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math/0003159
Suppose that MATH is an element of minimal length such that MATH and let MATH. We prove the lemma by induction on MATH. The case MATH is trivial. Initial step: MATH. If MATH then we have MATH. Hence MATH, MATH and MATH. So MATH by REF . Inductive step: if MATH then MATH. Let MATH with MATH and MATH, MATH, MATH. By induction MATH and, since MATH, MATH. If MATH then we can apply REF and we obtain MATH. If MATH then MATH, hence MATH. Moreover MATH and MATH so MATH and MATH. Hence MATH so MATH and MATH against the minimality of MATH.
math/0003159
Without loss of generality we can assume MATH. First step: MATH. This is clear since otherwise MATH. Second step: MATH and MATH for all MATH such that MATH. Let MATH and observe that MATH and MATH. Then as in first step we have MATH from which the claim follows. Let now MATH. If MATH then there exists MATH and MATH such that MATH and MATH where MATH. Since MATH we can assume MATH. Let now MATH, MATH and MATH, we have: MATH . Hence MATH and MATH against MATH.
math/0003159
By REF there exists a closed orbit MATH in MATH with trivial stabilizer. Then by REF there exists MATH such that MATH and by REF MATH is surjective. Now the surjectivity follows by homogeneity. The local triviality over MATH follows also from REF .
math/0003159
We prove that the map MATH is surjective. By REF this follows by the following two identities: MATH for each MATH-path MATH and for each admissible path MATH.
math/0003159
It's enough to prove that MATH is surjective. Let MATH and consider the sequence (see REF for the notation) : MATH . Since MATH and MATH we have that MATH is not surjective or that MATH is not injective. Suppose that MATH is not surjective, then up to the action of MATH we can assume that MATH. Then, for MATH consider MATH with MATH . Then CASE: MATH if MATH and MATH, since MATH, CASE: MATH if MATH and MATH . So MATH and it is clear that MATH and that MATH. If MATH is surjective and MATH is not injective the argument is similar.
math/0003159
We prove this proposition by induction on the order MATH on MATH. First step: MATH. If MATH we can take MATH and MATH. Inductive step. If MATH is not dominant then there exists MATH such that MATH. If MATH we observe that MATH (that is MATH and MATH) and that MATH and so we can apply the inductive hypothesis. If MATH we apply the previous lemma and the inductive hypothesis.
math/0003159
Observe that MATH so each irreducible component of MATH must have dimension at least MATH. Suppose now that MATH is dominant. By NAME 's theorem REF MATH is not empty. Observe also that by REF MATH is a smooth subset of MATH of dimension MATH. It is well known that MATH. Hence MATH . By the lemma above we have that if MATH and MATH. So MATH must be dense in MATH and MATH is a complete intersection. Moreover if MATH is regular we have that MATH so the singular locus has codimension at least two and normality and irreducibility follows. Finally by our discussion it is clear that if MATH is disconnected then MATH is not irreducible.
math/0003162
CASE: The condition MATH reads MATH this identity is proved for example, in CITE. CASE: Let MATH denote the angle function of MATH and MATH, defined by MATH; we then have MATH . Let MATH and MATH be the NAME forms of MATH, MATH; from REF applied to MATH, we infer MATH; similarly, we have MATH; putting together these two identities, we get MATH . This obviously implies MATH if MATH; the converse is also true, as the commutator MATH is invertible at each point where MATH.
math/0003162
Every self-dual NAME Hermitian REF-manifold MATH of non-constant curvature is conformally related (via REF ) to a self-dual NAME metric MATH of nowhere vanishing scalar curvature. A self-dual NAME metric is locally-symmetric if and only if its scalar curvature is constant CITE; thus, the one direction in the correspondence stated in the lemma follows by observing that MATH is locally-symmetric as soon as MATH is. Since the NAME tensor of a self-dual metric vanishes CITE, it follows from CITE that any self-dual NAME metric of nowhere vanishing scalar curvature gives rise to an NAME Hermitian metric in the same conformal class.
math/0003162
MATH. By REF , MATH has two distinct, non-constant eigenvalues at any point and there exists a positive, non-Kähler Hermitian structure MATH whose NAME form MATH generates the eigenspace of MATH corresponding to the simple eigenvalue. It follows that the Hermitian structure is preserved by the action of MATH, and therefore both functions MATH and MATH are constant along the orbits of MATH; in particular, MATH is colinear to MATH, at any point; this means that MATH; by REF , the vanishing of MATH is equivalent to the integrability of the negative almost Hermitian structure MATH. MATH. If MATH or, equivalently, if the negative almost Hermitian structure MATH is integrable, then, by REF , the NAME form MATH of MATH reads: MATH . According to REF we also have MATH and MATH; it follows that MATH; then, locally, MATH for a positive function MATH, that is, MATH is conformal to a NAME metric MATH. Since MATH, the NAME metric MATH is of zero scalar curvature. Clearly, the Killing field MATH preserves both MATH and MATH, hence, also, the NAME structure MATH. Two cases occur, according as MATH is homothetic or not to MATH. CASE: Suppose MATH is not homothetic to MATH; equivalently, the scalar curvature MATH of MATH does not vanishes; then, by CITE, MATH is a Killing vector field for MATH and MATH and is holomorphic with respect MATH. By the very definition of MATH we have that MATH; the Killing vector fields MATH and MATH are thus colinear everywhere (see REF ); it follows that MATH is a constant multiple of MATH. By considering MATH as a local coordinate on MATH and, by introducing a holomorphic coordinate MATH on the (locally defined) orbit-space for the holomorphic action of MATH on MATH, the metric MATH can be written in the following form: MATH where MATH is a smooth function satisfying the MATH . NAME field equation: MATH is a positive function given by MATH and MATH is a connection REF-form of the MATH-bundle MATH, whose curvature is given by MATH (see, for example, CITE). Moreover, the Killing field MATH is dual to MATH, and the (anti-self-dual) NAME form of the negative Hermitian structure MATH is given by MATH . By REF we have that MATH, so that the NAME form MATH of the positive Hermitian structure MATH is given by MATH . It is now easily seen that REF simultaneously define integrable almost complex structures if and only if MATH, or equivalently if and only if MATH. This means that MATH is a separable solution to the MATH . NAME field equation. Up to a change of the holomorphic coordinate MATH, it is explicitly given by CITE MATH for properly chosen constants MATH. Any such solution gives rise to a diagonal NAME self-dual NAME Hermitian metric pertaining to one of classes II, MATH, VIII and IX, depending on the choice of the constants MATH (see for example, CITE) for a common case of these metrics in the NAME IX case. CASE: If MATH is homothetic to MATH, that is, MATH is itself a NAME structure of zero scalar curvature, then MATH is locally hyperkähler and MATH is a Killing vector field preserving the NAME structure MATH. Then, one of the two following situations occurs: CASE: MATH is triholomorphic, that is, MATH preserves each NAME structure in the hyperkähler family: Then the quotient space, MATH, for the (real) action of MATH is flat and is endowed with a field of parallel straight lines. This situation is described by the CITE, and the metric MATH has the form: MATH for a positive harmonic function MATH on MATH and a REF-form MATH on MATH satisfying MATH . The Killing field MATH is dual to MATH and one may consider that the positive and negative Hermitian structures, MATH and MATH, correspond to REF-forms MATH respectively. We again conclude MATH, and therefore MATH. The case MATH corresponds to flat metrics in Class I, whereas, when MATH, by putting MATH, the metric becomes a diagonal NAME metric of Class II. CASE: MATH is not triholomorphic: Since, nevertheless, MATH preserves MATH, the metric MATH takes the form CITE MATH where MATH is a solution to the MATH . NAME field equation, MATH, MATH satisfies REF and MATH is a constant. Moreover, MATH is dual to MATH, and MATH is defined by the anti-self-dual form MATH . Similar arguments as above show that MATH, that is, MATH is a separable solution to the MATH . NAME field equation, and therefore our metric is again a diagonal NAME metric in one of the classes II, MATH, VIII or IX, compare CITE. The implication MATH is clear.
math/0003162
We first observe that REF can be equivalently written as MATH this shows that MATH is function of MATH, say MATH; from the above equality, we get REF , where MATH is a (still unknown) smooth function; in order to determine MATH, we differentiate REF by using REF and substitute into REF ; then, cancellations occur and REF eventually reduces to MATH the solutions of REF are given by REF .
math/0003162
By using REF , the right-hand side of MATH is easily computed; we thus obtain: MATH . Since MATH is self-dual and NAME, MATH, see REF . Then, by projecting REF to MATH, we get REF , whereas the projection of REF to MATH gives REF .
math/0003162
If we assume that MATH somewhere on MATH and that the anti-self-dual NAME tensor is identically zero, then, after contracting REF with MATH, we obtain MATH, which contradicts MATH.
math/0003162
MATH. According to REF , if the spectrum of MATH is everywhere degenerate, then either MATH vanishes identically (and therefore the hyperhermitian structure is closed by REF ) or MATH has two distinct eigenvalues MATH and MATH at any point. MATH. By REF , we know that a normalized generator MATH of the MATH-eigenspace of MATH is the NAME form of a positive Hermitian structure MATH. Let MATH be any self-dual REF-form orthogonal to MATH, with MATH; then, MATH and MATH are orthogonal, MATH-eigenforms of MATH; by substituting into REF , we get MATH . From the last two equalities, we get MATH, and by the first one we conclude that MATH. This shows that MATH is a multiple of MATH. It remains to prove that MATH does not vanish on MATH; by taking a two-fold cover of MATH if necessary, we may assume that the Hermitian structure MATH is globally defined on MATH; by REF , MATH is conformally NAME and NAME is the corresponding closed NAME form; but MATH is also a closed, self-dual REF-form, and a multiple of MATH, hence a constant (non zero) multiple of MATH. MATH. This is an immediate consequence of REF .
math/0003162
We first show that if MATH admits a non-closed hyperhermitian structure compatible with the negative orientation, then the corresponding NAME form MATH must be one of the forms MATH and MATH given in REF . From REF and the fact that MATH is a MATH-eigenform of MATH, we infer MATH . By differentiating REF and by using REF in order to compute MATH, we obtain MATH we infer: MATH . By substituting the above expression of MATH in REF , and by using REF again, we get MATH . Now, according to the above convention, by REF we end up with the following expression for MATH: MATH . This shows that every non-closed hyperhermitian structure is completely determined by the self-dual harmonic REF-form MATH. It remains to prove that MATH itself is determined, up to sign, by the metric MATH; then, the two possible values of MATH appearing in REF will only differ by conjugation of MATH or, equivalently, by substituting MATH to MATH. Notice that, according to our convention, at this stage we have the freedom to rescal the MATH-form MATH by a non-zero constant. In other words, by fixing one non-closed hyperhermitian structure and by following our convention, we know that any other non-closed hyperhermitian structure corresponds to a harmonic REF-form of the form MATH, where MATH is a non-zero constant. Our claim is that MATH; to see this, by using REF , we calculate MATH in the present situation, when MATH has degenerate spectrum, the norm of MATH is given by MATH; then, by REF , the above equality reduces itself to MATH it is readily checked that if the MATH-forms MATH and MATH simultaneously satisfy REF , then MATH. We now check that REF for either MATH or MATH are equivalent to REF . Keeping REF in mind, we see that REF can be equivalently re-written as MATH then, the equivalence `` REF " follows by a straightforward computation involving REF , and using REF ; REF-forms MATH and MATH thus correspond to two distinct, non-closed hyperhermitian structures MATH and MATH provided that REF holds, see Sec. CASE: As a final step, we have to prove that MATH is triholomorphic with respect to both hyperhermitian structures. For a general hyperhermitian structure MATH, with NAME form MATH, and for any Killing field MATH, we have MATH where MATH is the NAME derivative given by REF ; we thus only need to check that in our specific situation MATH commutes with MATH; by using REF , we get MATH the claim follows immediately.
math/0003162
Any such manifold is either a space of constant curvature, hence conformally flat, or a NAME manifold of constant holomorphic sectional curvature (see REF ). In the former case, the claim follows by REF , whereas in the latter case MATH; we then conclude by using REF .
math/0003162
Since MATH is Killing, equality REF MATH holds. For a quaternionic NAME manifold the curvature operator MATH acts on MATH by MATH, where MATH is a positive multiple of the scalar curvature, compare for example, CITE. Thus, projecting REF to MATH we get MATH .
math/0003162
We denote MATH any complex MATH -vector field of MATH and MATH the corresponding horizontal lift (considered as complex vector in MATH); then, MATH is integrable if and only if the following identity holds: MATH by the very definition of MATH we have MATH; moreover, since MATH belongs to MATH on MATH, the almost complex structure MATH (defined on MATH) commutes with MATH's for any trivialization MATH of MATH. Then, by REF it is easily seen that the integrability REF for MATH is the same as MATH . We now derive REF from the structure of the curvature tensor of the Riemannian symmetric spaces MATH and the corresponding non-compact duals, MATH and MATH (we refer to CITE for a general description of the curvature operator, MATH, of a Riemannian symmetric space). We first consider the simplest case of MATH (or its non-compact dual). The eigenspaces of MATH are then the simple factors MATH and MATH of the isotropy NAME sub-algebra MATH, and the orthogonal complement MATH of MATH in the space MATH of the skew-symmetric endomorphisms of MATH (note that MATH acts trivially on MATH); the decomposition MATH into eigenspaces of MATH then fits with the splitting REF ; MATH is thus identified to MATH, and MATH to MATH, whereas MATH corresponds to the kernel of MATH, the space MATH. This shows that the curvature operator acts on the first two factors in REF by multiplication with a non-zero constant (a certain multiple of the scalar curvature), and acts trivially on the third factor (therefore, MATH has thus three distinct eigenvalues, MATH and MATH); this observation also shows that any Killing field on MATH is necessarily quaternionic. As already observed, the almost complex structure MATH (defined on MATH) commutes with the MATH's, so that MATH is again a MATH -vector of MATH; we thus get MATH which means that MATH is an element of MATH. It then follows that MATH . But MATH is again a MATH -vector of MATH (see REF ), so that MATH; this implies REF . The same argument holds for the non-compact dual space MATH. The case of MATH (or its non-compact dual) is similar, but MATH is now a Hermitian symmetric space, whose canonical Hermitian structure MATH comutes with any MATH. The corresponding NAME form, MATH, then belongs to the space MATH and gives rise to a further splitting MATH where MATH is the orthogonal complement of MATH. Correspondingly, the eigenspaces of the curvature MATH are the bundles MATH, MATH, MATH, and MATH. Note that MATH acts trivially on MATH, whereas MATH is an eigenform of MATH corresponding to the simple eigenvalue; in particular, MATH must preserve MATH and MATH, so that MATH is of type MATH with respect to MATH; in other words, the almost complex structure MATH commutes with MATH, when acting on MATH. It follows that MATH belongs to MATH, and we conclude as in the case of MATH.
math/0003169
The hermitian metric MATH on MATH is defined by the integration along the fiber: for MATH and MATH, MATH . By CITE, the curvature form of MATH is NAME semipositive. Hence the smooth metric MATH on MATH induced from MATH has semipositive curvature form as well. Moreover, it is extended to a singular hermitian metric MATH over MATH. The multiplier ideal sheaf is trivial because the sections of MATH are MATH. In REF , the growth of the metric is logarithmic. So the NAME number vanishes, and the last statements follow from the regularization of positive currents CITE.
math/0003169
Let MATH be the minimal positive number such that MATH. We take a rational function MATH on MATH such that MATH. Let MATH be the normalization of MATH in the field MATH, and let MATH be a desingularization such that the composite morphism MATH is smooth over MATH. We have MATH . The NAME group MATH acts on MATH such that the above direct summands of MATH are eigenspaces with eigenvalues MATH. Since MATH is etale outside the support of MATH, MATH has only rational singularities, hence MATH. We apply REF to the sheaf MATH. By duality, we have MATH . By taking MATH (we may assume that MATH), we obtain our REF since MATH. For REF , we use the unipotent reduction theorem for the local monodromies of MATH CITE.
math/0003169
Let MATH be a log resolution for the pair MATH, and set MATH. The coefficients of MATH are less than MATH and negative coefficients appear only for exceptional divisors of MATH. We set MATH where MATH is an effective integral divisor and MATH is a MATH-divisor whose coefficients belong to the interval MATH. Since the support of MATH is exceptional for MATH, we have MATH. By applying the theorem to the pair MATH, we deduce that the sheaf MATH is numerically semipositive.
math/0003169
There exists an effective MATH-divisor MATH such that MATH is KLT and MATH.
math/0003169
Let MATH be the log resolution of the pair MATH. We have MATH for some MATH-divisor MATH. Then our assertion is proved in REF .
math/0003169
By the NAME lemma, there exists an effecive MATH-divisor MATH such that MATH is a MATH-divisor, the pair MATH is KLT, and that MATH is ample. Therefore, we may assume that MATH is a MATH-divisor and that MATH is ample. By the Base Point Free Theorem, there exists a proper surjective morphism MATH with connected fibers to a normal projective variety such that MATH for an ample NAME divisor MATH on MATH. We have MATH if and only if MATH. We shall show that there exists an effective MATH-divisor MATH on MATH such that MATH is KLT and MATH is ample. Since MATH is already assumed to be ample, we can write MATH with MATH and MATH being ample MATH-divisors. Since MATH is ample, there exists an effective MATH-divisor MATH such that MATH and that MATH is KLT. We set MATH. Then MATH for MATH. We construct birational morphisms MATH and MATH from smooth projective varieties such that MATH for a morphism MATH. We write MATH. If MATH and MATH are chosen suitably, then we may assume that the conditions of REF are satisfied for MATH and MATH. Then there exist MATH-divisors MATH and MATH on MATH such that MATH, MATH is effective, MATH and MATH is nef. Since MATH is ample, there exists a MATH-divisor MATH on MATH such that MATH is effective and MATH. Then we have MATH and MATH is KLT. Since MATH, we obtain our assertion.
math/0003169
We may assume that MATH by REF . Let MATH be the minimal resolution of singularities. Since MATH is effective, we can write MATH with MATH being KLT. Therefore, we may assume that MATH is smooth. By REF again, we may also assume that MATH is ample and that MATH is big. Assume first that MATH. Then the assertions follow immediately from the NAME theorem. We assume that MATH in the following. We prove REF . By the NAME theorem, MATH. Thus, if MATH, then MATH. Let us assume that MATH. Then there exists a surjective morphism MATH to a curve of genus MATH whose generic fiber is isomorphism to MATH. By REF , the vector bundle MATH is numerically semipositive. Since MATH is MATH-nef, it is MATH-generated, hence we have a surjective homomorphism MATH, and the latter sheaf is nef. Thus MATH. Since MATH, we have MATH. In order to prove REF , we take a general member MATH as a subscheme of MATH. We have an exact sequence MATH and MATH, hence it is sufficient to prove the freeness of MATH. Let MATH be any ideal sheaf of MATH of colength MATH. We shall prove that MATH. By duality, it is equivalent to MATH. Since MATH is even and MATH, we have MATH, and we have the desired vanishing.
math/0003169
By a crepant blowings-up, we may assume that MATH has only terminal singularities. Then we have MATH by CITE, and MATH is pseudo-effective by CITE (see also CITE). By the NAME theorem, we calculate MATH .
math/0003169
Assume that MATH is not LC. Let MATH be the LC threshold for MATH so that MATH and MATH is properly LC. Let MATH be a minimal center. By REF , for any positive rational number MATH, there exists an effective MATH-divisor MATH on MATH such that MATH and MATH is KLT. By the perturbation technique, we may assume that MATH is the only LC center for MATH and there exists only one LC place MATH above MATH if we replace MATH and MATH suitably. Therefore, there exists a birational morphism MATH from a smooth projective variety such that we can write MATH, where the support of MATH is a normal crossing divisor and the coefficients of MATH are strictly less than MATH. We consider an exact sequence MATH where MATH is the ideal sheaf for MATH. Since MATH is ample, we have MATH and MATH for MATH by the generalization of the NAME vanishing theorem. Since MATH, we obtain MATH. Hence the homomorphism MATH is surjective. We have MATH by REF . It follows that MATH is not contained in the base locus of MATH, a contradiction.
math/0003169
REF is proved in REF . We recall the proof for the convenience of the reader. We set MATH, MATH, MATH, and MATH for MATH. Since MATH and MATH, we can write MATH for some numbers MATH. Hence MATH . On the other hand, we have MATH . Therefore MATH . CASE: We may assume that MATH is ample by the base point free theorem. Assume that MATH is not PLT, and let MATH be the LC threshold so that MATH is properly LC. Let MATH be a minimal center. By CITE, for any positive rational number MATH, there exists an effective MATH-divisor MATH on MATH such that MATH and MATH is KLT. Then MATH for MATH. Since MATH can be arbitrarily small, we have MATH and MATH. If MATH, then MATH by REF . Otherwise, we have also MATH by REF . By the vanishing theorem, we have MATH as in the proof of REF . From an exact sequence MATH it follows that MATH is not contained in the base locus of MATH, a contradiction.
math/0003169
We shall prove REF simultaneously. Let MATH be the smallest positive integer such that MATH. We shall derive a contradiction from MATH. We take a general member MATH. Assume first that MATH is PLT and MATH. Then MATH is NAME canonical. We have an exact sequence MATH . We have MATH and MATH. On the other hand, we have MATH by CITE, a contradiction. Next assume that MATH is not PLT and MATH. Let MATH be the LC threshold so that MATH and MATH is properly LC. Let MATH be a minimal center. If MATH, then we have MATH. By CITE, for any positive rational number MATH, there exists an effective MATH-divisor MATH on MATH such that MATH and MATH is KLT. By the perturbation technique, we may assume that MATH is the only center if we replace MATH and MATH suitably. We consider an exact sequence MATH . Since MATH is ample, we have MATH by the vanishing theorem. Hence the homomorphism MATH is surjective. If MATH, then we have MATH by REF . We shall also prove that MATH in the case MATH. Then it follows that MATH is not contained in the base locus of MATH, a contradiction, and REF are proved. Assume that MATH. We set MATH so that MATH. Let MATH. If we set MATH and MATH, then MATH for some numbers MATH and MATH by the NAME theorem. By the vanishing theorem, we have MATH and MATH, because MATH. Then MATH . Thus we have MATH.
math/0003179
For REF , see CITE. For REF , suppose that MATH, then MATH as MATH and MATH by REF . Then, by the MATH-adic criterion CITE, MATH would be a MATH-order, a contradiction. For REF , see CITE.
math/0003179
MATH . Since the genus of a non - singular plane curve of degree MATH is MATH, REF follows from CITE. MATH . This is well known property of the Hermitian curve; see for example, CITE or CITE. MATH . If MATH, then from MATH and REF , either MATH or MATH. By hypothesis, MATH can only occur, and so, by REF , MATH is MATH-isomorphic to the Hermitian curve MATH. Then MATH; see CITE Let MATH. By REF , MATH. Suppose that MATH and let MATH be the MATH-Frobenius divisor associated to MATH. Then CITE MATH so that MATH, and hence MATH . Thus, as MATH, we would have MATH and hence MATH. If MATH, from REF we have that MATH; this contradicts CITE (compare REF ). MATH . By CITE, MATH for any MATH. Then REF follows as MATH and MATH by REF . MATH . Suppose that MATH. Then, by CITE, MATH for any MATH. Therefore MATH a contradiction as MATH. MATH . From CITE and the MATH-maximality of MATH we have MATH . Since MATH and MATH, REF follows.
math/0003179
The statement on the genus follows immediately from the upper bound on MATH. By REF we have that MATH. If MATH, then MATH and from REF MATH is MATH-Frobenius classical. In particular, REF holds true; that is, we have MATH. It is easy to see that MATH for MATH and that MATH. Moreover, MATH and MATH, and the result follows.
math/0003179
If REF holds, then MATH is classical by REF . For MATH, the hypothesis on MATH yields MATH and hence MATH. Thus MATH is classical by REF . Note that the following computations will provide another proof of this fact. For the rest of the proof we assume MATH to be non-classical, and we show that no one of REF ,, REF holds. From REF , MATH, where MATH for MATH, and MATH for MATH. Also, MATH by REF . Therefore, as MATH for each MATH CITE. From CITE we have that MATH for each MATH, where as before MATH denotes the MATH-Frobenius divisor associated to MATH. Thus, MATH or, equivalently, MATH . On the other hand, the discriminant of the above quadratic polynomial in MATH is MATH and hence MATH if and only if either MATH and MATH, or MATH and MATH. For these MATH's, the above inequality cannot actually hold, and hence MATH must be classical. Furthermore, if MATH, then MATH . It is easy to check that MATH, MATH for MATH, and that MATH for MATH; hence if REF holds, then MATH must be classical. Let MATH. If MATH, then MATH must be classical by REF . So we can suppose that MATH. It turns out that MATH and hence the result follows when REF is assumed to be true.
math/0003179
REF follows from REF . To prove REF , we first note that MATH (compare REF ), and that MATH is a power of MATH (see REF ). Now, with the same notation as in the proof of the previous proposition, we get MATH with MATH. So MATH. Furthermore, MATH and so MATH by REF . Hence MATH and REF follows.
math/0003179
Let MATH be the ramification divisor associated to MATH and suppose that MATH for each MATH. Then from CITE, MATH which is a contradiction as MATH and MATH.
math/0003179
By some computations we obtain that MATH is bigger than MATH in REF . So the curve MATH is classical for MATH. Let MATH be as in REF . Then the MATH-orders are MATH and MATH with MATH (compare CITE). Therefore, the MATH-orders are MATH and MATH with MATH. Since MATH, from REF , MATH, and hence MATH by the MATH-adic criterion CITE. Also, the MATH-Frobenius orders of MATH are MATH and MATH with MATH; see CITE. Suppose that MATH and keep up MATH to denote the MATH-Frobenius divisor associated to MATH. Then CITE MATH or equivalently MATH . The discriminant of this equation is MATH and it is positive for any MATH. Since MATH is the biggest root of the quadratic polynomial in MATH above, MATH, a contradiction. Finally, MATH divides MATH by CITE.
math/0003179
Suppose that MATH. By means of some computations, MATH and hence REF holds true. With the same notation as in the proof of that lemma, we can then use the following two facts: MATH, and MATH. Actually, we will improve the latter one. MATH is a power of MATH . Indeed, by MATH and the MATH-adic criterion CITE, a necessary and sufficient condition for MATH not to be a power of MATH is that MATH and MATH. If this occurs, one can argue as in the previous proof and obtain the following result: MATH . From this, MATH which is a contradiction as MATH. MATH . The claim is certainly true for MATH. So, MATH, with MATH. If MATH, by REF we have MATH. Thus, this fact together with MATH would yield MATH and hence MATH, a contradiction. Now from REF and CITE, we have MATH and REF follows from REF .
math/0003179
Let MATH and MATH. Then the image of the morphism MATH is the curve defined by MATH. This proves the lemma.
math/0003179
If REF holds, it is clear that MATH is MATH-covered by the Hermitian curve. This property extends to MATH via the previous lemma. For both curves, the MATH-maximality now follows from CITE.
math/0003179
Let MATH, MATH, and MATH. Then MATH and MATH so that MATH . This shows that MATH is contained in the NAME semigroup MATH at MATH. In particular, MATH. Since MATH (see CITE), the result follows.
math/0003179
If REF is satisfied, the result follows from REF . Now if MATH is MATH-maximal, then MATH is also MATH-maximal by REF and CITE. Then the corollary follows from REF .
math/0003179
REF follows from the identity MATH and REF . To show REF , it is enough to check that MATH. Recall that the NAME symbol MATH is defined by: MATH . In our case, since MATH, MATH. By the quadratic reciprocity low MATH from MATH and MATH we get MATH. Now, as MATH, we have that MATH. In other words, MATH viewed as an element in MATH is a non-square in MATH. Since MATH is as well a non-square in MATH, it follows then that MATH with MATH such that MATH. Hence MATH as, in particular, MATH is odd and as MATH.
math/0003179
The curve MATH is MATH-covered by MATH via the morphism MATH, where MATH and MATH.
math/0003179
Let MATH. It is not difficult to see that MATH and MATH. Hence, for MATH, MATH and hence MATH provided that MATH and MATH. Let MATH denote the set introduced in REF . Then MATH, and it is easily checked that MATH is a semigroup. By means of some computations we see that MATH, whence MATH follows.
math/0003179
The ``if" part follows from REF and here we do not use the hypothesis that MATH is prime. For the ``only if" part, we first notice that each MATH is MATH-rational. Now the case MATH and MATH in the proof of REF gives MATH. Therefore MATH because MATH CITE. As MATH and MATH is prime, the result follows.
math/0003179
Similar to the proof of REF .
math/0003180
Since MATH does not divide MATH, the equation has MATH solutions in MATH. If MATH is a solution, then MATH, as MATH, and hence MATH.
math/0003180
Suppose that MATH is not complete. Then there exists MATH such that for any MATH-rational line MATH through MATH, MATH . If MATH is a line such that REF holds and that MATH, then MATH . Let MATH. Then, as MATH is MATH-rational, MATH and thus, as MATH, MATH is the tangent line of MATH at MATH (see REF ). Therefore MATH since MATH; see CITE. CASE: MATH. From REF and NAME 's theorem there exists MATH such that MATH. Then, as MATH, MATH is the tangent line of MATH at MATH. We have that MATH and the claim follows from the fact that MATH CITE. Now there are at most MATH lines MATH such that MATH. For each of these lines, MATH; hence from REF we have MATH . Then MATH, a contradiction. This finishs the proof of REF .
math/0003185
We construct a functor MATH from the first category into the second category and investigate its properties. Let MATH be a NAME module over MATH. Suppose, MATH is already MATH-linearly and continuously embedded into a certain standard NAME module MATH for some NAME space MATH. Consider the intersection MATH of MATH with the norm-dense subset MATH. This intersection MATH is non-empty and MATH-invariant by construction. Moreover, for topological reasons it is complete with respect to the norm MATH, and the unit ball of MATH is complete with respect to the topology induced by the semi-norms MATH. By CITE MATH is a self-dual NAME W*-module over MATH, and the completion of it with respect to the norm MATH recovers MATH. Note, that MATH is a direct orthogonal summand of the NAME W*-module MATH CITE, and that the MATH-linear projection to it extends to the MATH-linear projection from MATH to MATH. Consider two MATH-linear continuous embeddings of MATH into MATH and MATH, respectively. The images of MATH in MATH and in MATH are linked by an isometric MATH-linear operator with carrier projections equal to the projections to the embedded copies of MATH in MATH and MATH, respectively. Repeating our construction we obtain two self-dual NAME W*-modules MATH and MATH which are MATH-linearly and isometrically isomorphic. By NAME 's theorem CITE they are unitarily isomorphic, too. Consequently, the functor MATH does not depend on the embedding of MATH as a NAME module over MATH. Let MATH be any NAME space. Then MATH is the NAME norm closure of the self-dual NAME MATH-module MATH with respect to the NAME space norm MATH. For general self-dual NAME MATH-modules MATH consider their canonical decomposition as MATH with MATH of MATH. Obviously, MATH is an orthogonal direct summand of the self-dual NAME MATH-module MATH for a certain NAME space MATH with MATH, where the orthogonal complement is the self-dual NAME MATH-module MATH. Hence, the completion of MATH with respect to the norm MATH is a NAME module over MATH by definition. This shows the functor MATH to be surjective. At the same time we constructed the inverse functor MATH. The respective sets of bounded MATH-linear morphisms of both these categories can be seen to coincide looking at the present construction. Since every bounded module map on a self-dual NAME MATH-module possesses an adjoint and since every bounded module map on a NAME module MATH over MATH preserves the subset MATH invariant, the coincidence of both the involutions comes to light.
math/0003185
Every isometric copy of MATH as a MATH-submodule of another NAME MATH-module MATH is an orthogonal summand of MATH since MATH is self-dual by assumption, compare CITE. In case MATH for some NAME space MATH we have the isometric algebraic embedding MATH, where MATH denotes the algebraic tensor product. Let us fix the standard center-valued trace MATH on MATH. The bounded module operators on MATH that admit a finite center-valued trace are contained in the C*-algebra of all 'compact' operators on MATH which is defined as the norm-closure of the linear hull of the operators MATH, since the trace class operators MATH are all compact operators. Therefore, projections onto a NAME C*-module MATH admit a finite value with respect to the fixed standard semifinite center-valued trace MATH if and only if they are 'compact' operators, that is, if and only if their image is a finitely generated NAME C*-module, CITE. By CITE the type MATH NAME algebra MATH may admit more then just one faithful semifinite normal extended center-valued trace, and an easy classification is available. However, if a projection MATH admits a finite center-valued trace value with respect to a certain such trace on MATH, then the NAME algebra MATH is of type MATH, that is, also the restriction of the standard faithful center-valued trace MATH to it would give only finite values in MATH. The converse obviously also holds. This implies our statement.
math/0003185
The equivalence of REF has been shown for countably generated NAME MATH-modules in CITE. The implication MATH can be shown to hold setting MATH for the existing unitary operator MATH and for MATH. The demonstration of the inverse implication requires slightly more work. For the given normalized tight (modular) frames MATH and MATH of MATH and MATH, respectively, we define a MATH-linear operator MATH by the rule MATH, MATH. For this operator MATH we have MATH for every MATH. Consequently, MATH for any MATH, and the operator MATH is unitary.
math/0003185
The set of algebraic generators MATH of MATH is a frame with respect to any MATH-valued inner product on MATH which turns MATH into a NAME MATH-module. That is the inequality MATH is satisfied for two finite positive real constants MATH and any MATH, see CITE. What is more, for any frame of MATH there exists another MATH-valued inner product MATH on MATH with respect to which it becomes normalized tight, that is MATH holds for any MATH. This inner product is unique, compare CITE, CITE. So REF of the previous theorem gives the complete statement.
math/0003185
Since both the modular NAME bases are sets of modular generators of MATH we obtain two MATH-valued (rectangular, without loss of generality) matrices MATH and MATH with MATH and MATH such that MATH . Combining these two sets of equalities in both the possible ways we obtain MATH for MATH, MATH. Now, since we deal with sets of coefficients MATH and MATH that are supposed to admit special carrier projections, the coefficients in front of the elements MATH and MATH at the right side can only take very specific values: MATH where MATH is the NAME symbol, MATH is the carrier projection of MATH and MATH is the carrier projection of MATH. So MATH and MATH are positive idempotent diagonal matrices with entries from MATH. The NAME relations MATH, MATH, MATH and MATH turn out to be fulfilled.
math/0003187
For a definition of MATH-equivalence and NAME forms, see for example CITE. It is well-known that MATH and MATH are MATH-equivalent iff they have isometric NAME forms; for an algebraic-topological proof of that combine NAME and NAME, CITE, or alternatively NAME, CITE. Thus REF is equivalent to REF . If MATH is obtained from MATH by a null move on a clasper MATH, then MATH lifts to the universal abelian cover MATH of the knot complement MATH. The lift MATH of MATH is a union of claspers translated by by the group MATH of deck transformations. Furthermore, MATH is obtained from MATH by surgery on MATH. Since clasper surgery preserves the homology and linking form, it follows that MATH and MATH have isometric NAME forms, thus REF implies REF . Suppose now that MATH is MATH-equivalent to MATH. Then, by NAME 's result, the integral homology REF-sphere MATH can be obtained from MATH by surgery on some null MATH. Moreover, if MATH is a knot in MATH that bounds a NAME surface MATH, we can always arrange (by an isotopy on MATH) that MATH is disjoint from MATH, thus MATH is MATH-null. Thus, if MATH and MATH are MATH-equivalent pairs, modulo null moves we can assume that MATH. In that case, Naik and Stanford show that MATH-equivalent pairs MATH and MATH are equivalent under a sequence of double MATH-elta-moves, that is null-moves where all the leaves of the clasper bound disjoint disks, CITE. Thus, REF implies REF and the Lemma follows.
math/0003187
The action will be defined in terms of MATH-moves. Let us recall surgery on a clasper with no trivalent vertices, a so-called MATH-move: MATH . A MATH-move can be thought of as a right-handed finger move, or a right-handed NAME twist, and has an inverse given by a left-handed finger move (indicated by a stroke in the opposite direction in CITE). We will call right-handed MATH-moves positive, and left-handed ones negative. To define the action, without loss of generality, we will assume that MATH is a collection of claspers of degree MATH. Further, we orient the flags of MATH outward, towards their univalent vertices. Consider a monomial MATH where the product is over the set of flags of MATH. For each flag MATH of MATH choose a segment MATH of MATH, such that segments of different flags are nonintersecting. For each flag MATH of MATH, choose a collection MATH of MATH-claspers such that when MATH, we have: MATH . Let MATH be the collection of MATH-claspers for all flags MATH of MATH, and let MATH denote the image of MATH after clasper surgery on MATH. This defines an element MATH. We claim that this element depends on MATH alone, and not on the intermediate choices of arcs and the MATH-claspers. Indeed, REF implies that if an edge of MATH passes through two oppositely oriented arcs of MATH, resulting in a clasper MATH, then MATH. This implies easily our previous claim. Thus, we can define the action of MATH on MATH by MATH. It is easy to see that the relations of the ring MATH hold.
math/0003187
If MATH is an oriented edge, let MATH denote the pair of flags such that MATH has the same orientation as MATH and opposite from MATH. This gives rise to a map MATH and MATH, which is an isomorphism.
math/0003187
It follows immediately using REF above.
math/0003187
We now claim that the proof of REF works without change, and proves REF . Indeed, REF work for MATH-null claspers as stated. Furthermore, if MATH is a MATH-null clasper, then each leaf of MATH lies in the commutator group MATH, where MATH. Since MATH, this implies that each leaf of MATH bounds a surface in MATH whose bands are null homologous links in MATH. The surfaces for different leaves may intersect each other. We can apply now the proof of REF exactly as stated, to conclude the proof of REF .
math/0003187
Using REF , we will rather describe a canonical isomorphism MATH. Recall the exact sequence MATH where MATH and MATH are the abelian groups of MATH-valued functions on the vertices and oriented edges of MATH. Let MATH. This gives rise to the element MATH where the product is taken over all edges of MATH. It is easy to see that this element depends on the image of MATH in MATH and gives rise to a map MATH. It is easy to see that this map is an isomorphism.
math/0003187
Our Sliding Lemma implies that an edge of MATH can slide past two arcs of MATH with opposite orientations. It can also slide past another edge. It is easy to see that this implies our result.
math/0003187
Since MATH, it suffices to consider the case of even MATH. Recalling REF , suppose that MATH (for MATH) is a collection of claspers in MATH each of degree MATH, and let MATH denote the set of arms of MATH. REF and its following discussion implies that MATH and that the nonzero contribution to the right hand side come from diagrams in MATH that touch all arms. Thus, contributing diagrams have at least MATH-colored legs, to be glued pairwise. Since pairwise gluing needs an even number number of univalent vertices, it follows that we need at least MATH-colored legs. Notice that MATH contains no struts. Thus, at most three MATH-colored legs meet at a vertex, and after gluing the MATH-colored legs we obtain trivalent graphs with at least MATH trivalent vertices, in other words of NAME degree at least MATH. Thus, MATH, which implies that MATH is a invariant of integral homology REF-spheres of type MATH with values in MATH.
math/0003187
It suffices to consider a collection MATH of claspers in MATH each of degree MATH. Let MATH denote the set of arms of MATH. The counting argument of the above Proposition shows that the contributions to MATH come from complete contractions of a disjoint union MATH of MATH vortices. A vortex is the diagram MATH, the next simplest unitrivalent graph after the strut. Furthermore, the MATH legs of MATH should touch all MATH arms of MATH. In other words, there is a REF correspondence between the legs of such MATH and the arms of MATH. Consider a leg MATH of MATH that touches an arm MATH of MATH. If MATH touches MATH, then due to the restriction of the negative inverse linking matrix of MATH (see REF), it needs to be contracted to another leg of MATH that touches MATH. But this is impossible, since the legs of MATH are in REF correspondence with the arms of MATH. Thus, each leg of MATH touches presicely one edge of MATH. In particular, each component MATH of MATH is colored by three edges of MATH. Notice that the set of edges of MATH is an algebraically split link. Given a vortex colored by three edges of MATH, the coefficient of it in the NAME integral equals to the triple NAME invariant, (as is easy to show, see for example CITE) and vanishes unless all three edges are part of a degree MATH clasper MATH. When the triple NAME invariant does not vanish, it equals to MATH using the orientation of the clasper MATH. Thus, the diagrams MATH that contribute are a disjoint union of MATH vortices MATH and these vortices are in REF correspondence with the set of claspers MATH, in such a way that the legs of each vortex MATH are colored by the edges of a unique clasper MATH. After we glue the legs of such MATH using the negative inverse linking matrix of MATH, the result follows.
math/0003187
Since MATH is an unlink and MATH is the disjoint union of an unknot MATH and the link MATH, it suffices to consider the case of a hairy strut MATH. In this case, MATH is a meridian of MATH. The formula for the NAME integral of the NAME Link of CITE (applied to MATH), together with the fact that MATH has linking number zero with MATH, imply that the hairy part MATH vanishes.
math/0003187
We have that MATH . In other words, the diagrams that contribute in MATH are those whose components either lie in MATH or are hairy struts of the shape MATH. Because of the restriction of the linking matrix REF, the pair of legs MATH of a hairy strut of the above shape must be glued to a pair of legs labeled by MATH as follows: MATH . The result of this gluing is equivalent to gluing pairs of MATH colored legs using the negative inverse hairy linking matrix. This concludes the proof of the lemma.
math/0003187
It is easy to see that linking numbers satisfy the above axioms. For Uniqueness, assume that MATH is another such function, and let MATH. consider a link MATH, and a surface MATH that bounds MATH. Using the Cutting Property, as in the proof of REF , it follows that MATH is a linear combination of MATH and MATH, and hence zero.
math/0003187
The Symmetry, Cutting and MATH-Sliding Properties follows immediately from REF . The Specialization Property follows from the well-known fact that given a covering space MATH and a cycle MATH in MATH that lifts to MATH in MATH, and a cycle MATH in MATH, then the intersection of MATH with MATH equals to the intersection of MATH with the push-forward of MATH in MATH. The Uniqueness statement follows from the proof of REF given in REF.
math/0003187
It will be more convenient to present the proof in the algebra MATH. Let MATH denote a change of scaling of a string-link MATH, as shown in the following figure in case MATH has three stands: MATH . The locality of the NAME integral implies that MATH for an associator MATH. Write MATH for an element MATH, where MATH is the free-Lie algebra MATH of two generators MATH. The following identity MATH (valid in a free NAME algebra of two generators) implies that MATH. If we project the above equality to the quotient MATH where connected diagrams with two legs either both on the first strand or both on the second strand vanish, then it follows that we can replace MATH by its image MATH. It is easy to see that MATH for any associator MATH. Now we can finish the proof of the lemma as follows. Consider a sting-link MATH with relative scaling obtained from a disk-basing of a special link MATH of three components and let MATH be the one obtained by a change of relative scaling of MATH. Projecting to MATH, and using the fact that MATH lies in the center of the NAME algebra MATH, it follows that MATH.
math/0003187
Consider the special link MATH, (this is an abbreviation for MATH) whose connected sum of the first two components gives MATH. How does the NAME integral of MATH determine that of MATH? The answer, though a bit complicated, is known by CITE. Following that notation, we have MATH where MATH is the operation that glues all MATH-colored legs of MATH to those of MATH (assuming that the number of legs of color MATH and of color MATH in MATH and MATH match; otherwise it is defined to be zero), and MATH where MATH is an (infinite) linear combination of of rooted trees with at least one trivalent vertex whose leaves are colored by MATH and whose root is colored by MATH. We will call such trees MATH-trees. The reader may consult CITE for the first few terms of MATH, which are given by any NAME formula, translated in terms of rooted trees. Upon projecting the answer to the quotient MATH, the above formula simplifies. Indeed, if we glue some disjoint union of trees of type MATH that contain at least one trivalent vertex to some hairy MATH-colored trees, the resulting connected graph will either have nontrivial homology, or at least two labels of MATH. Such graphs vanish in MATH. Thus, when projecting the above formula to MATH, we can assume that MATH. Using the fact of how the NAME integral of a link determines that of its sublinks and the above, it follows that MATH which, together with REF concludes the proof.
math/0003187
Consider the link MATH. Recall that a slide move is given by: In an artistic way, the next figure shows the result of a slide move (compare also with REF ) that replaces MATH by MATH, where MATH is the orientation reversed knot. MATH . As in the previous lemma, we have that MATH where MATH is the operation that replaces a MATH-colored leg to a MATH-colored one. Upon projecting to MATH, the above formula simplifies. Indeed, the MATH-colored hairy trees are (after removal of the hair) of two shapes MATH and MATH. When we glue, a MATH-colored hairy MATH vanishes in MATH. The remaining MATH-colored hairy trees are of the shapes MATH as well as the hairless MATH and MATH. When glued to MATH and MATH rooted trees, the ones of shapes MATH vanish in MATH, thus we remain it remains to consider only trees of shape MATH and the hairless MATH and MATH. When we glue these to MATH and MATH rooted trees, the only nonzero contribution comes from gluings like MATH . The above figure shows that each of the above gluings can be thought of as starting from a hairy graph of shape MATH, and adding to it some additional hair, first on the left and then on the right; each time multiplying the result by MATH (where MATH) and MATH is the number of left and right added hair. On the other hand, adding hair is the same as commuting with MATH (as follows by REF ). Translating from gluings back to NAME algebras, it follows that MATH where MATH means to replace the label MATH by MATH in the expression MATH. REF implies that MATH which concludes the proof.
math/0003187
This follows from REF once we show that MATH satisfies the Symmetry, Specialization, MATH-Sliding, Cutting and Initial Condition stated in that lemma. The symmetry follows by REF . Specialization follows from the fact that MATH coincides with the coefficient of a strut in the NAME integral, which equals to MATH. The MATH-Sliding property follows from REF , the Cutting property follows from REF . The Initial Condition property follows from the fact that the NAME integral is multiplicative for disjoint union of links, thus the only diagrams MATH that contribute to the NAME integral (and also in MATH) in this case is MATH, which contributes the linking number of MATH and MATH.
math/0003189
MATH . Let us recall three different moves on the set MATH of REF knots in MATH: CASE: Changing a crossing, that is, doing a MATH-modification in the language of CITE and CITE. CASE: Doing a MATH-move, that is, doing a MATH-modification in the language of CITE. CASE: Doing a null-move, in the language of CITE. These three moves lead in the usual way to three notions of finite type invariants and corresponding notions of MATH-equivalence, denoted by MATH and MATH respectively. Note that MATH implies that MATH for all invariants MATH of type MATH. It is a folk result (easily proven by the results of NAME and NAME) that MATH iff MATH. Furthermore, it is easy to see that if MATH, then MATH. Further, in CITE and CITE, itt was shown that MATH is a finite type invariant of type MATH with respect to the null move, where MATH takes values in an appropriate MATH-vector space. This discussion implies the following conclusion, for every fixed MATH. MATH . Thus, MATH is an additive (under connected sum) function on MATH (here MATH denotes the set of MATH-trivial knots with respect to the MATH-move). By REF and NAME it follows that MATH is a NAME invariant of degree at most MATH. We claim that MATH is MATH-valued of NAME degree MATH. Indeed, MATH is a vector space spanned by uni-trivalent graphs MATH with MATH vertices such that every component of MATH contains a trivalent vertex, modulo the MATH and MATH relations. Using the MATH relation, we can assume that distinct univalent vertices of MATH are joined to distinct trivalent vertices. It follows that the number of trivalent vertices of MATH is at least equal to the number of univalent vertices, and thus that MATH has at least MATH trivalent vertices, which gives rise to a null move of degree MATH, on which MATH vanishes. This implies that MATH is of NAME degree MATH. Furthermore, if MATH is a unitrivalent graph of degree MATH with more than MATH trivalent vertices, then MATH. The remaining graphs of degree MATH are a disjoint union of wheels that is, diagrams like MATH. This, together with a result of CITE implies that the degree MATH part of MATH lies in the algebra of NAME coefficients. This concludes the proof of the first REF . The second part follows easily since, up to a multiple, there is a unique MATH-valued NAME invariant of MATH-degree MATH.
math/0003189
MATH . Notice that MATH and that MATH. Thus, MATH, where MATH and MATH is a constant. In order to figure out MATH, we need a computation. Let MATH be a wheel with two legs attached to an unknot MATH. Consider the MATH-link of degree MATH (also denoted by MATH) in the complement of MATH. Let MATH for MATH and MATH denote the six leaves of MATH labeled as in REF . It follows from CITE (see also CITE) that MATH is the result of complete contractions of all pairs of leaves of MATH, where we label the edges by equivariant linking numbers of the leaves. This gives rise to a linear combination of trivalent graphs with two trivalent vertices and edges decorated by elements of MATH. The equivariant linking function MATH (see CITE) of the leaves is given as follows: MATH where MATH (respectively, MATH) if MATH (respectively, MATH), and the sign MATH depends on the sign of the clasp of the NAME double. The computation of the equivariant linking function is best seen by drawing the universal abelian cover of MATH and lifting MATH to it. Alternatively, one may use the genus MATH . NAME surface of MATH drawn above and identify the equivariant linking numbers in question with the equivariant linking numbers of links formed by meridians dual to the bands. In CITE NAME computes the matrix MATH of equivariant linking numbers of meridians MATH dual to the bands of a NAME surface by: MATH where MATH is the NAME matrix with respect to a basis consisting of bands, and MATH is the transpose of MATH. In our case, the NAME matrix is MATH and the corresponding matrix MATH is MATH . On the other hand, we have that MATH, since MATH where MATH is the coefficient of the degree MATH wheel MATH in the logarithm of the NAME integral, CITE. The result follows.
math/0003190
Let MATH, let MATH be homogeneous and let MATH. By NAME commutator formula, MATH . Then the first part follows immediately. Since MATH is a submodule of MATH, we have MATH. It is easy to see that MATH. This completes the proof.
math/0003190
We shall prove the first part by induction on MATH. From the definition of MATH, the lemma is true for MATH. Assume it is true for any MATH homogeneous vectors in MATH. Now let MATH be homogeneous and let MATH with MATH . Set MATH . Since MATH, in REF , we may take MATH. Let MATH be any nonnegative integer such that MATH. By REF , we have MATH . Notice that MATH . Thus MATH . Then it follows from the inductive hypothesis that MATH . This finishes the induction and concludes the proof.
math/0003190
Let MATH be the set defined by REF . We shall prove MATH . Let MATH and let MATH be homogeneous. Then for any MATH, MATH because REF MATH . This proves REF . Since MATH for MATH, REF implies MATH . By changing variable we get MATH . By REF , MATH and by REF MATH . Consequently, MATH . That is, MATH . Conversely, let MATH . Then REF holds for each homogeneous MATH. Recall REF with MATH: MATH . Applying MATH to REF , then using REF and the fundamental properties of delta function we get MATH . Furthermore, for any MATH, MATH . Thus MATH, hence MATH. This completes the proof.
math/0003190
Recall the conjugation REF of CITE: MATH . Because MATH is locally nilpotent on MATH, we may set MATH, so that we have MATH . Then the first part of the proposition follows immediately. By changing variable MATH we get MATH . This completes the proof.
math/0003190
From REF we easily get MATH . Using this and the fact that MATH, we get MATH . This proves REF . The second part follows from REF immediately.
math/0003190
First, using REF we get (CITE, REF ): MATH . Because MATH for MATH, MATH . Since for any homogeneous MATH, MATH REF , we have MATH noting that MATH for MATH. Using REF and all the above information we have MATH . Similarly, using the fact MATH we have MATH . This completes the proof.
math/0003190
Let MATH. For homogeneous MATH we set MATH . Then extend the definition by linearity. It follows from REF that MATH gives rise to a MATH-structure on MATH . In particular, MATH . The following arguments are classical and routine in nature. Let MATH be homogeneous and let MATH. For any MATH, because MATH in view of REF we have MATH . Since MATH is arbitrary, we must have MATH . Using REF we have MATH for every MATH. Consequently, MATH . Similarly, using the fact that MATH gives rise to the involution MATH of MATH REF we have MATH . Then the left action and right action of MATH on MATH give rise to a left action and right action of MATH on MATH. The rest can be proved similarly.
math/0003190
In view of REF we only need to prove MATH for MATH, MATH. Let us assume MATH is homogeneous. First, from the proof of REF we have MATH . Then using the fact MATH we get MATH . Using REF , we get MATH . (This argument is also similar to one in the proof REF .)
math/0003191
Write MATH and MATH. Then MATH and MATH. The result follows.
math/0003191
The NAME theorem tells us that MATH if and only if there exists a nonzero continuous linear functional on MATH which vanishes on MATH. The result now follows from REF .
math/0003191
Let MATH such that MATH and define a nonzero function MATH by MATH. For MATH, set MATH. Then MATH . Conversely suppose there exists MATH such that MATH. This means that MATH for all MATH, for all MATH except on a set MATH of measure zero. Also MATH for all MATH except on a set MATH of measure zero. Since MATH, we may choose MATH such that MATH for some MATH. Now define MATH. Then MATH and the calculation in REF shows that MATH.
math/0003191
Define a NAME space isomorphism MATH by MATH for MATH. We want to show that this isomorphism commutes with the action of MATH. Clearly it will be sufficient to show that MATH commutes with the action of MATH. If MATH, then MATH . Thus the action of MATH commutes with MATH. We deduce that for MATH, there exists MATH such that MATH if and only if there exists MATH such that MATH. The proposition now follows from REF .
math/0003191
We have MATH. By the above remarks, MATH is a sum of elements of the form MATH. Therefore we need only prove that MATH. But MATH if MATH and REF if MATH, and the result follows.
math/0003191
Let MATH. It was shown in CITE that MATH. Write MATH. If MATH and MATH, then by NAME 's inequality CITE applied to the function MATH for MATH, MATH consequently MATH . Therefore MATH is a continuous map from MATH into MATH for MATH. It is also continuous for MATH. The lemma follows because the map MATH is continuous; specifically MATH.
math/0003191
Let MATH and write MATH. Then MATH . Applying this in the case MATH, we obtain MATH. Using REF , we deduce that if MATH and MATH, then MATH. Since MATH, the result now follows from REF .
math/0003191
If MATH, then MATH and MATH by REF . Conversely suppose there exists MATH such that MATH. Then MATH for some MATH, so replacing MATH with MATH, we may assume that MATH. If MATH, then MATH and MATH. Using REF we see that MATH, and we deduce from REF that MATH for all MATH. It follows that MATH for all MATH, consequently MATH. Let MATH be a maximal ideal in MATH which contains MATH. By NAME theory there exists MATH such that MATH, so MATH.
math/0003191
Since MATH, we see from REF that MATH. Of course MATH. We now prove the stronger statement that MATH for all MATH. We have MATH . Therefore MATH and the result follows.
math/0003191
We shall use induction on MATH, so assume that the result is true with MATH in place of MATH. Let MATH denote the NAME graph of MATH with respect to the generators MATH. Thus the vertices of MATH are the elements of MATH, and MATH are joined by an edge if and only if MATH for some MATH. Also MATH acts by left multiplication on MATH. Suppose a nontrivial word in MATH is the identity, and choose such a word MATH with shortest possible length. Note that MATH must involve MATH, because MATH is the free product of the group generated by MATH and the group generated by MATH. By conjugating and taking inverses if necessary, we may assume without loss of generality that MATH ends with MATH. Write MATH, where MATH, and each of the MATH are one of the above MATH elements. Let us consider the path whose MATH-th vertex is MATH. Note that MATH, but MATH for MATH. Observe that the path of length REF from MATH to MATH cannot go through MATH (just go through the MATH possibilities for MATH, noting that MATH). Now remove the edge joining MATH and MATH. Since MATH is a tree CITE, the resulting graph will become two trees; one component MATH containing REF and the other component MATH containing MATH. Since the length REF path from MATH to MATH did not go though MATH, for MATH the path MATH remains in MATH at least until it passes through MATH again. Also the path must pass through MATH again in order to get back to REF. Since the paths MATH all have even length (all the MATH are words of length REF), it follows that MATH for some MATH, where MATH. We deduce that MATH, which contradicts the minimality of the length of MATH.
math/0003191
This follows immediately from REF : replace MATH with MATH for all MATH.
math/0003191
The above elements generate the subgroup generated by MATH . The result follows from REF .
math/0003191
Let MATH and let MATH be the free group on MATH. By REF there is a monomorphism MATH determined by the formula MATH . Note that MATH induces a NAME space monomorphism MATH. Set MATH. Since MATH is a MATH-zero divisor by REF , we see that MATH is a MATH-zero divisor, say MATH where MATH. Write MATH where MATH is a right transversal for MATH in MATH. Then MATH where MATH for all MATH. Also MATH for all MATH and MATH for some MATH. Define MATH by MATH. Then MATH and MATH as required.
math/0003191
Let MATH such that MATH and suppose MATH. Since MATH is a positive definite function by CITE, we can apply CITE to deduce that MATH. By REF MATH and the result is proven.
math/0003192
Given MATH and MATH, let MATH be the result of varying MATH by multiplying the MATH coordinate by MATH and adjusting the last coordinate so as to remain on MATH (that is, MATH). Differentiating the relation MATH implicitly with respect to MATH at REF yields MATH . By minimality of MATH, we know that the modulus of MATH has a minimum at MATH, hence MATH is purely imaginary. Plugging this into REF proves that MATH is real. If MATH then MATH has a tangent vector at MATH in the direction MATH, where MATH is the MATH coordinate vector. This contradicts minimality. Hence MATH.
math/0003192
For MATH, let MATH be the torus MATH shrunk in the last coordinate by MATH, that is, the set of MATH for which MATH, MATH and MATH. Write NAME 's formula as an iterated integral MATH . Here MATH is the circle of radius MATH. Let MATH be a compact set not containing MATH. For each fixed MATH, the function MATH has radius of convergence greater than MATH. Hence the inner integral in REF is MATH for some MATH. By continuity of the radius of convergence,we may integrate over MATH to see that MATH decreases exponentially. Thus if MATH is any neighborhood of MATH in MATH, the quantity MATH decreases exponentially. Thus we have reduced the problem to an integral over a neighborhood of MATH. Near MATH there is a parametrization MATH of MATH. Let MATH be the circle of radius MATH. Then when MATH is sufficiently small compared to MATH, the image of MATH under MATH is disjoint from MATH. Fix such a neighborhood. For any MATH, the function MATH has a single simple pole in the annulus bounded by MATH and MATH, occurring at MATH. The residue in the last variable of MATH at MATH is equal to MATH where MATH is defined in REF . Therefore, for each fixed MATH, MATH . But MATH is bounded by a constant multiple of MATH (the constant depending on the maximum of MATH on MATH) and hence MATH is exponentially decreasing, where MATH . Changing variables to MATH and MATH turns the quantity MATH into MATH and plugging in the definitions of MATH and MATH at REF above yields MATH which is none other than MATH.
math/0003192
The first statement is immediate. To prove the second, let MATH and see from the definition of MATH that MATH . By definition of MATH, the ratio MATH is some constant MATH independent of MATH, hence MATH . The right hand side of this is the derivative of MATH with respect to MATH at MATH. By definition of MATH this vanishes, and hence MATH. But MATH, so the gradient of MATH must vanish at MATH. Finally, observe that MATH. By strict minimality of MATH, the modulus of MATH is greater than MATH for any MATH.
math/0003192
Differentiate the equation MATH to get MATH which is the same as REF . Differentiate again to get MATH and use REF to eliminate MATH, giving REF . The formula for MATH follows from the definitions of MATH and of the partial derivative.
math/0003192
Changing variables to MATH, the integral becomes MATH the curve MATH is the image of MATH under MATH, so MATH and MATH remains in the right half plane, strictly except at REF. For MATH write MATH as MATH, where MATH is a polynomial of degree MATH and MATH is bounded; this can be done since MATH may be approximated by a degree MATH polynomial to within MATH at REF. First, evaluate MATH by moving the contour. Replace MATH by two line segments, the first of which goes along the positive real axis to some distance MATH and the second of which is strictly in the right half plane (we assumed MATH except at REF). The integral along the second segment is exponentially small since the integrand is. Hence the combined contribution is the series REF out to the MATH term. Next, bound MATH . With MATH representing different constants in different lines, we now observe that on MATH we have MATH. Thus, parametrizing MATH by arc-length, an upper bound is given by MATH . This is easily seen to be bounded above by MATH where MATH depends on the first MATH derivatives of MATH and MATH. Choosing MATH we have a remainder term that is MATH, proving the theorem.
math/0003192
The two-sided integral is the sum of two one-sided integrals on intervals MATH and MATH. The integral over MATH may be written as an integral over MATH of the function MATH. With MATH still denoting the coefficients resulting form the application of REF to the first integral, let MATH denote the coefficients when REF is applied to the second integral. In order to add the two integrals, we write MATH in terms of MATH by means of the following routine computation. Let MATH with MATH and MATH and define the analytic quantity MATH so that MATH . If MATH is odd, then then the hypothesis MATH implies that MATH is purely imaginary. We have MATH . Writing MATH for the inverse function to MATH and MATH for the inverse function to MATH we then have MATH . Hence, letting MATH denote the coefficient of MATH, MATH and thus MATH . When MATH is even, the computation is similar but easier, resulting in MATH . Now observe that if MATH is odd, hence MATH is purely imaginary, then MATH according to the sign of the argument of MATH. Setting MATH if MATH is even and MATH if MATH is odd, we recover the definition in REF and prove the Corollary.