paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9911206 | It suffices to show that the orbit of MATH becomes arbitrarily large as MATH increases. Since MATH is non-trivial, it contains MATH moving MATH to MATH for some MATH. Since MATH is spherically transitive, it contains MATH moving MATH to MATH. Consider now MATH: it belongs to MATH, and does not fix MATH, whence MATH's orbit contains at least MATH points. The argument applied to MATH shows that some MATH has least MATH points in its MATH-orbit, so at least MATH points in its MATH-orbit; and this process can be repeated an arbitrary number of times to produce vertices MATH with at least MATH points in their orbit. |
math/9911206 | Clearly if MATH is of infinite index in MATH then MATH is of infinite index in MATH, and is not trivial (MATH clearly cannot be abelian) so MATH is not just infinite. Conversely, assume MATH and let MATH. Let MATH; we will show that MATH is of finite index. Determine MATH such that MATH. Then there is a sequence MATH of length MATH such that MATH. Choose now two elements MATH of MATH. Because MATH is branched on MATH, it contains for MATH the elements MATH. Let MATH. It fixes all sequences except: those starting by MATH, upon which it acts as MATH, and possibly those starting by MATH for MATH. Consider MATH. Clearly MATH; as the commutator MATH was chosen arbitrarily, it follows that MATH contains MATH; and as MATH is normal, it contains MATH, the product having MATH factors. Now MATH is of finite index in MATH which is of finite index in MATH, so MATH is of finite index in MATH and the same holds for MATH. |
math/9911206 | Let MATH be just-non-solvable. Then MATH is a non-trivial normal subgroup, so MATH is solvable. Let now MATH be a non-trivial normal subgroup of MATH. It is shown in CITE that MATH contains MATH for some MATH, so it suffices to show that all MATH are solvable. Consider the chain MATH . The last group is abelian, hence solvable ; successive quotients in the sequence are also solvable because MATH and MATH is solvable by assumption. Also, MATH is solvable; therefore MATH is an extension of solvable groups, so is solvable. In case MATH is prime and MATH is a branch subgroup of MATH, the quotient MATH is a finite MATH-group, so is solvable. |
math/9911206 | Let MATH be a finite-index subgroup of MATH. By replacing MATH with its core MATH, still of finite index, we may suppose MATH is normal in MATH. By CITE, MATH contains MATH for some MATH, so MATH . |
math/9911206 | This follows again from REF. |
math/9911206 | Assume MATH is regular branch over MATH, and MATH. Recall that MATH contains a product of MATH copies of MATH at level MATH, and clearly MATH contains a product of MATH copies of MATH at level MATH, namely all but the one indexed by the vertex MATH. Take MATH. There is then a MATH such that MATH, so MATH contains the product of MATH copies of MATH at level MATH, hence is of finite index in MATH. |
math/9911206 | This follows from the above description of MATH. |
math/9911206 | Each of these subgroups MATH are normal finite-index subgroups of MATH. By the Quantitative Congruence Property, they are all contained in some MATH. It is therefore equivalent to study them directly or to study their images in MATH, which is a finite group. A computer algebra system, like CITE, can then be used to derive their structure. |
math/9911206 | The first three points are proven by NAME in CITE. To prove the fourth assertion, we apply REF to determine the structure of MATH for MATH, and note that MATH. For the last statement, note that MATH belongs to MATH, and that MATH giving, after conjugation, any generator of MATH in any position on any level MATH. |
math/9911206 | Define the following subgroups of MATH: MATH . Then the theorem follows from the following proposition. |
math/9911206 | A priori, MATH, as a subgroup of MATH, maps in MATH. Restricting to those pairs that fix MATH gives the result. Similarly, MATH, as a subgroup of MATH, maps in MATH, and MATH, as a subgroup of MATH, maps in MATH. |
math/9911206 | Consider the element MATH of MATH. Then MATH satisfies MATH, so MATH for all MATH; as MATH has only MATH-torsion, it follows that MATH is of infinite order. |
math/9911206 | The embedding is given by MATH, MATH, MATH, MATH. The index is infinite because the subgroups MATH and MATH do not intersect, one being torsion and the other torsion-free. |
math/9911206 | MATH is fractal by REF and the nature of the map MATH. As MATH is normal, MATH contains MATH and MATH, so by conjugation it contains MATH and MATH, so finally it contains MATH. |
math/9911206 | By direct computation, MATH. Apply REF . |
math/9911206 | Consider the groups MATH . Then MATH is a quotient of MATH, written MATH, via the map MATH, and the map MATH lifts to a map MATH. Define MATH where the notation implies that MATH. For any word MATH in MATH of length at least MATH representing an element of MATH, the corresponding words MATH will be strictly shorter; thus every MATH eventually gives MATH through iterated application of MATH, and thus MATH. We will obtain an explicit set of generators for MATH: let MATH and MATH . Then we claim that for all MATH . By direct application of the NAME algorithm CITE, we obtain the presentation MATH . Computation shows that MATH is of index MATH in MATH. From this we obtain, again using NAME, the presentation MATH . As a consequence, we can write MATH as a normal subgroup of MATH by keeping only those relators of MATH that do not appear in MATH and rewriting them in MATH, namely MATH . Then a direct computation shows that MATH for MATH. This proves that MATH . |
math/9911206 | For MATH, MATH is a subgroup of index MATH in MATH, so MATH is a subgroup of index MATH in MATH and MATH. For MATH one has MATH. |
math/9911206 | The first and third assertions can be checked on a computer. For the second, MATH contains MATH and MATH; these elements satisfy MATH and MATH. Then MATH contains MATH and MATH, so it contains MATH and MATH. |
math/9911206 | Let MATH and MATH be the minimum and maximum of the weight function MATH, and set MATH. By considering the series MATH, whose radius of convergence is MATH and comparing it with MATH, we obtain MATH so MATH is equivalent to MATH. |
math/9911206 | First, note that MATH cannot have polynomial growth, since it contains MATH whose growth function is greater than MATH CITE. Take as generators for MATH the set MATH; let MATH be strictly between the real root of the equation MATH and MATH, for instance MATH and let MATH be defined by MATH . Clearly any element MATH, when expressed as a minimal word in MATH, will have the form MATH, where the first and last MATH are optional and MATH. Indeed the function MATH satisfies the triangular inequalities MATH, etc. Choose once and for all a minimal expression for every element of MATH. Suppose now for contradiction that MATH. For some value MATH to be chosen later, partition MATH in two sets: MATH containing those elements MATH whose minimal expression MATH contains at least MATH occurrences of the generator MATH, and MATH the other elements. Define two generating series MATH . Clearly MATH. We will show that for an appropriate value of MATH both MATH and MATH will converge up to some MATH with MATH. We bound MATH by replacing MATH by a larger set, namely the set of all words MATH containing at least MATH occurrences of MATH. Then MATH . By NAME 's formula, MATH . Putting these together, we conclude that MATH converges up to any MATH if MATH and this will hold for MATH large enough, as both the first multiplicand and the denominator tend to MATH as MATH tends to MATH, while the second multiplicand tends to MATH. We then approximate MATH by considering the subset MATH of words MATH that either start or end by MATH, but not both; and further that contain an even number of MATH-s. The series MATH obtained this way will satisfy MATH. Now MATH injects in MATH through the map MATH, written MATH. We will compare MATH with MATH. Thanks to the choice of MATH, every generator MATH contributing MATH to MATH will contribute at most MATH to MATH, while every MATH contributing MATH to MATH will contribute MATH to MATH. We conclude that for all MATH we have MATH . This means every element of weight MATH in MATH can be written as a pair of elements of MATH with total weight at most MATH, or in formul MATH . The series MATH thus converges up to MATH; the same holds for MATH. Then the series MATH converges up to MATH, a contradiction. |
math/9911206 | The claims match those of REF , and are proved by restricting to elements preserving MATH the MATH and MATH in decompositions of the kind MATH. |
math/9911206 | First we prove MATH is normal in MATH, of index MATH, by writing MATH, MATH, MATH; similar relations hold for conjugates of MATH. A transversal of MATH in MATH is MATH. All subgroups in the diagram are then normal. Since MATH, we clearly have MATH. Now as MATH and MATH and MATH has index MATH, we must have MATH. Finally MATH, so MATH. Next MATH and similarly for MATH, so MATH and MATH. Also, MATH and MATH both belong to MATH, while MATH does not; so MATH is a proper subgroup of MATH, of index MATH (since MATH is the elementary abelian group MATH on MATH and MATH). Consider now MATH. It is in MATH since MATH. Also, MATH and similarly for other conjugates of MATH, so MATH, and MATH, so MATH. Finally MATH it is of index MATH in MATH (since MATH on MATH), and since MATH is of index MATH in MATH (with quotient MATH) we have all the claimed equalities. |
math/9911206 | MATH is fractal by REF and the nature of the map MATH. By direct computation, MATH, so MATH is branched on MATH. Then MATH, as is shown in CITE, so MATH has finite index and MATH is just-infinite by REF . MATH, so MATH has the congruence property. |
math/9911206 | For MATH, let MATH, the MATH-length of MATH, denote the minimal number of MATH's required to write MATH as a word over the alphabet MATH. We will show by induction on MATH that MATH is of infinite order. First, if MATH, that is, MATH, we conclude from MATH and MATH that MATH is of infinite order. Suppose now that MATH. If MATH, then MATH and it suffices to show that one of the MATH is of infinite order - this follows by induction since MATH and some MATH. We may thus suppose that MATH. Up to symmetry, it suffices also to consider elements MATH of the form MATH, MATH and MATH, for MATH. Write MATH. In the first case, we have MATH. It suffices to show that the first coordinate of this expression is non-trivial, as MATH contains at worst only MATH-torsion, being contained in the MATH-Sylow of MATH. Now map MATH to MATH, an elementary abelian group of order MATH. One checks that MATH in the abelian quotient, so the first coordinate maps to MATH in MATH. The second case is handled in the same way. Finally, if MATH, then MATH, so MATH (MATH factors); each factor has strictly smaller MATH-length than MATH, and as before the projection in one of the coordinates onto the abelian quotient gives some MATH. |
math/9911206 | Follows immediately from MATH and MATH. |
math/9911206 | First note that MATH, for every word in MATH and MATH whose number of MATH's is divisible by MATH can be written in these generators. Then compute MATH . The last assertion is also checked on this computation. |
math/9911206 | MATH is fractal by REF and the nature of the map MATH. The subgroup MATH described above has an infinite-index derived subgroup MATH (with infinite cyclic quotient), from which we conclude that MATH is not just-infinite; indeed MATH is normal in MATH and MATH is infinite. |
math/9911206 | For MATH, let MATH, the MATH-length of MATH, denote the minimal number of MATH's required to write MATH as a word over the alphabet MATH. We will show by induction on MATH that MATH is of infinite order. First, if MATH, that is, MATH, we conclude from MATH and MATH that MATH is of infinite order. Suppose now that MATH, and MATH. Then there is some sequence MATH of length MATH that is fixed by MATH and such that MATH. By REF , MATH, so it suffices to show that all MATH are of infinite order. Such a MATH can be written as MATH for some MATH and MATH; by symmetry let us suppose MATH. Then MATH, say. For any MATH, we have MATH, because all the components of MATH and MATH have MATH-length MATH. We distinguish three cases: CASE: MATH for some MATH. Then consider the image MATH of MATH in MATH. By REF , MATH, so MATH. But this is a contradiction, because MATH is elementary abelian of order MATH, generated by the independent images MATH and MATH. CASE: MATH for some MATH. Then by induction MATH is of infinite order, so MATH too, and MATH too. CASE: MATH for all MATH. We repeat the argument with MATH substituted for MATH. As there are finitely many elements MATH with MATH, we will eventually reach either an element of shorter length or an element already considered. In the latter case we obtain a relation of the form MATH from which MATH is seen to be of infinite order. |
math/9911206 | Define the following family of two-generated finite abelian groups: MATH . First suppose MATH; Consider the diagram of groups described above, and quotient all the groups by MATH. Then the quotient MATH is isomorphic to MATH, generated by MATH and MATH, and the quotient MATH is isomorphic to MATH, generated by MATH and MATH. As MATH, the index of MATH in MATH is MATH and the index of MATH is MATH. Then as MATH and MATH we deduce by induction that MATH and MATH, from which MATH follows. For MATH we have MATH. |
math/9911206 | Clearly MATH is normal of index MATH, being the kernel of the epimorphism MATH. Then MATH (as can be checked in the finite quotient MATH) but is of index at most MATH, so has precisely that index. Moreover, MATH is generated by the MATH: one has MATH, MATH, etc. MATH holds in all MATH-groups, and MATH has index MATH because it is MATH-generated MATH-step nilpotent. Now consider MATH. It is in MATH since MATH. Also, MATH and similarly for other conjugates, so MATH, and MATH, so MATH. Finally MATH it is of index MATH in MATH (since MATH on MATH), and since MATH is of index MATH in MATH (with quotient MATH) we have all the claimed equalities. |
math/9911206 | MATH is fractal by REF and the nature of the map MATH. By direct computation, MATH, so MATH is branched on MATH and is just-infinite by REF . |
math/9911206 | Take MATH, with MATH for some ray MATH; we will show that MATH is of infinite index in MATH. Let MATH be such that MATH. Then MATH, while by REF the index of MATH in MATH is infinite. |
math/9911206 | Since MATH, it follows that the MATH-space MATH is a subspace of MATH. The representation on a product of spaces is the tensor product of the representation on the spaces. |
math/9911206 | As MATH contains MATH, it follows by induction that MATH acts transitively on the sets MATH and MATH. As MATH contains MATH, it also permutes MATH and MATH, so it acts transitively on MATH. The same holds for MATH. The last assertion follows from REF . |
math/9911206 | Completely similar to REF. |
math/9911206 | As MATH contains MATH, it follows by induction that MATH acts transitively on the sets MATH, MATH and MATH. As MATH contains MATH, it also permutes MATH, MATH and MATH, so it acts transitively on MATH. The same holds for MATH and MATH. The last assertion follows from REF . |
math/9911206 | As MATH contains MATH, it follows by induction that MATH acts transitively on the sets MATH, MATH and MATH. As MATH contains MATH, it also permutes MATH, MATH and MATH, so it acts transitively on MATH. The same holds for MATH and MATH. The last assertion follows from REF . |
math/9911206 | Completely similar to REF. |
math/9911206 | Clearly MATH so MATH is abelian. Furthermore, MATH is a subgroup of MATH, and the natural map MATH induces a map MATH, so MATH is a direct summand of MATH, and their dimensions differ by MATH, which is MATH or MATH (recall MATH is the degree of the regular tree on which MATH acts). Now writing MATH we see that MATH is semi-simple and of dimension MATH. All such semisimple algebras are abelian, MATH, so MATH is abelian too. |
math/9911206 | By REF , the NAME algebra MATH is abelian, so it is isomorphic to MATH, where MATH is its dimension. This MATH in turn is equal to the number of double cosets MATH. These numbers MATH are computed in the corollaries in Subsection REF. By REF , the number of irreducible subrepresentations of MATH is MATH. Finally, MATH, where the MATH are irreducible representations. Since MATH and MATH is a power of MATH, the only possibility is that MATH for all MATH, and MATH . |
math/9911211 | CASE: By construction MATH is finite MATH-cyclic and thus by REF a reachability submodule. CASE: Since MATH is NAME, MATH is finitely generated and reachable. By REF this implies that MATH is finite MATH-cyclic. The foregoing discussion shows how to construct the desired MATH. |
math/9911211 | Let MATH and MATH. We have to show MATH. MATH is the sum of all reachability submodules of MATH. Since MATH is NAME, all reachability submodules MATH of MATH are finitely generated. By REF such modules MATH can be represented as MATH with some MATH. Since MATH with some MATH, we obtain MATH for an arbitrary reachability submodule MATH of MATH and thus MATH. The converse inclusion comes from the fact that by REF MATH is a reachability submodule of MATH and therefore contained in MATH for MATH. The latter implies: MATH. |
math/9911212 | If we expand, we obtain MATH . Taking the principal symbol of MATH as specified we obtain an error, in MATH . Now suppose we have chosen MATH such that the error, MATH is in MATH . Let MATH with MATH in MATH . Upon expanding we then obtain an extra term MATH with MATH of order MATH . Taking MATH . We have achieved an error one order better. Inducting and summing, we achieve an error in MATH . As the choice at each stage was forced, MATH is unique. |
math/9911212 | The Laplacian is defined by MATH and MATH where MATH is independent of the metric and MATH . The result follows simply by observing that in MATH and MATH terms not in MATH will vanish to order MATH at MATH unless MATH is applied to the MATH term. If MATH is applied once to such a term we obtain a first order differential operator vanishing to order MATH and if twice a zeroth order operator vanishing to order MATH . This is the result desired - we know there are no second order terms in MATH from our expression for the principal symbol. |
math/9911212 | Let MATH . As the principal symbols of MATH agree at MATH so we have that MATH is in MATH . Note that MATH . Expanding the factorizations for MATH and subtracting we have that, MATH . After cancelling the MATH we have that MATH . If MATH with MATH then we have that MATH and MATH . Taking the residue modulo MATH we have that MATH is congruent to zero modulo MATH . Recall that MATH with MATH . The only term of second order is MATH so we deduce that the principal symbol of MATH vanishes at MATH . Let MATH denote the principal symbol of MATH at MATH . We then have that MATH for each MATH. Iterating we conclude that MATH for each MATH which proves that MATH . Repeating, the result follows. |
math/9911212 | Let MATH be a volume form of MATH and MATH a volume form for MATH. We have by REF and MATH. We then have that MATH and similarly for MATH . It is now clear that MATH and MATH . For any MATH which does not involve MATH . |
math/9911212 | Let MATH be the volume form for MATH . Arguing as above we have that, MATH which modulo terms vanishing appropriately at MATH equals the bilinear form MATH which upon expanding modulo appropriately vanishing terms equals, MATH which does not involve MATH . |
math/9911214 | CASE: Suppose that MATH. Then we have MATH and MATH. Thus we get MATH or MATH. CASE: If MATH then MATH, and hence MATH. Thus we get MATH. Hence, by REF we have MATH. We next prove that MATH. Suppose that MATH satisfies MATH. Then we have MATH and MATH, which imply MATH. This contradicts to the assumption. Thus we get MATH. We next prove that MATH. If MATH then MATH, and hence MATH. Thus we get MATH. Therefore MATH. CASE: We first prove that MATH. Suppose that MATH satisfies MATH. Then we have MATH, which implies MATH. This contradicts to MATH. Thus we get MATH, and hence MATH by REF . CASE: By REF , we have MATH where the second inequality follows from REF . Thus we get MATH, and hence MATH. CASE: By REF , we get MATH. |
math/9911214 | REF - REF They are obvious. CASE: Set MATH. By the assumption on the total order MATH, the family MATH satisfies the sufficient condition in REF . Hence, MATH is a convex set in MATH. On the other hand, since MATH, MATH is a convex set in MATH by REF . Thus MATH is a biconvex set in MATH. Set MATH. Let MATH be the opposite order of MATH. Then MATH if MATH. Hence, MATH is a biconvex set in MATH. Thus MATH is a biconvex set in MATH since MATH. |
math/9911214 | The reflexive row and the symmetric row are obvious. To prove the transitive row, suppose that MATH for some MATH. For each MATH, choose MATH and MATH satisfying MATH and MATH. Then we have MATH . Thus we get MATH. Similarly, we see that for each MATH there exists MATH such that MATH. Therefore we get MATH. |
math/9911214 | CASE: We see that MATH if and only if MATH for all MATH. Hence the assertion follows from the fact that MATH if and only if MATH for MATH and MATH. CASE: This follows from REF . CASE: We see that MATH is an infinite set by REF . For each MATH, we have MATH. Thus we get MATH by REF . CASE: By REF , we see that the condition MATH is equivalent to the condition that for each MATH there exists MATH such that MATH and MATH. Thus MATH if and only if MATH. |
math/9911214 | CASE: Suppose that MATH. Since MATH for some MATH, we have MATH. Thus we get MATH. On the other hand, for each MATH, we have MATH and hence MATH for each MATH. Thus we get MATH by REF . CASE: Let MATH be an element of MATH constructed as in REF - REF . By the construction, we have MATH for some unique MATH. Since MATH for each MATH, we have MATH . By REF and the equality REF , we have MATH for each MATH since MATH. In addition, by REF we have MATH for each MATH. Thus we get MATH by REF . Moreover, by REF and the equality REF , we get MATH. CASE: Since MATH, we have MATH . On the other hand, since MATH, we have MATH . Therefore, by REF we get MATH. CASE: This is straightforward from REF . CASE: By the argument in the proof of REF , there exist an element MATH and MATH satisfying MATH for all MATH. Then we have MATH. Hence, by REF we have MATH. Moreover, since MATH for all MATH, we have MATH. Thus we get MATH. |
math/9911214 | This follows from REF . |
math/9911214 | CASE: This is straightforward from the definition. CASE: This follows immediately from REF . CASE: Let MATH be an element of MATH, and write MATH with MATH for all MATH and MATH for some MATH. Since MATH for each MATH, we have MATH with MATH, which implies that MATH since MATH. Thus MATH for each MATH, and hence MATH. CASE: Suppose that MATH and MATH with MATH. Then, by REF we have MATH. Conversely, suppose that MATH with MATH. Then we have MATH by REF , and hence MATH for all MATH since MATH. Moreover, MATH for all MATH since MATH. Thus MATH for all MATH, and hence MATH and MATH. Therefore MATH, which implies that MATH. |
math/9911214 | Suppose that there exist a symmetric subset MATH and a pointed subset MATH such that MATH. Then we have MATH . This proves the uniqueness of the decomposition. On the other hand, it is easy to see that the above subsets give the desired decomposition of MATH. In addition, we suppose that MATH is closed. Let MATH and MATH be elements of MATH such that MATH. Then we have MATH and MATH. Thus we get MATH, and hence MATH. Therefore MATH is closed. We next prove REF . Suppose that MATH for some MATH and MATH. Then MATH, since MATH is closed and MATH. This is a contradiction. Hence, REF is valid. Suppose that MATH for some MATH. Then, since MATH we have MATH by REF . This contradicts to MATH. Thus we get MATH for each MATH satisfying MATH. Therefore MATH is closed. |
math/9911214 | CASE: It is Clear. CASE: It is clear that MATH and MATH is a closed set. By REF , we have MATH where MATH (respectively, MATH) is the symmetric part of MATH (respectively, MATH) and MATH (respectively, MATH) is the pointed part of MATH (respectively, MATH). Then we have MATH . Indeed, if MATH then we have MATH and MATH by REF , and hence MATH by REF . Thus MATH. Similarly we have MATH. By REF , we have MATH. Moreover, we have MATH since MATH is pointed. Thus we get MATH by REF , and hence MATH is a parabolic set in MATH. CASE: By REF , there exist a subset MATH and an element MATH such that MATH. Then MATH since MATH, and hence MATH by REF . The uniqueness follows from REF . CASE: It is clear that MATH is pointed. By REF , we have MATH, and hence MATH is closed. Moreover, by REF we have MATH and MATH. In addition, MATH is closed. Thus MATH is closed, since MATH. |
math/9911214 | REF is straightforward from the definition. To prove REF , suppose that MATH. Then MATH by REF , hence MATH. Thus REF is valid. We prove REF . Write MATH with MATH. Then we have MATH for each MATH and MATH. Thus we get MATH for all MATH, and hence MATH. Thus REF is valid. CASE: It is obvious. |
math/9911214 | Thanks to REF , it suffices to show that the triplet MATH satisfies the six conditions R REF - R REF . The conditions R REF , R REF , and the first equality in R REF are obvious. By definition, we have MATH . Hence, to check the condition R REF , it suffices to show that each element of MATH can be written as MATH with MATH for all MATH. We have MATH for each MATH, and MATH for each MATH and MATH, where MATH such that MATH. If MATH satisfies MATH, then we have either MATH . In addition, we have MATH for each MATH. By REF - REF , we see that each element of MATH can be written as MATH with MATH. Hence, by induction on values of elements of MATH by the height function MATH, we see that each element of MATH can be written as a MATH-linear combination of MATH. Thus R REF and R REF are satisfied, and R REF is clear since MATH satisfies the required property in R REF . Finally, we check the second equality in R REF . Suppose that MATH with MATH. Since MATH, each element of MATH can be written as MATH with MATH for all MATH and MATH for some MATH. This fact implies that MATH. Thus the second equality in R REF is valid, since MATH fixes pointwise MATH. |
math/9911214 | Since MATH, REF follows from REF . REF follows from REF . |
math/9911214 | CASE: This follows from the fact that MATH is not empty if and only if MATH. CASE: By REF , we have MATH, which implies REF . By definition, we have MATH . Moreover, since MATH we have MATH, and hence MATH. Thus REF is valid. By definition, we have MATH . Thus REF is valid, since MATH. |
math/9911214 | CASE: Suppose that MATH with MATH. Then MATH and MATH. Since MATH is closed, we have MATH, and hence MATH by REF . Thus MATH is a convex set. CASE: It is clear that MATH. Suppose that MATH with MATH. Then MATH and MATH. If MATH then MATH. This contradicts to MATH. Thus we get MATH. Since MATH is closed, we have MATH, and hence MATH by REF . Therefore MATH is a real convex set. CASE: From REF , it follows that MATH is a real convex set. Since MATH is a biclosed set in MATH, the set MATH is a closed set. Thus, by REF , we see that MATH is a convex set, since MATH. |
math/9911214 | We have MATH. Since MATH is a closed set, REF follows from REF . From REF , it follows that MATH is a pointed biclosed set in MATH, hence REF follows from REF . |
math/9911214 | Suppose that MATH with MATH and MATH. Then we have MATH, MATH, and MATH. Thus we get MATH by REF , and hence MATH by REF . |
math/9911214 | CASE: From REF , it follows that MATH is a convex set in MATH. Thus the assertion follows from REF . CASE: By the equality REF , we have MATH . Since both MATH and MATH are convex sets in MATH, we see that both MATH and MATH are convex sets in MATH by REF , hence MATH is a biconvex set in MATH. To prove of the assertion for MATH, it suffices to exchange the sign MATH for MATH. |
math/9911214 | CASE: Suppose that there exists MATH such that MATH. Since MATH we have MATH. Thus we get MATH for all MATH by the convexity of MATH, and hence MATH. CASE: By REF , we have MATH. Suppose that MATH. Then there exists an element MATH such that MATH for some MATH. Moreover, we have MATH since MATH. By the convexity of MATH, we have MATH. This contradicts to MATH. Hence we have MATH. |
math/9911214 | It is clear that MATH. Suppose that MATH with MATH. By definition, there exist MATH such that MATH. By the convexity of MATH, we have MATH, and hence MATH. Thus MATH is a closed set. Suppose that MATH with MATH. By definition, we have MATH, and hence there exist MATH such that MATH and MATH for all MATH. By the convexity of MATH, we have MATH for all MATH. Thus we get MATH, and hence MATH. Therefore MATH is a closed set. Suppose that MATH. Then we have MATH. Hence we may assume that MATH. Then there exist MATH and MATH such that MATH. By the convexity of MATH, we have MATH. This contradicts to MATH. Thus we get MATH. Moreover, by definition, we have MATH, and hence MATH. Next we prove the second assertion. It suffices to show that MATH is a coclosed set in MATH. By the definition of MATH and REF , we see that MATH . Suppose that MATH with MATH. Then MATH by REF . By the convexity of MATH, we have MATH, and hence MATH. Thus MATH is a coclosed set in MATH. |
math/9911214 | CASE: From REF , it follows that MATH is a pointed closed subset of MATH. Hence there exists an element MATH such that MATH by REF . Then we have MATH . CASE: From REF , it follows for each MATH that MATH is a real biconvex set in MATH. In particular, MATH is a real convex set in MATH for each MATH. To prove the maximality of MATH, suppose that MATH for some real convex set MATH in MATH. By REF , there exists an element MATH such that MATH. Since MATH, it follows that MATH, which implies that MATH, and hence MATH. Therefore MATH is a maximal real convex set in MATH. Moreover, by the above argument, the injectivity of the mapping is obvious. Finally, we prove the surjectivity of the mapping. Let MATH be a maximal real convex set in MATH. By REF , there exists an element MATH such that MATH. The maximality of MATH implies that MATH. |
math/9911214 | From REF , it follows that MATH is a pointed biclosed subset of MATH. Hence, by REF , there exist a unique subset MATH and a unique element MATH such that MATH. By REF , we see that MATH for each MATH and that MATH for each MATH. Thus we get MATH and MATH. The second assertion follows from REF . |
math/9911214 | CASE: By the equality REF , we have MATH . The second assertion follows from REF . CASE: The assertion follows from REF , since REF is equivalent to the condition: MATH. |
math/9911214 | For each MATH, we see that MATH is a biconvex set in MATH, and hence MATH is a real biconvex set in MATH by REF . Thus the mapping MATH is well-defined. Moreover, by the second assertion in REF , we see that MATH and MATH. To prove the injectivity, suppose that MATH. By REF , we have MATH and MATH, and hence MATH since MATH. Thus we get MATH and MATH. By REF , we get MATH and MATH. Finally, we prove the surjectivity. Suppose that MATH. Then MATH. By REF , there exist a subset MATH and an element MATH such that MATH and MATH. Then MATH is a finite biconvex set in MATH, since MATH. By REF , there exists an element MATH such that MATH. Moreover, we have MATH by REF . Thus we get MATH and MATH by REF . |
math/9911214 | Since MATH we have MATH by REF . Hence REF follows from REF . Moreover, by REF , we have MATH . By the definition of MATH, we have MATH . Thus REF follows from REF . Moreover, by REF we have MATH and hence MATH. Thus we get MATH . |
math/9911214 | CASE: Put MATH. Then MATH. By REF , we have MATH where MATH. Since MATH we have MATH by REF . Thus, by REF we get MATH . By REF , and REF , we have MATH where the second equality follows from REF . By REF - REF with REF , we have MATH . Hence, by REF , we get MATH . CASE: By REF , we have MATH for each MATH. Hence MATH, which implies the surjectivity of MATH since MATH is bijective (see REF ). Moreover, since MATH is injective, MATH is bijective, so is MATH. CASE: Since MATH is surjective, we have MATH. Hence, it suffices to show that this union is disjoint. By REF , and the injectivity of MATH, we have the following equality: MATH for each MATH. Suppose that MATH for some MATH and MATH. By REF , we have MATH with a unique MATH. Thus we get MATH since MATH is injective. |
math/9911214 | We claim that if MATH then MATH. Indeed, if MATH for some MATH, then we have MATH by the convexity of MATH, and hence MATH for all MATH by the convexity of MATH, that is, MATH. Thus either MATH or MATH holds. |
math/9911214 | The ``if" part is obvious. Let us prove the ``only if" part. By REF , we have either MATH or MATH. If MATH and MATH, then MATH with MATH by REF . If MATH and MATH, then MATH with MATH by REF . If MATH, then MATH is a real biconvex set in MATH. Hence we have either MATH or MATH, that is, MATH or MATH, where MATH and MATH. |
math/9911215 | We need to show that MATH is surjective onto MATH if and only if MATH is surjective onto MATH. This follows from a general fact: if MATH and MATH are surjective linear maps between vector spaces, then MATH is surjective if and only if MATH. Clearly, this relation is symmetric in MATH and MATH, and we obtain the thesis. |
math/9911215 | We first consider the case that MATH is a smooth curve. Let us choose an arbitrary connection in MATH, an arbitrary Riemannian metric MATH on MATH and a smooth extension MATH of MATH, with MATH. Since the image of MATH is compact in MATH, there exists MATH which is a normal radius for all MATH, MATH. We define MATH to be the open set: MATH . Let now MATH be a referential of MATH along MATH; for instance, this referential can be chosen by parallel transport along MATH relative to the connection on MATH. Finally, we obtain a time-dependent local referential for MATH in MATH by setting, for MATH and for MATH, MATH equal to the parallel transport (relative to the connection of MATH) of MATH along the radial geodesic joining MATH and MATH. The general case of a continuous curve is easily obtained by a density argument. For, let MATH be continuous and let MATH be a totally normal radius for MATH, for all MATH. Let MATH be any smooth curve such that MATH for all MATH, where MATH is the distance induced by the Riemannian metric MATH on MATH. Then, if we repeat the above proof for the curve MATH, the open set MATH thus obtained will contain the graph of MATH, and we are done. |
math/9911215 | Let MATH be the subbundle of the cotangent bundle MATH given by the annihilator MATH of MATH. Apply REF to MATH and set MATH, where MATH is a time-dependent local referential of MATH defined in an open neighborhood of the graph of MATH. |
math/9911215 | Clearly MATH is smooth because MATH is smooth. To compute the differential of MATH we use the connection MATH adapted to the decomposition MATH introduced above. Let MATH be fixed and let MATH, that is, MATH is a vector field of class MATH along MATH with MATH. We write MATH with MATH and MATH for all MATH; using the properties of MATH we compute easily: MATH where MATH is the covariant derivative of MATH. Let now MATH be fixed; for the surjectivity of MATH we want to solve the equation in MATH: MATH. To this aim, we choose MATH, and we get: MATH . Since MATH is an isomorphism, REF is equivalent to a first order linear differential equation in MATH, that admits a unique solution satisfying MATH. Observe that since MATH, by REF we get that MATH is also of class MATH, and we are done. |
math/9911215 | It follows immediately from REF . |
math/9911215 | Simply consider a variation of MATH with variational vector field MATH and differentiate relation REF with respect to the variation parameter, using the local chart. |
math/9911215 | For MATH the tangent space MATH consists of those MATH vector fields MATH along MATH such that MATH. For a fixed MATH, REF is a first order linear differential equation for MATH; REF and standard results of existence and uniqueness of solutions of linear differential equations imply that the differential of MATH at any MATH maps the tangent space MATH isomorphically onto MATH. It follows from the inverse function theorem that MATH is a local diffeomorphism in MATH. Finally, by standard results on uniqueness of solutions of differential equations, we see that the restriction of MATH to MATH is injective. |
math/9911215 | The right side of REF vanishes at MATH, therefore, to conclude the proof, one only has to show that its representation in local coordinates satisfies the differential REF . This follows by direct computation, observing that the representation in local coordinates of the maps MATH is a solution of the homogeneous linear differential REF . |
math/9911215 | Follows directly from REF , observing that MATH is a submanifold chart for MATH, as remarked earlier. |
math/9911215 | The function MATH vanishes on MATH and therefore MATH vanishes on MATH. The conclusion follows by observing that, since MATH is nondegenerate, the MATH-orthogonal complement of MATH in MATH has dimension MATH. |
math/9911215 | For MATH, the kernel of the restriction of MATH to MATH is equal to the intersection of MATH with the MATH-orthogonal complement of MATH in MATH. By REF , it follows that the kernel of MATH projects by MATH into MATH, and therefore the projection of a characteristic is always horizontal. For the second part of the statement, observe that for MATH, MATH is a horizontal vector field in an open neighborhood of MATH whose value at MATH is MATH. Therefore MATH is MATH-orthogonal to MATH if and only if MATH. |
math/9911215 | Simply use REF and write the NAME equations of MATH in coordinates. |
math/9911215 | By REF and the observation above we get that MATH is a characteristic if and only if MATH is constant for every MATH. The conclusion follows. |
math/9911215 | By REF , we have: MATH . By REF , if MATH is a characteristic with MATH then MATH annihilates the right hand side of REF. Namely: MATH . We have to prove that if MATH annihilates the righthand side of REF then there exists a characteristic MATH with MATH and MATH. Define MATH by MATH for all MATH. By REF , we only have to prove that MATH. Computing as in REF, we see that, since MATH annihilates the righthand side of REF, then: MATH for any horizontal MATH vector field MATH along MATH. The conclusion follows. |
math/9911215 | By REF , the annihilator of the image of the differential of the restriction of the endpoint mapping to MATH is contained in the annihilator of MATH. The conclusion follows. |
math/9911215 | From REF, the fiber derivative MATH is easily computed as: MATH . For each MATH, the map MATH is clearly a diffeomorphism, whose inverse is given by: MATH . |
math/9911215 | A critical point of MATH is a curve satisfying the NAME - NAME REF associated to the Lagrangian MATH of REF. By REF , MATH is hyper-regular, hence the solutions of REF correspond, via MATH to the solutions of the associated Hamiltonian MATH, computed as follows. First, for MATH we have: MATH . Then, using REF, we compute: MATH . For the proof of the Proposition, we need to show that if MATH is an absolutely continuous curve in MATH, then MATH is horizontal and it is a solution for the NAME equations associated to the Hamiltonian MATH for some MATH if and only if it is a solution of the NAME equations associated to the sub-Riemannian Hamiltonian MATH of REF . The NAME equations of MATH are computed as follows: MATH . From the horizontality of MATH, using the first equation of REF we get: MATH and since MATH is an isomorphism, we get an explicit expression for the NAME multiplier MATH: MATH . Observe that, by a standard boot-strap argument, from REF it follows easily that MATH is smooth. We now write the NAME equations of the sub-Riemannian Hamiltonian and of MATH using a suitable time-dependent referential MATH of MATH. The choice of the referential is done as follows. Let MATH be a time-dependent referential of the annihilator MATH which is orthonormal with respect to MATH. For the orthogonality, it suffices to consider any referential of MATH and then to orthonormalize it by REF. Then, let MATH be the referential of MATH obtained by dualizing MATH. Finally, let MATH be any orthonormal referential of MATH, time-dependent or not. In the referential MATH, for MATH we have: MATH . We can rewrite REF as: MATH where MATH. On the other hand, the NAME equations for MATH are written as: MATH . Now, if MATH is horizontal and it satisfies REF for some MATH it follows that the second sum of the first equation in REF is zero, and therefore MATH satisfies also REF. Conversely, if MATH satisfies REF, then MATH is horizontal, and defining MATH by REF, it is easily seen that MATH is a solution of REF. |
math/9911215 | The fact that MATH is a smooth manifold follows easily from the transversality of MATH and REF . The proof of the second part of the statement is analogous to the proof of REF , keeping in mind that the critical points of the action functional associated to a hyper-regular Lagrangian in the space of curves joining MATH and MATH are the solutions of the NAME equations whose Hamiltonian lift vanishes on the tangent spaces of MATH and MATH. |
math/9911215 | By NAME - NAME inequality we have: MATH where the equality holds if and only if MATH is affinely parameterized. If MATH is affinely parameterized and it minimizes length, then, for any MATH, we have: MATH . Hence, MATH is a minimum of MATH. Conversely, suppose that MATH is a minimum of MATH. There exists an affinely parameterized horizontal curve MATH such that MATH is a reparameterization of MATH (see REF ). We have: MATH hence the above inequalities are indeed equalities, and MATH is affinely parameterized. Now, assume by contradiction that MATH connects MATH and MATH and satisfies MATH. By REF , we can assume that MATH is affinely parameterized, hence MATH. This is a contradiction, and we are done. |
math/9911215 | By contradiction, suppose that we can find a submanifold MATH through MATH which is transversal to MATH at MATH and disjoint from the open ball MATH. It follows that MATH is a length minimizer between the point MATH and the submanifold MATH, hence, by REF , MATH is a minimum point for the action functional in MATH. By possibly considering a small portion of MATH around MATH, we can assume that MATH is everywhere transversal to MATH. From REF it follows then that MATH is a normal geodesic, which is a contradiction. |
math/9911215 | It follows immediately from REF and the fact that affinely parameterized length minimizers are minima of the sub-Riemannian action functional. |
math/9911215 | Suppose that the pair MATH satisfying the thesis is found; then we obtain easily MATH . Since MATH is surjective, this proves the uniqueness of the pair. As to the existence, set MATH and define MATH as in REF. Obviously, MATH is absolutely continuous, nondecreasing and surjective. Suppose that MATH for some MATH, with MATH. Then, MATH, and therefore MATH. It follows that there exists a function MATH with MATH. The curve MATH is NAME continuous, hence absolutely continuous; for, if MATH, let MATH be such that MATH and MATH. Then, MATH . We are left with the proof that MATH almost everywhere. To see this, let MATH be chosen and let MATH be such that MATH. Then, we have: MATH . The conclusion follows by differentiating REF with respect to MATH. |
math/9911215 | Let MATH an absolutely continuous nondecreasing surjective map with MATH horizontal. Define: MATH . Clearly, MATH is horizontal outside MATH; to conclude the proof it suffices to show that MATH has null measure. To see this, observe that MATH has null measure and therefore MATH has null measure. Moreover, since MATH in MATH, it is not difficult to show that MATH has null measure, and we are done. |
math/9911215 | Let MATH be any Riemannian extension of MATH and apply REF . The curve MATH thus obtained is horizontal by REF . |
math/9911215 | We can assume without loss of generality that MATH. Let MATH be a codimension MATH submanifold containing a neighborhood of MATH in MATH and such that MATH. The existence of such a submanifold is easily proved using a coordinate system in MATH adapted to MATH around MATH. Observe that, by REF , we have MATH. Let MATH be a MATH-form in MATH along MATH such that MATH, MATH for all MATH and such that MATH. Let MATH be a sufficiently small open subset containing MATH and let MATH be sufficiently small. Consider the map MATH such that MATH is a solution of the sub-Riemannian Hamiltonian MATH defined in REF and MATH for all MATH. Let MATH, where MATH is the projection. By REF , MATH, which implies that MATH. It follows easily that the differential of MATH at MATH is an isomorphism, and by the Inverse Function Theorem, by possibly passing to smaller MATH and MATH, MATH is a diffeomorphism between MATH and an open neighborhood MATH of MATH in MATH. By possibly taking a smaller MATH, we can assume that MATH. We define a vector field MATH, a MATH-form MATH and a smooth map MATH on MATH by setting: MATH for all MATH. Since MATH does not depend on MATH, it follows easily that MATH . We prove next that MATH. To this aim, let MATH denote the flow of MATH, defined on an open subset of MATH; for MATH we set MATH. Clearly, MATH is an integral curve of MATH, and therefore we have MATH, hence MATH is invariant by the flow of MATH, that is, MATH . We show that MATH is also invariant by the flow of MATH; the equality MATH will follow from the fact that these two MATH-forms coincide on MATH. For the invariance of MATH, we argue as follows: let MATH, MATH and MATH; it suffices to prove that MATH is constant in MATH. In local coordinates MATH, MATH satisfies the following linear differential equation: MATH . For MATH fixed, let MATH be an orthonormal frame for MATH around MATH; by REF we have MATH, from which it follows: MATH . From REF it follows that MATH, and differentiating this expression we obtain: MATH . From REF, it follows: MATH . Using the second NAME equation in REF, we finally get: MATH . Using REF it is easily seen that MATH is constant in MATH, and MATH is invariant by the flow of MATH. The equality MATH is thus proven, and by REF we obtain: MATH . Let now MATH be a horizontal curve with MATH and MATH. Using REF, the length of MATH is estimated as follows: MATH . This implies that MATH is a length minimizer between MATH and MATH among all the horizontal curves with image in MATH. The conclusion of the proof will follow from the next Lemma, by possibly considering a smaller MATH. |
math/9911215 | We compare the sub-Riemannian metric MATH with the Euclidean metric relative to an arbitrary coordinate system around MATH. Let MATH be a coordinate system in MATH with MATH, MATH and MATH is an open neighborhood of MATH in MATH. Let MATH be the inverse image through MATH of a closed ball of radius MATH, MATH. For MATH and MATH, denote by MATH the Euclidean norm of the vector MATH. The set of vectors MATH that are tangent to the points of MATH with MATH form a compact subset of MATH, in which the continuous function MATH attains a positive minimum MATH. Observe that for all MATH tangent to some point of MATH, it is MATH. Take MATH. If MATH is a horizontal curve with MATH and MATH, then there exists MATH with MATH and MATH. Therefore, MATH where MATH denotes the Euclidean length of a curve. This concludes the proof. |
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