completeness... |
math/0008016 | Let MATH be a holomorphic lift of MATH and MATH. Using the basis REF of MATH, we write MATH, and define a MATH-valued MATH-form MATH on MATH. Let MATH be a conformal minimal immersion defined by MATH . Then the NAME map of MATH is, by definition, MATH where MATH is the compactification of MATH (see REF ). Since MATH is... |
math/0008016 | Suppose that the monodromy group is included in MATH. We shall make a contradiction. CASE: First, we shall describe the monodromy matrix. Write MATH and consider a system of ordinary differential equations MATH for MATH. Take a base point MATH. Let MATH REF such that the column vectors of MATH are a fundamental system ... |
math/0008016 | Since MATH is a subgroup of MATH, the NAME model of MATH is included in the set of Hermitian matrices. So each MATH is Hermitian: MATH . Since MATH is identity, we have MATH . Differentiating both sides of REF with respect to MATH at MATH, we have MATH . It follows that MATH . Using the fact that MATH is Hermitian, we ... |
math/0008016 | Let us take an arbitrary constant vector MATH. For a complex parameter MATH, we define MATH and MATH, MATH, inductively by MATH where MATH is the identity matrix. The right-hand side of REF implies that the MATH's are MATH-valued rational functions in the variable MATH, with a regular point MATH. We consider a formal p... |
math/0008016 | Let MATH be the first column of MATH and MATH the second column. Then we have MATH . So we have already seen in REF that MATH is a solution. Again, we consider the formal power series REF with coefficients REF. Differentiating REF with respect to MATH, we have MATH where we put MATH . In particular, MATH satisfies MATH... |
math/0008021 | REF is an evolution equation for the maps MATH, with initial condition MATH. As MATH is compact and everything is real analytic, the existence of a unique solution for MATH in MATH for some MATH follows from standard techniques in partial differential equations. For instance, one can prove it by applying the NAME - NAM... |
math/0008021 | Let MATH, and let MATH be the induced vector fields on MATH. Then, by definition of the moment map MATH, we see that MATH where MATH is the NAME derivative. As MATH is a MATH-orbit, the vector fields MATH for MATH are tangent to MATH, and generate its whole tangent space. Thus, MATH if and only if MATH on MATH for all ... |
math/0008021 | Let MATH. Then MATH is a closed REF-form on MATH, so there exists a smooth function MATH, unique up to addition of a constant, with MATH. Since MATH and MATH is tangent to MATH we have MATH, so MATH is constant on MATH, as MATH is connected. As the functions MATH are defined up to addition of a constant, we can choose ... |
math/0008021 | REF shows that MATH, and as MATH is a MATH-orbit it is a real analytic submanifold of MATH, with dimension MATH by assumption. Therefore REF shows that there exists a locally unique SL submanifold MATH in MATH containing MATH. But MATH is MATH-invariant and MATH, so MATH must also be (locally) MATH-invariant, by local ... |
math/0008021 | One can prove this quite simply by relating the tangent spaces MATH of MATH to those of MATH, and showing that each MATH is special Lagrangian. But we will instead give a proof using REF . Now MATH is a compact, nonsingular Riemannian MATH-manifold, with a natural orientation. Let MATH be the unique positive section of... |
math/0008021 | As by REF are not all zero with MATH and MATH, we see that MATH and MATH, so the interval MATH is well-defined. Since MATH for MATH, we have MATH and MATH. Therefore MATH is confined to MATH. Now MATH, so MATH has a turning point at REF. As the roots of MATH are MATH, they are all real, and there are none in MATH. So R... |
math/0008021 | From the proof of REF we know that MATH strictly increases from REF to REF in MATH, and so as MATH there is a unique MATH with MATH. Similarly, there is a unique MATH with MATH. Clearly MATH for MATH. Also, if MATH lies in MATH then MATH if and only if MATH. But we know from REF that if MATH are solutions to REF - REF ... |
math/0008021 | The only way for solutions of REF - REF to become singular is for some MATH to become zero, or for MATH. As neither of these can happen by REF , the solutions must exist for all MATH. Uniqueness of the solutions, with the given initial data, follows from standard results in differential equations. Since MATH we have MA... |
math/0008021 | Let MATH, for MATH positive integers with MATH. Then as MATH, MATH and MATH, we have MATH, MATH and MATH. Now REF of the SL MATH-fold MATH constructed in REF actually depends only on MATH and MATH rather than on MATH and MATH, as MATH are integers. Thus, adding an integer multiple of MATH to MATH makes no difference to... |
math/0008021 | It is obvious from REF and the definition of MATH and MATH that MATH is real analytic. As MATH, we have MATH and MATH. Also, as MATH the factor MATH in REF tends to zero, except near MATH and MATH. Hence, as MATH, the integrand in REF gets large near MATH and MATH, and very close to zero in between. So to understand MA... |
math/0008021 | From REF we see that MATH . Now MATH is conformal if and only if MATH and MATH are orthogonal and the same length. But MATH as MATH, so they are orthogonal. Also MATH . One can then prove from REF and MATH that MATH, and thus MATH is conformal. Now MATH is a cone in MATH, and is minimal because any calibrated submanifo... |
math/0008021 | We shall show that for each nonsingular point MATH in MATH, the tangent plane MATH is special Lagrangian. But as MATH is MATH-invariant and MATH, it is enough to verify this for one point in each orbit of MATH in MATH. Thus we can restrict our attention to MATH. Then the condition for MATH to be a nonsingular point of ... |
math/0008021 | For simplicity, we first suppose that MATH lies in the subgroup MATH of MATH, and treat MATH as a vector space rather than an affine space. Then MATH is an abelian NAME subalgebra of MATH, which we may regard as a vector space of commuting matrices. By standard results in linear algebra, we may decompose the complex ve... |
math/0008024 | By the definition of MATH, if MATH is a line connecting MATH and MATH, then MATH is a line connecting MATH and MATH. Therefore both MATH and MATH are contained in the plane MATH. The commutativity of the action of MATH and MATH is a direct consequence of the definition of MATH and MATH. |
math/0008024 | By REF , MATH is the maximal quotient on which the NAME involution acts as the MATH-multiplication. On the other hand, the NAME involution acts on MATH as the MATH-multiplication and we obtain the corollary. |
math/0008024 | Since the action of MATH is compatible, MATH is a homomorphism as MATH modules. We will prove that this is actually an isomorphism. Let MATH be the quotient of MATH by the involution MATH. Then by the NAME theorem, the genus MATH and MATH of MATH and MATH are MATH and MATH, respectively. Since the rank of MATH is MATH,... |
math/0008024 | By the definition of MATH and MATH, we have MATH . Since the involution MATH induces the MATH-multiplication on MATH we have MATH . REF yields the proposition. |
math/0008024 | Since MATH, it is enough to prove the injectivity of the map MATH. If MATH and MATH are collinear, then we showed that MATH and MATH are vertical to each other, therefore the images are different. If MATH and MATH are not collinear, then we may assume that MATH for MATH and MATH if MATH. Since two lines MATH and MATH a... |
math/0008024 | The automorphism of MATH induced by MATH (respectively, MATH) is the reflection with respect to the element MATH (respectively, MATH). By the definition of the isomorphism between MATH and MATH, MATH (respectively, MATH) is the reflection with respect to MATH (respectively, MATH). Using the compatibility of MATH and th... |
math/0008024 | Since MATH is equal to the representation matrix of MATH for the normalized REF forms MATH, we have MATH. By applying the transformation formula in REF for MATH in REF, we have MATH . By evaluating this equality at MATH, and MATH, we get MATH. Thus we get the proposition. |
math/0008024 | If we replace MATH by MATH with MATH, then MATH and MATH are replaced by MATH and MATH respectively, where MATH. By applying the quasi-periodicity of theta functions: MATH for theta constants to MATH, MATH, MATH and MATH, we have MATH . Since the entries of MATH are even, MATH is independent of the choice of MATH. |
math/0008024 | CASE: We choose MATH as a lifting of MATH. For a vector MATH, we have MATH for theta constants. Therefore we have MATH . We use the quasi-periodicity for theta constants. The equality MATH yields MATH. CASE: It is enough to prove that MATH is invariant under the action of generators of MATH. The group MATH is generated... |
math/0008024 | It is enough to prove the proposition for MATH. Note that if MATH (respectively, MATH) then MATH (respectively, MATH). Therefore we can check that the proposition by the definition of MATH. |
math/0008024 | In this proof, for two non zero functions MATH we write MATH if there exists a REF-th root of unity MATH such that MATH. To prove the equality REF , it enough to show this on the open set of MATH corresponding to cubic surfaces with no NAME points. We use the normal form of REF points as in REF; MATH . Since MATH for M... |
math/0008024 | By applying MATH to the equality MATH, we have the equality MATH. Therefore we have MATH . Thus we have the equality MATH, where MATH, MATH. For given MATH, we can choose MATH such that MATH by REF . Using these elements MATH, we have MATH and the similar equality for MATH. The equality MATH implies MATH and as a conse... |
math/0008024 | We can directly check the equality MATH . By REF , we have the equality MATH. As a consequence we have MATH for all MATH and get the commutativity of the diagram in the main theorem. We prove the last part of the theorem. By CITE, the map MATH is an isomorphism. Thus the map MATH is proper morphism and the image contai... |
math/0008027 | We use the following equality CITE. MATH . To prove the lemma, we put MATH. Then we have MATH . Putting MATH and MATH, the conclusion follows. |
math/0008041 | Straightforward calculations (most of them are done in CITE). |
math/0008041 | This follows from REF. |
math/0008041 | We prove REF by induction on MATH. If MATH there is nothing to show because MATH for MATH. Now let MATH and consider the exact sequence MATH . Since by induction hypothesis MATH and by the assumption MATH, we get that MATH. For MATH the exact sequence of the NAME homology together with REF yields MATH which proves REF ... |
math/0008041 | This follows from the fact that MATH. Induction on MATH proves that all MATH are cycles if MATH is one. |
math/0008041 | We proceed by induction on MATH to prove REF . If MATH, then MATH, and so MATH. Thus we choose MATH. Let MATH and assume that MATH. We see that MATH is a cycle because MATH . But MATH. Since MATH, it follows that MATH is a boundary for some element MATH. By the induction hypothesis we get MATH such that MATH. Note that... |
math/0008041 | We prove the assertion by induction on MATH. For MATH there is nothing to show. Let MATH. Since MATH and MATH we see that MATH is a cycle and therefore a boundary by the assumption that MATH. By REF we may assume that MATH. By the induction hypothesis we find the desired MATH in MATH. |
math/0008041 | Let MATH be an arbitrary basis of MATH. Since MATH there exists a cycle MATH with MATH. Furthermore MATH for MATH. In this situation we have MATH for some socle element MATH of MATH and we want to show that every equation MATH implies MATH for all MATH. Assume there is such an equation where not all MATH are zero. Afte... |
math/0008041 | We have MATH because MATH. Choose MATH. We prove by induction on MATH that we can find a basis MATH of MATH and a cycle MATH representing MATH such that every equation MATH implies MATH for all MATH. The cases MATH and MATH were shown in REF. Let MATH and MATH. Assume that there is a basis MATH and such an equation for... |
math/0008041 | This follows from REF and the fact that there are no non-trivial boundaries in MATH. |
math/0008041 | We have MATH . Hence there exists an integer MATH and a monomial MATH of MATH with MATH because all monomials have to cancel. Assume that MATH . Then MATH is a contradiction. Therefore MATH . |
math/0008041 | This follows from REF because the coefficients there have different leading terms. |
math/0008041 | Let MATH be a presentation of MATH such that MATH is free and MATH. We show that MATH . Since MATH, this will prove the theorem. After a suitable shift of the grading of MATH we may assume that MATH. Note that MATH is a submodule of a free module and we can apply our construction REF. Since MATH, we get a homogeneous c... |
math/0008041 | REF was shown in REF. In the proof of REF we proved in fact REF . Finally, REF follows from REF since MATH if MATH is the MATH-th syzygy module in the minimal graded free resolution of some module MATH. |
math/0008041 | The proof is essential the same as in CITE where the local case is treated. |
math/0008041 | According to REF the module MATH satisfies MATH if and only if MATH is a MATH-th syzygy module in a graded free resolution MATH of some graded MATH-module MATH. It is well-known (see for example CITE) that MATH as graded complexes where MATH is the minimal graded free resolution of MATH and MATH is split exact. Then MA... |
math/0008041 | We prove by induction on MATH that we find MATH and a set MATH such that the cycles MATH are MATH-linearly independent for MATH. The cases MATH and MATH follow from REF because if MATH, then all MATH have different degrees MATH and it suffices to show that these elements are not boundaries. Let MATH and assume that MAT... |
math/0008041 | We prove this by induction on MATH. The case MATH is trivial, so let MATH and without loss of generality we may assume that MATH. The set MATH is the disjoint union of the sets MATH and MATH. The induction hypothesis applied to MATH and MATH implies MATH . |
math/0008041 | Without loss of generality MATH. Since MATH, there exists a MATH-homogeneous cycle MATH such that MATH in MATH and MATH for some MATH. By the definition of MATH and REF we have MATH for MATH and MATH. We construct inductively MATH as above, as well as cycles MATH for each MATH such that MATH, MATH, MATH for MATH and su... |
math/0008041 | This follows from REF and the fact that MATH where MATH is the MATH-th syzygy module of a MATH-graded MATH-module MATH. |
math/0008041 | The assertion follows from REF with similar arguments as in the graded case. |
math/0008041 | Fix MATH and MATH. Let MATH be the number of monomials in MATH which are not monomials in MATH. We prove the statement by induction on MATH. If MATH, then MATH and there is nothing to show. Assume that MATH. Let MATH be the smallest monomial in MATH with respect to MATH which is not in MATH, and MATH be the largest mon... |
math/0008041 | It is well-known that MATH. Therefore MATH and MATH also has a MATH-linear resolution. CITE implies that a NAME monomial ideal, which is generated in degree MATH, has regularity MATH if and only if it is stable. Thus we get that MATH is a stable ideal, independent of the characteristic of MATH. Since MATH has a linear ... |
math/0008041 | This follows from REF. |
math/0008041 | By CITE we find a lex-ideal MATH with the same NAME function as MATH and MATH for all MATH. We see that MATH because these ideals share the same NAME function. It follows that MATH, and in particular MATH. Therefore MATH for all MATH where the last equality follows from MATH. |
math/0008043 | Multiplying REF by MATH and taking the expected value of both sides we get MATH. Since MATH, REF implies that MATH. Therefore MATH and the correlation coefficients MATH satisfy the recurrence MATH . For fixed MATH, this is a linear recurrence for MATH. REF implies that the characteristic equation of the recurrence has ... |
math/0008043 | We first show that for all MATH there are coefficients MATH and MATH such that MATH and MATH . We prove this by induction with respect to MATH. For MATH, this follows from REF with MATH, see REF . For a given MATH, suppose that MATH and both REF hold true for all MATH. We will prove that the same statement holds true f... |
math/0008043 | Applying MATH-times REF , we get MATH. This implies MATH. |
math/0008043 | Since MATH, from REF we get MATH . We now give another expression for the left hand side of REF . REF implies that MATH. Since MATH from REF we get MATH. By REF this implies MATH. Since MATH, combining the latter with REF we have MATH . We now substitute REF into REF as follows. Taking the conditional expectation MATH ... |
math/0008043 | CASE: Follows from the fact that the coefficients in REF must be non-negative. CASE: This is an immediate consequence of NAME 's criterion, see CITE. It is easy to see that MATH corresponds to the (unique) symmetric two-point distribution MATH. This is a degenerate case, often excluded from the general theory of orthog... |
math/0008043 | This follows from MATH. |
math/0008043 | The proof is by mathematical induction with respect to MATH. The result is trivially true for MATH (with MATH) and for MATH. By REF holds true for MATH. Since we assume that MATH, REF implies that REF holds true for MATH. Suppose REF holds true for MATH, and MATH. We will show that REF holds for MATH. The proof repeate... |
math/0008043 | All assumptions are symmetric with respect to time-reversal. Therefore, REF implies MATH. Since MATH, REF (used with MATH) proves that MATH are orthogonal with respect to MATH. Therefore, REF identifies uniquely the distribution MATH for both MATH and MATH cases. The distribution is not determined uniquely by the momen... |
math/0008043 | Clearly MATH proving REF . Similarly, MATH. Now MATH. Since MATH is MATH-measurable, we get MATH, which proves REF . |
math/0008043 | The case MATH is classical and the sum of the series is MATH. For MATH an explicit product representation for the series can be deduced from the facts collected in CITE, see also CITE. Namely, in the notation of CITE, MATH where MATH. Therefore, using CITE with MATH, MATH (see also CITE) we have MATH . Since the last e... |
math/0008043 | Applying NAME 's theorem to the function which by REF is non-negative we get MATH. Since MATH for all MATH and MATH, we get MATH. This proves stationarity. To prove REF , by the NAME property it suffices to show that MATH. Let MATH be an arbitrary bounded measurable function. We will verify that MATH . Since MATH is sq... |
math/0008043 | By the NAME property it suffices to show that MATH. To this end, as in the proof of REF , we fix a bounded measurable function MATH. We will verify that MATH by comparing the coefficients in the orthogonal expansions. From REF we get MATH, and MATH. Moreover, by orthogonality MATH except when MATH. Using these identiti... |
math/0008047 | It is well known that MATH is obtained from MATH by a pull-back of an automorphism MATH of MATH which preserves MATH. |
math/0008047 | It is easy to show them by the case check. |
math/0008047 | Suppose REF admits a generic solution. Then, MATH holds. Therefore, the LHS of REF equals to MATH. Then, the LHS of REF equals to MATH, which is positive. |
math/0008047 | Let us compute the ratio REF explicitly. MATH . In order to calculate MATH, it is convenient to evaluate MATH . Here MATH are given as follows: CASE: For MATH . CASE: For MATH, MATH where MATH, and MATH is MATH except for the cases: CASE: If MATH, MATH, MATH mod MATH, then MATH is MATH or MATH with MATH mod MATH. CASE:... |
math/0008047 | Owing to REF , we have MATH for any MATH. Since MATH, it suffices to show MATH for MATH. From REF both MATH and MATH for MATH are zero at MATH. Thus among MATH's the non-vanishing ones are only MATH, MATH, and MATH. Let MATH be the MATH-th row vector of the matrix MATH. In view of the above result, the linear dependenc... |
math/0008047 | By REF . Applying the NAME inversion formula CITE, we obtain REF . |
math/0008047 | By REF , it suffices to verify MATH. Actually, a stronger statement holds: Let us forget the relation REF , and regard MATH in REF as a nonnegative integer which is independent of MATH's. Then, MATH still holds. We prove the last statement by the double induction on MATH and the sum MATH. First, let MATH REF be arbitra... |
math/0008047 | By definition, MATH. Let MATH and MATH. From REF , we have MATH, and MATH when MATH. Thus, MATH. Therefore, MATH by REF . |
math/0008047 | If MATH, then the both hand sides of REF is REF. Suppose that MATH. By REF , we have MATH. By splitting the sum MATH of each diagonal element in MATH in REF , MATH is written as MATH . Then, REF follows from REF and the fact MATH where the last equality in REF is due to REF . |
math/0008047 | First, we consider MATH. For given MATH and MATH, let MATH be MATH, and MATH. Then, it is easy to check that MATH . Therefore, MATH converges for MATH. Next, consider MATH. By REF , MATH is a linear sum of the power series whose coefficient of MATH is MATH. Again, each series converges for MATH. |
math/0008047 | REF . Let MATH abbreviate MATH. By REF , we have MATH . In REF , we set MATH. After the substitution of REF and the change of the integration variable, we obtain MATH . Also, by REF , we have MATH . The equalities REF follows from REF , and the fact MATH if MATH. CASE: By using REF , the Right-hand side of REF is easil... |
math/0008047 | REF is immediately obtained from REF . REF follows from REF and the fact MATH. |
math/0008047 | It is enough to show that MATH for MATH and MATH which are related as MATH (MATH is defined in REF ). By REF , for any MATH it holds that MATH . By REF , it is easy to show MATH . With REF , we have MATH while for MATH, MATH . The equality REF follows from REF , and REF . |
math/0008047 | CASE: By rearranging the product indices, the LHS in REF becomes MATH . By REF , we have MATH. CASE: Using the same trick as above and the definition of MATH in REF , we have MATH . On the other hand, using (Q̃-I) and REF , we have MATH for MATH. REF is obtained by multiplying the above two equalities. |
math/0008047 | Let us regard REF as an equality of power series of MATH. Then, by substituting the series MATH for the variable MATH in REF and using REF , we obtain REF . |
math/0008047 | CASE: This is shown by the case check. CASE: This can be easily checked by REF . REF is equivalent to the matrix relation MATH. We have only to care that the matrix product in the LHS is well-defined. This is guaranteed by REF . The LHS of REF can be calculated in a similar way as follows: MATH where MATH is a sufficie... |
math/0008047 | We introduce another subset MATH REF of the index set MATH as MATH . Then, MATH and MATH. By (Q̃-I), MATH . For a given MATH, let MATH be a unique positive integer such that MATH. Then, with REF , it is easy to check that MATH for MATH. The claim now follows from REF by induction on MATH. |
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