paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010309 | We observe that MATH . That is MATH where MATH and MATH . Observe that MATH, where MATH is the NAME matrix MATH . Its determinant is given by the formula MATH. Therefore MATH, where MATH. Therefore, MATH is an invertible matrix, and one has MATH where MATH is a matrix with entries in MATH. Therefore MATH for MATH. NAME MATH as desired. |
math/0010309 | The proof uses REF and an analysis of the denominators of the relevent characteristic classes as discussed in Subsection REF By assumption, there is a rational cohomology class MATH such that MATH. Therefore there is a positive integer MATH such that MATH is an integral cohomology class. If MATH is a MATH-cycle on MATH, then the cohomology class of MATH is integral, where MATH. Note also that MATH is integral for all MATH. It follows that there is a line bundle MATH and connection MATH such that its curvature MATH, and also that the induced connection on the line bundle MATH has curvature MATH. In particular, we see that MATH . Observe that since MATH, it follows that MATH, where MATH denotes the cohomological dimension of MATH. Therefore, for all MATH, one has MATH . By REF , it follows that MATH where MATH. By REF , it follows that the NAME constant MATH. |
math/0010309 | Let MATH be a projection in MATH. Then MATH is also a projection in MATH, and one has MATH . Each term in the above equation is non-negative. By the Theorem B and by hypothesis, it follows that MATH belongs to the finite set MATH . Since MATH is an ICC group, therefore the enveloping NAME algebra of MATH is a factor, compare CITE. It follows that for each value of the trace, there is a unique projection up to NAME equivalence in the enveloping NAME algebra. Therefore there are at most MATH non-trivial projections in MATH, up to NAME equivalence in the enveloping NAME algebra. |
math/0010309 | To prove this, we will define an auxilliary algebra, which is a formal deformation quantization of MATH. Since MATH, there is a MATH-valued REF-cocycle MATH on MATH such that MATH. Define MATH as being the associative algebra of formal power series over the ring MATH of power series over MATH, where the multiplication in MATH is defined as MATH where MATH are MATH-bilinear maps MATH. Then we observe that MATH as algebras, where MATH and MATH denotes the ideal generated by MATH. Therefore the canonical projection map MATH is a homomorphism of algebras for all MATH. By adapting the proof in CITE, we see that the induced map in MATH-theory is an isomorphism, MATH . Then MATH is an isomorphism for all MATH. Setting MATH, we obtain the desired isomorphism. |
math/0010310 | Let MATH and MATH be the northern and southern hemispheres of MATH, that is, two disks in MATH such that MATH. Assume that MATH and MATH. Then MATH is MATH. We can extend any MATH to a diffeomorphism of the whole sphere by setting it to be the identity on MATH. Let MATH be the homomorphism defined in this way. This will be the rightmost map in our short exact sequence. First we show that the image of MATH is MATH. Let MATH be an element of MATH which fixes the puncture MATH. Note that MATH is a closed tubular neighborhood of MATH in MATH. By the uniqueness of tubular neighborhoods theorem, MATH is isotopic to the identity relative to MATH. This isotopy can be extended to an ambient isotopy of the MATH-times punctured sphere. We can therefore assume, without loss of generality, that MATH acts as the identity on MATH. Thus MATH. Now we show that the kernel of MATH is generated by MATH. Let MATH represent an element of the kernel of MATH. Let MATH be its extension to MATH which is the identity on MATH. Then there is an isotopy MATH such that MATH and MATH is the identity map. Now MATH restricted to MATH defines an element of the fundamental group of the space of all tubular neighborhoods of MATH. The proof of the uniqueness of tubular neighborhoods theorem CITE naturally extends to a proof that there is a homotopy equivalence between the space of tubular neighbourhoods of a point and MATH. Thus the fundamental group of the space of tubular neighbourhoods of a fixed point in MATH is MATH, generated by a rigid rotation through an angle of MATH. Consequently our family of diffeomorphisms MATH can be isotoped relative to endpoints so that its restriction to MATH is a rigid rotations by some multiple of MATH. Therefore MATH is isotopic to some power of MATH. |
math/0010310 | That the induced map is epic follows from NAME 's theorem CITE that that the genus two mapping class group is generated by five NAME twists, all of which happen to be MATH equivariant. See REF . This is the point where the analogous theorem fails for higher genus surfaces. That the induced map is one-to-one is more difficult. A proof can be found in CITE. |
math/0010310 | Onto is easy: Each of the five NAME twists shown in REF is sent to a half NAME twist around a curve separating two puncture points from the rest. Two examples are shown in REF . The definition of a half NAME twist is as illustrated in REF . These half NAME twists are the standard generators of the mapping class group of the MATH-times punctured sphere. That the kernel is MATH is an elementary exercise in (branched) covering space theory. |
math/0010321 | We have to prove REF for MATH. For MATH it is just MATH, for MATH it is MATH . But we have from REF : MATH and MATH for any MATH-morphism MATH. The general case is analogous. The statement of Lemma is true for any MATH-morphism MATH, not only for the NAME 's one. |
math/0010321 | We need to prove REF for MATH defined as above. The l.h.s.of REF has a form MATH where MATH is the set of admissible graphs with MATH vertices of the first type, MATH vertices of the second type, and MATH edges, and MATH. The numbers MATH are linear-quadratic expressions in the weights MATH,defined in REF, and the NAME weights MATH, defined in REF. We want to prove that MATH for any MATH. The idea arise to CITE: we identify MATH with the integral over the boundary MATH of the differential form of degree MATH, namely, with MATH. We have by the NAME formula: MATH . On the other hand, only the boundary strata of codimension MATH contributes in the l.h.s. of REF . We have: MATH where REF, and REF are the three types of the boundary strata of codimension MATH, listed in REF. The idea of the proof is to identify these summands with the summands contributed to MATH from REF . We consider the cases SREF - SREF CASE: The boundary stratum is MATH. It is clear that the integral over it factorizes in the product of two integrals: the integral over MATH and the integral over MATH. It is proved in REF, that the integral over MATH does not vanish only for MATH. The situation is as follows: two points move close to each other and far from MATH. This corresponds to the first summand of REF , containing the NAME - NAME bracket of polyvector fields. REF. CASE: Some points MATH move close to each other and close to MATH. The boundary stratum is MATH. It is clear that the integral factorizes into the product of two integrals, of an integral over MATH and of an integral over MATH (see, in particular, the following two subsections). Then, by REF , we have only the following possibilities: CASE: MATH and there is an edge from MATH to MATH; CASE: MATH and there are no edges connecting MATH and MATH. We analyze these strata separately below. It will be clear from our discussion how to apply REF to prove that other possible strata give zero contribution. REF . Let MATH be an edge starting at MATH, and let MATH be the angle between MATH and MATH (this is the angle MATH for MATH), MATH be the angle between MATH and MATH, and MATH be the angle between MATH and MATH, see REF . The integral over MATH should be the integral over MATH of the canonical MATH-form MATH on MATH. By our definition, MATH for MATH is the angle MATH. The MATH-form MATH is equal to MATH. In the limit MATH the angle MATH (see REF ). Therefore, we have two-form MATH. The angle MATH does not depend on the position of MATH. The ``new" edge will be MATH, and its angle is precisely MATH. The integral over MATH is MATH. In the general case, there are some edges starting at MATH, but no edge ends at MATH: in this case the angle of this edge would be equal to REF. These terms correspond to a summand of MATH (see REF ) and any they contribute to the third summand in REF . See REF for a further discussion. REF . We denote MATH by MATH. There no edges connecting MATH and MATH. See REF . In the limit when MATH tends to MATH and the points MATH (end-points of the edges starting from MATH) are far from MATH, the wedge product MATH tends to MATH. The integral MATH is the integral over MATH. The remaining part of the wedge product, MATH contributes to the integral over MATH. In the general case, if there exists any edge ending at MATH, an edge MATH, and MATH is far from MATH, the angle MATH tends to MATH when MATH tends to MATH. Therefore, no edge ends at MATH. These terms correspond to a summand of MATH and contribute to the third summand in REF . See REF for a further discussion. REF We claim that the total contribution of the two REF is equal exactly to MATH, where MATH. Here we consider two examples. MATH is a vector field, MATH is a function. Then MATH, and we have only the summand MATH, which corresponds exactly exactly to REF . Again, MATH is a vector field, MATH, but MATH is a MATH-form MATH. Then MATH . On the other hand, MATH . We see that in MATH the second summand in the right-hand side of REF cancells the second summand in the right-hand side of (MATH), whereas the first summand in the right-hand side of REF corresponds to REF , and the first summand in the right-hand side of REF corresponds to REF . In the general case the situation is analogous to this example. REF. CASE: Some points MATH of the first type and some points MATH of the second type move close to each other and close to MATH, MATH. The boundary stratum is MATH. The integral over MATH is exactly the NAME integral: the angle with MATH becomes the angle with MATH for MATH. We obtain the second summand in REF . REF is proven. |
math/0010321 | It is straightforward. |
math/0010321 | REF follows from REF , because MATH is a quasi-isomorphism by REF follow from the definitions. |
math/0010321 | CASE: In MATH we have: MATH . CASE: By definition, MATH. We have: MATH . |
math/0010321 | It is straightforward. |
math/0010321 | The numbers MATH, MATH in REF do not depend on the NAME algebra MATH. Therefore, one can suppose that MATH is semisimple. We proved in REF that if MATH is semisimple, that for MATH and MATH one has: MATH where MATH and MATH are both induced by the map MATH. Next, consider the isomorphism MATH, defined as MATH . It is proven in CITE that MATH. We obtain that MATH for MATH, MATH, and MATH, MATH are induced by the identity map MATH. Suppose now that for a map MATH we have MATH (MATH in our case). Then we have MATH. Indeed, set MATH and use the decompositions MATH which hold for a semisimple NAME algebra MATH. |
math/0010323 | Let MATH be a root of unity of order MATH. Let MATH be the subset of MATH of those MATH such that MATH. By CITE , the ideal in MATH of functions vanishing on MATH is MATH. Saying that MATH is regular is the same as saying that MATH. As MATH, this is the same as saying that MATH, which, using the isomorphism MATH, is equivalent to MATH. |
math/0010323 | CASE: Let MATH be a system of basic invariants. Let, as in the proof of the previous lemma, MATH be the subset of those MATH such that MATH does not divide MATH, and MATH the ideal generated by MATH. If MATH is monic in MATH, then MATH, so MATH is regular (by REF ). Now assume that MATH is regular, that is, that MATH. Let us construct from MATH a system of basic invariants MATH such that MATH is monic in MATH. Denote by MATH the complement of MATH (thus MATH). Let MATH be a family of complex numbers such that MATH. Let MATH. The requirements on MATH ensure that MATH is a system of basic invariants. By replacing MATH by MATH in MATH, one obtains the discriminant MATH. Let MATH be the image of MATH by the composition MATH . The polynomial MATH is non-zero (by REF ) and weighted homogeneous of weight MATH. There are two possibilities: CASE: Either MATH. By weighted homogeneity, MATH, with MATH. Once again by weighted homogeneity, MATH must be the only monomial of MATH of highest degree in MATH, and thus MATH is monic in MATH. CASE: Either MATH. A direct computation shows that the coefficient of MATH in MATH is the value of MATH evaluated at MATH and MATH. As MATH, this coefficient, seen as a polynomial function with variables MATH, is not constant. Thus it is possible to choose the MATH such that this coefficient is non-zero. By weighted homogeneity, the corresponding MATH will be monic in MATH. CASE: The additional assumption, in the notations of the proof of REF , is that MATH. We prove that the first assertion implies the last one, which is enough. Assume that MATH is regular, let MATH be a system of basic invariants. The proof of REF also proves that MATH is monic, once it has been noticed that, in the final discussion, MATH, since MATH. |
math/0010323 | If MATH is regular, then by REF we can find MATH such that the monomial MATH appears in MATH; this monomial must be a valuation monomial. Now assume that the valuation of the discriminant is MATH. Choose a system of basic invariants MATH. By weighted homogeneity, a valuation monomial in MATH can only involve those MATH for which MATH. Using REF , this implies that MATH is regular. |
math/0010323 | As MATH is obtained from MATH by removing complex codimension MATH subvarieties, MATH is surjective. In terms of paths, REF is the following: let MATH be a path in MATH from MATH to a divisor MATH; then in the MATH-homotopy class of MATH relatively to MATH, there exists a path avoiding MATH. This follows from standard general position arguments, as MATH has complex codimension MATH in MATH (whereas MATH has real dimension MATH), and MATH has complex codimension MATH in MATH (whereas the target point MATH has dimension MATH). REF is nothing more than an induction from REF. |
math/0010323 | The space MATH is trivial bundle of fiber MATH over MATH, so MATH can be lifted to a generator MATH of the monodromy around MATH in MATH (any lifting of a defining path suits). Let MATH be the set of irreducible components of MATH, let MATH the set of irreducible components of MATH which are not in MATH. The point REF applied to these MATH and MATH asserts that MATH is the image of a generator-of-the-monodromy MATH around MATH by the embedding morphism MATH. |
math/0010323 | We describe an inductive construction procedure for such generating sets. We keep the notations used in the previous subsection. Let MATH be such that MATH is MATH-dominant. If MATH, then MATH is the head monomial of MATH, MATH and REF gives generating sets with MATH generators-of-the-monodromy. Now assume MATH. Let MATH, let MATH, as in the previous subsection. We have MATH. The monomial MATH is dominant in MATH. Let MATH be the hypersurface in MATH defined by MATH, and let MATH. By the induction hypothesis, it is possible to generate MATH by a set MATH consisting of MATH generators-of-the-monodromy. As MATH, we have an exact sequence MATH . Using REF , lift MATH to a same cardinality set MATH of generators-of-the-monodromy around MATH in MATH. Let MATH be the set of all generators-of-the-monodromy around the irreducible components of MATH which are not in MATH. By REF , we have MATH. Thus MATH is generated by MATH. Using REF , lift MATH and MATH to same cardinality sets MATH and MATH of generators-of-the-monodromy around MATH in MATH. Choose MATH generic of direction MATH. The fibration argument used before still proves that MATH generates MATH. As MATH, the common irreducible components of MATH are irreducible components of the hypersurface MATH, thus, as MATH, they are also irreducible components of MATH. Using REF , this implies that elements of MATH are mapped to MATH in MATH. To generate MATH, it is enough to take the images in MATH of a generating set of MATH (of cardinal MATH) and of MATH (of cardinal MATH). |
math/0010323 | Note that MATH central implies MATH and that both being pure and being regular are invariant by conjugation properties. Hence the statement is independent from the choice of the starting leg MATH (only the cyclic order matters). This will allow us, later in the proof, to choose MATH at our convenience. Let us check that it is enough to prove the statement for only one basepoint chosen at our convenience. Let MATH be the basepoint in MATH we prefer to the imposed MATH. To compare the two situations, we can choose a path MATH from MATH to MATH in MATH. Along MATH we have isomorphisms MATH and MATH. We can drag our original spider along MATH to get a spider with center MATH: MATH . The isomorphism MATH maps the leg-generators of the original spider to the leg-generators of the new one. It maps central elements, pure elements, and elements whose image is regular to elements with the same property. If the statement is proven for spiders with basepoint MATH, then it follows for spiders with basepoint MATH. Consider the affine segment MATH in MATH. For each MATH, we denote by MATH the affine line of direction MATH passing through MATH (we therefore have MATH). As MATH is monic in MATH, MATH, so MATH is not generic, unless MATH. The lemma is obvious when MATH; we assume from now on that MATH. This implies that MATH and that the segment MATH has non-zero length and is transverse to MATH. Let MATH. The space MATH is a fiber bundle over MATH. The map MATH from MATH to the space of finite subsets of MATH is continuous, thus MATH is compact. Let MATH be large enough such that MATH (where MATH is the open ball of center MATH and radius MATH). Take MATH at distance MATH from MATH. We will prove the statement for spiders centered at MATH. MATH . To ease notations, let us fix affine coordinates MATH such that MATH and MATH. If the starting leg is well-chosen, the product MATH is represented by the loop MATH . Let MATH. The affine segment MATH yields an isomorphism MATH. As MATH is large enough, the outer surface of the cylinder of radius MATH around MATH does not intersect MATH, and the isomorphism MATH maps the product MATH to the element MATH represented by MATH (the MATH-action used in the formula is the linear one, not the weighted action; in terms of weighted action, the formula would of course be MATH). From that description of MATH, it is classical (see for example CITE, page REF) that MATH is central in MATH, that it is pure, and that the image MATH of MATH in MATH is MATH-regular. Thus MATH is central and pure in MATH, and MATH, which is conjugate to the image MATH of MATH in MATH, is a MATH-regular element. |
math/0010323 | It is enough to prove the proposition when MATH is irreductible: if MATH is reductible, it is a direct product of irreductible groups; except in a few degenerate and straightforward cases, MATH decomposes as a corresponding direct product; then MATH and MATH also decompose, and the reduction to the irreducible case follows. Choose MATH a regular degree for MATH (one can check that any irreducible reflection group admits at least one regular degree). We may assume that the discriminant MATH is monic in MATH. Indeed, by REF , there exists a system of basic invariants MATH of MATH such that the discriminant MATH of MATH is monic in a variable MATH. The system MATH is obtained from MATH by a certain sequence of algebraic substitutions. By performing the same substitutions among the corresponding elements of MATH (and leaving the remaining invariants unchanged), we get a new system MATH which satisfies the monicity assumption (in addition to the defining properties of MATH). The discrimant MATH of MATH is obtained from the discriminant MATH of MATH by ``forgetting" all monomials involving one or more of the inderminates MATH (this operation is the composition of MATH with MATH). As MATH is monic in MATH, and using weighted homogeneity, it is readily seen that MATH was already monic in MATH, and that MATH and MATH have the same degree in MATH. Let MATH be the hypersurface in MATH defined by MATH. Let MATH be the hypersurface in MATH defined by MATH. Let MATH be a MATH-generic line of direction MATH in MATH. The cardinal of MATH is MATH, thus MATH is generic relatively to MATH in MATH. By REF , the inclusion MATH is MATH-surjective. As it factors through MATH, the latter map is also MATH-surjective. |
nlin/0010040 | Consider the following commutative diagram: and let us set MATH, MATH, MATH, MATH, MATH, MATH, where MATH. Then the above diagram is infinitely continued to the left, while by setting MATH and passing to the inverse limit, we obtain REF with MATH, MATH, and MATH. |
nlin/0010040 | Let MATH and MATH. Then due to the definition of the NAME - NAME bracket one has MATH . The first summand vanishes, since MATH. On the other hand, MATH and consequently is a one-form on MATH. Hence, the second summand vanishes as well. |
nlin/0010040 | The family of coverings MATH is a deformation of MATH. Since we work with deformations which leave the equation structure unchanged, then, by REF , their infinitesimal parts are elements of MATH. Let MATH be such an element. Now, by REF , the deformation we are dealing with is to be trivial as a deformation of MATH endowed with the structure MATH. On the infinitesimal level, this means that the image of MATH in MATH should vanish. But by exactness of REF we see that MATH for some MATH. It now suffices to note that by construction of the connecting homomorphism, MATH. The family MATH allows us to find a shadow MATH explicitly. Namely, we obviously have MATH where MATH . Hence, infinitesimal action is given by the NAME - NAME bracket. In the coset MATH there exists a unique MATH- vertical representative MATH, and the corresponding element MATH is a required shadow. |
nlin/0010041 | It suffices to take the trivial bundle MATH for the bundles MATH and MATH and an arbitrary vector field MATH for the operator MATH. Flatness is a consequence of REF. |
nlin/0010045 | The NAME transform in REF establishes the correspondence between the terms in the asymptotic expansion of MATH and the terms of the small MATH expansion of MATH. This correspondence is MATH . Our plan is to modify the integrand in the definition of MATH, getting rid of the factor MATH, expand the integral in inverse powers of MATH and apply the correspondence REF to recover REF . First of all, as one can verify by direct substitution of the series for MATH, MATH and MATH . Applying REF , MATH where MATH and for simplicity we have omitted the argument MATH of the functions MATH, MATH and MATH. Similarly, using REF , we have MATH . Noticing the similarity between REF , we subtract the first from the second, with the appropriate factors, to obtain MATH where, as before, the argument MATH of MATH, MATH and MATH has been omitted. The right hand side of REF is exactly the integrand of REF if we perform the change of variables MATH and, therefore, MATH . The first term in the integral can be evaluated as follows, MATH where MATH . Since MATH we obtain MATH . Now we can expand the result in inverse powers of MATH and apply the correspondence REF . We obtain MATH . Next we need to expand the second part of the integrand in REF , MATH . Using the same notation as in REF , MATH where, as before, MATH is the MATH-th convolution of MATH with itself. Thus MATH . Finally we integrate against MATH to arrive at MATH . This is exactly the same as the MATH sum in REF with the exception of the extra MATH term in the summation above. For MATH we have MATH which, together with the terms MATH and MATH, gives the correct contribution MATH. |
nlin/0010045 | To show that the statement is true we need to prove that the function MATH is a nowhere decreasing function of MATH. On the line MATH we have MATH . Thus MATH and its derivative is, after simplification, MATH. |
nlin/0010058 | Any hydrodynamic conservation law MATH may be expanded, using REF , yielding MATH . The integrability condition for this is MATH . By differentiating this with respect to MATH the unity element, one finds that MATH satisfies the same equation, and hence is also conserved. These conserved densities are all homogeneous and may be labelled by their degree, so by NAME 's theorem MATH . They may also be normalised so that MATH . The basic relation REF may also be used to derive a recursion relation amongst the densities. Let MATH in REF , so, on using the unity relation, REF MATH . Multiplying by MATH and using NAME 's REF again yields MATH . Alternatively, from REF and NAME 's theorem one may derive MATH . The term in square brackets will not, in general, be zero, but for NAME manifolds it does vanish, while in the examples below it does not. Thus one has the following recursion scheme for these Hamiltonian densities: MATH . |
nlin/0010059 | The proof is obtained finding both the linear MATH and MATH satisfying REF . So doing, we introduce in REF the general form for MATH and MATH, namely MATH, MATH. The problem consits then in the evaluation of the coefficients of MATH and MATH by solving the algebraic system obtained identifying to zero the coefficients of the polynomial in MATH which results after this introduction of MATH and MATH in REF . The number of equations of this algebraic system is usely greater than the number of the unknown coefficients of MATH and MATH. In order to satisfy the entire system of algebraic equations, it is necessary then to introduce the conditions on the parameters of the differential system REF , which appear in the different statements of the theorem. |
nlin/0010059 | The cofactors of MATH are MATH, MATH, MATH respectively. REF use only these three algebraic solutions. For the other statements, it is easy to check, that under suitable assumptions, MATH is an algebraic solution of the LVREF, that is, that MATH verifies REF where MATH, the cofactor of MATH takes the values MATH for statement MATH where MATH, MATH for statement MATH, MATH for statements MATH and MATH, MATH for statement MATH, MATH for statement MATH, MATH for statements MATH. We note that MATH is the algebraic solution of REF for statement MATH, of REF for statement MATH, of REF for statements MATH and MATH of REF for statements MATH and REF , of REF for statement MATH, of REF for statement MATH, of REF for statements MATH. Hence to each statement corresponds a first integral MATH. |
nlin/0010059 | The first integral can be written as MATH with MATH. The cofactors of MATH are MATH, MATH, MATH respectively. Now it is easy to check that for REF - REF MATH and MATH are the algebraic solutions of REF , for REF MATH and MATH are the algebraic solutions of REF , for REF MATH and MATH are the algebraic solutions of REF and for REF MATH and MATH are the algebraic solutions of REF . Consequently MATH is an algebraic solution of the LVREF, that is, MATH verifies REF where MATH, the cofactor of MATH takes the value MATH for statement MATH, MATH for statement MATH, MATH for statement MATH, MATH for statement MATH, MATH for statement MATH, MATH for statement MATH, MATH for statement MATH and MATH for statement MATH. For each statement, a solution of system MATH. Hence to each statement corresponds a first integral MATH. |
quant-ph/0010031 | Note that we can write MATH where MATH. MATH is in MATH by definition. Since MATH contains MATH and hence MATH, it must also contain the identity matrix MATH. Hence, noting that MATH generates a one-dimensional NAME algebra isomorphic to MATH we have indeed MATH. |
quant-ph/0010031 | Using the relation MATH which follows from the definition of MATH, it can be verified by direct computation that the skew-Hermitian matrices MATH and MATH satisfy the equations MATH and a bit of algebra therefore shows that the MATH elements MATH and MATH for MATH generate the entire NAME algebra MATH. |
quant-ph/0010031 | If MATH then MATH where MATH. This leads to MATH . Since MATH by hypothesis it follows that MATH. Repeating this procedure MATH times shows that all the generators MATH and MATH for MATH are in MATH. Similarly, we can show that MATH and MATH in MATH implies that MATH contains all the generators MATH and MATH, MATH. |
quant-ph/0010031 | We evaluate MATH . Using MATH and MATH we obtain MATH, where MATH . Repeating the previous steps for MATH and MATH leads to MATH, where MATH . After MATH iterations, we have MATH where MATH . Since by REF this means MATH and noting that MATH we have MATH . The conclusion now follows from REF . |
quant-ph/0010031 | Let MATH. In this case the element MATH is in MATH and its sum and difference with MATH give rise to MATH which, along with their commutator MATH are in MATH. Starting with MATH and MATH, we have MATH . Since by REF for MATH, we have MATH. Similarly, starting with MATH and MATH, we can also prove MATH. This implies MATH and MATH are in MATH and hence the conclusion follows from REF . |
quant-ph/0010034 | We begin by deriving a more usable expression for MATH. MATH where we have used the fact that MATH is periodic of period MATH. Since MATH is one-to-one when restricted to its period MATH, all the kets MATH are mutually orthogonal. Hence, MATH . If MATH, then since MATH is a MATH-th root of unity, we have MATH . On the other hand, if MATH, then we can sum the geometric series to obtain MATH where we have used the fact that MATH is the primitive MATH-th root of unity given by MATH . The remaining part of the proposition is a consequence of the trigonometric identity MATH . |
quant-ph/0010034 | We begin by noting that MATH where we have made use of the inequalities MATH . It immediately follows that MATH . As a result, we can legitimately use the inequality MATH to simplify the expression for MATH. Thus, MATH . The remaining case, MATH is left to the reader. |
quant-ph/0010034 | Since MATH we know that MATH which can be rewritten as MATH . But, since MATH, it follows that MATH . Finally, since MATH and hence MATH, the above theorem can be applied. Thus, MATH is a convergent of the continued fraction expansion of MATH. |
quant-ph/0010034 | From the above theorem, we know that MATH where MATH is a monotone decreasing sequence of positive reals converging to zero. Thus, MATH . |
quant-ph/0010040 | CASE: The probability of error MATH of finding the hidden state MATH is given by MATH where MATH where MATH is the function that rounds to the nearest integer. Hence, MATH . Thus, MATH CASE: The computational cost of the NAME transform MATH is MATH single qubit operations. The transformations MATH and MATH each carry a computational cost of MATH. MATH REF is the computationally dominant step. In MATH REF there are MATH iterations. In each iteration, the NAME transform is applied twice. The transformations MATH and MATH are each applied once. Hence, each iteration comes with a computational cost of MATH, and so the total cost of MATH REF is MATH. |
quant-ph/0010101 | By REF , we need to compute MATH. For MATH, MATH and so MATH is the sign representation of MATH. Then MATH is one-dimensional and carries the trivial representation if MATH is even, or the sign representation if MATH is odd. Now for MATH even, we can easily compute a non-zero function MATH in MATH. For MATH is the top exterior power MATH. Thus MATH is a non-zero element of the one-dimensional subspace MATH of MATH. The tensor components of MATH are then given by MATH if for each MATH, the column MATH is a permutation MATH of MATH and MATH otherwise. Then we get MATH where MATH. The expression is very redundant, as each term appears MATH times. We remedy this by restricting the first permutation MATH to be MATH. This gives REF . |
quant-ph/0010101 | Let MATH. Then MATH where MATH is the character of MATH. If MATH acts trivially on MATH, then MATH. If MATH acts non-trivially, we claim MATH. To show this, it suffices to show that MATH has at least two distinct eigenvalues on MATH; this is because MATH is the sum of the MATH eigenvalues of MATH. Now the set MATH of MATH which act on MATH by a scalar is a normal subgroup of MATH. So if MATH, then MATH is MATH, the alternating group MATH or MATH. We can easily rule out the latter two possibilities, so MATH, which proves our claim. If MATH, then (since MATH and MATH), we have MATH, MATH and MATH. Our claim is clear here since MATH acts on MATH through the reflection representation of MATH on MATH. Therefore we have MATH as MATH where MATH is cardinality of the kernel of MATH. Our work in the previous paragraph computes MATH. |
quant-ph/0010101 | We have isomorphisms MATH since the representation of MATH on MATH factors through MATH. Thus the algebra MATH identifies with MATH. Now MATH is the algebra of MATH-invariant polynomial functions on traceless MATH diagonal matrices, and so is a polynomial algebra on the functions MATH and MATH. These invariants correspond (up to scaling) to MATH and MATH. The formula for the dimension follows easily. |
quant-ph/0010111 | We say the protocol has failed if many of the quantum states measured and published by NAME do not match what NAME sent or what NAME sent. If there are only very few cases with discrepancies the protocol is considered a success. If the protocol does not fail then the protocol is concealing to NAME. NAME has sent some substrings to NAME and as she could not know which substrings would be measured and published by NAME there are some substrings which she forwarded, but which were not tested. Hence NAME cannot measure the overall parity bit MATH even after getting to know the bases. NAME cannot measure before getting to know the bases as she will be detected cheating whenever NAME measures and publishes a quantum state she disturbed. If the protocol did not fail it is concealing to NAME unless NAME and NAME collude. This is clear as NAME does not have the complete quantum state, and can hence not measure the parity bit. If the protocol did not fail it is binding for NAME unless she colludes with NAME, which is impossible according to our assumption. If the protocol failed there are two cases to be considered. First, NAME published a lot of measurement results which do not match what NAME sent, but NAME was not complaining about NAME, then it is clear for NAME that NAME and NAME collude somehow, but as this is not possible according to our assumption this will not happen. The second case is that NAME complains about NAME, then NAME is identified as a cheater as every player complains about NAME. |
quant-ph/0010111 | The bit commitment used for forcing measurments need only be shortly binding. It need only be binding until the honest measurement is performed unreversibly. Hence the collusion MATH, which would be able to violate the binding condition but not the concealing condition of the above bit commitment, cannot cheat after the measurement is performed. |
quant-ph/0010111 | By REF we can have oblivious transfer from NAME to NAME and from NAME to NAME. As NAME cannot be in conflict with NAME or NAME (or a cheater can be identified) we can realize GCOT from NAME to any other player by the protocols of CITE with the security of the oblivious transfer channel of REF . The above protocol for GCOT between NAME and NAME is still concealing for NAME even if NAME and NAME collude, but the resulting commitments need not be binding any more. This is no problem as we allow NAME and NAME to collude only after the termination of the protocol (See comment after REF ). |
quant-ph/0010111 | If one looks at CITE one can see that the above theorem is proven there for all cases but one. The case left open is that two collusions MATH, MATH, covering MATH for a player MATH, are in conflict such that no player from MATH can use the oblivious transfer to any player of MATH (and vice versa). In this case we can proceed analogously to our three party protocols. All commitments are made via REF and equality of commitments is proven to NAME. This way we obtain GBCX in a way that whenever a player complains a cheater can be identified. The GCOT protocol can be realized analogously to the above three party protocol. A player NAME MATH runs subGCOT via NAME with a player NAME MATH such that NAME cannot distinguish data from NAME and data from NAME. Again NAME cannot complain without either being detected cheating or proving that NAME cheats. On top of subGCOT GCOT can easily be realized. With GBCX and GCOT we can realize all multi party protocols CITE if one keeps in mind that distributed bit commitments can be realized without problems even when conflicts are present CITE. The set MATH, for which MATH may be contained in MATH, can in the above case be chosen to be any set containing NAME. |
quant-ph/0010112 | If the players of two possible collusions MATH covering MATH cannot cooperate then it is not clear for MATH which collusion is cheating. To continue with the protocol all messages between players who are complaing about each other have to be exchanged over the broadcast channel or over secure channels via MATH. Obviously MATH learns all secrets or the protocol must be aborted. In both cases the protocol is not MATH-robust. |
quant-ph/0010112 | We show that it is impossible to realize a bit commitment for a party NAME MATH and NAME MATH. This is simple as we are almost in the two party scenario: Assume the collusion MATH can by no means measure the bit NAME committed to, then the collusion MATH can, by keeping every action at the quantum level cheat analogously to the two party situation, that is, there exists a unitary transform MATH which can change the bit NAME is committed to. The transform MATH must be jointly applied by all players in MATH . |
quant-ph/0010112 | Assume there exists a MATH with MATH for MATH. We would like to implement a bit commitment from a player from MATH to the player MATH. To prevent the players from MATH to jointly change the committed bit it must be possible for the players of MATH to measure the committed bit. Only the assumption that MATH does not collude with the players of MATH makes this attack impossible. If the player MATH is honest but curious and keeps everything at the quantum level, then he would be able to measure the committed bit if all players of the set MATH would together keep all their actions at the quantum level and later on give all their quantum information to the player MATH. NAME though the player MATH does not collude with the players from MATH we cannot keep the players from MATH from deviating from the protocol in giving away their secret data. |
quant-ph/0010112 | One can define a bit commitment protocol where NAME has to perform this measurement and send the information measured to NAME. As NAME can still not distinguish between the commitments zero and one the attack of NAME, Lo/NAME applies and there exists a unitary transform changing the bit. |
quant-ph/0010112 | If the honest but curious party remains independent of the two parties NAME and NAME bit commitment can be implemented by classical multiparty protocols. Now assume the honest but curious third party joins NAME or NAME after the commit phase is completed. The honest but curious third party will follow the protocol, but leave everything in superposition which need not be sent away as classical data. So the third party will perform some measurements. The third party can join NAME afterwards and NAME should still be unable to recover NAME bit. Hence the third party did only obtain measurement results which are of no use for NAME. Hence if the third party joins NAME we are in the situation of REF . NAME together with the third party can jointly perform a unitary transform which changes the bit NAME committed to. |
quant-ph/0010112 | Let MATH denote two sets of possibly colluding players and let MATH be an adversary structure containing the complements of MATH and MATH. We show that it is impossible to implement an oblivious transfer from NAME MATH to NAME MATH which is MATH-partially robust and MATH-secure after termination. Assume such an oblivious transfer were possible then we could with it implement a bit commitment from NAME to NAME which is MATH-partially robust during the commit phase and MATH-partially robust up to the unveil phase. This is easy to see as after termination of the commit phase security is the only critical issue. The data computed during the commit phase cannot be changed any more and fairness is not of interest until the unveil phase. It remains to be proven that a bit commitment from the player NAME MATH to the player NAME MATH is impossible. We look at the sets MATH, MATH and MATH and prove the impossibility analogously to REF . During the execution of the commit phase the protocol is MATH-partially robust hence we can assume that the players in MATH or the players in MATH collude and we still get a valid commitment from MATH to MATH. Now we assume the protocol to become MATH-partially robust afterwards. Then all players from MATH may join the players from MATH and the bit commitment remains concealing even if the players from MATH were colluding. Hence all quantum information in the posession of the players from MATH are of no use to NAME. According to REF the players from MATH can change the committed bit. Hence the bit commitment is not MATH-partially robust between commit and unveil. |
quant-ph/0010112 | We will let NAME commit to a bit string MATH. Commit via Secret Sharing REF CASE: NAME sends NAME a random string MATH. CASE: NAME shares the string MATH using a secret sharing scheme with access structure MATH. This protocol shares NAME secret MATH with the access structure MATH. If the receiver NAME is honest this protocol can be used to force measurements MATH-partially robustly REF and if NAME is not honest then we cannot prevent a dishonest sender from colluding and changing the committed bit together with the receiver of the bit commitment. No bit commitment scheme can. The unveil protocol is essentially a reconstruction of the shared secret. Unveil CASE: NAME announces the shares she sent. The players from MATH confirm the shares and NAME can then reconstruct MATH from his knowledge of MATH. |
quant-ph/0010112 | The partial robustness is the same as claimed by REF so we need to prove only the improved security. We have to see first that MATH is an adversary structure. The set MATH is an adversary structure and it contains all proper subsets of MATH. Hence the set MATH is an adversary structure, too. To obtain the higher security we choose for every pair of players one player who shall commit to the other. Let NAME and NAME be a pair of players for which either NAME, Bob-MATH, NAME, Bob-MATH or Alice-MATH and Bob-MATH. Otherwise exchange the names of the players. We will see that the bit commitment from REF between any two players NAME and NAME is MATH-secure if used in the above defined direction. As REF already proves the MATH-security we are left with proving the security against the possible collusion MATH. A collusion can only cheat if it is an authorized set able to recover a shared secret and if it contains the receiver of the bit commitment as the shared secret is encrypted by a key MATH only known to the sender and the receiver of the bit commitment. If the collusion contains the sender of the bit commitment then the committed bit can already be derived from the inputs of the colluding players and no security is lost. The direction of the bit commitment is chosen in a way that MATH either contains the sender of the bit commitment or it does not contain the receiver of the bit commitment and hence MATH is not able to reconstruct a secret bit in the bit commitment protocol. |
quant-ph/0010112 | The security of a bit commitment scheme is only relevant if the sender is honest. So we can assume throughout the proof that the secret is properly shared. Because no two collusions of MATH cover MATH every set of the access structure MATH contains at least two honest players. So even if all players of a collusion MATH leak their shares the secret remains shared among at least two honest players and no single honest but curious player gets to know a secret. |
quant-ph/0010112 | We are proving our claim using the above protocol. Whenever the receiver of the bit commitment complains about the sender then we need not be able to implement a bit commitment hence in the following we analyse only conflicts between the sender of the bit commitment and players other than the receiver. We consider two cases. First: NAME is honest, then every complaint about NAME comes from a cheater and NAME publishes the share the cheater complained about. So every complain about an honest NAME is equivalent to a leak of the share of the cheating player and we have seen that this does not harm the security of the secret sharing REF . The correctness of the secret sharing, which implies the binding property of the bit commitment, cannot be harmed by shares which become publicly known. Second: NAME is dishonest. Then NAME must be honest or we cannot expect a bit commitment to work. All shares the public ones as well as the only privately known ones pass the verifiable secret sharing test whenever the protocol has terminated and NAME has not been detected cheating. Hence every honest player is convinced that if NAME is honest then the secret is properly shared. This follows directly from the properties of the verifiable secret sharing scheme of CITE which was sketched in Subsection REF. The security of the protocol is not an issue if NAME is dishonest. |
cs/0011004 | Let MATH and MATH be two possible collusions covering MATH, then oblivious transfer cannot be implemented MATH-robustly between players of MATH and players of MATH. Between any two players NAME MATH and NAME MATH the oblivious transfer channel does not work, but it is not obvious for the player MATH who is refusing to cooperate. The player MATH must assist NAME and NAME. As no other player can assist we are in the three party situation with an oblivious transfer channel only between NAME and NAME and NAME and NAME. For each bit being transferred from NAME to NAME the player MATH knows either as much as NAME about this bit or he knows as much as NAME. The players NAME and NAME cannot agree on a bit known to both without MATH knowing it, too. Hence oblivious transfer from NAME to NAME becomes impossible without MATH having to learn a secret of NAME or a secret of NAME. |
cs/0011004 | We will not state a full proof here as it can be found in CITE. But we will restate the copying procedure as it is an important subprotocol of all of the following protocols. Suppose NAME is committed to NAME to a bit MATH and wants two instances of this commitment. Then NAME ceates MATH pairs of global bit commitments such that each pair NAME to MATH. Then all other player, by coin tossing, randomly partition these MATH pairs in three subsets of MATH pairs, thus obtaining three GBCX and ask NAME to prove the equality of the first new BCX with her GBCX for MATH. This destroys the old GBCX and one of the new GBCX, but an honest NAME can thereby convince all players that the two remaining GBCX both stand for the value MATH. |
cs/0011004 | Each player generates a GBCX and opens the commitment to NAME. In case of a conflict the player opens his GBCX publicly. Then NAME creates a GBCX such that the parity bit is the bit she wanted to create a DBC for. Only NAME knows how to open all commitments as she created one herself. |
cs/0011004 | We will essentially restate the GCOT protocol of CITE and see that with one-out-of-two UOT instead of one-out-of-two OT any conflict results in the identification of a cheater. MATH CASE: All participants together choose one decodable MATH linear code MATH with MATH and MATH for positive constants MATH, efficiently decoding MATH errors. CASE: NAME randomly picks MATH, committs to the bits MATH and MATH REF of the code words, and proves that the codewords fulfil the linear relations of MATH. CASE: NAME randomly picks MATH, with MATH and sets MATH for MATH and MATH for MATH. CASE: NAME runs MATH with NAME who gets MATH for MATH. NAME tells MATH to NAME who opens MATH for each MATH. CASE: NAME checks that MATH for MATH and MATH for MATH, sets MATH, for MATH and corrects MATH using MATH's decoding algorithm, commits to MATH for MATH, and proves that MATH. CASE: All players together randomly pick a subset MATH with MATH, MATH and NAME opens MATH and MATH for MATH. CASE: NAME proves that MATH for MATH. CASE: NAME randomly picks and announces a privacy amplification function MATH such that MATH and MATH and proves MATH and MATH. CASE: NAME sets MATH, commits to MATH and proves MATH. A conflict between NAME and NAME can only appear in connection with REF or REF . If these two steps would be performed honestly then all other steps can be checked by all other players and it becomes immediately clear who is cheating. In a conflict in connection with REF or REF NAME claims that NAME sent something inconsistent over the oblivious transfer channel or NAME accuses NAME to not have committed to what he received. In case of a conflict NAME opens all bits of MATH to which she is committed by the UOT also she opens her GBCX to these codewords, if she is not able to do it or unveils non code words or other inconsistent information she is detected cheating. The bits of MATH do not give away any secret as these are random code words. If NAME information is correctly unveiled and is consistent with all her past actions (proofs) then NAME was cheating if he did complain. If it was NAME complaining NAME has to prove zero knowledgly that the bit string MATH he is committed to equals MATH or equals MATH if he is able to convince all other players NAME is detected cheating (conflicts appearing during the proofs can be resolved easily as it is obvious for every player who is cheating). |
cs/0011004 | First we note that we do not need a broadcast channel as generating a GBC and unveiling it can be viewed as broadcasting. We now sketch the phases of a multi party protocol following CITE. To implement oblivious circuit evaluation to realize arbitrary functions we have to show the existence of an AND and a NOT function on DBCs and clearify how a protocol is initialized and how it is ended. Initialization Phase: All players have to agree on the function to be computed as well as on the circuit MATH to be used, they have to agree on an adversary structure MATH such that the protocol will be MATH robust and all players have to agree on the security parameters used and on a code MATH for the GCOT protocol. Then all players create DBCs to commit to their inputs. Computing Phase: The circuit is evaluated using AND and NOT gates on the input DBCs. An AND on commitments can be realized by the following protocol: NAME is committed to MATH and NAME is committed to MATH. Then NAME chooses a random bit MATH and runs MATH with NAME who gets MATH. We have MATH because for MATH we have MATH and hence MATH, for MATH we get MATH and MATH. To evaluate an AND on DBCs we observe that MATH. From this we can conclude that an AND operation on DBCs can be realized by MATH GPAND one for each pair of players and NAME operations for each player. To implement the NOT gate one player is picked who must invert his ``share". This players generates a new GBCX and proves that it is unequal to the GBCX he held before. Note that the GCOT within the AND protocol has to work only in one direction between every pair of players. Sometimes one needs several copies of a DBC. A DBC is copied by copying the GBCX it consists of. A GBCX can be copied by copying all its BCX with the procedure of REF . Revelation Phase: The result of a computation is hidden in DBCs. These have to be unveiled in a way to ensure the fairness of the protocol. Following CITE we use the techniques from CITE to gradually unveil the secret information such that no collusion can run off with an advantage of more than a fraction of a bit. Of course a MATH-secure protocol cannot be more than MATH-fair. |
cs/0011004 | Every player sends MATH times anonymously a random number to NAME. NAME sends her message MATH to every player together with MATH for all random numbers MATH . NAME received. Then every pair of players compares the message they received. Either they are all the same and the protocol was successful or two different messages show up (one might be the empty message). Now two cases can happen: CASE: The second message is correctly authenticated, then we have a high probability (depending on MATH) that the sender NAME was cheating or CASE: the second message is not correctly authenticated. In both cases we repeat the protocol until one of the following cases holds: CASE: The protocol was successful. CASE: NAME is in conflict with all other players and has to leave the protocol. CASE: The players complaining about NAME are always the same, then these must be cheating as NAME cannot know who sent which random number. CASE: Enough different correctly authenticated messages are found such that the probability that NAME is cheating is above a certain threshold and she is expelled from the protocol. |
cs/0011004 | To send a message anonymously one has to encode the message with an error correcting code to cope with the erasures of the AOT. For an anonymous broadcast NAME sends her message MATH anonymously to a player MATH. This player broadcasts the message. If he broadcasts something wrong NAME is in conflict with this player, complains about him using the authenticated broadcast, and picks another player MATH to start the procedure anew. Either the anonymous broadcast will eventually be successfull or NAME will leave the protocol as she is in conflict with all other players. |
cs/0011004 | For an anonymous broadcast with later identification NAME authenticates her message MATH with MATH random numbers which she sends anonymously to the players. Each player receives one random number. Then she anonymously broadcasts the thus authenticated message according to REF . No other player is later able to impersonate NAME as only she knows the secret random numbers of the honest players. |
cs/0011004 | We let all players create GBCX according to the protocol of CITE, but anonymously, using AOT and anonymous broadcast. Then after some time no new conflicts occur for MATH anonymous GBCX of each player (MATH is a security parameter which is polynomial in MATH). If a player NAME was unable to create a GBCX we will split the set of players in a way that one set contains all honest players and the other sets contain only cheaters. We explain this in more detail by the two cases which can occur: CASE: If NAME was honest then, as a cheater cannot distinguish between the honest players after some time if the cheater keeps complaining about NAME this cheater will be in conflict with all honest players. Furthermore all honest players will know it. Now we can seperate the set MATH of players several subsets such that all players in each subset are in conflict with the same players. Then we can be sure that one of the sets contains all honest players and every honest player knows it. CASE: If NAME was dishonest then we will also seperate the set MATH. NAME will be in one group with all players complaining about the same players as NAME did (these are all honest players if NAME were honest) all other players will be in the other sets. As NAME is a cheater and hence in conflict with an honest player all players in her group must be cheaters, too. |
cs/0011004 | NAME creates a GBCX following REF and NAME publishes positions of two substrings of the strings NAME sent to him. One substring where he knows all the bits and one substring where he knows nothing. The substrings must have approximately the same length. NAME publishes the bits of one of the substrings. Then NAME either learnt nothing new or he knows the bit NAME is committed to. We have realized UOT if we can show that no other player learns the bit NAME is committed to by the information published by NAME, but this is trivial as NAME sent different strings to different players. |
cs/0011004 | NAME sends, as a commitment, MATH bit strings of length MATH (MATH, MATH are security parameters which are polynomial in MATH) with parity MATH. Then the knowledge all other players have about MATH is negligible in MATH. Because the probability that a bit is received by at least one player is MATH and the probability that all players together have knowledge about all MATH is MATH which is negligible in MATH. If MATH strings are sent the probability remains negligible as MATH and MATH are polynomial in MATH. If NAME wanted to change the bit she committed to she has to change MATH bits. The probability that any single player does not detect this change is negligible in MATH. |
cs/0011004 | NAME creates a NAME and NAME publishes positions of two substrings of the strings NAME sent over the oblivious broadcast. One substring where he knows all the bits and one substring where he knows nothing. The substrings must have approximately the same length. NAME publishes the bits of one of the substrings. Then NAME either learnt nothing new or he knows the bit NAME is committed to. We have realized UOT if we can show that no other player learns the bit NAME is committed to by the information published by NAME. But as the substrings published are statistically independent of what the other players received this information just changes the probability of receiving a bit for each player. This change of probability can be coped with an suitable choice of the security parameters used in REF . |
cs/0011009 | We use induction on MATH; in the base case MATH, there is one (empty) maximal independent set, and for any MATH, MATH. Otherwise, we divide into cases according to the degrees of the vertices in MATH, as follows: CASE: If MATH contains a vertex MATH of degree three or more, then each maximal independent set MATH either contains MATH (in which case MATH is a maximal independent set of MATH) or it does not contain MATH (in which case MATH itself is a maximal independent set of MATH). Thus, by induction, the number of maximal independent sets of cardinality at most MATH is at most MATH as was to be proved. CASE: If G contains a degree-one vertex MATH, let its neighbor be MATH. Then each maximal independent set contains exactly one of MATH or MATH, and removing this vertex from the set produces a maximal independent set of either MATH or MATH. If the degree of MATH is MATH, this gives us by induction a bound of MATH on the number of maximal independent sets of cardinality at most MATH. CASE: If MATH contains an isolated vertex MATH, then each maximal independent set contains MATH, and the number of maximal independent sets of cardinality at most MATH is at most MATH . CASE: If MATH contains a chain MATH-MATH of degree two vertices, then each maximal independent set contains MATH, contains MATH, or does not contain MATH and contains MATH. Thus in this case the number of maximal independent sets of cardinality at most MATH is at most MATH . CASE: In the remaining case, MATH consists of a disjoint union of triangles, all maximal independent sets have exactly MATH vertices, and there are exactly MATH maximal independent sets. If MATH, then MATH . If MATH, there are no maximal independent sets of cardinality at most MATH. Thus in all cases the number of maximal independent sets is within the claimed bound. |
cs/0011009 | We use a recursive backtracking search, following the case analysis of REF : if there is a high-degree vertex, we try including it or not including it; if there is a degree-one vertex, we try including it or its neighbor; if there is a degree-zero vertex, we include it; and if all vertices form chains of degree-two vertices, we test whether the parameter MATH allows any small maximal independent sets, and if so we try including each of a chain of three adjacent vertices. The same case analysis shows that this algorithm performs MATH recursive calls. Each recursive call can easily be implemented in time polynomial in the size of the graph passed to the recursive call. Since our MATH bound is exponential in MATH, even when MATH, this polynomial overhead at the higher levels of the recursion is swamped by the time spent at lower levels of the recursion, and does not appear in our overall time bound. |
cs/0011009 | Clearly this is true of the initial values of MATH. Then for any MATH and any independent set MATH, we can color MATH by using a coloring of MATH and another color for each vertex in MATH, so MATH, and each step of our algorithm preserves the invariant. |
cs/0011009 | If we have any MATH-coloring of MATH, then the partition formed by separating the largest MATH color classes from the smallest color class satisfies the inequality MATH, so clearly this also is true when MATH is the partition maximizing MATH. If MATH were not maximal, due to the existence of another independent set MATH, then MATH would be a larger MATH-chromatic graph, violating the assumption of maximality of MATH. Similarly, suppose there were another MATH-chromatic set MATH. Then if MATH were empty, MATH would be a MATH-chromatic superset of MATH, violating the assumption of MATH's maximality. But if MATH were nonempty, MATH would be a better partition than MATH, so in either case we get a contradiction. |
cs/0011009 | Clearly, the initialization phase of the algorithm causes this to be true when MATH. Otherwise, let MATH be as in REF . By induction on MATH, MATH at the time we visit MATH. Then MATH, and MATH, so the inner loop for MATH will visit MATH and set MATH to MATH. |
cs/0011009 | MATH is itself a maximal MATH-chromatic subset of MATH, so REF shows that the algorithm correctly computes MATH. Clearly, the space is bounded by MATH. It remains to analyze the algorithm's time complexity. First, we consider the time spent initializing MATH. Since we perform a REF-coloring algorithm on each subset of MATH, this time is MATH . Finally, we bound the time in the main loop of the algorithm. We may possibly apply the algorithm of REF to generate small independent subsets of each set MATH. In the worst case, MATH and we can only limit the size of the generated independent sets to MATH. We spend constant time adjusting the value of MATH for each generated set. Thus, the total time can be bounded as MATH . This final term dominates the overall time bound. |
cs/0011015 | See REF. |
cs/0011015 | The correctness of NAME follows from REF . Below, we analyze the running time. We initialize a maximum heap CITE in MATH time to store the edges of MATH according to their weights. Let MATH be the running time of NAME excluding this initialization. Let MATH be the set of the heaviest edges in MATH. Then REF takes MATH time. In REF , we can compute MATH in MATH time CITE. From this matching, MATH can be found in MATH time CITE. Let MATH be the set of the edges of MATH adjacent to some node MATH with MATH; that is, MATH consists of the edges of MATH whose weights are reduced in MATH. Let MATH. REF updates every edge of MATH in the heap in MATH time. As MATH, REF to REF altogether use MATH time. Since the total weight of MATH is at most MATH, REF uses at most MATH time, where MATH is the maximum edge weight of MATH. In summary, for some positive integer MATH, MATH where MATH. By recursion, for some positive integers MATH with MATH and MATH, MATH . Since MATH is convex, by NAME 's Inequality CITE, MATH . Therefore, MATH . |
cs/0011015 | Suppose, for the sake of contradiction, that every minimum weight cover allows some bad node. Then we can obtain a contradiction by constructing another minimum weight cover with no bad node. Let MATH be a minimum weight cover of MATH with MATH as a bad node, that is, MATH. Recall that MATH is a minimum weight cover of MATH. Consider a maximum weight matching MATH of MATH. By REF , since MATH, MATH is matched by an edge in MATH, say, to a node MATH, and MATH. We call MATH the mate of MATH. Note that MATH cannot be a bad node; otherwise, MATH and a contradiction occurs. Since MATH is a cover of MATH, MATH. Thus, MATH. Define another cover MATH of MATH as follows. For each bad node defined by MATH, let MATH be the mate of MATH, define MATH and MATH. Note that MATH is not a bad node with respect to MATH, and neither is MATH since MATH. For all other nodes MATH, MATH is the same as MATH. Therefore, if MATH is a cover of MATH, MATH allows no bad node. Also, MATH. It remains to prove that MATH is a cover of MATH. By the definition of MATH, MATH if and only if MATH is the mate of a bad node with respect to MATH. Suppose MATH is not a cover of MATH. Then there exists an edge MATH such that MATH and MATH is the mate of a bad node. Recall that the latter implies that MATH. In other words, MATH . We can derive a contradiction as follows. CASE: MATH. Then MATH, which contradicts that MATH. CASE: MATH. Then MATH contains the edge MATH and MATH. Thus, MATH, which contradicts the fact that MATH allows no bad node. In conclusion, MATH is a cover of MATH. Together with the fact that MATH, we obtain the desired contradiction that MATH is a minimum weight cover of MATH with no bad node. REF follows. |
cs/0011015 | Consider any edge MATH of MATH. If MATH is not in MATH, then MATH. Assume that MATH is in MATH. Note that its weight in MATH is MATH. Since MATH is a cover, MATH. Thus, MATH. It follows that MATH is a cover of MATH. To show that MATH is a minimum weight one, we observe that MATH . By REF , MATH is minimum. |
cs/0011015 | The correctness of Compute-Min-Cover-MATH follows from REF . For the time complexity, the analysis is similar to that of REF . |
cs/0011015 | The running time of Recover-Max-Matching-MATH is dominated by the construction of MATH. Since MATH has at most MATH nodes and at most MATH edges, MATH can be constructed in MATH time using NAME algorithm CITE. It remains to show that MATH is a maximum weight matching of MATH. First, we argue that MATH has a perfect matching. Let MATH be a maximum weight matching of MATH. By REF , MATH for every edge MATH. Therefore, MATH is also a matching of MATH. Let MATH be the set of nodes in MATH unmatched by MATH. By REF , MATH for all MATH. Let MATH be MATH. Let MATH and MATH. Note that MATH forms a matching in MATH and every node in MATH is matched by either MATH, MATH or MATH. Thus, MATH has a perfect matching. Since MATH is a maximum cardinality matching of MATH, MATH must be a perfect matching. For every node MATH with MATH, MATH must be matched by MATH. Since there is no edge between MATH and any MATH in MATH, there exists some MATH with MATH. Thus, every node MATH with MATH must be matched by some edge in MATH. Therefore, MATH, and MATH is a maximum weight matching of MATH. |
cs/0011015 | REF follows from REF is proved as follows. Since MATH is a maximum weight matching of MATH, MATH. By REF , MATH if and only if MATH. We prove the latter as follows. Given a minimum weight cover MATH of MATH, we can obtain a cover MATH of MATH as follows. For any node MATH of MATH, MATH if MATH and MATH; otherwise, MATH. Note that MATH. Therefore, MATH and MATH. |
cs/0011015 | As MATH, let MATH be a shortest even-length alternating path for MATH where MATH. Based on MATH, we can construct an even-length alternating path MATH for MATH starting from MATH as follows. If MATH is not matched by MATH, MATH is simply a path of zero length. From now on, we assume that MATH is matched by MATH. As MATH is of even length, MATH is not matched by MATH. Then, by the definition of MATH, MATH is also not matched by MATH. Let MATH be the smallest integer in MATH such that MATH is not matched by MATH. Notice that, for all MATH, MATH is matched to MATH; furthermore, MATH contains an edge between MATH and MATH. Thus, MATH is an even-length alternating path for MATH. Similarly, for MATH, we can use MATH to define an even-length alternating path MATH for MATH starting from MATH. By construction, MATH are node-disjoint. |
cs/0011015 | The two statements are proved as follows. CASE: Let MATH be the largest integer such that MATH. By REF , MATH for all MATH, and MATH otherwise. Note that if MATH, MATH must be matched by MATH. Thus, MATH. Below, we prove the following two equalities: CASE: MATH. CASE: MATH. Then, by REF , MATH and MATH. Thus, MATH and REF follows. To show Equality REF , let MATH be the set of edges of MATH incident to MATH with MATH. Let MATH. Then, MATH. We claim that MATH is a maximum cardinality matching of MATH. Hence, MATH, and Equality REF follows. We prove the claim by contradiction. Suppose MATH is not a maximum cardinality matching of MATH. Then, there exists an alternating path MATH that can modify MATH to a larger matching of MATH CITE; in particular, the length of MATH must be odd and both of its endpoints are not matched by MATH. MATH must start from some node MATH with MATH and MATH; otherwise, MATH is alternating for MATH in MATH and MATH cannot be a maximum cardinality matching of MATH. Let MATH be a path formed by joining MATH with MATH. MATH is an even-length alternating path for MATH starting from MATH in MATH. This contradicts the fact that there is no even-length alternating path for MATH starting from MATH for MATH. To show Equality REF , we first note that MATH. It remains to prove the other direction. By REF , we can find MATH node-disjoint even-length alternating paths MATH for MATH, which start from MATH. MATH starts at MATH. Let MATH. Note that MATH and there are no edges in MATH incident to any of MATH. MATH is a matching of MATH and MATH of MATH. MATH. Since MATH by Equality REF , it follows that MATH. Therefore, Equality REF holds. CASE: We want to determine whether MATH for all nodes MATH in MATH time. By definition, MATH if and only if there is an even-length alternating path for MATH starting from MATH. Let us partition the nodes of MATH into two parts: MATH and MATH. Below, we give the details of computing MATH for all MATH. The case where MATH is symmetric. Let MATH be a directed graph over the node set MATH. MATH contains an edge MATH if there exists a node MATH such that MATH and MATH. Consider any node MATH of MATH that is unmatched by MATH. A directed path in MATH from MATH to a node MATH corresponds to a path in MATH, which is indeed an even-length alternating path for MATH starting from MATH. Therefore, for any MATH, MATH if and only if MATH is reachable from some node in MATH that is unmatched by MATH. We can identify all such MATH by using a depth-first search on MATH starting with all the nodes unmatched by MATH. The time required is MATH. As MATH, the lemma follows. |
cs/0011015 | Follows from REF . |
cs/0011018 | This theorem follows from REF . |
cs/0011018 | This lemma follows from REF and basics of linear programming CITE. |
cs/0011018 | CASE: By direct verification, MATH and MATH. Then, since MATH, by REF MATH and MATH. CASE: Note that MATH by REF . Then, by REF , MATH is an optimal mixed strategy of the online player. Thus, this statement follows from REF . As pointed out in REF , MATH. By direct evaluation and REF , MATH. Then, this statement follows from the fact that by definition, the MATH-th component of MATH equals MATH. CASE: To prove the uniqueness of MATH, by REF , it suffices to show that MATH has a unique element. By basics of linear programming CITE, MATH has only a finite number of extreme points, and any element in MATH is a finite convex combination of these extreme points. Thus, it suffices to show that MATH is the only extreme point of MATH as follows. Since MATH exists, MATH and MATH are extreme points of MATH and MATH by REF with MATH. On the other hand, let MATH be any extreme point of MATH. Since MATH is an extreme point of MATH, there is a square submatrix MATH of MATH such that MATH and MATH satisfy the five conditions in REF . Since MATH for MATH, MATH by REF . Since MATH is square, MATH and MATH. Then, by REF , MATH. Since MATH, we have MATH as desired. |
cs/0011018 | Straightforward. |
cs/0011018 | REF follows from the fact that MATH if MATH and MATH otherwise. REF follows from REF . For REF , we let MATH be the online player's MATH-th pure strategy; and let MATH be the the adversary's MATH-th pure strategy. As in REF, the payoff matrix MATH is defined by MATH. The statement then follows from the facts that MATH and that MATH if MATH or MATH otherwise. |
cs/0011018 | We use MATH to emphasize the dimension MATH of MATH. Let MATH be the submatrix of MATH obtained by deleting row MATH and column MATH. To expand MATH along the first row of MATH, observe that MATH. Furthermore, the first column of MATH equals MATH times that of MATH, while the other columns of MATH equal the corresponding ones of MATH; thus MATH. For MATH, MATH because in MATH, the first column equals MATH times the second column. Hence, MATH . The lemma immediately follows by induction on MATH. |
cs/0011018 | Let MATH, and MATH. By REF , MATH exists. Below we prove MATH and MATH. Then, REF follows from REF and the fact that MATH, MATH, and MATH. REF follows from REF and the fact that every component of MATH and MATH is greater than MATH. REF follows from REF To prove MATH and MATH, observe that MATH can be obtained by swapping MATH and MATH in MATH, and the MATH-th column MATH of MATH can be obtained from the MATH-th row of MATH by the same operation. Therefore, MATH if and only if MATH, and we only need to establish MATH. Since MATH if and only if MATH, we only show MATH for MATH as follows: MATH . |
cs/0011018 | Let MATH. By REF , MATH. By algebra, MATH is a decreasing function of MATH. Thus, MATH is a function of MATH whose minimum occurs at one end of the domain MATH. The lemma follows from this concavity. |
cs/0011023 | Since MATH knows MATH's bidding algorithm, MATH can perform at least as well as MATH by employing the same algorithm. Then this lemma follows from the fact that our auction is a zero-sum game. |
cs/0011023 | Let MATH be MATH's optimal bids for the MATH objects, respectively. MATH's probability of winning the MATH-th object is MATH. Since MATH's bids are within MATH, it is not to MATH's advantage to bid over MATH. Hence MATH and MATH. MATH's optimal bids maximize MATH as follows: MATH . |
cs/0011023 | Note that MATH. To show that MATH is a joint probability density function, we need only verify that the integral of MATH over MATH is MATH. Let MATH . Consider the case MATH. Then MATH . Hence if MATH, MATH which equals MATH . By symmetry, if MATH, MATH . It can be verified that MATH . Thus, MATH . |
cs/0011023 | Note that MATH, whether MATH is even or odd. CASE: This lemma is correct since if a random variable MATH is drawn from the uniform probability distribution on MATH, then MATH has the same probability distribution. CASE: The proof of REF shows that the marginal probability distribution of each MATH is MATH for MATH. It remains to show that MATH are also distributed the same way. Because these three random variables are symmetric to each other in REF , we only discuss MATH in detail. Let MATH. Since MATH and MATH are defined on MATH, MATH is defined on MATH. We have two cases: MATH and MATH. The two cases are symmetric, and we discuss only the latter. Let MATH denote the probability distribution of MATH. Then, MATH . Since MATH, MATH can take the form of REF , and we can obtain MATH. Since MATH, MATH is the probability distribution of MATH. |
cs/0011023 | REF gives an upper bound for MATH. REF give an upper bound for MATH, which in turn gives a matching lower bound for MATH because MATH. |
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