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hep-th/0003114
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Let MATH be the algebra of functions depending on MATH only. Any function from MATH is a pullback of some function on MATH to the entire extended phase space MATH REF. Let also MATH be a BRST observable. Then one can check that there exist functions MATH and MATH such that MATH . As a matter of REF can not depend on MATH as it has zero ghost number. Finally, it follows from REF that the NAME algebra of function on MATH is isomorphic with the NAME algebra of zero-ghost-number BRST invariant functions from MATH.
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hep-th/0003114
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REF evidently holds in the lowest order with respect to degree REF provided respective classical BRST charge MATH satisfies MATH. At the classical level the higher order terms in expansion of MATH with respect to MATH do not depend on the momenta MATH. Thus these terms belong to MATH. It is useful to assume that the same occurs at the quantum level: MATH . In the MATH-th MATH degree REF implies MATH where the quantity MATH is defined by MATH: MATH and MATH stands for MATH . Note that MATH is obviously nilpotent in MATH. It follows from the nilpotency of MATH that the compatibility condition for REF is MATH. In fact it is a sufficient condition for REF to admit a solution. Indeed, the cohomology of the differential MATH is trivial when evaluated on functions at least linear in MATH. To show it, we construct the ``contracting homotopy " MATH. Namely, let MATH be defined by its action on a homogeneous element MATH by MATH where MATH is inverse to MATH. For any MATH we have MATH . Since MATH is quadratic in MATH then REF-rd term vanishes and MATH implies MATH which in turn implies that REF admits a solution. Let us show that the necessary condition MATH is fulfilled. To this end assume MATH to satisfy REF for MATH. Thus the NAME identity MATH implies in the MATH-th degree that MATH. The particular solution to REF for MATH evidently reads as MATH . Iteratively applying this procedure one can construct a solution to REF at least locally. To show that REF admit a global solution we note that operators MATH as well as MATH are defined in a coordinate independent way. It implies that particular solution REF does not depend on the choice of the local coordinate system and thus it is a global solution. The quantum BRST charge constructed above obviously satisfies MATH which can be considered as an additional condition on the solution to REF . One can actually show that solution to REF is unique provided REF is imposed on MATH. Thus we have shown how to construct quantum BRST charge associated to the first class constraints MATH.
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hep-th/0003114
|
Consider an expansion of MATH in the homogeneous components MATH . The boundary REF implies MATH. REF obviously holds in the first degree. In the higher degrees we have MATH where MATH is given by MATH and MATH are terms of the expansion REF of MATH with respect to degree. Similarly to the proof of REF the necessary and sufficient condition for REF to admit solution is MATH. To show that the condition holds indeed assume that REF hold for all MATH. Then consider the identity MATH . Excluding contribution of degree MATH we arrive at MATH . The particular solution to REF for MATH is MATH . Iteratively applying this procedure one arrives at the particular solution to REF satisfying boundary condition MATH. Finally let us show the uniqueness. Taking into account that MATH belongs to MATH and MATH we conclude that MATH does not depend on MATH. Thus the general solution to REF is given by MATH . It is easy to see that the boundary condition requires MATH (recall that MATH.)
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hep-th/0003239
|
First of all the volume of MATH is infinite. It can be seen by calculating: MATH where we have used again the parameterization of the Euclidean NAME manifold given in the previous section. This implies that there are no MATH harmonic MATH- or equivalently MATH-forms. Now, as MATH is NAME and complete, REF implies that there are no MATH- and equivalently MATH-forms on MATH. It remains to show that any MATH harmonic MATH-form is a linear combination of MATH and MATH. For this we use a recent result of NAME, namely REF which we cite in full: Let MATH be a complete oriented Riemannian manifold and let MATH be a connected NAME group of isometries such that the Killing vector fields MATH it defines satisfy MATH . Then each MATH cohomology class is fixed by MATH. (Here MATH is the distance function of the Riemannian manifold.) We would like to apply this result to MATH with MATH acting on MATH by isometries of MATH. A glance at the metric REF assures us that the Killing fields of this action have indeed linear growth. Thus it is sufficient to find all MATH-invariant harmonic MATH-forms on MATH. Let MATH be such a form. In our coordinate chart MATH it must have the shape: MATH where MATH is a MATH-invariant function on MATH, moreover MATH and MATH are MATH invariant MATH-forms on MATH, and finally MATH is a MATH-invariant MATH-form on MATH. However there are very few MATH-invariant forms on MATH. Namely only the constant functions and constant times the volume form of the round MATH are MATH-invariant. It follows because MATH acts transitively showing that only the constant functions and equivalently constant multiples of the volume form are the MATH-invariant MATH- and MATH-forms respectively. Moreover there are no non-trivial MATH-invariant MATH-forms on MATH, which could be seen by looking at the dual vector field and seeing that the action of the MATH stabilizator of any point on the tangent space at that point has only the origin as its fixed point. It follows that our MATH-invariant MATH-form must have the form: MATH where MATH is the volume form of the unit MATH and MATH and MATH stand for a function on MATH depending only on MATH and MATH. Its NAME is given by MATH . In order that both MATH and MATH be closed we need that neither MATH nor MATH depend on MATH or MATH which means that MATH must have the form: MATH exactly as claimed. The result follows.
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hep-th/0003243
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Since the set of vectors MATH is a core fore MATH and MATH is closed, there is a sequence MATH such that MATH and MATH, strongly. Thus if MATH, one also has MATH and MATH. So the operator MATH, given by MATH is well defined. Its adjoint MATH also has the dense set of vectors MATH in its domain and MATH . This shows that MATH is closable (we use the symbol MATH also for its closure), and it establishes REF since MATH. For REF one makes use of the fact that for any vector MATH in the domain of MATH and MATH . Hence MATH, MATH. So MATH lies in the domain of MATH and MATH, MATH. An analogous statement holds for MATH, so the proof of the lemma is complete.
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hep-th/0003243
|
CASE: If MATH is affiliated with the wedge algebra MATH, say, the operators MATH are affiliated with MATH by covariance. Now for MATH varying in some open, bounded region MATH, there is a wedge MATH, hence the operators MATH, MATH, contain the common dense subspace MATH in their domains. Thus for MATH . Since MATH, the function MATH is a weak solution of the NAME - NAME equation, so one obtains from the preceding equation for any test function MATH with support in MATH where MATH is the complex conjugate of MATH. Making use of the temperateness assumption and the NAME - NAME property of MATH, this implies MATH where the integral is defined in the strong sense. Since MATH was arbitrary, the latter equation extends to all test functions MATH by continuity. CASE: The set of vectors of the form MATH, where MATH and MATH is any test function whose NAME transform MATH has compact support, has compact spectral support and it is dense in MATH since MATH is dense and MATH is continuous. By choosing the support of MATH properly, one obtains single particle states with spectral support in any given neighborhood of any point on the mass shell. It remains to be shown that these vectors belong to the domain of temperateness of MATH. There holds for any vector MATH in the domain of MATH for some polynomial MATH, depending only on MATH by the temperateness assumption. Hence MATH is an element of the domain of MATH, and the same holds true for MATH, where MATH, is the translated test function. The continuity of MATH and temperateness follow from the estimate MATH with the same polynomial MATH as above. Hence MATH is an element of the domain of temperateness MATH. The corresponding statements for MATH are established in the same manner.
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hep-th/0003243
|
Let MATH be any vector in the domain of temperateness of MATH with compact spectral support. Then MATH where MATH is defined as a strong integral. Now by the NAME - NAME property of MATH, there is for any MATH a MATH and a test function MATH whose NAME transform has support in any given neighbourhood of MATH such that MATH and MATH. In view of the latter fact, one may replace MATH in MATH by any member of the corresponding family of test functions MATH, defined in REF . Proceeding to the decomposition MATH and taking into account that MATH as MATH, it follows that MATH . Similarly, since MATH is a weak solution of the NAME - NAME equation according to REF , one may replace MATH in MATH by MATH. For MATH and the expression under the integral is a test function because of the support properties of MATH. Making use of the decomposition MATH and temperateness, which implies MATH for some polynomial MATH, one finds that MATH as MATH and MATH strongly. Combining these facts, one gets MATH . According to the choice of the test function MATH, its velocity support MATH is contained in a small neighborhood MATH of MATH, and consequently the operators MATH are localized in MATH. On the other hand, the operators MATH, appearing under the integral in REF , are affiliated with MATH, MATH. Because of locality, they commute with MATH on their respective domains if MATH is spacelike separated from MATH. In REF, there holds MATH and therefore also MATH if the respective neighborhoods are suitably chosen. Hence, according to the above geometrical considerations, the regions MATH and MATH are spacelike separated, and their spatial distance increases linearly with MATH. Because of the latter fact and since MATH is bounded, the two regions MATH and MATH are spacelike separated if MATH and MATH is sufficiently large. One can then reexpress the integral in REF according to MATH . In the latter expression, one can reverse now the passage from MATH to their respective asymptotically dominant parts, taking into account that, in the limit of asymptotic MATH, MATH, MATH, and MATH since MATH has compact spectral support CITE. Hence, by a straightforward estimate, one finds that MATH and MATH have the same limit as MATH. Plugging this information into relation REF and making use of the asymptotic REF , one arrives at MATH . In the resulting equation, one can replace MATH by MATH since MATH, where MATH was arbitrary, and MATH by the NAME structure of collision states. In view of relation REF , this completes the proof of REF. The proof of REF is similar, but now one has to take into account that the regions MATH and MATH are spacelike separated if MATH is sufficiently large. So in this case one arrives at an interpretation of the vectors MATH in terms of outgoing collision states.
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hep-th/0003243
|
The argument is very similar to the proof of the preceding lemma and it therefore suffices to indicate the main steps. In REF one has, in view of the fact that MATH, MATH, and the asymptotic relation REF , MATH where, in the last step, MATH have been replaced by their asymptotically dominant parts. Since MATH, the regions MATH and MATH are spacelike separated for MATH and MATH sufficiently large. Hence, by locality, MATH where, in the second equality, the transition from MATH to the asymptotically dominant parts has been reversed and, in the last step, the asymptotic relation REF has been used as well as the fact that MATH, MATH. This establishes REF . The proof of REF is analogous.
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hep-th/0003243
|
Let MATH be any local operator which is localized in MATH. Since MATH and MATH are solutions of the NAME - NAME equation of mass MATH, the commutator function MATH can be represented in the form MATH . Here the functions MATH are given by MATH where MATH is the momentum space wave function of the single particle vector MATH, and similarly for the other terms. Because of the localization properties of the operators MATH and MATH, the commutator function MATH and its time derivative vanish at time MATH in the half space MATH. In view of the representation REF , this implies that the functions MATH can be analytically continued in MATH into the lower half plane. As MATH is analytic in MATH in a strip about the origin, the functions MATH can likewise be analytically continued in MATH into some strip of the lower half plane. Now if MATH and MATH are open sets such that MATH vanishes for almost all spatial momenta MATH with MATH and MATH, the function MATH vanishes for these momenta as well. Being the boundary value of an analytic function with respect to MATH, MATH therefore vanishes for all MATH and MATH. Since MATH was arbitrary and the set of single particle states MATH, MATH, is dense in MATH, this implies MATH for MATH and MATH. Thus the complement of the support of MATH in momentum space MATH has the form MATH, where MATH is open. Hence, disregarding sets of measure MATH, the statement follows since MATH is different from zero.
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hep-th/0003243
|
The proof is based on induction in MATH. For MATH, one has MATH because of the support properties of MATH in momentum space. Assuming that the statement holds for MATH, let MATH be a test function whose NAME transform has support about points on the mass shell such that MATH. It then follows from relation REF that MATH . Hence, by relation REF and the induction hypothesis, one obtains MATH . Now let MATH be any local operator and MATH any test function such that MATH. There exists for given MATH a test function MATH as in the preceding step such that MATH. For the spectral support of MATH consists of the whole mass shell according to REF and consequently the set of vectors MATH is, for any compact set MATH, dense in the corresponding spectral subspace MATH of single particle states. As the collision states are continuous with respect to their single particle components, one can thus replace in REF the vector MATH by MATH, proving the statement.
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math-ph/0003001
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Obviously, MATH and MATH. To bound MATH we return to REF and note the bound (which holds in MATH) MATH . The second inequality holds because the convolution of two symmetric decreasing functions is symmetric decreasing. Next we turn to MATH, which is a bit more complicated. We split the integration region in REF into MATH and MATH. In region MATH we use MATH and hence the contribution to MATH from this region is bounded above by MATH for the same reason as in REF. In region MATH we use that MATH. We also note, for the integration over MATH, that we can take the mean of the integrand for MATH and MATH. In other words, we can bound this term as follows: MATH . Thus, we obtain the bound MATH .
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math-ph/0003001
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Thanks to REF we only need to establish the contraction property. We first note that for positive real numbers MATH where MATH is some point on the line between MATH and MATH. Thus we get MATH where MATH as in REF . In the last line we have simply noted that the integrals are smaller than the corresponding integral in REF.
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math-ph/0003001
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Take MATH and MATH so that MATH maps MATH into itself. Then, if MATH, the contraction condition will be satisfied.
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math-ph/0003001
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Define MATH to be the subset of MATH on which MATH for all MATH. If we can show that MATH also leaves MATH invariant, then we have shown the wanted REF on the solution in MATH (by the uniqueness of the solution on MATH) except for the possibility that equality also can occur. This is so, since we can apply the fixed point argument not only to MATH but also to MATH. Showing that REF holds is - according to REF - equivalent to showing that MATH holds. Now, using REF and the fact that the factors with the roots are monotone functions MATH and MATH respectively, it is enough to show that MATH holds. To proceed, we now compare the integrands pointwise. Using the explicit REF for MATH and MATH, and with MATH, we will have shown that the integrand of the left hand side pointwise majorizes the one on the right hand side of REF if MATH holds for all MATH. By symmetry, we only have to consider REF for MATH. To this end we exponentiate REF MATH and expand the exponential up to second order. (Note that this gives a lower bound on the exponential because the argument of the exponential function is positive.) That is, it suffices to show that MATH which follows by direct computation. Having established REF we see that REF indeed gives strict inequality in REF for the unique fixed point in MATH.
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math-ph/0003015
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The bundle MATH over NAME spacetime MATH is the trivial bundle MATH. Thus we can simplify our notation by identifying the spinor spaces over all points. Also, the covariant derivative coincides with the partial derivative of the individual components, and the curvature scalar vanishes. Therefore the functions MATH and MATH in REF of a NAME distribution have the form MATH, MATH with the corresponding scalar functions MATH and MATH. That is, the auxiliary two point function of any NAME state of the free NAME field on NAME spacetime is (up to a smooth part) a multiple of the unit matrix, and the nonvanishing components are two point functions of scalar NAME states. Since these are fixed up to a smooth part, the auxiliary two point function is of the form MATH with a scalar NAME distribution MATH (up to a smooth part), and we have MATH . From REF we know that the wave front set of MATH has the form REF , and its polarization set is obviously MATH . We obtain MATH from MATH by application of the operator MATH with principal symbol MATH. Following REF , this does not enlarge the wave front set. In addition, from REF we have the following restriction on the polarization set of MATH: MATH . Now MATH if MATH, and since the projection of the nontrivial part of the polarization set onto the cotangent bundle gives back the wave front set, we see that by the action of MATH the wave front set does not become smaller. Thus MATH has got the same wave front set as the scalar NAME distribution MATH. The equality for the polarization set in REF , however, does not follow from REF , since the operator MATH is characteristic just in the interesting points: we have MATH for MATH. Instead, we can give a direct calculation of the polarization vectors MATH over a point MATH in the wave front set. Because of REF , such vectors have to be a linear combination of the unit and gamma matrices, and we can set MATH. By definition, if MATH is the principal symbol of an operator MATH with MATH, a point MATH can only be contained in the polarization set of MATH if MATH. We know such an operator, since the two point distribution is a solution of the NAME equation MATH and we obtain MATH where MATH. Hence, since MATH, we must have MATH and MATH with MATH. This means that the polarization vectors over points MATH are indeed only the vectors of the form MATH. This completes the proof.
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math-ph/0003015
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A first (rather involved) proof for the NAME equation for spinor half densities was given by CITE. Here we give a simple proof that holds for the NAME equation on MATH. Once we haven chosen coordinate frames for the tangent and cotangent bundle, the operator MATH has principal and subprincipal symbols MATH . We choose MATH and see that MATH, thereby the operator is of real principal type. The characteristic curves of MATH are again the null geodesics. NAME 's connection acts on spinor fields MATH along the bicharacteristics that lie in the kernel of MATH for every point MATH, that is they satisfy MATH. Putting together the different terms in NAME 's connection and using MATH, NAME 's equations and the properties of the NAME symbols and the NAME matrices, yields MATH which is just the covariant derivative of MATH (as a vector field along the geodesic) along the geodesic line with respect to the spin connection. The proof for the adjoint operator is done in the same way.
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math-ph/0003020
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Choose MATH such that MATH. Then MATH and MATH annihilates MATH or vanishes identically. To prove independence, it is enough to notice that REF determining MATH, projected on eigenspaces of MATH reads MATH . Thus, MATH with MATH are linear in NAME variable of scaling degree MATH with successive terms involving variables of lower scaling weights and this proves independence.
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math-ph/0003020
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The order of extension does not depend on the basis chosen. In the basis of REF MATH. Then MATH as a straightforward consequence of REF.
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math-ph/0003020
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Let us take densities of Hamiltonians MATH, MATH. Their scaling degree are correspondingly MATH. These densities are gauge invariant, as follows from REF . Due to corollary to REF they are independent. Thus they may be taken as a part of coordinates on MATH. So MATH. In particular this implies that the multiplicity of MATH in MATH is less or equal to the one in MATH. However, as was shown in ref. CITE, for every MATH there exist such MATH that MATH is regular in MATH. So we have an opposite inequality of multiplicities. This completes the proof.
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math-ph/0003020
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Brute force checking of data presented in REF. We lack more elegant proof so far.
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math-ph/0003020
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As follows from scaling weight grading of NAME structures MATH . Since MATH . Recall that we have assumed ascending ordering of MATH, and so matrix MATH is lower triangular with respect to anti diagonal MATH. This proves that MATH equals to product of those constants. Since, by the choice of coordinates, we know that MATH is nondegenerate we conclude, that all those constants are nonzero.
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math-ph/0003020
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By the implicit function theorem one may resolve the system of polynomial equations MATH locally with respect to variables MATH. Consider the Hamiltonian flows in involution generated by MATH: MATH . Then one has MATH . Restricting on MATH and using the non degeneracy of MATH the result follows.
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math-ph/0003020
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Consider a ring of polynomial functions in MATH. It is obviously generated by elements of MATH. Using the adjoint action of nilpotent group with NAME algebra being MATH one may always reduce MATH to canonical form MATH (see REF ). Indeed, projecting REF on MATH eigenspaces we get MATH where MATH and MATH. This allows to resolve, starting from MATH and proceeding inductively, for MATH and MATH as polynomials in MATH. Recapitulating, MATH generate a ring of polynomials in MATH. The subring of gauge invariant polynomials is, then, obviously generated by MATH, that is, by NAME coordinates.
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math-ph/0003020
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Let MATH corresponds to group element fixing MATH, that is, MATH . Projecting on MATH eigenspaces we gain the system of REF . Since MATH, we have MATH . The rest of REF for MATH determines MATH. Given positive MATH there are MATH scalar equations. Due to injectivity of map MATH for negative MATH we may unambiguously solve for MATH unknowns contained in MATH and remain with MATH scalar equations for MATH, where MATH. Notice, that by definition, MATH iff MATH with MATH being the multiplicity of MATH in MATH. Starting with MATH case MATH we solve for MATH. Proceeding further with excluding MATH, MATH, we end up with MATH equations of order MATH, for each distinct MATH, to determine MATH unknowns contained in MATH. Notice, that the number of REF equals to the number of unknowns MATH and by NAME theorem we obtain no more than MATH solutions MATH, which determine that number of nonlinear transformations MATH of MATH. If only those MATH equations for MATH are independent we obtain an equality in REF .
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math-ph/0003020
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The action of NAME group MATH are inner in MATH and in MATH. For any root MATH, such that MATH, one has a reflection MATH in the hyperplane perpendicular to the root acting on MATH as follows: MATH . These reflections act canonically on NAME subalgebra MATH. Since all elements of MATH can be expressed as a product of these, we conclude that MATH stabilizes MATH. Take a root MATH, such that MATH, and consider the MATH orbit MATH that passes through it. Fix MATH and assume that MATH does not fix MATH. It suffices to prove that MATH is never a root. Since MATH stabilizes MATH, MATH is a linear combination of simple roots of MATH. For simply laced NAME algebras MATH, MATH itself is a root such that MATH. Assume that MATH is a root, then MATH . This is impossible for any two roots MATH and MATH of simply laced simple NAME algebra MATH.
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math-ph/0003020
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Consider the set MATH. Since MATH we have MATH. The set MATH is stable under the homomorphism MATH because MATH and MATH are stable. So MATH and, hence, MATH maps MATH in MATH. The surjectivity is obvious.
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math-ph/0003020
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By REF MATH generate the ring of gauge invariant polynomials in MATH. Restricting them on MATH and restricting adjoint group to MATH the proof follows.
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math-ph/0003020
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Indeed, let us choose another set of algebraically independent polynomials MATH . They are known to be of the same degrees. Using them we define new coordinates MATH: MATH . Starting from functions MATH of the lowest degree and proceeding up we conclude that MATH may be chosen to coincide with MATH.
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math-ph/0003020
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Indeed, NAME MATH, being invariant with respect to adjoint action of MATH, are not left invariant by action of group MATH. Let MATH. Then MATH . This implies that MATH results MATH invariant and hence depends on MATH only.
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math-ph/0003020
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Since MATH are MATH invariant polynomials and so is Killing metrics, we may make use of NAME theorem MATH stating that MATH is a polynomial in MATH invariant polynomials MATH. Now using definition of coordinates MATH REF and changing variables to MATH we obtain MATH . By another NAME theorem MATH are invariant with respect to NAME group of MATH. This means that MATH submatrix MATH is inverse of Killing metrics of rank MATH semisimple NAME algebra MATH restricted to NAME subalgebra. The latter coincides with Killing form of algebra MATH. Thus it is non degenerate due to celebrated NAME 's criterion of semisimplicity.
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math-ph/0003020
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Consider MATH. Rewrite REF as follows MATH . Since functions MATH are not altered we have the same pairing and due to REF MATH action preserves constant non degenerate submatrix of the matrix MATH. It is only possible if MATH acts linearly.
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math-ph/0003020
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Fix a MATH orbit MATH in MATH. It suffices to show that MATH permutes solutions of REF . Notice, that MATH invariant polynomials MATH, corresponding to abelian constituents of MATH are linear and may be chosen to be corresponding NAME variables as follows from REF. Then, due to REF MATH groups acts on them linearly in terms of MATH. Solving REF for MATH we get just MATH solutions as follows from NAME theorem. They correspond to different solutions of REF.
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math-ph/0003020
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Since MATH is a subgroup, there exist two transformations product of which is identity. Since they correspond to the same matrix MATH, the proof follows.
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math-ph/0003020
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NAME group MATH preserves metrics MATH, and is generated by reflections REF corresponding to simple roots of zero MATH grade. Group MATH, restricted to MATH, also preserves the metrics and is generated by reflections associated with simple roots of MATH grade REF, due to REF . We conclude that MATH is a finite group generated by MATH transformations associated to simple roots, that preserve MATH. Since it is an inverse of the Killing form of algebra MATH and MATH acts canonically REF, it follows that MATH is isomorphic to NAME group MATH.
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math-ph/0003020
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Due to REF, MATH is the momentum for the hierarchies in question, so their dynamical equations admit constant solutions, because of REF . Now we address the non degeneracy statement. MATH . Restricting on MATH and using REF and MATH we conclude MATH . Thus, due to assumption of REF , it suffices to prove the non degeneracy of MATH. The latter is obvious because MATH where MATH does not depend on MATH. Restriction on MATH does not alter the linear dependence of MATH on MATH because of REF . Due to REF MATH and is constant. It, thus, remains constant after restriction to MATH.
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math-ph/0003020
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Due to REF dispersionless limit of the second bracket remains linear in MATH. The result follows.
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math-ph/0003020
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MATH are NAME of finite dimensional NAME bracket REF and thus due to REF the result follows.
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math-ph/0003020
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Let us choose coordinates MATH as follows. Take the first MATH coordinates to be MATH and choose the rest to be canonical for finite dimensional bracket. The advantage of this choice is the simplicity of constraint REF MATH . Due to REF we have MATH . Both set of coordinates MATH and MATH are local coordinates on MATH, and so MATH does not degenerate at generic point of MATH. This proves non degeneracy. The following identity along with REF prove that MATH are indeed flat coordinates of metrics MATH: MATH . As a byproduct we conclude that brackets REF define the same geometry on MATH: MATH . This conclusion may be drawn also noting that both coordinates set are densities of mutually commuting integrals of corresponding exact brackets, the property that survives averaging.
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math-ph/0003020
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Indeed, due to REF MATH admits a restriction of MATH and preserves the latter. Hence, MATH were MATH invariant in MATH and so they remain restricted on MATH.
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math-ph/0003020
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Consider the following polynomial in MATH . When all MATH but MATH vanish it simplifies to MATH. At this point MATH and thus MATH is non degenerate. Indeed, let MATH be eigenvector of some representative, which always exists, of MATH in MATH with eigenvalue MATH. Then, due to homogeneity of MATH and from their MATH invariance, we conclude that only variables of degree MATH may differ from zero at this point of MATH. If there are more than one such variable, then eigenvalue MATH is degenerate and we can always choose MATH so that only MATH does not vanish. Due to regularity of MATH, vector MATH is not left fixed by any transformation from MATH, and thus the results follows.
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math-ph/0003020
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Let MATH be the metrics of the second NAME structure and let MATH - of the first. Introduce function MATH, and introduce the following vector fields MATH . Notice, that with this choice MATH and MATH and MATH as follows from considerations above, and where we have assumed MATH to be chosen anti diagonal with all nonzero entries being MATH. Then MATH, as follows from REF. One sees immediately that MATH. The second NAME structure is scaling weight graded and MATH is scaling weight NAME vector field. Thus MATH must be an eigenvector of MATH : MATH. In ref. CITE such flat pencils were called quasihomogeneous of degree MATH.
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math-ph/0003020
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Since obtained pencil of Hamiltonian structures satisfies NAME identity we have a flat pencil of metrics. It is quasihomogeneous as was shown in REF . Thus, following ref. CITE, it is enough to show that the degree of quasihomogeneity MATH. As was said in the proof of REF MATH. Due to quasihomogeneity of flat pencil we have MATH. But MATH . In the last line we have used the fact that MATH vanishes unless MATH, as follows from scaling weight grading of NAME structures. From this we obtain MATH . Since MATH is finite we obtain that MATH. Note that for the NAME conjugacy class MATH - NAME number, and we recover REF, obtained while constructing polynomial solutions to WDVV equations on the orbits of NAME groups CITE. Practically, we can find NAME potential MATH from the following relations MATH where MATH.
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math-ph/0003026
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Let MATH: MATH. This leads to MATH and, therefore, MATH must have the form MATH. Consequently, MATH and there exists MATH such that MATH and MATH. Therefore, in the symplectic decomposition MATH we obtain MATH . Moreover, MATH, so MATH and then MATH with MATH. Similarly, MATH with MATH. Since both MATH and MATH are effective, MATH. We can suppose, for example, that MATH. If MATH then the proof is finished. If not, since MATH, we can suppose that MATH. Denote then: MATH . Since MATH, MATH. Then necessarily MATH. Recall that MATH so we have MATH, that is, MATH and then, MATH . This implies that MATH.
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math-ph/0003026
|
For all MATH and MATH, MATH. Consequently, MATH, MATH. Recall that MATH is into. This leads to the relation MATH which holds for all MATH. Note that MATH since MATH is effective. Therefore, MATH, for all MATH and MATH holds for all MATH. Applying REF , we can deduce that there exists MATH such that MATH. Let then MATH be such that MATH. In the symplectic decomposition MATH we obtain MATH where MATH is an effective form on MATH. Moreover, MATH is parabolic on MATH, since MATH. From MATH one can conclude that there exists a symplectic basis MATH of MATH in which MATH .
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math-ph/0003026
|
MATH is non-degenerate on any subspace MATH of MATH such that MATH. Therefore, MATH must be even. Moreover, MATH is not zero, so MATH. Since MATH, MATH.
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math-ph/0003026
|
Let MATH be a vector satisfying MATH and let MATH be the symplectic decomposition of MATH in MATH. If MATH with MATH, we can write MATH, MATH. Then, if we assume that MATH for such a MATH, it is straightforward to obtain MATH and therefore, MATH. However, since MATH, MATH is non-degenerate on MATH and then MATH. It is impossible, so we can conclude that MATH.
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math-ph/0003026
|
We can assume that MATH is a polynomial map of degree less than MATH. Denote MATH. We assume first that there exists MATH such that MATH. Since MATH, MATH is an isomorphism and then MATH is a local diffeomorphism. Let MATH be defined by MATH . Denote MATH. MATH is a homogeneous polynom of degree MATH. It is easy to check that for any form MATH we have MATH where MATH. Then, one gets MATH . However, MATH, so MATH must be zero. This leads to MATH . Moreover, we have: MATH . Recall that MATH is a polynomial map with MATH. Therefore, MATH . Since MATH, it is easy to check that MATH . MATH is a hamiltonian vector field with the homogeneous coefficents of degree MATH. Consequently, there exists MATH for which MATH. Conversely, let us assume that there exists MATH with MATH in MATH such that MATH. Put MATH with MATH and MATH. Similar considerations lead to MATH . Consequently, one can conclude that MATH and MATH.
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math-ph/0003026
|
MATH is surjective. Moreover, MATH . Therefore, MATH is a local diffeomorphism.
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math-ph/0003026
|
Since MATH, MATH satisfies: MATH . Therefore, MATH and MATH . After observing that for any form MATH the relation MATH must hold (MATH), we get the result.
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math-ph/0003026
|
Suppose first that MATH. Then there exists MATH with MATH such that MATH. So, one has MATH . Therefore, MATH with MATH, MATH, MATH, MATH and MATH. Conversely, if there exist MATH and MATH such that MATH with MATH and MATH, then MATH. It means that MATH and, therefore, MATH with MATH, MATH.
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math-ph/0003027
|
In coordinates, any vector field MATH of MATH is of the type MATH, where MATH are functions. The time form writes as MATH. Hence, MATH. Therefore, MATH if and only if MATH and MATH. But the first condition on MATH means that the vector field MATH is projectable and, together with the second condition, that MATH is even constant. MATH .
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math-ph/0003027
|
If MATH where MATH, then, MATH. MATH .
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math-ph/0003027
|
In coordinates, we have MATH . Since MATH is projectable, this yields MATH. This does not depend on the coefficients MATH. The result turns out to be independent of the chart. MATH .
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math-ph/0003027
|
MATH is a (scaled) fibred morphism over MATH. Thus, REF yields the equality MATH where the components of MATH were given by MATH. MATH .
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math-ph/0003027
|
The coordinate expression shows that because of the projectabilty of MATH the vertical restriction does not contain coefficients of the type MATH. The result turns out to be independent of the chart. MATH .
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math-ph/0003027
|
The proof of MATH follows easily in virtue of the NAME rule and the fact that MATH is a symmetry of the contact maps. The proof of MATH can be obtained easily by considering the expressions of REF which are polynomial in the coordinates MATH or MATH. MATH .
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math-ph/0003027
|
REF yields, MATH. The proof of MATH can be seen directly because of the fact that MATH is a symmetry of MATH. For the proof of MATH we consider the coordinate expressions MATH and MATH, where we have set MATH to be the coefficient of MATH in REF and MATH the coefficient of MATH in REF . It can be easily seen that, if the sum of these expressions is zero, then, all MATH and all MATH have to be zero. MATH .
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math-ph/0003027
|
REF yields MATH. In analogy to the proof of REF this yields the equivalence between the condition MATH and the two conditions MATH, MATH. Clearly, MATH is equivalent to MATH. Thus, REF yield the result. MATH .
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math-ph/0003027
|
This follows directly from NAME 's formula using MATH and the closure of MATH. MATH .
|
math-ph/0003027
|
Let MATH be a second order connection such that MATH. This implies that there exists a local function MATH such that MATH. Using MATH, we get MATH. Let us set MATH; MATH is valued into MATH, and has coordinate expression MATH, with MATH. We have to show that a local function MATH only exists if MATH. Calculating in coordinates using MATH one gets the following system of equalities MATH, MATH, MATH. This systems implies that MATH. Thus, MATH. MATH .
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math-ph/0003027
|
MATH is locally equivalent to the closure of MATH. Hence, there is a local function MATH such that MATH. Therefore, MATH. MATH .
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math-ph/0003027
|
MATH. This is equivalent to the equation MATH. It follows directly from REF that MATH is a conserved quantity. MATH .
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math-ph/0003027
|
Both directions can be proved in analogy to the proof of the equivalence MATH. MATH .
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math-ph/0003027
|
By recalling that MATH we obtain that MATH. Hence, by observing that two vector fields MATH are equal if and only if MATH and MATH, we obtain the result. MATH .
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math-ph/0003027
|
The equalities MATH and MATH. yield the result. MATH .
|
math-ph/0003027
|
The first expression follows simply from the contact splitting of MATH and REF . The observer dependent expression of MATH follows simply from the splitting of MATH through the observer. The coordinate expression MATH with respect to a basis MATH yields the second expression. MATH .
|
math/0003002
|
Here is a well-known explicit description of the group MATH of points of order MATH on MATH. Let us denote by MATH the MATH-divisor MATH on MATH if MATH is odd and the MATH-divisor MATH if MATH is even. In both cases MATH is an effective divisor of degree MATH. Namely, let MATH be a subset of even cardinality. Then (CITE, Ch. NAME, REF ; CITE, pp. REF; see also CITE) the divisor MATH on MATH has degree MATH and MATH is principal. If MATH are two subsets of even cardinality in MATH then the divisors MATH and MATH are linearly equivalent if and only if either MATH or MATH. Also, if MATH then the divisor MATH is linearly equivalent to MATH. Hereafter we use the symbol MATH for the symmetric difference of two sets. Counting arguments imply easily that each point of MATH is the class of MATH for some MATH. We know that such a choice is not unique. However, in the case of odd MATH if we demand that MATH does not contain MATH then such a choice always exists and unique. This observation leads to a canonical group isomorphism MATH in the case of odd MATH. Here MATH stands for the linear equivalence class of a divisor. In the case of even MATH we are still able to define a canonical surjective group homomorphism MATH and one may easily check that the kernel of this map is the line generated by the set MATH, that is, the line generated by the constant function MATH. This gives rise to the injective homomorphism MATH which is an isomorphism, by counting arguments. So, in both (odd and even) cases we get a canonical isomorphism MATH, which obviously commutes with the actions of MATH. In other words, we constructed an isomorphism of MATH-modiles MATH and MATH. In light of REF , this ends the proof of REF .
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math/0003002
|
Let us put MATH . It follows from REF that the MATH-module MATH is very simple. Now the result follows readily from REF .
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math/0003002
|
Clearly, MATH . This implies easily that MATH where MATH. Since MATH is a proper subset of MATH, we have MATH . Recall that there exists MATH such that MATH lies in MATH and the exact multiplicative order of MATH is MATH. This implies that MATH. Since MATH, we conclude that MATH. Therefore MATH.
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math/0003002
|
Let us put MATH. We have MATH . Clearly, for each MATH one may find a MATH matrix with determinant MATH and trace MATH. This implies that MATH satisfies the conditions of REF . The construction described in REF allows us to construct a MATH-dimensional MATH-representation MATH of MATH for each subset MATH of of MATH. It is well-known (CITE, pp. REF) that MATH's exhaust the list of all absolutely irreducible MATH-representations of MATH and therefore MATH is isomorphic to MATH for some MATH. It follows from REF that either MATH is empty or MATH. The case of empty MATH corresponds to the trivial MATH-dimensional representation. Therefore MATH and MATH is MATH-dimensional.
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math/0003002
|
Let us put MATH. We know that MATH satisfies the conditions of REF . The construction described in REF allows us to construct a MATH-dimensional MATH-representation MATH of MATH for each subset MATH of of MATH. It is known (CITE, pp. REF) that MATH's exhaust the list of all absolutely irreducible MATH-representations of MATH and therefore MATH is isomorphic to MATH for some MATH. It follows from REF that either MATH is empty or MATH. The case of empty MATH corresponds to trivial MATH-dimensional representation. Therefore MATH and MATH is MATH-dimensional.
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math/0003002
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We may assume that MATH. Clearly, MATH is a faithful MATH-module and MATH . CASE: MATH is a semisimple MATH-module. Indeed, let MATH be a simple MATH-submodule. Then MATH is a non-zero MATH-stable subspace in MATH and therefore must coincide with MATH. On the other hand, each MATH is also a MATH-submodule in MATH, because MATH. In addition, if MATH is a MATH-submodule then MATH is a MATH-submodule in MATH, because MATH . Since MATH is simple, MATH or MATH. This implies that MATH is also simple. Hence MATH is a sum of simple MATH-modules and therefore is a semisimple MATH-module. CASE: The MATH-module MATH is either isotypic or induced. Indeed, let us split the semisimple MATH-module MATH into the direct sum MATH of its isotypic components. Dimension arguments imply that MATH. It follows easily from the arguments of the previous step that for each isotypic component MATH its image MATH is an isotypic MATH-submodule for each MATH and therefore is contained in some MATH. Similarly, MATH is an isotypic submodule obviously containing MATH. Since MATH is the isotypic component, MATH and therefore MATH. This means that MATH permutes the MATH; since MATH is MATH-simple, MATH permutes them transitively. This implies that all MATH have the same dimension MATH and therefore MATH divides MATH. Let MATH be the stabilizer of MATH in MATH, that is, MATH . The transitivity of the action of MATH on MATH-s implies that MATH. If MATH then MATH. This means that MATH for all MATH and MATH is isotypic. Assume that MATH and consider the MATH-module MATH. Clearly, MATH divides MATH and the MATH-module MATH is iduced by MATH. CASE: Assume that MATH and MATH. Then each MATH is one-dimensional and contains exactly one non-zero vector say, MATH. Then the sum MATH is a non-zero MATH-invariant vector which contradicts the simplicity of the MATH-module MATH.
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math/0003002
|
Since MATH is isotypic, there exist a simple MATH-module MATH, a positive integer MATH and an isomorphism MATH of MATH-modules. Let us put MATH . The isomorphism MATH gives rise to the isomorphism of MATH-vector spaces MATH . We have MATH . Clearly, MATH is isomorphic to the matrix algebra MATH of size MATH over MATH. Let us put MATH . Since MATH is simple, MATH is a finite-dimensional division algebra over MATH. Therefore MATH must be a finite field. We have MATH . Clearly, MATH divides MATH and therefore divides MATH. Clearly, MATH is always a cyclic group of order MATH and therefore has order dividing MATH. Clearly, MATH is stable under the adjoint action of MATH. This induces a homomorphism MATH . Since MATH is the center of MATH, it is stable under the action of MATH, that is, we get a homomorphism MATH, which must be trivial, since MATH is perfect and MATH is a cyclic group of order dividing MATH and therefore the kernel of the homomorphism must coincide with MATH. This implies that the center MATH of MATH commutes with MATH. Since MATH, we have MATH. This implies that MATH and one may rewrite MATH as MATH . It follows from the NAME density theorem that MATH with MATH. The adjoint action of MATH on MATH gives rise to a homomorphism MATH . Clearly, MATH and MATH provide MATH and MATH respectively with the structure of MATH-modules. Notice that MATH . Now our task boils down to comparison of the structures of MATH-module on MATH defined by MATH and MATH respectively. I claim that MATH . Indeed, notice that the conjugation by MATH in MATH leaves stable MATH and coincides on MATH with the conjugation by MATH. Since the centralizer of MATH in MATH coincides with MATH, there exists MATH such that MATH . Since the conjugation by MATH leaves stable the centralizer of MATH, that is, MATH and coincides on it with the conjugation by MATH, there exists a non-zero constant MATH such that MATH. This implies that MATH . Now one has only to recall that MATH and therefore MATH.
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math/0003002
|
We have MATH. Clearly, it suffices to check that the MATH-module MATH is very simple. First, notice that MATH acts doubly transitively on MATH. Indeed, each subgroup of MATH (except MATH itself) has index MATH (REF , p. CASE: This implies that MATH acts transitively on MATH. If the stabilizer MATH of a point MATH has index MATH then it follows easily from REF on p. REF) that MATH in conjugate to REF subgroup of upper-triangular matrices and therefore the MATH-set MATH is isomorphic to the projective line MATH with the standard action of MATH which is well-known to be doubly (and even triply) transitive. By REF , this implies that the MATH-module MATH is absolutely simple. Recall that MATH . By REF , there no absolutely simple nontrivial MATH-modules of dimension MATH. This implies that MATH is not isomorphic to a tensor product of absolutely simple MATH-modules of dimension MATH. Recall that all subgroups in MATH different from MATH have index MATH. It follows from REF that the MATH-module MATH is very simple. Since MATH, the MATH-module MATH is also very simple.
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math/0003002
|
We have MATH. First, notice that MATH acts doubly transitively on MATH. Indeed, the classification of subgroups of NAME groups REF , p. REF implies that each subgroup of MATH (except MATH itself) has index MATH. This implies that MATH acts transitively on MATH. If the stabilizer MATH of a point MATH has index MATH then it follows easily from the same classification that MATH is conjugate to the subgroup MATH generated by all MATH and MATH and therefore the MATH-set MATH is isomorphic to an ovoid MATH where the action of MATH is known to be doubly transitive REF on pp. REF of its proof on p. CASE: By REF , this implies that the MATH-module MATH is absolutely simple. Recall that MATH. By REF , there no absolutely simple nontrivial MATH-modules of dimension MATH. This implies that MATH is not isomorphic to a tensor product of absolutely simple MATH-modules of dimension MATH. Recall that all subgroups in MATH (except MATH itself) have index MATH. It follows from REF that the MATH-module MATH is very simple. Since MATH, the MATH-module MATH is also very simple.
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math/0003002
|
REF follow from REF respectively applied to MATH.
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math/0003002
|
Assume that MATH. Since MATH contains a subgroup isomorphic to MATH CITE, p. REF, it suffices to check the case of MATH, in light of REF . The group MATH has two conjugacy classes of maximal subgroups of index MATH and all other subgroups in MATH have index greater than MATH CITE, p. CASE: Therefore all subgroups in MATH (except MATH itself) have index greater than MATH and the action of MATH on the MATH-element set MATH is transitive. The permutation character (in both cases) is in notations of CITE, p. CASE: MATH, that is, MATH. The restriction of MATH to the set of MATH-regular elements coincides with absolutely irreducible NAME character MATH in notations of CITE, p. CASE: In particular, the corresponding MATH-module MATH is absolutely simple and has dimension MATH. Since MATH and MATH is a prime, the very simplicity of the MATH-module MATH follows from REF. This proves REF . Assume that MATH. Suppose MATH and the action of MATH on the MATH-element set MATH is transitive. All subgroups MATH in MATH of index MATH are isomorphic to MATH CITE, p. CASE: It follows from the already proven REF for MATH and REF that the MATH-module MATH is very simple. Suppose MATH. The action of MATH on the MATH-element MATH is transitive, since all subgroups in MATH (except MATH itself) have index MATH. All subgroups MATH in MATH of index MATH are isomorphic to MATH CITE, p. CASE: It follows from the already proven REF for MATH and REF that the MATH-module MATH is very simple.
|
math/0003008
|
The first formula is REF and the second one is established in the proof of CITE; compare also CITE.
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math/0003008
|
We will apply the Lemma with the roles of MATH and MATH interchanged. Since MATH holds for the counit MATH (for example, CITE), this requires replacing MATH by MATH. First assume that MATH is the centrally primitive idempotent of MATH corresponding to some irreducible MATH-module MATH. Then, by the Lemma, MATH. Thus, MATH . In general, MATH for certain irreducible MATH-modules MATH, and so MATH for the MATH-module MATH.
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math/0003010
|
Using only the facts that MATH and MATH define a measure (that is, not necessarily a probability measure), the proofs of REF will establish the equations: MATH . To get a recurrence relation for the MATH's, one simply plugs the second equation into the first. Similarly one obtains a recurrence relation for the MATH's. These recurrences allow one to solve for MATH in terms of MATH, implying that the formulas for MATH are proportional to the asserted values. Thus it is enough to prove that the asserted formulas for MATH satisfy the equation MATH. This follows readily from REF .
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math/0003010
|
The arguments are completely analogous to those for MATH in CITE REF , using the cycle index of the finite symplectic groups CITE.
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math/0003010
|
The most important observation (see CITE for a readable proof) is that the fixed space of an element MATH of MATH has isometry type MATH precisely when the partition corresponding to the polynomial MATH in the rational canonical form of MATH satisfies MATH and MATH. In other words, the partition has MATH parts and MATH's. Let MATH denote the coefficient of MATH in some polynomial MATH. Then one uses the cycle index for the unitary groups as in CITE to see that the sought probability for MATH is MATH . Using the fact that MATH this becomes MATH where the last equality is in the proof of REF. The formula for the MATH limit follows from the well-known identity MATH . Alternatively, it follows from the principle that the limit as MATH of MATH is MATH if MATH has a NAME expansion around MATH converging in a circle of radius MATH, together with the formula for MATH given in REF.
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math/0003010
|
The idea for all of the proofs is the same; hence we prove part two as follows: MATH as desired. Note that the meat of the lemma is the second equality, which follows from the formulas for MATH and MATH. The third equality is simply a relabelling of subscripts.
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math/0003010
|
The MATH probability of choosing a partition with MATH for all MATH is MATH . Since MATH is equal to MATH by definition, it it is enough to prove two claims: first that for every choice of MATH, MATH is equal to the asserted transition rule probability for moving from MATH to MATH, and second that the transition rule probabilities sum to one. The first claim follows from REF . For the second claim, observe that MATH because MATH is a measure and the columns of a partition are non-increasing in size as one moves to the right. Since MATH, it follows that MATH is a probability measure, as promised earlier.
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math/0003014
|
Integrating by parts, we obtain MATH . In view of REF , we have MATH . Now the lemma is proved by passing to the limit as MATH.
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math/0003014
|
The identity REF imply that MATH . Therefore, in view of REF , MATH . Taking into account REF and the second REF , we obtain MATH . REF implies that MATH and MATH satisfy the conditions of REF and that MATH. Obviously, REF follows from REF .
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math/0003014
|
In view of REF we have MATH . This estimates, REF imply that the functions MATH, MATH and MATH are uniformly bounded on MATH. Since MATH whenever MATH, integrating by parts with respect to MATH we obtain MATH . Now REF follows from REF .
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math/0003014
|
REF are obviously fulfilled; REF follow from the fact that MATH. Finally, REF holds true because the MATH-th derivative of the extended function MATH coincides with a linear combination of a MATH-function and two MATH-functions.
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math/0003014
|
Let MATH be the multiplication operator and MATH be the NAME multiplier generated by the characteristic function of the interval MATH. Then the NAME norm of the operator MATH acting in MATH is equal to MATH. Therefore MATH which implies that MATH .
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math/0003014
|
If MATH and MATH is the norm in MATH then MATH . One can easily see that MATH . Therefore REF implies the required estimate.
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math/0003014
|
According to REF , the NAME transform of MATH coincides on the interval MATH with the NAME transform of MATH . Since MATH is even, this implies that MATH for all MATH. Estimating MATH in the integrals on the right hand sides, we arrive at REF - REF .
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math/0003014
|
If we define MATH as in REF with MATH (see REF ) then REF follow from REF with MATH, REF .
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math/0003014
|
Since MATH, REF with MATH, MATH, MATH and MATH implies MATH . Let MATH be defined as in REF with MATH. Then REF follows from REF . Since MATH, REF imply that MATH . Estimating MATH with MATH and applying REF , we obtain REF .
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math/0003014
|
REF are proved by straightforward integration of REF . REF imply that MATH . Now, applying the first REF with MATH, we arrive at REF .
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math/0003032
|
A rank one algebraic factor has to have fibres of positive dimension. Hence the pre - image of the origin under the factor map is a union of finitely many rational tori of positive dimension and by REF MATH cannot be irreducible.
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math/0003032
|
By diagonalizing MATH over MATH and taking the real form of it, one immediately sees that the centralizer of MATH in MATH acts transitively on vectors with nonzero projections on all eigenspaces and thus has a single open and dense orbit. Since the centralizer over MATH is the closure of the centralizer over MATH, the MATH-linear span of the orbit of any integer or rational vector under the centralizer is an invariant rational subspace. Hence any integer point other than the origin belongs to the single open dense orbit of the centralizer of MATH in MATH. This implies the statement of the lemma.
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math/0003032
|
Choose MATH such that MATH. Let MATH be cyclic vectors for MATH and MATH, respectively. Now consider the integer vector MATH and find MATH commuting with MATH such that MATH. We have MATH. The conjugacy MATH maps bijectively the MATH - span of the MATH - orbit of MATH to MATH - span of the MATH - orbit of MATH. By cyclicity both spans coincide with MATH, and hence MATH.
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math/0003032
|
If MATH for MATH, then MATH has MATH as an eigenvalue, and hence has a rational subspace consisting of all invariant vectors. This subspace must be invariant under MATH which contradicts its irreducibility.
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math/0003032
|
MATH is a ring because MATH is a ring. As we pointed out above images of integer matrices are algebraic integers and images of matrices with determinant MATH are algebraic units. Hence MATH. Finally, for every polynomial MATH with integer coefficients, MATH is an integer matrix, hence MATH.
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math/0003032
|
By REF , MATH is isomorphic to the group of units in the order MATH, the statement follows from the NAME Unit Theorem (CITE, Ch. REF).
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