paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005001 | We shall just prove the inhomogeneous version REF; the homogeneous version easily follows from REF and a scaling and limiting argument based on REF . By REF and symmetry we may restrict to the region MATH. Since MATH, we thus have MATH. We now dyadically decompose the left-hand side of REF as MATH . The multiplier insi... |
math/0005001 | From REF we can estimate the left-hand side by MATH . For fixed MATH, the set of possible MATH ranges in an interval of length MATH, and vanishes unless MATH. The claim follows. |
math/0005001 | For brevity we shall denote the right-hand side of REF as MATH. The lower bound for MATH follows immediately from the Comparison principle, so it suffices to show the upper bound. The idea shall be to decompose MATH, MATH into dyadic shells, and then use REF to move MATH to be MATH. This may move MATH into another dyad... |
math/0005001 | In REF we have MATH by REF, from which the claimed bounds follow. It thus remains to consider REF . By symmetry we may assume that MATH. By REF , we have MATH for some MATH, MATH, MATH, MATH satisfying REF and MATH. To compute the right-hand side of this expression we shall use the identity MATH . We need only consider... |
math/0005001 | By NAME it suffices to show that MATH . From REF or REF it suffices to show that MATH and MATH for all MATH. This will be accomplished by REF and some tedious summation. Fix MATH. We first prove REF. We may assume REF. By REF we reduce to MATH . Estimating MATH and then performing the MATH summations, we reduce to MATH... |
math/0005001 | By duality and NAME it suffices to show that MATH . We estimate MATH by MATH. We then apply the inequality MATH and symmetry to reduce to MATH . We may minorize MATH by MATH. But then the estimate follows from REF and the MATH identity REF . |
math/0005001 | As this estimate is linear rather than multilinear we shall be able to apply some additional techniques such as NAME theory (and such mundane tools as the triangle inequality) in order to simplify the argument substantially (basically by preventing cross-terms when we finally pass to the bilinear setting). We first obs... |
math/0005001 | Applying duality and NAME as before, the estimate REF is equivalent to MATH . We may re-arrange the numerator, and write this more symmetrically as MATH . From the resonance identity REF we have MATH and hence MATH . Inserting this estimate into the above and using symmetry, we reduce to showing that MATH . We may of c... |
math/0005001 | By REF we may estimate the left-hand side of REF by MATH for some MATH, MATH. REF is now clear. To obtain REF, we observe from REF that MATH has a derivative of MATH in the MATH direction of MATH. In particular, MATH is monotone in the MATH direction, and for fixed values of MATH the MATH co-ordinate of MATH is constra... |
math/0005001 | To prove the claim it suffices by REF to show that MATH for all MATH-normalized MATH, where MATH is the coarse scale multiplier MATH . Fix the MATH. Decompose MATH, where MATH is the restriction of MATH to the box MATH. By REF, we may estimate the left-hand side of REF by MATH . We may estimate this by MATH where the f... |
math/0005001 | We apply the previous Lemma with MATH being the unit cube in MATH, and MATH being the lattice MATH. We observe that for any MATH we have MATH thanks to the Comparison principle, REF , and REF . The claim then follows from REF . |
math/0005001 | We treat the three cases of the Proposition separately. REF (MATH high-high interactions) MATH. The constraint MATH is clear, since MATH. To verify the second constraint, suppose MATH. Then MATH. Since MATH and MATH are positive, we must have MATH, hence MATH. This shows the second constraint. Now we prove REF. If REF ... |
math/0005001 | The symbols for MATH and MATH can both be majorized by MATH . Thus it suffices to show that MATH . By REF (splitting into three cases depending on which of the MATH is smallest) and the hypothesis MATH, we may replace the term MATH by MATH. By two applications of REF (and splitting into regions MATH and MATH) one can r... |
math/0005001 | We recommend reading this proof initially assuming the simplifying assumption MATH, as this case already contains the main idea of the argument. The reader may also wish to rescale MATH. First suppose that MATH. Then we estimate the left-hand side of REF by MATH . The claim then follows by REF with MATH. Note that this... |
math/0005001 | This will be a reprise of the proof of REF . (Alternatively, one could proceed using REF ). REF follows from REF, so we may assume that REF holds. We may also assume that MATH, hence MATH. By REF we have MATH for some MATH, MATH, MATH, MATH, MATH satisfying REF and MATH. From REF and the identity MATH it suffices to sh... |
math/0005001 | In addition to REF, we will need the algebraic identity MATH for all MATH. We first consider the case MATH. In this case we have MATH, so we may assume that MATH. REF now follows from REF, so we may assume REF. If MATH then we are essentially in the MATH case, since there is essentially no distinction between the const... |
math/0005001 | By duality and permutation of indices the estimate is equivalent to MATH . Thus we are in the MATH REF . By REF it suffices to show that MATH and MATH for all MATH. Since MATH, we can estimate MATH . Also, from the various cases of REF we have MATH whenever MATH. Combining these we obtain MATH . REF are now easily veri... |
math/0005005 | According to CITE or CITE there is an isomorphism of algebraic groups MATH induced by MATH. Letting MATH one has MATH . Applying MATH to this isomorphism we see that MATH in MATH, so MATH in MATH. We conclude that MATH. |
math/0005005 | REF with MATH implies that the subgroup MATH is normal, and for any two-sided tilting complex MATH the conjugation MATH is an automorphism of algebraic groups. Let us now switch to the notation MATH and MATH for the operations in MATH. Define an algebraic variety structure on each coset MATH using the multiplication ma... |
math/0005005 | It is known that there exist two-sided tilting complexes MATH; choose one. We obtain a group isomorphism MATH . By REF , MATH restricts to an isomorphism of algebraic groups MATH. So MATH is an isomorphism of locally algebraic groups. |
math/0005005 | Trivially MATH, and by CITE we have MATH. |
math/0005011 | CASE: We assume first that the system MATH is definite. Then the quotient map MATH is bijective. Indeed, the injectivity follows immediately from the definition of definiteness. To prove surjectivity, consider MATH. This means MATH and in view of REF there exists MATH such that MATH. Thus MATH. This proves surjectivity... |
math/0005011 | Let MATH with MATH in a neighborhood of MATH and MATH. Then MATH does the job. |
math/0005011 | By REF there are absolute continuous functions MATH with MATH, MATH, and MATH . Put MATH. We have MATH . Then MATH . Note that in view of REF the matrices MATH and MATH are invertible if MATH. Combining REF and the well - known estimate MATH we obtain MATH or MATH . Letting MATH we reach the conclusion. |
math/0005011 | By REF there are absolute continuous functions MATH, MATH, MATH, and MATH . Note that, again, MATH implies that MATH and MATH are invertible. In view of the regularity REF it suffices to show for MATH that MATH . By dominated convergence we have MATH . Integration by parts shows that MATH . Using REF this can be estima... |
math/0005011 | The essential self - adjointness of MATH follows, in view of REF, from REF with MATH and MATH. It is easy to see that, as in the case of a symmetric operator, the essential self - adjointness of the square of a s.l.r. in a NAME space implies the essential self - adjointness of the s.l.r. itself. However, the essential ... |
math/0005011 | This follows immediately from REF , and REF. |
math/0005011 | Consider MATH satisfying MATH . We have to show that MATH. REF translates into MATH . Note that REF implies that MATH and MATH are invertible on a set of positive NAME measure. Consequently, the systems MATH are definite. From REF we infer MATH. Hence MATH a.e. Since MATH is definite we infer from REF that MATH. In vie... |
math/0005011 | MATH follows easily by means of a NAME 's mollifier constructed in a neighborhood of the compact support of MATH. Next choose a cut - off function MATH with MATH in a neighborhood of MATH. Then, MATH and hence MATH since MATH. |
math/0005011 | Denote by MATH the distance function with respect to the metric MATH. Then fix MATH and put MATH. As in CITE one concludes MATH and MATH. Now choose a cut - off function MATH with MATH near MATH, and MATH. Then put MATH . MATH obviously has the desired properties. |
math/0005011 | Let MATH be a locally NAME function with compact support and put MATH. Using REF we find MATH . Using MATH the latter can be estimated MATH and thus MATH . We apply REF with MATH and obtain a sequence MATH of NAME functions MATH which satisfy REF with MATH. Putting MATH we have MATH and MATH . Since MATH as MATH we rea... |
math/0005011 | CASE: We note first that if MATH is elliptic (everywhere) then this is an easy consequence of elliptic regularity. Namely, if MATH then MATH and hence by elliptic regularity this implies MATH. That is, MATH is locally of NAME class MATH and hence in particular MATH . CASE: If MATH is not elliptic everywhere then we hav... |
math/0005011 | Since MATH REF implies MATH (compare REF), hence the symmetric operator REF satisfies NAME 's condition CITE. Thus all powers of MATH are essentially self - adjoint. Now, if MATH then MATH and hence MATH satisfies MATH. Consequently, Conjecture REF implies MATH and thus MATH. |
math/0005013 | We only prove REF . The proof is by induction on MATH . The statement is trivial if MATH or a limit ordinal. Suppose the statement is true for all ordinals not greater than MATH . Let MATH . If MATH is a neighborhood of MATH in MATH, then MATH is open in MATH . Thus there exist MATH such that MATH . Hence MATH . |
math/0005013 | We first consider the case MATH . Then MATH . For each MATH there exist an open neighborhood MATH of MATH and MATH such that whenever MATH for all MATH . By the compactness of MATH there exist MATH such that MATH . Let MATH . Then for all MATH and MATH we have MATH for some MATH . Since MATH . Taking limit as MATH we h... |
math/0005013 | For any MATH choose an open neighborhood MATH of MATH in MATH such that MATH and MATH . The collection MATH is an open cover of MATH. By REF , there exists a partition of unity MATH subordinated to MATH . If MATH for some MATH, let MATH if MATH, let MATH. Define MATH by MATH . The sum is well-defined since MATH is loca... |
math/0005013 | Let MATH be the function obtained from REF . If MATH, let MATH. Applying REF with MATH, MATH, and the function MATH yields a continuous function MATH such that MATH . Let MATH. Then MATH . Suppose that MATH and MATH is a compact subset of MATH . If MATH then there exists an open neighborhood MATH of MATH such that MATH... |
math/0005013 | The functions MATH are constructed inductively-MATH . By REF , there exists a continuous function MATH such that MATH. Extend MATH to a function on MATH by defining MATH to be MATH on MATH. Then REF hold. REF holds vacuously. Moreover, if MATH then there exists a neighborhood MATH of MATH in MATH such that MATH for all... |
math/0005013 | Let MATH be the sequence given by REF . Define MATH on MATH by MATH . Note that by REF , MATH converges uniformly on MATH . Hence MATH is well defined. Obviously, MATH . Claim. MATH for all MATH . Proof of Claim. Let MATH . We consider two cases. Suppose MATH . By REF , MATH is continuous on MATH for all MATH . Since M... |
math/0005013 | For MATH let MATH . If MATH let MATH be the MATH-neighborhood of MATH in MATH . For MATH it follows from REF that there exists MATH such that MATH and MATH for all compact subsets MATH of MATH . List the ordinals in MATH in a (possibly finite) sequence MATH . Here MATH or MATH . For each MATH let MATH . Then MATH is a ... |
math/0005013 | First we assume MATH is of the form MATH where MATH . By REF there exist a sequence MATH and a function MATH such that-MATH for all MATH converges pointwise to MATH and MATH . Then MATH by REF MATH . This implies that MATH by REF MATH . Hence there exist MATH and MATH such that MATH for all MATH, MATH converges pointwi... |
math/0005013 | The case MATH is trivial. Suppose the corollary holds for some MATH . If MATH it follows from REF that MATH is the pointwise limit of a bounded sequence MATH in MATH such that MATH . Since MATH by the inductive hypothesis, MATH . Conversely, if MATH then MATH is the pointwise limit of a sequence MATH in MATH with MATH ... |
math/0005013 | Write MATH as a sequence MATH . Without loss of generality, assume that MATH for all MATH . Since MATH is NAMEREF, there exists MATH such that MATH converges pointwise to MATH . Assume that MATH has been chosen so that MATH pointwise. If MATH, let MATH be the MATH-neighborhood of MATH in MATH and let MATH . For all MAT... |
math/0005013 | By REF , there exists MATH such that MATH . Then MATH is bounded in MATH pointwise and MATH. By the first statement in the proof of REF, there exists MATH such that MATH . Let MATH for all MATH. Then MATH and MATH for all MATH. It follows that MATH . |
math/0005013 | Let MATH . For each MATH and all MATH, let MATH be the MATH-neighborhood of MATH in MATH . Define MATH . Then MATH is a countable collection of compact subsets of MATH such that MATH . If MATH and MATH, by REF , there is a continuous function MATH on MATH such that MATH . Hence MATH for all MATH . By REF , there exists... |
math/0005013 | It is easy to see that for all MATH and MATH . If MATH,MATH then by REF there exist two sequences of continuous functions MATH and MATH converging pointwise to MATH and MATH respectively such that MATH and MATH . According to REF , MATH . Hence by REF , MATH . Finally, given MATH and MATH, choose MATH such that MATH . ... |
math/0005013 | By REF , there exists a sequence MATH converging to MATH pointwise such that MATH . For each MATH and MATH . By CITE (see the remark on CITE), MATH . REF implies that MATH . Since MATH converges to MATH pointwise, it follows from REF that MATH that is, MATH . |
math/0005013 | The proof is by induction on MATH . The result is clear if MATH or a limit ordinal. Assume that the lemma holds for some MATH . Suppose MATH and MATH are given. Let MATH . If MATH then MATH and we are done. Otherwise, assume that MATH . Then there exist a neighborhood MATH of MATH in MATH and MATH such that MATH for al... |
math/0005013 | From the above, we obtain a function MATH in MATH such that MATH and MATH . Since MATH it follows from REF that MATH . As MATH is bounded, we see that MATH . By REF , MATH . Also, MATH . Applying REF again gives MATH . |
math/0005013 | We may of course assume that neither MATH nor MATH is MATH and that MATH . The assumption on MATH yields a MATH-valued function MATH in MATH such that MATH . Denote MATH by MATH . Choose a sequence of ordinals MATH with MATH that strictly increases to MATH . Let MATH be any ordinal that is less than MATH . Fix a functi... |
math/0005015 | These maps are equivariant under the action of MATH. Hence, it remains to show injectivity and surjectivity. For this we may assume MATH. Let MATH be a line bundle on MATH. As MATH is smooth and MATH, there is a divisor MATH in MATH not meeting MATH such that MATH. Such a divisor MATH is a sum MATH. Moreover, such a MA... |
math/0005015 | Since MATH has no nontrivial characters, it follows from REF and from the proof of REF, that any line bundle on MATH admits a unique MATH-linearization. Assume that MATH and consider the induced action of MATH on the projectivization MATH. NAME 's fixed point theorem REF implies that there exists a nonzero section MATH... |
math/0005015 | By purity, it is sufficient to prove this when MATH is smooth in a neighborhood of MATH (since MATH is smooth, a function which is regular in the complement to a codimension REF subscheme is regular everywhere). Then, we can choose (étale) local coordinates MATH in MATH (that is, elements of MATH, MATH is free over MAT... |
math/0005015 | We only prove that MATH, because the sharper bound which is valid in the case of MATH won't be used below. We refer to REF for its proof. Let MATH be a basis of MATH. Then, MATH is a global section of the line bundle MATH. Moreover, MATH does not vanish on MATH. Therefore, we can write MATH for nonnegative integers MAT... |
math/0005015 | First we assume that MATH is effective. By REF , we have a nonzero invariant global section MATH. If MATH and MATH denote the action and the second projection MATH, respectively, endow the trivial line bundle MATH on MATH with the tensor-product metric. This is a locally constant metric on the trivial line bundle. The ... |
math/0005015 | It suffices to note that if MATH lies over the open subset MATH given by the definition of an adelic metric then the stabilizer of MATH equals MATH. |
math/0005015 | Let MATH be an ample line bundle on MATH and let MATH be a sufficiently large integer such that MATH is effective. It follows from the preceding section that MATH has a section MATH which does not vanish on MATH. This implies that the function MATH is bounded from above on MATH. Therefore, there exists a constant MATH ... |
math/0005015 | Fix a basis MATH of MATH consisting of (classes of) line bundles lying in the interior of the effective cone of MATH. Let MATH denote the open set of all linear combinations MATH such that for some MATH, MATH. Fix some ample line bundle MATH. It is well known that the height zeta function of MATH relative to MATH conve... |
math/0005015 | Since the natural action of MATH on MATH is trivial, for any MATH the function MATH is a height function for MATH, induced by a ``twisted" adelic metric. When MATH belongs to some compact subset MATH of MATH there exists a constant MATH such that for any MATH and any MATH, MATH . Indeed, the quotient MATH defines a bou... |
math/0005015 | NAME resolution of singularities (in char. MATH, see CITE or CITE) shows that there exists a composition of equivariant blow-up with smooth centers MATH such that CASE: MATH is a smooth projective equivariant compactification of MATH; CASE: MATH is equivariant (hence induces an isomorphism on the MATH); CASE: the bound... |
math/0005015 | Recall first NAME 's definition CITE of the NAME measure in this context, which extends NAME 's definition in the context of linear algebraic groups CITE. For any place MATH of MATH and any local non-vanishing differential form MATH on MATH, the self-dual measure on MATH induces a local measure MATH on MATH. These loca... |
math/0005015 | It is sufficient to prove the proposition when MATH as the integrals for different values of MATH are comparable. The NAME closure MATH of MATH in MATH need not be smooth. Nevertheless, we may introduce an equivariant proper modification MATH such that the NAME closure MATH of MATH is a smooth equivariant compactificat... |
math/0005015 | We prove this by induction on the codimension of MATH in MATH. If MATH, the adjunction formula shows that MATH . For MATH the nonnegative integer MATH is equal to MATH if and only if MATH. The lemma is proved. |
math/0005015 | Let MATH be the (nonrenormalized) NAME measure on MATH: by definition, on MATH, MATH. We have to estimate MATH . The integral in the limit is of the type studied in REF , but on the subgroup MATH. Denote therefore by primes MATH restrictions of objects from MATH to MATH. Thanks to the previous lemma, one has MATH, MATH... |
math/0005015 | Without loss of generality, we can assume that MATH. Now, using an analytic partition of unity on MATH and the assumption that the boundary MATH is a divisor with strict normal crossings, we see that it suffices to compute the integral over a relatively compact neighborhood of the origin in MATH (denoted by MATH) on wh... |
math/0005015 | We may assume MATH. Choose MATH such that the annulus MATH has positive measure in MATH and let MATH. Then, we have MATH . It follows that the integral over MATH converges if and only if the geometric series MATH converges, that is if MATH. |
math/0005015 | Let us assume for the moment that MATH. We shall write MATH, MATH and MATH. Then, in the domain MATH one has MATH . Now integrate by parts: for any MATH we have MATH and by induction MATH . For any MATH, let us define MATH; it is a MATH function on MATH. Let MATH be a point of MATH and let MATH be the set of MATH such ... |
math/0005015 | We split the integral along residue classes mod MATH. Let MATH and MATH so that MATH. We can introduce local (étale) coordinates MATH REF and MATH (MATH, MATH) around MATH such that locally, the divisor MATH is defined by the vanishing of MATH. Then, the local NAME measure identifies with the measure MATH on MATH. If M... |
math/0005015 | We use the fact that for any projective variety MATH of dimension MATH and degree MATH the number of MATH-points is estimated as MATH (see, for example, REF). Since MATH is projective, all MATH can be realized as subvarieties in some projective space. This proves the second part. To prove the first part, we apply NAME ... |
math/0005015 | Using the uniform estimates from REF we see that in the formula for MATH, each term with MATH is MATH, with uniform constants in MATH. Turning to the remaining terms, we get MATH . Finally, we have the desired estimate. |
math/0005015 | For any place MATH of MATH, let MATH . The previous proposition shows that in MATH, the NAME product MATH converges absoluetely to a holomorphic and bounded function MATH. For any MATH, one has MATH . It suffices to define MATH . The polynomial growth of MATH in vertical strips follows from the boundedness of MATH in M... |
math/0005015 | For MATH the local integrals converge absolutely in the domain under consideration and are bounded as in REF . For MATH, we have shown that there exists a constant MATH such that for all MATH and all MATH one has the estimate MATH . This implies that MATH is bounded independently of MATH, MATH and MATH. For any nonzero... |
math/0005015 | If MATH, the computation runs as follows MATH . For MATH, let MATH. Since we assumed MATH to be unramified over MATH, MATH. We will integrate over disks MATH for MATH suitably chosen. Indeed, if MATH and MATH, MATH hence, if MATH is chosen such that MATH . We can find such a MATH if and only if MATH, that is, MATH. If ... |
math/0005015 | Chose some element MATH in each orbit MATH and fix a local equation MATH for the corresponding MATH which is defined over the number field MATH. Then if MATH (for some MATH) is another element in the orbit of MATH, set MATH. This is well defined since we assumed that the equation MATH is invariant under MATH. Finally, ... |
math/0005018 | For convenience, we shall often write MATH. Let MATH . For the proof of the regularity of the functions MATH we shall make use of the following lemmas. The proof of the first lemma is trivial, using MATH. Let MATH and MATH. Assume that the real function MATH satisfies: For all MATH there exists constants MATH such that... |
math/0005023 | CASE: Suppose MATH is a normal connected meromorphic subgroup of MATH which is meromorphically isomorphic to a linear algebraic group. Then MATH meromorphically embeds in MATH. So MATH is both a complex torus and a linear algebraic group forcing it to be trivial. That is MATH is contained in MATH. CASE: As in REF. |
math/0005023 | CASE: MATH is commutative, so by REF , NAME meromorphic, thus by REF , meromorphically an extension of a complex torus MATH by a linear algebraic group MATH. As MATH is strongly minimal, MATH is either MATH or MATH. If MATH, then MATH. If MATH, then MATH has no proper infinite analytic subsets so is either an elliptic ... |
math/0005023 | CASE: Let MATH (=MATH). Suppose for the sake of contradiction that there is MATH such that MATH. Let MATH be the open unit disc in MATH and MATH be a coordinate function for any open neighborhood MATH of MATH in MATH where MATH is chosen such that MATH has dimension MATH. So if MATH, then MATH is an analytic subset of ... |
math/0005023 | CASE: Finding a meromorphic compactification. By assumption on MATH some open nonempty definable subset MATH of MATH is already a NAME open subset of a compact complex space MATH, which we may assume by resolution of singularities to be a manifold. Moreover by strong minimality of MATH, MATH is finite, say MATH. For MA... |
math/0005023 | Note that MATH being simple is almost strongly minimal. So if the lemma, as MATH has dimension MATH, MATH is semipluriminimal. By CITE, MATH is an almost direct product of pairwise orthogonal semiminimal groups. If MATH is already semiminimal, then as MATH is nonorthogonal to MATH via MATH, MATH must be algebraic, cont... |
math/0005023 | We prove the lemma by induction on MATH. It is clearly true for MATH. We may assume that the NAME degree of MATH is MATH. Let MATH be the (model-theoretic) stabilizer of MATH. If MATH is finite then by REF , as well as REF, MATH is, up to a set of NAME rank MATH, contained in a single translate of MATH. By induction hy... |
math/0005023 | As in the strongly minimal case we first find a compact complex manifold MATH containing MATH as a NAME open set, and then show that this gives MATH a NAME structure. CASE: Finding the compactification. Let MATH. By definition of MATH being a meromorphic group, let MATH be a definable subset (with NAME rank MATH) of MA... |
math/0005023 | Simplicity of MATH together with the the dichotomy theorem from CITE, implies that MATH is either nonorthogonal to MATH (namely has nontrivial algebraic reduction) or MATH is modular. In the first case, MATH is an algebraic group, so REF or REF hold. In the second case, every definable subset of MATH is a Boolean combi... |
math/0005023 | Note that if MATH satisfies the hypotheses of the lemma and MATH is a connected definable subgroup of MATH, or an image of MATH under a meromorphic homomorphism then MATH satisfies the hypotheses too (for suitable MATH). We will prove the lemma by induction on MATH. We consider various possibilities for MATH, Case I. M... |
math/0005023 | Suppose that MATH is neither trivial, nor an algebraic curve. Then MATH and by CITE and CITE, there is a strongly minimal group MATH definable in MATH. By REF MATH must be a complex torus, nonorthogonal to MATH. Nonorthogonality is witnessed by an analytic subset MATH of MATH which projects generically finite-to-one on... |
math/0005023 | REF : suppose MATH is nonorthogonal to MATH. Then clearly MATH. Suppose MATH is nontrivial and modular. Then MATH is nonorthogonal to the generic type of a strongly minimal modular torus MATH. Let MATH be a generic point of MATH (that is, a realization of MATH). Then there is generic MATH in MATH. Let MATH be the finit... |
math/0005025 | It follows from the definitions that MATH is MATH-stable. That it is a subspace of the same dimension as MATH also follows from the properness of the NAME. Moreover, it follows easily that the limit in the NAME is found by closing the bundle MATH over MATH. Given all this, the first statement follows from the second, o... |
math/0005025 | Since MATH is a MATH-module, MATH is a MATH-module contained in MATH which is isomorphic to MATH as a MATH-module. But this condition uniquely determines MATH. |
math/0005025 | Assume that MATH. We have to show that MATH is empty. First of all, since MATH is attractive, the image of MATH is contained in every connected open MATH-stable affine neighborhood of MATH, hence in MATH. Viewing MATH as a map to MATH, the fibre of MATH over MATH is finite, hence MATH is finite. Thus, MATH is a closed ... |
math/0005025 | Since MATH is attractive we may assume that MATH and MATH. Fix an equivariant projection MATH, and denote its restriction to MATH by MATH. Since MATH for all curves MATH, it follows that there is no MATH-curve in MATH, so by REF , MATH. This implies MATH is finite, since MATH is attractive. Let MATH be the ramification... |
math/0005025 | Let MATH be a MATH-stable line. Then by REF there is a MATH-line MATH, where MATH is the isotropy group of an arbitrary MATH, such that MATH. As MATH is nonsingular at MATH, there exists a MATH-stable curve MATH satisfying MATH. Setting MATH we obtain a MATH-stable surface, which contains MATH, and which satisfies MATH... |
math/0005025 | The first claim follows easily from the fact that MATH has a dense two dimensional MATH-orbit (compare CITE). Let MATH denote the two elements of MATH, and let MATH denote their weights. For any function MATH of weight MATH corresponding to a MATH-line MATH in MATH, there is a positive integer MATH such that MATH. Note... |
math/0005025 | Since MATH is generated by MATH-invariant surfaces and since MATH for all surfaces MATH which contain MATH, it is enough to show the proposition when MATH is a surface. If MATH is nonsingular at MATH, then MATH. Otherwise we know from REF that MATH is a cone over MATH, and for a cone, MATH. Since MATH is nonsingular at... |
math/0005025 | By a result of CITE, if MATH is rationally smooth at MATH, then MATH. By REF , we have MATH for every good MATH. Hence the result follows from REF . We will prove the last statement below. |
math/0005025 | Let MATH be the functions corresponding to the MATH-curves MATH in MATH. Since MATH, and since the MATH-curves are smooth and have non collinear weights, MATH equals the dimension of MATH. We may assume that MATH. Let MATH be the ideal generated by MATH. Then MATH is contained in the ideal of MATH. Now MATH equals the ... |
math/0005025 | The only facts we have to note are, firstly, that if MATH is short, then MATH and, secondly, if MATH is rationally smooth at MATH, then MATH (see CITE). |
math/0005025 | The existence of a long root MATH satisfying all the conditions in REF except possibly positivity follows from the Fundamental REF and the fact that MATH. To see MATH is positive suppose otherwise. It's then clear that MATH. If MATH also, then MATH since MATH, contradicting the assumption. Hence MATH. But as MATH is a ... |
math/0005025 | We will prove the following equivalent 'dual' statement: if MATH is the weight of a function corresponding to a MATH-stable line MATH, then there are MATH and MATH with MATH where MATH and MATH are the weights of functions corresponding to MATH-curves MATH and MATH, respectively. Let MATH be the MATH-eigenfunction corr... |
math/0005025 | The first assertion is a consequence of REF and the fact that MATH. For the second, use the fact that if a variety MATH is nonsingular at MATH, then MATH. Thus MATH . The final assertion follows from the fact that if MATH and MATH are both nonsingular at MATH, then MATH. |
math/0005025 | The first REF is standard. Moreover, MATH is surjective for all MATH and MATH maps the schematic tangent cone of MATH at MATH onto that of MATH at MATH. Consequently, it is also a surjection of the the associated reduced varieties. Since MATH is linear, REF is established. REF is an immediate consequence of the fact th... |
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