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math-ph/0103034 | First, we check that the definition is correct - indeed, since MATH and MATH are invertible elements from the center we might map it to the unit of the algebra. Next, we verify: MATH and, similarly, one can prove the relation for MATH. |
math-ph/0103034 | The algebra is described through the relations REF with the additional constraints MATH. Clearly, the projections MATH and MATH define two ideals, a commutative and a noncommutative one. When we restrict ourselves to the algebra of the noncommutative ideal, then MATH becomes a unit and the algebra of this ideal is generated by two MATH-commuting, MATH, unitaries, hence it is a complex matrix algebra of dimension MATH. |
math-ph/0103034 | First, let us notice that the commutation relations will be identical as for MATH: MATH . It only remains to prove whether the coproduct closes within MATH(we use the formal expressions within the multiplier of the tensor product): MATH where we use MATH to denote the operation modulo MATH and the last sum is over indices MATH, which run from MATH to MATH. Of course, a similar calculation could be done for MATH. |
math-ph/0103039 | The first observation we make is that for every probability measure MATH one has by property a., MATH . As a consequence of this and of property b., one has for every measurable set MATH the bound MATH . Define the constant MATH. An immediate consequence of REF is that for any probability measure MATH, one has MATH . Now take any two probability measures MATH and MATH. Denote by MATH the positive (respectively, negative) part of MATH. Since MATH and MATH are probability measures, one has MATH, say. Then, since MATH preserves probability, one has MATH . This completes the proof of REF . |
math-ph/0103039 | Fix once and for all MATH and MATH. By REF , there exist constants MATH and MATH such that the set MATH satisfies MATH for every MATH. By REF , we can apply REF to find MATH for some MATH and for MATH any integer. Since MATH preserves positivity and probability, one immediately gets the same estimate for arbitrary real times. By REF and the invariance of MATH, this yields for some constant MATH, MATH . The proof of REF is complete. |
math-ph/0103039 | We define the linear operator MATH and the stochastic convolution MATH . With these notations, the solution of REF reads MATH . In a first step, we show that for every couple of times MATH, there exists a constant MATH independent of the initial condition MATH such that MATH . For this purpose, we introduce the auxiliary process MATH defined by MATH. We have for MATH the equation MATH that is, MATH can be interpreted pathwise as the solution of the PDE MATH . If we denote by MATH the degree of MATH (remember that MATH), we have, thanks to the dissipativity of MATH, the inequality MATH where the MATH are some strictly positive constants and MATH denotes the left lower NAME derivative. An elementary computation allows to verify that the solutions of the ordinary differential equation MATH (with positive initial condition and MATH) satisfy the inequality MATH independently of the initial condition. Standard estimates on Gaussian processes show furthermore that for every MATH and every MATH, there exists a constant MATH such that MATH . Combining this with REF, we get REF. It remains to exploit the dissipativity of the linear operator MATH and the local boundedness of the nonlinearity to get the desired bound REF. We write for MATH the solution of REF as MATH . Since MATH, we have MATH where we used the fact that MATH is finite for every MATH, every MATH and every MATH. This technique can be iterated, using the fact that MATH, until one obtains the desired estimate REF. The proof of REF is complete. |
math-ph/0103039 | Recall that NAME 's theorem guarantees the existence of a probability measure MATH such that the transition probabilities MATH are all equivalent to MATH. This is a consequence of the NAME property and the irreducibility of MATH. By REF there exists a small set MATH such that MATH. For every MATH and every arbitrary MATH, we then have MATH for some MATH, MATH and a probability measure MATH. Since, by the NAME property, the function MATH is continuous and, by the accessibility of MATH, it is positive, there exists for every compact set MATH a constant MATH such that MATH . The proof of REF is complete. |
math-ph/0103039 | Let us denote by MATH a jointly measurable version of the densities of MATH with respect to MATH. We define for every MATH the sets MATH and MATH by MATH and the set MATH by MATH . Since MATH for every MATH, one has MATH for every MATH and therefore MATH, where MATH. Define the subset MATH of MATH by MATH . One has similarly MATH. Let us now define the sets MATH as above and define MATH. By REF with MATH and MATH, there exists a MATH-null set MATH such that for MATH one has MATH . Since on the other hand MATH, there exist a triple MATH and an integer MATH such that MATH, MATH, and e:covers MATH . This means that MATH covers simultaneously seven eights of the ``surfaces" of both sets MATH and MATH. (See REF for an illustration of this construction.) As a consequence of REF, the set MATH satisfies MATH. Similarly, the set MATH satisfies MATH. On the other hand, one has by the definitions of MATH and MATH that for MATH and MATH, MATH. Thus MATH for every MATH and every MATH. Defining a probability measure MATH by setting MATH, there exists MATH such that for every MATH, one has MATH and thus MATH is small. Since MATH, the proof of REF is complete. |
math-ph/0103041 | By induction in MATH. For MATH the claim is obvious. In the induction step MATH one may use the induction hypothesis and relations REF : MATH . |
math-ph/0103041 | We shall proceed by induction in MATH. If MATH then MATH and REF holds true since REF implies that MATH. The induction step MATH: according to REF , and owing to the fact that MATH is monotone, we have MATH . |
math-ph/0103041 | If MATH then MATH . Since MATH the sequence MATH is NAME in MATH and so MATH exists. Under REF we can apply the same reasoning to the sequence MATH to conclude that the limit MATH exists in MATH. Set MATH . If MATH then, owing to REF , we have MATH . REF implies that MATH is a NAME sequence in MATH and so MATH exists. To show REF let us first verify the inequality MATH valid for all MATH. Actually, using REF , we get MATH . To finish the estimate note that REF imply MATH . With the aid of an elementary identity, MATH we can derive from REF : if MATH then MATH . Set temporarily in this proof MATH . If MATH then MATH . Hence MATH . This shows that the sequence MATH is NAME and thus the limit on the Right-hand side of REF exists. We conclude that it holds true, in virtue of REF , that MATH . So equality REF has been verified as well. |
math-ph/0103041 | Choose an arbitrary vector MATH and set MATH . Then MATH and MATH as MATH. On the other hand, MATH . So the limit MATH exists. Consequently, since MATH is closed, MATH. But MATH has the same property and thus equality REF follows. Furthermore, the above computation also shows that MATH . Application of the following algebraic identity (easy to verify), MATH concludes the proof. |
math-ph/0103041 | We use notation of REF . From REF follows that, MATH, MATH, MATH . Since the set of vectors MATH is dense in MATH, we get, MATH, MATH, and hence MATH . Set MATH . REF implies that both sequences MATH and MATH are NAME in MATH and hence the limit REF exists in the operator norm, with MATH. Moreover, MATH, MATH . Next let us compute MATH. For MATH, set MATH. MATH doesn't depend on MATH since if MATH then MATH . We can apply REF to the operators MATH, MATH, MATH to conclude that MATH and MATH . On the other hand, MATH . Thus MATH . Consequently, MATH and MATH . Set MATH. According to REF , MATH. Now we can compute, using relation REF , a limit in MATH, MATH . So MATH. From the closeness of MATH, the equality MATH, and from the fact that the sequences MATH, MATH converge one deduces that MATH and hence, in fact, MATH. In addition, MATH . Combining REF one finds that MATH . To conclude the proof it suffices to apply the mapping MATH to equality REF . |
math-ph/0103041 | Taking into account the defining relations REF one finds that the constants MATH, MATH and MATH introduced in REF may be chosen as MATH . REF implies that MATH and so, according to the remark following REF holds true with MATH. Since MATH REF as well as all assumptions of REF are satisfied and so the conclusions of REF hold true. |
math-ph/0103041 | Set MATH . Since the Right-hand side of REF is in fact a matrix entry of MATH (compare REF ) this assumption may be rewritten as the equality MATH valid for all MATH. Since MATH is closed one easily derives from the last property that it holds true, MATH, MATH . Particularly, MATH. But MATH are mutually orthogonal eigenspaces of MATH. Consequently, if MATH, then the sequence MATH, MATH has the property: MATH and MATH, as MATH. Equality REF implies that MATH . Again owing to the fact that MATH is closed one concludes that MATH and MATH. |
math-ph/0103041 | The norm of MATH may be estimated as MATH . This way the assumptions of REF are satisfied and consequently, according to REF (and its proof), the same is true for REF (with MATH and MATH defined in REF and the constants MATH, MATH, MATH defined in REF ). Since it holds MATH (where MATH) and, by assumption, REF is satisfied with MATH we get MATH . This verifies REF ; the other assumptions of this proposition are verified as well as follows from REF . Note that, in virtue of REF , MATH coincides with the given operator MATH. Furthermore, MATH is a limit of Hermitian operators and so is itself Hermitian, and MATH is unitary. Equality REF holds true and this concludes the proof. |
math-ph/0103041 | Let MATH be the increasingly ordered set of eigenvalues of MATH, MATH. Set MATH . Then MATH . By assumption, if MATH is a non-critical index then MATH (for any MATH). Further notice that, due to the symmetry REF , MATH. According to NAME Theorem REF, for any MATH, MATH, MATH may be written as a convex combination (with non-negative coefficients) of eigenvalues of the operator MATH. Consequently, MATH . If MATH, MATH, then MATH is necessarily a critical index and MATH . This implies that MATH and so MATH . This immediately leads to the desired REF . |
math-ph/0103041 | We distinguish two cases. If MATH then MATH since, by assumption, MATH . So the distance may be estimated from below by MATH . If MATH then a lower bound to the distance is simply given by MATH . |
math-ph/0103041 | To estimate MATH we shall use relation REF. Note that MATH . If MATH is critical then we have, according to REF , MATH and consequently MATH . If MATH is non-critical and MATH then we have, according to REF , MATH and consequently MATH . In the case when MATH is non-critical and MATH one gets similarly MATH and MATH . |
math-ph/0103041 | CASE: If MATH then, after a usual limit procedure, we can choose for the integration path in REF the line which is parallel to the imaginary axis and intersects the real axis in the point MATH. So MATH . CASE: In the interlaced case we choose orthonormal bases in MATH and MATH so that MATH and MATH are diagonal, MATH and MATH. For brevity let us denote MATH by MATH. Then MATH, and we can use REF to estimate MATH . Symmetrically, MATH, and the result follows. |
math-ph/0103041 | We have to show that MATH . For brevity let us denote (in this proof) MATH . Here MATH is an ``inseparable" symbol (which this time doesn't have the meaning MATH of MATH). It holds MATH . Similarly, using REF , MATH . A combination of these two inequalities gives MATH . To obtain REF it suffices to apply MATH to this inequality. |
math-ph/0103041 | Notice that, if convenient, one can interchange MATH and MATH in MATH. It holds MATH . To finish the proof it suffices to apply MATH to this inequality. |
math-ph/0103047 | Use axioms NAME and NAME if MATH, axioms NAME and NAME if MATH. |
math-ph/0103047 | We may order the sequence of logical steps leading to MATH in such a way, that in the n'th step, establishing MATH, only the relations established in the former steps are needed. Going backwards from MATH, we see that MATH and each MATH has to be represented as the union of two sets, MATH and MATH, which are either equivalent to MATH or are ordered like MATH. Then we proceed by induction. The basis of the creative process is the axiom NAME. It is the only one, which does not need a pair of sets which are already in an order relation. So MATH. We assume that MATH for MATH has been established. Now either a similar procedure as in step one may occur in REF, or the existence of one or two of the former relations is needed. Checking the axioms it is obvious that REF propagates from the MATH to MATH in the process of establishing MATH. |
math-ph/0103047 | Axiom NAME assumes MATH and MATH. By REF , MATH is a union of sets MATH which are all eqivalent to MATH. Now define MATH, observe MATH by NAME, and use REF . |
math-ph/0103047 | We write the strong subadditivity of the entropy as a relation for the mean entropy: MATH where we denote MATH, as in REF, III and IV, and use MATH . Note that MATH and MATH. In case MATH, we define MATH, in accordance with the simple subadditivity of the entropy (which would be sufficient to check REF), MATH . Now we proceed in an inductive way along the process of creating the ordering, similar as we did in the proof of REF . We check each axiom, and assume that REF holds for the pairs of sets which are assumed to be in an order relation already. CASE: Equivalent sets have the same mean entropy, so the inequalities are invariant. CASE: We have MATH, MATH, hence MATH. CASE: Now MATH, MATH, so MATH. CASE: Here MATH, implying MATH, and MATH. CASE: Here we have MATH, MATH, MATH and MATH, which obviously implies MATH. CASE: Transitivity is obvious. |
math-ph/0103047 | One direction, assuming MATH, is easy. By REF , MATH is a union of some translates of MATH. So each boundary line of MATH contains the boundary line of at least one of these translated sets and can thus not be shorter. The proof of the assertion in the other logical direction fills the remaining part of this section. |
math-ph/0103047 | It is not difficult to see, that MATH is a convex set. (This would not be true, if MATH and MATH were oblique orthogonal lines, so we excluded this case; and we have no oblique rectangular boxes in MATH.) Consider a boundary line MATH of MATH and the corresponding boundary line MATH of MATH, f.e. the lower horizontal boundary lines, with lenghts MATH and MATH. (If its length happens to be zero, it consists of one point, f.e. the lowest point in the set.) Then MATH is the corresponding boundary line of MATH, and its length is MATH. |
math-ph/0103047 | If MATH is an atom, then MATH or MATH, and MATH by axiom NAME. For larger sets we have to treat different cases differently; but since the system is invariant under REF-degree rotations, it is sufficient to prove this proposition with the molecules MATH. With MATH we denote the boundary of MATH. Assume that the point MATH is the origin. CASE: MATH is the union of MATH with MATH, where MATH is MATH translated by one step to the right. We have to investigate, whether MATH is empty or whether it can be compared with MATH or with MATH. This depends on the form of MATH in the lowest and in the highest region: If MATH is not a vertical or oblique line, we say that the bottom of MATH is CASE: flat, if MATH; CASE: sharp, if MATH, and either MATH and MATH or vice versa; CASE: rectangular, if MATH, MATH and MATH. Analogously we denote the top, with MATH replaced by MATH. The following pictures show the sharp top of some MATH and the corresponding top of MATH with MATH inside marked with crosses, and the same for some MATH with a flat top. MATH . CASE: MATH is a vertical or oblique line: MATH and MATH by axiom NAME. CASE: Top and bottom are sharp: MATH is either a parallelepiped, or an equilateral triangle, where the vertical line is the ``base", or a trapezoid, also with the vertical line as the base. We have MATH and also MATH, all the nonvanishing adjacent border lines of MATH are shorter than the corresponding border lines of MATH, so MATH can be compared with MATH: MATH, which implies MATH. With axiom NAME we infer MATH. Example: MATH, with MATH inside. CASE: Bottom is flat, top is sharp, or vice versa: If MATH, then MATH; the top of MATH is sharp with adjacent border lines shorter than those at MATH, as in the above case. Other border lines of MATH have the same lengths as the corresponding lines of MATH. So MATH, and we use again axiom NAME. If MATH but MATH, the same argument is true, with top and bottom exchanged. Example: MATH, with MATH inside. CASE: Top and bottom are flat: (Horizontal lines and rectangles are here included.) Both MATH and MATH hold, with the other border lines of MATH with the same lengths as those of MATH. So again MATH holds and axiom NAME is applicable. Example: MATH, with MATH inside. Now, if MATH and MATH, the set MATH is an oblique rectangle, not in MATH. It is nevertheless true, that MATH, as is shown under the item ``Top and bottom are rectangular". If one of the extremal regions is rectangular, MATH can not be compared with MATH. Either MATH while MATH, or MATH while MATH, or both. The picture shows the rectangular top of some MATH, the top of MATH with MATH inside, MATH marked with crosses, and the top of a set MATH which is needed in the proof. MATH . So, to prove MATH one has to use REF, and establish the ordering MATH first. In any case one has, to establish MATH, to construct a set MATH, with MATH, MATH, and MATH, so that MATH by assumption (the induction hypothesis to prove the theorem). Then one has to use a set MATH, such that MATH and MATH (or MATH), which implies MATH by axiom NAME (or by axiom NAME). CASE: Top is rectangular, bottom flat (or vice versa): MATH is the union MATH, where MATH is MATH translated one unit to the left, MATH is MATH translated one unit to the right. So MATH, with MATH denoting the point left of MATH. We discuss the case with the rectangular top, the other case is analogous by reflection. We move MATH one unit upwards, denote this set as MATH, and observe MATH, see REF . The simplest case is just that of the example to REF. In this case MATH, so axiom NAME makes the proof of MATH complete. For any larger MATH also MATH is larger and MATH. By inspection of REF one sees that MATH, MATH, MATH. The other border lines have the same lengths as those of MATH. Since MATH, we have MATH, and MATH. Example: MATH inside-MATH. MATH, MATH inside. CASE: Top is rectangular, bottom sharp (or vice versa): MATH, where MATH (MATH denotes the point below MATH, etc.) is equivalent to the bottom of MATH: MATH since MATH and MATH. The remaining procedure is as above: We construct MATH, MATH, so that MATH, and observe MATH, MATH, MATH. The other border lines have the same lengths as those of MATH. Since MATH, we have MATH, and MATH. This implies MATH. Example: MATH inside-MATH, MATH, MATH inside. CASE: Top and bottom are rectangular: The bottom of MATH with MATH inside is like the bottom in REF upside down. MATH is defined accordingly as MATH. It follows that MATH, MATH. Again we use MATH, and perform the same procedures as in the last two cases. Example: MATH inside-MATH, MATH, MATH inside. Since it is needed at another place, we observe that the same procedure can be applied also if MATH is an oblique rectangle, because also in this case MATH and MATH. CASE: MATH is the union of MATH with MATH. The case MATH happens if MATH is a horizontal or a vertical line, or a narrow oblique strip perpendicular to MATH, where the border of MATH has MATH, and either MATH and MATH or MATH and MATH. In these cases axiom NAME applies. Note that using oblique lines rectangular to MATH is forbidden. (See the remark in the proof of REF .) If MATH is not the empty set, it is always a precursor of MATH in the ordering. To see this, one has to check the lower left end and the upper right end of MATH. We define the types for the lower left end, with obvious analogues for the upper right ends by symmetry. The pictures show again each type of end for some MATH, and the corresponding end of MATH with MATH inside marked with crosses. CASE: sharp ends: Either MATH, MATH, MATH, so that MATH, MATH, MATH or MATH, MATH, MATH, and MATH, MATH. The length MATH (or MATH) is also shorter than MATH (or n); how much shorter it is depends also on the other end. CASE: rectangular ends: MATH, MATH, MATH, so MATH, MATH, MATH. MATH . CASE: flat ends: MATH, so MATH, MATH, MATH. MATH . If no end is sharp, both MATH and MATH. Otherwise at least one of these border lines is shorter than those of MATH. In any case, MATH, so MATH, and axiom NAME applies, proving MATH. Example: MATH, MATH inside. CASE: Lines: If MATH is a vertical line, MATH, and axiom NAME applies, using MATH and MATH. For lines in other directions, the same argument applies, with the three points from MATH appearing in appropriate permutations. In the other cases, the types of the bottom and of the left side of MATH dictate the way of proof, whether axiom NAME can be used, or whether NAME is necessary. The types of the left side are defined in the same way as the types of the bottom, rotated by REF degrees: Flat means MATH, rectangular means MATH, MATH, MATH, and sharp means MATH with either MATH, MATH, MATH, or MATH, MATH, MATH. CASE: Left side is flat: MATH. Define MATH. By REF, since MATH. Observe MATH, with MATH. Then MATH has to be shown. If MATH, then MATH, otherwise MATH. On the right side: If MATH, then MATH; if MATH, MATH, then MATH; if MATH, then MATH, MATH. In all cases, MATH, so MATH and MATH. By symmetry of the system and of the molecule MATH under reflections at a diagonal, the proof for the cases with flat bottom, MATH, is analogous. The example for the axiom NAME is an example for this case. (What there has been termed MATH is here MATH:) CASE: Left side is sharp: Since the cases with MATH can be treated in the way described above, we assume MATH. Also the lines have alresdy been considered, so we have here MATH, MATH, MATH. Also here MATH, MATH and MATH defined as above. On the left side we observe MATH, on the right side the inequalities for different cases are as above. So again MATH holds, and axiom NAME applies. Example: MATH, with MATH inside. MATH. CASE: Left side and bottom are rectangular: It is necessary to use MATH. Also in this case MATH. We have to investigate on MATH. The left side and the bottom are: MATH, MATH, MATH, MATH, MATH. The other border lines have the same lengths as in MATH. Form MATH, with MATH. (The point MATH is marked with a cross.) Represent MATH, observe MATH. So, by axiom NAME, MATH, and axiom NAME implies MATH. The example for axiom NAME is an example for this case. CASE: Define MATH, MATH. So MATH. If MATH is a horizontal line, MATH and axiom NAME applies. For the other cases it is easy to see MATH: Observe MATH. For the left side, consider the sequence MATH of border lengths in MATH. The first one, which is not zero, is shorter by one in the set MATH. The analogue is true for the right side, considering MATH and MATH. Obviously MATH. So MATH, and axiom NAME applies. |
math/0103001 | The proof is based on CITE, which assumes that MATH is also a NAME algebra. Let MATH be a NAME homomorphism, and MATH satisfy MATH, the augmentation. Given a right integral MATH, we note that MATH whence MATH . Then MATH coincides with the set of all right integrals. Given MATH such that MATH, it follows that MATH whence MATH is a free MATH-module. Moreover, MATH is a direct MATH-summand in MATH since MATH defines a MATH-linear projection of MATH onto MATH. |
math/0103001 | We note that MATH by evaluating each side of REF on MATH. Then for each MATH, MATH . |
math/0103001 | The NAME theorem in CITE states that separable projective algebras are symmetric algebras. The result follows then from REF . |
math/0103001 | REF is proven by applying MATH to MATH, obtaining MATH for every MATH. REF is proven similarly. REF is easy. REF is proven by applying first MATH to MATH, obtaining that MATH if MATH is too. Next, if MATH, then MATH as well, which together with REF proves REF. REF is proven similarly. |
math/0103001 | Applying MATH to both sides of REF yields MATH for every MATH. It follows from the finite projectivity assumption on MATH that MATH is a NAME system. |
math/0103001 | It suffices by duality to establish the forward implication. Suppose MATH is an NAME for MATH and MATH a right norm. Now REF and the argument after it work for MATH and the right integrals MATH, MATH since MATH is the counit on MATH. It follows that MATH is an equation for the antipode in MATH. By taking MATH of both sides we see that MATH is a NAME system for MATH . Whence MATH is an NAME for MATH with right norm MATH. |
math/0103001 | The forward direction is obvious. For the converse, we use the fact that the MATH-submodule of integrals of an augmented NAME algebra is free of rank MATH compare CITE or REF . It follows that MATH. From NAME 's Theorem we obtain that the dual NAME algebra MATH is a NAME algebra. Whence MATH and MATH is an NAME. |
math/0103001 | Using the NAME coordinates MATH, we note that MATH . We make a computation as in CITE: MATH since MATH, MATH for every MATH and MATH. Whence MATH. Since MATH, it follows that MATH. It follows that MATH, so let the convolution inverse MATH act on both sides: MATH. Whence MATH, since MATH. |
math/0103001 | On the one hand, we have seen that MATH are NAME coordinates for MATH. On the other hand, the equation MATH for every MATH follows from REF and gives MATH . Then MATH is another NAME system for MATH. Since MATH and MATH are both dual bases to MATH, it follows that MATH. REF follows from applying MATH to both sides. |
math/0103001 | By REF , another NAME system for MATH is given by MATH, since MATH is an anti-automorphism. Then there exists a (derivative) MATH such that MATH . MATH is a left norm in MATH since MATH is a MATH-invariant anti-automorphism. Also MATH is a left integral in MATH by the following argument. For any MATH, we have MATH as MATH is group-like. Then for every MATH . Now both MATH and MATH equal MATH, since MATH, MATH and MATH is group-like. Since MATH is a scalar multiple of the norm MATH, it follows that MATH . Finally, MATH since MATH from REF , and MATH is nondegenerate. |
math/0103001 | Let MATH and denote the left norm MATH by MATH. Note that MATH since MATH. We note that MATH are NAME coordinates for MATH, since MATH is an anti-automorphism of MATH . Then the NAME automorphism MATH associated with MATH has inverse satisfying MATH whence MATH since MATH. It follows that MATH . From REF we have MATH, where MATH is the NAME automorphism of MATH. By REF , MATH since MATH and MATH are group-likes and MATH leaves MATH and MATH fixed. It follows that MATH for every MATH. REF follows. |
math/0103001 | First we note that MATH is unimodular and counimodular. Then it follows from the theorem above that MATH. Localizing with respect to the set MATH we may assume that MATH is invertible in MATH. Then MATH where MATH, respectively. We have to prove that MATH. It suffices to prove that MATH for any maximal ideal MATH in MATH. Since MATH is separable and coseparable over the field MATH, we deduce from the main theorem in CITE that MATH and therefore MATH. The desired result follows from the NAME Lemma because MATH is a direct summand in MATH. |
math/0103001 | First we note that MATH may be assumed invertible in MATH without loss of generality by localization with respect to powers of MATH. Let MATH in MATH be a maximal ideal. The characteristic of MATH is not MATH by our assumption. It is known that an algebra MATH is separable iff MATH is separable over MATH for every maximal ideal MATH CITE: whence MATH is MATH-separable. Furthermore note that if MATH is greater than MATH, then MATH . It then follows from CITE that MATH is MATH-separable for such MATH. If MATH and MATH denotes the algebraic closure of MATH, then MATH is either semisimple or isomorphic to the ring of dual numbers. But the latter is impossible since it is not a NAME algebra in characteristic different from MATH. Hence MATH is MATH-separable for all maximal ideal MATH. Hence MATH is MATH-separable by CITE. By REF then, MATH. |
math/0103001 | Since MATH is finite projective over MATH, it follows that MATH is a finite projective module. It remains to check that MATH, which we do below by using the NAME Relation and REF twice (for MATH and for MATH). Let MATH denote the restriction of MATH to MATH below. MATH . By sending MATH along the isomorphisms in the last set of equations, we compute that the NAME homomorphism MATH is given by REF . One may double check that MATH for every MATH by applying REF . |
math/0103001 | It will suffice to prove that MATH is free, the rest of the proof being entirely similar. First note that MATH is finitely generated since MATH is. If MATH is the maximal ideal of MATH, then the finite dimensional NAME algebra MATH is free over the NAME subalgebra MATH by purity and the freeness theorem in CITE. Suppose MATH is a MATH-linear isomorphism. Since MATH is finitely generated over MATH, MATH is contained in the radical of MATH. Now MATH lifts to a right MATH-homomorphism MATH with respect to the natural projections MATH and MATH. By NAME 's lemma, the homomorphism MATH is epi (compare CITE). Since MATH is finite projective, MATH is a MATH-split epi, which is bijective by NAME 's lemma applied to the underlying MATH-modules. Hence, MATH is free of finite rank. |
math/0103001 | We prove only that MATH is finite projective since the proof that MATH is entirely similar. First note that MATH is finitely generated. If MATH is a commutative ground ring, MATH is an epimorphism of MATH-modules, then it will suffice to show that the induced map MATH is epi too. Localizing at a maximal ideal MATH in MATH, we obtain a homomorphism denoted by MATH. By adapting a standard argument such as in CITE, we note that for every module MATH since MATH is finite projective. Then MATH maps MATH . By REF , MATH is free over MATH. It follows that MATH is epi for each maximal ideal MATH, whence MATH is epi. |
math/0103001 | The NAME automorphism MATH sends MATH into MATH by REF , since MATH is a NAME subalgebra of MATH. MATH is projective by REF . The conclusion follows then from REF . |
math/0103001 | The mapping MATH is clearly a bimodule homomorphism MATH. We compute for every MATH: MATH . |
math/0103001 | We work at first in the NAME algebra MATH in which MATH is embedded. Since MATH is a finite dimensional NAME algebra, there is a unique minimal polynomial of MATH, given by MATH. Since MATH is invertible, MATH and MATH. MATH is a NAME algebra with NAME MATH. Then MATH for every integer MATH, since each MATH is group-like. If MATH, then MATH, since otherwise MATH is root of MATH, a polynomial of degree less than MATH. Thus, MATH, but MATH since MATH on MATH. Then MATH, so that MATH. Since MATH, it follows that MATH. Clearly MATH is a NAME subalgebra of MATH of dimension MATH over MATH and it follows from the NAME theorem that MATH divides MATH. For MATH over an integral domain we arrive instead at MATH for some MATH. Since MATH is finite projective over an integral domain, it follows that MATH. |
math/0103001 | Let MATH be a prime ideal in MATH and MATH. Let MATH. Note that MATH is a finite projective NAME algebra over the domain MATH. By REF , there is an integer MATH such that MATH and MATH divides MATH. It follows that MATH for each prime ideal MATH of MATH. Since MATH is a finite projective over MATH, a standard argument gives MATH over all prime ideals, where the nilradical MATH is equal to the intersection of all prime ideals in MATH. Thus, MATH where MATH. Let MATH be integers such that MATH. Then MATH . It is clear that MATH is a monic polynomial with integer coefficients. In general, let MATH be the least number such that MATH: we will call MATH the characteristic of MATH. Clearly in this case MATH and MATH. Again by REF , MATH is a NAME algebra over MATH. Moreover, since MATH has no additive torsion elements, we conclude that MATH is a free module over MATH. Now by REF MATH is the minimal polynomial for MATH over MATH and therefore MATH divides MATH in MATH. Since MATH has the same roots in MATH as MATH it follows that a primitive MATH-root of unity is a MATH-root of unity and whence MATH divides MATH. |
math/0103001 | MATH is clearly a subring of MATH. Since we can choose a spanning set of MATH over MATH of the form MATH where the elements of the set are linearly independent modulo MATH, it follows that MATH generate a free module over MATH. We claim that MATH coincides with this module and therefore is free over MATH. First we need to prove that MATH. To do this, we observe that MATH and thus MATH. Then there is a canonical epimomorphism of algebras over MATH. Let MATH denote the image of MATH in MATH. Since MATH over MATH, we deduce that MATH over MATH from the fact that MATH is the prime subfield of MATH. Thus there is a canonical algebra epimorphism MATH, where MATH is a cyclic group of order MATH. Since MATH are linearly independent modulo MATH, it follows that MATH while MATH. Therefore all three MATH-algebras above have dimension MATH and MATH. The next step we need is to prove that MATH satisfies a monic polynomial equation of degree MATH over MATH. Clearly MATH and hence MATH with MATH. If all MATH then this is exactly what we need. Otherwise we can replace MATH by MATH. However new coefficients MATH are divisible by MATH. Continuing this process we will arrive at a monic polynomial in MATH of degree MATH because MATH and all the monomials of degree greater than MATH will be eliminated. Now it follows that MATH is a NAME algebra over MATH and a free module of rank MATH. Then MATH is a NAME algebra because MATH is a local ring. Let MATH be a NAME homomorphism for MATH. If MATH, it is clear that this is a NAME homomorphism for MATH. Since MATH, it follows that MATH is an invertible element of MATH. Hence, the relation MATH implies that MATH, which proves the theorem. |
math/0103001 | Let MATH be a NAME ring and MATH be the set of all torsion elements, that is, for any MATH there exists a positive integer MATH such that MATH. MATH is clearly an ideal of MATH. Since MATH is finitely generated over MATH, there exists a positive integer MATH such that MATH. Let MATH and MATH be canonical surjections. We claim that MATH is an embedding. Indeed, if MATH then MATH and MATH for some MATH. Obviously then MATH and MATH. Since MATH has no additive torsion and therefore is a NAME, it remains to prove that a NAME ring of a positive characteristic is a NAME. For that let us consider a multiplicatively closed set MATH consisting of all the non-divisors of zero of MATH. It is well-known that MATH is an embedding and MATH is a semi-local ring if MATH is NAME (see CITE). So, it is sufficient to prove that a semi-local ring MATH of positive characterestic is a NAME. Let MATH be the set of all maximal ideals of MATH and MATH be the corresponding localizations. Now let us consider a homomorphism MATH induced by canonical homomorphisms MATH. We claim that MATH is an embedding. Let in contrary MATH. Then MATH for any MATH and there exists MATH such that MATH. Let us consider the ideal MATH generated by all MATH. Clearly MATH. On the other hand MATH cannot belong to MATH because MATH is not in MATH. Therefore we get that MATH and consequently MATH. Since any MATH is a local ring of positive characteristic, REF implies the required result. |
math/0103001 | Note that MATH. |
math/0103001 | Localizing with respect to MATH we reduce the statement to a semi-local ring MATH. Then it is well-known (see CITE) that MATH and hence the statement follows from the NAME case. |
math/0103001 | Let MATH be an NAME with MATH a right norm. Then MATH is a left norm in MATH. Let MATH be the left distinguished group-like element in MATH satisfying MATH for every MATH. Moreover, note that MATH is a left norm in MATH. In this proof we denote elements of MATH as tensors in MATH. We claim that MATH is a left and right integral in MATH. We first show that it is a right integral. The transpose of REF is MATH. Applying MATH to both sides yields MATH. It follows easily that MATH . We next make a computation like that in CITE. Given a simple tensor MATH, note that in the second line below we use MATH for each MATH, and in the third line we use REF : MATH . In order to show that MATH is also a left integral, we note that REF applied to the right norm MATH in MATH is MATH. Apply MATH to obtain MATH . Applying MATH to both sides yields MATH. Whence MATH . Then MATH . Thus MATH is also a left integral. Next we note that MATH is an NAME for MATH, since MATH, the ordinary tensor product of algebras (recall that MATH is the ordinary tensor product of coalgebras MATH). This follows from MATH being a right integral in MATH on the one hand, while, on the other hand, MATH and MATH are NAME with NAME MATH and MATH. Since MATH is an NAME for MATH, it follows that MATH is a right norm in MATH. Since MATH is a left integral in MATH, it follows that it is a left norm too. Hence, MATH is unimodular. |
math/0103001 | Note that MATH is a right integral in MATH, since MATH and MATH are right integrals in MATH and MATH, respectively. Then MATH so that MATH is a right norm in MATH. By REF , MATH is an NAME with NAME MATH. |
math/0103001 | Since MATH is unimodular, REF shows that MATH has NAME automorphism MATH. Since MATH is almost commutative, a computation like NAME 's (compare CITE) shows that MATH of an almost commutative NAME algebra MATH is an inner automorphism: if MATH is the universal MATH-matrix satisfying REF , then MATH where MATH. Thus, the NAME automorphism is inner, and MATH is a symmetric algebra. |
math/0103008 | A point MATH is fixed by MATH if and only if there exists a one parameter subgroup MATH such that MATH . Let MATH (respectively, MATH) be the eigenspace of MATH with eigenvalue MATH (respectively, MATH). Let MATH be the sum of other eigenspaces. The above equation implies that CASE: MATH, MATH, MATH, CASE: MATH, MATH, CASE: MATH, MATH, MATH. The stability condition implies that MATH. Thus we have MATH, and MATH decomposes into a sum MATH and MATH, where MATH, MATH. Conversely MATH defines a fixed point. Thus we have a surjective morphism MATH . By the freeness of the MATH-action on MATH, this is injective. Let us identify the tangent bundle of MATH with MATH in REF. Its restriction to the fixed point set MATH decomposes as MATH where MATH, MATH are as in REF with MATH replaced by MATH (MATH), MATH, MATH are as above, and MATH, MATH are defined by exchanging the role of MATH, MATH and MATH, MATH. We let MATH acts on MATH (respectively, MATH) with weight MATH (respectively, MATH). Then this identification respects the MATH-action. Thus the tangent space of the fixed point component, which is the MATH-weight space of the whole tangent space, is identified with MATH. It is isomorphic to the tangent space of MATH. Therefore the map is isomorphism. |
math/0103008 | In CITE, the following was shown: varieties MATH, MATH and fixed point sets have MATH-partitions (see [CITE] for definition) such that each piece is an affine space bundle over a nonsingular projective manifold, which is a locally equivariant vector bundle. Moreover, each base manifold has a decomposable diagonal class as in [CITEEF]. Note that NAME homomorphisms are defined for arbitary varieties, and have obvious functorial properties. By the arguments in CITE, it is enough to show that the NAME homomorphism is an isomorphism for each base manifold. By the proof of [CITEEF], each base manifold has this property. |
math/0103008 | It is enough to show that MATH is a closed subvariety of MATH. By CITE the coordinate ring MATH (the ring of MATH-invariant polynomials in MATH) is generated by the following two types of functions: CASE: MATH, where MATH, , MATH is a cycle in our graph, that is, MATH, MATH, , MATH, MATH. CASE: MATH, where MATH, , MATH is a path in our graph, that is, MATH, MATH, , MATH, and MATH is a linear form on CITE MATH. Functions of the first type are invariant under the MATH-action. Functions of the second type are of weight MATH, MATH, MATH if MATH is the extension (by MATH) of a linear form of MATH, MATH, MATH respectively. Thus MATH is contained in MATH if and only if MATH maps MATH into MATH for any path MATH, , MATH. Similarly MATH is contained in MATH if and only if functions of the first type vanishes, and MATH maps MATH into MATH, and MATH to MATH for any path MATH, , MATH. Now the assertions are clear from these descriptions. |
math/0103008 | (Compare CITE.) REF The assertion follows from REF and results in CITE. CASE: We replace the group MATH by its maximal torus MATH. If we tensor the fraction field of MATH to the above homorphisms, it becomes isomorphisms by CITE since the MATH-fixed points are the same on all four varieties. Then the assertion for the torus follows from the freeness ( REF and CITE) of modules. Taking the NAME group invariant part, we get the assertion. |
math/0103008 | The morphism is equivariant under the MATH-action. Thus it is enough to check the assertion in a neighbourhood of MATH. The tangent bundles of these varieties, restricted to MATH, are both given by MATH and the differential of the morphism is the identity. Hence we have the assertion. |
math/0103008 | Recall that MATH is a finite union MATH. Hence both assertions will follow if we show that MATH is a lagrangian subvariety. Let MATH be the projection in REF. By REF, we have the following exact sequence of vector bundles over MATH: MATH (More precisely, we must restrict to the inverse image of the nonsingular locus of MATH.) By the definition of MATH, we have MATH and the induced bilinear form on MATH coincides with one induced from the symplectic form on MATH. Since MATH, MATH are lagrangian, the latter vanishes. It is also clear that the dimension of MATH is half of that of MATH. Just note that MATH and MATH are dual to each other with respect to the symplectic form. |
math/0103008 | If MATH for some MATH, then we have MATH. If MATH, then MATH. We continue this procedure successively. Since the total sum MATH of dimensions decreases under this, it will eventually stop at MATH with MATH for all MATH. |
math/0103008 | By REF MATH is generated by MATH. But this set consists of the single element MATH as shown in the proof of CITE. |
math/0103008 | By the definition, it is clear that the composite of homomorphisms is MATH. The exactness at the middle and the right terms are clear. We now prove the exactness at the left term. Suppose that MATH, considered as an element of MATH, is equal to MATH for some MATH. Since we have MATH, the injectivity of MATH implies that MATH. Then MATH implies that MATH. Thus we get the exactness. |
math/0103008 | CASE: By the assumption, MATH is surjective. It is true if and only if the following two statements hold: CASE: MATH is surjective, CASE: MATH is surjective. Then the second statement is equivalent to CASE: The homomorphism MATH is surjective. By REF we have MATH. Therefore MATH. By REF we have MATH . REF A subspace MATH of MATH defines subspaces MATH of MATH and MATH of MATH with an exact sequence MATH. For a generic MATH, we have MATH . Therefore we have the assertion. |
math/0103008 | Take MATH with MATH. Then we have MATH for some MATH. We have MATH. Hence REF implies MATH . First consider the case MATH. Then we have MATH . The first two inequalities give us MATH where we have used the assumption in the inequality. Thus we have checked REF in this case. Next consider the case MATH. We have REF except that the last line is replaced by MATH . REF is replaced by MATH . We also have REF in this case. Finally consider the case MATH. Then we have MATH . We have MATH where we have used the assumption in the inequality. Thus we have REF in this case. We have checked REF in all cases, and REF follows from MATH. |
math/0103008 | By the argument in CITE, the module MATH is integrable. Since the category of integrable MATH-modules is completely reducible, it is enough to show that characters of MATH, and MATH are equal. By the definition of MATH, we have MATH . Namely the weight of MATH as crystal is the same as the weight defined by the MATH-module structure. Thus REF imply the equality of the character. In fact, we do not need the full power of REF. We only need two things: REF there exists an isomorphism of sets MATH satisfying MATH, and REF MATH is isomorphic to MATH (MATH). Thus our proof follows from REF and CITE. |
math/0103008 | By REF, MATH is the total space of the vector bundle MATH over MATH (use MATH, MATH). Then MATH where the right hand side is the total space of the restriction of the vector bundle MATH. Since MATH is the total space of MATH, REF are clear. Let us describe the tangent bundles of submanifolds on the intersection. From now, restrictions or pull-backs of vector bundles are denoted by the same notation as original bundles. The tangent bundle of MATH appears in an exact sequence MATH . The tangent bundle of MATH appears as MATH . The tangent bundle of MATH appears as MATH . Now the transversality is clear. |
math/0103008 | CASE: The proof of CITE shows that MATH is generated by elements MATH with MATH for all MATH. By REF, we have MATH for all MATH. As we already used in the proof of REF, this implies that MATH is the highest weight vector MATH. Hence MATH is contained in MATH. CASE: The first statement follows from REF, that is, MATH. The rest of proof is based on the arguement in CITE. Let MATH be the inclusion. By the pull-back with support map with respect to MATH CITE, we have MATH . By CITE this map is compatible with the convolution product, that is, the following is a commutative diagram: MATH where the upper horizontal arrow is the convolution relative to MATH, and the lower horizontal arrow is the convolution relative to MATH. Furthermore, REF implies that MATH. By REF, we have an isomorphism MATH where MATH is the NAME isomorphism. By CITE, it is compatible with the convolution product, that is, the following is commutative: MATH where the horizotal arrows are the convolution product relative to MATH and MATH, and MATH is the NAME isomorphism REF. Moreover we have MATH by REF. Combining two commutative diagrams, we get the assertion. |
math/0103008 | The first equation follows from REF and MATH (a well-known property of the lowest weight vector). The last equation follows from REF and the compatibility of weight structures for the crystal and the MATH-module. Let us show REF. Assume that the left hand side is nonzero. Then MATH intersects with MATH, where MATH is the second projection. Take a point MATH in the image under MATH of a point in the intersection and consider the exact sequence REF. Then MATH is identified with the NAME variety of subspaces MATH in MATH with MATH by CITE or CITE. Since MATH is the lowest weight, we have MATH. Therefore, MATH . We must be careful with the second equality since it holds only for a generic MATH in general. However, MATH is isomorphic to MATH, which is a single point, as seen by the NAME group symmetry CITE. So the equality holds for our MATH. Thus we have MATH by the exactness of REF. But MATH . So the NAME variety is empty. This contradiction comes from our assumption. Hence we have REF. |
math/0103008 | We identify MATH with the tensor product MATH. By CITE the assignment MATH is a surjective homomorphism MATH with kernel MATH . Therefore REF implies that MATH factors through a MATH-module homomorphism MATH sending MATH to MATH. The uniqueness is clear. Since MATH, REF implies MATH . Together with REF, we have the surjectivity of the homomorphism. Now we compare the dimensions of the domain and the target. We have MATH . Thus the surjective homomorphism must be an isomorphism. The second statement follows from REF. |
math/0103008 | We have (compare CITE) MATH . By the argument in CITE, this implies that MATH is a simple MATH-module. Its MATH-th NAME polynomial is given by MATH . (Here NAME polynomials have values in MATH.) It is equal to the product of the NAME polynomials of MATH and MATH. Therefore, it is a subquotient of the tensor product module MATH . (A well-known argument based on REF.) By the localization theorem, REF and the NAME isomorphism REF , we know that the dimensions of the both hand sides are equal. (See also the remark below.) Hence we have the unique MATH-isomorphism MATH sending MATH to MATH. |
math/0103008 | Let MATH be the class represented by the structure sheaf of MATH, as in the previous section (the lowest weight vector). We have MATH for any MATH, MATH. Consider the element MATH of the Braid group corresponding to the longest element MATH of the NAME group (of MATH). It acts on MATH by CITE. Since MATH, MATH maps MATH to MATH, where MATH is an invertible element in MATH. We have MATH . Combining this with REF, we understand that it is enough to check the commutativity of the diagram for the element MATH. Let MATH be the image of MATH under the composition of REF. We want to show MATH . In the l - weight space decomposition REF, both hand sides of REF is contained in the summand MATH . Recall both MATH and MATH are a single point. Therefore the l - weight space is MATH-dimensional. Thus REF holds up to a nonzero constant in MATH. By CITE, we have MATH where MATH is a reduced expression of MATH, and MATH is an explicitly computable natural number. By REF we have MATH in MATH. Therefore, MATH . On the other hand, by REF we have MATH . We have used the commutativity of the diagram for the element MATH, which is obvious from the definition. Therefore the constant must be MATH. |
math/0103008 | Since MATH is generated by MATH, MATH is generated by MATH. Therefore, MATH is mapped to MATH under MATH. But it is equal to MATH by REF. |
math/0103008 | Take diagonal matices MATH, MATH whose entries are roots of MATH, MATH respectively. The evaluation at MATH defines a homomorphism MATH. Then we set MATH . The inclusion in REF induces a MATH-homomorphism MATH, sending MATH to MATH. If we denote by MATH the localization of a MATH-module MATH at the maximal ideal corresponding to MATH, the above equalities factor through as MATH . By the localization theorem CITE, we have MATH where MATH denotes the set of points fixed by MATH. So the result follows from the following. MATH. Let MATH be the eigenspace of MATH with eigenvalue MATH. Let MATH and MATH where MATH, , MATH is a path in our graph. Since MATH is fixed by MATH, MATH maps MATH to MATH. On the other hand, the condition MATH implies that MATH maps MATH to MATH and MATH to MATH as in the proof of REF. Therefore, MATH must be MATH by the assumption. Since the path is arbitrary, it means that MATH, that is, MATH. |
math/0103008 | The situation is almost the same as that studied in REF. So MATH is given by the same formula as in the tensor product crystal, if we know the codimension of MATH in MATH. In fact, it was given in CITE. We have MATH . (Although the MATH-action in CITE is different from our MATH-action, the argument works.) Now we repeat the argument in REF, where MATH (respectively, MATH) is replaced by MATH (respectively, MATH), and the order of the tensor product is reversed. Therefore, we get MATH and other formulas. |
math/0103008 | It is clear that MATH commutes with MATH. Let MATH be the map defined by omitting the first and last MATH from the above formula. Let MATH . We define MATH, MATH as in REF. If MATH, then MATH where we have used that the property MATH, MATH. Thus MATH commutes with MATH. From the formula MATH, it also commutes with MATH. Now we get (REF b, c) for MATH by REF and the preceding lemma. Then we get MATH where we have used MATH, MATH, which is clear from the definition. We have MATH . Since MATH, MATH commutes with MATH. Similarly, MATH commutes with MATH. |
math/0103010 | Let MATH be a smooth plane curve of degree MATH, and let MATH the plane containing it. Clearly, for reason of intersection number, the plane MATH can't lie in MATH. Since MATH then the general line MATH intersects MATH in at least REF points. But then MATH intersects also MATH in these points and hence it is contained in any quadric containing MATH, and hence in MATH. Since MATH is general then MATH implies that MATH, which is absurd. |
math/0103010 | Let MATH be a rank MATH quadric containing MATH. Then on MATH lie two MATH families of MATH-spaces; and let MATH be one of these MATH-spaces. Since MATH is a complete intersection of MATH and two quadrics, say MATH, and since MATH, then MATH. Since MATH and MATH are quadrics then any of the two cycles MATH and MATH is either MATH or a REF-dimensional quadric. But if some of these two cycles, say MATH, is MATH, then MATH will contain the cycle MATH. Therefore either MATH, and then MATH will be not irreducible, or MATH will be a quadric surface, and both cases are impossible. Therefore MATH and MATH are quadric surfaces in MATH, and MATH contains the intersection cycle MATH. As above MATH has not REF - dimensional components, and hence it is a REF - cycle. It is easy to see that for the general MATH the general such MATH is smooth, that is, MATH is a smooth elliptic quartic curve. Let, in turn, MATH be an elliptic quartic on MATH, that is, MATH is a complete intersection of two quadrics in REF-space MATH. Since MATH and MATH is a complete intersection of three quadrics in MATH, and since MATH is a complete intersection of two quadrics in MATH, then there exists a quadric MATH containing MATH such that MATH. Since the rank REF quadrics in MATH do not contain REF - spaces, then MATH, that is, MATH. |
math/0103010 | Let us denote by MATH. We will show that the general element of MATH, that is, the general pfaffian representation of MATH, corresponds to the general rank - two bundle MATH on MATH associated by the NAME - NAME correspondence to a curve MATH of genus MATH and degree MATH contained in MATH. To show this we may proceed in the following way (or as in CITE). To an element in MATH corresponds an unique pfaffian cubic REF-fold MATH such that MATH. Such a REF-fold contains (see REF for a proof) a quintic NAME surface MATH and MATH is a curve of degree MATH contained in MATH. Since MATH, by adjunction, MATH is an arithmetically normal curve of genus MATH and degree MATH contained in MATH. The bundle associated to MATH fits in the following exact sequence MATH it does not split being MATH and has no intermediate cohomology being MATH arithmetically normal. For this bundle we find, MATH, and MATH. |
math/0103011 | Let us consider the following commutative diagram: MATH . By an easy calculation, one can check that the linear map MATH is an isomorphism. Hence the result for the globular homology. The argument is similar for both branching and merging semi-cubical homology theories. |
math/0103011 | First let us make a short digression. Let MATH be a pair of MATH-functors MATH and MATH from a MATH-category MATH to a MATH-category MATH. Then the coequalizer MATH of MATH and MATH always exists in MATH and any element of MATH is a composite of elements of the form MATH where MATH runs over MATH (otherwise take the image of MATH in the coequalizer: this image still satisfies the universal property of the equalizer). Since the canonical MATH-functor MATH is the coequalizer of MATH and MATH, the previous remark does apply (the symbol MATH meaning the direct sum in MATH, which coincides with the disjoint union). But MATH identifies any element of MATH with an element of MATH therefore any element of MATH is a composite of elements of the form MATH with MATH running over MATH. |
math/0103011 | Let MATH be a MATH-functor such that for any MATH, MATH implies MATH in MATH. Then MATH induces a morphism of globular sets MATH and by hypothesis, MATH gives rise to a morphism of globular sets MATH such that MATH. Then the composite MATH yields a MATH-functor from MATH to MATH. Since MATH is a natural transformation, one gets the commutative diagram MATH so the equality MATH holds. Therefore MATH. One then obtains the commutative diagram MATH . Therefore there exists a unique natural transformation MATH such that MATH and such that MATH, that is, making the following diagram commutative: MATH . Suppose that there exists another MATH such that MATH. Proving that MATH is equivalent to proving that MATH by REF . So one is reduced to checking the equality MATH. The MATH-functor MATH is clearly surjective on the underlying sets. Therefore proving MATH is equivalent to proving MATH. But one has MATH, which concludes the proof. |
math/0103011 | By REF , any element of MATH is a composite of elements of the form MATH with MATH. Since MATH is clearly surjective on the underlying sets, any element of MATH is a composite of elements of the form MATH with MATH. So any element of MATH is a composite of elements of the form MATH with MATH running over MATH. The last part of the statement of the theorem is a consequence of the fact that MATH is a MATH-functor and of the universal property satisfied by MATH. |
math/0103011 | This is due to the fact that MATH for any MATH. |
math/0103011 | Any element MATH of MATH is a composite of elements MATH for MATH by REF . For a given MATH, let us call the smallest possible MATH the length of MATH. Now we check by induction on MATH the property MATH: ``for any MATH of length at most MATH, for any MATH, there exists MATH such that MATH and MATH". For MATH, this is an immediate consequence of the fact that MATH is a MATH-groupoid. Now suppose MATH proved for MATH with MATH. Let MATH be an element of MATH of length MATH. Then MATH for some MATH and with MATH and MATH of length at most MATH. Let MATH. Let MATH (respectively, MATH) be an inverse of MATH (respectively, MATH) for MATH. If MATH, then MATH so MATH is a solution. If MATH, then MATH, so MATH is now a solution. |
math/0103011 | Both MATH-categories MATH and MATH are freely generated by MATH-complexes whose atoms are clearly in bijection. Moreover by NAME 's formulae recalled in REF, the algebraic structure of MATH and MATH is completely characterized by the MATH and MATH operators, which are obviously preserved by the mapping. |
math/0103011 | Same proof as for REF . |
math/0103011 | If MATH is a MATH-dimensional face of MATH, then both sides of the equalities are equal to the empty set. Let us then suppose that MATH is at least of dimension MATH. If MATH does not belong to the image of MATH, then both sides are again empty. So suppose that MATH. Then MATH where MATH. Let MATH. Then MATH by definition of MATH in MATH and by definition of MATH where the notation MATH means that MATH is replaced by MATH. On the other hand, MATH. |
math/0103011 | We are going to prove by induction on MATH the statement MATH: ``for any valid expression MATH of MATH with at most MATH variables such that MATH, the expression MATH is valid in MATH." If MATH, then MATH for some face MATH of MATH. So MATH which is necessarily a valid expression in MATH. So MATH holds. Let us suppose that MATH is proved for MATH and let us consider a valid expression MATH of MATH with MATH variables. Then MATH with MATH and MATH being valid expressions having less than MATH variables. By the induction hypothesis, both MATH and MATH are valid expressions of MATH. If MATH, then by construction MATH and there is nothing to prove. Otherwise by construction again, MATH. Then by NAME 's formulae and by REF , MATH . Consequently, MATH, which proves that MATH is a valid expression of MATH. |
math/0103011 | The proof is threefold: REF one has to prove that MATH induces a MATH-functor MATH from MATH to MATH; REF afterwards we check that the underlying set map of MATH is injective; REF finally we prove that the image of MATH is the underlying set of MATH as a whole. CASE: We are going to prove MATH: ``the set map MATH from the set of faces of MATH to MATH induces a MATH-functor from MATH to MATH" by induction on MATH and using REF . The MATH-category MATH is the MATH-category MATH generated by one MATH-morphism. Therefore MATH holds. One has the commutative diagrams MATH for any MATH. Therefore by REF and the induction hypothesis, one sees that MATH induces a MATH-functor MATH from MATH to MATH. It remains to prove that MATH and that MATH to complete the proof. Let us check the first equality. By REF , MATH is a valid expression of MATH in which the MATH-dimensional faces MATH for MATH appear. In the MATH-complex MATH, that means that MATH contains not only the faces MATH, but also all their subfaces. Since MATH, then necessarily MATH. Because of the calculation rules in MATH described in REF , MATH is necessarily equal to MATH. And the desired equality is then proved. CASE: The previous paragraph shows that there is a well defined MATH-functor MATH from MATH to MATH characterized by CASE: if MATH (in particular MATH is a face of MATH), then MATH CASE: for MATH, if MATH exists for some MATH, then MATH. Moreover MATH and therefore MATH is injective. CASE: This is an immediate consequence of REF and of the construction of MATH. |
math/0103011 | By REF , the MATH-category MATH is graded by the vertices of MATH (however MATH). So this is a consequence of REF . |
math/0103011 | This is a remake of the proof that the natural morphism MATH from the globular to the branching semi-cubical nerve preserves the simplicial structure of both sides (see CITE). |
math/0103011 | See CITE. |
math/0103011 | See CITE. |
math/0103011 | This is an immediate corollary of REF . |
math/0103011 | Let MATH be a non-contracting MATH-category. Let MATH be the unique MATH-category such that MATH and with MATH, MATH and MATH. Then MATH for any MATH. So the regularity of the branching semi-globular nerve comes from that of the globular nerve CITE. |
math/0103011 | The group MATH is generated by the MATH for MATH. By REF , MATH where MATH is an expression using only the composition laws MATH for MATH and where MATH. By regularity of the branching semi-globular nerve, one deduces that MATH is generated by the MATH for MATH running over MATH. The remaining part of the statement is clear. |
math/0103011 | Let MATH and MATH be two morphisms of cuts from the branching semi-cubical nerve to the branching semi-globular nerve. For any non-contracting MATH-category MATH, MATH and MATH induce natural set maps from MATH to MATH. Since the only natural transformation from the identity functor of the category of sets to itself is the identity transformation (to see that, consider the case of a singleton), then MATH and MATH are both equal to the identity of MATH. If MATH is the free MATH-category MATH generated by one MATH-morphism MATH, then MATH is non-contracting. In that case, MATH. So in that case, MATH. For any non-contracting MATH-category MATH and any MATH, there exists a unique MATH-functor MATH from MATH to MATH such that MATH. So by naturality, MATH, and therefore MATH and MATH coincide everywhere. Notice that in general MATH so the reasoning of the MATH-th dimension does not apply to dimension MATH. Now suppose that we have proved that MATH for MATH and MATH. Then MATH and MATH are two MATH-functors from MATH to MATH which coincide on the MATH-dimensional faces of MATH and such that MATH. Then MATH and MATH induce the same labeling of the faces of MATH, therefore by freeness of MATH, MATH. Now let us prove the existence of this natural transformation. Let MATH. Then for any MATH, the morphism MATH cannot be MATH-dimensional otherwise MATH would be so as well. So the restriction of MATH to MATH gives rise to an element of MATH and therefore to elements of MATH and of MATH . The MATH-functors MATH and MATH for MATH are all non-contracting. Since the natural maps MATH for MATH arise from restrictions, then one obtains three natural morphisms of simplicial sets MATH which yield a cone based on the diagram MATH . Therefore one obtains a natural transformation MATH . For MATH, the corresponding element MATH is represented in REF . |
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