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math/0004054
We argue as follows: assume MATH; we recall estimate REF: MATH where MATH is given by REF and MATH satisfies REF. REF implies MATH, and thus REF simplifies as MATH . REF of MATH implies that MATH . Moreover, relation REF leads to MATH . This relation implies immediately that MATH and therefore, thanks to REF MATH . If MATH, the last relation implies immediately that for MATH large enough, MATH is at least equal to MATH. Assume now that MATH; now, the situation is more delicate, since none of the inequalities established so far implies an estimate on MATH. If MATH, we are done. Otherwise, we shall estimate MATH from below. Already, relation REF implies MATH or in other words MATH which implies MATH and therefore MATH . Thus, we have shown that MATH . If MATH, the relations MATH are contradictory for MATH large; therefore MATH . Thus, we have shown REF.
math/0004054
We go back to the original scales and time MATH; then MATH . Therefore, in coordinates MATH (see REF ), we have the relations MATH . We have also the estimate MATH . If MATH, REF is a consequence of REF. Otherwise, we observe that MATH belongs to MATH for all large enough MATH; therefore, we are able to use the equivalences REF, whence MATH which is valid because the dominant term in the right hand side of the above expression does not vanish; indeed, REF of MATH and REF of MATH, we can see that MATH which implies REF; in the original coordinates, REF translates as MATH . We infer from estimate REF that MATH . In the coordinates MATH and MATH, the system REF can be rewritten MATH . But the explicit solution of REF with initial data REF is given by MATH . If MATH is non negative, it is clear that MATH stays non negative for all time larger than MATH. If MATH is negative, we use REF: we estimate from below MATH by MATH, and after simplifications, we get MATH . Therefore, REF holds for all MATH. In particular, MATH and MATH which proves that in this case the limit of MATH and MATH is MATH, as MATH tends to infinity.
math/0004056
The spinor group MATH is completely defined in terms of the algebra MATH: MATH where MATH is a special NAME - NAME group, and MATH . Let MATH is a representation of the algebra MATH in a vector space MATH. The representation MATH induces via REF a representation of the group MATH, and also via REF a representation MATH of the group MATH. Further, in virtue of the decomposition REF an inverse transition MATH induces a transition MATH, and MATH induces a restriction MATH by means of mutually orthogonal idempotents (projection operators) MATH. At this point MATH commutes with MATH and anticommutes with MATH.
math/0004056
REF immediately follows from REF . Further, follows to CITE we see that for some point MATH and a vector field MATH the NAME formula with respect to a decomposition MATH gives MATH or MATH . Let MATH be a local orthonormal tangent frame of the submanifold MATH at the point MATH and let MATH be a local orthonormal frame on the normal bundle MATH at MATH. Then MATH is a local section of the tangent bundle MATH of the manifold MATH restricted to MATH. Now we can to write REF in the matrix form: MATH . Further, let MATH and MATH be respectively connection REF - forms for MATH and MATH lifted to MATH and MATH. If MATH is an usual double covering, then REF can be written as follows MATH . Using the standard formula CITE for MATH we obtain from REF MATH where MATH is a standard basis of the space MATH, MATH is a standard basis of the space MATH. Let MATH be a spinor bundle of the manifold MATH. In virtue of the decomposition MATH we have MATH, where MATH. Let MATH and MATH be NAME - NAME connections on the bundles MATH and MATH, respectively. Then MATH is a NAME - NAME product connection on MATH. At this point REF takes a form MATH where MATH is the NAME multiplication defined by REF . Whence in accordance with the definition of the spinor derivative REF and identifications MATH follows REF .
math/0004056
In the case of the Lorentzian manifold MATH we have the following immersions MATH and MATH. At this point in accordance with REF on the surfaces MATH and MATH there exist the spinor fields MATH and MATH, respectively. Let us find a NAME operator on the surface MATH. First of all, in accordance with REF and the formulae REF let consider the following two operators: MATH and MATH . The both operators act on the section of the spinor bundle MATH. Using REF we obtain MATH where MATH. Further, let MATH be the spinor field on the Lorentzian manifold MATH, then from REF follows MATH where MATH is a NAME operator of the surface MATH defined on the restriction MATH. We suppose now that the spinor field MATH on the manifold MATH is a real Killing spinor, that is, there exists such a number MATH that for any vector field MATH the derivative of MATH in the direction MATH equals MATH . Therefore, from REF for the restriction MATH we have MATH where MATH. Since MATH, then in virtue of the relation REF of REF from the last equation we obtain (recalling that MATH and MATH are the units of the pseudo - quaternion) MATH . Further, for the immersion of the time - like surface MATH the analogous calculations give (at this point MATH and MATH are the quaternion units) MATH .
math/0004061
Let MATH be an integral basis of MATH; complete it with integral vectors MATH to a basis of MATH. In a neighborhood of MATH the vectors MATH generate MATH independent, commuting, locally Hamiltonian vector fields MATH. The closed forms MATH define a family of MATH functions MATH. The commuting relations of such functions originate a constant skew-symmetric matrix. By a linear change of coordinates in MATH - hence not preserving integrality - we can replace the vectors MATH with a new family of vectors whose associated REF-forms have commuting relations that are represented by the rank MATH skew symmetric matrix MATH . We still denote MATH the new vectors and MATH the functions associated to them. The vector fields MATH commute, hence we can find functions MATH, MATH such that MATH for MATH. The MATH functions MATH are independent and, by a simple calculation MATH. Hence the commuting relations for these functions can be summarized by the matrix MATH which has blocks of dimension MATH and MATH. The construction we just did gives no informations on what the matrix MATH could be, but we can mimic a technique in CITE: consider the manifold MATH. MATH has an induced symplectic structure because its defining functions have non degenerate NAME brackets. Define the remaining MATH functions by choosing canonical (complex) coordinates in a neighborhood of MATH in MATH, MATH, and extending them to a neighborhood of MATH as constant along the flow of the symplectic gradient of the functions MATH. Finally we can replace the functions MATH with functions in involution. Note in fact that MATH are functions of MATH only, so we can obtain the new coordinates replacing the function MATH by the function MATH for appropriate MATH. This can be done in view of the fact that MATH is a closed REF-form, and hence it is locally the differential of a REF-form MATH; this defines the functions MATH. Finally define the functions MATH for MATH and MATH for MATH. It is easy to verify that MATH acts as the identity on the coordinates MATH. Hence the action of MATH has isotropy representation that can be reduced to a representation in MATH.
math/0004061
From REF it is obvious that the MATH belong to MATH. Let MATH, for MATH, be the list of real infinitesimal weights at MATH. Assume that the MATH are a set of MATH-generators of MATH. An element of MATH stabilizes all points of MATH if and only if MATH for all MATH (MATH means MATH) if and only if MATH for all MATH, if and only if MATH is an element of MATH, if and only if MATH is the identity of the torus. For the converse, if we assume that MATH is a strict subgroup of MATH, by eventually a MATH change of basis we can assume such subgroup is MATH with at least MATH not REF. The element of the torus MATH stabilizes all the points of the manifold without being REF. This contradicts effectiveness.
math/0004061
The notion of local maximum and local minimum makes perfect sense for closed REF-forms. A proof of this fact would require a long digression hence we omit it. A partial result (for rational REF-forms) is used in CITE. A full proof can be found in CITE.
math/0004061
Let MATH be a point fixed by the action. The real infinitesimal weights of the isotropy representation at MATH, MATH, must MATH-generate MATH and are exactly in number of MATH. By a MATH change of coordinates in MATH we can assume that MATH, where MATH is an integral basis for MATH and the MATH indicates the dual element (MATH). So the image of the local momentum map is the vertex at the origin of the standard tetrahedron. Choose now any MATH that is in the first octant of MATH, in other words choose a vector all whose coefficients with respect to the chosen basis are positive. The form MATH has a local minimum at MATH so, it must be exact. CASE: The subset of MATH associated to exact Hamiltonians is a subvector space and all the first quadrant is mapped to exact REF-forms. Hence the action is Hamiltonian.
math/0004061
Let MATH be a basis of the NAME algebra MATH. This basis is associated to MATH REF-forms MATH; to each of them is associated a finite dimensional MATH-vector space MATH. For any MATH-tuple of real numbers MATH such that the set MATH of MATH-subvector spaces of MATH is in direct sum, the form associated to the vector MATH has minimal kernel. In other words, given any such MATH-tuple, for any MATH in MATH . Observe that the MATH-tuples for which the family MATH is not composed of free summands are a countable subset of a continuum space.
math/0004061
Assume that the generic Hamiltonian has rationality degree MATH. Then we can choose a basis MATH of the free part of the first homology group of MATH so that MATH . This follows from the Theorem of Elementary Divisors and REF . If the generic Hamiltonian was less than MATH-rational then the matrix MATH would have less than MATH non zero lines. Hence there would be a relation of linear dependence among the lines. Such relation specifies a choice of an exact Hamiltonian, contradicting the hypothesis.
math/0004061
The thesis is equivalent to the fact that, if a quasi-periodic function generates a MATH-action, then MATH. The negation of this last statement can be proven to be equivalent to the following situation: a torus MATH, with MATH, acts on a closed symplectic manifold. This action is such that MATH for some MATH maximally irrational (that is, the one parametric subgroup generated by MATH is dense in MATH). Let MATH be a point that is a maximum for the Hamiltonian MATH; this point must exist given that the manifold is compact and MATH is exact; it is easy to prove that MATH is also a fixed point by the action of MATH. The fact that MATH is a maximum for MATH implies that the real infinitesimal weights of the isotropy representation at MATH, that we denoted by MATH, must satisfy the inequalities MATH . Some of these inequalities must be equalities, otherwise an argument as in REF would imply that the action is Hamiltonian. Call MATH the set of indices such that equality holds. We plan to show that MATH. If not there is a vector MATH (we are identifying MATH with MATH) such that MATH for all MATH. So, for MATH small, MATH . By REF this implies that the vectors MATH are associated to one-valued Hamiltonians, contradicting the hypothesis that MATH. We hence deduce that the space MATH is rational, and MATH cannot be maximally irrational.
math/0004061
In REF we proved that the smallest covering from which it is possible to define the momentum map as a function is that in which the pull back of a generic Hamiltonian is exact. Such covering is regular, with group of deck transformations which is free Abelian of rank equal to the rationality degree of the generic Hamiltonian. We denote the elements of the deck transformation of MATH by MATH, the projection in MATH of a point MATH will be denoted by MATH. Let MATH be a point of MATH. The image of MATH,the local momentum map at MATH, defines a convex wedge, MATH (that could well be MATH). For any point MATH in MATH above MATH the image of the momentum map MATH is contained in MATH . The convex wedge MATH is defined by some inequalities MATH . Each of such inequalities yields a local maximum at MATH for the Hamiltonian MATH. By REF the Hamiltonian MATH is exact and has unique maximum at MATH. This implies that for any MATH in MATH and for any MATH above MATH and this shows exactly that the image of MATH is bounded by the set of affine equations needed to conclude. We can now prove the convexity of MATH. Let MATH and MATH be two points in MATH, assume the interval MATH is not contained in the image of MATH. Let MATH be the first point in the segment not in the interior of MATH. We can find a sequence MATH in MATH such that MATH and MATH converges to MATH. The sequence MATH admits a subsequence MATH that converges to a point MATH in MATH. The local momentum map at MATH maps a neighborhood MATH of MATH into a convex wedge MATH. The points MATH belong to MATH for all MATH big enough. For every MATH big enough there is an unique open set MATH containing MATH and a point MATH above MATH. The point MATH must belong to all the convex wedges MATH otherwise it could not be in the closure of the image of MATH. Hence it is an attained value MATH. It is easy to be convinced that MATH is a proper wedge of MATH and that it does not contain the semi-line MATH . By REF we conclude that MATH cannot belong to the image of MATH. If the Hamiltonian map is such that MATH, then the image of the momentum map MATH is surjective. Let MATH a base for the cycles of the free part of MATH and call MATH . The MATH's are MATH vectors of MATH. The hypothesis implies that such vectors are a system of generators of MATH. In fact, if they were not, then the lines of MATH would be dependent and so MATH . This expresses a non zero, one-valued Hamiltonian. We can also think of MATH as a non zero vector orthogonal to all the MATH; such a vector can be found if and only if the MATH's do not span MATH. Since MATH must contain all the MATH-subgroup generated by the vectors MATH; in fact MATH. Hence, by the just proven convexity, the image of MATH must be MATH. Since the subalgebra MATH is rational we can choose a rational algebraic complement MATH so that not only MATH and MATH, but also MATH. The projection of the momentum map MATH on MATH factors through a function MATH. The image of MATH is a convex polytope as described in CITE and CITE. This means that there are critical manifolds fixed by the MATH action, MATH, and MATH. This implies that MATH. We are left to show that the image of MATH is exactly MATH. Let MATH be a point of MATH that projects on MATH, and let MATH be an element of MATH, the deck transformations of MATH. The element MATH has first component that is the same as that of MATH, while the second component is translated by a vector, associated to MATH, as described in REF . We can use, again, convexity of the image of MATH to show that MATH. This concludes the proof.
math/0004061
The NAME algebras of the two groups are the same and the fundamental vector fields are the same.
math/0004061
Assume the form MATH has a local maximum at the point MATH. We claim that MATH must then have a local maximum at MATH. In fact the form MATH is orthogonal to vectors of the form MATH, for any MATH (in other words MATH for any MATH); hence the Hessian of the form MATH, at the point MATH, has same signature and same rank as the Hessian of MATH at the same point.
math/0004061
In CITE was proven that the set MATH is a convex polytope MATH, and that MATH. We just have to show that MATH . We plan to use the fact that MATH is a MATH-manifold, and that the momentum map MATH, restricted to MATH, has values in MATH. Assume REF is false, let MATH be a point in MATH and MATH such that the line MATH is not all contained in MATH. The line in question must intersect MATH, since the projection on MATH of MATH is MATH. With a line of proof similar to that in REF there must be a point MATH and an attained value MATH that is not in MATH. Hence there must be a plane, defined by a linear equation MATH that bounds on one side the image of the local momentum map at MATH. Observe that this plane must be transverse to the line MATH, that is, MATH. In particular MATH must be non zero. The Hamiltonian associated to MATH has a local maximum at MATH, and hence, by REF must be exact in MATH. This is inconsistent with the hypotheses of the proposition, given that MATH is not zero.
math/0004061
Let MATH be a small perturbation of the given form MATH (MATH is a small closed REF-form). The theorem states that REF-form MATH is exact if and only if REF-form MATH is exact. There is one obvious implication in this statement: non exact Hamiltonians remain non exact Hamiltonians. If MATH is not exact then it must be true that MATH for some MATH non trivial loop in the fundamental group of the manifold. Any change in the symplectic form changes smoothly the form MATH, and so changes smoothly the integral that must hence be non zero for small deformations of MATH (non exactness is an open property). What is not obvious is local stability of exact Hamiltonians. In Subsection REF we proved that MATH is exact if and only if it has a minimum. This is equivalent to the existence of a stationary point MATH for MATH (that is, MATH) such that the Hessian matrix of MATH at MATH is positive definite; this is equivalent to the fact that the real infinitesimal weights of the isotropy representation at MATH, evaluated at MATH, give positive numbers or zero. Only the signs of the weights are dependent on the symplectic structure, and their dependence is continuous. So, the existence or not existence of stationary points with the above properties is independent of small equivariant changes of the symplectic REF-form; hence exactness is.
math/0004062
CASE: Since MATH is a morphism, we prove it for the generators. Let MATH, MATH. Then MATH, and we have MATH and analogously for MATH. CASE: As the previous item, we can prove it for generators MATH, MATH. We have MATH . CASE: Simply MATH . We remark that these equalities are easily understood with drawings. CASE: By induction. Notice first that MATH . For MATH we have MATH . Suppose the equality is true for MATH. Then MATH .
math/0004062
By induction, for MATH being MATH . If the equality holds for MATH, we have MATH .
math/0004062
Let MATH. We consider MATH, where MATH is the adjoint in MATH. It is easy to see by induction that this element is the action of the left hand side of REF on MATH. Now, the right hand side acts on MATH as a scalar, namely MATH . The assertion follows at once.
math/0004062
Let MATH, and let us define inductively MATH. Taking on MATH the bi-degree given by MATH, MATH, we have MATH. Furthermore, it is proved in REF that MATH and MATH. It is immediate to see by induction (it follows also from REF) that MATH, and MATH, where MATH. Thus, MATH if MATH. We shall prove that the set MATH is linearly independent. To see this, we compute MATH from where MATH if MATH. It is clear also that if MATH then MATH. Suppose inductively that for MATH, the set MATH is linearly independent. Then we have that if there were a linear combination MATH we would get, applying MATH, a linear combination MATH where the MATH's are the MATH's multiplied by non-zero factors. By the inductive hypothesis, we have that MATH. We apply now MATH and get that MATH. Continuing in this way, we get that MATH for all MATH, proving the inductive thesis. Now the assertion follows at once noting that this set lies in MATH, and then by REF the set MATH is linearly independent in MATH. Now, for the nilpotency order of MATH on MATH to be MATH it can happen that MATH or that MATH. In the first case we have MATH, and then MATH . In the second case it is easy to see using derivations that we also have MATH for MATH annihilates and MATH gives the same element in MATH when applied to both sides of the equality. We consider now the three cases in the statement. CASE: We have MATH, whence MATH. These equalities show how to write any monomial in MATH as a combination of the elements of the set MATH, from where MATH and the equality follows. CASE: We have MATH, and hence MATH . It can be seen, using derivations, that MATH. Furthermore, it can also be seen that MATH. We conclude now as in the previous case, since these equalities show how to write any monomial as a combination of those of REF. CASE: The same computations as in the previous case hold here, except for the relation between MATH and MATH. We are looking for the condition on the matrix for REF to be a basis of MATH. For this to happen, we must be able to write MATH as a linear combination of those elements; but now, the bi-degree of MATH is MATH and, moreover, it lies in the kernel of MATH. By REF , we have to write MATH as a combination of the elements in REF of bi-degree MATH which also lie in MATH, but there is only one such element: MATH. Hence we should have MATH. Taking MATH in both sides, we are led to the equations MATH for MATH. Since MATH and MATH, these equations are equivalent to MATH from where the conditions we stated are clear.
math/0004062
Let MATH be the space of the representation MATH. For MATH we denote also by MATH the element MATH. Let MATH be a basis of MATH. We have MATH, and hence MATH. We have an inclusion MATH (in fact, MATH is free as a right MATH-module by CITE) and hence MATH is finite dimensional. Now, MATH is of diagonal GT with matrix MATH, MATH, and thus it is of NAME type. Hence, the NAME matrix of this braiding has MATH in the main diagonal and MATH outside the main diagonal. Since by a result of NAME (see CITE) this matrix must correspond to a finite datum, the condition on MATH follows.
math/0004062
Since MATH has a finite number of simple modules up to isomorphism, it is enough to see that each simple module MATH can appear no more than MATH times (MATH depending on MATH) in a module MATH for it to be of finite NAME rank. Let then MATH be the space affording MATH, MATH and MATH such that MATH. Suppose MATH (MATH). Then we take MATH, where MATH is the element of the i-th copy of MATH corresponding to MATH. It is immediate to see that MATH is of diagonal GT with matrix MATH, MATH. Now, MATH since MATH is of finite NAME rank (since so is MATH), and MATH since MATH is of odd order. The same argument as in the previous lemma tells that MATH if MATH and MATH if MATH.
math/0004062
Straightforward (see CITE).
math/0004062
We have to prove that the group-likes associated with the basis MATH of MATH generate a finite group. These group-likes are nothing but MATH, MATH. Let MATH, let MATH be the least common multiple of the MATH's and let MATH be the group of MATH-roots of unity. Let MATH and MATH the symmetric group on the (finite) set MATH. It is clear then that there is an inclusion of the group generated by the MATH's into MATH.
math/0004070
We may assume that MATH, since otherwise MATH . But then actually MATH, since on MATH we have MATH, so that on this set MATH, which is integrable. Assume first that MATH. Fix MATH, and let MATH . Notice that MATH since MATH implies MATH. Thus for a very large MATH, we can break up MATH into convenient strings of terms as follows. There is maybe an initial string of MATH's during which MATH. Then there is a first time MATH when MATH, which initiates a string of no more than MATH terms, the sum of which is positive (using on each of these terms the fact that MATH). Beginning after the last term in this string, we repeat the previous analysis, finding maybe some MATH's until again some MATH initiates another string of no more than MATH terms and with positive sum. The full sum of MATH terms may end in the middle of either of these two kinds of strings (MATH's, or having positive sum). Thus we can find MATH such that MATH . Integrating both sides, dividing by MATH, and letting MATH gives MATH . Letting MATH and using the Dominated Convergence Theorem concludes the proof for the case MATH. To extend to the case MATH, for MATH let MATH, so that MATH and MATH a.e. and in MATH. Then for fixed MATH . Therefore MATH again by the Dominated Convergence Theorem. The full result follows by letting MATH.
math/0004070
It is enough to show that MATH . For then, letting MATH, applying this to MATH gives MATH so that MATH and hence MATH . Consider first MATH and its associated MATH, denoted by MATH. For any invariant function MATH such that MATH, for example MATH, we have MATH, so the Theorem gives MATH . Thus MATH is integrable (and, by a similar argument, so is MATH.) Now let MATH be arbitrary and apply the Theorem to MATH to conclude that MATH .
math/0004079
Without loss of generality, we may assume that MATH is a smooth point of the connection. Let MATH be a point of multiplicity MATH, and let MATH be a local coordinate at MATH. Note the effect of twisting (that is, replacing MATH by MATH) is to replace the local connection matrix MATH at MATH with respect to a basis MATH with MATH for the basis MATH. In particular, MATH will remain an admissible singularity for the new connection. After such a twist, we may assume MATH, with MATH. We argue by induction on MATH. If MATH, we replace MATH by MATH and argue as above. Assume MATH. Let MATH, and embed MATH in MATH via MATH. If MATH is a local parameter at MATH, and MATH is a local basis of MATH at MATH, then MATH is a local basis of MATH, and if MATH is the local block matrix of the connection MATH in the basis MATH, then MATH is the local block matrix of the connection in the basis MATH. If MATH has a local expansion MATH, then the polar part of this connection is MATH . Thus replacing now MATH by MATH, the local equation of the connection at MATH becomes MATH and therefore, is pseudo-admissible. On the other hand, MATH has decreased. We conclude by induction.
math/0004079
We show first that MATH is independent of the local bases. We work locally around a point MATH which is a singular point of the connection with multiplicity MATH. We assume first that MATH for a local coordinate, and that the connection matrix is MATH with MATH and MATH. (This includes both the admissible and pseudolog cases.) We take a gauge transformation of the form MATH with MATH. This amounts to MATH . Here MATH. We claim MATH is invariant. We compute (writing MATH, and computing modulo MATH) MATH (Note that MATH has entries in MATH, justifying replacing MATH by MATH.) To show invariance, it will suffice to show MATH . This expression can be written MATH . Verticality gives MATH . Multiplying through by MATH, the expression MATH above becomes MATH . It remains to show independence of MATH. Let MATH where MATH is meromorphic on MATH and invertible on MATH. Consider a diagram MATH . The classes of the rigidified bundles MATH and MATH differ by MATH, where MATH is the image of MATH. It follows that (with notation as above) MATH . Replacing MATH with MATH in REF, this gives the desired invariance.
math/0004079
We have MATH . The formulas in the lemma follow easily from this.
math/0004079
The connection MATH is vertical. Vanishing for curvature terms involving MATH for MATH implies MATH . The tactic is to eliminate first MATH from MATH, then MATH etc. One easily verifies matrix relations MATH . In particular, MATH . Write MATH . This yields MATH . Applying again now the relations REF yields MATH . Assume inductively that for some MATH, one can write MATH as follows: MATH . Applying REF to the first line, and isolating the terms in MATH and in MATH, one obtains MATH with MATH . It remains to arrange MATH. To this aim, write MATH . Now we group those terms differently. To a tuple MATH, with MATH, we associate the tuple MATH with MATH, MATH, and otherwise MATH for MATH. Using the first relation of REF again, this gives for those REF terms together MATH . This shows that the relation REF is true, with MATH replaced by MATH. As the last equation of REF for MATH is MATH, one obtains by induction that MATH vanishes on the variety defined by REF.
math/0004079
Straightforward.
math/0004079
The first two traces are equal by the diagram. For the third, note that since MATH, it follows that one has an exact sequence compatible with the endomorphism multiplication by MATH . Since MATH has no constant term in MATH, it acts nilpotently on the right.
math/0004079
Write MATH. Note that neither side of the identity changes if we replace MATH by MATH so we may assume MATH. Also, by linearity, we may assume MATH. The matrix for the action of MATH on MATH, the entries of which are themselves matrices, is MATH . The matrix for MATH is MATH. We write MATH for the naive trace, that is, the sum of the diagonal elements. For example, MATH. Then MATH . Also MATH . (The residue is computed in the ring MATH. Since MATH is in the center of this ring, we may move MATH past MATH under the residue.) It will therefore suffice to show MATH . Let MATH and write MATH with MATH. The assertion becomes MATH . Let MATH be a MATH matrix. Then MATH . In particular, taking MATH it follows that MATH . Write MATH. We have MATH, and MATH otherwise. Thus MATH . The conditions on the tuples MATH over which the right hand sums are taken become MATH . Replacing MATH by MATH, these become MATH proving the lemma. Write MATH. We must show MATH . This amounts to MATH . Suppose first MATH. With reference to REF, one can isolate the terms in MATH ending in MATH for MATH and write MATH . Here MATH . (Notice that since each MATH, terms with MATH are impossible. Also, since MATH, necessarily MATH.) The remainder MATH is given by the terms MATH in MATH. In the sum for MATH these terms arise when MATH and MATH, that is, MATH. There are MATH such terms: MATH . Finally, in REF we consider the terms with MATH. Writing MATH and replacing MATH with MATH for some MATH, it suffices to show MATH . We start with MATH . Note MATH is not possible because MATH. Again, by grouping together the terms ending with MATH we get MATH which is the desired equation.
math/0004079
By REF , we may replace MATH with MATH. By REF MATH is the sum of the MATH on MATH. By REF this is the same as MATH. Note the factor MATH on the right is because as a matrix MATH acts on MATH while the trace one wants is the action on MATH.
math/0004079
Define absolute forms MATH, MATH and MATH: MATH . The local term at MATH is MATH . Applying trace to MATH yields MATH . (Note here that MATH is not closed as an absolute form!) Also, modulo MATH, we have MATH because the residue of an exact form vanishes. The local term thus becomes MATH . We have MATH . Thus MATH . Expanding MATH and substituting on the right in REF MATH . Thus, REF becomes after summing over MATH . It remains to compare MATH and MATH. Recall MATH. We have MATH . From these equations it follows that MATH . Since this difference has no pole at MATH, and since MATH is invertible, we find MATH . Making this substitution in REF proves the proposition.
math/0004082
That the sequence MATH is nonincreasing, invariant under reordering, and nonnegative when MATH follows from the fact that it is a scaled limit of MATH (see below); it suffices therefore to show that MATH with probability REF. We first observe that MATH . Since decreasing MATH cannot decrease MATH, we conclude that it suffices to prove finiteness when MATH and MATH. But then REF below applies, expressing MATH as an infinite product. Moreover, the conditions on MATH suffice to force convergence of this product, and thus MATH. But this immediately implies MATH, as desired.
math/0004082
This follows immediately from the corresponding fact for MATH.
math/0004082
By REF above, MATH . Then MATH where the last step is valid since MATH and MATH. The theorem then follows by taking the limit MATH.
math/0004082
Let MATH. Then MATH, so we can extend MATH by adjoining MATH with multiplicity MATH of distribution MATH; denote the resulting random multiset by MATH. But then by REF , we can change the total ordering MATH so that MATH becomes maximal instead of minimal. We then find that MATH induces an increasing subsequence of MATH with respect to this new ordering; thus MATH . On the other hand, this is the only maximal increasing subsequence that passes through MATH; any other maximal increasing subsequence can have size at most MATH. We thus find MATH . But MATH is distributed as MATH since before the reordering every maximal increasing subsequence passes through MATH, and MATH is independent of MATH. So if we take the expectations and subtract/divide by the contribution of MATH, we find that we need only show MATH . Let MATH be a nonnegative-integer-valued random variable with finite first moment, and let MATH be an independent geometric random variable of parameter MATH. Then MATH . Similarly, for MATH, if MATH is finite, then MATH . MATH . Similarly, MATH and thus MATH . The theorem then follows from the following lemma, since MATH by assumption.
math/0004082
An increasing subsequence in MATH can pass through a point on a strict row or column at most once; thus MATH is unchanged if we remove any excess multiplicity in those rows and columns. Let MATH be the resulting multiset, then MATH . It will thus suffice to prove that MATH. But the moment generating function of MATH is MATH, where MATH . This product converges to an analytic function with no pole inside the open disc MATH, and thus the result follows.
math/0004083
We will prove REF by constructing a sign-reversing involution on the set of summands appearing on the right-hand side of REF , for which the condition MATH is violated. The argument will be first presented for the special case MATH, and then extended to a general MATH. The following general construction will be useful in the proof. Let MATH be an arbitrary walk, and let MATH be a vertex lying on its loop-erased part MATH. The edge sequence of MATH is canonically a subsequence of the edge sequence of MATH; in the latter sequence, let us identify the unique entry of the form MATH which is contained in the loop-erased subsequence. (The case MATH is an exception, to be kept in mind.) The walk MATH is now partitioned at the end of the entry MATH into two walks MATH such that CASE: the last entry in the edge sequence of MATH contributes to MATH; CASE: MATH does not visit any vertices which lie on MATH, except for MATH. REF - REF uniquely determine the partition MATH of the walk MATH; in the special case MATH, we set MATH to be the trivial path MATH, and let MATH. Let us get back to the proof. Assume MATH, and let the walks MATH and MATH be such that MATH and MATH pass through at least one common vertex. Among all such vertices, choose the one (call it MATH) which is closest to MATH along the self-avoiding walk MATH. We then partition MATH into MATH, following the rules above (see REF - REF ). Let us denote MATH. With this notation, REF can be restated as follows: CASE: MATH does not visit any vertices which lie on MATH, except for MATH. Let us now split MATH at the point of its first visit to MATH. More formally, we define the partition MATH of MATH by requiring that MATH does visit MATH before arriving at its endpoint. By the choice of MATH, CASE: MATH does not visit any vertices which lie on MATH, except for MATH; CASE: MATH does not visit any vertices which lie on MATH, except for ending at MATH. Everything is now ready for the path-switching argument. Let us create new walks MATH and MATH . The map MATH is the desired sign-reversing involution. The basic reason for this is the similarity of REF , which allows to interchange the portions MATH and MATH. It furthermore ensures that the new walk MATH splits into MATH and MATH. In particular, the loop-erased part of the initial segment of MATH remains invariant: MATH. Now REF show that MATH splits at MATH, as needed. As a result, applying the same procedure to MATH recovers MATH, and we are done. The case of an arbitrary MATH is proved by the same argument combined with a careful choice of the pair of paths to which it is applied. Take a term on the right-hand side of REF which corresponds to a MATH-tuple of walks MATH. Assume that this term does not appear in REF . Thus the set of triples MATH, MATH, MATH, such that the walks MATH and MATH pass through MATH is not empty. Among all such triples, let us choose lexicographically minimal, in the following order of priority: CASE: take the smallest possible value of MATH; CASE: for this MATH, let MATH to be as close as possible to MATH along MATH, among all intersections with MATH, for all MATH; CASE: for these MATH and MATH, find the smallest MATH such that MATH hits MATH. We then proceed exactly as before, working with the pair of walks MATH. Thus we partition the walk MATH into MATH, as prescribed by REF - REF above. Then locate CASE: the first visit of MATH to MATH, starting at MATH and split MATH accordingly into MATH. Exchanging the portions MATH and MATH of the walks MATH and MATH provides the desired sign-reversing involution. A straightforward verification is omitted.
math/0004083
Consider the right-hand side of REF . First we note that the loop-erased walks MATH are pairwise vertex-disjoint, and therefore MATH must be the identity permutation, under the assumptions of REF . It remains to show that the restrictions on the MATH may be relaxed as indicated. Assume that the walks MATH,, MATH satisfy the condition in REF . Suppose that the corresponding condition in REF is violated, that is, some walk MATH intersects MATH, for some MATH. We may assume that among all such violations, this one has the smallest value of MATH, which in particular means that MATH does not intersect MATH. We can then combine segments of MATH and MATH to create a walk MATH. By assumptions of REF , MATH must intersect the walk MATH. On the other hand, neither MATH nor MATH intersects MATH, a contradiction.
math/0004083
This theorem can be proved by a direct argument similar to the one used in the proof of REF . To save an effort, we will instead use a simple observation that will reduce REF to REF . Let us define a new network MATH by splitting every boundary vertex MATH into a source MATH and a sink MATH, converting all outgoing edges MATH into MATH, redirecting all incoming edges MATH into MATH, and keeping the edge weights intact. Let MATH and MATH be the sets of sources and sinks, respectively. Then the hitting matrix MATH of the original network MATH becomes a submatrix of the walk matrix MATH of the transformed network MATH; more precisely, MATH for MATH, while MATH for MATH and MATH. To complete the proof, it remains to carefully reformulate the statement of REF for MATH (with each MATH replaced by MATH and each MATH replaced by MATH) in terms of MATH.
math/0004084
Since the map MATH is differentiable for all MATH, there exists a vector field MATH on MATH satisfying REF. We need to check that MATH is MATH-vertical and right invariant. We have that MATH which proves that MATH is MATH-vertical, by definition. Also, MATH which proves that MATH is also right invariant.
math/0004084
For manifolds without corners, this is a result from CITE. For manifolds with corners the proof is the same.
math/0004084
If MATH is integrable, then REF is REF is Proposition REF , and REF is REF. The proof of REF in the general case are the same.
math/0004084
If MATH is a differentiable groupoid (not just a local one), this is REF. For local groupoids the proof is the same.
math/0004084
Because each MATH is a differential groupoid, REF follows from REF on each MATH. From this we obtain the desired relation everywhere on MATH. REF follows from the relation MATH valid on each MATH, and hence everywhere on MATH.
math/0004084
We begin by writing MATH. Then, using REF , we can find sections MATH of MATH, MATH, MATH, depending smoothly on MATH, such that MATH . To complete REF , we now use a local groupoid MATH integrating MATH. The existence of such a MATH is ensured by CITE. By replacing MATH with an open neighborhood of MATH, if necessary, we may identify MATH with a subset of MATH, using the exponential map. Then, for small MATH and a relatively compact neighborhood MATH of MATH, the maps MATH and MATH are diffeomorphisms onto neighborhoods of MATH in MATH, by REF . The desired map MATH is obtained from MATH. The proof of REF is similar, using the differentiability of the multiplication in a local groupoid.
math/0004084
Suppose first that the family MATH is a differentiable atlas. We begin by showing that the groupoid structure on MATH, induced from the groupoids MATH, is compatible with the differentiable structure defined by MATH. That is, we need to check that the structural morphisms are differentiable. We now check this. First, the domain map is differentiable and submersive because MATH, for all MATH, MATH, and MATH. Next, it is enough to check that the map MATH, defined on MATH is differentiable. Let MATH be such that MATH, and let MATH and MATH be two maps such that MATH and MATH, and such that their restrictions to some small set of the form MATH, MATH, is in the family MATH. We need to show that the induced map MATH is differentiable. It is enough to prove this in a small neighborhood of MATH. Because MATH forms an atlas, using REF , we see that we may assume MATH, eventually by changing MATH or MATH. The differentiability of MATH then follows by combining REF , and REF . This is enough to conclude that MATH is a differentiable groupoid whenever MATH is an atlas. Let MATH be the NAME algebroid of MATH corresponding to the differentiable structure defined by MATH. We need to check that MATH. We have that MATH, by construction, and hence we can identify as a set MATH with the disjoint union of the restrictions MATH. The two differentiable structures on MATH (the first induced from MATH and the second induced from MATH) are the same because the differential of the map MATH of REF , MATH, canonically identifies MATH and MATH, if MATH is as in that lemma. The NAME algebra structures on MATH and MATH also coincide because the two possible brackets of two vector fields coincide on each strata MATH, and hence they coincide everywhere. We have thus proved that, if MATH forms an atlas, then MATH is a differentiable groupoid with MATH. Conversely, suppose now that MATH is endowed with a differentiable structure, and let MATH. Since MATH is MATH-connected, we can choose vector fields MATH such that MATH for some MATH. If MATH are chosen to form a basis of MATH, as in REF , then the map MATH must be a diffeomorphism of a set of the form MATH onto an open neighborhood of MATH, for some open subset MATH. We obtain that if MATH is a differential groupoid such that MATH, then the family MATH is an atlas.
math/0004084
Consider the curve MATH where MATH and MATH is chosen such that MATH. By assumption, MATH is a closed curve on MATH, and hence, by the assumption that MATH is MATH - simply connected, we can continuously deform this curve to the constant curve MATH, within MATH, through closed curves based at MATH. More precisely, we can find MATH such that MATH and MATH, for all MATH. By an approximation argument, we can assume that MATH depends smoothly on MATH, for each integer MATH. Moreover, after replacing MATH by a large multiple MATH and each MATH, MATH, by MATH repeated MATH times, we can assume that there exist compactly supported sections MATH, depending smoothly on MATH, such that MATH, MATH, and MATH . The desired deformation is obtained by letting MATH . To see that MATH is a smooth family of admissible sections, we use REF for each MATH and the fact that each MATH is a smooth admissible section.
math/0004084
Let MATH be a local groupoid integrating MATH. Choose MATH and a neighborhood MATH of MATH small enough such that MATH is a diffeomorphism onto an open neighborhood MATH of MATH in MATH. Let MATH be a compact neighborhood of MATH. Because MATH is a compact subset of the open set MATH and MATH is continuous, MATH, we can find a neighborhood MATH of MATH such that MATH is defined in MATH whenever MATH, and MATH, and such that MATH . Then we can define MATH . Because MATH and MATH is a diffeomorphism, we obtain that MATH is smooth also.
math/0004084
By REF , we see that we it is enough to show that the family MATH, defined using the maps MATH of REF , is an atlas. Let MATH and MATH be two such maps, defined on MATH and, respectively, on MATH, such that their images intersect. Let MATH be an element of this intersection. As in the proof of REF , we can arrange that MATH. Then MATH and hence MATH so we may also assume that MATH (and hence MATH). By replacing MATH and MATH with some smaller, relatively compact neighborhoods, if necessary, we may further assume that the sections MATH are compactly supported. Let MATH . Since each MATH is MATH-simply connected, by REF we can find a smooth family MATH, MATH, such that MATH is the identity, MATH, the restriction to each groupoid MATH is differentiable. The smoothness of the family MATH in this case is reduces to showing that MATH is a smooth section of MATH over MATH. Let MATH be the map defined in REF using the admissible sections MATH, and let MATH. Then MATH in a small neighborhood of MATH. Since MATH is a local diffeomorphism for each fixed MATH, this proves the theorem.
math/0004084
Recall that MATH, the path groupoid of the foliation MATH, consists of fixed end point homotopy classes of paths MATH that are fully contained in a single leaf, with respect to homotopies within that leaf. As explained above, the morphism MATH defines a leafwise flat connection on MATH that preserves the NAME bracket. Let MATH be a MATH-simply connected groupoid that integrates MATH. We define then a groupoid MATH that integrates MATH as follows. As a smooth manifold, MATH . To define the multiplication, observe first that the leafwise flat connection on MATH defines a parallel transport map MATH which is a NAME algebra isomorphism for any path MATH fully contained in a leaf. Since MATH is simply connected for each MATH, we obtain by exponentiation a group morphism MATH . That is, the leafwise flat connection on MATH lifts to a leafwise flat connection on MATH that preserves the NAME group structure on the fibers. We are ready now to define the groupoid structure on MATH. Note first that MATH for MATH. Then MATH, MATH, and the product on MATH is given by the formula MATH where the composition of paths is given by concatenation. The flatness of MATH gives that MATH, which guarantees the associativity of the product.
math/0004084
We may assume that MATH is connected. Since semisimple NAME algebras are rigid, all fibers of MATH will be isomorphic NAME algebras. Fix one of these algebras and denote it by MATH. Also, let MATH be the group of automorphisms of MATH. Then, if we define MATH (the fibers are the sets of NAME algebra isomorphisms MATH), we obtain a MATH-principal bundle on MATH, which acquires by pull-back a foliation MATH of the same codimension as MATH. The path groupoid MATH of the foliation MATH has an induced free action of MATH, and we define the groupoid MATH by MATH. The composition of two paths in MATH is obtained by choosing composable liftings in MATH. Because the NAME algebra of MATH is MATH, we obtain that MATH integrates MATH.
math/0004084
Using REF or REF , we see that each of the algebroids MATH, obtained by restricting MATH to the stratum MATH, is integrable. By REF , the NAME algebroid MATH is then integrable.
math/0004084
The set of points satisfying REF at a face MATH is closed under the NAME bracket, by definition of a NAME flag. Since MATH is also closed under the NAME bracket, it follows that MATH is a NAME algebra. Fix a point MATH in the interior of a face MATH of codimension MATH with defining functions MATH. Let MATH be the corresponding hyperfaces, MATH, and let MATH. Also let MATH where the order of composition is not important because MATH, by the definition of boundary NAME datum. Let MATH, MATH, be a multi-index and MATH which, we recall, is a vector bundle on MATH, again by the definition of a boundary NAME datum. Also, let MATH and choose, arbitrarily, a complement MATH to MATH in MATH. Let MATH be the set of vector fields MATH defined using the diffeomorphism MATH for some small MATH. Then the restriction of MATH to MATH is MATH . Since this is a MATH-projective module - the module of sections of a bundle isomorphic to the direct sum of MATH and the trivial bundle generated by the set MATH (MATH), we obtain that MATH is a projective MATH - module, as desired.
math/0004084
Let MATH and MATH be the foliation and, respectively, the boundary NAME data defining MATH. Observe that the sections of MATH are vector fields that are tangent to all faces of MATH, and hence each open face of MATH is MATH-invariant. This means that the stratification of MATH by open faces is a MATH-invariant stratification of MATH. Fix a face MATH of codimension MATH with defining functions MATH, such that MATH. Let MATH be the interior of MATH, and denote MATH. For MATH, the intersection MATH is an integrable sub-bundle of MATH such that MATH maps MATH surjectively onto MATH. Moreover, the map MATH defines a splitting of MATH, that is, a NAME algebra morphism MATH. (This splitting is implicit in the description of MATH given in REF.) Consequently, MATH is integrable, by REF .
math/0004084
This follows from REF .
math/0004087
Let MATH, and let us consider the linear operator MATH associated to it defined by MATH . Let us suppose without loss of generality that MATH. Since MATH is completely continuous, we get that MATH .
math/0004087
If both MATH and MATH are scattered, then MATH is a NAME space and therefore MATH has the DPP. Now, let us suppose that one of them, say MATH, is not scattered. Since MATH is infinite, MATH is not NAME, and therefore there exist two sequences MATH and MATH such that MATH is weakly null and MATH for every MATH. Also, since MATH is not scattered, MATH contains an isomorphic copy of MATH, and therefore there exists a continuous surjective operator MATH CITE. Then let us consider the trilinear form MATH defined by MATH and let us consider the linear operator MATH canonically associated to it given by MATH . It is clear that MATH where MATH is given by MATH and MATH is the canonical linear isometry identifying both spaces. Since MATH, MATH and MATH are all of them weakly compact, so is MATH. So we now just have to see that MATH is not completely continuous. Let us consider a sequence of bounded functions MATH such that MATH, where MATH is the canonical basis of MATH. Then, according to REF , the sequence MATH weakly converges to zero, but, for each MATH, MATH a contradiction.
math/0004087
If MATH is scattered, then MATH is a NAME space for every MATH, and therefore MATH has the DPP. Now, if MATH is not scattered, we can consider the trilinear form MATH defined by MATH that is, the symmetrized respect to the two first variables of the trilinear form used in REF . Now we can apply analogous reasonings as before to conclude that the linear operator MATH defined by MATH is weakly compact and not completely continuous.
math/0004088
This follows immediately from CITE, where it is stated for the irreducible unitary representation with parameter MATH of the NAME group.
math/0004088
It is sufficient to observe that REF implies that the map MATH defines a continuous linear functional on MATH for MATH.
math/0004088
For MATH, we have MATH and therefore MATH .
math/0004088
For real MATH this is the infinitesimal version of the previous proposition, just differentiate MATH with respect to MATH and set MATH. For complex MATH it follows by linearity.
math/0004088
We will show the result for MATH, the general case can be shown similarly (see also the proof of REF ). Let MATH be a NAME function. Let MATH. Then we have MATH . Similarly, we get MATH and together these two inequalities imply MATH for MATH.
math/0004088
Let MATH be any sequence such that MATH and MATH for some MATH in the weak topology. To show that MATH is closable, we have to show that this implies MATH. Let us evaluate MATH between two exponential vectors MATH, MATH, MATH, then we get MATH and therefore MATH, as desired.
math/0004088
For MATH and MATH such that also MATH, we get MATH where we used REF . The first equality of the proposition now follows, since both MATH and MATH are derivations. The second equality follows immediately.
math/0004088
This formula is a consequence of the fact that MATH for all MATH, we get MATH .
math/0004088
This is obvious, since MATH is a derivation.
math/0004088
We have to show that for any sequence MATH in MATH with MATH and MATH, we get MATH. Let MATH. Set MATH, MATH, MATH, and MATH, then we have MATH and MATH. Thus we get MATH for all MATH, where MATH . But this implies MATH, since MATH is dense in MATH.
math/0004088
It is not difficult to check this directly on the NAME operators MATH, MATH. We get MATH and MATH . By linearity and continuity it therefore extends to all of MATH.
math/0004088
These properties can be deduced easily from the definition of the gradient and the properties of the derivation operator MATH and the inner product MATH.
math/0004088
We assume that such an operator MATH exists and show that this leads to a contradiction. Let MATH be the operator defined by MATH, it is easy to see that MATH and that MATH is given by MATH . Therefore, if MATH existed, we would have MATH which is clearly impossible.
math/0004088
From REF we get MATH . But since MATH and MATH commute with MATH, we can pull them out of the anti-commutator, and we get MATH .
math/0004088
Let MATH. Recalling the definition of MATH we get the following alternative expression for MATH, MATH . Evaluating this between two exponential vectors, we obtain MATH .
math/0004088
Let MATH be a sequence such that MATH and MATH in the weak topology. Then we get MATH for all MATH, and thus MATH.
math/0004088
CASE: Let MATH. We set MATH then we have MATH, and therefore MATH . On the other hand we have MATH, and MATH . Taking the difference of these two expressions, we get MATH . CASE: A straightforward computation gives MATH where we used that MATH defines a bounded operator, since MATH. REF can be shown similarly.
math/0004088
It is sufficient to check this for functionals of the form MATH, MATH. We get MATH .
math/0004088
Let MATH and set MATH then we have MATH for all NAME functions MATH. Therefore MATH for all MATH, since MATH and MATH for all MATH, and thus MATH . But this implies that the density of MATH is contained in the NAME spaces MATH for all MATH.
math/0004088
Since MATH, we have MATH for all polynomials MATH. We therefore get MATH . The hypotheses of the proposition assure MATH and therefore allow us to get the estimate MATH for all polynomials MATH. But this implies that MATH admits a bounded density.
math/0004088
To prove this, we show that the NAME integrals satisfy the same formula for the matrix elements between exponential vectors. Let MATH be such that its creation integral in the sense of NAME and NAME is defined with a domain containing the exponential vectors. Then we get MATH . For the annihilation integral we deduce the formula MATH .
math/0004089
REF follow from the greedy algorithm (see REF ). By REF of the exchange capacity, we have MATH. Since MATH, it follows from REF that MATH. Thus we obtain MATH.
math/0004089
If MATH, for each MATH the first MATH vertices in MATH must belong to MATH. Then it follows from REF that MATH. Since MATH and MATH, this implies MATH.
math/0004089
For any MATH we have MATH and MATH, and hence MATH.
math/0004089
When the MATH-scaling phase finishes with MATH, we have MATH for every MATH, which implies MATH as well as MATH. Similarly, when the MATH-scaling phase finishes with MATH, we have MATH for every MATH, which implies MATH as well as MATH. When the MATH-scaling phase ends with MATH due to REF , then MATH and MATH. By the definitions of MATH and MATH, we also have MATH for every MATH and MATH for every MATH. Therefore we have MATH as well as MATH.
math/0004089
By REF , the output MATH of the algorithm satisfies MATH. For any MATH, the weak duality in REF asserts MATH. Thus we have MATH, which implies by the integrality of MATH that MATH minimizes MATH.
math/0004089
The algorithm starts with MATH and ends with MATH, so the algorithm consists of MATH scaling phases. Each scaling phase finds MATH-augmenting paths. To find an augmenting path, we perform at most MATH pushes per extreme base. A saturating push requires MATH time while a nonsaturating one MATH time. Here, note that there are less than MATH nonsaturating pushes per augmenting path. Hence, the time spent in pushes per augmenting path is MATH. After each augmentation, we also update the expression MATH, which also takes MATH time per augmentation. Thus the overall complexity of MATH is MATH.
math/0004089
Let MATH be any minimizer of MATH. There exists a vector MATH with MATH such that MATH. Note that MATH for each MATH. By REF , there exists a subset MATH such that MATH. Then we have MATH. This implies MATH due to the assumption, and hence MATH.
math/0004089
Each time we call the procedure MATH, the algorithm adds a new arc to MATH or deletes a set of vertices. This can happen at most MATH times. Thus the overall running time of the algorithm is MATH, which is strongly polynomial.
math/0004091
MATH is a distance function. Indeed, MATH is symmetric and MATH if and only if MATH. We have MATH and MATH . Hence, MATH. Let MATH be a MATH - dispersed subspace of MATH. Since MATH is surjective, for any MATH, there is MATH such that MATH. Let MATH. By REF , MATH for all MATH. Thus MATH establishes an isometry between MATH and MATH.
math/0004091
Let MATH. We define MATH by MATH for MATH. (Since MATH,MATH.) We have, by the triangle inequality, MATH . On the other hand, MATH for MATH. Therefore, MATH for all MATH.
math/0004098
Immediate from REF , and REF .
math/0004098
Most of the details of the proof are contained in CITE and CITE, so we only sketch points not already covered there. The essential step (for the present applications) is REF , which shows that MATH intertwines the isometry MATH-with the restriction of the unitary operator MATH to the resolution subspace MATH. We have: MATH for all MATH, and all MATH. This proves REF .
math/0004098
The correspondence is MATH with MATH and in the reverse direction, MATH does the job, as can be checked by direct substitution.
math/0004098
It is clear that the operator MATH in REF is a projection if the family MATH consists of mutually commuting projections. We now prove the converse by induction starting with two given projections MATH such that MATH is given to be a projection. Then the commutator MATH satisfies MATH. Using that MATH we conclude that MATH, and therefore MATH; in other words, the two projections MATH commute. Suppose the lemma holds for fewer than MATH projections. If MATH is given as in REF , then MATH where MATH . Writing the operator MATH in matrix form relative to the two projections MATH and MATH, we get MATH with MATH, etc. But then REF - REF yield the conclusion: MATH and MATH imply MATH, and therefore MATH. As a result, the block matrix in REF reduces to MATH . A further calculation shows that MATH must then itself be a projection. From the definition of MATH in REF , and the induction hypothesis, we then conclude that the family MATH is indeed commutative.
math/0004098
We refer the reader to CITE.
math/0004098
The corollary is applied in REF below, so we postpone its proof to REF. The argument is in REF , and it is based on the ordered factorizations REF - REF .