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math/0105001 | We must show that MATH satisfies REF. Note that REF yields, in first order, MATH . Similarly, REF yields MATH . We finally note that REF implies that MATH . The difference of REF yields MATH . But, by REF , MATH. This implies that MATH proving that REF is satisfied. Now, switching MATH and MATH in REF , and subtracting it from REF (assuming MATH), we get MATH proving REF . |
math/0105001 | The fact that MATH is a closed bivector field was proven in CITE. Suppose MATH is a star product equivalent to MATH: MATH where MATH is an equivalence transformation. Assume MATH. We must show that, if MATH is the closed bivector given by the skew symmetric part of MATH, then MATH. The condition MATH implies that MATH. Hence MATH, for some vector field MATH. A simple computation just using the definitions shows that MATH and the result follows. |
math/0105001 | From REF we get, in second order, MATH . Similarly, from REF we get MATH . Finally, from REF we have MATH . Since we assume that MATH, subtracting REF from REF yields MATH . Using REF , we then get MATH . Taking the skew-symmetric part of this equation, and recalling that MATH, we finally have MATH . |
math/0105001 | The proof follows directly from REF . |
math/0105001 | Let MATH. A simple computation shows that if MATH, then MATH where MATH is defined by MATH. Thus MATH. |
math/0105001 | Since in our convention MATH, we use NAME 's explicit construction for the formal NAME structure MATH. REF 's star products in terms of the maps MATH is MATH . Since MATH is equivalent to MATH, and MATH is equivalent to MATH, by REF it suffices to compute MATH for MATH and MATH. It is clear from the expression just above that MATH. |
math/0105005 | Recall an equality due to NAME and NAME (see CITE). MATH . One easily verifies the following inclusions MATH . The rest follows from CITE. |
math/0105005 | The main ingredient of the proof is NAME 's sharp slope estimate CITE, which says that MATH if and only if MATH for all MATH. Suppose MATH. Then MATH for all MATH. By REF we have MATH . Conversely, suppose that MATH for all MATH. It suffices to show that MATH for all MATH. For MATH this statement is trivially true. We proceed by induction on MATH. Assume that MATH. By REF we have MATH. So MATH and MATH . From the hypothesis MATH, we have MATH. |
math/0105005 | The first statement follows from CITE. The second is precisely CITE. |
math/0105005 | For the special fibre MATH this follows immediately from REF. Let MATH be the field of fractions of MATH. Consider the generic fibre MATH. There are no points MATH in MATH with MATH, since for such MATH one can easily show that MATH is not integral over MATH, and MATH is integral over MATH. It follows that the affine part of MATH is nonsingular. The affine ramification points of the map MATH defined by MATH are those satisfy MATH. For each such MATH there are exactly MATH corresponding values of MATH, since MATH there. So there are MATH ramification points on the affine part. The function MATH in MATH and its first two derivatives have no common zeroes, so all affine ramification points are of index MATH. Let MATH be the ramification index at MATH. By NAME, we have MATH. It follows that MATH. But the genus of the special fibre MATH is MATH, hence the genus of MATH is MATH, and MATH. The differential form MATH has no poles at the affine part. The form MATH only has affine zeroes (of order REF) at points where the map MATH ramifies. At these points, MATH, so MATH has no affine zeroes. The degree of a differential form is MATH, hence MATH has a zero of order MATH at MATH. The function MATH has degree MATH and no poles at the affine part. Hence it has a pole of order MATH at MATH. Similarly, MATH has a pole of order MATH at MATH. So for MATH, MATH and MATH the form MATH is in MATH. From the assumption that MATH and MATH are coprime it follows that for MATH and MATH in this range the MATH have zeroes of different order at MATH, hence they are independent. From REF , it follows that the reduction of these differential forms modulo MATH form a basis for MATH, hence the MATH form a basis for MATH. |
math/0105005 | If MATH then MATH, which is zero in MATH. |
math/0105005 | It is easy to see that MATH if MATH. Thus MATH . Therefore, we have MATH . This proves the lemma. |
math/0105005 | We will prove the first part by induction. Suppose MATH and MATH . Note that this this trivially true if MATH. Write MATH. By CITE, we have MATH for some power series MATH. Thus the power series expansion of MATH is MATH . Apply the MATH action on the above first differential form. Since the MATH-action commutes with the restriction map (by REF ), we have MATH for some MATH. By the hypothesis, MATH divides MATH. For all MATH, by REF , MATH . For MATH coprime to MATH it follows from REF that MATH divides MATH. Thus MATH . Otherwise, except possibly when MATH and MATH, we have MATH . Therefore, MATH . For all integer MATH, by the hypothesis of the theorem, we obtain MATH . So, by REF , we have MATH. So MATH divides every sum of MATH except possibly the one on the first line. Combining this information with MATH, MATH and MATH yields for all MATH . Hence for such MATH . REF implies MATH so MATH which proves the induction hypothesis. If MATH then the above implies MATH lies in MATH . REF implies MATH lies in MATH. Hence, MATH lies in MATH. Now the theorem follows from MATH. |
math/0105005 | CASE: The hypotheses in REF are satisfied for all positive integers MATH and for all possible MATH and MATH. Thus our statement follows from REF . CASE: If MATH then MATH for all MATH in the range of REF by REF . This implies that for the particular MATH satisfying the hypothesis of REF we have MATH . This proves the NAME. |
math/0105005 | Note that MATH. Apply REF to this equation, we get MATH . We have MATH . Clearly, this is REF if MATH; otherwise, it is equal to MATH . Plugging this into REF yields the desired value for MATH. |
math/0105005 | Let MATH. Using the well-known identity MATH for all natural number MATH, one gets that MATH . If MATH then MATH . If MATH then MATH . If MATH then MATH . Substitute these back in MATH, we obtain the desired (in)equalities. |
math/0105005 | Take the identity MATH . Substitute the power series expansion of MATH in REF ; we get MATH . If MATH then MATH by REF ; hence MATH. This prove the first part of the lemma. If MATH then MATH and MATH. If MATH, MATH and MATH then, by REF , the term with minimal valuation in REF occurs at MATH. We have MATH . If MATH and MATH then the term with minimal valuation in REF occurs at MATH. We have MATH . This implies the second assertion. By MATH we have that MATH. Thus MATH . So the third assertion follows from this equality and the second assertion. |
math/0105005 | We will prove this lemma by induction on MATH. Let MATH. Suppose MATH. Note that all real numbers MATH and MATH satisfy MATH. So we have MATH . Suppose MATH and the equality holds. It reads MATH. Now suppose MATH. Let MATH. Let MATH for all MATH. One can find a natural number MATH such that MATH . Then MATH. We have MATH. So MATH . On the other hand, by induction hypothesis on MATH, one has MATH . Combining REF , one gets MATH where MATH. Using MATH one easily observes that MATH, and the first part of the lemma follows from REF . Now suppose MATH, MATH, the equality holds in REF and MATH. This can only happen if MATH. It follows by induction that MATH and MATH for MATH. From the equality in REF it follows that MATH. |
math/0105005 | Write MATH . Now we write MATH . Consider contributions of each factor of the product of REF in the coefficient of REF . Each MATH-th factor of REF contributes to the coefficients of MATH for some MATH, where MATH. When MATH then it has to contribute to the coefficient of MATH. Inductively for each MATH the MATH-th factor contributes precisely the coefficient to MATH. It is easy to see that MATH . Thus our assertion follows. |
math/0105005 | Let MATH. Let MATH where MATH, then MATH. Let MATH, write MATH for some MATH. From REF , it follows that MATH . Then by REF one easily verifies that REF holds. Assume REF holds. Then there is a MATH such that the equality in REF holds for MATH, which implies that MATH, MATH for MATH, MATH for MATH by REF . Thus MATH. So MATH. Those MATH which contribute terms in the sum REF with minimal valuation necessarily have MATH. By the identity MATH we have by REF MATH . This is equal to MATH if and only if MATH. By REF this is equivalent to MATH. Conversely, the conditions imply that the contribution of MATH with MATH to MATH in REF has valuation MATH. NAME from other MATH has higher valuation by the above arguments. Thus MATH. This finishes the proof of this lemma. |
math/0105005 | It suffices to prove the theorem for the case that MATH has constant coefficient MATH. On the one hand, MATH is independent of MATH; on the other hand, the curves MATH and MATH are isomorphic over MATH for any MATH, and hence have the same NAME polygon. With the assumption MATH we can use the results of REF. CASE: Set MATH. By the hypothesis on MATH, MATH and MATH it is elementary to check that for MATH and MATH in the range of REF we have MATH thus MATH for all MATH. By REF , we have MATH . Thus MATH by REF . CASE: Choose a value of MATH in the range of REF for MATH such that the following congruence has a solution for MATH, MATH . For any integer MATH define MATH . Note that MATH is monotonically decreasing as a function in MATH, and it converges to MATH as MATH approaches MATH. Suppose MATH, then there exists a positive integer MATH large enough such that MATH. Choose such a MATH, and such that MATH is a solution to the congruence above and such that MATH. For all MATH we have MATH . Thus, for all MATH and MATH we have by REF that MATH . On the other hand, since MATH by our assumption we have MATH . Hence, for all MATH one has by REF that MATH . Hence, the hypotheses of REF are satisfied. Again by REF , MATH where the equality holds if and only if MATH. In this case, we have MATH (by REF ), which contradicts our assumption that MATH. Therefore, we have MATH. |
math/0105005 | The denominator of the MATH function of MATH of (genus MATH) is MATH where MATH's are eigenvalues of the NAME endomorphism of MATH relative to MATH. We consider the MATH as elements of MATH. Extend the valuation MATH to MATH. Let MATH. Then MATH for all MATH (see CITE). But MATH so MATH. Thus MATH. Conversely, suppose MATH for every MATH. One easily derives from REF that MATH . Taking natural logarithm and then derivative at both sides, we get MATH . Then the left hand side of this power series converges MATH-adically for all MATH with MATH. Comparing to the right hand side series we have MATH for all MATH. Therefore, MATH. |
math/0105005 | Let MATH be a polynomial over MATH with MATH for some MATH. For any MATH, write MATH. Let MATH be a basis for the degree MATH extension MATH. For any MATH write MATH for some MATH. For any MATH, take its MATH-adic expansion MATH with MATH and some MATH. Then MATH . From this, one observes that MATH can be considered as a polynomial in MATH over MATH of total degree MATH . Write MATH for MATH, then MATH is a polynomial in MATH over MATH of total degree MATH . Then we observe that MATH . On the other hand, Ax's theorem CITE indicates MATH . Thus MATH . Applying REF , we have MATH. The second inequality is elementary. |
math/0105005 | We fix a natural number MATH. Define MATH as an element of the polynomial ring MATH in MATH variables. Let MATH be any integer with MATH and MATH. For every prime MATH and MATH, we write MATH for some integer MATH. So MATH. The following lemmas are suggested by NAME. We have MATH and it can be written as a polynomial in MATH with rational coefficients. Let MATH denote the evaluation of this polynomial at MATH. Then MATH is not the zero polynomial in MATH. For any MATH in MATH we have MATH . Define an auxiliary polynomial MATH. Note that MATH is equal to the MATH-coefficient of MATH . For fixed MATH the binomial coefficient MATH is a polynomial in MATH with rational coefficients. The first assertion follows. Consider MATH as a polynomial in MATH, its MATH-coefficient is equal to MATH by inspecting REF . Note MATH, so MATH is not the zero polynomial in MATH. Since MATH, we have MATH; hence MATH . This finishes the proof of REF . Let MATH be in MATH. The following statements are equivalent CASE: MATH for infinitely many primes MATH; CASE: there is a MATH with MATH and MATH, such that MATH for infinitely many primes MATH; CASE: there is a MATH with MATH and MATH, such that MATH for infinitely many prime MATH; CASE: there is a MATH with MATH and MATH, such that MATH. REF are clearly equivalent. REF are equivalent by REF are equivalent because MATH has to vanish if it vanishes modulo MATH for infinitely many primes MATH; conversely, since MATH, there are infinitely many prime MATH by NAME. This concludes REF . The complement MATH of MATH in MATH is the set of all MATH in MATH such that MATH for infinitely many prime MATH. Write MATH. By REF , MATH is equal to the set of all MATH in MATH such that MATH is a zero of a polynomial MATH in MATH. But MATH is not the zero polynomial by REF , so MATH is NAME closed. Therefore, MATH is NAME open and hence dense in MATH. This concludes the first part of the theorem. Now let MATH. Then there exists a natural number MATH such that for all MATH we have MATH and hence MATH by REF . Therefore, for every MATH we have MATH . This finishes the proof of REF . |
math/0105012 | By REF we may reduce to the case, where MATH. By REF , MATH. Now suppose by induction on MATH that MATH. Pick a simple reflection, such that MATH. By REF it follows that MATH. Also by REF one gets that MATH. By REF MATH is identified with the kernel of a non-zero homomorphism MATH. This implies that MATH (REF. The fact that MATH follows from REF ) and an easy induction on MATH (compare CITE). |
math/0105012 | By REF it suffices to prove uniqueness of a set of twisted NAME modules in MATH, where MATH. Let MATH be a reduced decomposition of MATH, then MATH by REF . |
math/0105012 | We go through the properties for the modules MATH. We may assume that MATH. REF implies that MATH and hence that MATH by REF . But MATH as MATH is simple. To verify REF , assume that MATH and MATH. Then write MATH for a suitable simple reflection MATH and therefore MATH. Suppose that MATH, then REF follows from applying REF and dualizing the short exact sequence MATH . REF are immediate using that translation commutes with duality MATH. By REF we get that MATH. |
math/0105012 | Let MATH denote the modules in the family of twisted NAME modules with only constant MATH-endomorphisms and MATH the modules in the other family in a regular block MATH. We will prove that MATH by induction on MATH. As MATH we get MATH, so that the morphism MATH in the relaxed REF has to be a constant multiple of the adjunction morphism. By the proof of REF , MATH (the morphism MATH in the relaxed REF ) has to be a constant multiple of the adjunction morphism. Now suppose that MATH and let MATH be a simple reflection with MATH. Then MATH and we have a commutative diagram MATH giving an isomorphism MATH. |
math/0105012 | This follows from the corresponding property MATH for NAME modules and the fact that the translation functor MATH becomes left translation of NAME modules under the equivalence MATH (see CITE). |
math/0105012 | This is REF . |
math/0105012 | This is REF . |
math/0105012 | On the level of sheaves MATH if MATH, since MATH is NAME and NAME irreducible of codimension MATH. Now one uses the local to global spectral sequence MATH and the higher cohomology vanishing MATH, which follows from the assumption that MATH is affinely embedded, to deduce the result. |
math/0105012 | We may lift MATH to a MATH-representation. On the level of MATH-sheaves we have an isomorphism MATH. This extends to an isomorphism of MATH-sheaves giving the desired result. |
math/0105012 | This is REF. |
math/0105012 | This is REF. |
math/0105012 | This follows from the long exact sequence and REF . |
math/0105012 | We use REF : MATH where MATH is the finite dimensional simple module with extremal weight MATH. Observe that MATH for arbitrary MATH and MATH. Now take a MATH-filtration MATH of MATH, such that MATH and MATH. Then use REF to get the desired result. |
math/0105012 | REF follows from the above by observing that MATH is free over MATH. Also REF follows from the above identification via REF and an easy induction on MATH. |
math/0105012 | Note that by REF we may reduce to the case where MATH for some simple reflection MATH. Since MATH what we need to prove is MATH. But this is clear from REF. The last statement follows from the first by noting that the tensor identity for induction ensures that we have an isomorphism MATH. |
math/0105012 | Let MATH and write MATH. Since MATH preserves MATH REF it follows that MATH. Combining this with REF we get the statement. |
math/0105012 | Note that MATH by REF . We shall verify REF leaving REF for later. The above definition gives MATH for all MATH. So REF is certainly satisfied. Via REF we see that REF is a consequence of the corresponding fact for ordinary NAME modules. Let now the notation and assumptions be as in REF. The well known effect of the wall crossing functor MATH on ordinary NAME modules gives the short exact sequence MATH . Applying the twisting functor MATH to this sequence we get (using REF on the middle term) MATH . The exactness of this sequence comes from REF. |
math/0105012 | The result is well known for ordinary NAME modules. It then follows for twisted NAME modules by the above proposition. In fact, we get MATH. |
math/0105014 | The proof of this proposition is exactly the same as the proof of REF . |
math/0105014 | The question can be reduced to the absolute case by the argument given in REF. Thus we will set MATH. Since the question is local (in étale topology), one may assume that MATH . In fact, it is enough to consider the local ring MATH with maximal ideal MATH. Denote MATH, then the two-term cut-off of cotangent complex is MATH . This induces a complex of MATH modules MATH . One may assume that MATH contains no linear terms in MATH, or otherwise the presentation of MATH could be changed to eliminate terms linear in MATH. That is, MATH and the above MATH is a zero map. This implies that the homology of the above complex is MATH . Because MATH is a perfect obstruction theory, MATH . Therefore MATH where the last equality is obvious. This shows, combining with the assumption, that MATH. However, MATH is obviously greater or equal to MATH. Thus MATH and MATH form a regular sequence. Therefore MATH is a local complete intersection. On the other hand, MATH implies that MATH and MATH is an isomorphism. Hence MATH, and virtual structure sheaf is the ordinary structure sheaf on MATH. |
math/0105014 | The proof of REF also proves this lemma: It is constructed there two principal NAME divisors MATH and MATH in MATH (with notations like REF). The proof there shows that MATH and MATH. By REF , MATH. |
math/0105014 | The (dual of the) obstruction theory MATH is quasi-isomorphic to MATH, which is a complex of locally free sheaves with zero differential. Since MATH is smooth, MATH. Now use the self-intersection formula REF. |
math/0105014 | It is obvious that the perfect obstruction theory of the product is the product of perfect obstruction theories on each component. |
math/0105014 | It is proved in CITE that the perfect obstruction theories of MATH and MATH are compatible with respect to MATH. It remains to apply REF . |
math/0105014 | The proof in CITE p. REF shows that MATH form a commutative diagram. Therefore, MATH . Now use the pull-back property REF MATH . This completes the proof. |
math/0105014 | The proof is the same as the proof in CITE pp. REF, with the (bivariant) cycle classes changed to (bivariant) MATH-classes. We summarize the spaces involved in the following diagram: MATH where the squares are cartesian and MATH is the universal curve. Since MATH, MATH and MATH are representable, proper and flat, they have natural orientations MATH (which are bivariant MATH-classes CITE). Since MATH is flat, MATH. Furthermore, since MATH is just blow-down of some rational curves, MATH. Therefore MATH. Then MATH where the last equality follows from REF . |
math/0105014 | Consider the following commutative diagram MATH . It induces a morphism MATH . We will show an analogue of REF in this setting. Namely, MATH . Note that the vertical maps only contracts rational components with two marked points (or less). That is the reason why only chains of rational curves are relevant. It is easy to see that MATH is a divisor MATH of normal crossing in MATH, and the ``smoothing map" MATH is a finite, unramified and surjective birational morphism. In other words, the smooth NAME stack MATH separates the normal crossing point. Although MATH is not equal to MATH, the difference is supported on the normal crossing subset. The following lemma shows how one could obtain the structure sheaf MATH by the inclusion-exclusion principle. Let MATH be a divisor with normal crossing, such that MATH are smooth, disjoint away from the origin. Furthermore locally at origin MATH is defined by MATH. Then MATH is an exact sequence. REF is equivalent to the study the exactness of the following sequence locally at the origin MATH . This equation holds because, for example, MATH, the following sequence MATH is exact. In the general case this follows from the inclusion-exclusion principle. For the exactness of REF at points away from the origin, where MATH divisors intersect (MATH), it follows from the induction as the divisor locally defined by the equation MATH. It remains to note that the normal crossing substack is stratified by the image of MATH with codimension MATH in MATH. This concludes the proof of REF. Now consider the following diagram MATH which defines a compatible perfect obstruction theory over MATH. Therefore by REF MATH . Proposition follows from the combination of REF . |
math/0105014 | First notice that the virtual structure sheaves in the above equations can be simultaneously erased from both sides by the Fundamental class and Forgetting tails axioms. First the case MATH. Let MATH be the divisors on MATH such that the generic curves have two components: one contains the MATH-th and MATH-th marked points, and the other contains the rest. It is well-known that MATH (see for example, CITE), where for the notational convenience we have used MATH and MATH for MATH-th universal cotangent line bundles on MATH and MATH respectively. It is easy to see that REF MATH for MATH. By projection formula MATH . It remains to compute MATH, which is a vector bundle because MATH. To compute MATH, we use some local arguments. The fibre MATH of MATH is a rational curve with MATH marked points MATH. An element of MATH is a rational function with poles of order no more than MATH at MATH. Therefore MATH is filtered by the degrees of poles at MATH: MATH . It is almost by definition that the graded piece MATH are isomorphic to MATH. Summarizing, MATH where ``MATH-theory" above means the equation is valid only in MATH-theory as we have used the graded pieces. It is now a matter of elementary computation (see REF) to deduce the genus zero string equation from REF. In the case MATH the NAME bundles MATH will naturally occur. Notice that MATH is actually equal to MATH, and this implies that the factor MATH commutes with MATH due to the projection formula. By NAME - NAME - NAME formula, REF is a rational function of MATH and MATH's. Therefore MATH . Now for MATH, we want to show MATH . The above equality can be proved using the same arguments as in the genus zero case. By REF and projection formula MATH . Because MATH for all MATH, MATH. MATH is a vector bundle. Now by NAME duality MATH, where MATH is the dualizing sheaf. NAME, MATH is the holomorphic differential with poles of order at most MATH at MATH. Thus we have a filtration MATH of degrees of poles at each marked point MATH as above and the graded bundles MATH is isomorphic to MATH and MATH. Therefore MATH . Notice again that the last equality holds only in MATH-theory (using graded objects). This means MATH which is equivalent to REF. |
math/0105014 | The proof is similar to the above proof of string equation. In the proof we again use MATH for MATH on MATH. We will use Forgetting tail axiom, REF and projection formula MATH . It is easy to see that this is equivalent to REF. |
math/0105014 | The deformation properties follow immediately from the definition. Setting MATH would make MATH the ordinary tensor product. The (super-)commutativity also follows from the definition. The associativity follows from the Splitting REF . It is easy to see that the MATH-theoretic WDVV equation MATH is equivalent to the associativity of the quantum MATH-ring. The WDVV equation follows from Splitting Axiom: Let MATH in REF. Different ways of splitting amount to the same invariants. |
math/0105014 | (see CITE) The first two are formal consequences of WDVV equation. The NAME connection of the metric MATH is MATH where MATH is the inverse matrix of MATH. Thus REF. holds. REF. is an obvious consequence of REF. |
math/0105014 | The proof relies on two ingredients: the string equation and WDVV equation. However, the WDVV equation here takes a slightly generalized form. Put MATH on the distinguished four marked points. The proof of WDVV equation goes through and we get MATH or equivalently, MATH . Put MATH. Since MATH, REF implies that MATH . Now by string equation MATH . |
math/0105015 | The flexible law can be expressed as MATH for all MATH. In an IP loop, this is equivalent to MATH, that is, MATH. Also, an easy calculation gives MATH in an IP loop. Thus REF are equivalent. Using the IP and REF , MATH is equivalent to MATH. Since the flexible law is just MATH, REF implies REF . Conversely, if REF holds, then taking MATH gives the flexible law, so that REF implies REF . Finally, REF are equivalent by the IP. |
math/0105015 | Applying REF twice, MATH. |
math/0105015 | REF are equivalent in IP loops. To prove WREF, start with MATH, which is MATH and set MATH and MATH, so that MATH. We get MATH . But MATH (see REF ), so MATH which is WREF. |
math/0105015 | Since the loop is alternative, the NAME property can be written as MATH. This gives us: MATH so MATH. Now, if MATH and MATH, we have MATH and hence MATH. |
math/0105015 | Let MATH denote MATH, so that MATH. Applying MATH to MATH gives MATH, and cancelling gives MATH. Replacing MATH with MATH yields MATH. In particular MATH, and so MATH. Next apply MATH to MATH to get MATH, and thus MATH. Cancelling yields MATH. Replacing MATH with MATH gives MATH and so MATH, which implies MATH. Replacing MATH with MATH gives MATH. Thus the loop satisfies the IP. Setting MATH in MATH yields the right alternative law MATH, and the right and left alternative laws are equivalent in IP loops. |
math/0105015 | The second equation is equivalent to the first in IP loops. So start with MATH. Set MATH and MATH (so MATH) to get MATH . Set MATH and MATH so MATH to get MATH . Putting these together, we have MATH . |
math/0105015 | Now we have MATH. If we let MATH and MATH, we get MATH, which (in flexible loops) is the NAME REF . |
math/0105015 | By REF and the preceding remarks, it is sufficient to prove that the ARIF loop MATH is MATH - PA for all odd MATH. So, for MATH, MATH, assume that MATH is MATH - PA. Setting MATH in REF , we get MATH. Now MATH and MATH, and so, by the MATH - PA, we have MATH. Applying this to MATH gives MATH, so MATH, which is the MATH - PA. |
math/0105015 | In a power alternative loop MATH, the subloop MATH generated by a given MATH induces a coset decomposition of MATH, and so if MATH is finite, the order of MATH must divide the order of MATH. Thus in a power alternative loop of odd order, each element is a square. Now apply REF . |
math/0105015 | We focus on MATH , since MATH are equivalent by the IP. Let MATH be a ARIF loop. On MATH, define three relations by MATH . Now MATH, and so MATH . It is clear that each of MATH is an equivalence relation. By REF and the fact that MATH is symmetric: MATH . Also, replacing MATH by MATH we have MATH . Replacing MATH by MATH we have MATH . So far, everything we have said holds in any IP power alternative loop. Our goal is now to prove MATH whenever MATH and either MATH is even or MATH is even. In the equations MATH set MATH and MATH. Then, by power alternativity, MATH so that MATH and MATH. We get: MATH . The first of these equations implies MATH, while the second implies MATH, so REF yields MATH. Set MATH and MATH to get MATH. Iterating this: MATH for every MATH. But by REF , we also have MATH, and iterating this we get: MATH for every MATH. Now, in view of REF , the lemma is equivalent to: MATH . We prove by induction on MATH that MATH holds for all MATH. By REF , it is sufficient to consider MATH, and the MATH case holds by the IP. Now, fix MATH. If MATH is even, we need to prove MATH for all MATH. Setting MATH in REF we get MATH, so it is sufficient to prove MATH whenever MATH. But since this is the same as MATH, it follows by applying MATH inductively to MATH, since MATH is even. If MATH is odd, we need to prove MATH for all MATH. Setting MATH in REF we get MATH, so it is sufficient to prove MATH whenever MATH. Now MATH is odd, so MATH. If MATH, then MATH (equivalently, MATH) follows by applying MATH inductively to MATH. If MATH, then induction gives us instead MATH, and hence (by MATH ) MATH. But also MATH (by REF with MATH), so MATH (by REF ), and hence MATH. |
math/0105015 | We first verify MATH . Applying REF twice, MATH. Let MATH and MATH, so MATH. Let MATH. Then MATH so MATH. Also, MATH, so MATH so MATH. Apply MATH to MATH to get MATH whenever MATH is even. Then apply MATH to get MATH whenever MATH are even. Now, the same argument starting from MATH results in MATH whenever MATH are even. Applying these with MATH, we get MATH for all MATH whenever MATH are even. But this implies that MATH whenever MATH is odd. Now apply MATH to MATH to get MATH whenever MATH is even. Then MATH yields MATH whenever MATH are even. The same argument starting from MATH results in MATH whenever MATH are even. Iterating as before, we get MATH for all MATH whenever MATH are even, which implies MATH whenever MATH is even. |
math/0105015 | We apply REF . MATH holds by power alternativity. But also MATH by the IP, and likewise MATH. |
math/0105015 | Fix MATH, and let MATH. By commutativity, MATH, which implies both that MATH is a subloop and that MATH is associative. |
math/0105015 | This is clear for MATH and MATH, so it is sufficient to prove it for MATH, which we do by induction on MATH. Assume it holds for MATH. By REF - associativity, MATH. Let MATH and MATH. Then MATH and MATH, so MATH. Hence, MATH. |
math/0105015 | Under the substitution MATH, MATH, REF reduces to: MATH which is an instance of MATH - associativity. |
math/0105015 | For example, say MATH and MATH are even, so we are considering REF . Let MATH and let MATH. Then the three special cases give us MATH, MATH, and MATH. But then REF yields REF for all MATH. |
math/0105015 | Let MATH and MATH. By REF , MATH and hence MATH for every MATH. By REF , MATH whenever MATH is even or MATH is even. Applying this to MATH and using (MATH), we have MATH . But REF also implies that MATH, so by the IP we have MATH . If MATH equals either MATH or MATH, then this yields MATH by (MATH). So, let MATH if MATH is even and let MATH if MATH is even. If MATH are both odd, then MATH is even and there is no restriction on MATH, so MATH can be either MATH or MATH. |
math/0105015 | First, consider REF. When MATH, this reduces to: MATH . If we let MATH and MATH, then this becomes MATH which is an instance of MATH - associativity. A similar argument works in REF but not in REF , where MATH is odd and MATH is even. To illustrate REF , consider MATH and MATH, MATH, or MATH. If MATH, we must verify MATH . This is no problem, since it is equivalent to MATH which is an instance of MATH - associativity. Likewise, the case MATH is no problem. But, when MATH, we must verify MATH . This is equivalent to MATH which requires MATH - associativity. However, this equation requires only MATH - associativity in the special case that MATH, and this case is sufficient by REF , applied with MATH, MATH, MATH, and MATH. The special case is condition REF , and conditions MATH are verified using MATH - associativity. The general situation is handled similarly. We must verify MATH where MATH is odd and MATH is even. By mirror symmetry, we may assume that MATH. Fix MATH. Let MATH be the assertion that this equation holds in the special case that MATH whenever MATH. So, we want to show MATH. Now, MATH holds for MATH large enough by MATH - associativity, and MATH holds by REF , so we are done. To be more specific, MATH asserts that MATH where MATH is MATH if MATH is odd and MATH if MATH is even. This is of form MATH, which is equivalent to MATH. Now, MATH has MATH blocks and MATH has MATH blocks, and MATH ends with MATH while MATH begins with MATH, so that the expression MATH has MATH blocks, so MATH follows from MATH - associativity whenever MATH, or MATH. Now, assume that MATH and assume that MATH holds. MATH is the special case of MATH with MATH. We conclude MATH by applying REF , with MATH, MATH, MATH, and MATH. Condition REF is MATH, and the conclusion, (MATH), is MATH. We must verify that conditions REF require only MATH - associativity. REF are easy. For (MATH), the expression MATH has no more than MATH blocks. Since MATH and MATH, we have MATH . |
math/0105015 | Say MATH is even; the argument for odd MATH is the same. Then we must prove MATH . We apply REF , with MATH, MATH, and MATH. Now MATH follows by MATH - associativity, and MATH follows by MATH - associativity plus REF . So, MATH follows by REF . |
math/0105015 | If MATH, then REF implies MATH. Thus, applying REF , the MATH (for MATH) are all the same. It follows by REF that MATH associates, so MATH is MATH - associative by REF . |
math/0105015 | Again, we must show that each MATH associates. Let MATH be the assertion that MATH associates for all MATH and MATH with MATH whenever MATH. So, our lemma is equivalent to MATH. Let MATH be the assertion that MATH associates for all MATH and MATH with MATH whenever MATH. So, our lemma is also equivalent to MATH. We shall in fact prove: CASE: MATH. CASE: MATH whenever MATH. CASE: MATH. CASE: MATH whenever MATH. Applying these items in order yields MATH and hence the lemma. First, note that, as in the proof of REF , MATH associates if MATH for some MATH with MATH. To prove MATH: Let MATH. We prove that MATH associates by showing that MATH; that is, MATH. Letting MATH so MATH, this reduces to MATH, which is true by MATH - associativity. To prove MATH when MATH and MATH is odd: MATH is now MATH, and we shall prove that MATH. Let MATH and MATH. We need to show that MATH. When MATH, this is true by MATH. But also MATH by REF . MATH now follows by REF . The proofs for MATH for MATH odd and for MATH are the same. Finally, we assume MATH and prove MATH. Let MATH. We prove that MATH associates by showing that MATH; that is, MATH. By REF applied twice, we have MATH. In particular, if MATH and MATH then MATH, so we get: MATH . We apply this equation to MATH: Now, MATH is a product of a word with MATH blocks, and by REF , this is equal to MATH . Thus, applying REF and power alternativity: MATH . Likewise, MATH is a product of a word with MATH blocks, of form MATH. This word associates by MATH, since it is the same as MATH. Thus, MATH the second MATH is obtained by applying REF , with MATH and MATH. We thus have MATH and hence MATH. |
math/0105015 | Consider the loop in REF . The nucleus is MATH, and all squares are in MATH, so that MATH is REF-element boolean group. This is not RIF because MATH, so that REF fails. |
math/0105015 | Consider the loop in REF . The nucleus is MATH, and all squares are in MATH, so that MATH is REF-element boolean group. This is not flexible because MATH. |
math/0105015 | Let MATH. Here, MATH is the identity element of the loop, so MATH and MATH by definition. Products MATH for distinct elements MATH of MATH are computed in the usual way from a NAME triple system MATH on MATH; that is, MATH is a set of REF subsets of MATH, and MATH, where MATH is the (unique) element of MATH such that MATH. For MATH, we take one of the standard examples of a triple system (see, for example, REF): MATH contains blocks of the form MATH and MATH, where MATH. So, for example MATH (since MATH), MATH (using MATH), MATH (using MATH,), and MATH (using MATH). Thus, MATH. However, MATH, since MATH (using MATH) and MATH (using MATH). |
math/0105017 | By REF , MATH and MATH are simple since they are tensor products of simple crystals. Let MATH be any sequence such that MATH. Then MATH is an isomorphism of simple MATH-crystals. But there is a unique MATH-crystal isomorphism MATH by REF . |
math/0105017 | We will prove the case MATH and MATH as the case MATH and MATH is similar. We have MATH. It must be shown that MATH. Let us fix MATH. For any isomorphism MATH that reorders MATH by a composition of local isomorphisms, define MATH to be the position in the threefold tensor MATH where MATH acts (see REF). Let MATH, MATH, MATH. Write MATH and let MATH (respectively, MATH) have value MATH (left) and MATH (right) on the two-fold tensor product MATH (respectively, MATH). Then the rows of the following table give all the possibilities: MATH . The columns MATH, MATH, and MATH agree with the defining conditions for MATH in REF . |
math/0105017 | REF holds by REF is proved by induction on MATH, by applying REF for MATH where MATH, MATH, and MATH. |
math/0105017 | Again REF follows by REF . The proof of REF follows by induction on MATH. The base of the induction REF is given by REF . For the induction step, REF are applied for MATH where MATH, MATH and MATH. |
math/0105017 | It must be shown that the tensor product construction for graded simple crystals is associative. This is true if the grading is ignored. It suffices to show that MATH . By REF we have MATH and MATH . In the second computation, REF is used. Also used is MATH, which holds because MATH acts on the rightmost tensor factor, which is not changed by MATH. One sees that REF are equal by REF. |
math/0105017 | The proof proceeds by induction on MATH. For MATH REF holds by definition. Suppose it holds for MATH tensor factors. Let MATH and MATH. Applying REF for MATH, induction, and REF , we have MATH which is evidently the right hand side of REF. |
math/0105017 | The proof immediately reduces to the case MATH. Let us prove the first assertion, as the second follows from it. Write MATH, MATH, MATH and MATH. By REF it follows that MATH where MATH. Therefore MATH as graded simple crystals. NAME both on the left by MATH and on the right by MATH, the result is a pair of isomorphic graded simple crystals MATH via the map MATH. In particular MATH. This argument implicitly uses REF . |
math/0105017 | It suffices to show that if MATH is one of the relations REF then MATH is also. The result is entirely straightforward unless MATH and MATH changes one of MATH. In this case (ignoring identical left and right factors) one has MATH. Applying MATH to both sides one obtains MATH. |
math/0105017 | It is easy to see that the map MATH given by MATH (where MATH is defined in REF) is a well-defined bijection that satisfies REF. For uniqueness, by definition the map MATH takes highest weight vectors to lowest weight vectors. But MATH has a unique highest weight vector and MATH has a unique lowest weight vector. |
math/0105017 | In the statement, MATH is computed with respect to a factorization into MATH column words, the last several of which may be empty. It suffices to show that if MATH are column words such that MATH with MATH one letter longer than MATH, then MATH. This reduction, together with the proof of this special case, can both be seen by considering a jeu de taquin on skew tableaux having at most two columns. |
math/0105017 | Observe that as MATH-crystals one has MATH using MATH columns. By the uniqueness in REF it follows that the crystals are also dual as MATH-crystals by the same map. |
math/0105017 | Consider the orientation-reversing automorphism of the NAME diagram of type MATH that sends MATH to MATH for all MATH with subscripts taken modulo MATH. The existence of this automorphism implies the existence of a map MATH satisfying REF. Since the NAME diagram automorphism is an involution, so is the induced map MATH. |
math/0105017 | REF is straightforward. For REF it suffices to show that if MATH is a relation of the form REF then so is MATH, but this is also straightforward. REF follows from REF , the connectedness of MATH, and the fact that MATH is a morphism of crystal graphs. |
math/0105017 | MATH and MATH obviously commute on sequences of column words. REF implies that MATH for all MATH. REF follows from REF . |
math/0105017 | Let MATH for MATH and MATH. Then MATH by the definition of MATH and REF . This suffices by the definition of MATH. |
math/0105017 | Let MATH be as before for MATH. MATH by the definition of MATH and REF . This suffices by the definition of MATH. |
math/0105017 | Let MATH, MATH be any reordering of the tensor factors of MATH, and MATH any composition of local isomorphisms. By REF , MATH . Using this one may reduce to the case MATH. The crystal MATH is connected (see REF). It then suffices to check REF for the single element MATH where MATH is the MATH-highest weight vector of MATH, by REF . Again by REF MATH is a MATH-highest weight vector of the same weight (namely MATH) as MATH where MATH is the unique MATH-highest weight vector in MATH. But there is only one such MATH-highest weight vector in MATH of this weight, so the two vectors must agree. It follows that both sides of REF evaluate to zero. |
math/0105017 | Let MATH and MATH be the tableaux on the left and right hand sides. Let MATH be obtained from MATH by replacing MATH by MATH and let MATH and MATH be the corresponding tableaux for MATH. By induction on the number of nonempty factors MATH. By definition the first MATH column words of MATH and MATH agree. To conclude MATH it suffices to show: CASE: MATH for MATH. CASE: MATH and MATH have the same shape. REF holds by the definition of MATH. For REF one has MATH by the fact that the MATH and MATH tableaux of a given word have the same shape, REF , the fact that MATH is shape-preserving, and the rule for how MATH and MATH change shapes. |
math/0105017 | Let MATH. Then MATH is a tableau of shape MATH by REF . It has the correct content to be MATH-LR by definition since the corresponding rectangles in MATH and MATH have the same heights. Since MATH is MATH-balanced, MATH is also, by REF. Thus the desired map is well-defined. Since MATH is an involution the map is bijective. |
math/0105017 | Let MATH. By REF there is a MATH such that MATH. Then MATH by REF . |
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