paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003068 | The simply connected MATH-manifolds MATH and MATH are non-spin and have the same invariants MATH; NAME 's classification CITE therefore tells us that they are homeomorphic. On the other hand, because MATH, these complex surfaces have NAME dimension MATH. REF therefore tells us that MATH whereas MATH . Thus these manifo... |
math/0003068 | First recall that, in terms of the decomposition MATH the curvature operator of an oriented Riemannian MATH-manifold can be expressed in the form MATH where MATH and MATH are trace-free, and where MATH exactly corresponds to MATH. On the other hand, every REF-form MATH on MATH can be uniquely written as MATH, where MAT... |
math/0003068 | By REF we know that MATH . On the other hand, MATH at all points by virtue of MATH by REF . Hence MATH and so MATH . Taking the square root of this inequality then yields the claim. |
math/0003068 | Let MATH be the minimal model of MATH. Then for every metric MATH on MATH there is a spin-MATH structure with MATH and such that the corresponding NAME invariant is non-zero for the polarization determined by MATH; in particular, the scalar curvature MATH of MATH is negative somewhere. Thus, REF tells us that MATH sati... |
math/0003068 | Suppose that MATH is a metric with MATH and total volume MATH. Thus, letting MATH denote the minimal model of MATH, and choosing a spin-MATH structure on MATH as in the proof of REF , MATH by virtue of REF . Since the spin-MATH structure is chosen so that MATH, equality must hold, it follows that MATH and MATH; in part... |
math/0003073 | The main idea of the proof relies on the identity MATH where MATH is the following coboundary operator: MATH . Observe that for MATH, MATH is a differential only for the MATH-reduction of the graded algebra of functionals. Using MATH is like working with a cosmological term, and, upon using the generalized NAME theorem... |
math/0003079 | Let MATH be a smooth function such that MATH for all MATH and define MATH . Then MATH and hence the tangent space of MATH at MATH is given by MATH . The tangent space to the MATH-orbit consists of all vector fields of the form MATH, where MATH. The map MATH identifies the quotient space MATH with the space of closed MA... |
math/0003079 | Choose a smooth path MATH such that MATH for every MATH and let MATH be defined by REF . Then MATH if and only if MATH. It follows from the definitions that this is equivalent to MATH for all MATH and all MATH. This means that there exists a smooth family of vector fields MATH such that MATH . Equivalently, MATH where ... |
math/0003079 | The ``if" part was proved in REF . Suppose that the path MATH is tangent to MATH. Choose a smooth function MATH such that MATH for every MATH and let MATH be defined by REF . By assumption, MATH is exact for every MATH. Fix a smooth path MATH and, for every MATH, choose MATH such that MATH . Then the function MATH is s... |
math/0003079 | Choose a lift MATH of MATH and denote MATH . Then MATH and hence MATH as claimed. MATH . |
math/0003079 | The parallel transport of MATH along a curve MATH is determined by the Hamiltonian functions MATH via REF . The functions MATH in REF correspond to the path MATH. By REF , the Hamiltonian isotopy determined by MATH preserves MATH if and only if MATH for every MATH. This shows that REF is equivalent to REF . To prove th... |
math/0003079 | Let MATH be defined by MATH . Since MATH is connected there exists a smooth path MATH such that, for every MATH, MATH . For example choose MATH in the interval MATH such that MATH for MATH and MATH for MATH. Then define MATH for MATH such that REF is satisfied. Let MATH be defined by MATH . By REF , the function MATH i... |
math/0003079 | Let MATH be an exact Lagrangian loop. Let MATH be the connection MATH-form defined by REF in the proof of REF , where the cutoff function MATH is chosen to be nondecreasing. Then MATH where MATH are given by REF and MATH. Taking the differential of this MATH-form on MATH we find MATH . Since MATH and MATH we obtain MAT... |
math/0003079 | Let MATH be given by REF with MATH replaced by MATH for MATH. Denote MATH and let MATH be defined by REF . Since MATH it follows from REF that there exists a function MATH such that MATH for every MATH. We shall prove that the required identity holds with MATH . To see this note that, by REF , MATH where MATH denotes t... |
math/0003079 | Let MATH be an exact Lagrangian loop and MATH be smooth functions such that the functions MATH defined by REF satisfy MATH for every MATH. For every smooth function MATH let MATH be given by REF . In particular, MATH is given by REF with MATH. We shall prove that MATH . To see this, note that MATH and hence MATH . This... |
math/0003079 | We prove that MATH. To see this, choose smooth functions MATH such that MATH and the map MATH defined by MATH is an embedding with MATH. Then MATH . Hence MATH where MATH. Hence MATH and this implies MATH . By REF in the appendix, two loops MATH and MATH, associated to two embedded discs MATH via REF , are Hamiltonian ... |
math/0003079 | We compute MATH . This proves the lemma. MATH . |
math/0003079 | Let MATH and MATH. Let MATH be given by REF . Then MATH is a MATH-holomorphic curve. By REF , MATH is tamed by MATH. Hence MATH . The infimum of the numbers on the right is MATH. This proves the lemma. MATH . |
math/0003079 | Let MATH and MATH. Let MATH be given by REF . Then MATH is a MATH-holomorphic curve. By REF , MATH is tamed by MATH. Hence MATH . The supremum of the numbers on the right is MATH. This proves the lemma. MATH . |
math/0003079 | REF . MATH . |
math/0003079 | REF . MATH . |
math/0003079 | If MATH for every MATH then MATH (with reversed orientation) and MATH form a sphere and the difference MATH is equal twice the first NAME number of this sphere. Hence the difference of the NAME numbers is an even multiple of MATH. This continues to hold whenever MATH is homotopic to MATH as a section of the bundle MATH... |
math/0003079 | In the case of MATH, consider the constant function MATH . Then a trivialization of the pullback tangent bundle MATH is determined by the coordinate chart MATH . In these coordinates the Hamiltonian flow is MATH . Since MATH we see that the NAME index of the loop MATH is equal to MATH. This proves the second equation i... |
math/0003079 | Let MATH be given by REF with MATH and MATH where MATH and MATH is given by REF . As in REF , MATH is a smooth nondecreasing cutoff function such that MATH for MATH near MATH and MATH for MATH near MATH. The formula MATH for MATH shows that MATH for MATH and MATH for MATH. By REF with MATH and MATH, MATH . The explicit... |
math/0003079 | The loop MATH, given by REF , is Hamiltonian isotopic to MATH and hence MATH . By REF , MATH and MATH . Hence, by REF , MATH . Here MATH denotes the connection MATH-form introduced in the proof of REF . Hence MATH . Since MATH the result follows from REF . MATH . |
math/0003079 | Let MATH be a smooth isotopy such that MATH and MATH. Define MATH for MATH. Then MATH and MATH for all MATH and MATH. Fix a Riemannian metric on MATH with volume form MATH and let MATH be defined by MATH . Choose MATH such that MATH and define MATH by MATH . Then MATH and MATH. Hence MATH for all MATH and MATH. Moreove... |
math/0003079 | The result follows again from NAME isotopy. We prove that there exists a MATH-form MATH such that MATH . To see this, choose any MATH-form MATH such that MATH. Then the integral of MATH over MATH vanishes and so MATH is exact. Hence there exists a smooth function MATH such that MATH and the MATH-form MATH satisfies REF... |
math/0003079 | Choose orientation preserving embeddings MATH such that MATH and MATH. We prove the result in four steps. CASE: There exists a diffeomorphism MATH that is isotopic to the identity and satisfies MATH. Choose a path MATH such that MATH and MATH. Next choose a smooth family of vector fields MATH such that MATH for every M... |
math/0003080 | It is clear from the definitions that the equivalence relation MATH is contained in MATH. For the converse, suppose MATH. Then there exist MATH and MATH, such that MATH. By splitting MATH and MATH into their component terms for MATH we obtain MATH for some MATH, MATH. It follows immediately from this that MATH. |
math/0003080 | All that remains to be verified is that NAME resulting from matches found in MATH can be added to MATH without altering MATH. We assume all polynomials in MATH to be monic (possible since MATH is a field). Now NAME result from two types of overlap. For the first case let MATH be polynomials in MATH such that MATH for s... |
math/0003081 | Assume for the block MATH the notations given in REF . Since MATH and MATH belong to different components of MATH, the MATH-edge MATH, together with its endpoints MATH, is a dipole of type REF in MATH. The cancellation of this dipole produces a dipole of type REF involving the MATH-coloured edge MATH. The sequence of c... |
math/0003081 | CASE: By deleting all dipoles involving the MATH- and MATH-edges connecting MATH with MATH we obtain the ``normal" crystallization of the lens space MATH (see CITE). CASE: MATH. |
math/0003081 | With the previous assumptions and notation, it suffices to show that the MATH-coloured graph MATH obtained by cancelling MATH in MATH is c.p.-isomorphic to MATH. First of all, exchange the names of the two cycles MATH and MATH, together with the first components in the labelling of their vertices. After this relabellin... |
math/0003081 | See REF . |
math/0003081 | The relations MATH, MATH, MATH and MATH hold. Therefore, we get for MATH the classical presentation of the NAME four group: MATH. To obtain the second result, define MATH and observe that both MATH and MATH commute with MATH. By NAME transformations, we have: MATH, MATH, MATH, MATH, which is a classical presentation of... |
math/0003081 | By MATH and MATH we can permute MATH in all possible way and therefore REF can be achieved. REF follow by a suitable application of MATH. REF follow by a combined application of the three maps. The unicity of such a MATH is straightforward. |
math/0003081 | By REF there exists MATH and MATH such that MATH. Hence, by REF : MATH. |
math/0003081 | MATH is MATH-equivalent to a MATH-tuple MATH verifying condition (MATH). By REF , MATH is MATH-equivalent to either MATH or MATH or MATH. Since MATH if and only if MATH, it is easy to see that the admissible MATH-tuples MATH and MATH both verify condition MATH and therefore the statement is achieved. |
math/0003081 | CASE: There is a finite number of traps of a fixed type. CASE: Trivial. |
math/0003081 | In one direction REF the statement is trivial since MATH are MATH-equivalent to MATH. To prove the converse, denote by MATH the graph whose vertex-set is the set MATH of all canonical MATH-tuples and whose edge-set is defined by the following rule: join two different vertices MATH and MATH by an edge iff there exist tw... |
math/0003081 | From REF we always get MATH and MATH. Moreover, MATH when either MATH or MATH or MATH, by REF . Now, if MATH then MATH, for MATH. Therefore MATH whenever MATH. On the other hand, it is easy to check that REF (respectively, REF ) includes all the minimal MATH-tuples MATH such that MATH and MATH (respectively, such that ... |
math/0003082 | Let MATH and MATH be a standard solution. If MATH and MATH is another solution of the conjugate equation, then MATH and MATH for some invertible MATH CITE, hence the conjugate of MATH with respect to MATH and MATH is given by MATH where MATH is the modular group of MATH , so that MATH because the minimal left inverse M... |
math/0003082 | We may suppose that MATH. Set MATH, where MATH is the minimal left inverse of MATH as before. Since the NAME cocycle MATH is a covariance cocycle for MATH CITE, there exists a one-dimensional character MATH of MATH such that MATH (both cocycles intertwine MATH and MATH). As shown in CITE, MATH . Moreover the NAME cocyc... |
math/0003082 | We have MATH CITE (the conjugate map is the one associated with MATH), hence the above REF applies, provided MATH is irreducible in End-MATH. If MATH is reducible in End-MATH we have MATH with MATH. By using the tracial property of MATH it is easy to check that MATH is a one-parameter group of unitaries in the finite-d... |
math/0003082 | See REF . |
math/0003082 | Let MATH be a cyclic separating vector such that MATH, MATH, and let MATH be the isometry of MATH given by MATH . The final projection of MATH is given by MATH thus MATH is a homomorphism of MATH onto MATH and MATH . Now the central support of MATH in MATH is MATH as MATH, hence if MATH there exists a unique MATH such ... |
math/0003082 | The map MATH is a functor of MATH tensor categories from MATH to a sub-tensor category of End-MATH, hence MATH. As MATH has a unitary braiding, every real object MATH is amenable REF , thus MATH, where MATH is the MATH norm of the fusion matrix MATH associated with MATH, therefore MATH, thus MATH. For any MATH, the obj... |
math/0003082 | Let MATH a holomorphic unitary covariance MATH-cocycle. As MATH is continuous, the map MATH is strongly continuous. To check this, note that by cocycle property it is enough to verify the continuity at MATH because then the strong limit MATH due to the normality of MATH. Let then MATH be a weak limit point of MATH as M... |
math/0003082 | By assumption and REF both MATH and MATH extend to MATH. As MATH and MATH are also intertwiners on MATH by weak continuity and the conjugate equation for MATH and MATH is obviously satisfied on MATH, the extension of MATH to MATH has finite index. |
math/0003082 | By REF MATH extends to MATH and has finite index. We are then in the case covered by REF . |
math/0003082 | As usual MATH is identified with MATH. If MATH, there exist two non-zero projections MATH with sum MATH. Thus MATH are different graded KMS functionals on MATH. This can be checked since both the extensions of MATH and MATH act trivially on MATH, and by usual approximation arguments. By the uniqueness assumption there ... |
math/0003082 | The holomorphic property of MATH is a direct consequence of the holomorphic property of the NAME cocycle because MATH is indeed a NAME cocycle, up to phase, with respect to two bounded positive normal functionals of MATH (see CITE and the previous section). Note now that, since the extension of MATH to MATH (still deno... |
math/0003082 | Because of additivity and wedge duality one can write MATH (intersection over all wedges containing MATH). If MATH is spacelike separated from MATH there is a wedge MATH with MATH and MATH, hence MATH is local. As MATH extends MATH and is local, wedge duality must hold for MATH too, therefore MATH for all wedges MATH. ... |
math/0003082 | Let MATH denote the sub-MATH-algebra of all elements with pointwise norm continuous orbit under the action of MATH of MATH, and set MATH. We have MATH where MATH is the MATH-weak closure of MATH. Indeed if MATH and MATH is an approximation of the identity in MATH by continuous functions with support in a ball of radius... |
math/0003082 | Let MATH be a NAME space of isometries implementing MATH on MATH and let MATH be an orthonormal basis of MATH, thus the MATH's are isometries in MATH with orthogonal final projections summing up to the identity and MATH, MATH. Then MATH where MATH and MATH. Therefore MATH. On the other hand MATH is a separating vector ... |
math/0003082 | Let MATH a KMS state of MATH extending MATH. As MATH is a separating state of MATH, also MATH is also separating as in the proof of REF , where MATH is the inner endomorphism of MATH implemented by MATH. As MATH extends MATH we have by REF that MATH, where the symbol MATH denotes quasi-equivalence. |
math/0003082 | The proof now follows by REF . |
math/0003082 | With MATH irreducible, we have to show that MATH is irreducible too. Let MATH, namely MATH and MATH for all MATH in MATH. As MATH is localized in a double cone MATH, MATH acts identically on MATH, hence MATH, thus MATH as desired. |
math/0003082 | As above noticed, MATH is a two-variable cocycle for the action of MATH on MATH, where MATH is the tensor category with conjugates generated by MATH (see CITE). Hence by CITE MATH . On the other hand, by the results in the previous section, we may extend MATH to the weak closure of MATH in the GNS representation of MAT... |
math/0003082 | Assume first that MATH is a half-line and let MATH be orthogonal to MATH; we have to show that MATH. Indeed if MATH is a half-line and MATH, then for all MATH for all MATH such that MATH. But, because of the KMS property, the function MATH is the boundary value of a function analytic in the strip MATH (as MATH and Dom-... |
math/0003082 | Clearly MATH for positive MATH. Setting MATH we have MATH therefore, by using the relation MATH, it follows that MATH . On the other hand MATH hence MATH showing the last part of the statement. |
math/0003082 | We first show that the triple MATH is a +hsm factorization with respect to MATH in the sense of CITE, namely these three algebras mutually commute and MATH, MATH and MATH are +half-sided modular inclusions. Now MATH and MATH commute by the isotony of MATH; for the same reason MATH commute with MATH, indeed MATH is MATH... |
math/0003082 | If MATH is strongly additive, then MATH is strongly additive (on the intervals of MATH), hence MATH is strongly additive and the above comment applies. |
math/0003082 | MATH follows by the above comments. On the other hand MATH because they are equivalent to the relative commutant property MATH for MATH and either MATH or MATH, which are indeed equivalent conditions in the conformal case CITE. |
math/0003082 | If MATH satisfies essential duality then, since MATH by MATH of the above proposition, we have MATH for MATH. On the other hand, by essential duality, we have MATH, hence MATH as desired. The case of arbitrary MATH is obtained by translation covariance. |
math/0003082 | By translation covariance there exists a unitary MATH such that MATH is localized in an interval contained in MATH, thus MATH on MATH for all MATH, MATH. It follows that MATH is a normal extension of MATH to MATH. |
math/0003082 | Setting MATH, MATH, we have MATH for large MATH. Moreover MATH and MATH converge increasingly respectively to MATH and MATH. Thus REF applies and gives MATH . |
math/0003082 | If MATH is KMS state, then MATH is a separating state. If MATH is a non-zero multiple of an isometry, then also MATH is a separating positive functional, which is quasi-equivalent to MATH because MATH is a separating vector for MATH. Take MATH, the element of MATH that implements MATH. If MATH we have MATH where MATH, ... |
math/0003082 | Let MATH be such that MATH if MATH. If MATH, write MATH with MATH. By the cocycle equation MATH we see that MATH. |
math/0003082 | We shall associate a covariant localized endomorphism MATH to a given MATH. Similarly as in REF , we set MATH where MATH is a vector in the time-zero hyperplane and MATH, with MATH the radius of the double cone in REF . We have to show that, for a fixed MATH, the above definition is independent of the choice of MATH, n... |
math/0003082 | Setting MATH, MATH, we trivially obtain a local net on MATH and MATH is a KMS with respect to translations. The conformal extension of MATH is then the conformal completion of MATH given by REF ; the additivity assumption is here unnecessary since MATH is independent of MATH. A detailed proof can be found in REF. |
math/0003082 | Note first that there is a conditional expectation MATH, MATH, given by MATH, MATH, where MATH is the orthogonal projection onto MATH. The case MATH is clearly a consequence of NAME 's theorem since MATH is globally invariant under the modular group of MATH. Now the translations on MATH (constructed as in REF ) restric... |
math/0003082 | If MATH is localized in the interval MATH REF of MATH, then clearly MATH restricts to the NAME algebra MATH for all MATH as it acts trivially on MATH and by duality for MATH. Hence MATH restricts to the MATH-algebra MATH, and then it extends to MATH by REF . Let MATH be also localized in the interval MATH. If MATH are ... |
math/0003082 | REF states that MATH . By the asymmetry of the chemical potential MATH, thus MATH . Summing up REF and setting MATH we obtain MATH . |
math/0003082 | Immediate. |
math/0003082 | Indeed MATH . Because of REF , one checks easily that MATH, MATH, where MATH and MATH is the one-parameter automorphism group. Since MATH commutes with MATH, it follows that MATH is an invariant subspace for MATH, thus the latter restricts to MATH on MATH and REF implies the statement. |
math/0003082 | To simplify notations we shall consider the tensor product MATH of the same net MATH by itself. The decomposition REF of the NAME space MATH of MATH gives a decomposition of MATH and we can define the operators MATH . The tensor product Hamiltonian MATH is equal to MATH. |
math/0003082 | Immediate by REF and the fact that inner automorphisms give the identity map in cyclic cohomology CITE. |
math/0003082 | Since MATH is the canonical endomorphism of MATH into MATH (see the appendix), the result is immediate. |
math/0003082 | Setting MATH we have (see REF and CITE) MATH, thus MATH which is a super-KMS functional with respect to the evolution Ad-MATH and the superderivation MATH. A direct verification shows that MATH is the JLO cocycle on MATH associated with MATH. |
math/0003082 | The proof could be given similarly to the one given in REF . However, it is easier to observe that one can consider sectors and conjugate sectors between different MATH-algebras. Then the canonical endomorphism MATH, as a map of MATH into MATH, is just the conjugate sector for the embedding MATH of MATH into MATH and t... |
math/0003082 | Let MATH be the operators in the conjugate equation for MATH; setting MATH and MATH REF holds true. |
math/0003087 | CASE: Let MATH the polar decomposition of MATH and MATH the spectral resolution of MATH. Then MATH and MATH, that is, MATH . Set now MATH with a unitary MATH. Since MATH is a trace vector we have MATH that is, MATH. Since every element of MATH is the linear combination of at most REF unitaries, it follows, that MATH. S... |
math/0003087 | Let MATH with MATH be a partial isometry and MATH, the polar decomposition of MATH. Then we have MATH since MATH is a partial isometry from MATH to MATH. Let now MATH be the spectral measure for MATH. Then MATH . From this we see, that MATH and, since MATH is separating for MATH and MATH, MATH. Since MATH is positive M... |
math/0003087 | Let MATH and MATH be two operators affiliated with MATH, such that MATH . Then MATH is closable, since MATH is finite (compare CITE), and its closure MATH is affiliated with MATH. Then it follows from REF that MATH, that is, MATH and MATH agree on the intersection of their domains. Since MATH is a core for both MATH an... |
math/0003087 | Let MATH be the polar decomposition of MATH and MATH the spectral projection of MATH for the interval MATH. Then MATH, such that MATH . Now MATH since MATH is a core for MATH and MATH. |
math/0003087 | CASE: Since MATH also MATH, where MATH is the polar decomposition of MATH. This is equivalent with MATH where MATH is the spectral measure of MATH and we have used the trace property of MATH and that MATH is in the final space of MATH. CASE: If MATH is cyclic there is an operator MATH, such that MATH and MATH . Now we ... |
math/0003087 | The existence and the asserted properties follow from REF and the uniqueness from REF . |
math/0003087 | The necessarity of the trace condition follows from REF and the sufficiency from REF. |
math/0003087 | Let MATH be the spectral projections of MATH corresponding to the interval MATH. Then MATH is in MATH with MATH, MATH as the polar decomposition. According to CITE MATH is a sequence of cyclic and separating vectors converging (MATH are invertible operators!) to MATH with modular objects: MATH where MATH is the conjuga... |
math/0003087 | CASE: Suppose that MATH are the modular objects corresponding to a cyclic and separating vector MATH. Since MATH is cyclic and separating there exists a non-singular operator MATH corresponding to MATH such that MATH and MATH (compare REF, and REF). Since the decomposition of MATH is unique up to a positive constant RE... |
math/0003087 | CASE: Let MATH be a type MATH factor (MATH), that is, it is isomorphic to MATH with a finite dimensional MATH. Since MATH is finite dimensional all linear operators are bounded and have finite trace, that is, REF is always satisfied. CASE: compare REF . |
math/0003087 | CASE: Let MATH. Then there exists a unitary MATH which commutes with MATH and MATH, such that MATH and MATH. Setting MATH we have MATH and MATH commutes with MATH (since MATH and MATH do). Then we can calculate MATH . Since MATH commutes with MATH we have MATH . With this, REF, and REF follows MATH . CASE: We can assum... |
math/0003087 | Let MATH be a unitary, such that MATH and MATH where MATH. Then from REF follows MATH . This shows MATH . |
math/0003087 | The relation defined in REF is an equivalence relation, since it is reflexive (choose MATH), symmetric REF and transitive: Let MATH and MATH; this means hat MATH and MATH, where MATH and MATH are unitaries with the properties described above. Then with MATH we have MATH and also MATH have the right properties, such tha... |
math/0003087 | Let MATH be two members of this class. Then there exist unitaries MATH, such that MATH, MATH commutes with MATH and MATH, and MATH (compare REF). Now define MATH . Then MATH and MATH such that the conditions of REF are fulfilled and MATH. |
math/0003087 | Since MATH and MATH are solutions of the inverse problem, there are two unitary operators MATH and MATH both commuting with MATH, such that MATH, MATH are the modular objects for MATH (MATH). CASE: Let MATH. Then there is a unitary MATH commuting with MATH and MATH, such that MATH. Setting MATH an easy calculation give... |
math/0003087 | Since MATH and MATH has no non-zero Abelian projection there exist for every MATH orthogonal families MATH and MATH of non-zero projections, such that MATH (compare CITE). Now MATH where the projections MATH are pairwise orthogonal projections, since MATH commutes with MATH, they are not MATH, since MATH is a factor (c... |
math/0003087 | The first assertion follows directly from REF and the fact, that MATH since MATH is a factor (compare CITE). The second assertion follows from MATH and MATH in the type MATH case and from REF in the type MATH case. |
math/0003087 | Since MATH are two selfadjoint operators having the same eigenvalues MATH REF we can write MATH where MATH are the corresponding (orthogonal) eigenprojections (and MATH) and the convergence is understood in the so-topology. Since MATH and MATH have the same multiplicities we have MATH (MATH), where MATH is the unique d... |
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