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math/9910003 | We see that MATH, which implies the first statement due to the expression MATH. The second statement is immediate from the first and REF. Since MATH, we have MATH. Then the last statement follows from the tables in CITE. |
math/9910003 | We have MATH and MATH for MATH, since MATH is the unique vertex connected to MATH. Then MATH and MATH commute, which implies MATH and thus REF. The form REF follows from the fact that MATH implies MATH and the exchange condition. |
math/9910003 | The former statement can be shown in the same way as REF , and the latter, directly. |
math/9910003 | The explicit form of MATH on MATH can be computed in a similar way to REF . Since MATH, we obtain the form REF. |
math/9910003 | Because MATH is MATH-invariant, it is sufficient to calculate the coefficients of the translations of antidominant weights. A translation MATH, MATH in the expansion of MATH appears as MATH where MATH and MATH . Then MATH, which implies MATH if MATH. Hence REF . |
math/9910003 | Consider MATH with MATH. Let MATH be the set of all the maximal antidominant weights in the expansion of MATH on MATH. Let MATH. Then we have MATH . Fix MATH. There exists MATH such that MATH and MATH. Suppose MATH for all MATH. Then MATH. Since MATH can be made arbitrary for suitable MATH, it follows that MATH and hence the result. |
math/9910003 | We note that MATH where MATH, MATH and MATH CITE. By using this relation and REF , we can check the behavior under the action of the NAME group (see the Appendix) and the holomorphy on the domain MATH. Then we see MATH. Since MATH fixes MATH, we have the proof. |
math/9910003 | First observe that if MATH, then MATH and MATH, where MATH for MATH and MATH. MATH . Then we have MATH . |
math/9910003 | We set MATH obtained in REF and set MATH . Because MATH is a reduced expression, we have for MATH . Let MATH. If MATH, then MATH and if MATH, then MATH. By using the identity MATH we arrive at REF. |
math/9910003 | We see MATH. Since MATH acts on MATH irreducibly, the statement follows from NAME 's lemma. |
math/9910003 | Let MATH be a linear map defined in the right hand side of REF. Owing to REF , we can evaluate the factor MATH at any element of MATH. Recall that every root system has a quasi-minuscule weight MATH, whose properties we have already investigated. CASE: MATH is not of type MATH where we have used REF . By applying MATH in the last equality, we obtain MATH . CASE: MATH is of type MATH . In a similar manner, we have MATH and consequently MATH where MATH. In any case, MATH and we have MATH, as required. |
math/9910008 | Suppose MATH is a MATH representation such that MATH has global coordinates MATH. Since MATH is a group, the representation MATH, with MATH is also a MATH representation. Moreover MATH has global coordinates MATH. The other permutations of coordinates are handled similarly. Suppose MATH is a MATH representation. Since MATH, the representation MATH, with MATH is also a MATH representation and the global coordinates of MATH is MATH. The other cases are handled similarly. |
math/9910008 | If MATH is a MATH representation, then MATH since MATH is abelian. Hence, MATH . If MATH, then MATH . This implies that the image of MATH is abelian, hence, is contained in MATH. |
math/9910008 | If MATH is a MATH representation and not a MATH representation, then at least two of the following MATH are in MATH. Since MATH implies MATH, at least two of the three global coordinates of MATH must be zero. Suppose two of the three global coordinates of MATH are zero, say MATH. One easily finds a MATH representation MATH such that the global coordinates of MATH are MATH: MATH where MATH. If MATH, then MATH and MATH is not a MATH representation. Since global coordinates are unique, MATH . Thus, MATH is a MATH representation but not a MATH representation. The proofs for the other cases are similar. |
math/9910008 | Suppose that MATH is an irrational multiple of MATH. Let MATH and MATH which does not correspond to a MATH representation class. Since MATH is an irrational multiple of MATH, MATH acts on the (transformed) subset of MATH, by an irrational rotation. By the compactness of MATH, there exists a MATH-value, MATH and MATH such that MATH is in the MATH-neighborhood of MATH and MATH . We first consider the special case where there exists an integer MATH such that the MATH-coordinate of MATH is strictly between MATH and MATH. Since MATH is an irrational multiple of MATH there are infinitely many integers MATH such that the MATH-coordinate of MATH is strictly between MATH and MATH. Choose MATH such that the MATH-coordinate of MATH is not in MATH. By the triangle inequality, there is some point on the MATH-orbit of MATH that is at most MATH from MATH . We now prove the proposition in general. Since MATH satisfies REF and is not a MATH representation class, MATH is a circle. Hence, we must have that MATH for some MATH. Therefore, there exists MATH such that MATH intersects MATH . Choose MATH. By the filtration MATH the set MATH contains all MATH-values that have the following properties: CASE: MATH intersects MATH CASE: There is a point in MATH whose MATH-orbit has at most MATH points with distinct MATH-coordinates between MATH and MATH. Note that the MATH-coordinate of MATH is the MATH-coordinate of MATH . Since MATH is an irrational multiple of MATH there is an infinite sequence of numbers MATH such that MATH, where MATH is the the MATH-coordinate of MATH. This forces MATH to intersect MATH. Of these, choose MATH such that MATH is not in MATH . Thus, the MATH-orbit of MATH has at least MATH points with distinct MATH-coordinates between MATH and MATH. Now at most MATH values of MATH yield MATH-orbits that are not MATH-dense. Thus, there exists a point MATH on the MATH-orbit of MATH such that MATH is between MATH and MATH, moreover, the MATH-orbit of MATH is MATH-dense. Since the MATH-neighborhood of MATH covers MATH, some point in the MATH-orbit of MATH comes within MATH of MATH . Finally, by REF , the set of MATH representation classes in MATH consists of six discrete points. This implies there is no loss of generality in assuming that MATH does not correspond to a MATH representation class. A symmetric argument holds if MATH is an irrational multiple of MATH . |
math/9910008 | Let MATH have infinite MATH-orbit and MATH which does not correspond to a MATH representation class. There are two cases. One possibility is that the MATH-orbit MATH has an infinite number of points on some circle MATH (respectively, MATH). Hence, there is an infinite number of points on MATH that have distinct MATH-coordinates. However, a priori, these points may not be dense in MATH . One uses the infinite number of points on MATH with distinct MATH-coordinates as in the proof of REF . The other possible case is that no circle MATH (or MATH) has an infinite number of points on MATH . As in the previous case, there are an infinite number of points on the MATH-orbit of MATH having distinct MATH-coordinates. The proof again follows similarly to the proof of REF . |
math/9910008 | Suppose MATH . Then MATH which implies that MATH or MATH a contradiction to the assumption that MATH are all non-zero. Suppose MATH. Recall that MATH satisfies REF . Thus, MATH . A similar argument applies in the case MATH . |
math/9910008 | By the assumption MATH, the only way that REF can have angles equal to an integer multiple of MATH is if MATH or MATH that is, MATH or MATH. Note that both cannot happen simultaneously. Suppose MATH. Then REF becomes MATH . REF and the assumption MATH lead to the following cases: CASE: Two terms in REF correspond to the first equation of REF , with one term equal to zero (angle MATH). Since MATH we must have that MATH so MATH . Now either MATH or MATH. If MATH then MATH hence MATH which yields the triples MATH and MATH . If MATH then MATH or MATH . The resulting triples MATH are MATH and MATH . Note that all of the above triples belong to a MATH-equivalence class appearing in REF . CASE: Two terms in REF correspond to the third equation of REF while the remaining term corresponds to the first equation in REF . The resulting triples are: MATH . Note that all above triples belong to a MATH-equivalence class appearing in REF . Suppose MATH. Then REF becomes MATH . REF and the assumption MATH lead to the following cases. CASE: Two terms in REF correspond to the first equation of REF , with one term equal to zero (angle MATH). The various possibilities lead to the triples: MATH . CASE: Two terms in REF correspond to the third equation of REF while the other corresponds to the first equation. The various possibilities lead to the triples: MATH . Again, the MATH-equivalence classes of these triples appear in REF . A similar argument holds if some angle in REF is an integer multiple of MATH, that is, MATH . |
math/9910013 | Assume first that MATH is non negative, and substitute MATH into REF; we obtain MATH so that MATH . Assume now that MATH is strictly negative. On one hand, REF implies the relation MATH the assumption on the sign of MATH is equivalent to MATH and therefore MATH . On the other hand, we subtract from the relation MATH and we infer that MATH . When we summarize REF, we find REF. |
math/9910013 | Let MATH be such that MATH is NAME continuous on MATH . Define MATH by MATH and let MATH be the NAME constant defined by MATH . Finally, let MATH be the NAME constant of MATH defined by MATH . There exists a function MATH which is bounded in a neighborhood of MATH such that for small positive values of MATH: MATH indeed, if MATH vanishes, or if MATH belongs to MATH, or if MATH belongs to MATH and the scalar product MATH is strictly positive, MATH vanishes; if MATH belongs to MATH and MATH vanishes, while MATH does not vanish, REF is a consequence of the smoothness of MATH in a neighborhood of MATH : for the values of MATH for which MATH belongs to MATH, MATH vanishes; for the values of MATH for which MATH does not belong to MATH, a NAME expansion shows that MATH hence REF. With the change of variable MATH, REF is equivalent to MATH where the function MATH is defined by MATH . Let us check that MATH is a strict contraction on MATH: if MATH, then MATH therefore, for MATH small enough, MATH is at most equal to MATH, and we can use the definitions of MATH and MATH: MATH . We estimate MATH as follows: by the triangle inequality, and the NAME condition on MATH, MATH . We apply REF, and we find MATH . Therefore, for MATH small enough, MATH maps MATH to itself. NAME, the NAME constant of MATH on this ball is at most equal to MATH. This proves that MATH has a fixed point in MATH for small enough values of MATH and completes the proof of the lemma. |
math/9910013 | Subtract REF for MATH from REF for MATH; with the change of variable MATH, we have to show the existence of a solution of MATH . If we denote by MATH the right hand side of the above equation, we have to choose a parameter MATH such that MATH will be a strict contraction of the ball MATH into itself. Let MATH be such that MATH is NAME continuous on MATH; denote by MATH the NAME constant of MATH over this ball, and define MATH let MATH be given by MATH . Denote finally by MATH the NAME constant of MATH defined as follows: MATH . Then, we have the estimate for MATH: MATH . It is straightforward that MATH . Therefore, if MATH satisfies the estimate MATH we may use the NAME continuity of MATH on the ball of radius MATH about MATH, and we find that if MATH belongs to MATH, MATH . We observe that MATH is at least equal to MATH, since MATH has eigenvectors relative to the eigenvalue MATH; therefore MATH thus, if MATH is so small that MATH maps MATH into itself; moreover, the NAME constant of MATH over this ball is at most equal to MATH; if MATH is a strict contraction from MATH to itself, which proves the lemma. |
math/9910013 | The theorem decomposes into an easy and a difficult part. The easy part is when MATH belongs to the interior of MATH. CASE: We choose MATH as in the proof of REF : the ball of center MATH and radius MATH is included in MATH. The number MATH defined by REF is equal to MATH, and the numbers MATH and MATH are given respectively by REF. We choose MATH and MATH. Assume that MATH satisfies the following inequalities: MATH . Then, if we write MATH we shall prove by induction that for small enough MATH, there exists a unique solution of REF, for MATH, which satisfies the estimate MATH . We claim that for MATH small enough, we can find a solution of MATH which satisfies the estimates MATH . In this construction, we seek a solution without considering the constraints, and we prove eventually that they are satisfied. It is clear that REF holds for MATH and that REF holds for MATH. Assume that it holds up to some exponent MATH. Thanks to REF, we have the estimates MATH . Therefore, we may apply REF with MATH: defining MATH, we can find MATH such that for MATH, there exists a unique MATH such that MATH . In particular, if MATH is defined by REF, with MATH replaced by MATH, we infer from REF that MATH therefore, with the help of the induction assumption, we have the estimate: MATH . If MATH satisfies the inequality MATH we can see that in fact MATH and therefore, instead of REF, we have MATH . Therefore, we have also MATH . Thus, REF hold. Let us prove that the vector MATH defined by REF belongs to MATH: since MATH we have the estimate MATH thus, if MATH and MATH satisfies MATH belongs to MATH, and the sequence MATH satisfies REF. This concludes the proof of the estimates in the first case. In particular, we can choose MATH. CASE: We define on MATH a norm denoted by MATH as follows: MATH . Pick MATH such that MATH is a diffeomorphism from an open neighborhood MATH to its image and such that MATH is included in an euclidean ball MATH such that MATH is NAME continuous on MATH; denote by MATH the NAME constant of MATH on MATH. Define MATH by MATH and MATH . A continuity argument shows taht the compact set MATH contains the ball of radius MATH about MATH. We will give now a description of the scheme REF in the new coordinates MATH. Assume therefore that MATH . We know that REF is equivalent to MATH . We map REF by MATH, and we calculate the NAME expansion of either side of REF around MATH. The left hand side of REF can be rewritten as MATH and therefore MATH where MATH . But MATH so that another NAME expansion gives MATH with MATH . Thus MATH . A similar calculation gives MATH with MATH . If we substitute REF into REF, we find that MATH where MATH and have the estimate MATH . Consider now the right hand side of REF. By definition of MATH, we have the identity MATH and a NAME expansion gives MATH with MATH . We substitute REF into REF, and we obtain MATH where MATH so that MATH . We have to estimate MATH; by elementary inequalities, MATH . The coefficient of MATH in the above bracket is MATH and since for MATH, MATH, this coefficient is at most equal to MATH. The coefficient of MATH in the same bracket is at most equal to MATH which is also at most equal to MATH. Therefore MATH . Thanks to the properties of MATH, MATH . Define MATH . In these new coordinates, we have MATH and MATH and MATH are given by MATH . Therefore, we have the estimates: MATH . We define MATH and MATH by MATH . Let now MATH be a number which satisfies MATH . Let MATH and let MATH and MATH be respectively as in REF. If we assume beyond REF that MATH we have the estimate MATH by elementary inequalities, MATH and therefore, if we define MATH we have shown that under REF , the following inequality holds: MATH . Let MATH be a number which satisfies the following inequalities: MATH . Assume that initially MATH . We will prove by induction that if MATH, then for all MATH . For MATH, REF guarantee that REF holds. The induction hypotheses imply that MATH and therefore, MATH, MATH and MATH belong to MATH. We may apply REF which guarantees the existence of MATH such that MATH . The ball of radius MATH about MATH is included in MATH; thus, if MATH also belongs to MATH. Similarly, MATH and if MATH belongs to MATH. Finally, thanks to REF of MATH, we have the inequality: MATH in virtue of REF of MATH and REF of MATH, we have MATH . Once again, if MATH belongs to MATH. Thus REF holds and we may apply the argument that followed. By definition of MATH and MATH, we have the inequalities MATH hence we infer that MATH and thanks to the induction hypothesis MATH . The equation in MATH has two distinct real roots if MATH is strictly positive; but this is always true if MATH and MATH . The smallest of the two roots of REF is inferior to MATH, since the substitution MATH in REF gives a negative left hand side; the largest of these two roots is at least equal to MATH; but relation REF implies MATH thus, if MATH relation REF implies MATH; if we substitute this inequality in the right hand side of REF, we find that the second inequality in REF holds for MATH; the first inequality in REF for MATH holds immediately, and the induction is proved. Thus, we can take as an upper bound of MATH the number MATH; we can take also MATH and MATH. |
math/9910013 | Let us check that MATH and MATH satisfy condition MATH. REF on MATH imply that MATH and MATH . Choose MATH; MATH and MATH satisfy condition MATH for small enough values of MATH. Then, it is clear that REF applies. |
math/9910013 | Any element MATH of MATH is included in an open ball MATH such that REF holds. We cover MATH by a finite number of balls MATH with associated numbers MATH, MATH, MATH and MATH. If we let MATH then any MATH belongs to a ball MATH, and in particular, MATH is included in MATH. If we take MATH it is immediate that the theorem holds, thanks to REF . |
math/9910013 | Let MATH be the compact set MATH and let MATH be as in REF ; cover MATH with a finite number of balls MATH; observe that, thanks to NAME - NAME 's theorem, the set MATH of functions MATH is relatively compact in MATH. The set of limit points of MATH as MATH tends to MATH is also a compact set, which we shall denote by MATH. There exists a finite subset MATH of MATH such that MATH . For each MATH, it is possible to find a finite increasing sequence of times MATH such that MATH . Thus, for all MATH, MATH . Therefore, we can decrease MATH so that MATH and thanks to REF and to REF, we can decrease MATH such that MATH . We simplify the notations by letting MATH and we take MATH be as in REF, where MATH is set equal to MATH, MATH is set equal to MATH and MATH is set equal to MATH. Now, we have to consider two cases: CASE: We have the inequality MATH hence, thanks to REF, we have the inequality MATH and therefore MATH . CASE: We observe that thanks to REF, we have the estimate MATH where MATH . The estimates on the first MATH components of the velocity in the straightened coordinates are immediate: MATH . In order to estimate the last coordinate, we partition MATH as follows: MATH . We write MATH as an union of discrete intervals: MATH . If MATH is defined as in REF, we observe that for MATH, MATH so that MATH . If MATH belongs to MATH, we observe that MATH and therefore, by the triangle inequality, MATH and using REF again, MATH . We observe that we have the elements of a telescoping sum: we sum REF for MATH varying from MATH to MATH, and we obtain MATH . Now, we sum REF from MATH to MATH, which yields MATH . But the terms MATH can be estimated, since they correspond to a summation over MATH: MATH . Therefore, we have proved that MATH . Summarizing this relation with REF, we can see that MATH . Relations REF do not depend on MATH; since we have only a finite number of these estimates, the theorem is proved. |
math/9910013 | Thanks to REF , we know that MATH is uniformly NAME continuous over MATH. Therefore, we may extract a subsequence, still denoted by MATH, such that REF hold. Thus MATH belongs to MATH, which means that MATH is a NAME continuous function CITE. For all MATH belonging to MATH, we have: MATH hence MATH. By definition of the scheme, we have REF , and thus MATH belongs to MATH. It follows that, for all MATH, the euclidean distance between MATH and MATH can be estimated as follows: MATH . Thanks to REF , we can see that for all MATH the euclidean distance between MATH and MATH is estimated by MATH. This allows us to pass to the limit when MATH tends to MATH and to conclude. |
math/9910013 | The measure MATH is a sum of NAME measures on MATH, more precisely, we have: MATH and the total variation of MATH on MATH is estimated by MATH . REF implies that MATH is a bounded family in MATH, the space of functions of bounded variation over MATH, with values in MATH. Using NAME 's theorem, we can extract another subsequence MATH which converges, except perhaps on a countable set of points, to a function of bounded variation. Hence MATH . Moreover, MATH NAME 's theorem implies that MATH converges to MATH in MATH. We extend MATH and MATH to MATH by MATH outside of MATH and still denote the respective extensions by MATH and MATH. The set MATH is a compact subset of MATH. The classical characterization of compact subsets of MATH CITE implies that MATH . Letting MATH, we can see that MATH converges to MATH in MATH. Let us define an approximate velocity MATH on MATH by MATH . The sequence MATH converges to MATH in MATH. Moreover, for all MATH and for all MATH, we have the identity MATH . We have immediately the following estimates for all MATH and all MATH: MATH . Let MATH be a continuous function over MATH with compact support included in MATH. For all small enough MATH, the support of MATH is included in MATH. The duality product MATH has the expression MATH . We wish to compare REF to MATH . We compare the right hand side of REF which is basically a numerical quadrature by the formula of rectangles to an appropriate integral. Let us rewrite the individual terms of the right hand side of REF as MATH . Consider the second term on the right hand side of REF: we have already proved (see relations REF) that there exists a constant MATH independent of MATH and MATH such that MATH . Denoting by MATH the modulus of continuity of MATH we can see that MATH . We consider now the first term on the right hand side of REF, which we would like to compare to REF . Thanks to the consistance REF have the following inequalities, for all MATH, and all MATH: MATH . For all MATH, let us define MATH . Denote by MATH the set MATH . Let MATH be the NAME constant of MATH restricted to MATH and let MATH be the modulus of continuity of MATH on MATH . With these notations, we can see that MATH . Since MATH is of class MATH in MATH, MATH converges strongly in MATH and MATH converges strongly to MATH in MATH and almost everywhere on MATH, the sequence MATH also converges strongly in MATH and almost everywhere on MATH to MATH. We see that MATH tends to MATH strongly in MATH and almost everywhere on MATH. We summarize relations REF together with the above convergence result, and we find that MATH which concludes the proof. |
math/9910013 | Define MATH is a sum of NAME measures on MATH; more precisely MATH . With all the previous results, we know that MATH converges to MATH weakly in MATH. Let us prove REF . Assume that MATH is a point of MATH such that MATH belongs to the interior of MATH. Then, by continuity of MATH, there exist MATH and MATH such that MATH . Since the sequence MATH converges uniformly to MATH as MATH tends to MATH, we can decrease MATH so that MATH . Relation REF implies the identity MATH . Relations REF imply that MATH . Possibly decreasing MATH, we have thus MATH . This proves that the support of MATH does not intersect the open set MATH, and therefore, relation REF holds. Assume now that MATH belongs to MATH, and let MATH be a ball having the properties of REF ; assume that the image of MATH by MATH and MATH is included in this ball for all small enough MATH. We rewrite REF as follows: for all continuous function MATH with compact support included in MATH and taking its values in MATH the following implication holds: MATH . In particular, if MATH vanishes for all MATH, then MATH also vanishes. The reader will check the equivalence of REF with REF. We infer from relation REF that MATH the above relation together with REF imply that there exists a constant MATH such that MATH . Since REF is local, it is enough to check it in the neighborhood of any MATH. Let MATH and MATH . We observe that if MATH belongs to MATH, then MATH . Therefore, we have the identity: MATH . We recall relation REF. Relation REF implies that MATH and therefore MATH . On the other hand, the definition of MATH is such that the MATH-th column of MATH is equal to MATH; therefore MATH . We infer from the above estimates that MATH and thus, there exists MATH such that for all MATH: MATH . We can see now that MATH which implies by a strightforward passage to the limit that MATH is non negative. This concludes the proof of the lemma. |
math/9910013 | Thanks to REF , we know that MATH and that there exists a nonnegative measure MATH such that MATH . We take the measure of the set MATH by the two sides of REF, and we find that MATH which implies immediately that MATH is parallel to MATH and proves the lemma. |
math/9910013 | We observe that MATH and that MATH . Moreover, for all MATH . Therefore, a straightforward integration yields MATH which implies MATH . We can write the analogous estimate on the interval MATH, which concludes the proof. |
math/9910013 | The idea of the proof is to find two succesive times MATH for which we can write down an estimate on the discrete velocities, and then to use REF to perform a discrete integration and to obtain a contradiction. We must deal with the fact that MATH does not converge uniformly to MATH. Without loss of generality, we may assume that MATH is continuous on the right and that for all MATH, MATH is also continuous from the right. According to NAME 's theorem, there exists a countable set MATH such that MATH . Assume that MATH vanishes; therefore, MATH is strictly positive. Choose MATH, and let MATH and MATH be such that MATH . An integration of REF on appropriate intervals yields MATH . Choose MATH and MATH; since MATH is a nonnegative measure, we have the following inequality for all MATH and all MATH: MATH . We integrate MATH on the interval MATH; since the measures MATH and MATH do not charge MATH and MATH, we find that MATH and therefore MATH . Choose now MATH; then, for MATH small enough, MATH and MATH belong to the interval MATH, and therefore, MATH where MATH tends to MATH as MATH tends to MATH. On the other hand, MATH tends to MATH and therefore, thanks to relation REF, there exists a family MATH such that MATH which is equivalent to MATH we infer from REF that MATH . Thus, for all MATH we infer from REF that MATH . Therefore, in the limit, for all MATH and for all MATH . On the other hand, relation REF implies that for all MATH, MATH . Under REF , relation REF contradicts relation REF. |
math/9910013 | Since MATH is strictly negative, there exists a real number MATH such that MATH is strictly positive on MATH. For all MATH, there exists MATH and MATH such that MATH . We prove now that there exists a maximal integer MATH such that MATH and denoting MATH the time MATH satisfies MATH . Let us first observe that for all small enough MATH and all MATH belonging to MATH we have MATH . Indeed, MATH and if MATH, we can see that REF holds. Therefore MATH exists and MATH . On the other hand, if there existed MATH such that for all MATH we had REF, then MATH would vanish on MATH, which contradicts REF . Therefore, we have shown that MATH that is, REF. We integrate discretely REF , and we find that for MATH . In the limit we have, MATH . The comparison of REF shows that MATH . Our purpose now is to obtain very precise estimates on the behavior of MATH beyond MATH. Thanks to the maximality of of MATH, we have the relation MATH let us estimate MATH: we substitute the value of MATH given by REF into this expression, and we also use REF with MATH replaced by MATH; we find MATH . We apply relation REF for MATH and we find that MATH . Therefore, we have MATH . On the other hand, if MATH is lesser than or equal to MATH, MATH if MATH is positive, then the sign condition on MATH implies that MATH . Thus, we have shown that MATH . If MATH is strictly positive, then for all small enough MATH, MATH . Let us estimate now the expression MATH: we have MATH . If MATH is non-negative, then MATH . We must estimate MATH: MATH and therefore MATH is non negative for all small enough MATH; the repetition of the argument shows that there exists MATH such that for all small enough MATH and all MATH, the expression MATH is non negative, and thus we have the relations MATH . On the other hand, if MATH is negative, we must have MATH and therefore MATH . These relations and the assumption on the sign of MATH imply that MATH which is strictly positive for MATH small enough. But now, we can see that MATH which is strictly positive for small enough MATH, and therefore MATH is strictly positive for MATH small enough, since MATH the same argument as above shows now that there exists MATH such that for all MATH, MATH . If we let MATH in the first case and MATH in the second case, we have now for MATH and MATH . Passing to the limit in REF, we can see that MATH . If we assume now that MATH vanishes, relation REF implies MATH . We observe that REF implies that for all MATH which implies immediately that for MATH which proves by a straightforward passage to the limit that MATH . This completes the proof of the lemma. |
math/9910013 | By uniform convergence of MATH to MATH, it is clear that MATH is equal to MATH. There remains to show that the initial condition on the impulsion is satisfied. Assume first that MATH belongs to the interior of MATH; then there exist MATH and MATH such that for all MATH and for all MATH . Then for all MATH belonging to MATH, MATH belongs to MATH for MATH small enough; we have indeed MATH which is strictly inferior to MATH for MATH small enough. Thus the constraints are not saturated for MATH and the convergence is a classical result. In the second case, MATH belongs to MATH; we have taken admissible initial conditions, so that MATH . We use the construction and notations of REF: MATH, MATH, MATH, MATH, MATH and MATH have the same signification as there. NAME 's formula yields MATH and REF of MATH gives MATH . Write MATH . Then the normal and tangential components of the impulsion are given by MATH . We wish to prove MATH which is equivalent to MATH . We recall relation REF. Relation REF implies that MATH and together with REF, we obtain in the limit MATH i. CASE: MATH . Let us show now that MATH considering two cases: MATH and MATH. When MATH vanishes, we have MATH and MATH . Thus, MATH and relation REF implies MATH therefore, a passage to the limit gives immediately MATH . If, on the other hand, MATH is strictly positive, then MATH which is strictly positive if MATH is small enough. Let MATH be the maximal interval such that MATH . Then, for all MATH, MATH which implies by discrete integration that MATH as long as MATH belongs to MATH. Moreover, if we choose any MATH and if MATH is at most equal to MATH, we can see that MATH for all small enough values of MATH. In particular, for all MATH, MATH which proves that MATH is at least equal to MATH. Therefore, MATH vanishes on the interval MATH; in the limit, MATH vanishes on MATH and therefore MATH which completes the proof of the lemma. |
math/9910013 | The measure MATH appearing in REF can be decomposed in the sum of an atomic part MATH and a diffuse part MATH. At each point of the support of MATH we have MATH thanks to relation REF. On any interval MATH which does not intersect the support of MATH, we multiply relation REF by MATH on the left, and we find that MATH . Define MATH . It is convenient to recall that MATH . Relations REF imply that in the sense of measures MATH . Our purpose now is to transform REF into a differential inequality. Let MATH be the norm of the bilinear mapping MATH . With this definition, MATH . We write now MATH . Define MATH and let MATH be the NAME constant of MATH for MATH and MATH; more precisely MATH . By construction, MATH is continuous and it is an increasing function of MATH and MATH. Fix MATH. If MATH and if MATH on MATH, we have the inequality MATH . But we can estimate MATH: MATH . Therefore we have the estimate MATH and we conclude that MATH satisfies the differential inequality MATH . Set MATH . While MATH and MATH, MATH satisfies the following differential inequality MATH . Let MATH be any positive number; consider the integrodifferential equation MATH with the initial condition MATH . It has a unique maximal solution which blows up in finite time, as soon as MATH is strictly positive and MATH is strictly positive. Let MATH be the largest time for which MATH . As MATH is a decreasing function of MATH, there exists a unique MATH such that MATH . Choose now MATH . Then we can compare the solution MATH of REF and the solution MATH of REF, and we find immediately that MATH . This concludes the proof of the lemma. |
math/9910013 | The only statement which deserves a proof is the last one; if it is not true, there exists MATH, a sequence of time steps still denoted by MATH and a sequence of integers MATH such that MATH . Without loss of generality, we may assume that MATH tends to MATH. First, MATH cannot be equal to MATH: we have learnt in REF that there exists a constant MATH and a time MATH such that for all MATH and all MATH, MATH . In particular, this estimate implies that MATH but MATH is at most equal to MATH, which contradicts REF. In the same fashion, we cannot have MATH; if it were the case, we could find an interval MATH containing MATH and MATH such that for all MATH, MATH is included in a ball of radius MATH about MATH included in the interior of MATH. But, in this case, MATH converges uniformly to MATH in MATH and this contradicts again REF. Thus, we assume that MATH is strictly positive and that MATH belongs to MATH. Choose a coordinate system such that the origin is at MATH; let MATH be the diffeomorphism defined at REF. In this case, MATH is given by REF. Define MATH . Let us compare MATH to MATH; it is convenient to define MATH then MATH . We observe that MATH and that MATH . These observations enable us to estimate the difference: there exists a constant MATH such that MATH . We infer from REF that there exists a constant MATH such that MATH . We use now REF: we can see that for all MATH, MATH so that MATH . If MATH is such that MATH and if MATH we can see that for all small enough MATH and all MATH the following estimate holds: MATH . But the function MATH defined by MATH converges almost everywhere on MATH to MATH; so does MATH. Therefore, in the limit, relation REF leads to MATH which is a contradiction. |
math/9910013 | Let Let MATH be the discrete time interval for which the numerical scheme REF has a solution; we know from REF that MATH . Assume that MATH . It is possible to extract from the sequence MATH a subsequence, still denoted by MATH, such that on all subinterval MATH included in MATH, MATH converges uniformly to MATH. In particular, thanks to REF we will have MATH and for MATH small enough we will have MATH . Thanks to REF , we can find MATH such that for all MATH, for all MATH and MATH satisfying condition MATH, it is possible to define a solution of the scheme for MATH, where MATH is independent of MATH. In particular, if we let MATH, MATH and MATH, we can extend the scheme up to MATH satisfying MATH which contradicts REF. This proves the desired result. |
math/9910015 | Suppose that MATH is a NAME function. Let MATH be a family of size MATH. Find a set MATH such that MATH. It follows that MATH. Similarly, if MATH is a basis for MATH, then MATH is a basis for MATH. |
math/9910015 | CASE: If MATH is a morphism between MATH and MATH, then MATH is a morphism between MATH and MATH. CASE: Suppose that MATH is such that MATH. Then MATH is cofinal in MATH. In other words, MATH. |
math/9910015 | CASE: Clearly MATH. To show the other inequality notice that there is a NAME mapping from MATH onto MATH. CASE: Suppose that MATH and let MATH be a NAME map such that MATH. It follows that MATH. Since MATH is a NAME mapping it follows that MATH. |
math/9910015 | CASE: Note that the set MATH has measure at most MATH. CASE: For an open set MATH let MATH . Note that MATH is a canonical representation of MATH as a union of disjoint basic intervals. Find open sets MATH covering MATH such that MATH. Let MATH be the lexicographic enumeration of MATH. Define for MATH, MATH and let MATH . Let MATH be such that MATH for each MATH. |
math/9910015 | CASE: Note that the set MATH is open and dense for every MATH. CASE: Let MATH be an increasing sequence of closed nowhere dense sets covering MATH. For each MATH let MATH . It is clear that MATH. |
math/9910015 | For MATH and MATH define MATH and MATH . We leave it to the reader to verify that these mappings have the required properties. |
math/9910015 | See CITE for REF for REF . |
math/9910015 | This is a special case of REF; see CITE |
math/9910015 | CASE: Let MATH be a family of NAME sets such that CASE: MATH, CASE: For every MATH, MATH is an open set of measure MATH. Look at the proof of REF to see that for each MATH, MATH and the mapping MATH is NAME. REF is obvious. CASE: By induction on complexity we show that for every MATH and a NAME set MATH there exists a NAME set MATH such that for every MATH, MATH is open and MATH. The only nontrivial part is to show that if the theorem holds for sets in MATH, then it holds for any set MATH. To see this write MATH where MATH is a descending sequence of sets in MATH. For each MATH let MATH be the set obtained from induction hypothesis for MATH and MATH. Let MATH. Each set MATH is NAME. Now define MATH . |
math/9910015 | CASE: By induction on the complexity we show that for any NAME set MATH there are NAME sets MATH and MATH such that CASE: MATH is open for every MATH, CASE: MATH is closed nowhere dense for every MATH and MATH, CASE: MATH . As before the nontrivial part is to show the theorem for the class MATH given that it holds for MATH. Suppose that MATH and MATH is the set obtained by applying the inductive hypothesis to MATH. Let MATH be an enumeration of basic sets in MATH. Define for MATH, MATH . Note that sets MATH are NAME. Let MATH . The vertical sections of the set MATH are closed nowhere dense and MATH, which ends the proof. CASE: For MATH let MATH . |
math/9910015 | Let MATH . For MATH define MATH if for all but finitely many MATH, MATH. MATH. To see that MATH define MATH and MATH such that for MATH and MATH we have MATH . For MATH put MATH, where MATH. If MATH let MATH be such that MATH . Verification that these mappings have the required properties is straightforward. To show that MATH we will find NAME functions MATH and MATH such that for MATH and MATH, MATH . Let MATH be a family of clopen probabilistically independent sets such that MATH. For MATH define MATH such that MATH . First consider MATH defined by MATH for MATH. Note that MATH is a NAME set and MATH has measure zero for every MATH. Fix a NAME isomorphism MATH and let MATH be defined as MATH for MATH. Apply REF to find a NAME mapping MATH such that MATH and define MATH. To define MATH we proceed as follows. Find a NAME set MATH such that CASE: MATH is a compact set of measure MATH for all MATH. CASE: MATH. CASE: For any basic open set MATH and MATH, if MATH then MATH has positive measure. First use REF to find a set MATH satisfying the first two conditions. Let MATH be an enumeration of basic open sets in MATH. For each MATH let MATH. By REF , the sets MATH are NAME for each MATH. Define MATH. For MATH define MATH . Note that MATH . Thus MATH . It follows that MATH so MATH for each MATH. Moreover, the mapping MATH is NAME (by REF ). Fix a NAME mapping from MATH to MATH such that MATH such that MATH . Finally define MATH by the formula: MATH . Suppose that for some MATH, MATH. It follows that, MATH . By the NAME category theorem, there is a basic open set MATH and MATH such that MATH but MATH . Therefore MATH which finishes the proof. MATH; there are NAME mappings MATH and MATH such that for any MATH and MATH, MATH . We will need the following lemma: There exists a good family MATH such that MATH . Fix MATH. Let MATH be an enumeration of all clopen sets. For MATH define MATH . Consider family MATH . We have to check that MATH satisfies conditions of REF . MATH . Let MATH be a dense open subset of MATH. Note that MATH for every MATH, by the density of MATH. Now define by induction a sequence MATH such that MATH and MATH for MATH. Clearly MATH . MATH . Suppose that MATH. For any MATH, MATH, where MATH for MATH. Order the sets MATH in such a way that MATH for MATH. It is easy to show by induction that MATH. Let MATH. For MATH define MATH. For MATH let MATH be such that MATH . Since MATH for all but finitely many MATH, by REF , MATH . Now suppose that MATH. This assumption means that there exists MATH such that MATH for MATH. It follows that MATH . REF follows immediately; compose the morphisms constructed in REF . |
math/9910015 | Let MATH be a MATH null set whose complement is meager. Use REF to find NAME functions MATH such that MATH . If MATH then MATH. It follows that MATH. |
math/9910015 | Let MATH be the family of clopen sets MATH such that there exists MATH and MATH such that MATH . Note that the family MATH is good (given the appropriate choice of the sequence MATH). For MATH let MATH if and only if MATH. For a strictly increasing function MATH define MATH such that MATH . Note that the image of MATH under MATH is rather small, MATH is not even cofinal in MATH. To finish the proof it is enough to show that if MATH for infinitely many MATH, then MATH . Find a sequence MATH such that for all MATH, MATH . Construct a real MATH such that MATH. Thus MATH but MATH for all MATH, so MATH. |
math/9910015 | Identify MATH with MATH and define MATH such that MATH and MATH. |
math/9910015 | Let MATH be any NAME function such that for MATH, MATH (see REF ) and let MATH for MATH. Verification that both functions have the required properties is straightforward. |
math/9910015 | Since every set MATH, MATH belongs to MATH, we get the second inequality. For the first inequality note that if MATH then MATH has size MATH. Thus any family that dominates MATH has to have a size at least MATH. |
math/9910015 | Clearly every subset of size MATH of MATH is unbounded. Moreover, for every MATH, MATH has size MATH. Thus MATH. |
math/9910015 | MATH. Suppose that MATH is a NAME mapping. Let MATH. Clearly MATH is a NAME set with all MATH meager and if MATH then MATH has required properties. MATH. We will need several lemmas. To avoid repetitions let us define: Suppose that MATH. MATH is nice if for every NAME function MATH there exists a function MATH such that MATH . Suppose that MATH is nice. Then for every NAME function MATH there exists MATH such that MATH . Suppose that a NAME mapping MATH is given. Let MATH denote the MATH-th element of MATH for MATH. For every MATH define a function MATH as follows: MATH . Since the mapping MATH is NAME and functions MATH can be coded as elements of MATH there is a function MATH such that MATH . Without loss of generality we can assume that MATH is a function from a MATH-element subset of MATH into MATH. Define MATH in the following way. Recursively choose MATH . Then let MATH be any function such that MATH for MATH. We show that the function MATH has the required properties. Suppose that MATH. Notice that the equality MATH implies that MATH . That finishes the proof since MATH for infinitely many MATH. Suppose that MATH is nice. Then for every NAME mapping MATH there exists an increasing sequence MATH such that MATH . Suppose that the lemma is not true and let MATH be a counterexample. Without loss of generality we can assume that MATH is increasing for all MATH. To get a contradiction we will define a NAME mapping MATH such that MATH is a dominating family. That will contradict the assumption that MATH is nice. Define MATH . Suppose that MATH is an increasing function. By the assumption there exist MATH and MATH such that MATH . In particular, MATH which finishes the proof. We now return to the proof that MATH implies MATH for REF. Let MATH be a NAME mapping. We want to show that MATH. Without loss of generality we can assume that MATH is the set built using the family from the proof of REF . For each MATH let MATH and MATH be such that MATH. By REF , there exists a sequence MATH such that CASE: MATH, for all MATH, CASE: MATH. For MATH let MATH. By REF , there exists a sequence MATH such that MATH . Without loss of generality we can assume that MATH, where MATH. Choose MATH such that MATH for all MATH. It follows that MATH for every MATH. |
math/9910015 | The inclusion MATH follows immediately from REF . Suppose that MATH. Let MATH be a NAME mapping. Since MATH there is a real MATH such that MATH. For MATH define for MATH, MATH . The mapping MATH is also NAME. Since MATH, it follows that there is an increasing function MATH such that MATH . Consider the set MATH . Clearly MATH is a dense MATH set. Moreover, for every MATH there is MATH such that MATH . It follows that MATH which finishes the proof. |
math/9910015 | We will establish the equivalence of REF . Suppose that MATH and MATH is a NAME mapping. Consider the morphism MATH witnessing that MATH. Let MATH be such that MATH. Then MATH is the object we are looking for. Suppose that MATH. Let MATH be a NAME mapping such that MATH for MATH. Consider the morphism MATH witnessing that MATH. It follows that there is no MATH such that MATH . Equivalence of REF follows from: MATH. To show that MATH define MATH as MATH . Similarly, define MATH is defined by: MATH . It is easy to see that these mappings have the required properties. To show that MATH identify MATH with MATH via functions MATH. For MATH let MATH . For MATH let MATH . The second part of REF follows readily from the first. |
math/9910015 | See CITE |
math/9910015 | Let MATH be an enumeration of clopen subsets of MATH. For MATH and MATH define MATH. The sets MATH are closed and MATH for every MATH. By the NAME Category theorem for each MATH there is a pair MATH such that MATH . Since MATH is a p-ideal, we can find MATH such that MATH . It follows that MATH. That finishes the proof of the first part. To prove the second part let MATH. Next, apply the above construction to MATH (which is closed) to get MATH, etc. |
math/9910015 | Consider MATH. Note that for MATH we have MATH . It follows that MATH is an analytic ideal. Moreover, MATH is a MATH-ideal; if MATH and MATH then MATH. To see this let MATH witness that MATH. Find MATH such that MATH for all MATH. Clearly, MATH for all MATH. Now the lemma follows immediately from the following: Let MATH be an analytic MATH-ideal of compact sets in a compact metrizable space MATH. Then MATH is actually MATH. See CITE, CITE or CITE. |
math/9910015 | Using REF represent MATH, where each MATH is closed. Apply REF to write for MATH, MATH, where MATH. It is clear that MATH generates MATH. |
math/9910015 | Implication MATH is obvious. MATH . We will use the following result. Suppose that MATH is an ideal containing all finite sets. The following conditions are equivalent: CASE: MATH has the NAME property, CASE: MATH is meager, CASE: there exists a partition MATH of MATH into disjoint intervals such that MATH . See CITE or CITE. Suppose that MATH. The ideal MATH is analytic and hence has the NAME property. By REF there exists a partition MATH of MATH into finite sets such that MATH . Choose MATH such that the set MATH . Let MATH be such that MATH. It follows that for every MATH, MATH which finishes the proof of REF. |
math/9910015 | Fix MATH and using the fact that MATH is a p-ideal find MATH such that MATH is cofinal in MATH. The set MATH . Let MATH be such that MATH. We have MATH . Choose MATH such that MATH is cofinal in MATH. The set MATH . Let MATH be such that MATH. It follows that MATH. |
math/9910015 | Use REF to find a descending sequence MATH generating MATH such that for each MATH, MATH . For MATH let MATH be an increasing sequence such that MATH . Identify MATH with MATH and define MATH and MATH such that MATH . Since MATH this will finish the proof. For MATH and MATH define MATH . Mapping MATH will be defined as follows. Suppose that MATH is given (with MATH). For MATH let MATH . Now define MATH . Note that MATH is a sum of at most MATH terms, each belonging to MATH. Thus, MATH for all MATH. The motivation for this definition is following: if MATH and MATH, then MATH . The requirements of the definition describe this situation and filter out ``background noise" coming with MATH. Finally define MATH . By the remarks above it is clear that if MATH and MATH then from the fact that MATH it follows that MATH. To finish the proof it remains to show that the range of MATH is contained in MATH. Let MATH be defined as above. For MATH, let MATH. Since MATH is finite for every MATH, by the lemma above, in order to show that MATH it would suffice to show that MATH. For each MATH, MATH . We prove this by induction on MATH. For each MATH, MATH so the lemma is true for MATH. Suppose it holds for MATH and all MATH. We have MATH which finishes the proof. Since sets MATH are closed, we conclude that MATH. In particular, by REF , MATH. |
math/9910015 | For MATH let MATH be defined as MATH. Let MATH. It is easy to see that MATH is a compact space. For MATH define MATH if there is a finite set MATH such that MATH or alternatively MATH . MATH. Define MATH and MATH as MATH and MATH. Suppose that MATH. That means that for any finite set MATH, MATH . It follows that MATH, hence MATH. Let MATH be the collection of compact subsets of MATH which are downward closed (with respect to MATH). Let MATH . As before, by REF , MATH is countably generated. Let MATH be any sequence generating MATH. We will show that MATH. We need functions MATH and MATH such that for MATH and MATH, MATH . Clearly, the dual morphism witnesses that MATH. Each set MATH is a set of branches of some tree. By taking the rightmost branch (towards the larger values) at every node of the tree, we produce a countable family MATH such that MATH . Since MATH is a p-ideal, for each MATH there is MATH such that MATH . Denote the first such MATH by MATH. Without loss of generality we can assume MATH for all MATH. We have the following two cases: CASE: There exists MATH such that CASE: MATH for every MATH. CASE: MATH for every MATH. In this case MATH is atomic. Note that REF implies that MATH . Thus by REF , MATH. REF , together with the fact that MATH is an ideal, implies that MATH for every MATH. CASE: Suppose that there is no MATH as in REF For MATH and MATH define MATH . For MATH let MATH . It is clear that these mappings have the required properties provided that they are correctly defined. Thus, the following lemma will complete the proof: For every MATH there exists a MATH such that MATH. Suppose not and let MATH be a strictly increasing function such that for every set MATH the set MATH is coinfinite. It follows that the family MATH generates a proper analytic ideal MATH. As before, MATH has the NAME property, hence by REF , there exists a partition MATH such that MATH . Let MATH for MATH and consider the function MATH . Clearly, MATH for all MATH. We will show that MATH for MATH, which will give the contradiction. Fix MATH. Suppose that MATH and MATH. Clearly, MATH and MATH. We have MATH . In particular, MATH . It follows that MATH. |
math/9910015 | Suppose that MATH. Let MATH be a NAME mapping. It is enough to find MATH such that MATH . Let MATH be a partition of MATH into infinitely many infinite pieces. For each MATH consider the mapping MATH and find MATH such that MATH . Then MATH is as required. |
math/9910015 | See CITE of CITE. |
math/9910015 | Let MATH be the collection of MATH such that CASE: MATH, CASE: MATH, CASE: MATH are pairwise different for MATH, CASE: MATH, CASE: MATH. From the assumption about the cardinal arithmetic in MATH it follows that MATH for MATH. In particular MATH. Moreover, since MATH we can find in MATH an enumeration MATH of MATH. Given MATH define MATH to be collection of all MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. MATH is filter-like for every MATH. Suppose that MATH. First we show that MATH is a function. Suppose that MATH. Consider the function MATH and note that exactly one of the following possibilities happens: CASE: there exists exactly one pair MATH such that MATH. In this case MATH agree on MATH with the value MATH, CASE: there exists exactly one MATH such that MATH so MATH . Now, put MATH and note that MATH has the required property. To finish the proof of REF note that MATH. Suppose that MATH. Let MATH and MATH be as in REF, and put MATH to be a countable set such that MATH and MATH are pairwise different. |
math/9910015 | We extend the measures using the NAME theorem and we only need to check that the requirements are consistent. Suppose that we have MATH-name MATH and MATH-name MATH such that MATH. A necessary and sufficient condition for both measures to have a common extension is that in such a case MATH. Let MATH and MATH be such that MATH . Let MATH . Set MATH belongs to MATH and MATH. Similarly MATH. In particular, MATH . |
math/9910015 | Without loss of generality we can assume that MATH and therefore we will drop the superscript MATH altogether. REF is clear. CASE: For a MATH-name MATH for a subset of MATH and MATH let MATH and MATH . Clearly, MATH . Note that if MATH then MATH for every MATH. It follows that MATH and MATH. Similarly, if MATH and MATH is a name for MATH then MATH and MATH. MATH. MATH is obvious. CASE: Suppose that MATH. Fix any real MATH and MATH. It follows that MATH is dense below MATH. Let MATH be a maximal antichain in MATH. We have MATH. It follows that MATH . Since MATH and MATH are pairwise disjoint, we get MATH . We conclude that MATH and since MATH was arbitrary, MATH. Now we show that MATH is a finitely additive measure. Suppose that MATH. Suppose that MATH. Let MATH and MATH be such that MATH. Then MATH and MATH - contradiction. Suppose that MATH and let MATH be a name for MATH. Let MATH, MATH be such that MATH and MATH. It follows that MATH and MATH. Thus MATH, so MATH . Suppose that MATH . There are reals MATH and MATH such that MATH, MATH and MATH. Use REF, to find MATH such that MATH and MATH. By REF, MATH. On the other hand, since MATH is additive, MATH - contradiction. CASE: Suppose that MATH is a MATH-name and for some MATH and MATH, MATH. That means that for every MATH, MATH . It follows that MATH . Since MATH is a measure, by looking at the complements we get, MATH, hence MATH . CASE: Suppose that MATH for MATH. Let MATH . If MATH is not dense in MATH, then the condition witnessing that has the required property. So suppose that MATH is dense and work towards a contradiction. Let MATH be a maximal antichain in MATH. Clearly MATH. Let MATH be such that MATH, which means that MATH . Similarly for MATH, MATH . Let MATH. We have MATH . This is a contradiction since MATH . |
math/9910015 | Induction on MATH. CASE: MATH. Let MATH be a generic filter and let MATH. Work in the model MATH. Since MATH there exists a NAME function MATH such that MATH . Since MATH is coded by a real, there exists a set MATH and a function MATH such that MATH . Function MATH and the set MATH are the objects we are looking for. CASE: MATH. Fix an increasing sequence MATH such that MATH and suppose that MATH is a MATH-name for a real number that is, a set of countably many antichains. Let MATH be a MATH-name for a real obtained by restriction conditions in these antichains to MATH. Note that MATH. Apply the induction hypothesis to MATH's to get NAME functions MATH and countable sets MATH. Let MATH and let MATH be defined as MATH . CASE: MATH. Since no reals are added at the step MATH there is nothing to prove. |
math/9910015 | Induction on MATH. |
math/9910015 | Suppose that MATH is a family of measure zero sets in MATH. Let MATH be a master set for MATH defined earlier. Without loss of generality we can assume that for some MATH, MATH, and let MATH be a MATH-name for MATH. As in REF, let MATH be the set such that MATH. Find MATH such that MATH. The random real added by MATH avoids all null sets coded in MATH, in particular, all MATH's. |
math/9910015 | Denote by MATH a MATH-name for MATH and let MATH be the least ordinal such that MATH is a MATH-name. We have the following two cases: CASE: MATH is a successor ordinal. Suppose first that MATH. Work in MATH and let MATH. For each MATH choose MATH and MATH such that MATH. Since MATH has a dense subset of size MATH, we can find MATH and MATH such that the set MATH is uncountable. Consider the set MATH in MATH. Observe that MATH is a closed set and if it was countable then all its elements would be in MATH. However, MATH and MATH. If MATH then the argument is identical except that we use MATH instead of MATH. In fact one can show that MATH . CASE: MATH is limit and MATH. In MATH we can find MATH and an uncountable set MATH such that MATH . Let MATH be a MATH-name for MATH. Suppose that MATH is a generic filter over MATH. For each MATH choose MATH and MATH such that MATH, where MATH is a MATH-name for the MATH-th element of MATH. There is an uncountable set MATH, and MATH such that MATH for MATH. Let MATH and let MATH be a MATH-name for MATH. As in the previous case, consider the set MATH in MATH. We see that MATH is uncountable because it contains an element which does not belong to MATH. |
math/9910015 | If MATH, then there is nothing to prove. Suppose that MATH and let MATH. Let MATH for MATH. Apply the induction hypothesis to get a condition MATH such that CASE: MATH, CASE: MATH, CASE: MATH CASE: MATH has at least MATH elements. Let MATH be a MATH-name for the set MATH . Let MATH be the list of all subsets of MATH of size at least MATH. Find a maximal antichain MATH below MATH such that MATH for MATH. We will need the following easy observation. Suppose that MATH is a family of subsets of MATH of measure MATH. Let MATH . Then MATH. Let MATH be the characteristic function of the set MATH for MATH. It follows that MATH. On the other hand, estimation of this integral yields, MATH and after simple computations we get MATH . It follows that we get the following estimates: MATH . Work in MATH and for each MATH apply REF to the family MATH and obtain a condition MATH such that MATH and MATH. Finally, define MATH to be a MATH-name such that for MATH, MATH. It is easy to see that MATH is as required. |
math/9910015 | Recall that for any tree MATH, MATH is an element of MATH such that for all MATH, MATH or MATH and for MATH, MATH. The Laver forcing MATH is the following forcing notion: MATH . For MATH, MATH if MATH. CASE: MATH. CASE: MATH. CASE: MATH does not add random reals. Moreover REF hold for the countable support iteration of Laver forcing as well. See CITE. Let MATH be a countable support iteration of length MATH the Laver forcing. It follows from REF that MATH in MATH, while both MATH and MATH are equal to MATH. |
math/9910015 | We will use forcing notion MATH defined below rather than MATH, it has a much simpler definition and has the required properties (the difficulties appear when unbounded reals are added). The infinitely equal forcing notion MATH is defined as follows: MATH if the following conditions are satisfied: CASE: MATH, CASE: MATH for all MATH, and CASE: MATH. For MATH we define MATH if MATH. Forcing MATH has the following properties: CASE: MATH. In fact, MATH where MATH is a generic real. CASE: MATH does not add random reals, CASE: MATH is MATH-bounding. See CITE. Let MATH be a countable support iteration such that for every MATH, CASE: MATH if MATH is even, and CASE: MATH if MATH is odd. Let MATH be a MATH-generic filter over MATH. It is clear that MATH. To see that MATH in the extension note that both forcing notions MATH and MATH are MATH-bounding and use REF. |
math/9910015 | The construction is canonical. Set the cardinal invariants corresponding to the families that MATH belongs to to MATH and the other ones to MATH. In our case MATH and MATH. Now consider the forcing notion that produces the model for the dual setup, that is, MATH and MATH. According to our table it is the iteration of NAME and random forcings, MATH. Let MATH be an increasing sequence of contable submodels of MATH such that CASE: MATH. CASE: for ever MATH, MATH is countable, CASE: MATH. For each MATH choose a pair MATH such that MATH is MATH over MATH. Note that such a pair will also be generic over MATH for MATH. Let MATH encode MATH as MATH . Let MATH. We will show that MATH has the required properties. To show that MATH fix a NAME function MATH and find MATH such that MATH is coded in MATH. Let MATH be any function which dominates MATH. For any MATH, MATH. Since MATH does not add dominating reals it follows that for every MATH there is a function MATH such that MATH. Since MATH is dominated by MATH we conclude that MATH for every MATH. To see that MATH let MATH. Observe that MATH is a continuous image of MATH. Moreover, if MATH is a meager set then MATH for MATH since MATH is a NAME real over MATH. The argument that MATH is identical. |
math/9910015 | Induction on MATH. Let MATH be a sequence of models having the required properties. Fix MATH and MATH. We will find MATH which is MATH-generic for MATH. If MATH then it is the usual proof that REF implies properness. If MATH then first find MATH which is MATH-generic for MATH and then use properness of MATH to get MATH which is MATH-generic. If MATH is limit then fix an increasing sequence MATH such that MATH. Use the induction hypothesis to find conditions MATH such that CASE: MATH, CASE: MATH is MATH-generic for MATH, CASE: MATH for each MATH. Let MATH be such that MATH for each MATH, it is the condition we are looking for. |
math/9910015 | In terms of cardinal invariants the statement of the theorem corresponds to the dual to the model for MATH and MATH, that is the one where MATH and MATH. The set we are looking for is defined using the forcing notion used to construct that model (compare REF ). Let MATH be an enumeration of MATH. Let MATH be a sequence of countable elementary submodels of MATH such that CASE: MATH, CASE: MATH, and MATH is countable, CASE: MATH for limit MATH. Note that from REF it follows that for every MATH, MATH ``MATH is countable." Let MATH be a sequence of reals such that CASE: MATH, CASE: MATH is MATH-generic over MATH for MATH, CASE: MATH is MATH-generic over MATH for MATH. For MATH define MATH . Let MATH. MATH. The set MATH is a NAME image of MATH. Given MATH find MATH such that MATH. Notice that MATH for MATH. In particular, no uncountable subset of MATH is in MATH. MATH. Consider the set MATH which is a NAME image of MATH. Let MATH. Let MATH . It is easy to see that MATH is a NAME set in MATH and MATH for every MATH. Suppose that MATH. Find MATH such that MATH and note that for MATH, MATH. It follows that no uncountable subset of MATH is in MATH. MATH . Let MATH be a NAME mapping. Find MATH such that MATH is coded in MATH. Let MATH be such that for every MATH, MATH. Since MATH is countable, such a MATH exists. Since both MATH and MATH are MATH-bounding (so is MATH) for every MATH, there exists MATH such that MATH. |
math/9910015 | Let MATH be a sequence of countable elementary submodels of MATH as in the previous proof. In this case we use the Laver forcing from REF . The only difference is that in order to ensure that the constructed set belongs to MATH we construct a set of witnesses for that. Let MATH be a sequence of reals such that CASE: MATH, CASE: MATH is MATH-generic over MATH for MATH, CASE: MATH is MATH-generic over MATH for all MATH. To meet REF we need the following result: Suppose that MATH is a countable model of MATH. Let MATH and let MATH be a random real over MATH. There exists MATH such that MATH is MATH-generic and MATH . See CITE, CITE or CITE. Let MATH. The difference between this and the previous construction is that we define the set of witnesses MATH that MATH. MATH. Let MATH be a NAME set with null sections. Find MATH such that MATH. Note that MATH since MATH is random over MATH for all MATH and MATH. MATH. Let MATH be a NAME mapping. Find MATH such that MATH is coded in MATH. Let MATH. Since MATH is countable, MATH is a null set. By REF for every MATH, MATH. MATH. This is obvious, by REF , for every MATH . |
math/9910015 | Fix MATH and let MATH be a maximal antichain of conditions deciding MATH. Use properness to find MATH such that each MATH is countable. By the assumption we can assume that elements of MATH are pairwise disjoint. Define MATH as MATH . Note that MATH is the function we are looking for. |
math/9910015 | For MATH and MATH we define MATH if MATH and first MATH elements of MATH and MATH are the same. For MATH and MATH let MATH if MATH and MATH. The forcing notions MATH, MATH (and the remaining ones as well) can be represented as the collections of perfect subsets of MATH (or MATH). This is not critical for the construction, but it makes it more natural. In case of MATH for MATH let MATH. Consider sets MATH of form MATH, where MATH satisfies the following conditions: CASE: MATH, CASE: for every MATH, MATH or MATH (so MATH), CASE: MATH. It is clear that every condition MATH corresponds to a set MATH as above and vice versa. Therefore from now on we identify MATH with these sets. Let MATH be the collection of subsets MATH such that CASE: MATH is NAME and NAME, CASE: MATH. The elements of MATH are MATH-names for the elements of MATH. Thus, the set MATH indeed corresponds to the iteration of MATH and MATH. For MATH and MATH let MATH mean that CASE: MATH, CASE: MATH . Note that MATH on MATH witnesses that it satisfies Axiom MATH. Let MATH be an enumeration of MATH, and MATH of MATH, and MATH an enumeration of MATH. We will build a MATH-tree MATH of elements of MATH. Let MATH denote the MATH-th level of MATH. The tree MATH satisfies the following inductive conditions: CASE: MATH . CASE: MATH, CASE: MATH, CASE: MATH, where MATH. CASE: MATH. We will describe how to build a set of immediate successors of an element MATH. Given MATH and MATH find MATH satisfying REF . By further shinking we can ensure that REF holds as well. REF is just the statement that the iteration of MATH and MATH is MATH-bounding. CASE: MATH is limit. Suppose that MATH for some MATH and that MATH is given. Fix an increasing sequence MATH such that MATH. Choose a sequence MATH such that CASE: MATH, CASE: for MATH, MATH, CASE: MATH. Use Axiom MATH to find MATH such that MATH. Level MATH will consist of elements selected in this way. Let MATH be a selector from elements of MATH. Note that MATH (by REF ), MATH (by REF ) and MATH (by REF ). |
math/9910015 | For every MATH and MATH define a node MATH in the following way: MATH and for MATH let MATH be the MATH-th element of MATH. For MATH and MATH define MATH if MATH . In particular, MATH is equivalent to MATH and MATH. It is easy to check that Laver forcing satisfies Axiom MATH. Suppose that CASE: MATH is an enumeration of MATH, CASE: MATH is an enumeration of MATH, CASE: MATH is an enumeration of MATH. We build a MATH-tree MATH satisfying the following inductive conditions: CASE: MATH CASE: MATH, (MATH adds a dominating real REF ) CASE: for every MATH, MATH has measure zero, (MATH does not add random reals REF ) CASE: MATH is uncountable MATH. (MATH preserves outer measure REF ), Next we want to chose a selector MATH from elements of MATH. REF will guarantee that MATH and REF that MATH. Unfortunately REF does not suffice to show that MATH. It is conceivable that MATH, because MATH is not a MATH-ideal. Therefore we need stronger property: CASE: For every NAME function MATH and a sequence MATH of conditions in MATH there exists an uncountable set MATH such that for each MATH-we can find sequence MATH such that MATH and MATH. REF is a translation of REF. Now we construct MATH along with MATH. On the step MATH we have MATH and MATH. Let MATH and pick MATH together with MATH as in REF . |
math/9910019 | The values at MATH follow from REF. For MATH, note that from the RHP REF, we have MATH. |
math/9910020 | We only need to give the formulae of the change of generators, which are MATH . |
math/9910022 | Let MATH be a smooth MATH-parameter family of metrics, and let MATH be a family of diffeomorphisms generated by vector fields MATH. Then we have MATH . Applying REF with MATH and MATH, we get MATH . Now define MATH . Applying REF with MATH and MATH, we obtain MATH . But evaluating REF at MATH shows that MATH . Hence MATH . So MATH is a solution of the NAME flow with MATH. Since solutions of the NAME flow are unique, it follows that MATH for as long as both solutions exist. |
math/9910022 | This is a straightforward computation using the identity MATH with REF , and REF . |
math/9910022 | This follows from MATH and all of REF - REF . |
math/9910022 | If MATH, the formula follows from REF . For the case MATH, we apply REF , and REF . If MATH and MATH, the formula follows from REF . And for the case MATH, we apply REF . |
math/9910022 | Using REF , we find that MATH which proves REF . Using REF , we get MATH where MATH is summed from MATH to MATH. REF then follows by applying REF - REF - REF . |
math/9910022 | If MATH, the standard formula MATH shows that REF holds. If MATH, the result is trivial by REF . If MATH and MATH, the observation MATH and identity REF together imply that MATH . If MATH and MATH, the observation MATH and identity REF imply MATH . |
math/9910022 | If MATH and MATH, this is just the contracted second NAME identity MATH . If MATH, MATH, and MATH is constant, this follows from REF and the calculation MATH . If MATH and MATH, one computes MATH . Finally, if MATH, one obtains MATH by a straightforward calculation. (See also REF and the remark after it in CITE). |
math/9910022 | This formula may be proved along the lines of CITE. Instead, we give an alternate proof using the space-time NAME and divergence identities. We note that taking the covariant derivative of identity REF yields MATH . So by using REF , substituting, and cancelling terms, we directly obtain MATH where MATH is the space-time Laplacian. Then using REF and the identity MATH, we conclude MATH . |
math/9910022 | If MATH, REF implies that MATH and MATH . Since MATH for all MATH, it is clear that REF holds if and only if both REF do. |
math/9910022 | For MATH, we use REF , and the fact that MATH to compute MATH . Hence REF is valid for all MATH if and only if REF holds. For MATH but MATH, we have MATH . So REF is valid for MATH and all MATH if and only if REF holds. Similarly, since MATH and MATH, we compute that MATH . It follows easily that REF is valid for MATH if and only if REF holds. Finally, we use REF to calculate MATH . Then noting that for MATH, MATH we compute MATH and collect terms to obtain MATH . So if REF holds, then REF is valid for MATH if and only if REF holds. |
math/9910022 | The metric MATH induces an inner product on MATH in the usual way; we shall abuse notation by writing MATH for MATH. If MATH is compact, take MATH, and otherwise let MATH be the function in the statement of the theorem. (By CITE and REF, such a function always exists if the time derivatives MATH of MATH and MATH of the NAME connection of MATH are bounded, and if MATH has positive sectional curvature.) By considering translates in time, it will suffice to prove there is MATH such that for every MATH, the quadratic form MATH is strictly positive on MATH, where MATH . Suppose MATH does not remain strictly positive, and let MATH denote the infimum of all MATH such that MATH for some MATH and MATH with MATH at MATH. We claim MATH. If not, there will be a sequence of compact sets MATH exhausting MATH, points MATH, and times MATH such that the first zero of MATH on MATH occurs at MATH. Since MATH on MATH and MATH if MATH is not compact, this is impossible. By the null eigenvector assumption, MATH . Define a tensor field MATH in a space-time neighborhood MATH of MATH by taking MATH at MATH and extending MATH by parallel transport along radial geodesics with respect to the connection MATH. (It suffices to extend MATH first radially along all MATH-geodesics which start tangent to the hypersurface MATH, and then along any curve with tangent MATH at MATH.) Notice that all symmetric space-like second covariant derivatives of MATH vanish at MATH. (Compare REF.) Indeed, with respect to a MATH-orthonormal frame MATH, one observes that for MATH, MATH . Hence for any MATH, we compute at MATH that MATH . Now consider the function MATH defined in MATH by MATH . NAME though MATH may not be compatible with the connection MATH, we still have MATH in a possibly smaller neighborhood MATH. Hence MATH attains its minimum in MATH at MATH, where we therefore have MATH and MATH for MATH, and MATH . To finish the proof, observe that there are constants MATH and MATH depending only on the bounds for MATH and MATH-on MATH such that MATH and MATH . There is MATH depending only on the NAME constant of MATH on MATH such that MATH and there is MATH depending only on the bounds for MATH, MATH, and MATH such that MATH . Combining these estimates with REF , we conclude that at MATH, MATH . Because MATH and the constants MATH cannot increase if MATH decreases, choosing MATH sufficiently small gives a contradiction. So MATH remains strictly positive on MATH, and the result follows by letting MATH. |
math/9910022 | Assume MATH, and denote the Right-hand side of REF by MATH. REF implies that the following identities are valid for all MATH: MATH . Thus for MATH, we have MATH . On the other hand, we use REF - REF with REF to compute directly that MATH . In the same way, we compute for MATH that MATH and MATH . Then by using the divergence identity MATH we can write MATH . Cancelling terms yields MATH . Recalling REF , we conclude that the special case MATH of REF holds if and only if MATH hence if and only if REF holds. Now if MATH, identities REF - REF let us write MATH . On the other hand, REF - REF imply that MATH and MATH . Noticing that MATH by the second NAME identity, we write the diffusion term in the form MATH and cancel terms to obtain MATH . Thus the special case MATH of REF holds if and only if MATH hence if and only if REF holds. Finally, the equivalence of REF and the case MATH of REF is clear when we observe that MATH and MATH . |
math/9910022 | For MATH, the equation MATH is equivalent to MATH since MATH is the unique torsion-free connection compatible with MATH; this is REF . Assuming REF , the equation MATH is valid for MATH if and only if MATH . This says that when we lower an index, MATH is a MATH-form. The equation MATH is valid for MATH and MATH if and only if MATH holds for all MATH and MATH; this is REF . The identity MATH is satisfied automatically for all MATH. |
math/9910022 | Identities REF follow easily from REF . To derive REF , we use REF to compute MATH . To derive REF , we recall that MATH and calculate MATH . Finally, to derive REF , we use REF to compute MATH . |
math/9910022 | The first two equations are easy. For the third, we substitute the formula MATH into the calculation MATH and cancel terms. |
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