paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0104199 | At time MATH for all MATH we have MATH when MATH has sidelength MATH. Thus at time REF, the lemma is satisfied. Let MATH be the first time at which the lemma fails and MATH be one of the cubes for which it fails. It must be the case by hypothesis that MATH does not intersect MATH. Then we have MATH . Let MATH be the last time before MATH when MATH . Then we have MATH . However on the time interval MATH the lemma is satisfied. We will invoke REF . Now for any MATH we have MATH . This is because MATH. Further for any ancestor MATH with MATH having sidelength MATH, with MATH, MATH . Further for any follower MATH (that is a cube which contributes to MATH and is in particular contained in MATH) of MATH with MATH having sidelength MATH with MATH, we have MATH . Applying REF, we conclude that MATH. Thus we conclude MATH by counting the elements of MATH by MATH. Calculating, we find MATH which since MATH cannot possibly account for REF. Now applying REF to REF , we observe that for any MATH between MATH and MATH, we have dissipation at MATH of MATH. Thus to reach a contradiction, it suffices to show that on this time interval MATH . For this reason we can also ignore the ultra-low term MATH in REF . Using REF, we observe that for any MATH in MATH we have the estimates MATH as well as MATH by the lower bound on MATH. Now we calculate MATH . Thus MATH cannot contribute to the growth. Now to estimate MATH, we observe that for MATH we have the estimate MATH while for ancestor MATH of sidelength MATH, we apply REF (as well as the hypotheses of the lemma for squares larger than MATH) MATH since of course MATH. Now we just estimate MATH . Thus MATH could not have grown which is a contradiction. |
math/0104200 | For any reduced, irreducible projective curve MATH, not necessarily smooth, define MATH . Let MATH be a fibration of smooth projective surface over a smooth projective curve, all defined over a number field MATH. Suppose the fibration has good reduction at MATH. For any MATH, write MATH for the fiber of MATH at MATH. Define MATH . Denote by MATH the MATH-components of the fiber MATH. Define MATH . Note that MATH coincides with the usual definition when MATH is a genus one fibration (this follows from the case-by-case analysis in CITE), or when MATH is a smooth fiber (in which case MATH is the trace of the geometric NAME at MATH acting on MATH). Moreover, for every MATH, we have MATH which is the analog for our fibration of CITE. Consequently, MATH . On the other hand, since MATH has good reduction at MATH, the NAME conjecture gives MATH . Combine REF together with REF , we get MATH . Recall that the submodule MATH of the MATH-module MATH is generated by a fixed multisection MATH on MATH along with all components of all fibers of MATH over MATH. The geometric NAME MATH fixes MATH and acts non-trivially on any non-MATH-rational components of each fiber of MATH. Furthermore, in light of the NAME conjecture, MATH has at least one smooth fiber over MATH for all but finitely many MATH, and the image in MATH of any two such multisections differ by a non-zero MATH-multiple. Consequently, for all but finitely many MATH, MATH whence MATH . Denote by MATH the MATH-function associated to the MATH-module MATH. Following CITE, we have, for MATH, MATH . Combine REF , we get MATH . MATH is a finite dimensional MATH-module, so MATH is an NAME MATH-function. In particular, the second term on the right side of REF is MATH. Under NAME 's conjecture, the first term on the right side of REF is MATH. Invoke the NAME isomorphism REF and we get REF . |
math/0104200 | Let MATH be a semistable elliptic fibration over a number field MATH. Fix an integer MATH. Then MATH where MATH denotes the fiber product of MATH copies of MATH with itself over MATH. Denote by MATH . NAME 's desingularization of MATH CITE via a canonical sequence of blowups. See CITE for a description of the desingularization. Write MATH . Since MATH is smooth, we have MATH. For MATH the singular locus consists of isolated ordinary double points. Write MATH for the usual big-MATH notation with the constant depending on MATH only (and not on MATH); as usual the constant could change from line to line. Then MATH . Substitute this into REF and expand MATH using the binomial theorem, we get MATH . Set MATH. Then MATH by the NAME conjecture, so REF becomes MATH . Suppose MATH has a section. Then for every MATH we have MATH. The semistable fibration MATH induces a fibration MATH. CITE shows that MATH induces an isomorphism of fundamental groups MATH when MATH; his argument applies mutatis mutandis for MATH as well. By the compatibility of MATH-adic cohomology with NAME cohomology, MATH is then a MATH-equivariant isomorphism for MATH. When MATH, NAME 's exact sequence REF implies that MATH is also an isomorphism. The Lemma then follows. Invoke the Lemma and we see that MATH is equal to MATH since MATH for MATH. Thus MATH . Apply NAME 's conjecture and we get REF . |
math/0104200 | We can assume that MATH is relatively minimal. Write MATH for the generic fiber. Fix an integer MATH. Then MATH and hence MATH, are both divisors on the elliptic curve MATH over the function field MATH of MATH. Add up the points in MATH and we see that MATH is the divisor of a function on MATH. Since MATH is relatively minimal, by the NAME property MATH extends to a divisor MATH on MATH which is disjoint from MATH, and the Lemma follows by taking MATH. |
math/0104200 | It suffices to assume that MATH is an odd-prime power MATH, and MATH in NAME 's Theorem is equal to MATH . The two cases MATH and MATH are immediate consequence of the first part of the following Lemma. Let MATH be monic and quadratic, and let MATH be a prime. CASE: Suppose MATH. Then for any integer MATH, MATH . Moreover, there exists a unique MATH such that MATH. CASE: If MATH, then MATH has a unique solution MATH and has no solution MATH for any MATH. Write MATH. Since MATH, MATH by hypothesis. Furthermore, the quadratic formula shows that MATH is a double root of MATH, whence MATH. Thus for any MATH and any MATH, MATH . Thus MATH if and only if MATH. This gives the first part of REF . To get the second part, note that if MATH, then combining MATH and MATH with REF , we get MATH if and only if MATH. Repeat this argument finitely many times and we get MATH if and only if MATH, as desired. As for REF , MATH implies that MATH is also a repeated root MATH, whence MATH as well, and REF shows that MATH has at most one solution. But MATH is exactly divisible by MATH, by hypothesis. Returning to the proof of the last case MATH of REF , denote by MATH the maximal order in MATH, and by MATH the minimal polynomial of MATH. Then MATH for some integers MATH with MATH, so with the change of variable MATH, MATH . Thus MATH . If MATH, then the lifting argument in the proof of NAME 's Lemma shows that the number of solutions MATH of MATH is either MATH or MATH, depending on whether MATH is split or inert in MATH. This gives REF when MATH and MATH. The last subcase MATH is handled by REF . This completes the proof of REF . |
math/0104200 | Denote by MATH the elliptic modular surface associated to MATH. Both MATH and MATH are defined over MATH. CITE shows that MATH is a semistable fibration, and that its group of section is finite even after we extend the base field to MATH. To simplify the notation, write MATH and MATH for MATH and MATH, respectively. Then MATH since MATH for every bad fiber MATH. The number of MATH with MATH or MATH is at most MATH, so MATH . The supersingular fibers do not contribute to the sum, so for the rest of this section, we will assume that MATH . Thanks to REF , the inner-most sum in REF is simply MATH where MATH is any curve with MATH, and MATH as before. As we run through all MATH with MATH and MATH, by REF we run through precisely all integers MATH such that MATH . The number of MATH with MATH and with MATH equals to the order in MATH with discriminant MATH, is equal to MATH, where MATH . Combine all these together, invoke REF and we can rewrite REF as MATH where MATH runs through all integers as in REF and MATH as in the last Section. Since MATH, if MATH is odd then the double sum in REF is zero, whence the entire REF is MATH. Substitute this back into REF and we get, for odd MATH, MATH . REF then follows. To handle even MATH, we rewrite REF in terms of the NAME trace formula for NAME operators on cusp forms for MATH; REF then follows from the NAME conjecture estimate for these traces. For any fixed MATH, the number of pairs MATH with MATH or MATH is at most four. Since MATH is bounded from the above in terms of MATH, with MATH and recall that MATH, for even MATH we can further rewrite REF as MATH where the MATH and MATH are as in the end of the last Section. The NAME trace formula says that the trace of the NAME operator MATH on the space of weight MATH modular forms on MATH with MATH, is given by CITE MATH . This allows us to rewrite REF as MATH since the NAME conjecture gives MATH while REF gives MATH. Substitute these back into REF and we get REF for even MATH. |
math/0104200 | Part of the following calculation is inspired by CITE. From the spectral sequence for MATH we get a MATH-equivariant filtration on MATH with MATH where MATH. Thus MATH is the product of the MATH-functions associated to MATH for MATH. Moreover, MATH . To prove the Lemma it then suffices to show that for every MATH, MATH . We begin with MATH. Since MATH, we have MATH. In particular, MATH. Thus MATH has a simple pole at MATH while MATH, so both sides of REF are MATH when MATH. Denote by MATH the singular locus of MATH; it is also the singular locus for MATH. Denote by MATH the closed immersion, and by MATH the open embedding of MATH in MATH. For any MATH-adic sheaf MATH on MATH we have an exact sequence MATH . This becomes a short exact sequence for MATH, by the local invariant cycle theorem. Moreover, MATH where MATH is the pull-back to MATH of MATH via MATH and MATH the corresponding map. Consequently, MATH . Since MATH is a skyscraper sheaf, for MATH the NAME formula gives MATH . CITE shows that MATH is canonically identified with the space of weight MATH modular forms on MATH plus its complex conjugate. There are no weight MATH modular forms on MATH, so for MATH both sides of REF are zero. Finally, consider the case MATH. The NAME formula gives MATH so REF contributes MATH to both sides of REF . As for REF , MATH where MATH runs through the closed points of MATH and MATH denotes the closed immersion. From the theory of modular curves we see that MATH is a subfield of the cyclotomic field MATH. The fibers of MATH are also defined over some subfields of MATH, and hence the same holds for the fibers of MATH. That means as MATH-modules, MATH is a direct sum of Abelian characters MATH's. The MATH-function associated to MATH is therefore a product of Abelian MATH-series MATH. It is classical that at MATH, MATH either has a pole or has neither a pole nor a zero; we pick up a pole precisely when MATH is a MATH-dimensional MATH-fixed space of MATH. Thus the contribution from REF to the two sides of REF are equal as well, and the Lemma follows. |
math/0104200 | Let MATH be a non-split, isotrivial elliptic fibration over MATH. Then we can find an integer MATH, a smooth curve MATH of genus one and a non-constant function MATH such that, for all but finite many points MATH, the fiber MATH is the MATH-th order twist of MATH by MATH; in other words, MATH is isomorphic to MATH over the extension MATH. We begin with MATH, that is, a quadratic twist family. Then MATH for every smooth fiber MATH, whence by the NAME conjecture, MATH . Note that up to a term that is holomorphic for MATH, the first term on the right side is the logarithmic derivative of the NAME convolution of the MATH-series of the Jacobian MATH of MATH. Since MATH is a modular elliptic curve, the convolution MATH-series has a simple pole at MATH and MATH and is holomorphic for MATH CITE, whence the residue of REF at MATH is exactly one. This gives REF for quadratic twist families. Next, consider the case MATH. Then MATH is a (possibly trivial) MATH-torsor of MATH, and MATH is not a MATH-th power. Every smooth curve of genus one over any finite field MATH has a MATH-rational point, so MATH is MATH-isomorphic to MATH for some element MATH. Consequently, except for a finite number of MATH (the number of which is bounded from the above independent of MATH), MATH is isomorphic to MATH. It is classical CITE that for MATH sufficiently large, MATH where for MATH, MATH is a character of order MATH. Consequently, MATH . Denote by MATH the double cover of MATH given by the (possibly singular) model MATH. Then the inner sum in REF is simply MATH where the MATH-constant is independent of MATH. This is MATH by the NAME conjecture, so the residue at MATH of the left side of REF is zero. The same argument applies mutatis mutandis to the case MATH and MATH. |
math/0104201 | Obvious. |
math/0104201 | Obvious. |
math/0104201 | It follows from CITE that if MATH or MATH is not wild then any of its non-semisimple blocks is NAME equivalent to one of the algebras in REF. We show that all algebras in this list with the exception of MATH, MATH and MATH are derived wild. Let MATH be one of the algebras from REF and let MATH be one of the algebras MATH. Since MATH is wild, there exists MATH-bimodule MATH such that the functor MATH preserves indecomposability and isomorphism classes. We set MATH and denote by MATH the matrix corresponding to the map MATH with respect to some fixed basis. CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: Set MATH for MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH. CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH, MATH and let MATH be the following complex of MATH-bimodules: MATH where MATH . CASE: Let MATH. Since box MATH is wild, there exists MATH-bimodule MATH such that the functor MATH preserves indecomposability and isomorphism classes. Denote by MATH the following complex of MATH-bimodules. Set MATH for MATH, MATH for MATH, MATH, MATH . It is not difficult to verify that the functor MATH, which acts from the category of finite-dimensional MATH-modules to the category MATH, where MATH is one of the algebras from REF , preserves indecomposability and isomorphism classes. So, MATH is derived wild. It follows from CITE that algebra MATH is derived tame. It follows from CITE that the algebra MATH is derived tame. The derived tameness of the algebra MATH follows from REF (see REF). |
math/0104201 | The proof follows from CITE. |
math/0104201 | Consider the functor MATH defined as follows. The modules are MATH and the differential maps are MATH where MATH is the matrix corresponding to the map MATH in some fixed basis. It is easy to see that MATH is a representation equivalence. |
math/0104201 | Let MATH be the box corresponding to MATH (see above). Let us consider the sub-box MATH of MATH, where MATH, MATH and MATH. It is easy to see that MATH, where MATH is the principal box corresponding to the hereditary algebra MATH (that is, MATH). Consider the trivial category MATH with the set of vertices MATH and the functor MATH which maps: MATH . Construct the amalgamation MATH, or, the same, the couniversal square: MATH . Consider now the box MATH, where MATH. We say that the box MATH is obtained from MATH by reducing the sub-box MATH. By CITE, MATH induces the representation equivalence MATH. Straightforward calculation, which we omit, shows that MATH and there exists the sub-bimodule MATH of the bimodule MATH such that MATH, MATH for each MATH and MATH for each MATH. Hence it follows from the definition of the representations of a box that MATH. Then it is ease to see that MATH for each indecomposable representation MATH of MATH, where MATH is as in REF . |
math/0104201 | It is ease to see that MATH, MATH, and MATH has the following minimal projective resolution: MATH where MATH . Straightforward calculation, which we omit, shows that MATH . Hence the theorem follows from REF . |
math/0104206 | CASE: The equivalence of REF as well as the implication MATH are straightforward. Now assume MATH exists and MATH. If MATH, then MATH according to REF and therefore MATH, which is impossible (again by REF ). If MATH then MATH and, hence, there is an element MATH. But then the point MATH is outside MATH - a contradiction. CASE: Assume MATH. According to REF we have to show that exactly one of the inequalities MATH and MATH holds. The opposite (strict) inequalities are excluded by the same reasons as in the proof of REF . Next we exclude that simultaneously MATH and MATH. Assume that MATH and MATH. In particular MATH and, therefore, one of these base facets differs from MATH. If MATH, then there is a point MATH and by the equality MATH we get a contradiction because MATH, that is, MATH is outside MATH. It only remains to show that MATH and MATH can not hold simultaneously. But if this were the case, then for every lattice point MATH we would have MATH . This is a contradiction, because MATH for MATH. CASE: The proof is straightforward. CASE: Suppose MATH. Then REF is satisfied for MATH, MATH: note that MATH and MATH for MATH. Similarly REF holds with MATH. If REF holds, then MATH is clearly a column vector with base facet MATH: MATH for all MATH. If REF holds, then MATH for all facets MATH. The only vector satisfying this condition is MATH. CASE: Let MATH, MATH. One has MATH, and there exists a facet MATH with MATH. If MATH, then MATH, and this is impossible for MATH. Thus MATH, and since MATH, it follows that MATH, MATH. Now consider the facet MATH. As seen already, it is different from MATH and MATH. So MATH forces MATH. Moreover, for all facets MATH we must have MATH. In view of REF both MATH and MATH are invertible column vectors. |
math/0104206 | CASE: That MATH exists follows from the definition of the product and the equality MATH. Thus, by REF we only need to show that MATH. But the same proposition implies MATH. CASE: That the existence of MATH and MATH implies that of MATH follows from MATH. The example (see REF ) MATH completes the proof because MATH and MATH exist and MATH does not exist. |
math/0104206 | We define the mapping MATH as follows. For every lattice point MATH consider the recursively given sequence of nonnegative integers: MATH . We also have the lattice point MATH . To the data MATH and MATH we now associate the lattice point MATH where the sequence MATH is related to MATH in the same way as MATH to MATH. It is straightforward to see that we do not get outside the source polytope and that the mapping MATH is an affine isomorphism respecting the lattice structures. |
math/0104206 | Pick MATH. First notice that it is impossible to have simultaneously MATH, MATH and MATH. In fact, if these inequalities held, then the product MATH existed because MATH would be the base facet for MATH, and REF applies. But similar arguments show that the product MATH existed as well - a contradiction with REF . By symmetry the inequalities MATH, MATH for MATH are also impossible. It only remains to exclude the case in which MATH and MATH. Then we can conclude that MATH is parallel to either MATH or MATH, as claimed. But if MATH and MATH then the product MATH exists and, simultaneously, MATH which is impossible because MATH is balanced. |
math/0104206 | First consider the case when MATH exists. As a lattice polytope MATH decomposes into lattice polytopal layers parallel to the affine plane spanned by MATH, MATH and MATH, i. e. we consider the maximal lattice polytopes in the sections of MATH with REF-planes parallel to MATH. Clearly, the automorphisms MATH and MATH restrict to automorphisms of MATH for each of these layers MATH and we have to check that MATH and MATH are the same layer by layer. If a layer MATH has dimension MATH, then at least two of the vectors MATH, MATH and MATH do not belong to MATH. On the other hand, if MATH or MATH, then one easily deduces from REF that all the three vectors MATH, MATH and MATH belong to MATH. Therefore, MATH. But this means that MATH, MATH. In particular, the automorphisms MATH and MATH restrict to the identity automorphism of MATH and the commutator equality becomes trivial on the layer MATH. So we only need to consider REF-dimensional layers MATH. Then MATH and MATH exists as a product in MATH. Since MATH we have MATH. Without loss of generality we can therefore assume MATH and MATH. By a suitable affine integral unimodular change of the coordinates in MATH one can achieve that MATH, MATH is an edge of MATH, and MATH, MATH. Moreover, by a parallel shift we can also assume that MATH is the lowest right vertex of MATH. Consider the triangle MATH (see REF ). We have MATH. On the other hand MATH is point-wise fixed under automorphisms of type MATH, MATH and MATH (MATH). Therefore we only need to check that the two automorphisms MATH coincide at the vertex MATH of MATH. Observe that all the automorphisms involved restrict to elements of MATH. Therefore, after the natural identification MATH in such a way that MATH and MATH for MATH the problem has been reduced to the equation MATH the verification of which is accomplished by a routine computation. The case when MATH exists is just a consequence of the previous case. Now assume MATH. In this situation, too, we could check the desired equality layer by layer. In fact, using REF it is easy to show that MATH for any of the layers MATH as above. But the arguments below do not get simplified by the consideration of layers. By REF none of the products MATH and MATH exists. By REF there are only two possibilities: either MATH and the claim follows from REF or MATH, what we will assume. Since MATH is parallel to MATH, we get MATH. Then by REF there is exactly one facet MATH such that MATH and for all the other facets MATH we have MATH (the pairing is considered on MATH). Now MATH extends to a unique facet MATH, MATH, and we see that MATH. Again by REF MATH. Using similar arguments for MATH and MATH we arrive at the conclusion that the two facets MATH meet along a common facet (a codimension REF face of MATH) and, moreover, MATH is a base facet both for MATH and MATH. Assume for the moment that MATH contains a lattice point MATH in its relative interior. Then MATH and MATH. Assume in addition that MATH is in the relative interior of MATH. Then MATH. The automorphism MATH fixes pointwise both MATH and MATH. On the other hand MATH is generated by MATH and MATH (or, similarly, by MATH and MATH). Therefore, it suffices to show that the elementary automorphisms MATH and MATH commute at MATH. It is clear that both automorphisms restrict to an automorphism of the subalgebra MATH where MATH is the lattice REF dimensional parallelepiped with vertices MATH, MATH, MATH, MATH. (Clearly, MATH is a unit lattice square up to unimodular transformation.) Now the desired commutativity follows from the easily checked equation MATH where the semigroup operation is written multiplicatively. Consider the general case. Since MATH consists of non-zero-divisors it is enough to show that MATH and MATH coincide on NAME subalgebras MATH for all sufficiently big natural numbers MATH. It is in general not true that MATH (unless MATH is a normal polytope, that is the semigroup MATH is normal, see CITE for a detailed study of these properties). But the monomials of degree MATH in MATH constitute a subset of MATH such that all the arguments above that we have used for MATH apply to it. It only remains to notice that for MATH big enough the existence of the desired lattice points in the relative interiors as above is guaranteed. |
math/0104206 | The desired infinite series of decompositions is derived as follows. First fix some MATH such that MATH and then choose MATH recursively such that MATH . The vector MATH gets decomposed at each of the indices MATH because MATH and MATH share the base facet in MATH. |
math/0104206 | Consider the `infinite lattice polytopes' MATH where we mean the filtered unions of polytopes and their ambient Euclidean spaces. It is enough to show that there is a global affine (i. e. linear + constant) transformation MATH such that MATH and MATH. The existence of such MATH is established as follows. By REF (and REF ) for every index MATH there exists MATH such that MATH is a subset of MATH. Consider the minimal subpolytope MATH for which MATH and MATH. It is clear from REF that MATH and MATH are naturally isomorphic as lattice polytopes and, moreover, MATH is a face of MATH. In particular we obtain an isomorphism between MATH and a face of MATH. Let MATH denote the affine continuation of this isomorphism to the ambient Euclidean spaces. Then the MATH constitute a compatible system and, therefore, induce a global affine transformation MATH. It is an easy exercise to show that MATH satisfies all the desired conditions. |
math/0104206 | Choose MATH. There is no loss of generality in assuming MATH for some MATH, MATH and MATH. In particular, MATH restricts to an automorphism of each of the MATH, MATH. CASE: We claim that for every MATH there exists MATH such that MATH. (MATH denotes to the multiplicative group of invertible elements.) Let MATH be a facet. It is enough to show that every element MATH satisfies the condition MATH for some MATH and MATH with MATH. In fact, by running MATH through MATH we prove the claim. In the argument below we consider the polytopes MATH for all facets MATH. Clearly, if MATH is a not a base facet of an element of MATH, then MATH does not show up in a doubling spectrum starting with MATH - we use this doubled polytope only as an auxiliary object. If MATH for some MATH, then the two automorphisms MATH commute. If MATH is not the base facet of any of the column vectors, then none of the products MATH and MATH exists for any MATH. In fact, we have MATH (by REF) and MATH. Since MATH, REF implies that MATH and MATH again commute. For simplicity put MATH. Consider the point MATH that corresponds to MATH. We have MATH for some MATH and MATH with MATH, where MATH is considered in MATH. It is easily seen that MATH where MATH. Clearly, MATH. On the other hand MATH . Since MATH and MATH commute, we get what we want by comparing these two expressions. CASE: Let MATH be a polytope, MATH and MATH be a MATH-automorphism of MATH of type MATH, MATH for all MATH. Put MATH where MATH and the operation in MATH is written multiplicatively. The ratio is in fact independent of the choice of MATH because MATH. Then MATH . (See also REF .) Returning to our central automorphism MATH we see that MATH whenever the points MATH differ by a vector from MATH. Since we can replace MATH by some polytope obtained by successive doublings at base facets of column vectors, it is enough to apply the following lemma: clearly MATH for the lattice point MATH constructed there, and thus MATH since MATH and MATH differ by a sum of column vectors. |
math/0104206 | We construct a chain MATH of lattice points MATH such that MATH is a multiple of MATH and MATH for MATH. Let MATH. If MATH, then we choose MATH. Otherwise MATH for MATH, and we set MATH. Since MATH is parallel to the extension of MATH to MATH for MATH (by MATH) and MATH, we have reached a lattice point with the desired properties. |
math/0104206 | As remarked, we have fixed a doubling spectrum MATH. For every MATH consider the subsets MATH of MATH. (Unlike the classical situation the two sets MATH and MATH are not completely similar.) We have the subgroups MATH generated correspondingly by the MATH, MATH and MATH, MATH (MATH). Assume MATH and MATH. Claim. The mappings MATH are group isomorphisms, where the free MATH-modules MATH and MATH are viewed as additive abelian groups. Moreover, the canonical surjection MATH is injective both on MATH and MATH. First one observes that MATH does not exist for any pair MATH . REF implies this for the pairs from MATH without the condition that MATH is balanced, whereas this condition is needed for the pairs from MATH: if MATH existed then for some MATH then MATH, contradicting the condition that MATH is balanced. As a result of this observation all elements MATH and MATH have presentations MATH . For all points MATH and MATH with MATH we have MATH where MATH is a MATH-linear combination of the points from MATH with height above MATH at least REF and MATH for MATH. (The semigroup operation is written multiplicatively). It is immediate from the second equation that the presentation REF is uniquely determined by MATH. Observe that for each MATH there exists MATH such that the corresponding MATH is REF. Therefore, by running MATH through MATH we see that likewise MATH uniquely determines the presentation REF . Now the claim follows. Since MATH is balanced, for all vectors MATH, MATH and MATH and elements MATH we have: MATH and MATH . The listed cases exhaust all possibilities since MATH, MATH and the product MATH does not exist - we have MATH and REF applies. (It is not difficult to find examples for which MATH exists.) Since MATH, MATH and MATH whenever the corresponding product exists, we arrive at the inclusions: MATH for all MATH. Changing MATH by MATH we get MATH for all MATH. Now we are ready to prove that MATH . Choose an element MATH is such that MATH. We want to show that MATH for all MATH and MATH. Since MATH is the inductive limit of the groups MATH and MATH restricts to an automorphism of MATH for MATH, there exist MATH mapping to MATH provided MATH and such that MATH is the image of MATH in MATH. We can also assume MATH and that MATH gets decomposed in MATH at MATH. We will show that MATH maps to the center MATH. Clearly, this implies that MATH is central in MATH. For the elements of MATH we have MATH where MATH is the column vector corresponding to MATH. The claim above and REF imply that MATH commutes both with elements of MATH and MATH. Hence it also commutes with MATH. We have shown the inclusion MATH. The other inclusion follows from REF . |
math/0104206 | We have MATH for MATH, MATH and MATH. The product table for MATH consists of only the equation MATH (see REF ). We now describe the group MATH explicitly. First we make the following observation: if MATH is a polytope and MATH then MATH is naturally isomorphic to MATH so that the points of MATH are mapped to themselves under this isomorphism. It follows from this observation and REF that for producing a sequence of polytopes isomorphic to a doubling spectrum (starting with MATH) one only needs to decompose at each step one of the original column vectors of MATH or one of the vectors of type either MATH or MATH. (See REF .) Fix a doubling spectrum MATH. Consider the points MATH, MATH and MATH from MATH. We have MATH and MATH. Based on the aforementioned general observation one easily sees that for every MATH the set MATH looks as follows. There is certain system of points in MATH such that MATH where MATH . Moreover, the numbers MATH, MATH and MATH can be arbitrarily big if MATH is big enough. Now MATH admits the following description. Let MATH and MATH denote two disjoint copies of MATH and let MATH be an ``origin", MATH. Then MATH is generated by symbols MATH where MATH, MATH and MATH and these symbols are subject to the standard NAME relations: MATH . In REF we have tried to visualize these data. It only remains to notice that the proof of REF goes through for MATH without any change. It is of course important that for a pair MATH from our index set we can always find MATH such that MATH and MATH are also in the index set. |
math/0104206 | The second assertion is REF . Suppose MATH is a central extension, i. CASE: MATH. We want to construct a splitting homomorphism MATH. According to CITE this will show the universality of the central extension MATH. Pick MATH. Since the extension MATH is central, the commutators of type MATH with MATH and MATH coincide. Therefore we can use the notation MATH for this common value. For every generator MATH of MATH with MATH and MATH we choose an index MATH such that the column vector MATH is decomposed at MATH and put MATH . (Here MATH is the corresponding vector.) All we have to show is that CASE: the MATH are independent of the choices of MATH, CASE: the MATH are subject to the same relations as the MATH. To this end we make the following observation. Let MATH be the quadrangle considered in REF , and MATH be its column vectors. For some MATH . Consider MATH such that MATH exists. Then the assignments MATH, MATH, MATH REF give rise to a group homomorphism MATH. This observation follows from the definition of polytopal NAME groups and REF : after the identifications MATH, MATH and MATH the higher members of the doubling spectrum MATH of MATH only admit column vectors that also show up in MATH. It is also crucial that the defining relations for MATH are preserved by the corresponding elements of MATH because both MATH and MATH are balanced. By REF every central extension of MATH splits. On the other hand the homomorphism MATH in the pull back diagram MATH is central since MATH is central. A splitting of MATH yields the commutative triangle MATH which simultaneously shows several things: CASE: the definition of MATH is in fact independent of the choice of MATH, CASE: MATH for all MATH, CASE: for all MATH and all MATH for which MATH exists one has MATH. In fact, by lifting MATH and MATH to the appropriate vectors in MATH one observes that MATH for all MATH. Applying this formula to the column vectors MATH, MATH, and MATH (playing the roles of MATH, MATH and MATH) we prove the independence of MATH of the choice of MATH. The other two properties of the MATH follow similarly by using appropriate mappings MATH: for REF we use the same mapping MATH, MATH, MATH and for REF we use the mapping MATH, MATH, MATH, MATH. Only one NAME relation remains to be checked. Consider a pair of column vectors MATH such that MATH. We want to show that MATH for all MATH. Like in the proof of REF one must consider relations of this type separately. Let MATH be as in REF . Then MATH . Assume we have shown that none of the products MATH exists. Then, since the corresponding sums are non-zero vectors, we get MATH . It follows that MATH . It remains to show that the products REF do not exist. (Notice that this difficulty is absent in the case MATH.) For simplicity put MATH. Thus MATH and MATH have the same base facet in MATH, namely MATH. (The arguments below use REF several times.) First observe that the inequality MATH is impossible because otherwise MATH would exist. This already excludes the existence of MATH. Also, the product MATH does not exist because MATH. By the equations MATH and MATH we have MATH for every facet MATH different from MATH. Since MATH the base facet MATH is different from MATH. Therefore MATH, and MATH does not exist. Finally we have to exclude the existence of MATH. There are two cases - either MATH or MATH. If MATH we have MATH by REF , and we are done. If MATH and MATH existed, then its image in MATH under the MATH-rotation would be MATH - a contradiction because we have assumed that MATH does not exist. |
math/0104206 | CASE: The mapping MATH extends to doubling spectra of MATH and MATH and induces a map MATH as follows. If MATH is doubled with respect to MATH, then we also double MATH along MATH and extend MATH by setting MATH, MATH, MATH, MATH. If MATH is doubled with respect to a facet MATH that is not of type MATH for some MATH, then we put MATH. Using REF one checks easily that the extension of MATH again satisfies REF implies that the product MATH exists if and only if MATH exists. Furthermore it follows from REF that MATH if and only if MATH. In conjunction with REF this guarantees the compatibility of the assignment MATH with the NAME relations. CASE: We must check that MATH maps the center of MATH to the center of MATH. This certainly holds if MATH for MATH commutes with every MATH for all MATH. This is somewhat tedious, but, in view of REF , straightforward to check. Hence the claim on induced homomorphisms follows from REF , and that on surjectivity is then trivial. REF is obvious in view of REF . |
math/0104206 | It follows immediately from REF that the polytopes in each class are MATH-equivalent, and that polytopes from different classes are not MATH-equivalent. Now let MATH be a balanced polygon. CASE: There are column vectors MATH such that MATH exists. We claim that in this case MATH. There exists a facet MATH of MATH such that MATH. Thus we have the following table whose entries are given by the ``scalar products" of the vectors MATH with the facets MATH: MATH . Note that MATH have non-negative heights with respect to all other facets. Evidently MATH and MATH form a basis of the lattice MATH, and every column vector MATH is a linear combination of MATH and MATH, MATH. Since MATH, and at most one of these numbers can be negative, one easily checks that MATH and the following implications hold: MATH . But MATH, MATH is impossible, since MATH in this case. We have now shown that indeed MATH. If MATH are the only column vectors, then we are in REF and can stop the discussion, since this case is possible: see the proof of REF , where we have considered the polygon MATH . So suppose at least one of MATH is a further column vector. Assume that MATH. Then clearly MATH, and MATH is parallel to MATH. After a change of coordinates we can assume MATH, MATH, so that MATH. Since MATH is parallel to MATH, a parallel translation moves MATH into a position where the line segment MATH is contained in MATH and MATH is a vertex. The point MATH must also lie in MATH, which is parallel to MATH. Since MATH is parallel to MATH, the lower end point of MATH must belong to MATH, and we can sketch part of MATH as in REF . The ``upper" end point of MATH must lie in MATH and so MATH has exactly the three facets MATH. It follows that MATH is a multiple of the lattice unit triangle, and therefore MATH. This is REF . If MATH is a column vector, then MATH, and consequently MATH, MATH. Moreover, MATH and, hence, MATH have height MATH with respect to all the other facets. Then MATH is a column vector, and we are again in REF . Therefore, in addition to MATH only MATH can be a column vector (unless we are in REF ). Then MATH is the base facet of MATH, and this implies MATH, MATH (since MATH is balanced), so that MATH is again parallel to MATH. In this case MATH is projectively equivalent to the polytope MATH, since all the facets except MATH and MATH must be parallel to MATH. (See REF where we have sketched an affine-integrally equivalent polytope.) REF . None of the products MATH REF exists. There are two possibilities: either all the column vectors share the same base edge or there are MATH with MATH. In the first situation it is clear that MATH can have an arbitrary number of column vectors: just consider the balanced quadrangles MATH . In this situation MATH. Now assume MATH for some MATH. By REF we have MATH. We claim that there are only two possibilities: either MATH or MATH. First we show that any element of MATH is parallel to either MATH or to MATH. In fact, if there is MATH which is not parallel to one of these edges then either MATH or MATH (for otherwise MATH would share the base edges with both MATH and MATH which is impossible by REF ). But then REF implies that one of the products MATH or MATH exists - a contradiction. In particular we see that MATH. One only needs to show that if MATH then MATH as well. In fact, if MATH, then all the facets of MATH different from MATH and MATH are parallel to MATH, and MATH and MATH are parallel to MATH. So MATH is a parallelogram. Since MATH and MATH span the lattice MATH, the parallelogram MATH is projectively equivalent to the unit square (up to affine-integral equivalence). In this case MATH. That the case MATH is also possible is shown by the balanced pentagon MATH whose only column vectors are MATH and MATH. |
math/0104206 | The first equation is just the definition of MATH. It is enough to prove the second, third and fourth equations - the last two follow from them since the unit square is the direct product of two line segments, and since MATH is MATH-equivalent to MATH. Let MATH, MATH and MATH denote the groups on the right hand side of the three equations. By MATH and MATH we denote two mutually disjoint copies of MATH, which are also disjoint from the original MATH, and introduce the following index sets: MATH . By identifying MATH and MATH with free MATH-bases of two copies of MATH, say MATH and MATH, and MATH with that of MATH, we can view the groups MATH, MATH, and MATH as the corresponding subgroups of MATH, MATH and MATH in an obvious way. In particular, these groups consist of elementary automorphisms of free MATH-modules with respect to their distinguished bases. For each of the three cases we denote by MATH, MATH the associated elementary automorphism, MATH belonging to the corresponding index set. They are subject to the standard NAME relations. Similar arguments as in the proof of REF (and REF ) show that MATH and MATH can be assumed to be generated by symbols MATH where MATH, MATH and, respectively, MATH or MATH. (REF itself is about MATH.) Moreover, these symbols are subject to the same NAME relations as the MATH. Therefore we have natural surjective group homomorphisms MATH . By REF we only need to show the equations MATH . First we show that MATH (MATH is for the center). Since MATH, the centers MATH, MATH and MATH are in the subgroups generated by MATH, MATH, where the MATH are respectively from MATH, MATH and MATH. We have the group epimorphism MATH given by deleting the last column and row. Therefore, if MATH then MATH is actually concentrated in the smaller subgroup, generated by MATH with MATH. Now we are done by the observation that for arbitrary homomorphisms MATH, MATH and MATH the equations MATH and MATH simultaneously for all MATH imply that MATH, MATH, MATH and MATH are actually zero homomorphisms. It follows that MATH, MATH and MATH. To derive the opposite inclusions we apply the same arguments as in REF . They go through once we show that the analogue of the second part of the Claim in that proof remains true when we change MATH there by the unstable versions of each of the homomorphisms MATH, MATH and MATH. More precisely, for each natural number MATH we introduce the sets MATH, MATH, MATH which are finite versions of MATH, MATH and MATH, defined via the sets MATH and MATH. They give rise to the corresponding unstable subgroups MATH, MATH, MATH. We have surjective group homomorphisms MATH whose sources are the appropriate unstable NAME groups. One should notice that for successive indices MATH these NAME groups may not correspond to successive members in the fixed doubling spectra - there may be big intervals of intermediate members. Assume MATH. We let MATH denote the subgroups which are the same for these unstable NAME groups as MATH and MATH for MATH in the proof of REF . What we want to show is that the restrictions MATH are all injective group homomorphisms. We have the natural embeddings MATH . Under these embeddings the generators of MATH (MATH, MATH) are sent to standard elementary matrices MATH with the same MATH. Similarly, the generators of MATH (MATH, MATH) are sent to standard elementary matrices MATH with the same MATH. This, clearly, gives the result. |
math/0104207 | CASE: The curve MATH is isomorphic, as a curve with MATH action, to MATH for any permutation MATH of the indices MATH. CASE: Conjugation by MATH defines an isomorphism MATH between MATH and MATH. There is clearly a natural MATH equivariant isomorphism between the curves MATH and MATH, and from this the result follows. |
math/0104207 | The subbundle MATH is the MATH-invariant part of MATH; since the group MATH is finite, it is a direct summand and MATH as MATH-bundle. On the other hand, MATH is zero, since MATH is equal to MATH which is zero. |
math/0104207 | Let MATH. It is enough to show that the fiber MATH of MATH at MATH has dimension given by REF . The MATH-equivariant bundle MATH on MATH is defined by a representation of MATH. By the principal theorem of CITE it follows that the parabolic degree MATH is MATH. Applying the NAME theorem to REF gives MATH . The result follows by observing that MATH, and by REF. |
math/0104207 | The bundle MATH splits, as a vector bundle with MATH-action, as direct sum MATH. Pulling back to MATH and applying MATH both respect this direct sum decomposition; this proves the result. |
math/0104207 | Basically this follows from REF . MATH is the direct sum, over MATH and over connected components MATH of MATH, of MATH. It is therefore enough to prove the following: given two elements MATH in MATH, connected components MATH in MATH and elements MATH, for every connected component MATH of MATH the element MATH is in MATH. Here MATH is the natural inclusion. As restriction preserves the grading, and MATH raises the degree by MATH, the only thing left to prove is indeed REF . |
math/0104207 | Write MATH for the natural inclusion. Using the excess intersection formula, it is a straightforward computation to check that the product MATH is equal to MATH where MATH is MATH . The only thing one needs to remember is that the assumption that all the MATH have even degree allows one to move them around in the product freely. Analogously, MATH is equal to MATH where MATH is MATH . |
math/0104207 | The only thing left to prove is associativity: we will check that the sufficient condition of REF is verified. We will construct on MATH two vector bundles MATH and MATH such that the left hand side of REF is MATH REF and the right hand side of REF is MATH REF ; then REF we will prove that MATH and MATH are isomorphic. |
math/0104207 | The left hand side is the top NAME class of the bundle MATH. As equivalent vector bundles in the NAME group have the same NAME classes the result follows. |
math/0104207 | Let MATH. We have an exact sequence MATH as all sheaves involved are MATH-sheaves and the maps are MATH-equivariant, the long exact sequence of cohomology is an exact sequence of finite dimensional MATH representations. As a MATH representation, MATH is equal to MATH, since MATH is connected. As the connected component MATH of MATH has stabilizer MATH, the space MATH is equal to MATH (see the appendix for details); analogously MATH is the representation MATH and MATH is MATH. Moreover MATH is equal to MATH and analogously for MATH. |
math/0104207 | For any MATH bundle MATH, any subgroup MATH of MATH and any representation MATH of MATH, we have MATH. The result follows immediately by applying this remark to the induced representations in the previous Lemma, and taking MATH. |
math/0104207 | It is of course enough to prove that MATH is isomorphic as representation of MATH to MATH. Consider the natural map MATH. It is easy to prove that it is surjective. As deformations of MATH are unobstructed, this proves that there exists a smoothing of MATH preserving the MATH action; therefore there is a flat family of curves MATH over a small disk such that MATH acts fiberwise, the central fiber is MATH and the other fibers are smooth. Hence the general fiber MATH of MATH is isomorphic to the NAME MATH cover MATH of MATH branched over four points with stabilizers MATH, MATH, MATH and MATH (defined in the appendix); the MATH representation MATH does not depend on the particular branch points chosen. On the other hand, since MATH is a MATH vector bundle with fiber the MATH of the fiber, MATH is isomorphic as a representation to MATH. The same argument also applies to MATH, completing the proof. |
math/0104207 | By REF for suitable MATH, so MATH if MATH. By REF combined with REF we get that MATH has rank MATH and therefore MATH. Therefore MATH, where MATH is the inclusion and MATH is the cup product on MATH. Therefore MATH. This shows REF . REF follows immediately from REF . |
math/0104207 | Let MATH; choose MATH and MATH. Define MATH such classes MATH generate MATH for MATH even and MATH for MATH odd. We will prove that MATH; this proves the stronger statement that every element of MATH skew commutes with every element of MATH. It is enough to check that, for any fixed MATH, MATH for MATH. Let MATH, MATH, and MATH. Using this new notation, we have to check that MATH . Let MATH. By REF where MATH is the natural inclusion; analogously MATH . Note that MATH in view of REF ; on the other hand, MATH, and this completes the proof. |
math/0104207 | This follows immediately from REF . |
math/0104207 | Both vector spaces are defined as direct sums over the elements of MATH, so it is enough to define MATH componentwise. As already remarked, the fixed locus of MATH on MATH is naturally isomorphic to the submanifold MATH of MATH induced by the natural surjection MATH. This isomorphism determines MATH. The grading is preserved, since in both cases it is chosen so as to have the graded pieces distributed symmetrical around zero: in MATH because MATH is equal to half the real codimension of MATH in MATH, and in MATH because in CITE the grading of each summand of MATH is centered around zero. The morphism MATH is MATH-equivariant by comparing REF with paragraph REF We see that MATH also preserves the duality, by comparing REF with REF. |
math/0104207 | We use the notation of Construction REF. Let MATH be the representation of MATH on MATH induced by the natural action of MATH. As a MATH-equivariant vector bundle, MATH is isomorphic to the tensor product of MATH (with the trivial MATH-action) and of MATH. This implies that MATH is isomorphic to MATH, where MATH is MATH. |
math/0104207 | Let MATH, and write MATH. We have to prove that for every MATH and MATH . The fixed locus of MATH is naturally isomorphic to MATH: for every MATH, let MATH be the natural projection. Comparing with the definition of the product in REF, we see that we have to prove that MATH where MATH is the graph defect defined in REF Note that MATH coincides with MATH as defined in CITE. Because of the splitting REF , MATH where MATH viewed as a bundle on MATH (canonically isomorphic to MATH). Hence is it is enough to prove REF in the case where MATH acts transitively on MATH; write MATH for MATH (the unique orbit of MATH). In this case the rank MATH of MATH is equal, by comparing REF with the definition of the graph defect, to MATH. This completes the proof, by REF , since MATH for every MATH. |
math/0104207 | Let MATH be the NAME algebra MATH with the degree map given by MATH. Replace also the degree map on MATH by MATH. Then in CITE a ring isomorphism MATH compatible with the duality pairing is constructed. In the definition of the product MATH on MATH the degree is used in two places. First, for the definition of the pushforward MATH where MATH is the inclusion. With the new definition of the degree the pushforward changes by a sign MATH. Second, in the definition of the class MATH in REF. In follows from the definition that MATH is changed from MATH to MATH. Let MATH. Then, for the products to be the same, MATH in REF has to be replaced by MATH. Putting this together, we have to change MATH by a factor of MATH, where by REF MATH that is, MATH is replaced by MATH. The explicit formula in terms of the NAME operators follows from the definition of MATH in CITE directly before REF . |
math/0104207 | As all the NAME classes of MATH vanish, we get MATH if an only if the rank of MATH is zero, in which case MATH. By REF the rank of MATH is MATH. |
math/0104207 | The proof will occupy the rest of this section. We use some of the ideas of CITE p. CASE: Let MATH be a subgroup of MATH and assume MATH. Then MATH is connected and there is a canonical isomorphism MATH. There is a MATH-equivariant morphism MATH given on points by MATH; for any subgroup MATH of MATH its restriction is a morphism MATH. The action of MATH on MATH by MATH commutes with the MATH-action and the map MATH is just the quotient map for this action. Let MATH with MATH. As in CITE p. REF one shows that MATH is isomorphic to MATH (in particular MATH is connected) and that the action of MATH on MATH is trivial. Therefore MATH is a natural isomorphism MATH. Let MATH be a subgroup of MATH, and let MATH. We identify MATH with MATH. Note that the restriction of MATH to MATH is given by sending MATH to MATH. As all MATH are divisible by MATH, we can define MATH and for MATH we define MATH. By definition we get MATH. The proof of the next lemma shows that this is the decomposition of MATH into connected components. There is a canonical isomorphism MATH. MATH is identified with MATH . For any MATH with MATH, we get an isomorphism MATH. Grouping the elements of MATH in sets of MATH elements, each of which contained in an orbit of MATH, defines a surjection MATH, giving isomorphisms MATH and MATH for a subgroup MATH of MATH with MATH. Furthermore with the same proof as in CITE p. REF the induced isomorphism MATH is independent of the choice of MATH. As MATH, there is a canonical isomorphism MATH . Putting everything together we get a canonical isomorphism MATH which by definition commutes with the MATH-action. This shows REF . We note that the action of MATH on MATH is just the restriction of the action on MATH. Furthermore for any subgroup MATH, the normal bundle of MATH in MATH is just the restriction of the the normal bundle of MATH in MATH. In particular the age MATH of an element MATH is the same for both actions, and the bundle MATH on MATH is the restriction of the corresponding bundle on MATH(in view of REF ). Therefore the ring structure on MATH is determined in the same way as for MATH: MATH and if we write MATH for a class MATH in the summand corresponding to MATH, then MATH with MATH given by REF with the restriction to MATH replaced by that over MATH. Here MATH is the embedding of MATH into MATH. We need to determine how the product is distributed over the connected components MATH. We write MATH for a class MATH and MATH for a class MATH. Let MATH, MATH, let MATH, MATH. Then MATH if MATH. Otherwise MATH . Here in the second line the inner sum is over all MATH such that MATH and MATH. In the last row we require in addition that MATH. Note that this is equivalent to MATH, MATH, MATH. Finally MATH, MATH, MATH are the inclusions. From the definitions it is also obvious that, if MATH and thus also MATH, then the isomorphisms MATH and MATH commute with the pullbacks induced by the inclusions. Therefore REF follows from REF , thus completing the proof of REF . |
math/0104211 | Let MATH be a square root of MATH (that is, MATH). First, note that MATH, for any line bundle MATH such that MATH; moreover if MATH is NAME, so is MATH, by the remarks after REF . Therefore, if the result is true for some square root of MATH, then it is also true for any other square root. If MATH, and MATH is the trivial line bundle, MATH is NAME 's bundle MATH (see REF), so the result follows from REF below. So, assuming MATH, consider the following diagram, induced by the quotient homomorphism MATH, whose commutativity is easily established, and whose vertical sequences are principal fibrations: MATH . Here MATH denotes the connected component which contains the image of the inclusion MATH, and similarly for MATH. Using the techniques of NAME CITE, we see that the marked NAME group MATH, of REF , gives rise to a flat projective NAME bundle MATH, which by construction is in the image of the natural map from the space of all projective structures on MATH to MATH. Then, By REF , there is a flat vector bundle MATH such that MATH and MATH has divisor order MATH, so that it can be given as an extension MATH, with MATH. MATH is indecomposable because of NAME 's theorem, thus MATH, and by uniqueness, MATH. On the other hand, if MATH is such that MATH, then MATH is NAME by definition, and the commutativity of the diagram implies MATH. Thus, by exactness, MATH and MATH, differ by tensoring with a MATH-bundle MATH (MATH), so MATH is NAME as required. |
math/0104211 | . Since MATH has only scalar commutants, the action of MATH by conjugation is free and given MATH in the same orbit, there is a unique MATH such that MATH. The same arguments used in REF, prove that MATH (the center acts trivially) is a complex manifold, and that the tangent spaces agree with the required cohomology groups. Moreover, MATH is an analytic submanifold of MATH, since the inclusion MATH is complex linear when MATH is NAME. Since MATH is a free group, MATH, for any MATH-module MATH. Therefore, the symplectic form vanishes on any two tangent vectors to MATH. |
math/0104211 | . We may suppose that MATH (see the appendix). Let MATH be MATH distinct complex numbers of modulus REF. Let MATH, and MATH be the permutation matrix MATH, for a canonical basis MATH. It is easy to see that MATH form an irreducible set of unitary matrices (that is, no subspace of MATH is preserved by both). Hence, the representation of MATH given by MATH, is unitary, NAME and irreducible. Stability is clear from NAME 's theorem. |
math/0104211 | . In any holomorphic parameter space, stable bundles form the complement of an analytic subset REF . Since MATH is non-empty (irreducible unitary representations give stable bundles, for MATH), the existence of the family MATH implies that MATH is open and dense in MATH. The same applies to MATH, since stable NAME bundles exist, by REF . |
math/0104211 | . This follows from REF, where it is shown that the space of isomorphism classes of simple vector bundles with degree REF and rank MATH, MATH, verifies the universal property of a coarse moduli space in the holomorphic category. In other words, for every holomorphic family MATH of simple vector bundles over MATH, parametrized by a complex manifold MATH, the ``universal map" MATH sending MATH to the equivalence class of MATH is holomorphic. Since our maps MATH and MATH are actually the ``universal maps" for the holomorphic families MATH and MATH constructed above (consisting of stable, hence simple, bundles), and MATH is an open subset of MATH, we have the coarse moduli space universal property for MATH as well. |
math/0104211 | . By definition of NAME group, every MATH is loxodromic, and any two generators of MATH have distinct fixed points. Since two non trivial NAME transformations commute if and only if they have the same fixed points, only the identity commutes with all elements in MATH. Therefore, every representation MATH has only scalar commutants. |
math/0104211 | . To prove MATH, note that an isomorphism between MATH and MATH is a holomorphic global section of MATH, consisting of invertible matrices. So it corresponds to a holomorphic MATH such that MATH. To prove MATH, let MATH be as in REF and put MATH where MATH. We obtain MATH, MATH and MATH belongs to MATH, since MATH is equivalent to MATH . Conversely, if we have REF , then clearly MATH verifies REF . |
math/0104211 | . CASE: First note that both maps MATH and MATH are defined from MATH to MATH. Let MATH and MATH. For small MATH, we can expand MATH as: MATH. Let MATH denote MATH, for brevity. Discarding second order terms, we find: MATH . The derivative of the curve of representations MATH is given by MATH, so the differential at MATH in the MATH direction is finally given by: CASE: For any MATH and MATH, we have MATH (by REF ). Letting MATH, we see that the differential of MATH, at the origin is zero: MATH. Hence MATH. Conversely, if MATH, then MATH is tangent to the fiber of the map MATH at MATH, which means tangent to the image of MATH at REF, so that there is a MATH, such that MATH. |
math/0104211 | Since MATH and MATH have the same dimension, MATH REF , we just need to show that MATH has trivial kernel. Since MATH is the composition MATH, the differential at MATH is the composition: MATH, and so, (by REF ): MATH . |
math/0104211 | . Let MATH be a cocycle representative of MATH and similarly for MATH. Let MATH, so that MATH and MATH, for all MATH. By NAME 's theorem, we have: MATH where the curves MATH are the MATH sides of the boundary of the polygon MATH, whose vertices can be ordered as MATH. Half of the MATH sides give (using the notation MATH): MATH . A similar computation for the remaining MATH sides gives the desired formula. |
math/0104211 | . Since MATH is a free group, MATH, for any MATH-module MATH. Therefore, if MATH, then MATH and by REF , MATH. Therefore, MATH. |
math/0104211 | When MATH, REF , togheter imply that MATH is invertible. The set MATH is a closed analytic subset of MATH which is not the whole set. Since MATH is a holomorphic map between the complex manifolds MATH and MATH, on the complement MATH, MATH is a local diffeomorphism, and therefore, it is an open map. So, we can take MATH. |
math/0104211 | . Write MATH, where MATH act by MATH and MATH. Consider the NAME representation MATH given by assigning to MATH the MATH matrix MATH whose entries are all zero except for ones on the principal diagonal and on the diagonal above it (MATH are the nonzero entries). We claim that MATH. Clearly MATH, so assume that MATH. By construction, MATH is an extension of the form MATH. and so, by REF , the lemma can be proved by showing that MATH. Sections of MATH over MATH correspond to holomorphic functions MATH satisfying MATH. This means that MATH is a constant (being an entire doubly periodic function), and that the other components of MATH have REF as period, and verify MATH. Since an abelian differential with zero ``MATH-periods" (in this case MATH) has to be zero, we get MATH, and MATH constant. Repeating this argument we get MATH, for all MATH, and MATH is a constant and so, MATH. |
math/0104211 | . By NAME 's theorem, we may assume that MATH is indecomposable of degree REF. Then by REF , MATH. Since MATH is NAME and NAME is a line bundle, we conclude that MATH is also NAME (see the remarks after REF ). |
math/0104211 | . If MATH for some MATH, then MATH is a non-conjugate NAME representation, for all MATH. Moreover MATH, by REF . |
math/0104215 | Let MATH be an analytic NAME symmetry of order MATH. Therefore the relation MATH holds. MATH has already been diagonalized and hence MATH. From the analyticity of MATH, we can expand MATH into the power series with respect to MATH as MATH where MATH and MATH are homogeneous polynomials of order MATH. Since the lowest order terms of MATH must be zero, we obtain the relation MATH . Let us explicitly write MATH as MATH . Straightforward computation gives MATH . Since MATH is of order MATH, at least one coefficient MATH is not zero. This implies REF . |
math/0104215 | Set MATH and MATH. Hence REF can be rewritten as MATH . The l.h.s. of REF can be expanded into the NAME series with respect to MATH from the analyticity of MATH. This implies MATH must be an integer. |
math/0104215 | We show that MATH is an eigenvalue of MATH defined as REF and MATH is an eigenvector belonging to MATH. First by differentiating REF with respect to MATH and substituting MATH and MATH, one obtains MATH . On the other hand, MATH . By substituting REF into REF , we have MATH . Furthermore, since MATH, MATH . The lemma is proved. |
math/0104215 | Let the vector field MATH be a quasihomogeneous NAME symmetry of degree MATH. The quasihomogeneous system REF is transformed into the autonomous system REF by the transformation REF. Applying the transformation REF to the NAME symmetry REF, we have MATH . Putting MATH, we have Thus MATH which is analytic with respect to MATH. Then the new independent variable MATH satisfies a linear ordinary differential equation MATH . A straightforward computation tells that the transformed vector field REF is a NAME symmetry of the extended autonomous system MATH . Therefore, applying REF , we have a resonance relation MATH where MATH. Multiplying MATH to the both hands of the first equation in REF and rewriting MATH and MATH, we obtain REF . This completes the proof. |
math/0104215 | Applying the scale transformation REF to the system REF, we have MATH where MATH is a formal NAME series with respect to MATH (respectively, MATH) without constant terms. On the other hand, an analytic (respectively, polynomial) NAME symmetry MATH of the system REF is transformed into the form MATH (respectively, MATH) for some MATH, where MATH are quasihomogeneous vector fields of degree MATH. Hence MATH is an analytic (respectively, polynomial) NAME symmetry of the system REF. The system REF approaches the truncated system REF as MATH (respectively, MATH). On the other hand, the NAME symmetry REF (respectively, REF) approaches MATH simultaneously. Thus the truncated system REF possesses a NAME symmetry MATH. |
math/0104217 | The last part is an easy consequence of the first statements, compare CITE. The first part can be proven exactly as in CITE provided that one is still able to construct a derivation of weight -REF on MATH: By assumption there is an effective divisor MATH and a vector field MATH such that MATH. Let MATH be the subgroup of automorphisms fixing MATH. The existence of MATH implies that MATH is nontrivial. Furthermore the linear representation of MATH on the MATH-jets of a point MATH is faithful if MATH. Therefore MATH is linear algebraic and contains a linear algebraic one parameter group MATH, that is, MATH or MATH. Since MATH stabilizes the divisor MATH there is a MATH-linearization of the line bundle MATH and a dual MATH-action MATH on MATH (compare CITE). If MATH there are semi-invariant elements MATH of MATH, that is, MATH which generate MATH as a MATH-algebra ( MATH is the weight of the semi-invariant element MATH). Let MATH correspond to MATH. Let MATH be the homogeneous coordinate ring of MATH. Since MATH acts trivial on MATH via MATH it acts semi-trivial on MATH with weight MATH, and the quotients MATH are equal for all MATH. This implies MATH and MATH since otherwise MATH would act trivial on all of MATH. Now one can twist the MATH-linearization: Let MATH act on MATH via MATH for all homogeneous MATH. Hence for an arbitrary polynomial MATH: MATH and MATH. It follows MATH where MATH is the NAME corresponding to MATH. One can divide MATH by MATH to get a MATH derivation MATH. If MATH let MATH still be homogeneous generators of MATH, the element MATH corresponding to MATH. Since the unipotent group MATH fixes MATH the group acts trivial on MATH. This implies for homogeneous MATH that MATH . Once more one can divide the corresponding derivation MATH by MATH and gets a MATH derivation MATH on MATH. Now one constructs an element MATH as in CITE with MATH . Then MATH with MATH and MATH (compare CITE), and the theorem follows. |
math/0104217 | Let MATH be the connected and nontrivial algebraic subgroup of the automorphism group of MATH which fixes the zero locus MATH of MATH. Because the action of MATH on the vector space of MATH-jets at a fixed point will be faithful for MATH, the group MATH is linear algebraic. Let MATH be the minimal algebraic subgroup whose NAME algebra contains MATH. A representation of MATH in MATH shows that MATH CITE. Because MATH is commutative, the fixed point locus MATH is contained in MATH and MATH stabilizes MATH. By composing the various MATH- and MATH-actions one can move the divisor MATH along orbits to a linearly equivalent divisor MATH stabilized by MATH. MATH implies MATH, linear equivalence means MATH, consequently MATH. And MATH is reduced and irreducible because MATH is. |
math/0104217 | By construction MATH is linearly equivalent to MATH, hence very ample. The corresponding embedding MATH maps MATH into a hyperplane, and the cone MATH consists of the lines through the cone vertex MATH (not in the hyperplane) and points MATH. Such a line MATH cuts MATH transversally in one point, consequently (in MATH): MATH . MATH implies the hypothesis. |
math/0104217 | Since MATH is very ample, by NAME there exist two smooth hyperplane sections MATH with MATH. Then MATH does not contain any line MATH from MATH to a point MATH. As in the proof of the previous claim the MATH intersect every line MATH from MATH to a point MATH exactly in MATH with intersection multiplicity MATH, that is, transversal. Therefore the strikt transforms MATH do not intersect at all. Since the MATH are smooth, the intersection multiplicities of the MATH in MATH are MATH. Then, by CITE MATH . |
math/0104217 | By the adjunction formulas one gets MATH and MATH . The nonnormal locus MATH on MATH is given by the conductor ideal of the normalization MATH, and its support consists of whole fibers of the MATH bundle MATH. From the subadjunction formula for normalization (and the formula for the canonical bundle on smooth ruled surfaces), it follows MATH where MATH is the section with negative self intersection on the smooth ruled surface MATH and MATH is a fiber. Let MATH be a fiber of MATH and MATH a general fiber of MATH. Then MATH. Furthermore, MATH, where MATH is a line from the cone vertex MATH to a point MATH. Since MATH is a section of the projective line bundle MATH over MATH, the intersection multiplicity MATH. Therefore: MATH . Let MATH be defined by MATH. As in the proof of REF , MATH. Consequently: MATH and this implies MATH. But the index MATH of the NAME threefold MATH is MATH, hence MATH and MATH. |
math/0104217 | Set MATH. First consider the structure sheaf sequence of MATH tensorized by the tangent bundle MATH, MATH and the beginning of the corresponding long exact sequence, MATH . The NAME sequence MATH and MATH for MATH, MATH imply MATH. Therefore, MATH is surjective. By the normal sequence MATH one finally has MATH. |
math/0104217 | The polynomial MATH may be written as MATH . Since the intersection with MATH is supposed to be a cone over a basis MATH with vertex MATH, this cone MATH is given by the equation MATH, and MATH. Since the hypersurface is smooth, the rest follows from computing the gradient of MATH: the coefficient of MATH must not be MATH because otherwise the vertex MATH will be a singularity in MATH, too. Similarly, the coefficient of at least one of the monomials MATH, MATH, must not vanish, because otherwise the point MATH lies in MATH and will be not smooth. |
math/0104217 | Since MATH, the curve MATH is no line. Since vector fields on projective spaces always vanish on complete linear subspaces, MATH vanishes on the plane MATH. On the other hand, MATH does not vanish on the hyperplane MATH, because it contains the cone MATH, and MATH should be non trivial on MATH. Since every vector field on a cone vanishes in the vertex, MATH vanishes in MATH. This implies the hypothesis, because the zero locus of vector fields on MATH consists of the eigenspaces of the transposed corresponding matrix. |
math/0104219 | This can be done by cutting and pasting MATH along some disks. Note that such operations do not have any effect on REF if we take a suitable choice of MATH. |
math/0104219 | Suppose that there exists an arc MATH of MATH such that MATH. By exchanging MATH if necessary, we may assume that MATH is outermost in MATH, that is, MATH. If MATH connects different vertices, then a MATH-compression of MATH along MATH reduces MATH. Otherwise, MATH incidents a single vertex, say MATH. We perform a MATH-compression of MATH along MATH, and obtain an annulus MATH consisting of the disk MATH and the resultant band MATH. Since we chose an outermost arc MATH and MATH, there exists a compressing disk for MATH in MATH. By retaking MATH along the compressing disk, we can reduce MATH. |
math/0104219 | Suppose that there is a face MATH such that MATH can be oriented. Then, since no corner of MATH intersects MATH, and by REF , MATH has non-zero intersection number with MATH on MATH as illustrated in REF . This is a contradiction. |
math/0104219 | This can be done by an isotopy of MATH since REF assures us that MATH is irreducible. |
math/0104219 | This can be done by the same argument to REF . |
math/0104219 | Suppose that there is a vertex MATH with valency REF. Then only one edge MATH incident to MATH, and hence exactly one of MATH and MATH is attached to MATH or contained in MATH. Thus MATH intersects MATH in two points. Since MATH is prime, MATH bounds a disk MATH in MATH which intersects MATH in an unknotted arc. In the former case, MATH lies under a subarc of MATH by the minimality of the number of bridges MATH. Then by an isotopy of MATH along REF-ball which is bounded by MATH, we can reduce MATH. See REF . In the later case, MATH intersects MATH in one point, and MATH bounds a pair of a REF-ball and an unknotted subarc of MATH by the minimality of MATH. Then an isotopy of MATH along the pair can reduce MATH. See REF . |
math/0104219 | Suppose there exists a face MATH as REF . Then MATH consists of an edge MATH of MATH and a subarc MATH of the boundary of a vertex MATH of MATH. By REF , MATH intersects MATH. Moreover, since the loop MATH bounds a disk MATH in MATH, MATH and MATH meets exactly one of MATH and MATH, say MATH. Thus a loop MATH intersects MATH in two points. Since MATH is prime, MATH intersects MATH in an embedded arc. Then, there are two posibilities for MATH, MATH or MATH. In the formar case, MATH bounds a pair of a REF-ball and an unknotted arc, and an isotopy of MATH along the pair eliminates MATH. In the later case, MATH bounds a REF-ball , and an isotopy of MATH along REF-ball eliminates MATH. These contradict the minimality of MATH. |
math/0104219 | If all corners of MATH do not meet MATH, then this is same to REF . If exactly one corner of MATH meets MATH or MATH at one point, then MATH and some MATH have the intersection number MATH, or a vertex which meets MATH along the corner intersects some MATH in one point. Since MATH and MATH must have the intersection number zero, MATH is bounded by a loop of MATH consisting of a vertex and an edge MATH, and MATH intersects MATH in one point. Then REF gives the conclusion. If some corners of MATH meet both MATH and MATH, then the corners of MATH have the intersection number zero with MATH because MATH and MATH have the intersection number zero. In such a situation, we have a contradiction same as the proof of REF . |
math/0104224 | If MATH then MATH is homeomorphic to the connected sum of MATH copies of MATH, and therefore the statement is straightforward. If MATH, the presentation becomes MATH . REF shows an NAME which induces (MATH), and so, by CITE, this presentation is geometric. If MATH, the presentation becomes MATH . Therefore, if we replace MATH with MATH, REF also gives an NAME inducing (MATH). |
math/0104224 | MATH is obtained by NAME surgery on MATH, with coefficients MATH and MATH, along the trivial link with two components MATH. |
math/0104224 | Both the link MATH and the surgery coefficients defining MATH are invariant with respect to the rotation MATH of MATH, which sends the MATH-th component of MATH onto the MATH-th component (mod MATH). Let MATH be the cyclic group of order MATH generated by MATH. Observe that the fixed-point set of the action of MATH on MATH is a trivial knot disjoint from MATH. Therefore, we have an action of MATH on MATH, with a knot MATH as fixed-point set. The quotient MATH is precisely the manifold MATH, which is homeomorphic to MATH by REF , and MATH is obviously a knot MATH, which only depends on MATH and MATH. Moreover, MATH is the Borromean rings, as showed in REF . This proves the statement. |
math/0104224 | The link MATH admits an invertible involution MATH, whose axis intersects each component in two points (see the dashed line of REF ), and the rotation symmetry MATH of order MATH which was discussed in REF . These symmetries induce symmetries (also denoted by MATH and MATH) on the periodic NAME manifold MATH, such that MATH. We have MATH (see CITE and CITE) and MATH (see REF ). It is immediate to see that MATH induces a symmetry (also denoted by MATH) on the orbifold MATH, and MATH is the orbifold MATH. As we see from REF , MATH induces a strongly invertible involution (also denoted by MATH) on the link MATH. Using the Montesinos algorithm we see that MATH. This concludes the proof. |
math/0104224 | From REF , MATH is the MATH-fold cyclic covering of MATH, branched over a knot MATH which does not depend on MATH. By isotopy and NAME moves it is easy to obtain (see REF ) a diagram of MATH, which is a NAME 's normal form of type MATH. This proves the statement. |
math/0104226 | Defining as usual the annihilator of MATH by MATH one has that denseness of MATH is equivalent to MATH . Since MATH the proof is concluded if the range of MATH is closed. This follows from the closed range theorem since the range of MATH is closed by the surjectivity hypothesis. |
math/0104226 | Here we just give the main steps of the proof refering to REF, for the details. One starts writing the presumed resolvent of an extension MATH of MATH as MATH where MATH has to be determined. NAME requires MATH or, equivalently, MATH . Therefore posing MATH, where MATH, REF is equivalent to MATH . The resolvent identity MATH is then equivalent to MATH . Suppose now that there exist a (necessarily closed) operator MATH and an open set MATH, invariant with respect to complex conjugation, such that MATH . Then REF forces MATH to satisfy the relation MATH whereas REF , at least in the case MATH is densely defined, and has a bounded inverse given by MATH as we are pretending, is equivalent to MATH . By REF , for any self-adjoint MATH, the linear operator MATH satisfies REF and, by REF , has a bounded inverse for any MATH (at this point REF is used). Therefore (see the proof of REF) MATH is the resolvent of a self-adjoint operator MATH (here REF is needed). For any MATH one has MATH the definition of MATH being MATH-independent thanks to resolvent identity REF . Being MATH injective, REF imply MATH and so the definition MATH is MATH-independent. Therefore any MATH can be equivalently re-written as MATH where MATH and MATH . This implies, for any MATH, MATH . Conversely any MATH, MATH, MATH, admits the decomposition MATH, where MATH . Note that MATH by REF and MATH. |
math/0104226 | Being MATH injective, by continuity and density the thesis follows from the identity MATH . |
math/0104226 | By the definition of MATH, MATH and MATH one has, for any MATH, MATH . The proof is then concluded by REF . |
math/0104226 | By the definition of MATH one has MATH and so, since MATH and MATH in conclusion there follows MATH if and only if Range-MATH is closed. By the closed range theorem Range-MATH is closed if and only if the Range-MATH is closed, and this is equivalent to the range of MATH being closed. Being MATH surjective, MATH is injective with a closed range and so MATH is a bijection. By NAME 's theory we know that any MATH can be univocally decomposed as MATH that is, MATH . The above decomposition can be then rearranged as MATH . By REF one has MATH . Since the scalar product of MATH can be equivalently written as MATH one has MATH . This implies, since Range-MATH is closed, MATH . Thus, being MATH, the vector MATH is a generic element of MATH and we have shown that MATH. It is then straighforward to check that MATH. By REF one has MATH . This implies MATH thus MATH is isometric if and only if MATH where MATH. By using the identities MATH and MATH one has MATH and so MATH is an isometry. By again using identity REF one can check that MATH has an inverse defined by MATH . Thus MATH is unitary. Let us now take MATH. Then MATH and so MATH and MATH . |
math/0104226 | By REF one has MATH . Thus, by inverting the relation MATH given in the previous theorem, one obtains MATH . Since MATH and MATH if and only if MATH (see for example, REF ), the range of MATH is dense and thus MATH is densely defined as MATH is a continuos bijection. By REF one has MATH and so, since MATH and MATH, MATH is self-adjoint if and only if MATH . Such an equality is then an immediate conseguence of the unitarity of both MATH and MATH. |
math/0104227 | Since MATH is compact, at a minimum of the solution MATH we have MATH with MATH positive semidefinite. From REF , we then have, at the minimum point, MATH is in MATH. Therefore since the cones are connected, by continuity we have MATH is positive MATH-admissible. A similar argument holds in the negative MATH-admissible case. If we make the conformal change of metric MATH, then for any function h, MATH where MATH is the Hessian of MATH with respect to the metric MATH, and MATH is taken with respect to MATH. We have for the NAME symbols (see CITE) MATH . Therefore MATH . We let MATH . From REF , we see that the linearization at the solution MATH in the direction MATH is given by MATH . Since MATH is in MATH, from REF , we are done. |
math/0104227 | By positive MATH-admissible, we mean that the matrix MATH is in MATH. The multiple of MATH is irrelevant, since MATH. Letting MATH, we must show that MATH, that is, MATH is elliptic at MATH for MATH. We have MATH . From REF , the first two terms together are in MATH. The last term is a non-negative multiple of the identity, so again using REF , we are done. |
math/0104227 | Assume we have a solution MATH of REF , with MATH. We let MATH . Then letting MATH, the function MATH satisfies MATH where MATH is a linear elliptic operator (this follows from REF , see REF), so by the maximum principle, we have MATH, that is, MATH. The proof of the strict upper inequality is similar. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.