paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904014 | Let MATH be the set of all associated semisimple pairs. Obviously, MATH. It follows from REF that MATH is unipotent and therefore MATH is closed in MATH. Since MATH is finite, MATH. On the other hand, MATH. Therefore MATH. Recall that MATH is the commuting variety. Hence, the assertion is equivalent to that MATH is irr... |
math/9904014 | From REF , it follows that MATH is a semi-direct product of the unipotent group MATH and a finite group MATH. CASE: Take an arbitrary MATH. It is a semisimple element of finite order. Since MATH is an associated semisimple pair for MATH, it follows from REF that MATH for some MATH. Hence MATH. By REF , one may assume t... |
math/9904014 | CASE: Suppose a MATH-triple MATH commutes with MATH. Then we may choose a MATH-triple containing MATH inside of the reductive algebra MATH. CASE: This readily follows from the NAME theory. CASE: In this case MATH satisfy commutator relations REF . From REF , we then conclude that MATH is MATH-conjugate to MATH. |
math/9904014 | The proof is much the same as for the previous assertion. Take a nilpotent element MATH. It then follows from REF that MATH . If MATH, then we must have MATH and MATH for ``parity" reason. Thus MATH is regular in MATH and hence in MATH. This argument can be reversed. |
math/9904014 | CASE: The space MATH possesses the MATH-filtration and MATH. It follows from the definition of MATH-limit that MATH. Furthermore, MATH. Under our assumption, this means that MATH and the eigenvalues of MATH on MATH are nonnegative integers. Assume that MATH for some MATH. Since MATH is killed by some power of MATH, we ... |
math/9904014 | Since MATH and MATH, we have MATH and MATH. That is, the property of being a dual pair is equivalent to that MATH. We first prove that MATH commutes with MATH. REF says that MATH commutes with MATH. Therefore the subalgebras generated by MATH, MATH and MATH commute. By REF , the subalgebra generated by MATH and MATH is... |
math/9904014 | CASE: Since MATH is Abelian in both cases, REF is satisfied. By REF , the other hypotheses are satisfied, too. The centre of MATH is equal to MATH. CASE: Clearly, MATH is reductive if and only if MATH is reductive. If MATH is reductive, then it contains a suitable MATH-triple together with MATH. The opposite implicatio... |
math/9904014 | CASE: The argument is close to that in REF. By definition, MATH, where MATH consists of nilpotent elements. As MATH, we have MATH. Thus, MATH. In the rectangular case, MATH is reductive. Hence MATH is reductive, too. This clearly forces that MATH. CASE: Since MATH is semisimple, the previous equality means MATH is a NA... |
math/9904014 | CASE: Since MATH, taking the double centralizer gives MATH. Whence MATH is NAME in MATH. Next, MATH, which means MATH is regular in MATH. The centralizer of MATH in MATH is equal to MATH, the centre of MATH. That is, MATH is distinguished in MATH. Since any distinguished element is even (see for example, CITE), the ass... |
math/9904014 | Note first that MATH is the centre of MATH for any MATH. CASE: By definition, the linear space MATH is generated by all elements of the form MATH REF that lie in MATH. Let MATH and MATH. If MATH and MATH, then MATH. CASE: If MATH is even, then MATH. Since MATH, we conclude that MATH. Hence MATH, by the first claim. |
math/9904014 | CASE: Let MATH be the centre of MATH. Then MATH and therefore MATH. In case MATH is even, MATH is generated by MATH, and MATH as NAME algebra. Hence MATH is in the centre of MATH, that is, MATH. CASE: Since MATH and MATH, the assertion follows from the previous Lemma. |
math/9904014 | CASE: Since MATH, the affine space MATH is transversal to the orbit MATH at MATH. Consider the subspace MATH. Since MATH is a principal MATH-triple in MATH (see REF), MATH is a section of the open sheet in MATH. This is a classical result of CITE. Therefore almost all elements in MATH are semisimple and MATH-conjugate ... |
math/9904014 | By REF , each member of a rectangular pn-pair is excellent. |
math/9904016 | Write MATH where MATH and MATH are just maps of the complex plane. Since MATH corresponds to a complete global analytic function MATH, we need to verify that MATH corresponds to some other complete global analytic function MATH. From our definitions we see that MATH is obtained from MATH by substituting each function e... |
math/9904016 | Without loss of generality let MATH and MATH lie on the real axes of, respectively, MATH and MATH. The functions MATH of the function elements MATH, which represent MATH locally, will then have power series on the real axis (of MATH) with real coefficients. Since a real power series when analytically continued along th... |
math/9904016 | Near the point of intersection MATH is represented by a function element MATH where MATH is conformal. The equality of angles, formed by a pair of lines in MATH and their images by MATH in MATH, is simply the geometrical statement that MATH is conformal. |
math/9904016 | Suppose MATH and MATH intersect with angle MATH on a nontrivial NAME surface MATH; for convenience, let MATH be the point of intersection. These curves are fixed by NAME reflections MATH and MATH respectively, and MATH is an isometry of MATH. The neighborhood of the point of intersection is the graph MATH where MATH is... |
math/9904016 | Suppose MATH contains a real curve and call its completion MATH. We recall that MATH, and its closure in MATH, MATH, are straight lines and MATH cannot have an endpoint within MATH REF . The possible geometrical relationships between MATH and MATH are diagrammed in REF . Since MATH is nontrivial, at least two vertices ... |
math/9904016 | We use the real curves to decompose MATH into a set of tiles REF MATH. REF tells us that MATH, so that MATH, where MATH is the edge group of MATH. On the other hand, if MATH is an isometry of MATH, then MATH, where MATH for some MATH. But since MATH has no nontrivial isometries, the map MATH must be the identity and MA... |
math/9904016 | If MATH is finite we can view MATH as a finite cell complex. We can relate the number of REF, MATH, and the number of REF, MATH, in this complex to the number of REF, MATH. Since every REF-cell is bounded by three REF-cells, each of which bounds exactly one other REF-cell, MATH. Similarly, the boundary of each REF-cell... |
math/9904016 | If MATH is not discrete we can find a sequence of MATH such that both MATH and MATH. Let MATH be a regular point of MATH, then MATH. Near MATH we can represent MATH by the graph MATH where MATH is a neighborhood of MATH in MATH, MATH is conformal in MATH, and MATH. Since MATH is an isometry, MATH. From MATH we see that... |
math/9904016 | Let MATH be an open ball of radius MATH centered at an arbitrary point MATH. Consider the piece of the NAME surface within this ball, MATH, and the projection MATH. MATH is finitely discrete if there is a uniform upper bound on the number of preimages MATH. MATH is covered by the orbit of closed graphs, MATH, where MAT... |
math/9904016 | MATH is the union of countably many closed graphs MATH given by the orbit of MATH under the action of the edge group. Since each graph has at most three singular points, MATH has countably many branch points. If there were uncountably many self-intersection points then uncountably many must arise from one pair of disti... |
math/9904016 | We arrive at MATH by composing bijections MATH such that MATH and MATH are (correspondingly) uniformly bounded. The Lemma then follows by application of the triangle inequality. Since MATH is crystallographic, we can partition MATH into translates of a bounded fundamental region, MATH, of its lattice MATH. Thus for any... |
math/9904016 | The expression MATH indicates there exist constants MATH and MATH (independent of MATH) such that for sufficiently large MATH, MATH. Let MATH be a disk of radius MATH centered at the origin and let MATH . We first obtain a bound on the difference, MATH, when neither MATH nor MATH is a crossing point or a branch point o... |
math/9904016 | First consider the case MATH; then MATH . Now consider the group MATH, where MATH is normal in MATH. Clearly MATH. Moreover, one easily verifies MATH, where MATH is any of the three generators of MATH. These two facts together show MATH; MATH is clearly the lattice group of MATH. Next consider the case MATH. For the ge... |
math/9904016 | By REF , MATH has a discrete lattice, that is, the orbit MATH is discrete in MATH. For right triangles, REF gives us MATH if MATH, MATH otherwise. Thus discreteness of MATH implies discreteness of the star MATH in MATH. Since MATH, MATH is just the orbit of MATH under action of the group MATH generated by MATH. Clearly... |
math/9904016 | Let MATH be the isometry group of MATH, MATH its derived point group, and MATH its lattice group. Since conjugation by MATH leaves MATH invariant, consider the automorphisms MATH given by MATH. If MATH has rank MATH, then the homomorphism MATH, where MATH, induces a representation of MATH by integral MATH matrices of d... |
math/9904016 | Suppose MATH divides MATH and MATH. We may assume that MATH does not divide MATH since otherwise MATH, MATH and MATH would have MATH as a common divisor. Let MATH; then MATH and MATH is a subgroup of MATH of order MATH. By looking in MATH (or MATH), we can find a translation MATH such that MATH and MATH. Now consider t... |
math/9904016 | The properties listed in REF are simple consequences of general results. Since MATH we see that MATH is isomorphic to an abstract group generated by an element of order REF, MATH, and an element MATH, which has order MATH, if MATH is odd (and therefore MATH) or MATH is even and just one of MATH and MATH is odd (since t... |
math/9904036 | We proceed by induction on MATH, assuming in addition that the index of MATH is MATH. We just did it for MATH. Assume the construction is done for MATH. Let MATH be an integer as in the proposition, and set MATH . Since MATH, the integer MATH is positive. Also, MATH because MATH and MATH. It implies MATH hence there ex... |
math/9904044 | We first replace MATH by its double-adjoint so that we can assume that MATH is closed (it is easy to check that REF remain valid). The problem is to show that it is self-adjoint. Let MATH be the range of the operator MATH. It is a closed subspace of MATH (as MATH, and MATH is closed). Let MATH be the bounded operator o... |
math/9904044 | We give the proof for completeness. The commutation with MATH-translations is clear. Then MATH contains (the inverse NAME transform of) MATH and all its translates. Hence if MATH is orthogonal to MATH then the function MATH on MATH belongs to MATH and has a vanishing ``inverse NAME transform", hence MATH (almost everyw... |
math/9904044 | Let us first assume that MATH is bounded. We use the (inverse NAME transform of the) function MATH and define MATH to be MATH. Let us consider the domain MATH consisting of all finite linear combinations of translates of MATH. It is dense by the argument using unicity of NAME transform in MATH we have used previously. ... |
math/9904044 | One just checks that MATH intertwines MATH with MATH, and also MATH with MATH and that the inversion MATH also intertwines MATH with MATH, and MATH with MATH. |
math/9904044 | Let us take MATH and consider the linear operator on MATH: MATH . It commutes with the action of MATH hence stabilizes each MATH and is a multiple MATH of the identity there. On the other hand, if we choose MATH and MATH in MATH and consider MATH we obtain a bounded operator MATH on MATH commuting with dilations and su... |
math/9904044 | One applies REF . |
math/9904044 | If the function MATH on MATH is chosen smooth with compact support (so that MATH is entire) then, for any MATH the function MATH, viewed in the additive picture, is smooth on MATH, has compact support, and vanishes identically in a neighborhood of the origin. So its image under the inversion also belongs to the NAME cl... |
math/9904044 | We have to check the identity: MATH for all NAME functions MATH with NAME Transform MATH. Both integrals are analytic in MATH, hence both sides are smooth (bounded) functions of MATH. It will be enough to prove the identity after integrating against MATH with an arbitrary NAME function MATH. With the notations of REF ,... |
math/9904044 | We have to check an identity: MATH for all NAME functions MATH with NAME Transform MATH. Both integrals are analytic in the strip MATH, their ratio is thus a meromorphic function, which depends neither on MATH nor on MATH as it equals MATH on the critical line. Furthermore for any given MATH we can choose MATH, with MA... |
math/9904044 | Let MATH and MATH. One checks that MATH (MATH). We choose as homogeneous function MATH. For these choices the identity of REF becomes MATH . Adding a suitable linear combinations of these identities for MATH gives MATH hence the result after evaluating the integrals in terms of MATH. |
math/9904044 | We have to show that MATH is a multiplier of the NAME class. Let MATH. Using REF and the partial fraction expansion of the logarithmic derivative of MATH (as in CITE for the real and complex NAME Gamma functions), or NAME 's formula, or any other means, one finds MATH, MATH, so that MATH. |
math/9904044 | Indeed, it commutes with MATH by REF and it commutes with MATH by construction. |
math/9904044 | Assuming the validity of the estimates we see that both sides of the identity are analytic functions of MATH, so it is enough to prove the identity on the critical line: MATH . As in the proof of REF , it is enough to prove it after integrating against an arbitrary NAME function MATH. With MATH, and using REF this beco... |
math/9904044 | We have used the notation MATH for the NAME 's constant (MATH). Let MATH for MATH be the homogeneous distribution MATH on MATH. It is a tempered distribution. The formula MATH defines its analytic continuation to MATH, with a simple pole at MATH. Using MATH for MATH, MATH, and expanding in MATH gives MATH . As MATH the... |
math/9904049 | Denote the space obtained at stage MATH by MATH and organize the projections of the fiber squares of all stages as MATH . Then MATH and MATH is the proper transform of MATH in MATH if MATH, while MATH is the component of the exceptional divisor over MATH. The statement will follow once it has been shown that the stated... |
math/9904049 | This has been obtained while proving the theorem, and is formulated separately only for the ease of future reference. |
math/9904049 | In a blowup MATH of a smooth algebraic variety MATH along a smooth center MATH, if MATH is the proper transform of a smooth variety MATH, then in terms of ideal sheaves MATH. Applied at each step, this equality yields MATH, where MATH denotes the composition of the stated blowups. |
math/9904049 | We concentrate on the normal crossing property, which implies the other claims. By construction, the proper transform of every polydiagonal MATH under MATH is a smooth divisor; it will be denoted by MATH. The proper transforms of MATH and MATH become disjoint when that of their intersection MATH is blown up, unless one... |
math/9904049 | Straightforwardly from the construction of MATH and REF , MATH where MATH is the fiber of the normal bundle to MATH in MATH, which is MATH by an easy dimension count. REF converts this formula into MATH . Since MATH and MATH, there results a recurrence relation MATH and both claims immediately follow. |
math/9904049 | By induction on MATH, each MATH is the closure of MATH in MATH . The basis is clear: MATH. Then, MATH is the blowup of MATH along MATH or in other terms, along MATH . This ideal sheaf becomes MATH upon multiplying by an invertible ideal sheaf, and blowing up MATH is equivalent to taking the closure of the graph of the ... |
math/9904049 | Start with notation for the products from REF : MATH where MATH is the number of essential blocks in a partition MATH. If MATH is the only essential block of MATH, then MATH, so MATH can indeed be used for MATH. MATH . Now take the left of these two diagrams, where MATH and MATH are rational maps defined on MATH, and n... |
math/9904049 | The normal space MATH at a point MATH to MATH is independent of MATH (assumed positive): its dimension is equal to the cardinality of the nest MATH . The nest alone determines the iterated blowup of MATH induced from REF , and the preimage of the origin under it is isomorphic to MATH. |
math/9904049 | Look at the equations of the large diagonals containing MATH, that is, MATH for all pairs of MATH and MATH belonging to the same block of MATH. Equations coming from different blocks of MATH are independent of each other, leading to the product decomposition. |
math/9904049 | The complement to MATH in MATH is MATH, the configuration space of MATH distinct labeled points in MATH modulo translations. Since MATH is the complement in MATH to the projectivization of the arrangement MATH, it follows that MATH is the orbit space of the diagonal action of MATH on this product by dilations. Separate... |
math/9904049 | CASE: The direct factor MATH is the small diagonal MATH. The essential shape of the bottom partition of MATH is the integer MATH, thus by definition, there is a map MATH. The bundle MATH is the pullback by MATH of the tautologial line bundle over MATH; since MATH is an iterated blowup, REF (formulated below) has to be ... |
math/9904049 | The normal bundle MATH is the pullback MATH. |
math/9904049 | REF follows directly from the definition. CASE: Nothing to be done when MATH and MATH are disjoint. When MATH, denote by MATH the exceptional divisor of MATH, and by MATH the projection MATH, then MATH and the claim follows. |
math/9904049 | Similarly to REF , the definition of MATH implies REF can be checked by induction on MATH, where the inductive step follows by applying the case MATH. Thus, it is enough to show that each divisor MATH is isomorphic to MATH, where MATH and MATH. The argument is based on REF . Every partition from MATH belongs to one of ... |
math/9904049 | REF gives MATH, where MATH. By REF , the arrangements MATH transform isomorphically from the normal spaces to MATH in MATH into the normal spaces to MATH in MATH. At the next stage, MATH is a bundle over MATH with fibers isomorphic to MATH. The relevant blowup centers of the subsequent stages are its subbundles; their ... |
math/9904049 | Put together REF . |
math/9904049 | Fix a partition MATH of MATH into two-element blocks and let the nest MATH be the set MATH of blocks of MATH. The map MATH takes the divisor MATH of MATH into the stratum MATH of MATH. The divisor is isomorphic to MATH by REF and the stratum is isomorphic to MATH. Since MATH, it follows that MATH maps to MATH. Tracing ... |
math/9904049 | CASE: It is enough to show this for MATH and MATH. The required sublattice of MATH is generated by the union MATH, where the only essential block of MATH REF is MATH (respectively, MATH). CASE: It is enough to consider the same MATH and MATH as in REF and then write explicitly the equations for the large diagonals. CAS... |
math/9904049 | First, reduce to the case of all MATH points colliding at the same point in MATH. Suppose a collision MATH occurs at MATH. If it could be studied near each MATH independently of the other points, as for MATH, the isotropy subgroup would have been MATH, where MATH is the isotropy subgroup of the collision near MATH. It ... |
math/9904050 | That MATH is a NAME pair follows from REF . Since REF hold, the difference MATH can be represented as a sum of a contraction (with norm strictly less than one) and a finite-rank operator. Thus MATH is a NAME pair by REF . Since REF holds one can apply REF to conclude that MATH . By REF one gets MATH . Combining REF, an... |
math/9904050 | By REF one concludes that MATH and hence by REF , the pair MATH is a NAME pair. In particular, REF implies MATH . Hence, REF yields MATH proving REF. |
math/9904050 | It suffices to combine REF . |
math/9904050 | By hypothesis, there exist orthogonal projections MATH in MATH such that MATH and the pairs MATH and MATH, MATH are NAME pairs of projections. Since MATH one infers MATH, MATH and hence MATH, MATH are NAME pairs. Moreover, MATH. Thus, MATH by repeatedly using REF. |
math/9904050 | Since MATH and MATH have a bounded inverse, MATH and MATH are well-defined and (compare CITE) MATH . Using standard estimates for resolvents MATH and MATH of the dissipative operators MATH and MATH entering REF (compare REF) one concludes MATH if MATH. Thus, MATH by REF. |
math/9904050 | Since MATH for some MATH and MATH, one infers that MATH. Thus, there exists a closed interval MATH, MATH, such that MATH for all MATH. Hence, for given MATH, one can find a clockwise oriented bounded contour MATH encircling MATH such that MATH and MATH . The second resolvent identity for MATH and MATH then implies MATH... |
math/9904050 | REF follows from REF . Next one notes that for MATH sufficiently small, MATH by REF , and hence MATH . Since by REF (replacing MATH by MATH), MATH one obtains REF. |
math/9904050 | Since by REF for some MATH, REF implies REF. Due to the fact that MATH (compare REF ) one can represent the spectral projections MATH and MATH by the NAME integrals REF and arguing as in the proof of REF then yields MATH . Thus, by REF , the pair MATH is a NAME pair of orthogonal projections, which together with REF pr... |
math/9904050 | Define two orthogonal projections MATH and MATH such that MATH and MATH where MATH (taking into account that MATH). Next, we note MATH where the clockwise oriented contour MATH encircles MATH. On the other hand, MATH using REF. Since MATH, the last term in REF is a trace class operator and, hence, MATH if and only if M... |
math/9904050 | By hypothesis, MATH for some MATH. Since MATH and MATH as MATH, the operators MATH are also invertible for MATH, MATH sufficiently large. Since the operator-valued logarithm is a continuous function of its (dissipative) argument in the MATH-topology, MATH and hence by REF, MATH . By REF , MATH and thus MATH and MATH . ... |
math/9904050 | Fix a MATH, MATH. Denote by MATH the eigenvalues of MATH different from MATH with corresponding multiplicities MATH. First we prove the following representation, MATH with MATH . Since MATH, there exists a MATH such that MATH . For sufficiently small values of MATH, MATH, the self-adjoint operator MATH is well-defined ... |
math/9904050 | First one notes that MATH, MATH is invertible. For MATH this holds by hypothesis and for MATH this holds since MATH. Thus, MATH and MATH are well-defined. In the following let MATH. Since MATH, the representation MATH and estimates for the resolvents MATH and MATH, MATH, analogous to those in the proof of REF yield REF... |
math/9904050 | Pick MATH in REF . Taking the imaginary part of both sides of REF, an explicit computation of the integrals in REF yields MATH . |
math/9904050 | Under REF one can apply REF for MATH. Thus, MATH and therefore REF holds due to REF and the fact that the left-hand side of REF is real. |
math/9904050 | Since by hypothesis, MATH, REF implies that MATH for all MATH except possibly at a finite number of points MATH. Introducing the notation MATH one obtains for MATH sufficiently small, MATH . By REF , MATH and by REF (for MATH sufficiently small), MATH while MATH . Combining REF - REF proves REF. Setting MATH one gets b... |
math/9904050 | By REF , MATH for all MATH sufficiently small. By REF , MATH . By REF one concludes that MATH and from standard properties of the operator logarithm (compare CITE) one also infers MATH . Combining REF - REF one obtains MATH and by REF one infers MATH where MATH . By REF , MATH and therefore, MATH . Since by REF, MATH, ... |
math/9904050 | Let MATH. Then MATH and MATH prove REF. Next, let MATH. Then MATH and MATH prove REF. |
math/9904050 | By REF the pairs MATH and MATH have a trindex and hence the pair MATH has a generalized trace and MATH . Moreover, the following representations hold MATH . Under REF , MATH, MATH and hence, by REF , MATH . Combining REF - REF proves REF. |
math/9904050 | Introduce the family of trace class operators MATH . Then, MATH and hence the NAME determinant of MATH is well-defined. By the analytic NAME theorem the set of MATH such that MATH does not have a bounded inverse is discrete and therefore there exists a MATH such that MATH is well-defined. By REF , the operator MATH has... |
math/9904050 | Differentiating MATH with respect to MATH (compare , CITE) MATH one obtains by REF MATH since MATH. However, MATH can be computed explicitly, MATH iterating the second resolvent identity. Combining REF - REF one infers MATH . Taking traces in REF one gets MATH and thus, differentiating REF with respect to MATH, MATH . ... |
math/9904050 | First of all, one notes that the boundary values MATH and MATH exist MATH a.e. in the topology MATH for every MATH (but in general not for MATH,) CITE, CITE (see also CITE, CITE, CITE). By REF, the operator MATH has a bounded inverse for a.e. MATH. Moreover (see, for example, CITE), MATH . Thus, there exists a set MATH... |
math/9904050 | Since REF have been proven in CITE, we focus on REF. Let MATH be an increasing family of orthogonal projections of rank MATH, that is, MATH, MATH, with MATH . Combining the norm continuity of the logarithm of bounded dissipative operators as discussed in REF with the exponential NAME representation for MATH (that is, t... |
math/9904050 | Since MATH and MATH, the spectrum of MATH is a discrete set with only possible accumulation points at MATH. Next, the normal boundary values MATH exist in norm for all MATH. Moreover, MATH which can be seen as follows: suppose that REF is false, then by compactness of MATH there exists a MATH such that MATH . Multiplyi... |
math/9904050 | The assertion is a direct consequence of REF . |
math/9904053 | If MATH is a sphere or torus, MATH, so that there is nothing to prove. We may thus assume MATH. Let MATH be the bundle projection, MATH the NAME class of the tangent bundle along the fibers, and MATH the fundamental class dual to the orientation class MATH. Let MATH denote integration along the fiber. Then MATH . Denot... |
math/9904053 | This follows directly from REF, the equality MATH for closed hyperbolic NAME surfaces MATH, and the multiplicativity of the NAME characteristic in fiber bundles. |
math/9904053 | If MATH, then MATH or MATH must be a circle, so that the right-hand-side of REF vanishes. Similarly, if MATH and either MATH or MATH is a circle the right-hand-side of REF vanishes and there is nothing to prove. Thus the only interesting case is that of a surface bundle, for which we appeal to REF . |
math/9904053 | As MATH is an NAME manifold, we have NAME 's inequality MATH, with equality only if MATH is flat. Flat manifolds are finitely covered by the torus, and so their simplicial volumes vanish and there is nothing to prove in that case. Thus we can assume MATH. This means that neither MATH nor MATH can be a circle or a REF-t... |
math/9904058 | (outlined by NAME): It is known that MATH is a MATH bundle over MATH with monodromy MATH CITE. By standard MATH-manifold theory MATH can be isotoped to a fiber preserving isotopy. By composing with obvious diffeomorphism that extends, we can assume that the fiber orientation is preserved. Since MATH has to commute with... |
math/9904058 | By replacing one of the fishtails in MATH by MATH we obtain a homotopy MATH-sphere, which can be easily checked to be MATH REF . Similarly by replacing one of the cusps in MATH we obtain a homotopy MATH which can be checked to be the standard MATH. Unlike the previous case, this check is surprisingly difficult (it requ... |
math/9904067 | For completeness we present a short proof based on the following classical characterization of meager sets in MATH. MATH . Suppose that MATH is a meager set. Without loss of generality we can assume that MATH for some partition MATH and real MATH. Let MATH for every MATH. Define MATH . Suppose that MATH. Define MATH. I... |
math/9904067 | We start with the following easy observation: Suppose that MATH is a finite set and MATH are such that CASE: MATH and MATH, CASE: MATH. There exist MATH such that MATH . Let MATH . Check that for every MATH, MATH . Thus MATH for all MATH. By NAME theorem there are MATH such that MATH . In particular, MATH . Fix a parti... |
math/9904079 | By REF MATH . Hence, MATH . Since MATH and since MATH and MATH are irreducible, we have MATH . Theorem is proved. |
math/9904079 | The natural homomorphism MATH is, clearly, a homomorphism of MATH-modules. It is not difficult to verify that MATH . Let MATH be a fixed MATH-tableau, MATH and MATH two MATH-semistandard sequences with elements from MATH and MATH, respectively. Then by REF the vectors MATH form a basis of a subspace MATH which is also ... |
math/9904079 | Let MATH be a supercommutative superalgebra, MATH a MATH-module; let MATH and MATH. The elements of MATH may be considered as functions on MATH with values in MATH. Let MATH. Therefore, MATH determines a homomorphism MATH. Set MATH . Observe that MATH naturally acts on MATH and on the algebra of functions on MATH. |
math/9904079 | By REF MATH and MATH hence, MATH. Since homomorphism REF is a homomorphism of MATH-modules, its kernel coinsides with MATH. Let MATH be a MATH rectangle. The condition MATH means that MATH and by REF it suffices to demonstrate that MATH, where MATH is a fixed standard rectangular tableau of size MATH, belongs to the id... |
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