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math/0104289 | Let MATH be an H-M rep. of the form MATH. Then, MATH and MATH lie over the same element of MATH. In each width REF MATH orbit, there are four elements to which MATH gives an element with MATH or REF. Apply the shift to the remaining (from REF) elements in the orbits containing H-M and near H-M reps. that lie over MATH with MATH. There are MATH elements in the width REF cusps with shifts of this type. The remaining elements with MATH REF must have lifting invariant -REF. A representative has the form MATH with MATH as in REF . With MATH the centralizer of MATH in MATH, the MATH orbit of MATH includes MATH . An analog REF proof shows that if MATH, then MATH. As there, this argument uses the special presentation of MATH from REF . |
math/0104289 | REF account for all of the intersections of MATH applied to orbits with MATH with MATH. There are four such intersections. REF shows that MATH applied to orbits with MATH contributes eight intersections with MATH. Since there are REF elements in MATH, MATH orbits MATH account for all the remaining intersections with MATH. Now we show why MATH give the indicated intersections. Such intersections lie above corresponding level REF intersections. REF gives the formula MATH acting on reduced inner NAME classes. Suppose MATH lies over MATH and MATH has middle product REF. For example, this would hold if MATH is an H-M rep. Then REF , MATH lies over MATH, on an H-M or near H-M rep. orbit, and its shift also has middle product REF. So MATH, MATH applied to MATH is an intersection of MATH and MATH. If we use instead MATH, this gives another intersection of MATH and MATH: MATH. If MATH is the H-M rep. on MATH, then MATH is an H-M rep. Suppose MATH (MATH; the complement of an H-M rep), then MATH is a near H-M rep. as given by REF . Similarly, replacing MATH by MATH and MATH replaces MATH by the complement of MATH and the other near H-M rep. over MATH. |
math/0104289 | Since MATH is a NAME cover with kernel of exponent MATH, MATH factors through MATH. There are two types of generators of MATH: MATH and MATH with MATH. Let MATH. By assumption, there are MATH with MATH and MATH. From MATH, MATH . Similarly, MATH. Conclude: MATH acts trivially on MATH. Finally, suppose MATH. Since MATH is a pro-free, pro-MATH group, MATH is nontrivial if and only if MATH is not in MATH. This translates the last sentence. |
math/0104289 | Since MATH is a NAME cover of MATH it is a quotient of MATH. This defines MATH as the maximal quotient of MATH on which MATH acts trivially. So, there is NAME cover MATH, a central extension of MATH with kernel MATH. Each MATH element of MATH has a unique MATH lift to MATH. Suppose MATH and MATH. Then, the unique lifts of the entries of MATH to MATH determine the image of MATH in MATH. Further, this element only depends on the MATH orbit of MATH. The two given generators in the groups MATH, MATH, or MATH have product of order REF. The resulting groups have order REF, the NAME REF-group as a quotient and a pair of generators with one of order REF and one of order REF. Only the group MATH has these properties: elements of order REF don't generate the dihedral group, and the quaternion group's only element of order REF is in its center. |
math/0104289 | If MATH has order REF, and MATH, both MATH and MATH are order REF lifts to MATH of MATH. So, they are equal. The first computation is almost trivial: MATH . The lifting invariant is a braid invariant. The formulas of REF come to computing MATH when without the hats it is REF. Each element in the pullback MATH of MATH in MATH has order REF. Therefore, MATH is an elementary REF-group and a MATH module. Suppose MATH has order REF. For MATH, MATH is the trivial element if and only if MATH. So, as MATH varies, MATH (or MATH) runs over a MATH module MATH of MATH mapping to MATH by the canonical map MATH. Further, MATH has the same image as MATH, where MATH and MATH differ by the element generating the centralizer of MATH. In particular, MATH also maps to MATH under this map. Now suppose MATH. Then, there are MATH with MATH having the same image in MATH. Conclude: There exists MATH with MATH if and only if MATH. If, however, MATH, then MATH is a proper MATH module of MATH mapping surjectively to MATH. So, as a MATH module, MATH is the direct sum of MATH and MATH. Consider MATH. REF shows this is a MATH module. Then, MATH is a MATH module of order REF. Thus, it is either MATH or MATH. Suppose MATH, and MATH acts nontrivially here. This would force MATH to have a nontrivial MATH quotient, a contradiction to MATH being REF-perfect. Now assume MATH and MATH acts trivially. Then, MATH is a central NAME cover with kernel MATH. Exclude this because the NAME multiplier of MATH is MATH. So MATH holds. We may choose MATH to map trivially and MATH to map nontrivially by MATH to MATH. Thus, as an abelian group we can can write MATH as MATH. Since MATH maps surjectively to MATH and has the same order, It is isomorphic to MATH. Contrary to how we formed MATH, each element in MATH would have order REF. This concludes showing MATH: Both MATH orbits are nonempty. We show there are eight values of MATH with MATH (respectively, -REF). Consider pairs MATH with MATH over the trivial element in MATH. These form a MATH module MATH consisting of the two cosets of the submodule MATH where MATH. So, each of the two invariants contribute eight elements. Now we show the distinguishing conditions of REF hold. CASE: If MATH is in an H-M rep. orbit, then MATH . CASE: Otherwise MATH and MATH. Toward the former, consider MATH when MATH is an H-M, a near H-M rep. or a complement of such. Let MATH centralize MATH, MATH. Then, MATH. Let MATH centralize MATH. For the complement MATH of MATH, MATH. Next, suppose MATH where MATH centralizes MATH. As in REF , this is a near H-M rep. Here MATH. Let MATH be the MATH for the complement MATH of MATH. To show MATH, use REF proof. Assume MATH is a lift of MATH and MATH is a lift of MATH. As usual write MATH as MATH: MATH (respectively, MATH) permutes the coordinates as MATH (respectively, MATH). Since MATH (respectively, MATH), MATH and MATH. Also, MATH and MATH. Then, MATH and MATH is MATH . Let MATH be the subgroup mapping isomorphically to MATH by the action of MATH. To simplify exponents replace MATH by MATH. Similarly, form MATH. The above shows that for all MATH, MATH and MATH are in the same conjugacy class of MATH. To finish it suffices that the remaining eight pairs, with MATH, MATH are in different conjugacy classes in MATH. Do an explicit calculation using MATH and MATH, MATH and MATH. Elements in MATH: MATH . Apply MATH (respectively, MATH) to this list, to get two new lists MATH, MATH. For example, the REFth item in MATH (respectively, the REFth item in List MATH) is MATH (respectively, MATH). Treat these two lists as giving a permutation MATH of MATH, where MATH fixes MATH, MATH. The rules for MATH: MATH if the ith element of MATH equals the MATH-th element of MATH. The result: MATH . The final step is to check the collection of pairs MATH where MATH. Eight of these pairs are in MATH or in MATH. Examples: MATH while MATH. |
math/0104289 | Elements of odd order in MATH have a presentation as a product of an even number of disjoint REF-cycles, so they are always in MATH. If MATH is split, then any element MATH has a lift MATH having the same order. Since MATH is a central extension with order REF kernel, the order of a lift of an element of even order is independent of the lift. This concludes the proof. |
math/0104289 | Apply REF . Elements of order REF generate MATH. Now suppose MATH lifts to an element of order REF. Then, the nonsplit central extension MATH must be the quotient of the universal central extension of the perfect group MATH that characterizes orders of lifts from MATH in REF . The criterion for this is that MATH is a product of MATH disjoint REF-cycles with MATH odd REF . |
math/0104289 | If MATH, then MATH, and MATH. From our previous notation MATH conjugates MATH into MATH. Right cosets of MATH in MATH are the MATH cosets that MATH fixes. Suppose MATH conjugates MATH to MATH. Running over MATH gives elements whose inverses conjugate MATH into MATH. If MATH conjugates MATH to MATH, then MATH with MATH. So, the cosets with representatives conjugating MATH to MATH are in one-one correspondence with MATH. The test for spin separation does not depend the choice of MATH. If MATH contains no MATH, then MATH contains REF-Sylow of MATH and MATH fixes no cosets: MATH. This implies MATH with MATH odd. Also, REF-Sylow of MATH can't contain MATH, or else the representation won't be faithful. So REF-Sylow MATH of MATH has order at most REF. This implies MATH is divisible by MATH, a contradiction to REF . Now assume REF-Sylow of MATH has order at most REF. As above, let MATH be the number of cosets MATH fixes. Then, MATH and MATH with MATH odd, implies MATH with MATH odd. The centralizer of MATH has a REF-Sylow of order MATH, and this centralizer is in MATH. This implies MATH, a contradiction. Now suppose REF-Sylow MATH of MATH has order at least MATH. Assume first that MATH surjects onto a REF-Sylow MATH of MATH. Then MATH is the whole REF-Sylow MATH of MATH, for MATH is a NAME cover. If, however, MATH does not surject onto a REF-Sylow of MATH, then the kernel of the map has order MATH. So, MATH contains MATH, and MATH is not faithful. Conclude: MATH. |
math/0104289 | Use the notation for MATH of order REF in the statement. To be explicit assume MATH - MATH-up to conjugacy MATH - MATH lifts MATH. Given one lift MATH, all others have the shape MATH. Map MATH by MATH. REF says the kernel of MATH has rank REF, so the image MATH of MATH is a homogeneous space for the squares of lifts of MATH. Conclude: MATH and MATH determine MATH; and MATH runs over the subset MATH of MATH that MATH fixes. Suppose MATH has order REF and MATH is spin separating. Then, some MATH has order REF, MATH and MATH stabilizes MATH. Write MATH. Eight divides the degree of the representation. As in REF , spin separation is equivalent to MATH being four times an odd number. CASE: Suppose MATH is trivial on the NAME REF-group MATH. REF says MATH contains exactly two conjugates of MATH, and all of MATH centralizes MATH. Let MATH be REF (respectively, REF) if MATH (respectively, MATH). So, MATH fixes MATH integers. Therefore, REF divides MATH, and MATH is not spin separating. CASE: Suppose MATH is nontrivial on MATH. If MATH is the only element in MATH in its conjugacy class, then the computation just concluded gives MATH . So, REF exactly divides MATH and MATH is spin separating. If three elements in MATH are conjugate to MATH, then MATH. Now suppose MATH. As MATH is nontrivial on MATH, conjugates (under MATH) of MATH in MATH fall in two MATH orbits. Let MATH be a representative of one of these orbits. Then, MATH, so MATH. The argument is similar for REF conjugates of MATH in MATH. Now we count subgroups MATH, up to conjugacy in MATH, giving spin separating representations of degree REF. Giving MATH is equivalent to giving pairs MATH with MATH subject to these conditions. MATH . Note: MATH and MATH are in the same conjugacy class of MATH; this asks exactly that MATH and MATH are in different conjugacy classes of MATH. NAME elements of MATH from the proof of REF . With the cosets of MATH labeled as there, MATH represents MATH. Thus, its centralizer in MATH is MATH . Calculate: MATH. Uniquely determine a representative for MATH by taking MATH and MATH. The elements MATH run over all choices MATH. So, MATH equals the number of pairs MATH and MATH with one having three nonzero entries, and the other not. Given any pair MATH, there are two choices for MATH satisfying this condition. Conjugate the set of groups MATH attached to these choices by any lift of MATH (centralizing MATH). The result is a distinct subgroup. Conclude there are four such subgroups, falling in pairs according to MATH or MATH. Now, suppose MATH. As above, for MATH, MATH fixes twelve integers. Running over the ten elements of MATH, transitivity of the representation guarantees some element of MATH fixes any given integer in MATH. As MATH, these sets of fixed integers are disjoint, and they form a set of imprimitivity. With no loss assume MATH, so MATH. Elements of MATH are exactly those permuting the fixed integers for MATH. Acting on MATH cosets gives the degree REF representation of MATH on pairs of distinct integers from MATH. Continue the hypothesis MATH. Suppose MATH has order REF. Then, MATH has order REF, and MATH fixes no integers. On the other hand, MATH. From above, MATH fixes exactly sixty integers. Conclude: MATH is a product of twelve REF-cycles and six REF-cycles. For MATH of order REF, similarly deduce MATH is a product of four REF-cycles and eighteen REF-cycles. |
math/0104289 | If MATH, as in REF , the conditions on MATH are that it centralizes this element of MATH and it acts on MATH in the standard NAME REF-group representation. For this case, MATH (respectively, REF) if MATH (respectively, MATH and MATH). |
math/0104289 | Assume for MATH, MATH gives a faithful spin separation. According to REF-Sylow MATH of MATH has order MATH or MATH and it contains an element MATH. REF lists all cases where MATH is REF or REF. The only degrees left as dangling possibilities are where MATH has order divisible by REF or where it has order MATH. In the former case, since MATH contains nontrivial elements of MATH, it must contain all of MATH. So it is not faithful. Similarly for the latter case: The action of an element of order REF on MATH forces this to be all of MATH. |
math/0104289 | The group MATH has only one faithful transitive permutation representation as a subgroup of MATH. In this representation elements of order REF have the shape MATH. So these generators are in MATH: MATH. This is the only embedding we must test to check if MATH has a spin separating representation of any kind. Form a central extension of MATH by mapping MATH to MATH and MATH to MATH. In the extension, the square of MATH is cleaved away from the squares MATH (they remain equal). Let MATH be the trivial MATH module. Recall: MATH has rank MATH and order MATH. So, the REFst characteristic REF-Frattini module MATH (notation of REF) of MATH has dimension MATH CITE: NAME 's formula for the number of generators of a subgroup of a free group of rank MATH and index MATH. As a MATH module it is indecomposable CITE. The maximal quotient of it on which MATH acts trivially is MATH. We ouline our computation that MATH has dimension two. Use the notation of CITE. The augmentation map gives MATH. Denote its kernel by MATH. The version of NAME 's Theorem (CITE or CITE) in CITE gives an effective tool for computing the NAME layers of any MATH-group ring using the NAME basis of its universal enveloping algebra. The NAME layer at the top of MATH (the second radical layer in MATH) is MATH. The projective module MATH maps naturally and surjectively to MATH extending the direct sum of two augmentation maps MATH. The kernel of this map is MATH CITE. The explicit basis of MATH from the group elements produces a natural interpretation of MATH as a matrix. Row reduce and compute explicitly a basis of MATH from that of MATH. This gives MATH a MATH module structure. Rational canonical form of the matrix action MATH of MATH (respectively, MATH and MATH) on this module determines the maximal quotient on which MATH (respectively, MATH) acts like the identity matrix, giving the maximal quotient on which MATH acts like the identity. We let GAP do this computation. All non-split extensions of MATH are isomorphic as groups though not as extensions. As in REF, the isomorphisms between them come from MATH which is isomorphic to a NAME REF-group. Extensions correspond to which order REF subgroup of MATH cleaves from the other order REF subgroups. The proof of REF reviews the generators and relations for MATH. In that notation, symbols MATH, MATH (in the multiplicative group of units in the NAME algebra), generate subject to the relations MATH and MATH. The map MATH appears from MATH. Finally, MATH. Like MATH, MATH has no non-trivial coreless subgroup. So, only its regular representation is faithful and its generators MATH and MATH, of order REF, have the shape MATH; they are in MATH. Involutions in MATH are the products of REF disjoint two cycles. So REF implies they lift to the same order on pullback in MATH. Now we show MATH (the pullback of MATH to MATH) does not split off MATH. If the extension splits, the lifts would retain the relation MATH. So, it suffices that the square of any lift of MATH differs from the square of any lift of MATH. With no loss: MATH . Write each REF-cycle's lift to MATH as the product of three obvious generators. Example: MATH lifts to MATH. Square the lifts of MATH and MATH using the relations from the previous paragraph. Then multiply the two squares together to get -REF, the generator of the kernel of MATH. This also shows -REF is in the NAME subgroup, so the extension is NAME, and therefore nonsplit. The same procedure for MATH shows the squares of the lifts of MATH, MATH, and MATH are the same. Every non-split extension of MATH has one of these cleaved off from the rest. This shows the spin extension of MATH splits. |
math/0104289 | Write MATH as MATH with MATH (respectively, MATH) the disjoint cycles in MATH that don't (respectively, do) appear in MATH. Similarly, write MATH as MATH. Since MATH and MATH commute, so do MATH and MATH. Proving the lemma amounts to extracting a MATH-pair MATH that appears in MATH. Then, an induction characterizes MATH as given by products of disjoint MATH-pairs. Suppose MATH appears in a disjoint cycle MATH in MATH. By assumption, MATH appears in a disjoint cycle MATH in MATH. Since we have extracted MATH, MATH. Suppose MATH is not in a disjoint cycle of MATH. Then, MATH applied to MATH has the effect MATH. Thus, MATH has order larger than REF. So, contrary to a previous deduction, MATH and MATH don't commute. Therefore, MATH appears in a disjoint cycle MATH of MATH. Calculate: MATH applied to MATH has the effect MATH. Now compute the effect of MATH on MATH to conclude MATH contains the disjoint cycle MATH. That is, MATH appears in MATH. |
math/0104289 | The number of integers in the support of REF from MATH is the same as in REF-cycles of MATH. From REF this means there are twenty such integers, and therefore two REF-cycles in MATH. Since this element fixes no integers there are also four REF-cycles in MATH. Without loss, take REF cosets of MATH above the cosets MATH, MATH, of MATH to be MATH, with MATH, MATH, MATH, and MATH. The MATH-s are four representatives of MATH cosets in MATH. First assume MATH. For each MATH, MATH cycles MATH, MATH (use that MATH commutes with MATH). The argument is the same if MATH: Write the distinct cosets above as MATH with MATH, MATH, MATH and MATH. Now inspect the relation between two elements of MATH. There are six elements in MATH. An element MATH of order REF that stabilizes MATH is transitive on the remaining five elements of MATH. Since MATH, MATH and MATH, these values are independent of their arguments. Use that MATH and MATH. Thus, MATH, and MATH, REF or REF. If MATH and MATH have eight common fixed integers, then they also move eight common integers. So we show the former and conclude MATH and MATH. With no loss, assume MATH and find the cosets from among MATH fixed by MATH. Since MATH, as MATH varies, MATH runs over all five other elements of MATH. Use MATH to see MATH fixes exactly two of these cosets as MATH varies, and eight cosets in all. |
math/0104289 | Suppose at each level MATH there are lifts MATH to MATH as above, where MATH has the same order as MATH. The sets MATH of such lifts in MATH form a projective system of nonempty closed subsets. So, there is an element in the projective limit of the MATH-s giving the desired lift MATH. Now consider the relation between lifts to MATH to that of lifts to MATH. Apply REF to the pullback MATH of MATH in MATH. Then, MATH is a MATH-projective cover of MATH. So, it maps surjectively to the MATH-Frattini cover of MATH. A lifting MATH to MATH, as in the statement where MATH has the same order as MATH, will map to such a lifting in the image. Conversely, the universal MATH-Frattini cover of MATH has a map back through MATH over MATH. Again, a lifting of the type in the lemma to MATH produces one in MATH. |
math/0104289 | The effective procedure for computing the monodromy is CITE. For MATH, consider MATH the corresponding elements in MATH, so and MATH and MATH are the corresponding MATH factors. Then, MATH . So, mod MATH, the MATH-s are factors of automorphy. From CITE, however, finding factors of automorphy without modding out by MATH improves if we know MATH is in the level REF congruence subgroup of MATH. Then, the REFth power of the MATH-null is an automorphic function. |
math/0104289 | Restrict the argument to any geometrically connected component of MATH. Assume the result is false and MATH is the constant image of the moduli map. So, with MATH running over MATH, there are an infinite number of inequivalent NAME covers MATH with group MATH (isomorphic to MATH, though not necessarily equal to it). Consider the group MATH these MATH-s generate in the automorphism group of MATH. Since MATH, this is a finite group. So there are only finitely many covers up to equivalence in this class. The family is connected: There is only one cover up to equivalence, contradicting the above. |
math/0104289 | NAME shows MATH has genus REF. CITE shows for any MATH, the general cyclic degree REF triple cover with MATH branch points has a simple Jacobian. This result applies to available for MATH covers no matter the number of branch points. The proof considers MATH, the function field MATH of the branch point locus, with the field of definition for the generic cover MATH and its group of automorphisms adjoined. Then, for MATH a prime, MATH acts on the MATH-division points MATH according to this recipe: CASE: If MATH then MATH with MATH and MATH a standard isotropic decomposition with respect to the NAME pairing, MATH fixing MATH and MATH, inducing a group between MATH and MATH; and CASE: if MATH, then MATH makes MATH a MATH vector space, with MATH inducing a group between MATH and MATH. Suppose MATH is not simple. Let MATH be a finite extension over which MATH is isogenous to a product of abelian varieties, so MATH acts through a reducible normal subgroup of one of the simple groups occurring in REF. Then, the finite group MATH would contain infinitely many nonisomorphic simple groups, according to the composition factors appearing in the cases of REF. Finally, the cover MATH is unramified, and for each copy MATH of a MATH in MATH, MATH is a NAME cover. So, MATH has a degree REF isogeny to MATH with MATH an elliptic curve (REF and CITE). If MATH and MATH generate MATH, then their associated elliptic curves MATH, MATH, give the curves appearing in the statement of the lemma. These isogeny factors account for MATH up to isogeny. |
math/0104289 | Since the moduli space MATH is a family of MATH branch point covers, REF shows it does not satisfy the moduli interpretation for factoring through MATH. As the monodromy group has MATH as a quotient, this means MATH is not a quotient of a modular curve. |
nlin/0104009 | The eigenvalues of MATH are, in general, complex. It would be more convenient to deal with nonnegative real eigenvalues. The trick is to consider the matrix MATH which, due to the semigroup property is also doubly stochastic. In addition, it is symmetric (all eigenvalues are real) and positive definite (that is, all eigenvalues are nonnegative). To relate the eigenvalues of the matrices MATH and MATH we write for the second largest eigenvalue of MATH, MATH where MATH is the eigenvector of both MATH and MATH with the eigenvalue REF, and MATH is the eigenvector of MATH corresponding to MATH. From REF and the inequality MATH we infer that to prove the Theorem it is enough to show that the mean of MATH converges to zero as MATH. The simplest way to estimate the second largest eigenvalue of MATH is to compute its trace. Since MATH is positively defined, we have MATH . Thus if we can find such MATH that MATH, it will imply that MATH. It turns out that it is enough to take MATH. However let us consider MATH too. MATH . And calculating the mean MATH where due to the invariance of measure of MATH the means of the different matrix elements are equal. To calculate MATH one can either integrate over the measure of MATH, as was done in CITE, or apply various invariance considerations CITE. The result is MATH and therefore MATH as MATH. This agrees with the numerical observation that MATH: the eigenvalues of MATH are then of order MATH and there are MATH of them (not counting REF). In the case MATH we have MATH . Applying the averaging we again find that due to the invariance of the measure all contributions from the first sum are the same and there are MATH of them; there are MATH contributions from the second and the third sum (the contributions are equal) and MATH contributions from the last. Counting the number of terms in each sum, we write MATH where the averages can be calculated using CITE, MATH . Bringing everything together, we obtain MATH which effectively finishes the proof. |
nlin/0104031 | Since MATH, we have for MATH and each MATH where we used REF for the last inequality. Choosing MATH such that MATH it follows that MATH. |
nlin/0104031 | For the reader's convenience we recall the brief argument from CITE which deduces the estimate on the essential spectral radius from NAME 's formula CITE. Let MATH. Since MATH is relatively compact in MATH, there are MATH such that MATH, where MATH. For MATH, REF implies that MATH . Choosing MATH we can conclude that for each MATH the set MATH can be covered by a finite number of MATH - balls of radius MATH, MATH. This is precisely what is needed to apply NAME 's formula, which states that the essential spectral radius of MATH is at most MATH. |
nlin/0104031 | Since MATH, there are MATH with MATH and MATH. Hence MATH so that MATH belongs to the MATH - closure of MATH. But since MATH is closed in MATH, this implies that MATH. |
nlin/0104031 | Let us start by considering only MATH. Recall from REF that MATH where the Jacobian MATH of MATH is of class MATH. Then, for MATH, holds MATH . To continue we need to note the following. For each MATH there is MATH such that if MATH (MATH from the definition of MATH in REF ), then MATH for each MATH. Obviously, MATH. To investigate the NAME norm of MATH it is more convenient to introduce normal coordinates. Let us be more precise. Given an arbitrary point MATH let us consider a neighbourhood MATH so small to be contained in the domain of injectivity of the exponential map. We assume that MATH is sufficiently small that the ball of centre MATH and radius MATH is always contained in MATH. We consider an isometric identification of MATH with MATH and then consider the chart MATH given by the exponential map. On the torus MATH the situation is rather trivial. Here all tangent spaces MATH are canonically identified with each other and with MATH and MATH. Since there is no danger of confusion, we will even skip the ``mod MATH" henceforth. (It suffices to choose MATH.) Then, for each MATH, we can identify MATH with a MATH matrix MATH, so MATH. Since MATH is of class MATH there are MATH such that MATH . It follows that MATH . For manifolds other than MATH more care has to be taken to derive an analogous estimate. To continue we need to recall that the vectors MATH are adapted to the unstable foliation. Given a vector MATH in MATH and a point MATH we can write MATH as MATH, with MATH and MATH. Therefore, for each MATH, an unstable subspace MATH can be represented by an operator MATH; the subspace is just MATH. Clearly MATH. In terms of the operators MATH the MATH-Hölder continuity of the unstable distribution implies CITE: There is a constant MATH, independent of MATH and MATH, such that MATH where MATH. For MATH, let MATH, MATH. Denote MATH. Then, from REF we have MATH . As MATH we find for sufficiently small MATH (such that MATH) that MATH and hence MATH. So we conclude MATH provided MATH is chosen small enough. This finishes the proof of the sub-lemma. It is convenient to introduce the distortion MATH . We can then write MATH . Since the Jacobian MATH is of class MATH by assumption and since MATH, the following estimate is immediate: For each MATH there exists a constant MATH such that, for each MATH, MATH. Thus MATH . To estimate the stable norm we write MATH . But MATH and MATH. Applied with MATH instead of MATH this yields at once MATH . To obtain strict contraction for the MATH-norm, we must approximate MATH by a suitable function MATH. There are MATH and MATH which do not depend on the parameter MATH from the definition of MATH and MATH such that for any choice of MATH and for each MATH there exists MATH such that MATH . We postpone the proof of the above result to the end of the section and finish the proof of REF before. So we choose MATH and fix an integer MATH such that MATH . Then we choose MATH so small that MATH and MATH. The sub-lemma implies now MATH . Using such a decomposition and denoting MATH, MATH so that MATH . Combined with REF this yields for MATH with a sufficiently large MATH . Combining and iterating the last two estimates (and observing REF ) finally yields MATH for a suitable constant MATH that depends on MATH. To conclude the proof of the Lemma note that REF extend by continuity to MATH and MATH respectively. The boundedness of MATH is then obvious. |
nlin/0104031 | The basic idea is to define MATH as an average of MATH on a small ``ball", centred at MATH, in MATH. Since only uniform NAME continuity along stable leaves is demanded from the approximating function MATH (except for global measurability, of course), its construction can be carried out ``leafwise". We fix a stable leaf MATH. Since MATH is a stable leaf, it is a MATH-dimensional MATH manifold with the Riemannian structure inherited from MATH (as smooth as the map MATH). In order to simplify the exposition we restrict again to the case MATH where all tangent spaces MATH can be naturally immersed into MATH, yet the general case can be treated in essentially the same way. For each MATH we will define an isometric identification MATH of the respective tangent spaces. In addition, for each MATH we consider a chart MATH where MATH with a radius MATH that can be chosen the same for all MATH and all MATH. We will use MATH to define ``balls centred at MATH" in MATH. The charts MATH enjoy the following useful properties: There exists a constant MATH (also independent of MATH and of MATH) such that for MATH and sufficiently small MATH holds: CASE: For all MATH and all MATH, MATH where MATH denotes the Jacobian determinant of MATH. CASE: For all MATH with MATH and all MATH, MATH and MATH . CASE: For all MATH with MATH, MATH . Before defining the maps MATH let us see how such maps allow to construct the wanted smooth approximation of MATH. Let MATH, denote by MATH the Riemannian volume on MATH and define for MATH and MATH where MATH. Then MATH because MATH maps the ball MATH onto the ball MATH isometrically. It follows MATH . We estimate the NAME constant of MATH. To this end consider MATH with MATH. Then MATH . Choose now MATH. Then this is the desired estimate on MATH. In order to estimate MATH we proceed differently. To this end denote MATH. Then MATH . Hence MATH with a constant that depends only on MATH, because MATH and MATH. It follows that MATH. To conclude we need to finally define the charts MATH and the identifications MATH and prove REF . Throughout the rest of this proof various constants MATH will appear that we do not specify precisely, but that all depend at most on the first and second derivatives of the stable manifolds MATH which in turn are uniformly bounded, see the appendix for a reference. In addition, they depend on a suitably small choice of the radius MATH. To begin with, we define for each pair MATH a linear map MATH by MATH for each MATH. Clearly MATH where MATH depends only on the second derivative (curvature) of MATH. In addition, locally we can view MATH as a graph over MATH. Let MATH such that MATH for each MATH, and define MATH, MATH . It is easy to see that MATH and hence MATH so that MATH, see also REF . Similarly, MATH. This implies in particular that MATH is a diffeomorphism close to the identity. Note also that MATH . The last ingredient needed is a way to identify the tangent space of MATH at different points. Choose a basis MATH of MATH, then MATH is a basis of MATH, unfortunately it may not be orthonormal. Yet, MATH . This means that setting MATH the new basis MATH is orthonormal. We then choose the isometric identification MATH. We are now in the position to prove REF . REFThis is nothing else than REF . REFConsider MATH and let MATH such that MATH. Consider the map MATH defined by MATH . We can apply the orthogonal projection MATH on MATH to both sides of the above equation to obtain MATH . Note that if we consider MATH as a map in the ambient space, as above, then for MATH we have MATH and thus it immediately follows MATH, which implies MATH because MATH is a MATH function. By using the above facts and recalling the definition of MATH in REF we can compute MATH where all the constants MATH depend only on the curvature of MATH. To obtain the estimate on the Jacobian we need to differentiate REF which yields as before (observing that MATH) MATH . The wanted estimate follows now from MATH. REFAgain we use REF to have MATH this means that MATH from which REF immediately follows. |
nlin/0104031 | Since MATH is the transfer operator of MATH, we have MATH for each MATH and MATH. In order to estimate MATH let MATH with MATH and MATH. Since MATH expands MATH, MATH may be as large as MATH. Therefore let MATH be the smallest integer such that MATH and fix a chain of points MATH such that MATH for all MATH. Since MATH, it follows that MATH and hence MATH . It follows that MATH. Exactly the same estimate holds also for the MATH - norm. For the unstable norm one proceeds as in REF and in NAME REF, taking the same care when estimating the NAME constant of MATH along stable fibres that we took above in the case of MATH. For sufficiently small MATH this leads to the estimate MATH and for large enough MATH we obtain with arguments analogous to those leading to REF that MATH . Iterating this estimate for fixed MATH and recalling that MATH leads to the estimate MATH. Since we can choose MATH as large as we like, MATH. |
nlin/0104031 | Let us start with MATH. For MATH, and MATH, MATH . Hence MATH, so that MATH extends to a continuous linear functional on MATH, and by NAME 's theorem there exists MATH such that MATH for each MATH. The equality for all the functions in MATH follows by an approximation argument. The idea is to approximate a function in MATH by a smooth function but keeping under control the regularity along the stable direction. To this end it is convenient to introduce the following singular convolution: MATH where MATH is the intersection of the local stable manifold with the ball of radius MATH centred at MATH, MATH is the Riemannian measure restricted to the stable manifold, and MATH is chosen so that MATH. Obviously, MATH . There exist constants MATH such that for each MATH we have MATH. Clearly MATH is bounded with respect to the sup norm. If MATH then, by the uniform smoothness of the stable manifold it follows that there exists MATH such that, for MATH small enough MATH . From the above equation, remembering that the stable manifolds have uniformly bounded curvature, it follows that there exists MATH such that MATH. The continuity of MATH is an immediate consequence of the continuity of the stable foliation. Notice that MATH and MATH are measures. By NAME 's theorem, each MATH can be approximated MATH-almost everywhere by a sequence MATH, MATH, and in view of NAME 's dominated convergence theorem also the MATH - and MATH - integrals of MATH tend to zero. Now fix MATH and MATH. Because of REF there is MATH such that MATH . Hence, recalling REF and NAME REF, we have MATH . The second and the fourth expression tend to MATH as MATH for fixed MATH and MATH. Hence MATH . Hence MATH for all MATH and all MATH. If MATH and MATH, then MATH and MATH by REF . It follows that MATH is a positive measure and MATH is an eigenvalue of MATH. Finally REF implies that for MATH . It follows that MATH is absolutely continuous with respect to MATH and MATH-almost surely. Observe also that, because of REF , each MATH can be represented as MATH for some MATH. Next, if MATH, then MATH follows since MATH for all MATH. The last statement follows immediately from the observation MATH, see REF . |
nlin/0104031 | CASE: For each MATH, MATH . CASE: Let MATH. Recall from REF that MATH is a bounded linear operator with MATH. Introduce the map MATH by the relation MATH with MATH and MATH as in REF . Then CASE: MATH is linear by inspection of the definition CASE: MATH is injective, because MATH, and since MATH, that is, MATH, it follows from REF that also MATH and hence MATH. CASE: MATH is surjective, as we show below. Hence MATH is the number of ergodic components of MATH. Proof of the surjectivity of MATH: Let MATH and choose MATH such that MATH. Then, for each MATH, MATH so that MATH when MATH. It follows that MATH exists. To show that MATH, consider MATH . It follows that MATH. Hence MATH for all MATH, so that MATH. Since MATH, this space has a basis consisting of indicator functions of pairwise disjoint sets MATH. Let MATH and MATH. Then MATH. CASE: The third statement follows from NAME 's Theorem. Indeed, if MATH there exists a measurable MATH-invariant set MATH, MATH, such that MATH . The characteristic function MATH of MATH is a measurable function, moreover, if MATH and MATH, MATH, then MATH hence MATH. Accordingly, MATH. Consequently MATH where we have used REF . We can then define, for each MATH, the sets MATH. Clearly these are invariant sets as well and their characteristic functions belong to MATH, hence we have MATH. But, in view of what we just proved in REF, there can be only finitely many invariant sets of positive MATH measure. If MATH is ergodic it follows that MATH for each MATH so that MATH for MATH-almost all MATH. CASE: The proof of this part depends heavily on REF and its proof. So we postpone it to REF. |
nlin/0104031 | Since the stable and unstable distributions are MATH (see REF in the appendix) the regularity statement follows like REF from the main result of CITE which says in our context that since MATH is MATH in each of its two real variables separately, it is jointly MATH in both variables. The third assertion is a restatement of REF which implies that MATH and MATH are the stable and unstable manifold of MATH. In addition, it is always possible to reparametrise the two coordinate axis so that REF is satisfied. |
nlin/0104031 | Given MATH let us consider a neighbourhood MATH that is foliated on the one hand by the leaves MATH, MATH, and, on the other hand, by the leaves MATH, MATH. For each MATH, define the two functions MATH by MATH where MATH denotes the unique point in MATH, see REF . The map MATH is injective and there is a constant MATH, which can be chosen the same for all MATH and MATH, such that for all MATH . In addition MATH is absolutely continuous and its Jacobian MATH satisfies MATH . The injectivity of MATH is a consequence of the transversality of the defining foliations MATH and MATH. The uniformity of all constants is due to the compactness of MATH. Because of REF , in a suitable neighbourhood MATH of MATH, both foliations can be straightened simultaneously by some MATH-diffeomorphism MATH from MATH onto MATH. This means in particular that for any MATH . Without loss of generality we can assume that there is a constant MATH such that MATH for all MATH. Now let MATH and MATH. Then MATH . It follows that MATH . Let MATH. There are two ways to estimate MATH, and we will need them both. Since MATH we have MATH because MATH is of class MATH. Similarly, MATH . This yields REF . We turn to the estimate for the Jacobian MATH. First note that MATH is continuously differentiable. Indeed, the implicit function theorem gives in view of REF for all MATH from which it follows that MATH by the previous estimate and since MATH. We return to the proof of REF . In our estimates, in order to compare MATH and MATH, we will have to compare MATH to MATH. But since we have control over MATH only along stable manifolds, we will compare MATH to MATH and care for the remainder by estimating the difference of MATH and MATH. We start by noticing that MATH in view of REF so that MATH . Similarly, denoting MATH we have MATH where we used REF , the second estimate in REF and the fact that MATH. Accordingly, MATH . Now we can estimate the NAME constant of MATH: MATH where we used in the last step that MATH if MATH is small enough. Therefore MATH where we used the first estimate in REF , estimate REF and also REF . This proves the lemma. |
nlin/0104031 | Let MATH be an admissible stochastic kernel family. We draw first conclusions for this family from REF . It is immediately seen that MATH; this is the first assertion of REF . Combining REF with NAME REF we see that there is a constant MATH such that for each MATH there is MATH with MATH and MATH . Our first conclusion is that for sufficiently small MATH and MATH, MATH . Since MATH it follows that MATH . We turn to the estimate for MATH. Remember that, in dimension two, the stable and unstable distributions are of class MATH, (see REF in the appendix). Therefore they can be described by MATH fields of unit tangent vectors MATH and MATH. Each vector field MATH can be written as MATH where MATH is measurable, MATH and MATH. One just has to make sure by choosing MATH sufficiently small that the angle between MATH and MATH is small enough if MATH. So let MATH. Then MATH where MATH . Observe that we interpret MATH as a vector so that the dot-products under the integral signs are inner products. Let MATH. Then MATH and MATH . Since this integral extends only over points MATH, we have MATH for sufficiently small MATH. In order to estimate MATH we apply REF to the MATH-valued function MATH. Since MATH is MATH in both coordinates and since MATH is admissible, MATH is MATH-dominated by the same MATH that dominates MATH. Hence MATH for sufficiently small MATH and MATH. So there exists a constant MATH such that MATH. We turn to the estimate of MATH. MATH . Since MATH, it follows from REF that MATH for a suitable constant MATH that depends on the MATH-norm of MATH. Since MATH is MATH-dominated by by MATH (see the definition of admissible kernels in REF ), we see as above that MATH. In order to collect the last estimates we write MATH . Observe that MATH when MATH is small enough because MATH and MATH. We invoke REF once more and conclude that MATH when MATH and MATH are sufficiently small. Note also that MATH. Hence MATH so that MATH . Combined with the estimate REF for the stable norm this yields for sufficiently large MATH which finishes the proof of REF . |
nlin/0104031 | We start by proving the lemma not for MATH but for MATH where MATH is the iterate from REF for which MATH. The reduction to the case MATH is then accomplished in REF . Let MATH. Since MATH, it follows from REF that MATH whence, for sufficiently small MATH, MATH . By induction, it follows MATH . (Observe that MATH.) Hence there is MATH such that MATH for all MATH so that MATH . We turn to the second estimate from REF . Again we consider first MATH instead of MATH. For each MATH we have MATH where we used REF for the first, REF for the second, and REF for the third inequality. For MATH this leads, by induction on MATH and observing the choice of MATH, to MATH . It remains to carry over these results for MATH to MATH. For this we use repeatedly the next two lemmas to show that there exists a smooth averaging operator MATH such that MATH with MATH for some MATH. Applying REF and the last estimate to this operator instead of MATH and using estimate REF for MATH at most MATH times then finishes the proof of REF . |
nlin/0104031 | We define MATH. Then MATH with MATH . It follows immediately that MATH if MATH, thus MATH is a kernel family. Clearly, MATH, so MATH is a stochastic kernel family. Moreover, MATH . Observe that CASE: MATH is NAME continuous, CASE: MATH, MATH and MATH are bounded and NAME continuous, because MATH is of class MATH. Therefore the verification of REF - REF for the kernel MATH and for the terms MATH, MATH, is straightforward. The term MATH needs some more care. First note that MATH . This controls the MATH-contribution to REF and reduces the NAME estimates for MATH in REF to the corresponding estimates for MATH. But MATH for a suitable MATH so that this term is bounded by MATH, and finally MATH so that this term is bounded by MATH. Hence MATH is an admissible kernel family. |
nlin/0104031 | Let MATH. Obviously MATH if MATH. The associated kernel MATH is MATH . It follows that MATH and REF - REF for MATH, MATH and MATH can be deduced easily from the corresponding properties for MATH, MATH and MATH. |
nlin/0104031 | The last step in the discussion of smooth random perturbations consists in showing that there exists a constant MATH such that MATH. To this end consider MATH. Let MATH. Denoting as before MATH we have MATH . Our goal is to show that MATH. We denote by MATH the flow at unit speed along the unstable manifolds. Let MATH be the signed distance between MATH and MATH within MATH. The direction is chosen such that MATH (see REF ). We first look at MATH, a term which involves a difference of MATH-values along an unstable fibre. It is natural to try to estimate this difference in terms of MATH where MATH, as in the preceding proof, denotes the unit tangent field in unstable direction. MATH . Changing variables to MATH and setting MATH we get MATH where MATH denotes the Jacobian of MATH, MATH, and MATH . To proceed we will show that MATH is a test vector field in MATH for a suitable MATH. To this end it suffices to check that MATH for some MATH. It is easy to estimate MATH. Because of the uniform transversality of the stable and unstable foliation, MATH so that, for each MATH with MATH, the set MATH is contained in the interval MATH. Hence MATH with a constant MATH which depends only on the stable and unstable foliation. It remains to estimate MATH. Various points and corresponding stable and unstable fibres which we use in the following computation are shown schematically in REF . Here are some estimates on distances between related points. We abbreviate MATH as MATH etc. Then, if MATH is sufficiently small, MATH for suitable constants that depend on the lower bound of the angles of intersection of the stable and unstable foliation and other quantities associated with them. We turn to the NAME estimate for MATH. MATH . Notice that according to our assumptions MATH is MATH-close to REF. The MATH-part in REF is most easily estimated, since MATH by REF , and since MATH. This implies MATH . Denote the expression in curly brackets in MATH by MATH. For each fixed MATH, the NAME measure of the set of the points MATH that belong to exactly one of the two indicator sets is of order MATH where MATH is a local MATH diffeomorphisms that straightens the stable and unstable foliation simultaneously, see REF . Therefore the corresponding part of the integral can be estimated as MATH. Now consider for fixed MATH the set MATH of those MATH which belong to both indicator sets. Then MATH, and for MATH we have MATH . MATH and MATH are easy to estimate: MATH where we have used MATH. It follows that MATH . To estimate MATH we shift the difference in MATH to the kernel MATH using the following change of variables: MATH, MATH. The map MATH is injective and there is a constant MATH, which can be chosen the same for all MATH, MATH and MATH, such that MATH . In addition MATH is absolutely continuous and its Jacobian MATH satisfies MATH . The proof is very similar to that of NAME REF. The rôle of the points MATH and MATH in that NAME is played by the points MATH and MATH here, that of MATH and MATH by MATH and MATH. The only difference is that MATH in that proof, while in the present situation MATH. Therefore the estimate REF does not change at all (except for the constant) whereas the estimate in REF is replaced by the weaker one MATH. This leads to MATH . Armed with these estimates we continue to estimate MATH from REF . Observe first that MATH and that, as in REF , MATH . This implies MATH where we used REF for the last step. Combining this with REF we find that, for MATH, MATH so that MATH for sufficiently small MATH. It follows that MATH. Therefore, MATH see REF to recall the meaning of MATH. It remains to estimate the integral MATH see REF . The strategy is similar to that for the estimate of MATH. For each MATH, we use the coordinate change MATH. The map MATH is injective and there is a constant MATH, which can be chosen the same for all MATH and MATH and MATH, such that MATH . In addition MATH is continuously differentiable and its Jacobian MATH satisfies MATH . We postpone the proof of this NAME to the end of this section. Using the coordinate change MATH, MATH and we have to estimate MATH and MATH in order to bound MATH in terms of MATH. We start with the estimate for MATH. In view of REF and of REF we have MATH . So it suffices to show that, for each fixed MATH with MATH, the supremum and the MATH - NAME constant of MATH are bounded (uniformly in MATH) by MATH. The relevant points are sketched in REF . Recall that MATH. Then MATH because of REF . Similarly, invoking REF , MATH . To analyse the NAME constant of MATH we consider two different cases which may happen for points MATH and MATH. First, if MATH, we use REF to estimate MATH (Observe that MATH.) Otherwise, if MATH we proceed as follows: MATH . In any case, MATH . The NAME constant of MATH is estimated along the same lines using REF - the details are left to the reader. We turn to the estimates for MATH. Because of REF we have MATH . In order to estimate the NAME constant of MATH we consider the cases MATH and MATH separately as in the estimate for MATH. This yields MATH . |
nlin/0104031 | Let MATH be a MATH-diffeomorphism as in the proof of NAME REF. This means that MATH . Hence MATH for a suitable MATH. This implies REF . To proceed further, we need the estimate MATH . This is proved as follows: Since MATH, we have MATH and MATH since the manifold MATH is compact and since MATH is determined by the implicit function theorem. Now we apply REF to prove estimate REF . Since MATH, it suffices to estimate MATH. To this end let MATH and assume without loss of generality that MATH. For MATH denote MATH. Then MATH, MATH, and MATH . We turn to the proof of REF . Obviously it suffices to estimate MATH where MATH denotes the identity matrix. MATH where we used REF for the last step. The proof of estimate REF is more subtle. To simplify the notation denote by MATH and MATH the maps MATH and MATH, respectively. Then MATH and hence MATH so that MATH . Since MATH, it follows that MATH. Now consider again MATH and MATH with MATH. For such points MATH . Multiplying this equation with the row vector MATH from the left we obtain MATH because MATH. Observing now that also MATH we can conclude that MATH for suitable MATH, where we used the boundedness of MATH, the estimate MATH and the fact that MATH in the last step. So we proved MATH . Interchanging the rôles of MATH and MATH we obtain in the same way MATH with the same constants. Therefore we can add both inequalities and arrive at MATH provided MATH is sufficiently small. Hence MATH where we used REF in the last step. From this estimate REF follows immediately. |
nlin/0104031 | By using a smooth partition of unity we can always restrict ourselves to a situation in which the functions are all supported in small balls of radius MATH. In such a ball one can consider two changes of variables MATH and MATH that straighten the foliations of the maps MATH and MATH, respectively. As seen in the previous section these are MATH. Accordingly, setting MATH, MATH is MATH. More can be said. There exist constants MATH and MATH such that MATH . Since the above result is hardly surprising and its proof is a bit technical we postpone it to the end of the present section. For each MATH and MATH holds MATH . Clearly there exists MATH such that MATH, hence NAME REF implies MATH . Analogously, MATH . Since MATH sends the stable and unstable manifolds of MATH to the corresponding ones of MATH and is MATH it follows that there exists a MATH-Hölder function MATH such that MATH. Accordingly, MATH, for some constant MATH. Therefore the following estimate finishes the proof: MATH where the last inequality follows from NAME REF. |
nlin/0104031 | For MATH, MATH, so MATH. In order to estimate MATH, let MATH. Observing that MATH is a convolution kernel, one verifies easily that for MATH and MATH. Hence MATH, and since the same arguments apply to the MATH norm, this finishes the proof of REF . Next we estimate MATH. To this end consider MATH with MATH. Then MATH . As MATH is the supremum of all such integrals, this proves REF . Again the same estimate applies to the MATH norm. The proof of REF is similar to that of REF . Consider MATH with MATH. Then MATH and the argument is concluded as before. Estimate REF is proved in the same way. We turn to REF . To this end we consider MATH and MATH as maps from MATH to MATH and denote MATH. For MATH define MATH so MATH and MATH and MATH, MATH are the NAME of MATH and MATH, respectively. Therefore, not writing the variable MATH explicitly and extending MATH periodically to the covering MATH of MATH, we have MATH . Fix MATH such that MATH for all MATH. Then MATH . Clearly, MATH, and a routine calculation shows that MATH provided this distance is small enough to make the denominator of the fraction not smaller than MATH, say. |
nlin/0104031 | From REF follows MATH and the first inequality follows by REF . The second inequality is proven in the same way since the rôles of MATH and MATH are perfectly symmetric. |
nlin/0104031 | We fix MATH and MATH and consider only MATH with MATH and MATH. Several constants introduced in the sequel will depend on MATH, MATH and MATH in various ways, but we will not precisely keep track of them in this section. So MATH denotes an arbitrary constant that depends in particular on MATH, but not on MATH. Recall from REF that REF applies also to MATH and hence gives estimates on MATH. In particular, MATH. Therefore MATH so that MATH provided that MATH. Since MATH, it follows that MATH . Observe next that MATH . Then it follows from REF and from REF that MATH . The next step is to apply REF to MATH and MATH with the rôles of these two operators interchanged, that is, interpreting MATH as a perturbation of MATH. This is possible, because of the following two observations: CASE: The operators MATH satisfy the NAME type inequality from REF with uniform constants that are close to those for MATH and MATH, compare REF . (The dependence of these constants on the map MATH is continuous under MATH changes of MATH.) CASE: In view of REF and our convention on the use of MATH, MATH is bounded in terms of quantities depending only on MATH and on the kernel MATH. Therefore REF guarantees that MATH with constants that do not depend on MATH and on MATH if MATH and MATH are small enough. Estimate REF is exactly REF . For REF of this proposition - as discussed in REF - we must bound the three terms MATH where MATH, MATH, MATH and MATH denote the resolvents of MATH, MATH, MATH and MATH, respectively. Because of REF , the first and the third term are bounded by MATH. It remains to estimate second one. Observing the resolvent estimate REF , the norm comparisons REF , we have MATH . Collecting the estimates for the three terms in REF and setting MATH we conclude that MATH . This is REF . |
nlin/0104031 | If MATH and MATH are the unitary vectors fields in the unstable directions for the two maps, it is clear that there exists MATH such that MATH . The estimate MATH is an immediate consequence of REF and the definition of MATH. The second is more subtle and requires some work. We start by studying the derivative of the map MATH. From the definition of MATH is follows immediately that MATH where MATH is the Jacobian of the holonomy MATH from the unstable fibre MATH to the fibre MATH at the point MATH, and MATH is the Jacobian of the holonomy MATH from the unstable fibre MATH to the fibre MATH at the point MATH. While MATH are the stable and unstable unit vector fields respectively. The above formulae imply that MATH . As the same formulae apply to the change of coordinates MATH associated to the map MATH, we have MATH where the tilde designate the quantities relative to MATH. It is immediate to see that MATH are MATH matrix valued functions. Moreover, the fact that the distributions are uniformly transversal, REF and the fact that MATH show that there exists a constant MATH such that MATH . Hence, to see that MATH is close to unity we need only to control the ratio between the NAME of the corresponding holonomies. Since the stable and the unstable one are treated exactly in the same way we will consider only the unstable ones. We want to prove that there exist MATH such that MATH . The above fact is a ready consequence of the well known formula for the Jacobian of holonomies CITE: MATH where MATH, MATH and MATH, MATH. Since MATH and MATH are on the same unstable fibre, MATH, so that there is some constant MATH such that, for each MATH, MATH . Accordingly, letting MATH, MATH . Denote by MATH the maximum of the derivatives of the maps in the neighbourhood under consideration. Since we know already that MATH, it follows that MATH, and the same estimate holds for MATH. Remembering REF and the fact that the foliations are MATH this yields immediately MATH . The result follows then by choosing MATH, which yields MATH. |
nlin/0104031 | Our starting point is the formal identity MATH . We show that MATH is indeed invertible: For MATH we have MATH where we used REF . Using REF the second summand is further estimated by MATH where we also used REF . Combining both estimates we see that MATH . This proves that MATH is invertible with MATH if MATH is sufficiently small. Hence MATH where we used REF . This proves REF . We turn to the proof of REF . MATH . This is even a bit stronger than REF . |
nlin/0104031 | A direct computation yields, for each MATH, MATH, MATH, MATH . From this the lemma follows immediately with MATH. |
nlin/0104031 | Let us set MATH. The idea of the proof is to represent the embedding MATH as a composition of compact and continuous linear maps: MATH . The first embedding is continuous, because MATH, MATH. The compactness of the second embedding is equivalent to the compact embedding of MATH into MATH, and this is well known because MATH. The continuity of the last embedding is an immediate consequence of the next lemma. |
nlin/0104031 | REF is obvious, REF follows from NAME 's theorem, because MATH is a skew product over the identity, REF is a consequence of REF , and REF is again obvious, since, as already remarked, MATH parametrises an unstable manifold. |
nlin/0104031 | The result follows by using polar coordinates around the point MATH and observing that MATH outside the ball MATH: MATH where MATH is the measure on the MATH dimensional spherical surface MATH. |
nlin/0104031 | Let MATH be a smooth partition of unity such that each MATH is contained in an open set MATH where MATH satisfy the smallness requirements previously discussed for the generic local neighbourhood MATH. In addition MATH and MATH are chosen such that, if MATH denotes the chart associated to MATH so that MATH, then MATH. Let us start by considering any MATH and consider MATH, MATH. Applying the coordinate change from REF to the function MATH and using the formula from REF we have MATH where MATH are the canonical unit vectors in MATH. This suggests to define MATH . Notice that REF imply MATH because MATH, and MATH . Hence, for each MATH, MATH since MATH and MATH . Let MATH. There exists MATH such that, for each MATH, MATH, MATH and MATH . REF is clearly the local version of what we aim at; we postpone its proof to the end of the section and use it beforehand to conclude the argument. To go to the global level the first step is to define globally MATH and MATH. Let MATH be such that MATH on MATH and MATH outside MATH. Then MATH and MATH are well defined on all of MATH, MATH, MATH for suitable MATH, and since MATH is supported in MATH it follows from REF that MATH . We can now conclude the argument by going back to our partition of unity. Given MATH and MATH for each MATH we have MATH such that MATH where MATH . |
nlin/0104031 | From REF it is clear that the supremum norms of the above test functions are bounded, more care is needed to control the NAME norms. First of all, as stated in REF , MATH and MATH are MATH-Hölder in MATH and the same for MATH. We discuss MATH first. The idea is to see how the function varies when moving along the unstable manifold and when varying MATH. Since MATH it follows MATH where we have used REF and the fact that the Jacobian of the holonomy is close to the unity when the two transversal manifolds are close, see REF . Next, given MATH, we can define a map MATH by the stable holonomy MATH between the unstable manifold of MATH and of MATH. Namely, for each point MATH of the first unstable manifold we define MATH such that MATH . The reader can look at REF to have a quick pictorial idea of all the quantities defined so far. Accordingly, MATH . But MATH since the holonomy is close to unity and the distance between the two manifolds is bounded by MATH. By the same reason, since the stable foliation is uniformly transversal to the unstable one, MATH. Using REF one shows that MATH. Then it follows, for each MATH with MATH, MATH . This clearly implies that the test function MATH is MATH-Hölder with NAME norm bounded uniformly in MATH. The study of the vectors MATH is completely similar apart for the presence of the singular kernel MATH. To control it, we will use the following result. There exists a constant MATH such that, for each MATH, MATH . An explicit computation yields MATH for some fixed constant MATH. The estimate is then done by dividing the integral into two parts. In the first inequality one considers separately the integral in a ball of radius MATH around MATH and the rest (where MATH). This yields MATH from which the wanted result follows. The other inequality is done in the same way by separating a ball of radius MATH around MATH with MATH large enough. The regularity of MATH follows then trivially from the same arguments used to study MATH and by NAME REF. |
nlin/0104031 | In a sufficiently small neighbourhood MATH we can use the coordinates MATH introduced in the proof of REF , that is, MATH. Recall from REF that MATH. We can then construct a vector field basis of the type requested in the lemma as MATH. Note that, according to REF and the regularity of the unstable distribution stated in REF , the vector fields MATH are in MATH for a suitable MATH. Consider now some MATH with MATH and MATH, MATH which is MATH when restricted to any unstable manifold, and let MATH. Then MATH for some MATH that does not depend on MATH and, denoting MATH, we have MATH . In view of REF the functions MATH are NAME in the variable MATH, and in view of REF also MATH is NAME with respect to MATH. This allows to integrate by parts obtaining MATH where MATH . In particular, MATH . In view of REF , not only MATH, but also MATH. Finally, if MATH is a different basis then there must exist functions MATH, but MATH when restricted to any unstable manifold, such that MATH. Then the test vector field MATH can be written as MATH with MATH. Thus MATH which yields the same formula only with different functions MATH. |
nlin/0104031 | Let MATH. Without loss of generality we can assume that MATH. We must show that the measure MATH, locally conditioned to the unstable foliation, is absolutely continuous on MATH-almost every fibre with respect to the Riemannian measure on the fibre with a density of bounded variation. So let MATH be a small neighbourhood on which, as in the last proof, we can use the change of coordinates MATH from REF , and recall that MATH. Since the maps MATH for each fixed MATH are absolutely continuous with NAME close to MATH which are NAME in the variable MATH (see REF ), it suffices to prove the same property for the measure MATH and its conditional measures on the subspaces MATH. Hence we must show that there exists a constant MATH such that for each MATH with MATH, MATH, MATH where MATH. We will shortly see that this estimate implies for MATH-almost every MATH where MATH denotes the conditional measure of MATH on MATH. This implies the announced absolute continuity of MATH with respect to NAME measure on MATH with a density of bounded variation. To obtain REF from REF it is convenient to introduce some natural notations: let MATH be the MATH-algebra generated by the subspaces MATH; MATH the measure restricted to the MATH-algebra MATH (in our case this is simply a marginal of MATH); MATH be the MATH functions that are MATH measurable; and by MATH let us designate the expectation with respect to the measure MATH. Clearly, in this coordinate free language, MATH corresponds to the conditional expectation MATH. Finally, let us designate by MATH the norms in the spaces MATH. Then MATH where the third equality is due to the fact that the MATH measurable functions are simply functions of only one coordinate (hence the obvious denseness of MATH in the corresponding MATH), and the last inequality follows by REF . This proves that for each function MATH there exists a set MATH, MATH, such that if MATH, REF applies. Since MATH is separable, REF follows by standard approximation arguments. To conclude we are left with the proof of estimate REF . Let MATH. Because of the continuity properties of MATH the vector field MATH is of the type considered in the previous proof, and we define MATH as there. Then MATH. Since MATH is defined in terms of MATH (see REF ), we start with the following observation: For MATH recall from NAME REF that MATH for MATH where MATH was chosen such that MATH. Hence, applying identity REF repeatedly (for MATH), we get MATH where MATH denotes the distortion defined in REF and where MATH for all MATH because MATH and MATH. In the rest of the proof we will, for simplifying the estimate, assume that MATH and MATH so that each integer MATH is trivially a multiple of MATH and write MATH instead of MATH. Observing REF , it follows that MATH where, for interchanging the sum and the limit, we made use of the fact that MATH . Since MATH with a constant that depends only on the Jacobian MATH of MATH and on the vector fields MATH, it follows that the first integral is bounded by MATH. (Observe that MATH has bounded density with respect to MATH and that MATH is MATH-invariant.) For the second integral the corresponding bound follows from the boundedness of the MATH. This proves estimate REF and concludes the proof of REF . |
nlin/0104031 | As usual it suffices to consider MATH. By definition there are MATH such that MATH . Since MATH is dense in MATH, we may assume that MATH. Our first goal is to estimate MATH in terms of MATH. To this end we approximate a given MATH by functions MATH, MATH, MATH, which are such that MATH for some MATH and MATH. Hence MATH so that, letting first MATH and then MATH, we see that MATH. Next we have to estimate MATH in terms of MATH. We start with MATH so that MATH for some MATH. Then MATH from which MATH. If we can extend this estimate to all MATH, it follows that MATH and hence MATH. As we will have to work in local coordinates we end up with a weaker estimate MATH where MATH is a constant determined by a suitable smooth partition of unity, compare REF and the remarks thereafter. Since we are in dimension MATH, there is a local MATH change of coordinates that straightens both foliations simultaneously. Therefore it suffices to show that each vector field MATH with MATH can be approximated in the MATH-sense by MATH-vector fields MATH with MATH. But this can be done by smoothening MATH with convolution kernels in both coordinate directions independently. |
nlin/0104031 | Considering instead of MATH a sufficiently high iterate, we may assume that the constant MATH in REF , that is, that MATH expands and contracts strictly along the unstable respectively, stable leaves. Let MATH. As already mentioned MATH is just the Jacobian of the holonomy between MATH and MATH. Calling MATH the tangent space to MATH, and setting MATH, the following formula holds, CITE, MATH where, for convenience, we have set MATH and MATH. Since MATH and MATH belong to the same unstable manifold it follows that MATH. Thus the problem of estimating the size of the Jacobian is reduced to the one of understanding the ``distance" between the tangent spaces MATH and MATH. This requires a little preparation. First of all, as usual in the construction of stable and unstable manifolds, it is convenient to choose appropriate coordinates in the tangent spaces at the points MATH. We will choose coordinates such that MATH is the unstable subspace and MATH the stable one. Clearly, by using such adapted coordinates, the maps MATH are all represented by matrices of the form MATH where MATH and MATH. Since we want to compare nearby matrices it is convenient to use coordinates related to MATH also for the quantities associated to MATH. To do so just use the exponential map to extend the above coordinates in the tangent space of MATH to coordinates in a neighbourhood of MATH. Clearly, such neighbourhoods can be chosen all of uniform size, hence if MATH is small enough, MATH always belongs to the corresponding neighbourhood. Since the above coordinates are uniformly smooth and the map MATH is MATH it follows that the derivatives MATH are all represented by matrices of the form MATH where MATH. Using the above coordinates the spaces MATH and MATH can be represented by MATH matrices, namely MATH and MATH. The basic estimate on the above subspaces is given in the next lemma the proof of which is postponed to the end of the appendix. There exists a constant MATH such that MATH . By this sub-lemma it follows that there exists a constant MATH such that MATH from which the first statement of REF follows. In addition, since each term of the sum in REF is clearly MATH, the above estimate on the NAME constant is indeed an estimate on the corresponding derivative. In order to show MATH-Hölder continuity of this derivative we note first the formula MATH where MATH. Since the maps MATH are all uniformly invertible (because of hyperbolicity), all the derivatives appearing in the above formula are well defined. Varying MATH the points MATH move along the unstable manifold, by definition. It follows that MATH and, together with NAME REF, this implies that the above series converges exponentially fast. Next, let MATH, MATH, MATH and, as before, MATH, MATH. Accordingly MATH . Now each term in REF is clearly NAME with respect to the variables MATH with NAME constant bounded by MATH. It follows MATH where we have chosen MATH appropriately. |
nlin/0104031 | To start we need to obtain a recursive equation for the MATH. Since MATH it follows that the space MATH is sent into the space MATH with MATH, that is, MATH. Accordingly, MATH where MATH. From this follows immediately (recall that we assumed MATH) MATH and, observing REF , MATH . |
nlin/0104070 | We first show that MATH defined in REF is a NAME transform of MATH and of MATH. To do that we have to check that the function MATH, which due to REF must be of the form MATH does satisfy REF . This can be verified directly using REF applied to MATH and MATH. Now, since MATH exists and describes an asymptotic lattice, it is enough to show that REF with the correct constant and sign apply also for the pair (MATH, MATH) and for the pair (MATH, MATH), what can be done by direct calculation using REF. |
quant-ph/0104100 | Given a quantum MATH cell probe solution to the static data structure problem MATH, we can get a MATH safe coinless quantum protocol for the corresponding communication problem by just simulating the cell probe solution. If in addition, the query scheme is address-only, the messages from NAME to NAME need consist only of the `address' part. This can be seen as follows. Let the state vector of the data qubits before the MATH-th query be MATH. MATH is independent of the query element and the stored data. NAME keeps MATH special ancilla registers in states MATH at the start of the protocol MATH. These special ancilla registers are in tensor with the rest of the qubits of NAME and NAME at the start of MATH. Protocol MATH simulates the cell probe solution, but with the following modification. To simulate the MATH-th query of the cell probe solution, NAME prepares her `address' and `data' qubits as in the query scheme, but sends the `address' qubits only. NAME treats those `address' qubits together with MATH in the MATH-th special ancilla register as NAME 's query, and performs the oracle table transformation on them. He then sends these qubits (both the `address' as well as the MATH-th special register qubits) to NAME. NAME exchanges the contents of the MATH-th special register with her `data' qubits (that is, exchanges the basis states), and proceeds with the simulation of the query scheme. This gives us a MATH safe coinless quantum protocol with the same error probability as that of the cell probe query scheme. |
quant-ph/0104100 | By REF , MATH. But since the density matrix of MATH is independent of MATH, MATH. Hence, by again using REF , we get that MATH. |
quant-ph/0104100 | The proof is by induction on MATH, using REF repeatedly. We also use the fact that MATH for MATH, since MATH are independent classical random variables. |
quant-ph/0104100 | Let MATH be the density matrix of MATH when MATH. Let MATH be the (marginal) probability that MATH and MATH the (conditional) probability that MATH given MATH. Define MATH. We now have MATH . |
quant-ph/0104100 | Suppose the given protocol for MATH has worst case error MATH. Define MATH. To prove the quantum round elimination lemma it suffices to give, by the harder direction of the minimax lemma REF , for any probability distribution MATH on MATH, a MATH safe public coin quantum protocol MATH for MATH with average distributional error MATH. To this end, we will first construct a probability distribution MATH on MATH as follows. Choose MATH uniformly at random. Choose independently, for each MATH, MATH according to distribution MATH. Set MATH and throw away MATH. By the easier direction of the minimax lemma REF , we get a MATH safe coinless quantum protocol MATH for MATH with distributional error, MATH. In MATH, NAME gets MATH, NAME gets MATH, MATH and MATH. We shall construct the desired protocol MATH from the protocol MATH. Let MATH be the first message of NAME in MATH. By the definition of a safe protocol, MATH has two parts: MATH qubits long, and the ``safe" overhead MATH, MATH qubits long. Let the input to NAME be denoted by the classical random variable MATH where MATH is the classical random variable corresponding to the MATH-th input to NAME. Let the classical random variable MATH denote the input MATH of NAME. Define MATH to be the average error of MATH under distribution MATH when MATH is fixed and MATH are fixed to MATH. Using REF and the fact that under distribution MATH, MATH are independent classical random variables, we get that MATH . Also MATH . The expectations above are under distribution MATH. For any MATH, MATH, define the MATH safe coinless quantum protocol MATH for the function MATH as follows. NAME is given MATH and NAME is given MATH. NAME sets MATH to the given value, and both NAME and NAME set MATH to the values MATH. NAME puts an independent copy of a pure state MATH (defined below) for each of the inputs MATH. NAME sets MATH and NAME sets MATH. Then they run protocol MATH on these inputs. Here MATH, where MATH is the (marginal) probability of MATH under distribution MATH. Since MATH is a safe coinless quantum protocol, so is MATH. Because MATH is a secure protocol, the probability that MATH makes an error for an input MATH, MATH, is the average probability of error of MATH under distribution MATH when MATH is fixed to the given value, MATH are fixed to MATH, and MATH are fixed to MATH. Hence, the average probability of error of MATH under distribution MATH . Let MATH denote the first message of MATH and MATH denote the register MATH holding the input MATH to NAME. Because of the ``secureness" of MATH, the density matrix of MATH in protocol MATH is the same as the density matrix of MATH in protocol MATH when MATH are set to MATH. Hence MATH . Using REF , we get a MATH safe coinless quantum protocol MATH for MATH with MATH . We now construct a MATH safe public coin quantum protocol MATH for MATH, which is nothing but a probability distribution (under MATH) over the safe coinless quantum protocols MATH, MATH, MATH. For protocol MATH, we get (note that the expectations below are under distribution MATH) MATH . The first inequality follows from REF , the second inequality follows from the concavity of the fourth root function and the last inequality from from REF . This completes the proof of the quantum round elimination lemma. |
quant-ph/0104100 | The proof is similar to that of REF , but using REF instead of REF . Suppose the given protocol for MATH has worst case error MATH. Define MATH. To prove the classical round elimination lemma it suffices to give, by the harder direction of the minimax lemma REF , for any probability distribution MATH on MATH, a MATH public coin classical randomised protocol MATH for MATH with average distributional error MATH. To this end, we construct the probability distribution MATH on MATH as before. By the easier direction of the minimax lemma REF , we get a MATH classical deterministic protocol MATH for MATH with distributional error, MATH. In MATH, NAME gets MATH, NAME gets MATH, MATH and a copy of MATH. We shall construct the desired protocol MATH from the protocol MATH. Let MATH be the first message of NAME in MATH. Let the input to NAME be denoted by the classical random variable MATH where MATH is the classical random variable corresponding to the MATH-th input to NAME. Let the classical random variable MATH denote the input MATH of NAME. Define MATH to be the average error of MATH under distribution MATH when MATH is fixed and MATH are fixed to MATH. Arguing as before, we get MATH . Also MATH . The expectations above are under distribution MATH. For any MATH, MATH, define the MATH private coin classical randomised protocol MATH for the function MATH as follows. NAME is given MATH and NAME is given MATH. NAME sets MATH to the given value, and both NAME and NAME set MATH to the values MATH. NAME tosses her private coin to choose MATH, where each MATH is chosen independently according to the (marginal) distribution on MATH induced by MATH. NAME sets MATH and NAME sets MATH. Then they run protocol MATH on these inputs. The probability that MATH makes an error for an input MATH, MATH, is the average probability of error of MATH under distribution MATH when MATH is fixed to the given value, MATH are fixed to MATH, and MATH are fixed to MATH. Hence, the average probability of error of MATH under distribution MATH . Let MATH denote the first message of MATH and MATH denote the register MATH holding the input MATH to NAME. Then MATH . Using REF and arguing as before, we can complete the proof of the classical round elimination lemma. |
quant-ph/0104100 | Consider the static rank parity data structure problem where the storage scheme has to store a set MATH, MATH, and the query scheme, given a query MATH, has to decide whether the rank of MATH in MATH is odd or even. CITE have shown the existence of two-level perfect hash tables containing, for each member MATH of the stored subset MATH, MATH's rank in MATH, and using MATH cells of word size MATH and requiring only MATH classical deterministic cell probes. Combining a MATH address-only quantum cell probe solution to the static predecessor problem with such a perfect hash table, gives us a MATH address-only quantum cell probe solution to the static rank parity problem. The error probability of the cell probe scheme for the rank parity problem is the same as the error probability of the cell probe scheme for the predecessor problem. By REF , we get a MATH safe coinless quantum protocol for the rank parity communication game MATH. The error probability of the communication protocol is the same as that of the cell probe scheme for the predecessor problem. |
quant-ph/0104100 | Consider the problem MATH. NAME, who is given MATH, computes the concatenation MATH. NAME, who is given MATH, MATH and MATH, computes MATH NAME and NAME then run the protocol for MATH on the inputs MATH, MATH to solve the problem MATH. |
quant-ph/0104100 | Consider the problem MATH. NAME, given MATH and MATH, computes MATH. NAME, given MATH, computes the sets MATH where MATH . Above, the integers MATH are to be thought of as bit strings of length MATH. NAME also computes MATH. NAME and NAME then run the protocol for MATH on inputs MATH, MATH to solve the problem MATH. |
quant-ph/0104100 | We basically imitate the proof of CITE, but in our quantum setting. By REF , it suffices to consider communication protocols for the rank parity communication game MATH. Let MATH. Let MATH. For any given constants MATH, define MATH . We shall show that the rank parity communication game MATH does not have bounded error MATH safe public coin quantum protocols, thus proving the desired lower bounds on the query complexity of static rank parity (and hence, static predecessor) by REF . Given a MATH safe public coin quantum protocol for MATH with error probability MATH (MATH), we get a MATH safe public coin quantum protocol for MATH with the same error probability MATH, by REF . Using the quantum round elimination lemma REF , we get a MATH safe public coin quantum protocol for MATH but the error probability increases to at most MATH. Using the reduction of REF , we get a MATH safe public coin quantum protocol for MATH with error probability at most MATH. From the given values of the parameters, we see that MATH . This implies that we also have a MATH safe public coin quantum protocol for MATH with error probability at most MATH. Using the quantum round elimination lemma REF again, we get a MATH safe public coin quantum protocol for MATH but the error probability increases to at most MATH. We do the above steps repeatedly. After applying the above steps MATH times, we get a MATH safe public coin quantum protocol for MATH with error probability at most MATH. By applying the above steps MATH times, we finally get a MATH safe public coin quantum protocol for MATH with error probability at most MATH. From the given values of the parameters, we see that MATH . Thus we get a zero round protocol for a rank parity problem on a non-trivial domain with error probability less than MATH, which is a contradiction. In the above proof, we are tacitly ignoring ``rounding off" problems. We remark that this does not affect the correctness of the proof. |
quant-ph/0104100 | We recall the following reduction from MATH to MATH (see CITE): In MATH, NAME is given MATH, NAME is given MATH, MATH, and copies of MATH, and they have to communicate and decide if MATH. To reduce MATH to MATH, NAME constructs MATH by concatenating MATH, NAME constructs MATH by concatenating MATH. It is easy to see that MATH iff MATH. Suppose MATH has a MATH safe public coin quantum protocol with worst case error probability less than MATH. Suppose MATH where MATH. For MATH, define MATH . Also define MATH and MATH. Then MATH and MATH . We now apply the above self-reduction and the quantum round elimination lemma REF alternately. Before the MATH-th stage, we have a MATH safe public coin quantum protocol for MATH with worst case error probability less than MATH. Here MATH if MATH is odd, MATH otherwise. For the MATH-th stage, we apply the self-reduction with MATH. This gives us a MATH safe public coin quantum protocol for MATH with the same error probability. We now apply the quantum round elimination lemma REF to get a MATH safe public coin quantum protocol for MATH with worst case error probability less than MATH. Here MATH if MATH and MATH if MATH. This completes the MATH-th stage. Applying the self-reduction and the round elimination lemma alternately MATH times gives us a zero round quantum protocol for the `greater-than' problem on a domain of size MATH with worst case error probability less than MATH, which is a contradiction. In the above proof, we are tacitly ignoring ``rounding off" problems. We remark that this does not affect the correctness of the proof. This proves the quantum lower bound of MATH on the message complexity. Using the classical round elimination lemma REF instead of the quantum one, and treating a classical randomised protocol with complexity MATH as a MATH protocol, we get the stronger classical lower bound of MATH. |
quant-ph/0104100 | The proof follows by modifying NAME 's lower bound proof for quantum ordered searching CITE. There, it was shown that if MATH is stored in sorted order in a table MATH then, given any query element MATH, MATH probes are required by any quantum search strategy to find out the smallest index MATH, MATH, such that MATH. We observe that the lemma above does not follow directly from the result of NAME, since we only need to decide if MATH is present in the table or not, and this is a weaker requirement. To prove the lemma, we follow the adversary strategy of CITE with some minor changes. We study the behaviour of the quantum query scheme with query element MATH. The proof of NAME is based on a clever strategy of subdividing ``intervals" (an interval is a contiguous set of locations in the sorted table). We work instead with ``logical intervals", where a logical interval denotes the set of locations in the table where elements contiguous in the natural ordering are stored (as determined by the fixed storing order). After this definition, one can easily show that the same subdivision strategy as in CITE goes through. In NAME 's proof, the adversary constructs inputs by padding with zeros from the beginning up to the left of an interval, and with ones from the end up to the right of the interval. Instead, we pad with small numbers REF from the logical beginning up to the logical left of a logical interval, and with large numbers REF from the logical end up to the logical right of the logical interval. We store the appropriate `pointer values' in the `pointer locations' (predetermined by the storing strategy). After doing this, one can easily show that the same error analysis of CITE goes through. Thus, the adversary finally can produce two inputs, one of them containing MATH and the other not, such that the behaviour of the query scheme is very similar on both. This is a contradiction. |
quant-ph/0104100 | Our proof follows from the NAME theoretic arguments of CITE together with REF . The details are omitted. |
quant-ph/0104100 | We first give an overview of the plan of the proof, before getting down to the details. The proof proceeds in stages. CASE: Starting from MATH, we construct a MATH private coin protocol MATH, where the first message is independent of NAME 's input, and MATH. The important idea in this step is to first generate NAME 's message using a new private coin without ``looking" at her input, and after that, to adjust NAME 's old private coin in a suitable manner so as to be consistent with her message and input. CASE: Suppose the coin tosses in MATH were done in public. Then NAME can generate the first message of MATH himself, as it is independent of NAME 's input. Doing this gives us a MATH public coin protocol MATH, such that MATH for every MATH. The protocol MATH of s our desired MATH public coin classical randomised protocol for MATH. We have MATH . We now give the details of the proof. Let MATH be the probability distribution of the first message MATH of protocol MATH when NAME 's input MATH. Let MATH denote NAME 's input register. Define MATH, where MATH is the (marginal) probability of MATH under distribution MATH. MATH is the probability distribution of the average first message under distribution MATH. By REF , we get that MATH . For MATH and an instance MATH of the first message of NAME, let MATH denote the (conditional) probability that the private coin toss of NAME results in MATH, given that NAME 's input is MATH and her first message in protocol MATH is MATH. Let MATH denote the probability that the first message of NAME in MATH is MATH, given that her input is MATH. Let MATH denote the probability of MATH occurring in the average first message of NAME. Then, MATH. CASE: We construct a MATH private coin classical randomised protocol MATH for MATH with average error under distribution MATH, MATH, and where the probability distribution of the first message is independent of the input to NAME. Suppose NAME is given MATH and NAME is given MATH. NAME tosses a fresh private coin to pick MATH with probability MATH. NAME then sets her old private coin to MATH with probability MATH. (If in MATH, message MATH cannot occur when NAME 's input is MATH, we say that protocol MATH gives an error if such a thing happens.) After this, NAME and NAME behave as in protocol MATH (henceforth, NAME ignores the new private coin which she had tossed to generate her first message MATH). Hence in MATH, the probability distribution of the first message is independent of NAME 's input. Let us now compare the situations in protocols MATH and MATH when NAME 's input is MATH, NAME 's input is MATH, NAME has finished tossing her private coins, but no communication has taken place as yet. In protocol MATH, the probability that NAME 's private coin toss results in MATH is MATH . In protocol MATH, the probability that NAME 's (old) private coin toss results in MATH is MATH . Thus, the MATH distance between the probability distributions on NAME 's (old) private coin toss is MATH . Hence, the error probability of MATH on input MATH . Let MATH be the probability that MATH under distribution MATH. Then, the average error of MATH under distribution MATH, MATH, is bounded by MATH . The last inequality follows from the ``average encoding theorem" REF . CASE: We now construct our desired MATH public coin classical randomised protocol MATH for MATH with MATH. Suppose all the coin tosses of NAME and NAME in MATH were done publicly before any communication takes place. Now there is no need for the first message from NAME to NAME, because NAME can reconstruct the message by looking at the public coin tosses. This gives us the protocol MATH, and trivially MATH . This completes the proof of REF . |
quant-ph/0104100 | We first give an overview of the plan of the proof, before getting down to the details. The proof proceeds in stages. We remark on the similarities between the stages in the quantum proof, and the stages in the classical proof REF . Stages REFA and REFB of the quantum proof together correspond to f the classical proof, and Stages REFA and REFB of the quantum proof together correspond to f the classical proof. CASE: Starting from the MATH safe coinless protocol MATH, we construct a MATH safe coinless protocol MATH with MATH for every MATH. MATH contains an extra ``secure" copy of NAME 's input MATH, but is otherwise the same as MATH. CASE: Starting from MATH, we construct a MATH safe coinless protocol MATH, where the first message is independent of NAME 's input, and MATH. The important idea in this step is to first generate NAME 's average message (which is independent of her input), and after that, use the extra ``secure" copy of NAME 's input MATH to apply a unitary transformation MATH on some of her qubits without touching her message. MATH is used to adjust NAME 's state in a suitable manner so as to be consistent with her input and message. This ``adjustment" step requires the use of the ``local transition theorem" REF . CASE: Since in MATH the first message is independent of NAME 's input, NAME can generate it himself. But it is also necessary to achieve the correct entanglement between NAME 's qubits and the first message. NAME does this by first sending a safe message of MATH qubits. NAME then applies a unitary transformation MATH on some of her qubits, using the extra ``secure" copy of her input MATH, to achieve the correct entanglement. The existence of such a MATH follows from REF . Doing all this gives us a MATH safe coinless protocol MATH, such that MATH for every MATH. CASE: Since the first message of NAME in MATH is zero qubits long, NAME can concatenate his first two messages, giving us a MATH safe coinless protocol MATH, such that MATH for every MATH. The technical reason behind this is that unitary transformations on disjoint sets of qubits commute. The protocol MATH of s our desired MATH safe coinless quantum protocol for MATH. We have MATH . We now give the details of the proof. Let MATH be the density matrix of the first message MATH of protocol MATH when NAME 's input MATH. Let MATH denote NAME 's input register. Define MATH, where MATH is the (marginal) probability of MATH under distribution MATH. MATH is the density matrix of the average first message under distribution MATH. By the ``secureness" of MATH, MATH is also the density matrix of the first message when MATH is fed to NAME 's input register MATH, where MATH. By REF , we get that MATH CASE: We first construct a MATH safe coinless quantum protocol MATH for MATH such that MATH, for every MATH. Let MATH be NAME 's input register in MATH. In MATH, NAME has an additional register MATH, and the input MATH to NAME is fed to register MATH, instead of MATH. MATH is initialised to MATH in MATH. In protocol MATH, NAME first copies the contents of MATH to MATH. After that, things in MATH proceed as in MATH. Register MATH is not touched henceforth, and thus, MATH holds an extra ``secure" copy of MATH throughout the run of protocol MATH. CASE: We now construct a MATH safe coinless quantum protocol MATH for MATH with average error under distribution MATH, MATH, and where the density matrix of the first message is independent of the input MATH to NAME. NAME is given MATH and NAME is given MATH. Consider the situation in MATH after the first message has been prepared by NAME, but before it is sent to NAME. Let register MATH denote NAME 's qubits excluding the message qubits MATH and the qubits of the ``secure" copy MATH (in particular, MATH includes the qubits of register MATH). Without loss of generality, one can assume that register MATH has at least MATH qubits, because one can initially pad up MATH with ancilla qubits set to MATH. Let MATH be the state vector of MATH in MATH at this point, where the subscripts denote the registers. MATH is a purification of MATH. We note that MATH is also the state vector of MATH in protocol MATH at this point. MATH is similar to MATH except for the following. NAME puts MATH in register MATH (instead of copying MATH to MATH as in MATH) to create the first message in register MATH with density matrix MATH. MATH now contains a purification MATH of MATH. Then NAME applies a unitary transformation MATH depending upon MATH (which is available ``securely" in register MATH) on MATH, so that MATH is ``close" to MATH. Here MATH stands for the identity transformation on MATH. REF tells us that there exists a unitary transformation MATH on MATH such that MATH . Thus, MATH is the state vector of MATH in MATH after the application of MATH. NAME then sends register MATH to NAME and after this, NAME and NAME behave as in MATH. Application of MATH does not affect the density matrix of register MATH, which continues to be MATH. Hence in MATH, the density matrix of the first message is independent of NAME 's input. Let us now compare the situations in protocols MATH and MATH when NAME 's input is MATH, NAME 's input is MATH, NAME has prepared her first message, but no communication has taken place as yet. At this point, in both protocols MATH and MATH, the state vector of NAME 's qubits is the same, and in tensor with the state vector of NAME 's qubits. Let MATH denote the register of NAME 's qubits (including his input qubits MATH) and let MATH denote the state vector of MATH at this point. Hence the global state of protocol MATH at this point is MATH, and the global state of MATH is MATH. Therefore, the global states of protocols MATH and MATH at this point differ in trace distance by the quantity MATH . Using REF , we see that the error probability of MATH on input MATH . Let MATH be the probability that MATH under distribution MATH. Then, the average error of MATH under distribution MATH, MATH, is bounded by MATH . For the second inequality above, we use the concavity of the square root function. The last inequality follows from the ``average encoding theorem" REF . CASE: We now construct a MATH safe coinless quantum protocol MATH for MATH with MATH, for all MATH. NAME is given MATH and NAME is given MATH. The protocol MATH will be constructed from MATH. The input MATH is fed to register MATH of NAME, and the input MATH is fed to register MATH of NAME. Let register MATH denote all the qubits of register MATH, except the last MATH qubits. In protocol MATH the registers initially in NAME 's possession are MATH and MATH, and the registers initially in NAME 's possession are MATH, MATH, and a new register MATH, where MATH is MATH qubits long. The qubits of MATH are initially set to MATH. NAME first prepares the state vector MATH in register MATH as in protocol MATH. He then constructs a canonical purification of MATH in registers MATH. The density matrix of MATH is MATH. NAME then sends MATH to NAME. The density matrix of MATH is independent of the inputs MATH (in fact, if the canonical purification in MATH is the NAME purification, then the density matrix of MATH is also MATH). After receiving MATH, NAME treats MATH as the register MATH in the remainder of the protocol. MATH now contains a purification of MATH. NAME applies a unitary transformation MATH depending upon MATH (which is available ``securely" in register MATH) on MATH, so that the state vector of MATH becomes MATH. The existence of such a MATH follows from REF . At this point, the global state vector (over all the qubits of NAME and NAME) in MATH is the same as the global state vector in MATH viz. MATH. NAME now treats register MATH as if it were the first message of NAME in MATH, and proceeds to compute his response MATH of length MATH. NAME sends MATH to NAME and after this protocol MATH proceeds as in MATH. In MATH . NAME starts the communication, the communication goes on for MATH rounds, the first message of NAME of length MATH (that is, register MATH) is a safe message, and the first message of NAME is zero qubits long. CASE: We finally construct a MATH safe coinless quantum protocol MATH for MATH with MATH, for all MATH. In protocol MATH, NAME (after doing the same computations as in MATH) first sends as a single message register MATH of length MATH, and after that NAME applies MATH on MATH followed by her appropriate unitary transformation on MATH (the unitary transformation of NAME in MATH on her qubits MATH after she has received the first two messages of NAME). At this point, the global state vector (over all the qubits of NAME and NAME) in MATH is the same as the global state vector in MATH, since unitary transformations on disjoint sets of qubits commute. After this, things in MATH proceed as in MATH. In protocol MATH . NAME starts the communication, the communication goes on for MATH rounds, and the first message of NAME of length MATH contains a safe overhead (the register MATH) of MATH qubits. This completes the proof of REF . |
cs/0105013 | Processor MATH REF that copies from the output register of MATH must continually rewrite its output register for MATH - otherwise there can be a deadlock where the value written by MATH is different from the value MATH reads from MATH. Similarly, MATH must repeatedly write, otherwise there can be a deadlock where all the registers have the same value, and the MATH program counter is past the first write to its register that incremented this value. Therefore, processors continually write into their output registers. Since all processors repeatedly write their output registers, we can construct an execution where reads are concurrent with writes and obtain arbitrary values. This construction can be used to show that the protocol does not converge (and also that it is not stable). |
cs/0105013 | There are two proof obligations, stability (closure) from legitimate configurations and convergence from arbitrary configurations to legitimate ones. Closure. It is straightforward to verify that in any processor cycle from a legitimate configuration, a processor writes to at most two registers as it changes the counter value. Thus when the neighbor reads these registers, at most two reads can have incorrect values due to concurrent writing. If both have correct values, the token passes correctly (a subsequent read by the process can still obtain an incorrect value, but only by getting REF for all reads, which causes no harm). If both have incorrect values, then the reader observes no change in counter values. If just one returns an incorrect value, then the reader observes parity of zero, which is harmless. This reasoning shows that the protocol is stable. Convergence. The remaining task is to verify that the protocol guarantees to reach a legitimate configuration in any execution. Suppose all processors have completed at least one cycle of REF. In the subsequent execution, a processor only writes a register if that register requires change to agree with the processor's counter. Note that by standard arguments, no deadlock is possible in this system and that MATH increments its counter infinitely many times in an execution. It is still possible that one processor can read more than two incorrect values due to concurrent writes (consider an initial state with many counter values; as these values are propagated to some MATH, it could be that MATH happens to read many registers concurrent with MATH writing to them). Since the counter range is MATH and there are MATH processors, it follows that at least one counter value MATH is not present in the system. By the arguments given for the proof of closure, no processor incorrectly reads input registers to get the value MATH in such a configuration. Because MATH increments MATH infinitely, we can suppose MATH but no other processor or register encoding equals MATH, and by standard arguments (and the propagation of values observed in the proof of closure), a legitimate configuration eventually is reached. |
cs/0105013 | The closure argument is the same as given in the proof of REF , inspecting each of the four cases of reading overlapping with writing of the two bits that change when a processor changes its counter and writes the one new NAME code bit and the parity bit. In each case, the neighbor processor either reads the old value, or ignores the values it reads (because parity is incorrect), or obtains the new counter value. The change from old to new counter value is essentially atomic. Proof of convergence requires new arguments. Consider some configuration of an execution prior to which each processor has completed at least two cycles of REF , so that output registers agree with counter values (unless the processor has read a new value and updated its counter). Observe that thereafter, if processor MATH successively reads two different NAME code values from its input registers, each with correct parity, then MATH concurrently wrote at least once to its output registers. Moreover, if MATH successively reads MATH different NAME code values with correct parity, then MATH wrote at least MATH times a new counter value and read at least MATH times from its own input registers, by the structure of the loop REF . A consequence of these observations is that if MATH successively reads MATH different counter values with correct parity, then MATH wrote at least one new counter value in the same period. In particular, if MATH successively reads MATH different counter values, then we may assert that MATH read MATH's output registers and wrote a new counter in the same period. By the standard argument refuting deadlock, processor MATH increments its counter infinitely often in any execution. Therefore we can consider an execution suffix starting with MATH. In the typical reflected NAME code, the high-order bit starting from MATH does not change until the counter has incremented MATH times. Therefore, until MATH has incremented MATH at least MATH times, any read by MATH obtains a value with zero in the high-order bit. The observations above imply that, before MATH changes at the high-order bit, each processor has copied some counter value obtained via MATH - such counter values may be inaccurate due to reads overlapping writes or more than one write (bit change) for one scan of a set of registers, however the value for the high-order bit stabilizes to zero in this execution fragment. In a configuration where no counter or register set has REF in the high-order bit, the event of MATH changing the high-order bit creates a unique occurrence of REF in that position. Since MATH does not again change its counter until observing the same value from MATH, convergence is guaranteed. |
cs/0105015 | Both hyper-arc consistency and range consistency verify all values of all domains. But hyper-arc consistency verifies the constraints with respect to the exact domains MATH, while range consistency verifies the constraints with respect to intervals that include the domains: MATH. A constraint that holds on a domain MATH also holds on the interval MATH since MATH. The converse is not true, see REF . Hence MATH. Both range consistency and bound consistency verify the constraints with respect to intervals that include the domains. But bound consistency only considers MATH and MATH for a domain MATH, while range consistency considers all values in MATH. Since MATH, MATH. REF shows that MATH cannot be discarded. |
cs/0105015 | Since the definition of arc consistency and hyper-arc consistency is equivalent for binary constraints, we only need to consider the filtering of the alldifferentconstraint. Consider the constraint MATH: MATH and the corresponding decomposition in terms of disequalities, denoted by MATH. If a value MATH is not arc consistent with respect to the set MATH, then it is also not hyper-arc consistent with respect to MATH. Indeed, when MATH is not arc consistent with respect to MATH, then we cannot find a MATH for some variable MATH such that MATH. But then we also cannot find a MATH-uple MATH, since we cannot find a value MATH such that MATH. Therefore, MATH. The converse is not true, as illustrated in REF . |
cs/0105015 | By construction we have that the alldifferent constraint is equivalent to the simultaneous consideration of the sequence of corresponding disequalities. The number of disequalities is precisely MATH. If we consider only MATH disequalities simultaneously (MATH), there are MATH unconstrained relations between variables, and the corresponding variables could take the same value when a certain instantiation is considered. Therefore, we really need to take all MATH constraints into consideration, which corresponds to the relational MATH-consistency. |
cs/0105017 | Assume that we are trying to find a vertex MATH in zonotope MATH extreme in direction MATH, that is, that maximizes the dot product MATH. Assume that MATH is the NAME sum of line segments of the form MATH, where MATH. We simply set each MATH independently to MATH or MATH, depending upon whether the projection of MATH onto MATH is negative or positive. |
cs/0105017 | Assume that we are trying to find a vertex MATH in zonotope MATH extreme in direction MATH. Order the MATH's with MATH according to their projection onto vector MATH, breaking ties arbitrarily. In decreasing order by projection along MATH, set the corresponding MATH's to MATH until doing so would violate the constraint that MATH. Set MATH for this ``transitional" vector to the maximum value allowed by this constraint, and finally set the remaining MATH's to MATH. Then MATH maximizes MATH. |
cs/0105017 | As in CITE, consider the polytope MATH that is the NAME sum of MATH and MATH, that is, MATH. We are trying to minimize over MATH the convex quadratic objective function MATH, that is, the length of a line segment between MATH and MATH. For a given direction MATH, we can find the solution MATH to the linear optimization problem for MATH by using REF to find the MATH optimizing MATH over MATH and the MATH optimizing MATH over MATH. Now given a point MATH, we can use this observation and the polynomial-time equivalence between separation and optimization to solve the separation problem for MATH and MATH in time linear in MATH and polynomial in MATH and MATH. We can use this solution to the separation problem for MATH as a subroutine for the ellipsoid method (see CITE) in order to optimize MATH over MATH. Given an optimizing choice of MATH, it is easy to find the best pair of parallel hyperplanes and a decision boundary, either the MATH-SVM decision boundary or some other reasonable choice within the parallel family. |
cs/0105017 | Given a point MATH and zonotope MATH, MATH, we can use REF and the polynomial-time equivalence between separation and optimization to solve the separation problem for MATH and MATH in time linear in MATH and polynomial in MATH, MATH and MATH. We can solve the separation problem for the intersection of zonotopes MATH simply by solving it separately for each zonotope. We now use the equivalence between separation and optimization in the other direction to conclude that we can also solve the optimization problem for an intersection of zonotopes. |
cs/0105024 | CASE: Suppose MATH is closed under MATH. Then all values in domains of variables in MATH are supported. CASE: Take some MATH. MATH is closed under (arr-MATH), thus also MATH. Then there exists some MATH with MATH. This index and MATH support MATH. CASE: For some MATH consider the necessarily failing condition of (arr-MATH). Thus a value MATH exists in both MATH and MATH, for some MATH with MATH. Assigning the MATH to the MATH, and MATH and MATH, is a solution supporting MATH. CASE: Consider a value MATH, and the following cases: CASE: MATH . The index MATH can not select the variable MATH; however, MATH remains satisfiable. Therefore, there is a solution for MATH that is indifferent to the value of MATH, and so supports MATH. MATH . Here the condition of (arr-MATH) is fulfilled, its consequence holds and with it MATH. A supporting solution is therefore MATH, MATH, MATH for all MATH. REF some MATH contains more than one element Consider some index MATH with MATH that also fulfills MATH. Such an index exists because otherwise (arr-MATH) would be applicable. Choose a MATH and instantiate MATH, MATH, MATH for all MATH. This solution to MATH does not assign to MATH and hence supports MATH. CASE: Suppose here that MATH is not closed under MATH. Then domains of some variables in MATH contain unsupported values. CASE: Assume (arr-MATH) is applicable, that is, MATH. Then there is some value MATH for all MATH. Clearly, MATH is not part of any solution. CASE: Suppose some MATH could be removed by (arr-MATH). From the condition of (arr-MATH) it follows that with MATH no index MATH can be found that allows the same value for MATH and MATH. CASE: For a singleton index domain and so a possible application of (arr-MATH)consider MATH with MATH. Such a MATH can not be supported by MATH. |
cs/0105024 | Suppose MATH and MATH are arc-consistent. Any solution for MATH assigns a value to MATH that is also supported by a solution to MATH, and vice versa. Due to the conditions on variables, such solutions do not assign to the same variables. Therefore there union is also a solution for MATH. Thus, a supporting solution for any domain value of a variable in MATH, and MATH, can be extended to a supporting solution for MATH. Hence, MATH is arc-consistent. |
cs/0105027 | The NAME assumption in pomdps warrants that MATH . MATH is the probability of the part of the history, dependent on the environment, that is unknown to the agent and can be only sampled. MATH is the probability of the part of the history, dependent on the agent, that is known to the agent and can be computed (and differentiated). Therefore we can compute MATH . |
cs/0105027 | In our setup, MATH, and MATH; and MATH by REF . According to REF MATH. So we can use NAME 's inequality and we get the following deviation bound for a policy MATH: MATH . After solving for MATH, we get the statement of REF . |
cs/0105027 | Given a class of policies MATH with finite covering number MATH, the upper bound MATH on the likelihood ratio, and MATH, MATH . Note the relationship to REF . The only essential difference is in the covering number, which takes into account the extension from a single policy MATH to the class MATH. This requires the sample size MATH to increase accordingly to achieve the given confidence level. The derivation is similar to uniform convergence result of CITE(see pages REF), using NAME 's inequality instead of NAME 's. Solving for MATH gives us the statement of the theorem. |
cs/0105029 | (of REF for MATH) Let MATH be a vector REF-coloring of MATH. Let MATH be the average degree of MATH. Let MATH. (This is slightly different from the choice made by CITE. They choose MATH. It is the only change that we make to their algorithm.) Choose a random vector MATH according to the standard MATH-dimensional normal distribution. Let MATH. Let MATH be size of MATH and let MATH be the number of edges contained in MATH. An independent set MATH of size MATH is then easily obtained by removing one vertex from each edge contained in MATH. We show that the expected size of MATH is MATH. Let MATH, where MATH, denote the tail of the standard normal distribution. It is well known that MATH, for every MATH. It is also known that if MATH is an arbitrary unit vector in MATH, and MATH is a random vector chosen according to the standard MATH-dimensional normal distribution, then the inner product MATH is distributed according to the standard one dimensional normal distribution. Furthermore, if MATH and MATH are orthogonal unit vectors then the two random variables MATH and MATH are independent. It is easy to see, then, that: MATH where MATH and MATH are two unit vectors such that MATH, and MATH and MATH, respectively, are the number of vertices and edges in the graph. It is not difficult to see that the probability MATH is a monotone increasing function of the angle between MATH and MATH. As we would like to obtain an upper bound on the probability, we may assume, therefore, that MATH. CITE argue that MATH where the rightmost equality follows from the fact that MATH is also a unit vector. We obtain a slightly sharper upper bound on this probability: If MATH and MATH are unit vectors such that MATH then MATH . Let MATH and MATH be two unit vectors such that MATH. Note that MATH and MATH form an angle of MATH. Let MATH be the tip of MATH. Draw a line perpendicular to MATH that passes through MATH. Similarly, draw a line perpendicular to MATH that passes through the tip of MATH, as shown in REF . It is easy to see that these two lines intersect at the point MATH which is MATH. (This follows from the fact that MATH so that MATH and the fact that MATH. Note that MATH is also a unit vector.) The projection of a standard MATH-dimensional normal vector MATH on the plane spanned by MATH and MATH is a standard REF-dimensional normal vector which we denote by MATH. Note that MATH and MATH. The probability that we have to bound is therefore the probability that the random vector MATH falls into the wedge defined by the angle MATH. CITE bound this probability by the probability that MATH falls to the right of the vertical line that passes through MATH, which is MATH. Let MATH and MATH be unit vectors in the plane spanned by MATH and MATH such that the angle formed by them and MATH is MATH (see REF ). Draw a line through MATH which is perpendicular to MATH. Similarly, draw a line through MATH which is perpendicular to MATH. Let MATH be the point on the first line in the direction of MATH. A simple calculation shows that MATH. We bound the probability that MATH falls into the wedge formed by MATH by the probability that it falls into the wedge formed by MATH. This probability is just MATH. As MATH, the events MATH and MATH are independent. Thus, this probability is just MATH. Using a more complicated analysis, presented in REF, we can show that MATH. Thus, the bound given in REF is asymptotically tight. We are now back in the proof of REF . As MATH, we get that MATH . With MATH we have MATH and therefore MATH . Thus, MATH and the proof of the theorem (for MATH) is completed. |
cs/0105029 | Assume that MATH and consider the natural semidefinite programming relaxation of the maximum independent set problem: MATH . An almost optimal solution MATH of this semidefinite program can be found in polynomial time. As MATH is assumed to contain an independent set of size at least MATH, we may assume that MATH . (The MATH term comes from the fact that MATH is only an almost optimal solution of the program. We can make this term much smaller if we wish, but MATH is small enough for our purposes.) We now use the following simple facts: If MATH and MATH, for every MATH, then for any MATH, at least MATH of the MATH's satisfy MATH. Indeed, if the claim is not satisfied then MATH, a contradiction. It is easy to check that MATH, if MATH. Using this fact with MATH, MATH, and MATH, we get that for at least MATH of the vectors satisfy MATH. Thus, if MATH, then MATH. Let MATH and MATH be unit vectors such that MATH, MATH and MATH. Let MATH and MATH, respectively, be the normalized projections of MATH and MATH on the space orthogonal to MATH. Then MATH . Let MATH and MATH. Then MATH . (Recall that MATH is a unit vector so MATH.) . Thus, MATH . As MATH, we get that MATH, and the numerator of the expression given above for MATH can be simplified as follows: MATH and the claim follows. We continue now with the proof of REF . Recall that MATH, where MATH, and that MATH. We may assume that MATH, for every MATH. Otherwise, we can very slightly perturb MATH. (Recall that the vectors MATH form, in any case, only an almost optimal solution of the semidefinite program.) Suppose now that MATH and MATH. Thus MATH, MATH and MATH. Let MATH and MATH be the normalized projections of MATH and MATH on the space orthogonal to MATH. The expression given for MATH in REF is decreasing in both MATH and MATH. Thus, MATH . We obtained, therefore, a vector MATH-coloring of MATH. This completes the proof. |
cs/0105029 | The proof is by induction on MATH. Assume at first that MATH. It is easy to see that a graph is vector MATH-colorable, for some MATH, if and only if the graph contains no edges. Thus, MATH is an independent set of size MATH. Assume, therefore, that MATH. Let MATH be the maximum degree of MATH. We describe two ways of finding independent sets of MATH. Using REF , we can find, in polynomial time, an independent set of MATH of size MATH. Alternatively, let MATH be a vertex of MATH of degree MATH and let MATH be the set of its neighbors. It follows from REF that the subgraph MATH induced by MATH is vector MATH-colorable. By the induction hypothesis, we can find in MATH, in polynomial time, an independent set of size MATH. This independent set is also an independent set of MATH. Taking the larger of these two independent sets, we obtain an independent set of MATH of size MATH as required. It is easy to verify, by induction, that MATH, where MATH. We omit the straightforward details. This completes the proof of the lemma. |
cs/0105029 | Suppose that MATH contains an independent set of size MATH. By REF , we can find, in polynomial time, a subset MATH of size MATH and a vector MATH-coloring of MATH, where MATH. By REF , we can find, in polynomial time, an independent set of MATH of size MATH. As MATH, and MATH, we get that the size of this independent set, which is also an independent set of MATH, is MATH, as required. |
cs/0105029 | Let MATH. Consulting REF , we see that MATH where MATH and MATH . Moving to polar coordinates, we get that MATH where MATH is the region MATH expressed in polar coordinates. Using the sine theorem, we get that MATH and thus MATH . Putting all this together, we get that MATH . Next, we change the variable of integration. Let MATH, so that MATH. We get that MATH . Using integration by parts we finally get the concise formula: MATH . Let us now consider the integral MATH . Since MATH is an increasing function for MATH, we get that MATH . On the other hand, by integrating by parts, we get that MATH . Letting MATH, we get that MATH and thus MATH as claimed. |
cs/0105029 | Let MATH. It is easy to see that MATH and therefore, there is at least one MATH, such that MATH, and the claim follows. |
cs/0105029 | Assume, for contradiction, that REF does not hold for any red vertex. If we sum up over all red vertices, we get that MATH . Now, MATH . Combining this with REF and the NAME inequality, we get that MATH a contradiction. |
cs/0105029 | Let MATH, and let MATH, for MATH . By REF , with MATH and MATH, at least one such set MATH satisfies MATH . We now remove from the graph all the red vertices MATH, for which MATH is small. More formally, we remove all vertices MATH for which MATH. We let MATH be the remaining set of red vertices. It is easy to see, by REF , that in the remaining graph we have MATH, and thus, we can apply REF , with MATH, MATH, MATH, and we get a set MATH, such that MATH and MATH. For every MATH, we know that MATH, and therefore MATH. If all vertices in MATH have the same degree into MATH, then clearly MATH, and we are done. We therefore partition the vertices of MATH into sets of vertices with roughly the same degree into MATH, MATH. By REF , with MATH and MATH, there is at least one set MATH, such that MATH . For every MATH, we know that MATH, and therefore MATH. In MATH, the degrees into MATH are roughly the same, and thus, MATH. Thus, we proved that in the collection MATH, whose size is MATH, there is at least one set MATH that satisfies the required properties. |
cs/0105036 | We check in MATH that MATH is not an answer set of MATH as follows. Guess a subset MATH of MATH, and verify that: CASE: MATH is not a model for MATH, or REF MATH is a model for MATH and MATH. The construction of MATH (see REF ) is feasible in polynomial time, and the tasks REF are clearly tractable. Thus, deciding whether MATH is not an answer set for MATH is in MATH, and, consequently, deciding whether MATH is an answer set for MATH is in MATH. |
cs/0105036 | Given a ground MATH program MATH and a ground literal MATH, we verify that MATH is a brave consequence of MATH as follows. Guess a set MATH of ground literals, check that REF MATH is an answer set for MATH, and REF MATH is true with respect to MATH. Task REF is clearly polynomial; while REF is in MATH, by virtue of REF . The problem therefore lies in MATH. MATH-hardness follows from REF and the results in CITE. |
cs/0105036 | Given a ground MATH program MATH and a ground literal MATH, we verify that MATH is not a cautious consequence of MATH as follows. Guess a set MATH of ground literals, check that REF MATH is an answer set for MATH, and REF MATH is not true with respect to MATH. Task REF is clearly polynomial; while REF is in MATH, by virtue of REF . Therefore, the complement of cautious reasoning is in MATH, and cautious reasoning is in MATH. MATH-hardness follows from REF and the results in CITE. |
cs/0105036 | First we show that MATH is equal to MATH (as defined in CITE): Deletion REF never applies, since for every literal L and any two rules MATH, MATH, MATH holds, thus violating REF and therefore no rule can be overridden. It is evident that the deletion REF are equal to deletion REF of MATH in REF, respectively. The first ones delete rules, where some NAF literal is contained in MATH, while the second ones delete all NAF literals of the remaining rules. Next, we show that the criteria for a consistent set MATH of literals being an answer set of a positive (that is, NAF free) program (as in CITE) is equal to the notion of satisfaction: Since the set is consistent, REF does not apply. REF says: MATH (the body is true) implies that the head is true. This is logically equivalent to ``The body is not true or the head is true", which is the definition of rule satisfaction. In total we have that the minimal models of MATH are equal to the consistent answer sets of MATH, since answer sets are minimal by definition. Additionally, we require in REF that MATH is also a model of MATH, while in CITE there is no such requirement. However, all minimal models of MATH are also models of MATH: All rules in MATH are satisfied, and only the deletion REF have been applied (as shown above). So, for any rule MATH, which has been deleted by REF , some literal in MATH is in MATH, so MATH's body is not true, and thus MATH is satisfied in MATH. If a rule MATH, which has been transformed by REF , is satisfied without MATH, then either MATH is true or MATH is not true, so adding any NAF part to it does not change its satisfaction status. |
cs/0105036 | First we show that given an answer set MATH for MATH there exists a consistent answer set MATH for MATH such that MATH. We proceed by constructing the model MATH. Let MATH. Let MATH be the set of ground literals MATH such that there exists a (defeasible) rule MATH with MATH such that MATH is overridden in MATH and MATH is the tuple of arguments appearing in the head of MATH. Let MATH be the set of ground literals MATH such that there exist two rules MATH such that MATH, MATH is defeasible and MATH overrides MATH in MATH. Let denote by MATH the collection of sets of ground literals such that each element MATH satisfies the following properties: CASE: for each literal MATH there is a literal MATH in MATH, for some object identifier MATH, CASE: for each MATH such that the body of MATH is true in MATH, for at least one literal MATH of the head of MATH a corresponding literal MATH occurs in MATH, CASE: MATH, CASE: MATH is a consistent set of literals. First observe that the family MATH is not empty (that is, there is at least a set of consistent sets of literals satisfying REF above). This immediately follows from the fact that MATH is an answer set of the program MATH. Let MATH, for a generic MATH. It is easy to show that MATH is independent on which MATH is chosen. Indeed, no literals from MATH appear in the NAF part of the rules in MATH. Now we examine which rules the program MATH contains (for any set MATH). Both the rules with head predicate MATH and MATH and the rules of the form MATH (added by REF ) appear unchanged in MATH. Indeed, these rules do not contain a NAF part (recall that the GL transformation can modify only rules in which a NAF part occurs). Each constraint of MATH (added by REF), that is a rule of the form MATH (where MATH is a literal not occurring in MATH) is translated into the rule MATH. The other rules in MATH originate from rules of MATH obtained by rewriting rules of MATH (see REF). Thus, consider a rule MATH of MATH. If MATH is defeasible and overridden in MATH then the corresponding rule in MATH (generated by REF) contains a NAF part not satisfied in MATH, by construction of MATH and MATH. Hence, such a rule appears neither in MATH nor in MATH. The other case we have to consider is that the rule MATH is either a strict rule or a defeasible rule not overridden in MATH. First suppose that MATH is a strict rule or is a defeasible rule not threatened in MATH (recall that a rule not threatened in MATH is certainly not overridden in MATH). In this case, the corresponding rule, say MATH in MATH (generated by REF) has the same body of MATH and the head modified by renaming predicates (from MATH to MATH) and by adding the object MATH (from which the rule MATH comes) as first argument in each head literal. Since the body of MATH does not contain literals from MATH and MATH, and further MATH (for each MATH), MATH is eliminated by the GL transformation with respect to MATH if and only if MATH is eliminated by the GL transformation with respect to MATH. Moreover, in case MATH is not eliminated by the GL transformation with respect to MATH, since the body of MATH does not contain literals from MATH and MATH, the GL transformation with respect to MATH modifies the body of MATH in the same way the GL transformation with respect to MATH modifies the body of MATH. Thus, each rule MATH in MATH has a corresponding rule in MATH with the same body and a rewritten head. Now suppose that MATH is a threatened defeasible rule that is not overridden in MATH. In this case, the corresponding rule, say MATH in MATH (generated by REF) has the head modified by renaming predicates (from MATH to MATH) and by adding the object MATH as first argument in each head literal and a body obtained by adding to the body of MATH a literal of the form MATH, where MATH is the object from which MATH comes, and MATH represents the tuple of terms appearing in the head literals of MATH. Since the rule MATH is not overridden in MATH, the literal MATH cannot belong to MATH and hence cannot belong to MATH. Thus, the GL transformation with respect to MATH eliminates the NAF part of the rule MATH. As a consequence, also in this case, each rule in MATH has a corresponding rule in MATH with Now we prove that, for any MATH, MATH is a model for MATH. Indeed, rules with head predicate MATH are clearly satisfied. Further, rules with head predicate MATH are satisfied by construction of set MATH, MATH and MATH. Moreover, rules of the form MATH are satisfied since MATH and by construction of MATH. Rules of MATH of the form MATH, originated by the translation of the constraints, are satisfied since MATH is a consistent set of literals. Consider now the remaining rules (those corresponding to rules of MATH). Let MATH be a rule of MATH and MATH the rule of MATH corresponding to MATH. As shown earlier, the two rules have the same body. Thus, if the body of MATH is true with respect to MATH, the body of MATH is true with respect to MATH, since no literal of MATH can appear in the body of the rule MATH (and hence of the rule MATH). As a consequence, by REF of the collection MATH to which the set MATH belongs, the head of the rule MATH is true in MATH. Thus, MATH is a model for MATH. Now we prove the following claim: CASE: Let MATH be a model for MATH. Then, MATH is a model for MATH. By contradiction suppose that MATH is not a model for MATH. Thus, there exists a rule MATH with body true in MATH and head false in MATH. Since, as shown earlier, the rule MATH has a corresponding rule MATH in MATH with the same body of MATH and the head obtained by replacing each literal MATH of the head of MATH by the corresponding literal MATH, where MATH is the object from which MATH comes. Since MATH is a model for MATH, at least one of the "MATH literal" of the head of MATH must be true in MATH. Then, due to the presence of the rules of the form MATH in MATH, MATH must contain also the "MATH corresponding literal" belonging to the head of MATH (contradiction) Moreover we prove that each model for MATH CASE: REF contains MATH, and CASE: REF belongs to MATH. To prove REF , suppose by contradiction MATH is a model for MATH such that MATH. Thus, MATH. On the other hand, by REF , MATH is a model for MATH. But since MATH is an answer set for MATH and then a minimal model for MATH, a contradiction arises. Now we have to prove REF above. First observe that REF . and REF. of the family MATH are trivially verified by the models of MATH. Thus, by contradiction suppose there exists a model MATH of MATH such that it does not satisfy one of REF . or REF. characterizing the family MATH. First suppose that REF . is not satisfied by MATH, that is, there is a literal MATH in MATH such that no corresponding literal MATH occurs in MATH, for some object identifier MATH. Let MATH be the set obtained by MATH by eliminating all such literals MATH. It is easy to see that MATH is still a model for MATH. Indeed, rules of the form MATH are satisfied since literals MATH dropped from MATH do not have corresponding "MATH literals" by hypothesis. Further, no other rule in MATH contains a literal from MATH in the head. On the other hand, by REF , MATH is a model for MATH. But this is a contradiction, since MATH and MATH is an answer set for MATH. Suppose now that REF . is not satisfied by MATH, that is, there is a rule MATH such that the body of MATH is true in MATH and no "MATH literals" corresponding to literals of the head occur in MATH. Since, MATH (see REF above), the corresponding rule of MATH is not satisfied. But this implies that MATH can not be a model for MATH (contradiction). A consequence of the fact that every model of MATH contains MATH is that every model of MATH must contain the sets MATH, MATH and MATH. because of the rules added by REF. Thus, the model MATH for MATH such that no set MATH exists such that MATH is a minimal model for MATH, that is, a consistent answer set for MATH. Hence, the first part of the proof is concluded, since MATH. Now, we prove that given a consistent answer set MATH for MATH, MATH is an answer set for MATH. First we prove that a literal MATH belongs to MATH if and only if there exits a literal MATH in MATH, for some object identifier MATH. Indeed, MATH implies that MATH. But since MATH is a minimal model for MATH, there must exits a rule in MATH with head containing the literal MATH and body true with respect to MATH (otherwise the literal MATH could be dropped from MATH without invalidate any rule of MATH and thus MATH would not be minimal). Conversely, if MATH, for some object identifier MATH, the literal MATH belongs to MATH, since MATH is a model for MATH and the rule MATH belongs to MATH. Thus, MATH. Moreover, we prove that every rule of MATH has a corresponding rule in MATH with same body and a head obtained by replacing the MATH literals with the MATH corresponding ones and by eliminating the object argument from these literals. Indeed, from the above result, the GL transformation deletes a rule MATH from MATH if either the corresponding rule belonging to MATH is overridden in MATH (due the the literal MATH occurring in the body of MATH) or some negated (by negation MATH) literal is false in MATH. But this literal is false in MATH if and only if it is false in MATH. On the other hand, in case the rule is not deleted by the GL transformation, its body is rewritten in the same way of the corresponding rule appearing in MATH. As a consequence, MATH is a model for MATH. Indeed, if the body of a rule MATH of MATH is true with respect to MATH, the corresponding rule MATH of MATH has the body true with respect to MATH. Hence, at least one of the head literals of MATH must be true in MATH. Let MATH such a literal. As shown earlier, this implies that MATH belongs to MATH. But MATH appears in the head of MATH and hence MATH is satisfied in MATH. Now we prove that MATH is minimal. By contradiction, suppose that MATH is a model for MATH. Consider the literals belonging to the set MATH. Because of the correspondence between the rules of MATH and the rules of MATH, the set of literals obtained from MATH by eliminating all the literals MATH belonging to the set MATH as well as the corresponding MATH literals is still a model for MATH. But this is a contradiction, since MATH is a consistent answer set for MATH. Since MATH is a model for MATH and is minimal, MATH is an answer set for the program MATH. Hence the proof is concluded. |
gr-qc/0105066 | Since MATH, antisymmetrising over MATH indices gives an identity MATH . Since the tensor MATH is trace-free, we get MATH . Absorbing the deltas gives the theorem. |
gr-qc/0105066 | The proof is analogous to the proof of REF but this time the antisymmetrisation is over MATH indices. Starting with MATH since the tensor is trace-free, we get MATH . Absorbing the deltas gives the theorem. |
gr-qc/0105066 | The proof is analogous to the proof of REF . This time we antisymmetrise over MATH indices to get the identity. MATH . Since the tensor MATH is trace-free, we get MATH . Absorbing the deltas gives the result. |
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