paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
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 that are all well-defined on MATH and are MATH-reducible. |
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 of the indicial equation of REF satisfy MATH. Then MATH . Let MATH . Then by REF, the log-term coefficient MATH of REF is given by MATH, where MATH and MATH for MATH. Hence MATH is a polynomial in MATH of order MATH. We now define MATH, and we define MATH and MATH for MATH by the relations MATH . It follows that MATH, and hence MATH. Then, defining MATH, we also have MATH for MATH. Since MATH, we have MATH . If the only roots of the polynomial MATH with respect to MATH are MATH and MATH, then it follows that MATH would be nonnegative. However, MATH for all MATH, hence this polynomial must have some root MATH, and then MATH. For this MATH and MATH, we have MATH, and thus we have at least one surface for each MATH. Since MATH is a polynomial of degree MATH in MATH, there are at most MATH roots, and hence at most MATH surfaces. For MATH, one can check by explicitly computing the polynomial for MATH that there is always at least one root 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 the latter part. Let MATH be another lift on MATH and denote MATH, MATH. The lifts MATH and MATH are related by MATH for some function MATH. It follows that MATH . Since MATH, we have MATH . Therefore MATH. This implies that MATH . |
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 initial condition. Then MATH differs from MATH only by MATH for some MATH. Therefore, MATH satisfies MATH, which implies that MATH is congruent to MATH by the isometric action of MATH. |
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. Next we shall prove the converse. Let MATH be a conformal immersion with holomorphic right NAME map. Let us choose a lift MATH of MATH arbitrarily. We set MATH and assume MATH and MATH has a local expression MATH, MATH for a local complex coordinate MATH. The integrability REF is written as MATH the conformality REF is MATH and the condition of nondegeneracy REF is MATH . On the other hand, we have MATH . Hence the right NAME map MATH is holomorphic if and only if there exists a holomorphic map MATH and a non-vanishing function MATH such that MATH . Differentiating REF with respect to MATH, we have MATH . It follows from REF that MATH . Since the left-hand side of the above equation is Hermitian, so is the right-hand side, namely, MATH . Hence, MATH . This implies from REF that MATH, that is, MATH must be holomorphic. Therefore MATH is a holomorphic MATH-form on MATH. Moreover, by conformality REF, MATH is null. Next, let MATH be a holomorphic map obtained by solving the differential equation MATH for MATH as in REF under the initial condition MATH for a base point MATH. We want to show that MATH is also a lift of MATH, that is, MATH. For this, we have only to prove that MATH differs from MATH by a MATH-valued function, namely, that MATH takes values in MATH. In fact, MATH satisfies MATH and the right-hand side MATH is a MATH-form which has values in MATH, that is, MATH. Therefore MATH remains in MATH. Thus MATH admits a holomorphic lift MATH if 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 MATH. Hence the function MATH is subharmonic, which must be constant since MATH is compact. By an isometry in MATH, we may assume that MATH, where MATH is the identity element of MATH. Then MATH is identically MATH. On the other hand, MATH is identically MATH. This implies that the eigenvalues MATH of MATH satisfy MATH . Here, MATH, because MATH is Hermitian. Moreover, they are positive around MATH because all of them are MATH at MATH. Therefore, MATH holds. However, since both sides are MATH, the equality is attained in REF. Thus, MATH . Since MATH is Hermitian, this implies that MATH is equal to the identity matrix, a contradiction. |
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 curvature, which yields the assertion by the same argument as in minimal surface theory. (See CITE.) |
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 of MATH that MATH is bounded in MATH. The compactness of the fiber MATH of the fiber bundle MATH implies that MATH is bounded in MATH. Hence MATH is also bounded in MATH because it is the image of MATH under the diffeomorphism MATH of MATH. If we write MATH, then MATH holds for some constant MATH. Since MATH we have MATH for some constant MATH. Thus, we have MATH . |
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 single-valued on MATH by REF , MATH is a single-valued map defined on MATH. Since the induced metric MATH of MATH is given by MATH (see REF), we have MATH. Hence the dual total curvature MATH is the total curvature of the minimal immersion MATH. As in the proof of REF, we have MATH where MATH is the homology degree of MATH. Let MATH be the ends of MATH. By REF , MATH has a pole at each end. Let MATH denote the maximum order of the pole of MATH at the end MATH. Then, as in the proof of REF, we have MATH where MATH is the number of ends. It should be remarked that to prove REF, we do not need well-definedness of MATH on MATH, but only the fact that the NAME map MATH is single-valued on MATH. It can be easily checked that the order of MATH at MATH is equal to MATH. By REF , we have MATH . By REF - REF, we have MATH . This proves the theorem. |
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 of solutions of REF. Let us denote the monodromy representation by MATH, that is, MATH is determined by MATH for any covering transformation MATH of MATH. By our assumption, MATH, in particular, eigenvalues of MATH are of the form MATH (MATH). Since MATH is non-singular at each point of MATH, an arbitrary fundamental system MATH of solutions to REF can be written as MATH . We let MATH the monodromy matrix determined by MATH. Then we can write MATH . Hence, eigenvalues of MATH are also of the form MATH REF and MATH is diagonalizable. CASE: Since we assumed that MATH has a pole of order MATH at the origin, MATH has the NAME expansion MATH . We shall show that all eigenvalues of MATH vanish. Let MATH be an eigenvalue of MATH. First we shall prove that MATH is a real number. We have only to prove it for an eigenvalue MATH such that MATH is not an eigenvalue for any positive integer MATH. (In fact, whenever both MATH and MATH are eigenvalues for an integer MATH, MATH is real if and only if MATH is also.) By REF in Appendix, there exists a solution MATH to REF which is expanded as MATH . Denote by MATH the covering transformation corresponding to a loop going once around the origin in MATH in the positive direction. Then MATH satisfies MATH . If we take a fundamental system MATH of solutions to REF so that the first column vector MATH is equal to MATH, then MATH . Hence, MATH is an eigenvalue of the monodromy matrix MATH of REF. Then it follows from REF that MATH must be real. Thus we can conclude that all the eigenvalues of MATH are real. Moreover, the nullity condition MATH gives MATH in particular, MATH. Namely, the square-sum of eigenvalues of MATH is zero. Hence, all the eigenvalues of MATH must be zero. CASE: Finally, we shall make a contradiction. Suppose MATH is diagonalizable. Then MATH vanishes, which contradicts the completeness around the origin. Hence, MATH is not diagonalizable, and its NAME normal form is MATH where MATH denotes MATH or MATH. By REF of Appendix, there exist linearly independent solutions MATH, MATH of REF such that MATH . For the covering transformation MATH as above, MATH . If we take a fundamental system MATH of solutions so that MATH the monodromy matrix is given by MATH . This means that the monodromy matrix MATH of REF is not diagonalizable, which contradicts the conclusion of REF . |
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 obtain MATH . |
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 power series MATH . The ordinary differential REF has MATH linearly independent solutions in the space of formal power series of MATH, and by REF , we know that these formal power series converge. So we may treat the solution as a formal power series. Differentiating REF, we have MATH . By our assumption, MATH is a nonsingular matrix. Hence MATH defined in REF are regular values (that is, MATH). Moreover, if MATH is a MATH-eigenvector of MATH, then MATH is a solution to REF , because MATH. |
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 because MATH. In REF, if we put MATH then MATH where MATH is a holomorphic function in MATH. Hence, it can be written as MATH . On the other hand, if we set MATH we have MATH from REF. Therefore, MATH is a solution to REF. Moreover, MATH is of the form MATH, by REF. |
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 - NAME Theorem CITE to an evolution equation for MATH derived from REF . Thus the family MATH exists, and it remains to prove that MATH is special Lagrangian. Now by REF , as MATH and MATH is real analytic, there is a locally unique real analytic special Lagrangian submanifold MATH in MATH containing MATH. We shall show that MATH locally. To do this, observe that REF also makes sense as an evolution equation for submanifolds of MATH. That is, we could look for a family MATH of real analytic maps MATH with MATH, satisfying the equation MATH using the index notation for tensors on MATH. It follows as above that for some MATH there exists a unique solution to this problem. Let MATH and MATH, and set MATH, so that MATH. Treating MATH and MATH as real vector spaces, we have orthogonal direct sums MATH and MATH, where MATH is the perpendicular subspace to MATH. This induces a splitting MATH. Now MATH, and MATH is calibrated with respect to MATH. This implies that the component of MATH in MATH is zero, because this measures the change in MATH under small variations of the subspace MATH, but MATH is maximum and therefore stationary. Since MATH lies in MATH, it follows that MATH as the component in MATH comes from the component of MATH in MATH, which is zero. Therefore MATH . Because the splitting MATH is orthogonal we have MATH for some MATH. Contracting this with REF shows that MATH at MATH, as the right-hand side of REF lies in MATH, so its contraction with MATH is zero. REF holds for all MATH and MATH. Therefore by REF the MATH also satisfy REF , and so MATH by uniqueness. Hence MATH maps MATH to MATH, and MATH maps MATH to MATH, if MATH is sufficiently small. Finally, suppose MATH is an embedding. Then MATH is also an embedding for small MATH. But MATH is a normal vector field to MATH in MATH, with length MATH. As MATH is nonvanishing, this vector field is nonzero, so MATH is an embedding for small MATH, with MATH an open subset of MATH, which is therefore special Lagrangian. If MATH is an immersion, then MATH is a special Lagrangian immersion, in a similar way. |
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. By REF this holds if and only if MATH on MATH for all MATH. Since MATH is connected, this is true if and only if MATH is constant on MATH, so that MATH for some MATH. But MATH is MATH-invariant, so MATH is MATH-invariant, and MATH. |
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 uniquely by requiring that MATH on MATH. Clearly MATH is linear in MATH. Hence there is a unique map MATH with MATH for all MATH, and thus MATH as in REF above. The MATH-equivariance in REF follows because MATH is MATH-invariant. Thus we have constructed a particular moment map MATH for MATH, with the property that MATH on MATH. Hence MATH admits a moment map. If MATH is any moment map for MATH then MATH is a constant MATH, and so MATH. |
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 uniqueness. Hence MATH, by REF . We can also construct MATH by an evolution equation, as in REF . Choose MATH, and let MATH be the stabilizer of MATH in MATH. Set MATH, and define MATH by MATH for all MATH. Then MATH is an immersion, with MATH. Clearly MATH and MATH are real analytic, and MATH on MATH. As MATH is oriented and MATH acts on it by isometries, we can choose a nonvanishing, MATH-invariant section MATH of MATH. Suppose for the moment that MATH is compact. Then REF applies to give an embedding MATH, whose image is an open subset of MATH. As MATH is MATH-invariant and MATH is MATH-equivariant, we see by uniqueness that MATH is equivariant under the actions of MATH on MATH and MATH. Thus MATH is a MATH-orbit in MATH isomorphic to MATH for MATH, and MATH is fibred by a smooth REF-parameter family of MATH-orbits isomorphic to MATH near MATH. This completes the proof, except that we assumed MATH was compact to apply REF . This assumption is in fact unnecessary. As in the discussion after REF , for any MATH the MATH exist near MATH for MATH and some MATH. We can then use MATH-equivariance to extend MATH uniquely to all of MATH. |
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 with MATH. It follows easily from REF that MATH and MATH are real analytic. Consider a REF-parameter family MATH of maps MATH defined by MATH, where MATH is a differentiable function. Calculation shows that the MATH satisfy the evolution REF if and only if MATH satisfies the o.d.e. MATH . But then MATH, which is real. So MATH is constant along the integral curves of REF . In particular, if MATH then MATH, MATH is an integral curve of REF , and the result then follows quickly from REF . |
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 REF is the only turning point of MATH in this interval. Thus MATH in MATH, as MATH. REF shows that MATH for MATH. As MATH, this gives MATH, so MATH. |
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 then MATH is confined to MATH, and MATH. Thus MATH for all MATH for which the solutions exist, as we have to prove. As MATH is confined to MATH, there exists MATH with MATH for MATH and all MATH for which the solution exists. Thus MATH so that MATH for MATH. |
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 MATH by REF . By REF , MATH if and only if MATH, that is, if and only if MATH. By REF this happens exactly when MATH. But MATH is confined to MATH by REF , where MATH and MATH for MATH. Hence MATH if and only if MATH or MATH. So we see that MATH must cycle up and down between MATH and MATH, turning only at MATH and MATH. Using the ideas of REF, we see from REF that the time taken for MATH to increase from MATH to MATH is MATH which is finite, as MATH has only single roots at MATH. Similarly, the time for MATH to decrease from MATH to MATH is also MATH. Thus, the time taken for MATH to start at MATH, increase to MATH, and decrease back to MATH, is MATH. That is, MATH undergoes periodic oscillations with period MATH. REF shows that MATH is also periodic with period MATH, and also that MATH, as MATH. It is then easy to see that MATH for all MATH and MATH given by MATH. |
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. This means that the REF-parameter family of MATH-orbits MATH which make up MATH satisfies MATH for all MATH, and is periodic, with period MATH. The definition of MATH given in REF differs from REF in that MATH rather than MATH, and MATH rather than MATH. We allow MATH to add the point REF to MATH, which makes MATH closed. We allow MATH because MATH and MATH exist for all MATH by REF . In fact, as MATH has period MATH, we would get the same MATH if we replaced MATH by MATH. So MATH is a closed loop of MATH-orbits MATH, together with the point zero. It is clear that MATH is closed and, at least as an immersed submanifold, it is topologically a cone on MATH, with just one singular point at REF. So we need only show that MATH is embedded. If two MATH-orbits MATH intersect, then they are the same. But from REF , a MATH-orbit locally determines MATH uniquely. Thus, our loop of MATH-orbits cannot cross itself, and must be embedded. |
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 MATH as MATH, it is enough to study the integral REF near MATH and MATH. We shall model it at MATH. Suppose MATH, so that MATH has multiplicity MATH. Then near MATH we have MATH . Since MATH this gives MATH, so that MATH. Therefore, when MATH we have MATH taking only the highest-order terms. Thus, when MATH is small we see that MATH . Changing variables to MATH, after some surprising cancellations we get MATH where we have approximated the second integral by replacing the upper limit MATH by MATH. Hence, for small MATH we have MATH and similarly MATH so that MATH as MATH. Next consider the behaviour of MATH as MATH. When MATH is close to REF, MATH is small and MATH is close to MATH. So write MATH, for MATH small. Then, setting MATH, MATH taking only the highest order terms, REF - REF become MATH . The first two of these equations show that MATH and MATH undergo approximately simple harmonic oscillations with period MATH. Then the third equation shows that MATH, as MATH is approximately constant, and so MATH as MATH. |
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 submanifold is minimal. Therefore the intersection MATH is minimal in MATH. But MATH is the image of MATH. Hence MATH is a conformal map with minimal image. But it is well known in the field of harmonic maps that a conformal map from a NAME surface is harmonic if and only if it has minimal image. Thus MATH is harmonic. |
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 is that MATH is a nonsingular point of MATH, and the vector fields MATH are linearly independent at MATH. Suppose these conditions hold. Choose a basis MATH of MATH such that MATH are orthonormal, which is possible by linear independence of MATH at MATH. Now MATH is normal to MATH, which is a special Lagrangian plane in MATH. Therefore MATH lies in MATH, where MATH is the complex structure on MATH. Hence MATH are orthonormal in MATH. Extend them to an orthonormal basis of MATH with vectors MATH, so that MATH . Now the level sets of the moment map MATH of MATH are orthogonal to MATH for all MATH. Hence MATH is the subspace of MATH orthogonal to MATH for MATH. Thus MATH. But MATH, and so MATH is the span of MATH and MATH. Hence MATH . Comparing REF and remembering that the bases are orthonormal, we see that in effect we have orthogonal direct sums MATH and MATH. It is easy to see that as MATH is an SL plane with phase REF, this implies that MATH is an SL plane with phase MATH or MATH, depending on the orientation chosen for MATH. Thus, if MATH is even then MATH is special Lagrangian with phase REF, and if MATH is odd then MATH is special Lagrangian with phase MATH, with the appropriate orientation. |
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 vector space MATH into a direct sum of eigenspaces of the action of MATH. Actually, one usually considers the eigenspaces of a single matrix, rather than of a vector space of commuting matrices. But the eigenspace decomposition of MATH under a generic element of MATH is the same as its decomposition under MATH, so the two points of view are equivalent. Let us write the eigenspace decomposition as MATH where MATH are nonzero eigenspaces of MATH in MATH with distinct, nonzero eigenvalues in MATH, and MATH is the zero eigenspace of MATH. The decomposition REF is unique up to the order of the subspaces MATH, and is orthogonal as MATH. Each MATH acts on MATH by multiplication by MATH for some MATH, and is zero on MATH. Now MATH is connected and abelian, so MATH, and MATH is surjective. Hence we can write each MATH as MATH for MATH, and so MATH acts on MATH by multiplication by MATH, and as the identity on MATH. Putting MATH and MATH, we have shown that REF satisfies REF and the second part of REF . Now consider the general case with MATH. By projecting MATH from MATH to MATH we can reduce to the previous case, and decompose MATH into eigenspaces. However, we now have to allow for MATH to act by translations in each factor, as well as by MATH rotations. Since the MATH part of MATH acts on MATH with nonzero eigenvalue, and MATH is abelian, by moving the origin in MATH we can eliminate the translation part, so that MATH acts on MATH by multiplication by MATH for MATH, as in REF . Moving the origin is allowed, as we seek only an affine isomorphism MATH, rather than a vector space isomorphism. However, moving the origin in MATH in REF has no effect on the MATH translation-component of the action of MATH, because this is the zero eigenspace of the MATH part of MATH. So we cannot eliminate translations in the MATH directions by choosing the origin appropriately. Instead, define MATH to be the complex vector subspace of MATH generated by the MATH translation-components of MATH, and let MATH be the orthogonal complement to MATH in MATH. Then we have an affine isomorphism MATH such that each MATH acts on MATH by multiplication by MATH for MATH, MATH acts by translations on MATH, and MATH acts trivially on MATH, so that REF and parts of REF are satisfied. Now we have put the MATH-action in a standard form, we prove the `only if' part of the theorem. Suppose MATH is special Lagrangian in MATH, and MATH is normal to MATH at MATH for every MATH in MATH. The key idea we shall use is that for each MATH in MATH, as MATH is normal to MATH and MATH is Lagrangian, the vector field MATH is tangent to MATH at MATH. By exponentiating MATH we get a REF-parameter family of diffeomorphisms of MATH, which locally preserve MATH. That is, for each MATH in the interior of MATH, there exists MATH such that MATH for MATH. As MATH is abelian and the vector fields MATH are holomorphic, we see that MATH for all MATH. Thus the MATH form an abelian NAME group MATH of diffeomorphisms of MATH, isomorphic to MATH. We use this to extend MATH to a globally invariant submanifold MATH. Define MATH . Then it is not difficult to show that MATH is a special Lagrangian submanifold of MATH containing MATH, invariant under MATH. One way to prove this is to use real analyticity, and the results of REF. Each connected component MATH of the interior of MATH contains a connected component MATH of the interior of MATH. As MATH is real analytic and MATH is the orbit of MATH under a NAME group, MATH is also real analytic. So as the special Lagrangian condition holds on a nonempty open subset MATH of MATH, it holds on all of MATH. As MATH is Lagrangian we have MATH. But MATH for MATH. Hence MATH . Write MATH, where MATH is the projection of MATH to MATH. Let MATH act on MATH by multiplication by MATH for MATH. Then MATH . Combining REF for all MATH, and remembering that the eigenvalues of MATH on MATH are distinct and nonzero, we see that MATH . By considering the tangent spaces of MATH we find that MATH admits a local product structure, and deduce that MATH, where MATH is a Lagrangian submanifold of MATH for MATH, and MATH is Lagrangian in MATH. Let the MATH and MATH be as small as possible such that MATH. This defines the MATH and MATH uniquely. As MATH is special Lagrangian, it follows that MATH and MATH are actually special Lagrangian in MATH and MATH with some phases; we can fix the phases to be REF by choosing the holomorphic volume forms on MATH appropriately. Since MATH and MATH are defined uniquely using MATH, which is invariant under MATH, the MATH and MATH must also be invariant under MATH. But MATH multiplies by MATH in MATH for some MATH for MATH, and MATH can take any value in MATH as MATH varies. Therefore MATH is invariant under all dilations of MATH for MATH, and is a cone by REF . This proves REF. Similarly, MATH is invariant under the action of MATH on MATH. Let MATH be an orbit of a point in MATH under MATH. Now MATH acts by translations on MATH, and trivially on MATH. Thus MATH, where MATH is an affine subspace of MATH and MATH. By definition these translations on MATH generate MATH over MATH, which forces MATH. However, MATH and MATH is Lagrangian, so that MATH. Therefore MATH, and MATH is a Lagrangian plane in MATH. By choosing the phase of the holomorphic volume form on MATH appropriately, we can assume that MATH is an SL plane. Thus MATH is fibred by orbits MATH of the form MATH, where MATH is an SL plane in MATH and MATH. It easily follows that MATH, where MATH in MATH, and MATH is an SL submanifold of MATH. As MATH acts on MATH by translations in the direction of MATH, it follows that MATH acts on MATH by translations in the direction of MATH. This proves the `only if' part of REF . But the `if' part follows very easily, given the discussion above, so the proof is complete. |
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, it is enough to prove the surjectivity. We use the same notations MATH, MATH as in the last paragraph. Let MATH be a small deformation of the cubic threefold MATH over MATH such that the generic fibers MATH at MATH are sufficiently generic. We can extend a line MATH to a family of lines MATH contained in MATH. The family of incidental subvarieties MATH is a family of curves on MATH. By CITE, the generic fiber MATH of MATH at MATH is a smooth curve of genus REF. Moreover the family of the NAME involutions on the fibers comes to be an involution MATH of MATH preserving each fiber. Thus we have the relative cylinder map MATH and it factors through the maximal quotient MATH of MATH on which the involution MATH acts as the MATH-multiplication: MATH . Since MATH is a proper smooth morphism of cubic threefolds, MATH is a smooth MATH-sheaf. By considering the rank of fibers and the specialization map, we get the smoothness of the sheaf MATH. On the other hand, by REF, for MATH the homomorphism MATH is surjective. Therefore MATH is surjective. Since MATH is a curve with one node, whose normalization MATH is a curve of genus REF, we have MATH and we get the theorem. |
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 are disjoint, there are exactly REF lines which intersect both MATH and MATH. Two of them are MATH and MATH. The rest of them are written as MATH, MATH, MATH. Then we can find MATH such that MATH is a tritangent. We denote MATH, MATH. Then MATH are mutually disjoint lines. Moreover, there exists a unique line MATH such that MATH are mutually disjoint lines. By blowing them down, we obtain a marking of MATH. In particular, MATH, MATH, MATH and MATH. Note that the action of MATH on the set of tritangents and MATH are equivariant. The action of the transposition MATH is trivial on MATH and non-trivial on MATH, therefore MATH and MATH are different. |
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 the intersection form on MATH, we get the reflection vector for MATH (respectively, MATH). |
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 by complex reflections, (compare CITE) and the invariance under reflections can be proved similarly as in REF . We apply the transformation formula to compute MATH, where MATH. We use the notations MATH of REF By the definition of the embedding MATH given in REF , we have MATH . Since we can choose MATH as a lifting of MATH, MATH is equal to MATH, where MATH . To use REF , we write MATH, MATH. Then we have MATH . Since MATH ( mod MATH), MATH is an integer if MATH ( mod MATH). |
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 MATH, MATH and MATH, we have MATH . For a point MATH, we define a MATH multi-valued function MATH. As we mentioned, MATH depends on the path connecting MATH and MATH. We choose a lifting MATH of MATH in MATH. Then MATH and MATH. By the quasi-periodicity for theta functions, the multivalued meromorphic function MATH on MATH becomes a single valued rational function on MATH. Using the table of REF, we have the following table on the order of zero of the numerator and the denominator at MATH of REF : MATH . As a consequence, the rational function MATH is equal to MATH, where MATH is a constant independent of MATH. First we evaluate the function MATH at MATH. The value MATH is equal to MATH . Here we used the relation between theta functions and theta constants: MATH . Next we consider the limit MATH. To compute the limit of theta functions, we choose a path from MATH to MATH such that MATH tends to MATH if MATH tends to MATH. Using this path, MATH is a function on MATH defined on a neighborhood of MATH. We can choose a local parameter MATH of MATH at MATH such that MATH and MATH. Since the order of zero of MATH at MATH is one, we have MATH. Therefore we have MATH where MATH. Put MATH and use the equality MATH for MATH, then we have MATH where MATH. Therefore the limit REF is equal to MATH . Multiplying the equality REF for MATH, we have MATH . By the equality REF , MATH is equal to the right hand side of REF . |
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 consequence, we have the equality MATH for MATH. In the same way by applying MATH to the equality MATH, we get the equality MATH. By taking quotient, we have MATH, where MATH, MATH. Again using REF and the same technic as MATH, we get MATH for MATH. |
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 contains an open dense subset and it is surjective. By the commutativity of diagram, since MATH is injective, it is isomorphism. Thus we get the last statement of the main theorem. |
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 . We show REF by induction on MATH. The case MATH is trivial and for MATH and MATH the assertion is true by assumption. Now let MATH, MATH and consider MATH . By REF and the induction on MATH we get that MATH. Hence MATH. |
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 . We define MATH . Then MATH and this proves REF . If MATH, we see that MATH for some MATH and therefore MATH if and only if MATH. We prove REF by induction on MATH. For MATH there is nothing to show. Let MATH and assume MATH. By the same argument as in the proof of REF we get MATH for some MATH. The induction hypothesis implies MATH, and then MATH. |
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. After a base change we may assume that MATH. We get MATH contradicting to REF . |
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 a cycle MATH with MATH where not all MATH are zero. After a base change of MATH we may assume that MATH. Then MATH and therefore MATH for some element MATH. By REF we can find an element MATH such that MATH and MATH. Now REF guarantees that we can apply our induction hypothesis to MATH and we find a base change MATH of MATH, MATH in MATH (with respect to the new basis) such that MATH are MATH-linearly independent for MATH. By REF we have MATH for MATH. Then MATH is the desired cycle because MATH in MATH. |
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 cycle MATH in MATH. There are no boundaries except for zero in MATH and therefore we only have to construct enough MATH-linearly independent cycles in MATH to prove the assertion. Assume that MATH and construct the numbers MATH for MATH by REF. Let MATH and set MATH. Consider the (by REF) cycles MATH with MATH where MATH . We have MATH and therefore MATH . If we show that the cycles MATH are MATH-linearly independent, the assertion follows. Take MATH, MATH for some MATH. It is easy to see by the construction REF that MATH . It is enough to show that the initial terms of the cycles are MATH-linearly independent. If cycles have different initial monomials in the MATH, there is nothing to show. Take MATH and assume that the corresponding cycles have the same initial monomial in the MATH. We have to consider two cases. If MATH, then MATH and the cycles are the same. For MATH the construction implies MATH which proves the MATH-linearly independence. |
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 MATH splits as a graded module into MATH where MATH is a graded free MATH-module. If MATH, there is nothing to show. For MATH it follows that MATH. Then REF applied to MATH proves the corollary, since MATH for all integers MATH. |
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 MATH. Again it suffices to show that the cycles MATH are not boundaries for a suitable subset MATH. If such a set exists, then nothing is to prove. Otherwise there exists a MATH with MATH and we may assume MATH. By REF we find MATH such that MATH in MATH and MATH. By REF we can apply our induction hypothesis and assume that MATH has the desired properties in MATH. Again by REF we have MATH and MATH is the desired element. |
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 suitable MATH. Furthermore, MATH is an element of the NAME complex with respect to the variables MATH with MATH. For MATH we take all cycles MATH with MATH which have different MATH-degree. They are not zero and therefore MATH-linearly independent in homology. By REF there are at least MATH of them and this concludes the proof. Let MATH. By REF we can choose MATH in a way such that MATH for MATH and some MATH. Choose MATH and MATH. If MATH and MATH for MATH are constructed, then define MATH with MATH and the given MATH for MATH. For MATH with MATH, re-choose MATH by REF in such a way that MATH for MATH and some MATH. Note that since MATH has no monomial which is divided by some MATH for MATH, we can use REF to avoid these MATH in the construction of MATH again. By REF the cycles MATH are also not zero in MATH. Clearly MATH and the assertion follows. |
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 monomial which is in MATH, but not in MATH. Define the ideal MATH by MATH. Then MATH, and MATH is also stable. Thus, by the induction hypothesis MATH . Since MATH, the revlex order implies MATH . Therefore MATH yields MATH which proves the assertion. |
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 resolution, we obtain by the main result in CITE that MATH for all MATH. |
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 two distinct real roots and their product is MATH. There is therefore only one root, MATH, in the interval MATH and MATH. This shows that MATH. In particular, MATH, and MATH for MATH. |
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 for MATH. Conditioning on additional variable MATH and using the induction assumption we get MATH . Now adding MATH to the condition in MATH and using the induction assumption we get MATH . Combining these two expressions we get the linear equation MATH for unknown random variable MATH. Notice that since MATH, if MATH then MATH are linearly related. Therefore, MATH, contradicting REF . (Case MATH requires separate verification: MATH by REF ). Thus MATH and MATH is determined uniquely as the linear function of MATH. This proves REF . By symmetry (or by similar reasoning) REF holds true. The equation gives MATH which shows that MATH. This completes the induction proof of REF . Since MATH for MATH, the coefficients MATH in a linear regression are determined uniquely. A calculation gives MATH . Using REF we have MATH as MATH. Passing to the limit as MATH in REF we get REF . |
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 of both sides of REF with MATH we get MATH . Replacing MATH in the above expression by the right hand side of REF we get REF . If REF holds true then identity REF follows by a simple calculation, since by REF and the assumption MATH we have 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 orthogonal polynomials because MATH for all MATH. CASE: This is the consequence of the classical result that normal distribution is determined uniquely by its moments. CASE: This follows from NAME 's result as reported in addendum REF (page REF) of CITE. Notice that the latter deals with normalized rather than monic polynomials, see REF below. Since MATH, NAME 's theorem is used with MATH and the condition MATH holds true because MATH is concave when MATH. Several explicit weight functions for MATH-Hermite polynomials with MATH are given in CITE. |
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 repeatedly uses conditional expectations MATH. Consider MATH. We will first obtain two REF which allow us to determine conditional moments MATH and MATH. The first equation is obtained as follows. By induction assumption, MATH. On the other hand, REF gives MATH . Therefore MATH . To obtain the second equation, consider MATH. By induction assumption, MATH. On the other hand, REF with MATH give MATH . (We will use relations REF later on.) Thus, using the induction assumption, we obtain the second equation MATH . Polynomials MATH satisfy second order recurrence of the form MATH. We use it to rewrite REF as follows. MATH . Now use REF to eliminate MATH from this equation. Since MATH this gives MATH . By REF , identity MATH (recall that MATH is given by REF ) and elementary calculation we now get MATH . Since MATH, and by induction assumption MATH, this implies MATH . Since REF means that MATH, this ends the proof. |
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 moments when MATH which corresponds to MATH. Finally, when the distribution is determined uniquely, the odd-order moments MATH by REF . Therefore the distribution of MATH is symmetric. |
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 expression is a product of positive factors, this ends the proof when MATH. |
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 square integrable, and MATH are orthogonal in MATH we can write MATH, where MATH is orthogonal to all polynomials in variables MATH. In the case MATH we have MATH because MATH are an orthogonal basis of MATH. Notice that the joint distribution of MATH is given by MATH . From REF and orthogonality we get MATH . Moreover, MATH for MATH. Therefore MATH . Similar calculation gives MATH . Since by symmetry a similar formula holds true for the integral of MATH instead of MATH, we get MATH . Comparing the coefficients in the expansions we verify that MATH proving REF . |
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 identities and the expansion MATH we see that MATH . We now turn to the right hand side of REF . Since MATH a calculation based on REF yields MATH . By symmetry, MATH . Another elementary calculation using REF gives MATH . Since MATH, we can now verify that REF holds true by verifying the relation MATH . Comparing the coefficients in the expansions, the latter reduces to the following. CASE: Coefficients at MATH match when MATH (this holds true by REF ) CASE: Coefficients at MATH for MATH match when MATH (this holds true after a longer calculation using MATH, REF , and REF ) CASE: Coefficients at MATH and at MATH for MATH match when MATH (this holds true by REF ) This implies that REF holds true. |
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: If MATH, MATH, MATH mod MATH, then MATH is MATH or MATH with MATH mod MATH. CASE: For MATH, MATH . The factors MATH and MATH are all nonzero for a generic string solution MATH. Thus they are canceled in the ratio in the Right-hand side of REF , and we find MATH . From REF we obtain REF . |
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 dependence MATH can possibly hold only when MATH is independent of MATH. Consequently we consider the equation MATH. The MATH-th component of the vector MATH is given by MATH where we have taken into account REF and MATH. Due to REF the last expression is equal to MATH. Therefore the equation MATH is equivalent to MATH for any MATH. This admits only the trivial solution for MATH if MATH. |
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 arbitrary, and suppose MATH (that is, MATH for any MATH). Then MATH is equivalent to the positivity of MATH, which is a principal minor of the tensor product of two positive-definite matrices, MATH and MATH with MATH and MATH some integers. Therefore, the claim is true. Next, suppose that MATH (that is, MATH) and MATH is any nonnegative integer. Then MATH. Finally, let MATH REF and MATH be arbitrary. Then there exists some MATH such that MATH. Set MATH if MATH, MATH otherwise, and MATH. Then, one can split the determinant as MATH. By the induction hypothesis, the Right-hand side is positive. |
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 easily calculated as REF . CASE: In REF , replace MATH and MATH in the both hand sides by MATH and MATH. Accordingly, MATH in the Right-hand side in REF should be also replaced by MATH (compare REF ). Then, using REF , we obtain MATH . The equality REF follows from REF , and the identity MATH . |
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 sufficiently large number. REF immediately follows from REF and the relation MATH. |
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. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.