paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006078 | Note that MATH is semisimple by REF . We only need to prove the invertibility of the matrix formed by MATH where MATH are as above, MATH is a basis in the space MATH of characters of MATH (we used above REF for the quantum trace). Observe that the linear map MATH takes any element of the form MATH (that is, MATH), wher... |
math/0006078 | First let us show that MATH, equipped with a natural involution MATH where MATH, is a MATH-subalgebra of the tensor product MATH-algebra MATH. For this it suffices to show that MATH, that is, MATH, MATH for MATH. For instance, one computes: MATH for all MATH. The right-hand sides of the above equations are equal since ... |
math/0006078 | Since MATH also implements MATH REF , MATH is central, therefore MATH is also central. Clearly, MATH must commute with MATH. The same Proposition gives MATH, which allows us to compute MATH . REF and the trace property imply that MATH . Since MATH and MATH is central, the above relation means that MATH and, therefore, ... |
math/0006078 | Relations REF are obvious, REF follows from REF . Let us prove relations REF . On the one hand, for all MATH we have, using the definitions of MATH, REF and the notation MATH for a linear operator MATH and two vectors MATH of a NAME space: MATH . And, on the other hand, using the definition of MATH, we compute : MATH w... |
math/0006078 | The proof follows from REF . |
math/0006078 | CASE: The definition of counit implies : MATH hence MATH. We used here the identity MATH and REF . Similarly, MATH. Using the antipode property, we compute MATH . CASE: Set MATH. Then MATH is central and invertible, MATH, and from REF we conclude that MATH. Next, MATH . Applying the counit to both sides of the last equ... |
math/0006078 | One verifies that MATH is a homomorphism exactly as in CITE. The properties of MATH and MATH follow directly from definitions. For the counit axiom we have, using the properties of counital maps, REF , and REF : MATH . REF can be verified by a direct computation. Next, we observe that MATH and MATH. The antipode axiom ... |
math/0006081 | Every compact MATH-space MATH has a MATH-invariant point. Indeed, there exists a MATH-map MATH, and since MATH is MATH-invariant, so is MATH. Let MATH be the compact space of all maximal chains of closed subsets of MATH. We saw that MATH is a compact MATH-space. Thus MATH has a MATH-invariant point. |
math/0006081 | NAME 's lemma implies that there exists a minimal element MATH in the set of all closed non-empty subsemigroups of MATH. Fix MATH. We claim that MATH (and hence MATH is a singleton). The set MATH, being a closed subsemigroup of MATH, is equal to MATH. It follows that the closed subsemigroup MATH is non-empty. Hence MAT... |
math/0006081 | The composition MATH is a MATH-map of MATH into itself, hence it has the form MATH, where MATH. Since MATH, we have MATH. |
math/0006081 | According to REF , there is MATH such that MATH for all MATH. Since MATH is a compact MATH-space contained in MATH, we have MATH by the minimality of MATH. Thus there exists MATH such that MATH. Let MATH be the MATH-map defined by MATH. Then MATH for every MATH, therefore MATH (the identity map of MATH). We have proved... |
math/0006082 | There is a bijective correspondence on the level of comples tori, which restricts to abelian varieties. Given a complex torus MATH and a basis in MATH, that is, an isomorphism MATH, from the exact sequence REF a representation MATH is deduced, and this determines a complex torus MATH together with a covering map MATH, ... |
math/0006082 | The NAME space MATH may be viewed as a quotient: the space MATH of lattice bases MATH in MATH, subject to the locally closed condition that the alternating form defined by MATH is the imaginary part of some positive hermitian form (a condition that the NAME relations express in terms of the NAME coordinates of MATH), d... |
math/0006082 | The isogeny is a sum of embeddings if and only if the inclusion MATH is given by a pair of inclusions MATH. If the condition is satisfied, as in the beginning of the section, calling MATH the quotient varieties, the product isogeny MATH factors as MATH . Introducing bases also in the homology of the varieties MATH and ... |
math/0006082 | To an embedding is associated a pair of varieties, as we have seen. Conversely, given two abelian varieties MATH, with symplectic bases for polarizations of types MATH, by means of the isomorphism MATH, from the exact sequence REF an inclusion MATH is deduced. Define the torus MATH and call MATH the quotient isogeny. T... |
math/0006082 | Take MATH, the connected component of REF in the kernel, take MATH, let MATH and MATH be the complementary abelian subvarieties, and let MATH be the restriction of MATH. Uniqueness of the choice is a consequence of uniqueness in the classical reducibility theorem. |
math/0006082 | By hypothesis the matrices of the isogenies satisfy the condition of REF and there are inclusions MATH and MATH as is seen in the proof of the lemma. Using these inclusions it is easily seen that there exists MATH if and only if MATH sends MATH and this happens if and only if MATH sends MATH. If this happens then there... |
math/0006082 | Assume that we are given abelian varieties MATH with symplectic bases for polarizations of types MATH, and an isogeny MATH of type MATH. Because of REF for MATH there are inclusions MATH. Then define the torus MATH. By means of REF an isomorphism MATH is obtained. Because the types MATH are related, this gives on MATH ... |
math/0006083 | Every element in the image of MATH is made by contracting copies of the structure constants of the NAME algebra, which are invariant under MATH. |
math/0006083 | Given a MATH-circus MATH, consider the MATH-circus MATH with an additional double lasso which encircles two knot strands and/or handles of double lassos. The resolution of MATH along the new double lasso is equal to the difference between MATH and a variant MATH in which the two strands have crosses each other. Since M... |
math/0006083 | A multiple crossing can be done step by step: MATH . (MATH is the operation of resolving the double lasso.) Modulo Relation REF, we can forget the extra little hooks. |
math/0006083 | MATH . |
math/0006083 | The map from MATH to MATH defined above is clearly surjective. To see that it is injective, we need to check that diagrams differing by cyclic permutations are already equal in MATH. It suffices to check that two NAME diagrams based on an interval differing by moving a single leg from the beginning to the end of the in... |
math/0006083 | This is a general property of the holonomy of a connection with values in a NAME algebra in which the connection form is primitive. (The usual example is the holonomy of a NAME valued connection, which takes values in the the NAME group or, alternatively, the grouplike elements inside the universal enveloping algebra.) |
math/0006083 | By the definition of a universal finite type invariant MATH, MATH where MATH is the MATH-circus MATH in the complement of MATH. By the Splitting relation, this is MATH. |
math/0006083 | By the above computations, the invariant of the connected cable of a knot MATH is MATH, multiplied by MATH, and closed up with a twist. The conjugating elements MATH and MATH can be swept through the knot and cancel each other. The factor MATH in MATH can be combined with MATH so that we apply MATH to MATH. The twisted... |
math/0006083 | The glued diagrams are the same on the two sides; we either combine the legs of MATH and MATH into one set and then glue with MATH, or we split the legs of MATH into two pieces which are then glued with MATH and MATH. (Note that there are no combinatorial factors to worry about: in both cases, we take the sum over all ... |
math/0006083 | As before, the diagrams are the same on both sides. |
math/0006083 | We use a sliding argument similar to the one in the proof of REF . Link relations in MATH can be slid over diagrams in MATH, as shown in REF . |
math/0006083 | As advertised, we use the equality of links MATH. Let us see what this equality of links says about the NAME integral of the NAME link. On the MATH side, we see the connected sum of two open NAME links; by REF, the invariant of the connected sum is the connected sum of the invariants. To write this conveniently, let MA... |
math/0006083 | Using REF and noting that gluing with MATH takes the legs of a diagram in MATH and averages over all ways of ordering them, as in the definition of MATH, we see that MATH . |
math/0006083 | MATH . In the second equality, we use REF . This is allowed, since the contraction descends to MATH by the argument of REF . Note that it is crucial that MATH is invariant for this argument. |
math/0006083 | First note that any vacuum diagrams that appear in the MATH's pass through unchanged to the result; let us assume that there are none, so that we can use the vacuum projection MATH without changing the result. By REF , MATH . Let MATH. Each MATH has at least one leg, since if the MATH of MATH eats all the legs of MATH,... |
math/0006083 | MATH . |
math/0006083 | We need to check that after resolving each vertex like MATH we are left with a collection of MATH double lassos that are trivial in MATH once you forget the knot. The internal vertices can be pulled apart one by one in the specified order: at each stage, at least one side of the encircled lasso ends in a loop with noth... |
math/0006083 | Because MATH is boundary connected, there is a good ordering on the internal vertices. (Order the vertices from the external vertices on the knot inward.) Every good ordering has a comptabile routing. |
math/0006083 | The minimal vertex in the first ordering must have an external vertex as a neighbor. Therefore, there is no obstruction to moving this vertex to the first position in the second ordering by a series of transpositions. We can repeat this for each vertex in turn. |
math/0006083 | We need to check that the two possibilities for the routing at each internal vertex are equivalent. This follows from two applications of the vertex relation: MATH where MATH is younger than MATH and the order of the two vertices in the middle sum is chosen depending on which neighbor of MATH is younger, or is irreleva... |
math/0006083 | Pick a good ordering for each routing. We can adjust the routing while keeping the ordering fixed by REF , so we just need to check that we can change the ordering. The two orderings are related by a chain of good orderings related by adjacent transpositions by REF . At each transposition, the two vertices involved mus... |
math/0006083 | Proceed by induction on MATH. The result is trivial for MATH. MATH . |
math/0006083 | The first relation says that MATH is REF-dimensional. For the second relation, note that both sides, considered as elements of MATH, are multiples of the projection onto the antisymmetric part MATH (which is REF-dimensional). A little computation fixes the constant. (Note that the constant depends on the metric. Here w... |
math/0006083 | Proceed by induction. This is a straightforward computation for MATH. For MATH, compute as follows: MATH . |
math/0006083 | Proceed by induction. The statement is trivial for MATH. For MATH, the two ends of the first strut on the left hand side can either connect to the two ends of a single right hand strut or they can connect to two different struts. These happen in MATH and MATH ways, respectively. (Note that there are MATH ways in all of... |
math/0006083 | By REF , we find MATH . Set MATH. Then by REF , MATH so MATH . |
math/0006085 | Consider the map MATH . It is a smooth fibration with fiber MATH, where MATH. Since the base MATH is contractible, we obtain that the inclusion MATH is a homotopy equivalence. Hence, the integral cohomology ring of MATH coincides with MATH, which we calculate below. REF from CITE describes algebra MATH, where MATH is a... |
math/0006085 | Let MATH be an orthonormal base. We may assume that MATH. We want to calculate the Hessian of function MATH at a billiard trajectory MATH, where MATH and MATH . Let MATH denote the orthogonal vector to MATH lying in the MATH-plane, that is, MATH . Any tangent vector MATH is determined by numbers MATH, where MATH and MA... |
math/0006085 | Let MATH be the equatorial sphere consisting of unit vectors orthogonal to MATH. Any point MATH and an angle REF determine a critical submanifold MATH. Fix an eigenvalue MATH (given by REF ), such that REF is negative. Consider the subbundle MATH of the normal bundle MATH consisting of eigen vectors of the Hessian with... |
math/0006085 | Note that the critical value MATH equals MATH . Hence for MATH we have MATH. Choose constants MATH such that MATH . Each MATH is a compact manifold with boundary and we obtain a filtration MATH and the inclusion MATH is a homotopy equivalence (as follows easily from REF). Using REF and the NAME isomorphism we obtain MA... |
math/0006085 | We will give here a simple proof working for MATH. The case MATH will follow from REF below. Consider filtration MATH as in the proof of REF . Let MATH denote MATH. We obtain a filtration MATH such that the inclusion MATH is a homotopy equivalence and MATH (using the NAME isomorphism). Hence MATH is nonzero (and one-di... |
math/0006085 | As in the proof of REF we obtain that the negative normal bundle MATH splits as a direct sum of MATH vector bundles MATH of rank MATH, one for each negative eigenvalue MATH of the Hessian. Here MATH. Let MATH denote the tangent bundle of MATH. Let MATH be a rank MATH vector bundle over MATH such that its fiber over a l... |
math/0006085 | By REF the first NAME class of MATH is MATH. This implies our statement. |
math/0006085 | Consider filtration MATH (compare proof of REF ) and the associated spectral sequence MATH . MATH contains a single critical submanifold MATH with index MATH. The normal bundle to MATH is orientable if MATH is even and it is non-orientable if MATH is odd. The NAME isomorphism gives MATH . Here MATH denotes the nontrivi... |
math/0006085 | Consider the universal MATH-bundle MATH and the associated fibration MATH, having MATH as the fiber. The total space MATH is homotopy equivalent to MATH. The NAME spectral sequence of this fibration converges to the cohomology algebra MATH. The initial term is MATH where MATH, the cohomology of the fiber, is understood... |
math/0006085 | Let MATH denote the space of orbits MATH. Function MATH determines a continuous function MATH. We want to show that any orbit MATH, representing points MATH with MATH, is not a critical point of MATH in the sense of REF . This would imply that the number of critical orbits of MATH is at least the number of critical poi... |
math/0006085 | Consider fibration MATH where the image of a cyclic configuration MATH under projection MATH is given by MATH. The fiber of MATH is the configuration space MATH. Consider the NAME 's spectral sequence of this fibration. The cohomology of the fiber MATH is described by REF ; it has generators MATH, which multiply accord... |
math/0006085 | First we will assume that MATH; the case MATH will be treated separately later. We will describe the additive structure of MATH, using approach of the NAME theory. Let MATH be the unit sphere. Consider the total length function MATH where for MATH we have MATH . The critical points of MATH are MATH-periodic billiard tr... |
math/0006086 | Suppose that MATH and that MATH vanishes on all the elements in the claim. Since MATH vanishes on MATH for MATH, it can be written as a linear combination of the MATH. Since the subdiagram of the NAME diagram for MATH spanned by the vertices of MATH is the NAME diagram of a semisimple group, it follows that MATH is zer... |
math/0006086 | Let MATH. Let MATH be the saturation of MATH in MATH, and define MATH similarly. Since MATH is a direct summand of MATH, there is an induced injection MATH. Suppose that MATH. Then MATH vanishes on MATH. Of course, it takes rational values on MATH and integral values on MATH. Thus, MATH determines a homomorphism MATH t... |
math/0006086 | Since every MATH bundle has a semistable holomorphic structure, this is clear from the previous lemma and REF . |
math/0006086 | REF shows that the parabolic subgroup and the topological type of the MATH-bundle MATH are determined by MATH and the topological type of MATH. Given two connections MATH, MATH on MATH which define semistable holomorphic structures, an open dense set of the line in MATH joining them will also define semistable holomorp... |
math/0006086 | Since the NAME group acts trivially on MATH, it suffices to divide out by MATH. We may thus assume that MATH is semisimple. By CITE, for MATH, MATH if and only if MATH for all MATH. But the simplicial cone MATH is spanned over MATH by the elements MATH, MATH, and MATH for some positive constant MATH. Thus MATH if and o... |
math/0006086 | Suppose that MATH is a point of NAME type with MATH. Let MATH be a MATH-connection in MATH. Let MATH be the holomorphic MATH-bundle determined by MATH. According to REF there is an arbitrarily small deformation of MATH to a holomorphic MATH-bundle MATH contained in MATH. We can extend MATH to a MATH-connection on this ... |
math/0006086 | As we have seen, MATH is superharmonic at MATH if and only if MATH. Consequently, MATH is superharmonic if and only if MATH. |
math/0006086 | Since the simplicial cone MATH is spanned by MATH, the first statement is clear. Since MATH for all MATH, by for example, CITE, it follows that, if MATH is superharmonic, then MATH. Finally, if MATH is harmonic, then MATH and MATH are both superharmonic, so that MATH. |
math/0006086 | Since the negative of a harmonic function is harmonic and the negative of a superharmonic function is subharmonic, the result for a subharmonic function MATH with MATH follows from that for a superharmonic function MATH with MATH. After replacing MATH by MATH, we may assume that MATH is superharmonic except at MATH and... |
math/0006086 | By REF appplied to the dual root system, the set MATH is a basis for MATH. Thus if MATH for MATH, then MATH is uniquely determined by the conditions MATH for MATH and MATH for MATH. The positivity statement is the special case of the previous lemma, applied to the inequality MATH on MATH and viewing MATH as superharmon... |
math/0006086 | The first statement is immediate from REF . Suppose MATH. Then MATH in the NAME ordering if and only if MATH, if and only if MATH. Since MATH, it follows from the first statement that this holds if and only if MATH which is clearly equivalent to the statement MATH. |
math/0006086 | The condition MATH is equivalent to the condition that for each character MATH of MATH we have MATH. The condition that MATH is harmonic at MATH means that MATH. The condition that MATH is equivalent to MATH. Thus, by REF , MATH is of NAME type for MATH. The final statement follows from REF . |
math/0006086 | There are only finitely many points MATH of NAME type for MATH with MATH. Thus, it suffices to show that if MATH, and if there is no point MATH of NAME type for MATH with MATH, then a finite sequence of the moves listed can be applied to MATH, each one not increasing MATH, to produce MATH. First suppose that MATH. Then... |
math/0006086 | We shall consider each type of move separately. We begin with two general lemmas. Let MATH and MATH be connected linear algebraic groups over MATH. Suppose that MATH is a surjective homomorphism. Let MATH be a holomorphic MATH-bundle over a curve MATH of arbitrary genus and let MATH be the holomorphic MATH bundle MATH.... |
math/0006086 | The proof given above for moves of Types REF applies equally well for curves of higher genus. For the case of moves of Type REF we further divided into the case when MATH was strictly subharmonic at MATH and the case when MATH was harmonic at MATH. The proof in this case when MATH is harmonic reduces to the following e... |
math/0006086 | It suffices to show that, for all characters MATH of MATH, we have MATH, where MATH is the integer such that MATH. But clearly MATH, and we are reduced to the case of a line bundle, that is, MATH and MATH. In this case, it is easy to see that, if MATH is the line bundle over MATH corresponding to MATH, then the line bu... |
math/0006086 | We may assume that the cover MATH was chosen so that each MATH, MATH, lifts to MATH. By hypothesis, MATH lifts to MATH. Thus, we have lifted the transition functions MATH of MATH to a collection MATH. By definition, MATH is the coboundary of MATH, viewed as an element of MATH, and similarly for MATH. But clearly, since... |
math/0006086 | By REF , there is an arbitrarily small deformation of MATH which is semistable. If MATH are the transition functions for MATH as defined above, this means that we can find transition functions MATH, depending on MATH in a small disk about MATH in MATH, such that MATH, MATH, and such that the bundle whose transition fun... |
math/0006086 | That we can arrange REF after an arbitrarily small deformation of MATH follows from REF . Since MATH lifts to MATH, REF follows from REF follows from REF . |
math/0006086 | Since there are no nonempty triple intersections involving MATH and MATH, MATH is vacuously a MATH-cocycle. To prove the rest of the result, we shall show: for all MATH, there exists MATH and, for all MATH, there exists MATH such that MATH . For this says that the MATH-cocycles MATH and MATH are cohomologous. Clearly, ... |
math/0006086 | By definition, MATH, and so it suffices to show that MATH. Choose MATH, and let MATH be define by MATH and MATH is harmonic outside of MATH. By REF , MATH and MATH. Since MATH is harmonic except at MATH it follows from REF that MATH is superharmonic. Clearly, by REF MATH for some point MATH of NAME type for MATH and MA... |
math/0006086 | Clearly, if MATH, then MATH. To prove the converse, we use our previous results on harmonic and superharmonic functions. The NAME diagram of MATH is a union of MATH connected components MATH. For MATH, let MATH be the set of the vertices of MATH together with MATH, and assume that MATH. For MATH, both MATH and MATH are... |
math/0006086 | The assumption that MATH is long implies that MATH for every simple coroot MATH which is not orthogonal to MATH. Thus the condition that MATH is harmonic except at MATH and MATH implies that MATH for all MATH with MATH. Hence MATH is constant. If in addition MATH, then MATH and so MATH for all MATH. Thus MATH. Hence, f... |
math/0006086 | It suffices to prove this for a superharmonic function MATH which is harmonic except at a single coroot MATH. If MATH, then we have seen that MATH is linear and increasing toward MATH on each subset MATH with its natural ordering. Now suppose that MATH. Then in particular MATH does not correspond to a trivalent vertex ... |
math/0006086 | In case MATH is not of type MATH, the first statement follows from REF and the second from REF . In case MATH is of type MATH, the first statement is trivially true since every vertex is special, and the second again follows from REF . |
math/0006086 | Since MATH is the highest root, MATH for all MATH, and so MATH is superharmonic. The result is then immediate from the previous corollary. |
math/0006086 | It suffices to show that MATH. By REF applied to the superharmonic function MATH, we see that MATH. Thus by REF , MATH. Since MATH, in fact MATH. |
math/0006086 | By REF any point MATH which is of NAME type for the trivial bundle and indexes a minimally unstable stratum must be of the form MATH for some MATH. By REF , only MATH for MATH special can index a minimally unstable stratum. |
math/0006089 | We proceed by induction, the case MATH being true by inspection. For MATH we surely have MATH . Notice that from REF of the MATH, MATH and therefore MATH . Now using the fact (easily proven by your favorite calculus student) that MATH for MATH, we conclude that MATH as desired. |
math/0006089 | To show that the product REF converges, we must show that the corresponding sum MATH converges. But by REF we certainly have MATH for MATH, and therefore MATH . Similarly, using the fact that MATH for any real numbers MATH, we see that for MATH . REF for MATH follow from those for MATH, as it is easily verified by hand... |
math/0006089 | Writing MATH, we have from the binomial theorem MATH . Since all of these binomial coefficients MATH are integers, each term in this last sum is visibly divisible by MATH. |
math/0006089 | If MATH is any prime power dividing MATH, a MATH-fold application of REF shows us that MATH is divisible by MATH. Then, since MATH we see that MATH is also divisible by MATH. In particular, since MATH was an arbitrary prime power dividing MATH, we see that MATH is divisible by every prime power that divides MATH. This ... |
math/0006089 | We proceed by induction on MATH, the cases MATH and MATH being evident by inspection. For the inductive step, suppose (as our induction hypothesis) that MATH is indeed an integer for a given MATH. We may write MATH by REF of MATH and MATH, respectively. The second factor MATH is an integer by the induction hypothesis. ... |
math/0006089 | We shall prove the contrapositive, that no algebraic number can satisfy the NAME REF for all positive MATH. Suppose that MATH is algebraic. Without loss of generality, we may suppose that MATH by adding an appropriate integer. Let MATH be the minimal polynomial for MATH, where the coefficients MATH are integers. Now su... |
math/0006089 | It is immediate that MATH where we have used REF for the second inequality. |
math/0006089 | We can easily show show that the MATH provide the very close rational approximations needed in REF to make MATH a NAME number. Indeed, MATH can be written as a fraction whose denominator is MATH by REF , while REF tells us that for MATH where the last inequality is by REF . Since MATH tends to infinity with MATH, this ... |
math/0006089 | We have just shown in REF that MATH is irrational. As for proving that MATH is absolutely abnormal, the idea is that for every integer base MATH, the number MATH is just a tiny bit less than the MATH-adic fraction MATH. Since the MATH-ary expansion of MATH terminates in an infinite string of zeros, the slightly smaller... |
math/0006093 | Let hence REF generate solutions of REF and let for a second TLM system REF the deflection MATH be defined by REF . Then for every sequence of states MATH incident at a cell the total states of the second system are MATH . This substituted for MATH in REF yields MATH . By virtue of the additivity of MATH, MATH follows ... |
math/0006094 | Define the vector MATH in the MATH plane as the shift of the first collision point. By a direct computation one finds MATH . Since all the incoming shock of the linearly degenerate family have the same speed MATH, by simple geometrical considerations it follows that the vector MATH is constant during all interactions R... |
math/0006094 | If MATH, MATH are the shift rates before interaction, and MATH, MATH after interaction, then REF follows easily from the conservation relation MATH because by assumptions no waves of other families are generated. By REF the conclusion follows. |
math/0006094 | We consider only the case of linearly degenerate family MATH, since in the other case the proof is exactly the one given in CITE. Let MATH, MATH, be the position of the shock MATH of the MATH-th family in involution, and let MATH be the value of the NAME coordinate at MATH. For MATH, define MATH as the projection of MA... |
math/0006094 | The theorem will be proved outside the times of interaction, because the NAME dependence in MATH of the approximate semigroup implies the validity of REF for all MATH. If is sufficient to show that MATH is piecewise constant, with jumps only at the points MATH where MATH has a shock MATH, and the following relation hol... |
math/0006094 | For genuinely nonlinear fields, the proof is the same as in CITE. We then restrict the proof to the case of a linearly degenerate fields MATH. Assume that there exists two jumps MATH, MATH of the MATH-th family, with positions MATH, such that MATH . For definiteness, assume MATH, and the following conditions is satisfi... |
math/0006094 | The first inequality is an easy consequence of the MATH continuous dependence for front tracking solutions, see CITE. For the second one, note that all the shocks different from MATH have size uniformly bigger than MATH, so that their position is shifted of the order MATH. Thus the second inequality follows by standard... |
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