paper stringlengths 9 16 | proof stringlengths 0 131k |
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cond-mat/0002219 | If a set of bosonic fields with four-species has the same commutation relation as REF , the kernel MATH whose dual fields are replaced with those fields is equivalent to the original kernel. We put c-numbers MATH and MATH into the dual fields. Under a simultaneous replacement MATH and MATH the commutation relation REF ... |
cond-mat/0002219 | By REF we have MATH where MATH. Due to the orthogonality MATH and MATH, MATH it follows that MATH . Using the relation MATH we obtain the lemma. |
cs/0002008 | Suppose, by way of contradiction, this were not true. Let MATH be a state of MATH such that a deadlock MATH is reachable from MATH, but not by an initial nontrivial motion which is trivial in MATH. Let MATH be a behaviour of MATH with initial state MATH and reaching the deadlock. Write MATH where MATH and MATH. We clai... |
cs/0002008 | Using the algebra of designs, organize the system as a composite of the given subsystem and its complement: MATH . Now apply REF to the composite of the evaluation of the two subsystems. |
cs/0002012 | Let MATH be the number of mismatches between MATH and MATH at the positions in MATH. Let MATH . First, we prove the following claim. For any MATH such that MATH, where MATH is the constant in the algorithm closestString, there are indices MATH such that for any MATH. MATH . Consider indices MATH such that MATH. Then fo... |
cs/0002012 | CASE: Let MATH. By REF , MATH-where the last inequality is because MATH and that MATH is increasing for MATH. It is easy to verify that for MATH, MATH . Therefore, REF is proved. CASE: Let MATH. By REF is proved. MATH . |
cs/0002012 | Let MATH. Then, for any two strings MATH and MATH of length MATH, we have MATH. Thus for any MATH, MATH . Therefore, the following optimization problem MATH has a solution with cost MATH. Suppose that the optimization problem has an optimal solution MATH such that MATH. Then MATH . Now we solve REF approximately. Simil... |
cs/0002012 | Given an instance of Closest String, suppose MATH is an optimal solution and the optimal cost is MATH, that is, MATH for all MATH. Let MATH be defined as REF of Algorithm closestString. Since for every position in MATH, at least one of the MATH strings MATH conflict the optimal center string MATH, so we have MATH. As f... |
cs/0002012 | Obviously, the size of MATH in REF is at most MATH. REF takes MATH time. Other steps take less than that time. Thus, the total time required is MATH, which is polynomial in term of input size for any constant MATH. From REF , the performance ratio of the algorithm is MATH. MATH . |
cs/0002012 | Let MATH be an optimal center string and MATH be the length-MATH substring of MATH that is the closest to MATH. Let MATH. Let MATH be any small positive number and MATH be any fixed integer. Let MATH. If MATH, then clearly we can find a solution MATH within ratio MATH in REF . So, we assume that MATH from now on. By RE... |
gr-qc/0002016 | Suppose the MATH- and MATH-scalars are related as in REF , that is, MATH. Then the above calculations prove that REF are equivalent. Since REF is equivalent to the vanishing of MATH and since REF is equivalent to MATH being a MATH-potential of a NAME potential of the NAME spinor, the theorem follows. |
gr-qc/0002034 | From REF , MATH if MATH so MATH cannot cross MATH from below. Since MATH for small MATH, the lemma follows. |
gr-qc/0002034 | If MATH for MATH, then MATH exists (because MATH and MATH) so MATH exists as well. We will show that the orbit can be continued beyond MATH provided that MATH. Since MATH, the only obstruction to extending the solution is the possibility that MATH,MATH, or MATH might be unbounded. To see that MATH is bounded we note th... |
gr-qc/0002034 | A simple calculation yields MATH . REF follows immediately from REF . To prove REF note that MATH if MATH, so MATH cannot cross zero from above. |
gr-qc/0002034 | CASE: We note that MATH, so MATH if MATH. CASE: If MATH and MATH, then MATH by REF . Next, we note that MATH if MATH and MATH. If MATH then MATH but MATH since MATH when MATH by REF . CASE: The function MATH changes sign at each zero for which MATH. From REF , MATH changes sign at most once. Thus, for MATH, MATH and at... |
gr-qc/0002034 | The proof is standard so we just outline it. We introduce new variables MATH, MATH, MATH, and rewrite REF as the first order system MATH . We will use the sup norm throughout this discussion: MATH means the MATH. Consider the space MATH of quadruples of functions MATH where MATH, and MATH and each of the four functions... |
gr-qc/0002034 | Let MATH. Then, REF become MATH with the behavior at the origin MATH . For MATH (decoupling of gravity) REF with REF have constant flat-spactime solutions MATH, MATH. Inserting these solutions into REF gives the NAME equation MATH whose unique solution satisfying REF is MATH . This solution has infinite rotation as MAT... |
gr-qc/0002034 | Note that MATH so integrating gives MATH. Therefore, MATH for MATH. To show that MATH assume MATH. Then MATH so if MATH we have MATH or MATH so MATH is bounded. Since MATH, MATH is bounded below. Thus, by integrating one concludes that MATH. But MATH and MATH so MATH. This contradicts MATH so we must have MATH. |
gr-qc/0002034 | In order to prove that MATH becomes positive at some point MATH, we will show that MATH. By REF we have MATH, so we must show that MATH. The proof of this fact is divided into two cases: CASE: MATH, and REF MATH. Before considering these cases we list some useful properties of the function MATH. We have: CASE: MATH; CA... |
gr-qc/0002034 | Note that MATH. Let MATH be the first zero of MATH, that is, MATH and MATH. If MATH then MATH. Thus MATH can have a zero for MATH only if MATH. Then, from REF , MATH for all MATH. Define a function MATH. A calculation gives MATH so by integrating we get MATH. On the other hand we have MATH, so MATH; contradiction. |
gr-qc/0002034 | If MATH is bounded above then there is an integer MATH such that MATH for all MATH but MATH for some MATH and hence, by REF for all MATH, MATH. We next show that there is a MATH such that for all MATH, MATH (that is, the orbit is ultimately in QREF or QREF). Note that, by REF the orbit must satisfy either MATH or MATH,... |
gr-qc/0002034 | Suppose that MATH for all MATH. We claim that this implies MATH. To see this, suppose that MATH for some MATH. Let MATH which exists because MATH. Note that MATH for all MATH. Choose a MATH such that MATH. If MATH for some MATH, then by REF MATH for MATH, where the last but one inequality follows from the fact that MAT... |
gr-qc/0002034 | From the previous lemma we know that there exists a MATH such that MATH for MATH. Let MATH for MATH. A calculation shows that MATH so MATH if MATH. Multiplying REF by MATH we obtain MATH . The right hand side is positive for MATH so MATH is negative and increasing, hence it has a finite non-positive limit. This implies... |
gr-qc/0002034 | Suppose that MATH for all MATH, so MATH for all MATH. We have from REF that MATH. Integrating this inequality from some MATH to some MATH, we obtain MATH which implies (by the NAME inequality) that MATH is bounded. Next, MATH, MATH, implies that MATH; moreover, by REF MATH, hence MATH is not integrable near MATH. Since... |
gr-qc/0002034 | CASE: Suppose the MATH-orbit crashes in QREF or QREF. By REF , MATH for some MATH with MATH; hence, for MATH sufficiently near MATH we have MATH with MATH. By REF , MATH for all MATH. CASE: This case is much more difficult and will require several auxiliary results. It follows from REF that nearby orbits have rotation ... |
gr-qc/0002034 | We set MATH and then compute that MATH . Note that MATH and all terms on the right side of REF are negative except for the last two. If MATH, we combine the term MATH with MATH; clearly, MATH. Next, we combine the term MATH with MATH to get MATH where MATH; the maximum value of this expression occurs when MATH and that... |
gr-qc/0002034 | The assumption on MATH tells us that the orbit lies in QREF or QREF for MATH near MATH. For simplicity of exposition we only discuss the case of REF, that is, MATH. In particular, MATH is a monotone function and hence has a limit at MATH. Thus, MATH is continuous; in particular, if we suppose that MATH, then MATH for M... |
gr-qc/0002034 | If the MATH-orbit crashes at MATH with rotation MATH, then there is a MATH such that MATH. Let MATH. By REF , if MATH is sufficiently close to MATH, MATH (along the MATH-orbit) is bounded in QREF by MATH; MATH is a uniform bound on MATH in QREF for all MATH sufficiently near MATH. Next, choose MATH such that MATH and s... |
gr-qc/0002034 | From REF we have MATH. Moreover, since MATH and MATH, so if MATH then MATH. Since MATH, we have MATH if MATH. Thus, if MATH, and MATH in the interval MATH; if MATH, then MATH for all MATH in the interval MATH because MATH cannot cross that value from above. |
gr-qc/0002034 | To show that MATH goes to MATH, we note that MATH for all MATH near MATH and MATH by REF . Hence, MATH. By REF , MATH, hence MATH for MATH. Thus, MATH. Let MATH be chosen so that MATH. Then, if MATH is sufficiently close to MATH, MATH so MATH. For such MATH we then have MATH and MATH because MATH; thus, MATH. |
gr-qc/0002034 | Suppose that the MATH-orbit crashes at MATH with MATH for all MATH and MATH for MATH near MATH. For MATH near MATH there is a MATH with MATH by REF . Since MATH, MATH, that is, MATH, so the MATH-orbit crashes, MATH, or exits QREF to REF (or QREF to REF), MATH, never to return. In either case, the MATH-orbit has rotatio... |
gr-qc/0002034 | Let MATH for all MATH for which the a-orbit is defined. Note that MATH and MATH by REF and hence, MATH. Also note that MATH is a lower bound for MATH by REF ; hence, MATH has a greatest lower bound MATH. We will show that the MATH-orbit is a globally regular solution and MATH for large MATH. We first show that MATH, th... |
hep-th/0002169 | From REF the left hand side is equal to MATH . By performing the sum over MATH we obtain MATH . However, the last expression is equal to the right hand side of REF. |
hep-th/0002169 | This is a direct consequence of REF. |
hep-th/0002169 | If MATH, we find that MATH . As for the case MATH, we have MATH where we used MATH. Thus REF shows that MATH . This completes the proof of REF. |
hep-th/0002169 | We see that MATH by REF . Then REF follows from REF . |
hep-th/0002169 | See CITE. |
hep-th/0002169 | REF is an easy consequence of REF while REF follows from REF , MATH and NAME. |
hep-th/0002169 | Recall that MATH since MATH for the inclusion MATH. |
hep-th/0002169 | If MATH the only relevant part for the calculation of MATH is the term MATH in REF. Then use the series expansion REF for MATH. |
hep-th/0002169 | Let MATH be a point of MATH. By He CITE, the NAME tangent space of MATH at MATH is given by MATH the obstruction of infinitesimal liftings belong to the kernel of the composition of homomorphisms MATH and MATH where MATH is the hypercohomology associated to the complex MATH. Moreover there is an exact sequence MATH whe... |
hep-th/0002169 | For a coherent system MATH belonging to MATH, our assumptions and CITE imply that CASE: MATH is surjective in codimension REF (and hence MATH) and MATH is a (slope) stable sheaf, or CASE: MATH is injective and MATH is a (slope) stable sheaf according as REF MATH or REF MATH. For the second case, MATH is also genericall... |
hep-th/0002169 | Let MATH be an element of MATH. Since MATH for MATH, we obtain MATH . Since MATH is a stable sheaf of positive degree, NAME duality and REF imply that MATH . By using the canonical exact sequence MATH we see that MATH . |
hep-th/0002169 | Let MATH be a family of coherent systems parametrized by a scheme MATH such that MATH for all MATH, where MATH is a line bundle on MATH. Replacing MATH by MATH, we may assume that MATH. We consider a locally-free resolution REF MATH . Then MATH is injective and it defines an effective NAME divisor MATH on MATH. MATH is... |
hep-th/0002169 | That the assertions hold is guaranteed by CITE unless MATH and MATH is not ample. Hence we may assume that MATH and MATH satisfies MATH. The following argument is very similar to the last part of the proof of CITE. Let MATH be an ample line bundle in REF. Replacing MATH by MATH, MATH, we may assume that the evaluation ... |
hep-th/0002169 | See CITE. |
hep-th/0002169 | We may assume that MATH is smooth by applying REF successively. Since MATH is a projective morphism, the NAME spectral sequence for MATH degenerates. Moreover we obtain MATH, and hence MATH for MATH. Since MATH is the NAME dual of MATH, we obtain our claim. |
hep-th/0002169 | By REF , MATH is isomorphic to MATH. Hence we shall prove that MATH is smooth. Let MATH be an element of MATH. Then condition (MATH-REF) implies that MATH is injective and MATH is a rank-REF torsion-free sheaf when restricted to its support MATH. In order to prove the smoothness of MATH at MATH, it is sufficient to pro... |
hep-th/0002169 | It is sufficient to prove the deformation equivalence of MATH. By the connectedness of the moduli space of polarized MATH surfaces, there is a family of polarized MATH surfaces MATH such that MATH is irreducible and there are two points MATH which satisfy MATH. Then there is a family of moduli spaces of coherent system... |
hep-th/0002169 | Using REF the left hand side can be rewritten as MATH . Then the integrand of each factor becomes MATH where we used REF. By performing the integrals using REF we thus obtain MATH which can be cast in the desired form thanks to the identity MATH . |
hep-th/0002169 | Using REF we see that the left hand side is equal to MATH . |
hep-th/0002169 | The calculation is similar to that in the proof of REF . The left hand side is equal to MATH . By performing the Gaussian integrals we obtain that MATH . This readily leads to the desired result. |
hep-th/0002193 | Obvious. |
hep-th/0002193 | The proposition holds for the exterior algebra, and therefore also for the algebra MATH since the latter is a ``small" deformation of the exterior algebra. |
hep-th/0002193 | For MATH this is a well-known fact about the symmetric algebra MATH. Since the augmented NAME complex is exact for MATH it is also exact for small MATH and consequently for all MATH. |
hep-th/0002193 | We have to check that MATH vanishes. Note that MATH and that the differentials MATH and MATH restricted to MATH coincide with the multiplication maps MATH and MATH respectively. Thus we have the following commutative diagram: MATH . Now the Lemma follows because MATH (associativity) obviously annihilates MATH. |
hep-th/0002193 | The MATH - bigraded component of MATH is equal to MATH hence the MATH - bigraded component of the bicomplex MATH vanishes for MATH or MATH. Thus the MATH - bigraded component of the bicomplex MATH is bounded. Therefore both spectral sequences of the bicomplex MATH converge to the cohomology of the total complex MATH. T... |
hep-th/0002193 | This follows immediately from the definition of MATH and MATH. |
hep-th/0002193 | By ampleness a sheaf MATH can be covered by a finite sum of sheaves MATH. Now the statement follows from the Proposition, because MATH for all MATH. |
hep-th/0002193 | The group MATH coincides with MATH. Let MATH be the maximal integer (it exists because the global dimension is finite) such that for some MATH there exists arbitrarily large MATH such that MATH. Assume that MATH. Choose an epimorphism MATH. Let MATH denote its kernel. Then for MATH we have MATH hence MATH implies MATH.... |
hep-th/0002193 | To prove this we only need to show that multiplication by MATH is a monomorphism. If MATH is a bundle, it can be embedded into a direct sum MATH because by ampleness the dual bundle MATH is covered by a direct sum of line bundles. Now, since the morphism MATH is mono for any MATH the same is true for the morphism MATH. |
hep-th/0002193 | We have the following exact sequence in the category MATH: MATH . Since MATH we have MATH. Assume that MATH has a nontrivial section. Then MATH has a nontrivial section too. For the same reason MATH has a nontrivial section, and so on. Thus for any MATH the bundle MATH has a nontrivial section. By REF we have isomorphi... |
hep-th/0002193 | Let MATH be a finitely generated graded MATH-module such that MATH. Consider an exact sequence: MATH . Since MATH the module MATH is finite dimensional. This implies that for MATH the map MATH is surjective. Moreover, these maps are isomorphisms for MATH because all MATH are finite dimensional vector spaces. Let us ide... |
hep-th/0002193 | The existence of such a monad was proved above. The uniqueness follows from REF . The equality of dimensions of MATH and MATH follows immediately from the exact sequence REF . |
hep-th/0002193 | Let MATH be a bundle on MATH trivial on the line MATH. We showed above that any such bundle comes from a monad unique up to an isomorphism. Conversely, suppose we have a monad MATH with MATH such that its restriction to the line MATH is a monad with the cohomology MATH. Then the cohomology of this monad is a bundle on ... |
hep-th/0002193 | Obvious. |
hep-th/0002193 | By the preceding theorem, it is sufficient to show that the NAME transform of every multiplier is a convolute, and vice versa. The former fact is proved in REF, ch. CASE: Let us prove the converse. By REF , every convolute has the form MATH for some MATH and rapidly decreasing continuous functions MATH. Let MATH be the... |
hep-th/0002193 | We will prove that the MATH - product of two convolutes of MATH is well-defined, and is again a convolute of MATH. The rest is obvious. It is sufficient to consider the case when MATH . Then, integrating by parts, we may rewrite the MATH - product in the following form: MATH . Derivatives acting on the exponential brin... |
hep-th/0002193 | The first statement is an immediate consequence of REF . The second statement follows from a simple computation. |
math-ph/0002003 | CASE: The time evolution of MATH is unitary and thus MATH. The stated analyticity is an immediate consequence of the elementary properties of the NAME transform . The asymptotic behavior follows then from the NAME lemma. CASE: We have in MATH, MATH . For MATH we push the integration contour through the upper half plane... |
math-ph/0002003 | To get a contradiction, assume MATH is a MATH solution of REF . Multiplying REF by MATH, and summing with respect to MATH from MATH to MATH we get MATH . For MATH the imaginary part of MATH is nonpositive, by REF , and is strictly negative for MATH large enough. Thus if for some such MATH, MATH is nonzero then the last... |
math-ph/0002003 | CASE: We look at MATH, the case of MATH being similar. Dropping the MATH superscript we have from REF MATH . To find the analytic properties of the solution MATH it is convenient to regard REF as a contractive equation in the space MATH of sequences MATH in the norm MATH. Let MATH be large. The map MATH defined by MATH... |
math-ph/0002003 | By analyticity and continuity MATH does not vanish for any MATH and MATH. By REF the function MATH defined through MATH, where MATH and MATH has the required properties. Since no solution of the homogeneous equation is bounded on MATH, MATH is the unique solution with the desired properties. |
math-ph/0002003 | Since MATH is analytic in MATH, continuous and nonzero in MATH, MATH is bounded below in compact sets in MATH. Then, the smoothness properties of MATH derive easily from those of MATH on which we concentrate now. CASE: For MATH we write, using REF , MATH . The last term in parenthesis can be rewritten, using also REF ,... |
math-ph/0002003 | Indeed, MATH . |
math-ph/0002003 | Consider MATH endowed with the norm MATH, where MATH. If MATH is continuous and MATH, a straightforward calculation shows that MATH where the last relation follows from the NAME lemma. The integral REF can be written as MATH . Since MATH is locally in MATH and bounded for large MATH it is clear that for large enough MA... |
math-ph/0002014 | Let us define MATH for all MATH by REF, and let us extend MATH to all MATH by setting MATH when MATH. To prove REF Note that MATH, which implies that MATH is subharmonic (we use MATH and MATH, by REF ). Set MATH with MATH and small. Obviously, MATH is subharmonic on the open set MATH because MATH is harmonic there. Cle... |
math-ph/0002014 | In polar coordinates, MATH, one has MATH. Therefore, it suffices to prove that for each angle MATH, and with MATH denoted simply by MATH, MATH where MATH denotes the distance of the origin to the boundary of MATH along the ray MATH. If MATH then REF is trivial because the right side is zero while the left side is evide... |
math-ph/0002021 | By REF is continuous, hence we can find a seminorm MATH so that MATH for all MATH. Thus there are positive constants MATH and MATH so that MATH holds for all MATH. In the first equality, the invariance of MATH was used, and in the second, REF was applied. Thus, for any NAME MATH, and any MATH, MATH in MATH, one obtains... |
math-ph/0002021 | Select a local trivialization of MATH over MATH. With respect to it, there are smooth MATH-valued functions MATH of MATH such that MATH . Now suppose that MATH is not in MATH, so that MATH isn't contained in any of the MATH. Then, making use of the fact that the wavefront set of a scalar distribution may be characteriz... |
math-ph/0002021 | Let MATH be any point in MATH. Then there is a coordinate chart MATH around MATH so that, for small MATH, MATH holds on a neighbourhood of MATH, where MATH . In such a coordinate system, we can also define ``spatial" translations MATH for MATH in a sufficiently small neighbourhood MATH of the origin in MATH. Let MATH b... |
math-ph/0002027 | Since the eigenbasis MATH of MATH is an orthonormal basis for MATH, it suffices to show that for each MATH, MATH. But MATH . |
math-ph/0002027 | The proof is by induction on MATH. The formula clearly holds when MATH. For MATH, the determinant is a rational function of MATH with a double pole at MATH; we can write MATH . The coefficient MATH is zero since the determinant is even under the exchange of MATH and MATH (exchange the first two rows and exchange the fi... |
math-ph/0002027 | When MATH is the upper half plane this follows by combining REF with REF and the calculation REF . For arbitrary MATH, REF shows that MATH (where MATH is the NAME 's function on MATH) by conformal invariance of the height moments and of MATH. This completes the proof. |
math-ph/0002038 | It follows by a standard averaging argument CITE that the set MATH of invariant probability measures supported on MATH is non-empty: for any probability measure MATH supported on MATH, the set of weak* limit points of the set of finite time averages MATH gives at least one non-trivial element of MATH. Since MATH consis... |
math-ph/0002038 | In the case of a Hamiltonian flow with NAME vector field MATH, this is a case of NAME 's theorem CITE. In fact, MATH where MATH. In the case of MATH actions, we can apply NAME 's theorem to any one parameter subgroup. |
math-ph/0002038 | By REF no such invariant foliation exists unless MATH, so we may assume this is the case. The NAME metric MATH is then a well-defined metric on MATH. We denote by MATH the associated homogeneous Hamiltonian (length squared of a covector). Since the sets MATH and MATH are the same, the latter carries a NAME foliation by... |
math-ph/0002038 | The assumption implies that MATH is flat for all MATH in the interval. Let MATH denote the curvature tensor of MATH . It is clearly a real analytic function of MATH. Since MATH in MATH, it must vanish identically. Therefore the NAME 's flow MATH on MATH has no conjugate points. By REF, it follows that MATH is flat and ... |
math-ph/0002038 | CASE: MATH: By definition, MATH where MATH is a cutoff to MATH . Since MATH is a finite set for each MATH, there exists MATH such that MATH . We form the sequence MATH and then choose a sub-ladder MATH with a unique quantum limit. Then MATH . CASE: MATH: It is clear that for each ladder MATH we have MATH . Therefore th... |
math-ph/0002038 | Let MATH be a MATH-invariant neighbourhood of MATH. Let MATH be a sequence of regular points such that MATH. For each MATH and sufficiently large MATH, there exists at least one component MATH of MATH such that MATH. By REF , MATH is charged by an amount MATH. We now break up the discussion into two cases: CASE: All MA... |
math-ph/0002038 | Existence of a compact singular orbit contradicts the the uniform boundedness of eigenfunctions assumption. Indeed, it follows from REF that, for any MATH, there exist a compact, singular orbit MATH and MATH-normalized joint eigenfunctions MATH such that for some MATH . However, the estimate in REF cannot hold since by... |
math-ph/0002038 | We fix an energy level MATH and consider eigenvalues of MATH lying in MATH for some fixed MATH . The eigenfunctions we consider are the joint eigenfunctions of MATH with joint eigenvalues MATH respectively, satisfying MATH for some MATH . We recall that MATH and MATH corresponds to the energy shell MATH of the classica... |
math-ph/0002039 | The surjectivity of MATH guarantees that the normal bundle MATH is disjoint from the wave front set of MATH and hence MATH is well-defined (see CITE). Thus we can define MATH . To verify the equation of the lemma, it suffices by the continuity of MATH to consider the case where MATH. In this case, MATH, and hence MATH ... |
math-ph/0002039 | The proof is by induction on MATH. From REF , MATH where the summation goes over all partitions MATH with at least two elements in the partition (that is, MATH). From REF , we have MATH where the sum is taken over all pairs MATH of partitions of the set MATH. If we use the inductive assumption that REF holds when the n... |
math-ph/0002039 | By the determinant formula, MATH where the sum goes over all MATH-tuples MATH of permutations of MATH. We claim that MATH where the summation on the right goes over the set of MATH-tuples MATH such that no proper subset of MATH is invariant under the group generated by the MATH. (Each such MATH corresponds to a connect... |
math-ph/0002039 | We have the identity MATH . The proof of this identity is the same as that of REF . Indeed, the connected functions of MATH are equal to REF (except that REF-point connected function equals REF); hence REF follows. The identity REF can be rewritten as MATH where the summation on the right goes over all mutually connect... |
math-ph/0002039 | We must show that MATH . Assume without loss of generality that MATH. Let MATH be an oriented connected graph with zero boundary as in the definition of MATH. Since MATH and MATH are connected by a chain of loops in MATH, we can choose disjoint sets of edges MATH such that MATH forms a path starting at MATH and ending ... |
math-ph/0002045 | The existence of continuous sections can be proved applying NAME 's Lemma to constant sections in the (trivial) part of the vector bundle over one chart MATH, getting continuous sections of the whole vector bundle supported in one chart MATH over which the vector bundle is trivial. Any bundle homomorphism MATH maps sec... |
math-ph/0002045 | MATH: MATH . Assume MATH by NAME 's theorem. Let MATH be the fibrewise projection, MATH. Define MATH by MATH for MATH, MATH. Then MATH is a correctly defined surjective bundle homomorphism. Let MATH be the induced MATH-module map, a projection onto MATH. Note, that MATH for every MATH. Therefore, MATH by construction. ... |
math-ph/0002045 | Because of the categorical equivalence by NAME and NAME it is sufficient to indicate the existence and the properties of MATH-valued inner products on finitely generated projective MATH- or MATH-modules. For MATH the MATH-valued inner product is defined as MATH. For direct summands MATH of MATH we reduce this MATH-valu... |
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