paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0102030 | The induction is on the set of REF. The comments at the start of this section apply therefore to MATH instead of MATH and thus we let MATH so that we have MATH . |
gr-qc/0102061 | A static solution MATH obeys MATH where MATH is the Laplacian in D dimensions. This equation clearly correspond to the extremum condition MATH for the static energy functional MATH where the functionals MATH and MATH stand for the two terms on the right-hand side. Note that not only W but also MATH and MATH are non-negative. Now, let MATH be a static solution. Consider the one-parameter family of configurations MATH . It is easy to check that MATH . Since MATH is an extremum of MATH, it must in particular make MATH satationary with respect to variations in MATH; that is, MATH . Differentiating REF using REF gives us MATH . Since MATH and MATH are non-negative REF cannot be satisfied for MATH unless MATH. This means that MATH has to be space-independent and equal to one of the zeros of MATH. This is just a trivial solution and the theorem precludes non-trivial space-dependent solutions. QED . |
hep-th/0102039 | It suffices to prove that the inclusions MATH hold for any pair of indices MATH, where MATH . The action of these operators on an arbitrary monomial MATH yields a polynomial, since MATH is a multiple of MATH. The homogeneous degree of this polynomial is either MATH (if MATH) or MATH (if MATH). Therefore, if the original monomial belongs to MATH the resulting polynomial is also in MATH for MATH, but lies outside MATH for MATH. On the other hand, the degrees of the variables MATH in the resulting polynomial remain equal to MATH if MATH, while the degrees of MATH and MATH satisfy MATH where MATH . Therefore, if the original monomial belongs to the space MATH so does the resulting polynomial in all three cases MATH. |
hep-th/0102039 | The statement follows from REF and the fact that the differential parts of MATH and MATH preserve the modules MATH and MATH for any non-negative integer MATH, whereas the differential part of MATH preserves the module MATH only for MATH. |
hep-th/0102039 | The gauge transformation REF with gauge factor MATH together with the change of variables MATH map the gauge spin Hamiltonian MATH to a matrix NAME operator REF, with potential MATH . |
hep-th/0102162 | The internal propagators are symmetric in MATH and MATH. An external MATH-leg goes into a MATH-leg by amputation (their propagators are inverses as integral kernels), and vice versa. |
hep-th/0102175 | By virtue of the above mentioned abstract NAME theorem CITE, there is an isomorphism between the cohomology of the complex REF and the cohomology MATH of the paracompact space MATH with coefficients in the constant sheaf MATH. Since MATH is a strong deformation retract of MATH CITE, the cohomology MATH is isomorphic to the cohomology MATH of MATH with coefficients in the constant sheaf MATH CITE and, consequently, to the NAME cohomology MATH of MATH. |
hep-th/0102190 | We will prove that the prepotential satisfies this system by showing existence of the following: CASE: an `invertible metric' MATH CASE: an associative algebra with structure constants MATH CASE: a relation between the third order derivatives of MATH and the structure constants: MATH . If conditions MATH are met, then MATH satisfies the generalized WDVV system. Associativity of the algebra can be expressed through an identity on the structure constants MATH which due to REF also reads MATH and multiplying by MATH from the right gives the desired result. The rest of the proof deals with a discussion of REF. It is well-known CITE that the right hand side of REF equals the NAME superpotential associated with the corresponding NAME algebra. Using this connection, we can define the primary fields MATH and The chiral ring is an associative algebra defined by MATH . Instead of using the MATH as coordinates on the part of the moduli space we're interested in, we want to use the MATH. For the chiral ring this implies that in the new coordinates MATH which again is an associative algebra, but with different structure constants MATH. This is the algebra we will use in the rest of the proof. For the relation REF we turn to another aspect of NAME theory: the NAME equations (see for example, CITE and references therein). These form a coupled set of first order partial differential equations which express how the integrals of holomorphic differentials over homology cycles of a NAME surface in a family depend on the moduli. Flat coordinates of the NAME theory are a set of coordinates MATH on moduli space such that MATH where MATH is given by MATH . The fact that such coordinates indeed exist will not be discussed here. In terms of these coordinates the following set of NAME equations hold CITE MATH for any cycle MATH. These equations were derived by making use of the chiral ring, expressed in the flat coordinates, and therefore the structure constants MATH appear. Making a change of coordinates to the MATH and using the fact that the MATH satisfy REF one finds MATH . Taking MATH we get MATH which is the intended relation REF . The only thing that is left to do, is to prove that MATH is invertible. This will not be discussed in the present paper. |
hep-th/0102190 | Starting from the multiplication structure in REF we find (see REF ) MATH we will rewrite it in such a way that it becomes of the form MATH . As a first step, we use REF : MATH . The notation MATH stands for the vector with components MATH and we used a matrix notation for the structure constants. The proof becomes somewhat technical, so let us first give a general outline of it. The strategy will be to get rid of the second term of REF by cancelling it with part of the third term, since we want an algebra in which the first term gives the structure constants. For this cancelling we'll use equation MATH in combination with the following relation which expresses the fact that MATH is a graded function MATH . Cancelling is possible at the expense of introducing yet another term which then has to be canceled etcetera. This recursive process does come to an end however, and by performing it we automatically calculate modulo MATH instead of MATH. We rewrite MATH by splitting up the third term and rewriting one part of it using MATH: MATH . Now we use REF to work out the product MATH and the result is: MATH . We now use REF again to rewrite the second term in the first line: MATH . Note that by cancelling the one term, we automatically calculate modulo MATH. The expression between brackets in the first line seems to spoil our achievement but it doesn't: until now we rewrote MATH and we can now rewrite using the same procedure MATH . This is a recursive process. If it stops at some point, then we get a multiplication structure MATH for some polynomial MATH and the theorem is proven. To see that the process indeed stops, we refer to the lemma below. |
hep-th/0102190 | One finds CITE that the degree of MATH is MATH . Dividing this by MATH, we get an object of degree MATH. The MATH are defined through MATH-and if we can show that for MATH we can't divide MATH by MATH, we have shown that MATH is nilpotent since it is strictly upper triangular. Since MATH we find that indeed for MATH the degree of MATH is bigger than the degree of MATH and since they are both polynomials we can't divide the two. This finishes the proof of the lemma. |
math-ph/0102009 | Let MATH be the components of MATH and MATH the components of MATH. Then there is a disjoint union MATH such that MATH . Let MATH be the set of those MATH for which MATH consists of a single element MATH. These MATH belong to components MATH that are large enough and survive the MATH applications of MATH without having to merge with other components. It follows from REF that MATH, that is, MATH, since otherwise, the number of elements of the set MATH would be greater than the number MATH of elements of MATH. Of course, we have MATH. Combining these, we have MATH . With MATH, we have MATH. |
math-ph/0102009 | Let us apply the last lemma repeatedly with MATH where MATH is the number of components of MATH, and MATH. We get MATH, hence the number of components decreases to REF fast. The times MATH form, at the same time, approximately a geometric series in which even the largest term, obtained with MATH, is at most MATH. Therefore the sum of this series is still MATH for an appropriate constant MATH. |
math-ph/0102009 | Let MATH be a connected subset of MATH with the property that MATH is not empty. We have to give a lower bound on the set MATH. It is easy to verify the following commutation property of the rules MATH and MATH: MATH . It follows that MATH . If MATH is not empty then MATH. It follows from REF that MATH is not empty. The set MATH then contains a full triangle of span MATH, which contains MATH elements. |
math-ph/0102009 | Proof of REF . Let MATH and MATH be two points in MATH. Let MATH be a point of MATH in MATH, and MATH a point of MATH in MATH. These points are connected in MATH by a path. Each edge of the path is contained in exactly one tile held by MATH. We have obtained a path of tiles connecting the tile with center MATH to the tile with center MATH. The centers of these tiles form a path connecting MATH and MATH in MATH. Proof of REF . Let MATH be two points in MATH. We have to find a path in MATH connecting them. Since the set MATH is connected it is enough to find such a path when MATH are in two neighboring tiles, and then work step-by-step. If the intersection point of the two neighbor tiles is in MATH then MATH are clearly connected through it. Otherwise, MATH contains the edge in both tiles opposite the intersection. It is easy to see from REF that these two edges have an edge of MATH connecting them. |
math-ph/0102009 | Suppose that the first relation does not hold. Without loss of generality, let us assume that MATH . Let MATH be arbitrary. Let MATH. Then MATH. Let MATH, MATH. The triple MATH is a connected closed cut of MATH. The connectedness follows immediately from the definition. To show that it is a closed cut, we have to show MATH. The relation MATH implies MATH, and hence, since MATH is a component of MATH, we have MATH. It follows from the fact that MATH is a connected cut and from MATH that MATH . This, together with REF, implies MATH . |
math-ph/0102009 | Let MATH be a set with MATH. We will estimate MATH. Let MATH be a closed cut of MATH with MATH. Let us break MATH into components MATH, and MATH similarly into components MATH. Then, REF says that we have either MATH for all MATH, or MATH for all MATH. Without loss of generality, assume that REF holds. Then we have MATH . |
math-ph/0102009 | We apply the above theorems to MATH,MATH and MATH consecutively, with MATH throughout, but with MATH in the three stages. |
math-ph/0102009 | Let MATH. Let MATH be a connected cut of MATH with MATH. Without loss of generality, we can assume that it is a closed cut. Our goal is to estimate MATH. We will find a certain cut MATH of MATH. For each element MATH of MATH, we define an element MATH in MATH, and set MATH. To define MATH, remember the notation MATH from REF. Let us group the neighbors of MATH in three connected pairs MATH where MATH . The pair MATH consists of the centers of those tiles containing the corner MATH. For each MATH, the pair MATH may intersect one of the sets MATH. It cannot intersect both since MATH and MATH are separated by MATH. CASE: Suppose that only one pair, say MATH, is intersected by MATH, and MATH. Then let MATH. CASE: Suppose that two pairs are intersected by MATH, and the third one, say MATH, is not, and MATH. Then let MATH. CASE: In all other cases, we choose MATH arbitrarily from the set MATH. Now let MATH . The triple MATH is a cut. It is enough to prove that if there is a path between some elements MATH for MATH then this path passes through an element of MATH. Let MATH be such a path. For both MATH, the element MATH is contained in a tile MATH for some MATH. Let MATH be the last MATH such that MATH for some MATH in MATH. Let MATH be the tile containing the pair MATH. Then MATH, since MATH is a closed cut. It is easy to see from the definition above that MATH is either MATH or MATH. Let us complete the proof of REF . We replace the cut MATH with the closed cut MATH where MATH. Let MATH be the components of MATH, and MATH the components of MATH. It follows from REF that either MATH for all MATH, or MATH for all MATH. Let us suppose without loss of generality that REF holds. It follows from the definition of MATH and MATH that the tile MATH intersects MATH for all MATH. Let MATH . Then MATH, MATH. It follows from the connectedness of MATH that MATH for a triangle MATH such that MATH. Then the triangle MATH contains MATH, and the triangle MATH contains MATH. Let MATH, that is, the blowup of MATH by MATH. Let us call the sets MATH, MATH neighbors if they either intersect or have neighboring elements. It follows from the connectedness or MATH that the set MATH is connected under this neighbor relation. Indeed, we constructed MATH in such a way that MATH. Therefore if MATH and MATH are neighbors then MATH and MATH intersect. Let us call two triangles MATH, MATH neighbors if they intersect. Then from the fact that the set MATH is connected under the neighbor relation, it follows that the set MATH is also connected under its neighbor relation. According to REF , if triangles MATH intersect then there is a triangle containing their union whose span is MATH. It follows that there is a triangle MATH containing MATH such that MATH. As we know, MATH for any nonnegative MATH. It follows from REF that MATH . We have therefore MATH . Finally, MATH where we used the assumption MATH to imply that the coefficient of MATH is not positive, therefore we can replace MATH with REF. |
math-ph/0102009 | For a subset MATH of MATH, let MATH . Suppose that MATH is a connected cut of MATH with MATH. Without loss of generality, we can assume that it is a closed cut. Our goal is to estimate MATH. From the fact that MATH are separated by a cut, it follows that the sets MATH are disjoint. Let MATH . We have MATH for MATH. The first relation follows immediately from the definition. For the second relation, note that MATH which is contained in MATH by the closedness of the cut MATH. Now we proceed similarly to the proof of REF . However, we are trying to make the new cutting set MATH smaller than the old one. Let us use the notation introduced above. There is an element MATH of MATH, and a mapping MATH defined on MATH such that we have MATH, and with MATH the triple MATH is a cut of MATH. The proof of this lemma is left to the next section. Now we conclude the proof of REF analogously to the end of the proof of REF . Let MATH be closed cut such that MATH. Let MATH be the components of MATH, and MATH the components of MATH. It follows from REF that either MATH for all MATH, or MATH for all MATH. Let us suppose without loss of generality that REF holds. Let MATH. Let us remember the superfluous element MATH, and define MATH. It follows from our construction that MATH . It follows from the connectedness of MATH that MATH for a triangle MATH such that MATH. Then MATH. Let MATH for MATH, and MATH for MATH. Just as in the proof of REF , we can conclude that there is a triangle MATH containing MATH such that MATH. It follows from REF that, for MATH, MATH . We have therefore MATH . Finally, MATH where we used the assumption MATH to imply that the coefficient of MATH is not positive, therefore we can replace MATH with REF. |
math-ph/0102009 | In later parts of the proof, we will give an algorithm for the definition of the distinct elements MATH, the number MATH with MATH, and the sets MATH . Let MATH. Let MATH. Assume that MATH and MATH have already been defined. First we see that, given MATH, what conditions must be satisfied by MATH and MATH to make MATH a cut of MATH. The element MATH is contained in three tiles MATH for MATH. They are numbered in such a way that MATH . Let us write MATH. We say that MATH is superfluous if one of the MATH does not intersect the set MATH. We will choose MATH later in such a way that there is a MATH such that MATH is superfluous. The point MATH is the first superfluous element of the sequence MATH. If MATH is not superfluous then there is a MATH and MATH such that MATH . Such a MATH is called eligible for MATH. Let MATH be the set of those (one or two) elements of MATH that are eligible for MATH. If MATH is not superfluous then MATH. If REF are satisfied then MATH is a cut of MATH. Suppose that there is a path MATH going from MATH to MATH in MATH. Let MATH be the first element of the path that is not in MATH. We will prove that it is in MATH. The point MATH in the intersection of MATH and MATH is the neighbor of an element of MATH, since it is in MATH. If it is an element of MATH itself then MATH. Since MATH, it follows that MATH and we are done. Suppose therefore that MATH. Then MATH, since MATH is a closed cut. Let MATH be such that MATH. Then MATH. If MATH then MATH, by the definition of MATH. Then MATH is not superfluous, and by REF , MATH is either MATH or MATH. CASE: After REF , what is left from REF to prove is that the sequences MATH can be chosen satisfying REF in such a way that one of the MATH is superfluous. The construction will contain an appropriately chosen constant MATH or REF. If MATH then we say that a forward choice is made at time MATH. In this case, MATH is in corner MATH of the tile containing both MATH and MATH. We call this tile the backward tile. The value of the linear function MATH is greater on MATH than on MATH. Let us call the two other tiles containing MATH the forward tiles. The set MATH is the set of the centers of one or two forward tiles for MATH. In case of a forward choice, the corner MATH of one of the forward tiles is chosen for MATH. Suppose that there is a MATH in MATH satisfying MATH . Then choosing MATH as such a MATH would make a strong forward choice. If, in addition to REF, we also have MATH then we say that a strong forward choice is made. Suppose that there is a MATH in MATH satisfying REF. Then MATH is such a MATH, and with MATH a strong forward choice is made. REF are the only ones restricting the choice of MATH and MATH for MATH. Otherwise, the choice is arbitrary. Suppose that no superfluous MATH was found for MATH, all earlier choices (if any) were forward, and MATH . Then there is a MATH in MATH satisfying REF and therefore a forward choice can be made. If there is a MATH in MATH satisfying REF then all choices beginning with MATH are strongly forward, until a superfluous node is found. By REF , the elements of MATH are contained in two different sets MATH. It follows from REF that the two forward tiles are contained in different sets MATH. There is an edge between the corners MATH of the two forward tiles. Since MATH separates MATH, it must contain one of these points MATH. Since all our earlier choices were forward, the function MATH is strictly decreasing on the sequence MATH. Therefore it is not possible that MATH is equal to one of the earlier elements of the sequence, and hence REF is satisfied. If a MATH in MATH can be found satisfying REF then according to REF , the strong forward choice MATH, MATH is made. From MATH, it follows that either MATH is superfluous or MATH. In the latter case, the conditions of the present lemma are satisfied for MATH, implying that the next choice is also strong forward, etc. CASE: If MATH can be chosen superfluous then it is chosen so. CASE: If MATH cannot be chosen superfluous but it can be chosen to make MATH then it is chosen so. In this case, MATH is chosen to make MATH. If the second case of the above condition occurs then all conditions of REF are satisfied with MATH. Suppose that none of the choices of REF are possible, and MATH can be chosen to either make MATH superfluous or to satisfy the conditions of REF with MATH. Then they are chosen so. The elements MATH can always be chosen in such a way that either REF or REF applies. Before giving the proof of this lemma, let us finish, with its help, the proof of REF . The complete algorithm of choosing MATH is as follows. Choose MATH to satisfy REF . If the second part applies then choose MATH accordingly. If REF applies then choose MATH to satisfy REF . From now on, choose MATH to satisfy REF . A superfluous MATH will always found. Indeed, if the first part of REF applies then MATH is superfluous. If the second part applies then the conditions of REF are satisfied with MATH. If REF applies then they are satisfied with MATH. From this time on, strong forward choices can be made until a superfluous MATH is found. This is unavoidable since MATH is finite and hence we cannot go on making strong forward choices forever. |
math-ph/0102009 | Suppose that the statement of the lemma does not hold. We will arrive at a contradiction. Choose MATH arbitrarily. We have MATH. We can choose MATH to get MATH, MATH. We will show that we can then make a forward choice (not strong) for each MATH and recreate REF indefinitely. This is the desired contradiction since our set is finite. Assume that we succeeded until MATH. By REF , there is a MATH in MATH such that REF holds. If MATH then with the choice MATH, MATH, MATH . REF would apply, and we assumed this is impossible. Therefore MATH. Without loss of generality, let us assume MATH . Then MATH. From MATH, it follows that MATH, hence MATH. Since MATH is not superfluous, the assumption MATH implies MATH . Let us show MATH. It is easy to check that the two tiles MATH and MATH intersect in MATH. If MATH belonged to MATH then, by REF , the tile MATH would be contained in MATH while for similar reason, the tile MATH is contained in MATH. Then the intersection point MATH would have to belong to MATH. But then we could satisfy REF with MATH. We have MATH. Indeed, if it belonged to MATH then the choice MATH, MATH would again satisfy all conditions of REF which we supposed is impossible. We found that the neighborhood of MATH is just a shift of the neighborhood of MATH. This could continue indefinitely. |
math-ph/0102018 | Pick any region MATH and use the existence of a unitary partial intertwiner MATH following from the above definition MATH where the second line is the definition of a representation which is equivalent to MATH and is identical to MATH in its restriction to MATH . Therefore for all regions MATH the range of MATH is according to NAME duality contained in that of MATH . This is so because MATH for MATH . This relation together with NAME duality then tells us that MATH from which one concludes that MATH defines an endomorphism of MATH. |
math-ph/0102018 | The previous theorem is then a consequence for MATH and the fact that there exists a dense set of one particle states in the domain of MATH (that is, the analytically continued NAME) which is identical to the domain of MATH . Although the PFG's for wedge regions always exist, their use for the construction of the wedge algebras from the wedge localized subspace is presently limited to their ``temperedness". A polarization-free generator G is called tempered if there exists a dense subspace MATH (domain of temperedness) of its domain which is stable under translations such that for any MATH the function MATH is strongly continuous and polynomial bounded in norm for large x, and the same holds also true for MATH . |
math-ph/0102029 | Let MATH solve REF with MATH and satisfy the condition MATH . The solution MATH is an entire function of MATH and of MATH (see CITE, CITE). Since MATH and MATH satisfy the first REF , one has: MATH where MATH does not depend on MATH. Thus MATH . Note that MATH may be not defined for some MATH, namely for some MATH, namely for MATH, where MATH are the eigenvalues of the problem MATH . Since MATH, the condition MATH can be satisfies only if MATH. There are at most finitely many negative eigenvalues of the selfadjoint NAME operator MATH in MATH. For REF -REF to be solvable, when MATH it is necessary and sufficient that the appropriate orthogonaltiy conditions are satisfied. Namely one finds MATH . For this series to be defined at MATH it is necessary and sufficient that MATH. Note that MATH by the uniqueness of the solution to the NAME problem (see REF ). Since we have assumed MATH for MATH, the function MATH is an entire function of MATH on the complex MATH-plane. Therefore MATH is well-defined as a meromorphic function of the parameter MATH with values in MATH. Note that REF -REF is always solvable, but if the operator MATH has negative eigenvalues, then the solution to REF may grow exponentially as MATH. From REF one concludes MATH since MATH. The zeros of the function MATH are precisely the NAME eigenvalues MATH of MATH, while the zeros of the function MATH are precisely the eigenvalues of the problem MATH . It is well known (see for example, CITE) that the knowledge of MATH and MATH for all MATH determines MATH uniquely because two spectra of MATH with the same homogeneous boundary condition at MATH and two different homogeneous boundary condition at MATH, determine MATH uniquely. The zeros of MATH are the numbers MATH and only these numbers, while its poles are the numbers MATH and only these numbers. REF is proved. |
math-ph/0102029 | From REF it follows that MATH . Thus MATH . The poles of the function REF are the eigenvalues MATH, and this is the only information one can get from REF . The knowledge of one spectrum MATH of MATH determines, roughly speaking, ``half of the potential": namely, if MATH is known on the interval MATH, then the data MATH known for all MATH determine MATH on MATH uniquely (see CITE, CITE, CITE). By the same reason if MATH then MATH is uniquely determined on MATH by the set MATH known for all MATH. REF is proved. |
math-ph/0102030 | This result is known: the density of the domain of definition of the symmetric operator MATH mentioned in REF and the existence of a selfadjoint extension are proved in CITE. The defect indices of MATH are MATH or MATH , so that by NAME extension theory MATH has selfadjoint extensions (see CITE). Actually we assume in the Appendix that MATH, in which case the conclusion of REF is obvious: MATH is the linear dense subset in MATH on which MATH is defined. |
math-ph/0102030 | Let MATH and MATH, so MATH . Let us prove that: MATH where MATH is an arbitrary fixed point, MATH, MATH, MATH. If REF is proved, then MATH . From REF the desired conclusion REF follows immediately by the NAME theorem about linear functionals in MATH. To complete the proof, one has to prove estimate REF. This estimate follows from the inequality: MATH where MATH, MATH, MATH, MATH is a strictly inner open subinterval of MATH. Indeed, since MATH is selfadjoint, REF implies: MATH . Moreover MATH so, using REF, one gets: MATH . From REF one gets REF. Let us finish the proof by proving REF. In fact, REF is a particular case of the well-known elliptic estimates (see for example, [REF, pp. REF]), but an elementary proof of REF is given below in the Appendix. REF is proved. |
math/0102001 | CASE: Let MATH be the map MATH. Then MATH. Hence, MATH . CASE: For MATH, MATH where MATH is conjugation by MATH. Then MATH . |
math/0102001 | CASE: By REF , the vector field MATH is right-invariant under MATH. Thus, for MATH and MATH, MATH . CASE: MATH . |
math/0102001 | MATH . |
math/0102001 | CASE: Applying MATH to both sides of MATH we get MATH . This last expression is precisely the covariant derivative MATH of MATH induced by the connection MATH, where we view MATH as a map from MATH to MATH. CASE: MATH . |
math/0102001 | Since MATH and MATH are MATH-valued and MATH and MATH are scalar-valued, MATH is a MATH-valued MATH-form on MATH. CASE: If MATH and MATH is the fundamental vector field on MATH associated to MATH under the MATH-action, then MATH is the fundamental vector field on MATH associated to MATH. Hence, MATH . CASE: NAME: For MATH, MATH . Hence, MATH is a connection on MATH. To prove that MATH is basic with respect to the MATH-action, we check the horizontality condition MATH and the invariance condition MATH. CASE: Horizontality: MATH . CASE: Invariance: MATH . In this sum, MATH and MATH because both connections MATH and MATH are MATH-invariant REF ; moreover, by REF , MATH . Hence, MATH. |
math/0102001 | MATH . To compute a bracket such as MATH, first choose a basis MATH for MATH and write MATH with MATH being ordinary forms on MATH. Then MATH . Therefore, MATH . Since MATH is basic with respect to the MATH-action, it follows easily that MATH and MATH. Hence, MATH is also basic with respect to the MATH-action. |
math/0102001 | To check that MATH is basic with respect to the MATH-action, we compute for MATH: MATH . To check that MATH is basic with respect to the MATH-action, let MATH. Since MATH and MATH are both horizontal with respect to MATH, MATH. If MATH, then by the MATH-equivariance of MATH and MATH, MATH . Hence, MATH . Since MATH is MATH-invariant, MATH for any MATH. |
math/0102001 | Since MATH, MATH . Hence, MATH . If MATH, then MATH. |
math/0102001 | First we check that MATH is a connection on MATH. CASE: Let MATH and let MATH be its fundamental vector field on MATH. Then the fundamental vector field of MATH on MATH is MATH, because MATH acts trivially on the first factor. Hence, MATH . CASE: NAME: MATH . Next we check that MATH is basic with respect to the MATH-action. CASE: Horizontality: MATH . CASE: MATH-invariance: MATH . |
math/0102001 | MATH . In the last term, if we sum over only MATH such that MATH, instead of all MATH, it becomes MATH . Hence, MATH . The final formula now follows from REF . |
math/0102005 | We have to show how to reconstruct MATH from the functor MATH. Choose an affine open covering MATH and let MATH, MATH be the inclusion maps. Assume that MATH is a right exact functor commuting with direct sums. Then MATH will be a quasi-coherent sheaf on MATH with a MATH structure. There is a corresponding quasi-coherent sheaf MATH on MATH. In a similar way we find quasi-coherent sheaves MATH on MATH together with maps MATH. We define MATH. It is easy to see that if MATH then MATH. |
math/0102005 | As above we may assume MATH, MATH where MATH is a coherent MATH-module for finite maps MATH, MATH. Then MATH. Here MATH is a locally free MATH-module. Thus we have to show that if MATH is a finite map and MATH are coherent MATH-modules such that MATH is locally free and MATH is locally free then MATH is locally free. Since the question is local on MATH we may reduce to the case that MATH is affine. Then MATH is affine as well and hence MATH is a direct summand of a free MATH-module. So we reduce to the case MATH which is obvious. It is sufficient to prove the assertion on the rank for all pullbacks MATH for MATH algebraically closed. Hence we may assume that MATH with MATH algebraically closed. Now let MATH, MATH be respectively the left rank of MATH and MATH. We have to show that MATH for all closed points MATH. Since MATH is an extension of MATH objects of the form MATH for some MATH, this is clear. |
math/0102005 | Let MATH be the cokernel of MATH. By REF is free over MATH. Choose an isomorphism MATH and lift this to a map MATH. Let MATH its cokernel. NAME with MATH yields MATH. Since MATH we obtain MATH by NAME 's lemma. Now factor MATH through a map MATH and let MATH be the pullback of MATH and MATH. Thus we have an exact sequence: MATH . Since MATH is flat over MATH this sequence remains exact if we tensor with MATH. Since MATH is isomorphic to coker MATH we deduce that MATH. By NAME we obtain MATH. This clearly implies what we want. |
math/0102005 | Assume that MATH is locally free on the left. We will show that it is also locally free on the right. First consider the case that MATH. Then MATH and MATH are regular of the same dimension. As above we may assume that MATH for finite maps MATH, MATH. We then have the following chain of implications: MATH . The last implication follows from the fact that MATH is regular. Now consider the case where MATH is general. From the hypotheses that MATH is locally free over MATH we obtain that MATH is flat over MATH and hence MATH is also flat over MATH (since MATH is finite). Thus MATH is flat over MATH. Since MATH is finite the formation of MATH commutes with base change. By the above discussion we know that for every MATH we have that MATH is locally free over MATH. Then REF with MATH shows that MATH itself is locally free. |
math/0102005 | The author learned this beautiful formula from notes by CITE where it is shown that it holds more generally in the setting of derived categories. In our current setting it follows trivially from REF. |
math/0102005 | According to REF the left structure of MATH is given by the ordinary vector bundle dual of the right structure of MATH. Thus the right rank of MATH equals the left rank of MATH. In the same way we find that the right rank of MATH equals the left rank of MATH. Now from REF we easily obtain that the left rank of MATH equals the left rank of MATH which finishes the proof. |
math/0102005 | If MATH is a locally free coherent sheaf-bimodule on MATH and we have a base extension MATH then using REF we see that there is at least a map of sheaf-bimodules MATH. Then by looking at the left or right structure we see that this map is an isomorphism. |
math/0102005 | We only consider REF . REF is similar. With a similar method as the one that was used in the proof of REF it suffices to prove this in the case that MATH. If we restrict to this case then it is sufficient to prove that for all closed points MATH the map MATH is non-zero. Now this map is obtained by adjointness from the identity map MATH . Since this map is obviously non-zero we are done. |
math/0102005 | This is a direct computation. Let MATH, MATH where MATH, MATH are sections of MATH and MATH respectively. We have MATH . Thus we have MATH . If we look at the following pullback diagram: MATH then we find MATH . We now compute MATH . To prove the claim about preservation of epimorphisms one simply checks that epimorphisms are preserved in each individual step. |
math/0102005 | By the definition of a point module we have a surjective map MATH which according to REF corresponds to a surjective map MATH . Thus the image of MATH lies inside MATH. It follows that the image of MATH lies inside MATH. This proves what we want. |
math/0102005 | The map MATH is the composition MATH. The first map is a section and so it is a closed immersion. In particular it is proper. The second map is also proper since it is the base extension of a proper map. Thus MATH is also proper. Now we can verify REF . Since MATH is proper it is sufficient to verify that the image of MATH is finite on the left and right. This is clear since by the previous lemma this image is contained in the support of MATH and MATH was coherent by hypotheses. |
math/0102005 | If MATH is a sheaf-MATH-graded algebra on MATH then we define the MATH-graded sheaf algebra MATH by MATH . It is clear that we have MATH. Furthermore it is also clear that MATH is a non-commutative symmetric algebra with MATH and MATH for all MATH. Since such a non-commutative symmetric algebra is obtained by twisting from MATH we are done. |
math/0102005 | Both claims are similar so we only consider the second one. Since we may shift MATH we may without loss of generality assume that MATH. In that case MATH is described by a triple MATH where MATH, MATH are locally free of rank one over MATH and MATH is a surjective map. We have to extend this triple to a quintuple MATH where MATH is also locally free of rank one over MATH and MATH is another surjective map. The entries in such a quintuple are not arbitrary since the relation MATH has to be satisfied. To clarify this restriction we note that point modules and truncated point modules are preserved under twisting (see REF). Hence we may without loss of generality assume that MATH is in standard form, that is, MATH for some sheaf-bimodule MATH which is locally free of rank two on both sides. In order for MATH to define an object in MATH module we need that the composition MATH is equal to zero since this composition represents the action of MATH. From REF below it follows that this composition may be described in the following alternative way: MATH where MATH is obtained from MATH by adjointness. Thus the pair MATH is a quotient of MATH. If we now show that MATH is itself locally free of rank one then we are done. This last fact follows from REF below. |
math/0102005 | This is standard. |
math/0102005 | This is a direct consequence of REF if we view MATH as a MATH-bimodule. |
math/0102005 | Using REF it suffices to prove this in the case that MATH. But then it is sufficient to show that MATH is not zero (as MATH has rank two by REF ). Since MATH is not zero this is clear. |
math/0102005 | All claims are local on MATH so we may and we will assume that MATH is affine. In addition we may replace MATH by the scheme-theoretic support of MATH, that is, we may assume that MATH is finite. It follows that MATH is also affine, say MATH. Therefore MATH is obtained from a finitely generated MATH module MATH and MATH, MATH. The map MATH is obtained from the obvious map MATH. To prove that MATH is a closed immersion we simply remark that MATH is surjective in degree MATH. Now we make the additional hypotheses on our data, that is, MATH is a smooth connected curve over MATH and MATH is locally free of rank two over MATH. To prove our claim we may now make the additional simplifying assumption that MATH where MATH is a discrete valuation ring. The fact that MATH is NAME implies that MATH has no embedded components. So MATH is free of rank one or two over MATH and MATH embeds in MATH. If MATH is free of rank one then MATH and hence MATH is an isomorphism. So assume that MATH has rank two. Thus MATH where MATH satisfies a monic quadratic equation over MATH. We now have to show that the kernel MATH of MATH is generated by one element. Let MATH. Then MATH is generated by MATH, MATH and MATH. Write MATH, MATH with MATH. Then MATH . Thus MATH is indeed generated by a single quadratic element. |
math/0102005 | In view of the above discussion it is clearly sufficient to prove this for MATH. We will start by giving an alternative description of MATH. Without loss of generality we may assume that MATH. An object in MATH has a unique representative of the form MATH where MATH and MATH is an epimorphism. There exist sections MATH of MATH and an element MATH of MATH such that MATH and MATH. According to REF MATH corresponds to an epimorphism MATH and furthermore MATH. Since MATH contains all information to reconstruct MATH and MATH we conclude that MATH is in one-one correspondence with the set of quotients of MATH on MATH which are of rank one over MATH. If we apply this the discussion before the statement of the theorem with MATH, MATH, MATH then we find MATH . Since this bijection is obviously compatible with base extension we find that the functor MATH is represented by MATH. This finishes the proof. |
math/0102005 | This is not an immediate consequence of semi-continuity since we are not assuming that MATH is flat over MATH. We use the theorem on formal functions. For MATH let MATH where MATH is the maximal ideal corresponding to MATH. In addition let MATH be the restriction of MATH to MATH. Then one has CITE MATH . Thus in order to show that MATH for MATH it is sufficient to show that CASE: MATH for all MATH and all MATH. Similarly it is easy to see that for MATH to be surjective it is sufficient that the condition CASE: MATH is surjective holds for all MATH and all MATH. Our proof will be by induction on MATH. It follows from the hypotheses that MATH and MATH are satisfied. Assume now that MATH and MATH are satisfied. We have an exact sequence MATH . Thus MATH is the quotient of a sheaf with vanishing higher cohomology, and since we are in dimension one it follows that MATH itself has vanishing higher cohomology. Thus it follows that MATH is exact, and furthermore the induction hypotheses imply that MATH for MATH. So this proves MATH. In order to prove MATH we use the following commutative diagram with exact rows: MATH . Since the outermost vertical maps are surjective the same holds for the middle one. This proves MATH. |
math/0102005 | Since MATH is strongly graded and MATH it is clear that MATH and hence MATH is noetherian. So we need only verify REF . and REF. from REF . Let MATH. We compute MATH . Since MATH and MATH is finite, it is sufficient to prove the analogues of REF for MATH. According to REF we have to show that MATH when restricted to the MATH fibers of MATH becomes eventually generated by global sections. This follows from the fact that according to REF the MATH-fibers are preserved under the MATH's and the fact that MATH when restricted to a MATH-fiber of MATH is equal to MATH. |
math/0102005 | Without loss of generality we may assume that MATH is algebraically closed. As far as REF is concerned we will only consider the left structure of MATH. The statement about the right structure follows by symmetry. Assume that we have shown that MATH is locally free on the left of rank MATH for MATH. We tensor REF on the left with MATH. Since MATH and MATH we obtain that MATH and MATH is an extension of MATH and MATH for some MATH. This yields MATH . On the other hand we have from REF MATH . Combining these two inequalities yields MATH for all MATH. Since MATH this yields that MATH is locally free of rank MATH on the left. By induction we obtain the corresponding statement for all MATH. From this we easily obtain that REF are exact on the left. |
math/0102005 | We have maps of MATH-objects induced by the multiplication in MATH which are surjective in degree MATH. Since MATH is generated by global sections on the right these may be turned into maps MATH for certain MATH which are still surjective in degree MATH. Let MATH with MATH noetherian. Then there is some MATH such that MATH is generated in degree one. Hence there is some MATH, which we will take MATH, such that there is an epimorphism MATH which using the the maps given in REF may be turned into epimorphisms MATH . This implies REF . We now compute MATH . According to CITE and CITE the map MATH is an isomorphism in high degree. Hence for MATH: MATH . Thus MATH has finite cohomological dimension. To prove REF we may then assume that MATH for MATH large. Since in that case MATH we are done. |
math/0102005 | We first observe that MATH is in fact coherent. To this end it is sufficient to show that the diagonal submodule MATH is a finitely generated MATH-module. This may be verified after tensoring with MATH. From REF one obtains MATH . Thus MATH . The righthand side of REF is the graded-MATH-module associated to the coherent MATH-module MATH (for the ample line bundle given by MATH). Hence this graded module is finitely generated. From the computation in the previous paragraph we also learn that MATH is indeed given by the sheaf MATH supported on the diagonal. We claim that the support of MATH is finite over both factors of MATH. Again it is clearly sufficient to check this over MATH but then it follows from the explicit form of MATH given above. As indicated above MATH is flat over MATH. Hence the same is true for MATH. Since MATH is locally free over both factors it follows from REF that MATH is locally free on the left and on the right. By tensoring with MATH we deduce that the left and right rank of MATH are equal to MATH. |
math/0102005 | It is sufficient to prove that for every MATH we have that MATH lies in MATH. Since MATH is a finitely generated MATH-module in every degree we may prove this after specialization. We compute MATH . Thus MATH is up to finite length modules the graded MATH-module associated to the coherent MATH-module MATH. Hence it is finitely generated. |
math/0102005 | We first have to construct a natural transformation MATH . Taking into account the equivalences MATH this diagram may be rewritten as MATH . Here MATH is MATH applied to the left grading and similarly for MATH. The natural transformation is now obtained by functoriality from the canonical map MATH . We claim this natural transformation is an isomorphism. Both branches of REF represent right exact functors so it is sufficient to consider the value on the projective generators MATH of MATH. This verification may be done after specialization. We find MATH where have used REF. An easy verification shows that MATH from which we deduce that MATH is finite dimensional for any MATH. This implies that the natural transformation in REF is in fact a natural isomorphism. |
math/0102005 | According to CITE MATH is MATH-finite. Therefore, according to REF it is sufficient to prove that MATH has a strongly ample sequence. To this end we verify the conditions for REF . It is standard that these conditions lift from MATH to MATH and hence we may check them over MATH. Over MATH they follow from the explicit description of MATH given in REF . |
math/0102005 | We first discuss the first statement. Given REF it is sufficient to check that MATH satisfies the conditions of REF . It is easy to see that MATH is compatible with base change and is MATH-flat. One may then invoke the explicit description of MATH given in REF. To prove the last statement we note that is is true over MATH by REF. We may then invoke NAME 's lemma for MATH (given that everything is compatible with base change as we have shown above). |
math/0102005 | Given our preparatory work it is sufficient to prove MATH. By REF we obtain that MATH is an ample sequence in MATH. Given REF and the MATH-algebra version of the NAME theorem CITE it is sufficient to prove that MATH forms a strongly ample sequence in MATH. Using REF together with REF this may be checked over MATH. Then we invoke again the explicit description of MATH given in REF. |
math/0102007 | Let MATH. If MATH, then MATH satisfies both inequalities. Thus we may assume that MATH. Similarly, if MATH, then MATH satisfies both inequalities and we may assume that MATH. Consider function MATH. We have MATH . It follows that MATH is negative in some neighborhood of MATH and positive in some neighborhood of MATH. Hence the set MATH is a nonempty closed subset of MATH. Let MATH. Then MATH and MATH, that is, MATH and MATH. We have MATH and MATH . |
math/0102007 | Let MATH for MATH. By REF , there is MATH satisfying REF . Let MATH. We have MATH and MATH . |
math/0102008 | In order to see that MATH is bounded let MATH. We can write MATH as MATH, where MATH is a block sequence in MATH so that MATH and MATH. Thus MATH. If MATH, then MATH . If MATH for some MATH, MATH, then it follows from our REF that MATH . In order to show that MATH is strictly singular we consider an arbitrary infinite dimensional subspace MATH of MATH, and let MATH. MATH contains an element MATH for which there exists a MATH so that MATH. Indeed, as it was shown in CITE, we find for any MATH and MATH a normalized block MATH in MATH which is MATH-equivalent to the MATH-unit vector basis, in particular it follows that MATH . On the other hand it was shown in CITE that given MATH we can choose MATH big enough so that for any MATH, MATH, it follows that MATH, with MATH, for MATH. Since clearly MATH there exists a MATH so that MATH. Now assume that MATH is an infinite dimensional subspace of MATH on which MATH acts as an isomorphism. For any MATH we can choose a MATH, MATH, so that MATH, with MATH. As before we write MATH as MATH where MATH is a block sequence with MATH and MATH for MATH. Then MATH. Since MATH was arbitrary and since MATH for MATH we arrive to a contradiction to the assumption that MATH in an isomorphism. Finally, MATH is not compact since MATH is a seminormalized block sequence. |
math/0102008 | For REF we have to show that the function MATH, MATH is increasing. By taking derivatives we need to show that MATH. An easy computation shows that: MATH . In order to show REF let MATH and choose MATH such that MATH . Secondly choose a MATH so that MATH. Then MATH . |
math/0102008 | Using REF we can write for any MATH and MATH as MATH where MATH and MATH is associated to MATH, for MATH, with MATH, and where MATH is a block with respect to MATH. Assume MATH, MATH, with MATH and MATH, and let MATH. Without loss of generality we can assume that MATH, for MATH. By recursion we will choose for each MATH a node MATH, with MATH and MATH, so that MATH . For MATH, and since MATH, the claim follows from the assumptions on MATH. Assume MATH, with MATH, have been chosen. Then it follows from REF that MATH and by the induction hypothesis that MATH . Thus there exists a MATH so that for MATH it follows that MATH . This finishes the proof of the claim. For MATH let MATH (recall that MATH, with MATH, is the associated vector to a tree of length at least MATH). Since MATH it follows that on the MATH, and, then, that MATH . In the last of above inequalities we used REF , the remark after REF , and REF . |
math/0102008 | We choose MATH so that MATH, where MATH is chosen as in REF , and so that MATH, the first element of the previously chosen sequence MATH. For MATH with MATH we consider the sequence MATH, with MATH and MATH and observe that the two series MATH and MATH are finite (note that MATH and that MATH is summable). By REF also the series MATH is finite. Therefore MATH is uniformly bounded on MATH. By induction on MATH we will show that MATH . This would imply the statement of the Lemma if we choose MATH. If MATH the claim is trivial (note that MATH). Assume that for some MATH and all MATH, MATH, REF is true and assume MATH, MATH, and MATH. By REF we can choose MATH, MATH, MATH, MATH, MATH, and MATH so that (put MATH and MATH) MATH . We can assume that MATH for MATH. If MATH then the cardinality of the support of MATH is smaller than the cardinality of supp-MATH. If MATH and if, say, MATH then we could split MATH into MATH and MATH, choose MATH and observe that above inequalities still hold. Thus we can assume that for all MATH, the support of MATH is of lower cardinality than the support of MATH. Thus we can assume that our induction hypothesis applies to MATH, MATH. We distinguish now between two cases. If MATH, then it follows from REF (applied to MATH and MATH) that MATH . If MATH we deduce that MATH . Thus in both cases we conclude that MATH which finishes the proof of REF . |
math/0102008 | Note that for MATH . Therefore it will be enough to show that for MATH, MATH . Using REF we can easily prove by induction on MATH that MATH and (by applying MATH) it follows from REF that (recall that MATH) MATH . We distinguish between two kinds of MATH's in MATH. If MATH is such that MATH we deduce that MATH . Thus MATH . To estimate MATH, with MATH, we put MATH and MATH and observe for MATH that MATH . As in REF we observe that MATH which implies by REF that MATH and thus (by REF ) MATH . REF imply that MATH . Finally REF imply the claimed REF . |
math/0102008 | Since MATH is close to a logarithmic function we can choose a MATH so that for MATH and MATH . Assuming furthermore that MATH we deduce for MATH that MATH . Let MATH, let MATH, such that MATH, and MATH. Choose MATH and MATH so that MATH. For MATH let MATH so that MATH. For two numbers MATH, MATH, we let MATH and we choose for MATH, MATH, with MATH whenever MATH so that MATH. Now we observe that MATH is well ordered by MATH and its cardinality is MATH which is at least MATH and at most MATH. Thus we deduce MATH . Solving these inequalities for MATH we obtain that MATH . Choosing for the MATH the values MATH, MATH, MATH and MATH we deduce for all choices MATH. This implies that MATH and thus that MATH . Now let MATH and choose for MATH a MATH so that MATH, and reorder MATH and MATH into MATH and MATH. Letting MATH we obtain the claimed inequality: MATH . |
math/0102008 | We proceed by induction on MATH (assume that MATH). For MATH, MATH for some MATH, MATH and some MATH. Then MATH and MATH. The inductive step, from MATH to MATH proceeds as follows: By the definition of MATH, we separate three cases: CASE: Assume that MATH where MATH is an interval, MATH for all MATH and MATH. Let MATH . Then MATH and by the induction hypothesis we have MATH . Set MATH and after noting that MATH, for MATH set MATH where the sum over an empty set of indices is zero, and MATH if MATH. It is easy to see that the above choices of MATH and MATH, MATH, satisfy the conclusion of the lemma. CASE: Assume that MATH where MATH is an interval and MATH in MATH. By the induction hypothesis we have that MATH . Set MATH and after noting that MATH (since MATH is injective) for MATH, and MATH, for MATH set MATH and MATH if MATH. It is easy to see that the conclusion of the lemma is satisfied. CASE: Assume that MATH and MATH where MATH for MATH and MATH, MATH, MATH, MATH has rational coordinates, and MATH, for MATH. Let MATH and MATH . If MATH then we proceed as in REF . Therefore without loss of generality, we assume that MATH. Let MATH and MATH . By the induction hypothesis we have MATH . Set MATH and after noting that MATH and that the sets MATH, MATH (for MATH), MATH (for MATH and MATH), MATH (for MATH) are mutually disjoint (by the injectivity of MATH), set MATH . It is easy to see that the conclusion of the lemma is satisfied. |
math/0102010 | We compute using REF : MATH . |
math/0102010 | We note that: for all MATH which implies that MATH for all MATH. |
math/0102010 | Most of the proof is found in CITE or CITE: we need only establish the symmetric NAME condition as well as the characterization above. Let MATH. Note that MATH is anti-isomorphic to MATH as algebras, via the map MATH which has inverse given by MATH. Clearly then MATH. Note also that MATH as a consequence of REF . We compute that MATH for all MATH and all MATH: MATH while MATH . Suppose MATH is an algebra with idempotent MATH and conditional expectation MATH satisfying the conditions above. Since MATH and MATH for each MATH, there is a surjective mapping of MATH. If MATH for some MATH, then MATH; applying MATH and REF , we see that MATH. It follows that the mapping MATH is an algebra isomorphism forming a commutative triangle with MATH and MATH. |
math/0102010 | Let MATH. We compute using the symmetric product assumption: MATH . |
math/0102010 | From the first of the depth MATH conditions, we see that MATH for all MATH. Since MATH and MATH, we readily see that MATH is a strongly separable extension of MATH with NAME trace of index MATH. Similarly, MATH is a strongly separable extension with MATH as conditional expectation, dual bases MATH, MATH and index MATH. In particular, MATH is a separable extension of the separable algebra MATH, and is itself a separable algebra CITE. Similarly, MATH is MATH-separable. MATH is a non-degenerate trace on MATH since it is a NAME homomorphism by transitivity: MATH by the NAME property. Similarly, MATH is a non-degenerate trace. |
math/0102010 | The inverse mapping is given by MATH. We note that MATH for MATH, since MATH. The second statement can be proven similarly. |
math/0102010 | Let MATH, MATH denote dual bases in MATH for MATH restricted thereon. It follows from REF that the elements MATH, MATH are dual bases for the trace MATH restricted to MATH. Define MATH by MATH . Since MATH, MATH are dual bases for MATH by transitivity, it follows that MATH and MATH for every MATH. The left MATH-module property of MATH follows from: for all MATH, MATH since MATH by REF . Next, MATH since MATH. Finally, let MATH and use the NAME relations as well as the expression for MATH below REF to compute: MATH . |
math/0102010 | We compute: for each MATH, MATH while MATH by the NAME property MATH. |
math/0102010 | We note that MATH and MATH. The proposition follows easily from the dual bases equations and the depth MATH assumption: MATH for all MATH. |
math/0102010 | Now MATH and MATH follow from the usual NAME REF for MATH. At a point below in this proof, we will need to know that MATH which follows from MATH for by the basic construction theorem MATH in MATH. Note that MATH. We compute: for each MATH, MATH . Thus, MATH. The computation MATH proceeds similarly, where MATH clearly defines a bimodule projection of MATH onto MATH (compare REF ). As a result, we have MATH. We will show that MATH by showing that MATH and noting that MATH by a computation very similar to that for MATH above. Using the braid-like relations and REF , we compute: MATH . |
math/0102010 | If MATH for some MATH, then for all MATH we have MATH, since MATH (depth MATH property). Taking MATH and using the braid-like relation between NAME idempotents, and NAME property of MATH we have MATH therefore MATH. Similarly, one proves that MATH implies MATH. |
math/0102010 | The formula for MATH follows from the identity MATH and MATH: MATH . Then the second equation follows from the computation : MATH . To establish the third formula, we use the NAME property and commuting square condition to compute: for all MATH, MATH . |
math/0102010 | We obtain this formula by multiplying both sides of REF by MATH on the left and taking MATH from both sides. |
math/0102010 | Let us take MATH, then using REF , the commuting square condition, and the NAME property we have MATH . Therefore, MATH and since MATH as vector spaces, we have MATH. |
math/0102010 | Using non-degeneracy of the duality form and definition of MATH we compute for all MATH: MATH whence the formula follows. |
math/0102010 | We use the non-degeneracy of the duality form: MATH therefore, we have MATH. Using REF we conclude that MATH . We replace MATH by MATH to obtain the result. |
math/0102010 | Let MATH then MATH . |
math/0102010 | We have seen that MATH, therefore MATH and MATH commute. By REF , MATH . |
math/0102010 | First, one can define a coalgebra structure on MATH using the duality form from REF and show that MATH in a way similar to how it was shown above for the comultiplication MATH of MATH that MATH. Then we compute: MATH . Note that in the third line MATH commutes with each of the elements in MATH, so that MATH is also equal to MATH. |
math/0102010 | CASE: We have MATH, whence the result follows by non-degeneracy of the bilinear pairing MATH. CASE: We compute, using REF and the anti-multiplicativity of MATH: MATH . CASE: Since both sides of the given equation belong to MATH, it suffices to evaluate them against MATH for all MATH: MATH where we used REF . CASE: We evaluate both sides against elements MATH (note that MATH commutes with MATH): MATH . CASE: From REF , properties of MATH, and properties of the separability element MATH we have MATH . |
math/0102010 | CASE: Let MATH then MATH . CASE: From Lemma (REF c) and REF we have MATH . |
math/0102010 | First, let us note that for all MATH we have MATH if and only if MATH for all MATH. Indeed, if MATH and MATH for all MATH then MATH for all MATH. But since MATH by REF and MATH is non-degenerate, we conclude that MATH. Let MATH and MATH. We compute, using REF : MATH whence the result follows. |
math/0102010 | This follows from the fact that every MATH can be written as MATH, where MATH commute with MATH. |
math/0102010 | This is obtained from REF by replacing MATH with MATH, multiplying both sides by MATH on the left, and taking MATH from both sides. |
math/0102010 | Let MATH then using REF and Lemma (REF d) we compute: MATH since MATH, whence the proposition follows from non-degeneracy of MATH and bijectivity of MATH. |
math/0102010 | Using the definition and properties of MATH and REF for all MATH we have: MATH and using REF and bijectivity of MATH we obtain: MATH . Next, using the duality form we have: for MATH, MATH as required. |
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