paper
stringlengths 9
16
| proof
stringlengths 0
131k
|
|---|---|
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 inside the summation has essentially disjoint MATH and MATH supports as MATH varies dyadically. By REF , we can thus estimate the above by MATH . By the triangle inequality and Comparison principle we may estimate this by MATH . By the NAME lemma we have MATH . The claim then follows from REF and a simple computation.
|
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 dyadic shell, depending on the relative sizes of MATH and MATH. By REF we may assume that MATH. We then split REF into three pieces determined by the regions MATH . In the first case we dyadically decompose MATH and use the triangle inequality to estimate the left-hand side of REF by MATH . We subdivide MATH into MATH regions of the form MATH for integers MATH. By REF we can thus estimate the previous by MATH . By REF and the Comparison principle the expression inside the sup is MATH. Since MATH, the claim follows for this case. In the second case we repeat the above arguments, eventually estimating this contribution by MATH . If we use REF to shift MATH down by MATH and MATH up by MATH, and use the crude estimate MATH we can estimate the expression inside the sup by MATH. Since MATH, the claim then follows for this case. It remains to consider the third case. We can dyadically decompose into pieces MATH for MATH. By NAME 's test REF it suffices to control the contribution of a single MATH, which we now fix. The quantity MATH fluctuates between MATH and MATH. We then dyadically decompose in this quantity and use the triangle inequality to estimate the contribution of this case by MATH . We now decompose the condition MATH into MATH for MATH as before. Using NAME 's test we can estimate the previous by MATH . By REF the expression inside the supremum is MATH. Since MATH and MATH, the claim then follows.
|
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 three cases: MATH, MATH, and MATH. (The case MATH then follows by symmetry). If MATH, we see from REF that MATH variable is contained in the union of two intervals of length MATH at worst, and REF follows. If MATH, we must have MATH, so REF shows that MATH is contained in the union of two intervals of length MATH, and REF follows. If MATH, then we must have MATH, so REF shows that MATH is contained in the union of two intervals of length MATH. But MATH is also contained in an interval of length MATH. REF follows.
|
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 which is certainly true (with about MATH to spare). Now we show REF. We may assume MATH. We first deal with the contribution where REF holds. In this case we have MATH, so we reduce to MATH . But this is easily verified. Now we deal with the cases where REF applies. We do not have perfect symmetry and must consider the cases MATH separately. In the first case we reduce by REF to MATH . Performing the MATH summation we reduce to MATH which is easily verified. To unify the second and third cases we replace MATH by MATH. By asymmetry it suffices now to show the second case. We simplify using the first half of REF to MATH . We may assume MATH since the inner sum vanishes otherwise. Performing the MATH summation we reduce to MATH which is easily verified (with about MATH to spare). To finish the proof of REF it remains to deal with the cases where REF holds. This reduces to MATH . Performing the MATH summations, we reduce to MATH which is easily verified (with about MATH to spare).
|
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 observe that if MATH is constant in MATH, then this estimate follows immediately from the one-dimensional NAME embedding MATH. Thus we may subtract off the mean and assume that MATH is identically zero. By dividing the spatial frequency into regions MATH and MATH and using symmetry of the cubic MATH we may assume further that MATH is supported on the half-plane MATH. For each dyadic MATH, let MATH be a smooth localization to the frequency range MATH. Since we have MATH and the NAME estimate MATH it suffices to show that MATH uniformly in MATH. (This trick works in general for all NAME estimates; see for example, CITE). Fix MATH. Squaring the above estimate and using duality, we reduce to showing that MATH . By NAME, this will follow if we can show that MATH . Since MATH are positive and comparable to MATH, we see that MATH is negative and comparable to MATH. We may assume REF. By a dyadic decomposition it thus suffices to show that MATH . We may assume MATH as the other cases are similar or better. First consider the contribution when MATH. In this case we use REF to estimate the above by MATH and this easily sums to MATH as desired. By REF it remains only to consider the case when MATH. We then apply REF to reduce to MATH . But this is easily verified also.
|
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 course replace MATH by MATH. The above estimate is then equivalent to the trilinear estimate MATH . But this is an immediate consequence of REF and NAME.
|
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 constrained inside an interval of length MATH. REF (for MATH) then follows by NAME 's theorem; the claim for MATH follows by symmetry.
|
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 functions MATH on MATH are defined by MATH . The claim then follows since MATH .
|
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 holds then the claim follows from REF. In light of this, REF, and the preceding discussion, the only remaining possibility is that MATH. By symmetry we may assume that MATH. In this case MATH and MATH are near opposite sides of the upper light cone, while MATH is near the time axis. The intersection of one light cone with a translated inverse of the other will basically be a graph over an ellipsoid of bounded eccentricity. The variable MATH can be localized to a ball of radius MATH, so one can localize MATH and MATH to similar balls by REF . We thus have MATH where MATH are such that MATH. We may of course assume that MATH. By REF we obtain MATH . Since MATH and MATH, we that the above set is within MATH of an ellipsoid of bounded eccentricity and principal radii MATH. Since the set is also contained in a ball of radius MATH, it must have measure MATH, and the claim follows. REF (NAME interactions) MATH; MATH. We either have MATH or MATH. REF : (MATH case) MATH. By repeating REF argument we have MATH where MATH are such that MATH. We may of course assume that MATH. From REF we have MATH. This relation motivates the following partition. Let MATH be a large number, let MATH be a maximal MATH-separated subset of the sphere MATH, and for each MATH let MATH denote the cone MATH . We then partition the MATH and MATH variable into these cones: MATH . The contribution of a single term vanishes unless MATH. Thus each cone MATH interacts with at most MATH cones MATH, and conversely. By REF we thus have MATH for one such pair MATH of cones. Fix MATH, MATH. Let MATH be a unit vector which is coplanar with MATH, MATH, is perpendicular with MATH, and obeys MATH. (This latter condition is redundant unless MATH). Observe that the conditions of REF apply with MATH, and MATH. Since MATH, we thus have from REF that MATH . Also, from REF and the fact that MATH we have MATH . Combining the two estimates we obtain REF . REF : (MATH interactions) MATH. We modify REF argument as follows. We begin with MATH where the MATH are as before. Since MATH and MATH, the sine rule (see REF ) shows that MATH. We then introduce the cones MATH as before and partition MATH . The contribution of a single term vanishes unless MATH. By NAME 's test REF we reduce to MATH . One then applies REF as before; the function MATH is MATH instead of MATH, and we have MATH rather than MATH, but the two changes essentially cancel each other out, and we obtain the same bounds. We omit the details. REF (MATH high-high interactions) MATH; MATH. We begin with MATH where the MATH are as before. Since MATH and MATH, the sine rule shows that MATH. We can then introduce cones as before, except with angular width MATH rather than MATH. We then apply REF with MATH and MATH. REF follows.
|
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 replace MATH by MATH for MATH. We have thus reduced to MATH . We may assume REF as usual. We can break this up dyadically as MATH . We may assume that two of the MATH is positive and the third is negative. We may assume REF, so in particular MATH. Suppose for the moment that we are in the situation REF. Then from the estimate MATH and REF we have MATH . Also, from the permutation invariance of MATH we have MATH . By symmetry we thus have MATH for all choices of MATH. We thus reduce to MATH . By NAME 's test we may assume that MATH for some fixed MATH. Since MATH and MATH we have MATH . Also, from the various cases of REF and the hypothesis MATH we have MATH . From these estimates and the hypothesis MATH we easily see that REF holds uniformly in MATH.
|
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 argument also gives the MATH case with MATH. Henceforth we will assume MATH. We now give the more elementary MATH argument, in which MATH. We remark that this argument is essentially in CITE. Divide MATH into sectors MATH of angular width MATH for some large constant MATH, with vertex at the origin. We can estimate REF as MATH . The summands vanish unless MATH. Thus each MATH interacts with at most MATH sectors MATH, and conversely. By NAME 's test REF it thus suffices to show that MATH . But this follows from REF and elementary geometry. We now give the general MATH argument. We shall assume that REF has already been proven for some MATH, and show that this implies REF with MATH replaced by MATH. Since we have already proven REF for MATH, the claim then follows by iteration. Roughly speaking, the point is as follows. The condition MATH has too much curvature in the MATH case to be usefully decomposed into boxes or sectors. However, at scales MATH and below, the condition no longer depends on MATH of the dimensions, and we can use REF to reduce to the MATH case. One then uses REF and the iteration REF to handle the coarse scales, thus introducing the MATH loss. We turn to the details. Fix MATH, MATH, MATH, MATH. We may localize MATH to a ball of the form MATH for some MATH. By REF we can thus localize MATH to a similar ball MATH. By a mild rotation and scaling we may assume that MATH. We thus have to show that MATH where MATH . Let MATH denote the box centered at the origin with sides parallel to the axes, and all side-lengths equal to MATH except for the MATH side-length, which is MATH. Let MATH be the canonical tiling lattice of MATH, so that MATH is thus a box covering. We wish to apply REF . We begin by proving the fine-scale estimate MATH for all MATH. We shall achieve this by reducing to the MATH case already proven; alternatively one can modify the MATH argument to prove this estimate directly. Fix MATH, MATH, MATH. In order for the left-hand side of REF to be non-zero, there must exist MATH and MATH such that MATH. From this and elementary geometry we see that MATH . For any MATH, write MATH, where MATH is the orthogonal projection to MATH and MATH is the complement of this projection. We observe that when MATH for MATH, we have the estimates MATH which implies that MATH . From this and REF we see that the condition MATH implies that MATH. In other words, the components MATH and MATH do not significantly affect the angle. We can thus estimate the left-hand side of REF by MATH . Recall that the variables MATH, MATH are constrained to a ball of radius MATH. Applying REF we can therefore estimate the previous by MATH and REF then follows from the MATH case of REF with MATH, which we have proven previously. From REF we have MATH where MATH whenever REF holds, and MATH otherwise. Applying REF with MATH replaced by MATH (the geometric mean of MATH and MATH) we thus obtain MATH which simplifies to REF with MATH replaced by MATH, as desired.
|
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 show that MATH with the right-hand side replaced by MATH in the exceptional case MATH, MATH. As in REF , we need only consider three cases: MATH, MATH, and MATH. Suppose MATH. Define the radius MATH by MATH . If MATH, then the set in REF is contained in an annulus of radius MATH and thickness MATH. If MATH, then the above set is contained in a ball of radius MATH. In either case the claim follows (checking the MATH and MATH cases separately), noting that we must have MATH and MATH. Now suppose MATH. We can assume that MATH, since the set in REF vanishes otherwise. But then this set is contained in an annulus of thickness MATH while simultaneously being contained in a ball of radius MATH, and REF follows. Now suppose MATH. We can assume that MATH. But then this set is contained in an annulus of thickness MATH and simultaneously in a ball of radius MATH, and REF follows.
|
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 constraints MATH and MATH when MATH. Thus by symmetry of the first and second variable it suffices to consider the case MATH . By some permutation of REF we thus have MATH for some MATH obeying REF and MATH. Since MATH, the measure of this set is MATH. REF follows. (Alternatively, one can argue using REF ). Having disposed of the case MATH, it remains by symmetry to deal with the case MATH . First suppose that MATH, which forces MATH. Then we can repeat the MATH arguments for this case. Thus we may assume that MATH. We first consider REF . By REF we have MATH . The desired bound REF then follows from REF . It remains to consider REF . The desired bound REF has now simplified to MATH though in the exceptional REF our task is still to show REF. We estimate REF by MATH . The main difficulty with REF is that there are two separate phenomena that need to be exploited to obtain a sharp bound. The first is the transversality between the surfaces MATH. The other is the angular constraint REF. To be able to exploit both simultaneously we again use REF . The key is to cover MATH by boxes which are large enough to capture an optimum amount of transversality subject to the condition that the boxes stay small enough so as not to disturb the angular constraint. We turn to the details. Let MATH be a large number (MATH will do) and define MATH, MATH. Let MATH be the box MATH . Let MATH be the canonical tiling lattice for MATH, so that MATH is a box covering. Let MATH be the multiplier defined in REF. It is clear that MATH unless MATH, MATH, MATH, MATH, and MATH. We now claim that MATH except in the exceptional REF , in which case one must place an additional factor of MATH on the right-hand side. Assuming REF for the moment, we see from REF that MATH . From REF (for instance) we have MATH so by the tensor product lemma REF the above simplifies to MATH . Applying REF we thus obtain MATH which is REF as desired. In the exceptional REF we repeat the above argument but acquire an extra factor of MATH, which ultimately yields REF instead of REF. It remains to prove REF. Fix MATH with MATH, MATH, and MATH. It suffices to show that MATH with an additional factor of MATH in the right-hand side in the exceptional REF . We may assume that MATH, since the above expression vanishes otherwise. We shall use REF and elementary geometry. Let MATH be such that MATH. Discarding the restrictions on MATH and MATH, and using REF we reduce to showing MATH for all MATH, with an additional factor of MATH when REF holds. Fix MATH; we may assume that MATH since the set vanishes otherwise. In particular we have MATH. We now split into three cases depending on the value of MATH. (The estimate REF is symmetric with respect to interchanging MATH and MATH). First suppose that MATH. Using REF we can estimate the left-hand side of REF by MATH . To finish the proof of REF in this case it thus suffices to show that MATH. Since MATH and MATH, we are done unless MATH. But in this case we have MATH since we have assumed MATH. Since MATH, MATH are within MATH of MATH and MATH respectively, REF follows. Now suppose that MATH. Using REF we can estimate the left-hand side of REF by MATH . Since MATH, we thus see that this set is contained in a MATH neighbourhood of a hyperplane, and also in a ball of radius MATH. REF follows. Finally, suppose that MATH. Using REF as before, the left-hand side of REF becomes MATH . First suppose that MATH, so in particular MATH. Since MATH, the above set is contained in a MATH neighbourhood of a hyperplane, and also in a ball of radius MATH. One can then bound the measure of the above set as MATH which is REF with the additional factor of MATH. Now suppose MATH, so that MATH. From this and the estimates MATH we see that MATH and MATH. From this we see that the size of the above set is at most MATH. When MATH, then MATH and MATH, and the above estimate simplifies to REF as desired. When MATH then MATH and the above estimate simplifies to MATH which is REF with the additional factor of MATH.
|
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 verified using the fact that MATH. Note that the MATH, MATH, MATH summations may give powers of MATH, but this acceptable because of MATH factor and the assumption MATH.
|
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 map MATH, MATH. Since MATH is an automorphism of algebraic groups, the variety structure is independent of the representative MATH. Let us prove that MATH is a locally algebraic group. For MATH and MATH, multiplication is the morphism MATH . Similarly for the inverse: MATH .
|
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. Now we have MATH and in view of REF we reach the conclusion. CASE: If MATH is not definite but MATH we replace MATH by MATH, where MATH is the characteristic function of an interval MATH. The system MATH is definite on MATH and REF. applies. To complete the proof it remains to note that we obtain a linear isomorphism, MATH, from MATH onto MATH as follows: for MATH let MATH be the solution of the differential equation MATH with MATH.
|
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 estimated by MATH and we reach the conclusion.
|
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 self - adjointness of MATH can easily be seen directly: According to REF let MATH with MATH, and MATH . For MATH we choose, according to REF , representatives MATH and put MATH. Since MATH vanishes if MATH is not invertible the function MATH is well - defined. Moreover MATH hence MATH lies in MATH and it converges to MATH in MATH. Finally, we calculate MATH . Thus MATH and MATH and the claim is proved.
|
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 view of REF and MATH a.e. we may apply the definiteness of MATH to conclude that MATH.
|
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 reach the conclusion by invoking the dominated convergence theorem.
|
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 have to invoke the hyperbolic equation method as presented for example, by CITE. As in REF. we have MATH and hence, by elliptic regularity, MATH is locally of NAME class MATH, in particular MATH is locally square integrable. We now show that MATH is square integrable. Choose a large compact set MATH and let MATH be a first order differential operator which coincides with MATH over MATH and which vanishes outside a large compact set MATH. Now consider the operator MATH . MATH is a formally symmetric differential operator which vanishes outside a compact set. Hence, MATH has bounded propagation speed, in particular it satisfies NAME 's condition MATH. Thus by the hyperbolic equation method CITE all powers of MATH are essentially self - adjoint. Next choose a cut - off function MATH with MATH in a neighborhood of MATH and MATH. Then the commutator MATH is a first order differential operator which is supported in MATH. In particular MATH and hence MATH. Then MATH is square integrable. Since MATH is essentially self - adjoint, this implies that MATH hence MATH is square integrable. This implies that MATH is square integrable.
|
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 have MATH . Using REF , it is easy to verify by induction that MATH for all MATH . In particular, MATH . Hence MATH . Suppose the assertion is true for some MATH . Let MATH be a sequence in MATH that converges pointwise to a function MATH . Suppose there exists MATH such that MATH for all MATH and MATH . We need to show MATH . Since MATH we have MATH . For each MATH, let MATH denote the MATH-neighborhood of MATH . Denote MATH by MATH . From REF , for each MATH and MATH. By the inductive hypothesis, we see that MATH . From this and applying REF with MATH for all MATH, we see that MATH . Let MATH . Then MATH . Thus MATH . Therefore MATH . Hence MATH . Suppose MATH is a limit ordinal and the statement holds for all ordinals MATH . Let MATH be a sequence that converges pointwise to a function MATH and let MATH be given-MATH . Suppose that MATH for all MATH, and MATH . Then MATH and MATH .
|
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 locally finite. Let MATH . Then MATH is a finite set, MATH for all MATH and MATH . If MATH then MATH hence MATH. Therefore, MATH for some MATH . But then MATH implies that MATH . It follows that MATH . This shows that MATH . Finally, if MATH is a point in MATH, there exists an open neighborhood MATH of MATH in MATH such that MATH and MATH is finite. Now MATH . Hence MATH is continuous on MATH, since it is a finite linear combination of continuous functions. In particular, MATH is continuous at MATH. As MATH is arbitrary, MATH is continuous on MATH.
|
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 . Note that MATH . By REF , MATH by the continuity of MATH . In particular, MATH . It follows that MATH . Repeating the argument inductively yields that MATH . Hence MATH, as required.
|
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 . Note that since MATH is continuous at MATH . Hence there exists a neighborhood MATH of MATH in MATH such that MATH for all MATH . Let MATH . Then MATH is a neighborhood of MATH in MATH . For all MATH . Hence MATH . This proves REF . Suppose that MATH have been chosen. By REF , there exists a continuous function MATH such that MATH . Define MATH on MATH by MATH . Then MATH is continuous on MATH . By REF , MATH . Hence MATH is defined and continuous on MATH . Moreover, it follows from REF that MATH . From REF and the definition of MATH we have MATH . Therefore, MATH . Now define MATH . Then MATH is continuous on MATH . This proves REF . Furthermore, MATH . This proves REF . Also, MATH by the definition of MATH . This proves REF . Finally, suppose MATH . Assume that MATH . Then MATH . Thus there exists a neighborhood MATH of MATH such that MATH whenever MATH . Note that since MATH is continuous at MATH . Therefore, there exists a neighborhood MATH of MATH in MATH such that MATH for all MATH . Let MATH . Then MATH is a neighborhood of MATH in MATH such that MATH whenever MATH . Hence MATH . This proves REF .
|
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 MATH converges uniformly to MATH on MATH and MATH is an open subset of MATH is continuous at MATH . Hence MATH . Now suppose MATH . Then MATH . There is a neighborhood MATH of MATH in MATH such that MATH for all MATH . Also, for MATH . Since MATH is continuous on MATH is continuous on MATH for all MATH . Similarly, MATH and MATH is continuous on MATH by REF ; thus, MATH is continuous on MATH . Hence there exists a neighborhood MATH of MATH in MATH such that MATH and MATH . Let MATH . Then MATH is a neighborhood of MATH in MATH. If MATH then MATH . Thus MATH . If MATH then MATH . Thus MATH if MATH . Hence MATH whenever MATH . Therefore MATH . This proves the claim. It follows by induction that MATH for MATH . Indeed, the Claim yields the assertion for MATH . If the inclusion holds for some MATH let MATH . Then MATH . Hence the inclusion holds for MATH . If the inclusion holds for all MATH where MATH is a limit ordinal, then MATH . This proves the inclusion for MATH . In particular, if MATH then MATH . Thus MATH for all MATH . Hence MATH . Of course, since MATH . Therefore MATH .
|
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 closed subset of MATH . It is also easy to see that MATH and MATH are disjoint if MATH . Thus MATH is a partition of MATH into clopen (in MATH) subsets. Now define MATH to be MATH . Since MATH is a compact subset of MATH . From the clopeness of the partition MATH it follows readily that MATH . By REF , there exists a function MATH on MATH such that MATH and MATH . Finally, define MATH to be MATH and MATH to be MATH . It follows from the choices of the MATH's that MATH . Since MATH pointwise on MATH . Suppose MATH is a compact subset of MATH . Then MATH . To complete the proof, we claim that for any MATH and any MATH . The proof of this is by induction on MATH . The case MATH and the limit case is trivial. Now assume that the claim holds for some MATH . Let MATH . Choose MATH such that MATH and MATH where MATH is the metric on MATH . Denote MATH by MATH and the MATH-ball in MATH centered at MATH by MATH where MATH . Note that MATH by the inductive hypothesis: For all MATH . This implies that MATH for all MATH . Thus MATH . By REF , MATH . In particular, MATH . Since MATH is arbitrary, this shows that MATH .
|
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 pointwise to MATH and MATH . We may assume that MATH for all MATH for otherwise, simply replace MATH by MATH . Continuing, we obtain MATH and MATH for each MATH such that CASE: MATH CASE: MATH for all MATH CASE: MATH for all MATH CASE: MATH (pointwise) for all MATH and CASE: MATH converges uniformly to MATH on MATH . Let MATH and MATH . By REF , MATH for all MATH . Given MATH there exists MATH such that for all MATH . Then MATH . Therefore MATH . By REF , MATH for all MATH . Therefore, MATH by REF . Moreover, MATH . This proves the theorem in case MATH, with MATH in place of MATH . For a general nonzero countable ordinal MATH write MATH in NAME normal form as MATH where MATH . If MATH then MATH . By the previous case, there exists MATH such that MATH and MATH converges pointwise to MATH . If MATH take MATH for all MATH . Then MATH for all MATH, MATH and MATH converges pointwise to MATH.
|
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 . Since MATH by REF MATH . Thus MATH . Now assume that the corollary holds for all MATH where MATH is a countable limit ordinal. Let MATH . By the inductive hypothesis, MATH for MATH . Hence MATH is the uniform limit of a sequence in MATH, and thus belongs to MATH . Conversely, assume that MATH . For every MATH there exists MATH such that MATH . By REF , the exists MATH such that MATH . Thus MATH as required.
|
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 MATH the function MATH is continuous on MATH . Let MATH be a continuous extension of the function onto MATH . Then MATH . If MATH then MATH since MATH . If MATH then there exists MATH such that MATH thus MATH for all MATH . Therefore MATH for all MATH . Hence MATH . Thus MATH pointwise-MATH . Now for each MATH let MATH . Since MATH for all MATH, on MATH we have MATH . Thus MATH on MATH for all MATH . In particular, on the set MATH since MATH . As MATH we see that MATH pointwise-MATH . Also, for each MATH is bounded (by MATH) in MATH thus MATH is bounded in MATH .
|
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 such that MATH converges pointwise to MATH and MATH is bounded in MATH for all MATH. List the elements of MATH in a sequence MATH . Take MATH if MATH is of the form MATH for some MATH . Let MATH . Suppose MATH has been chosen. Then MATH converges to MATH pointwise, MATH is a bounded sequence in MATH and MATH . By REF , there exists MATH such that MATH . Let MATH for all MATH. Then MATH . Therefore MATH and MATH converges pointwise to MATH. We claim that for all MATH and for all MATH . We prove the claim by induction on MATH . The claim is trivial if MATH or MATH is a limit ordinal. Assume that MATH is a successor ordinal and that the claim holds for MATH . Let MATH . Then MATH . If MATH there exists MATH such that MATH . Choose MATH such that MATH . Then MATH and MATH since MATH . By REF , MATH . Thus MATH . But since MATH there exists an open set MATH in MATH such that MATH . By REF , MATH . Therefore MATH a contradiction. This proves the claim. From the claim, MATH for all MATH . Therefore MATH . Since MATH by CITE, (or REF ), MATH .
|
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 . Then MATH . Thus 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 all MATH . Since MATH on MATH and MATH is continuous on MATH there exists a neighborhood MATH of MATH in MATH such that MATH and MATH for all MATH . Set MATH . Then MATH is a neighborhood of MATH . Claim. MATH . Note that if MATH then MATH . Hence there exists MATH such that MATH . Also, MATH implies that MATH . Thus MATH by the inductive hypothesis. Since MATH as required. Now if MATH is a neighborhood of MATH in MATH there exist MATH such that MATH where, in the last inequality, MATH since MATH by the claim. Therefore, MATH . By the claim, MATH . Since MATH is an arbitrary neighborhood of MATH this shows that MATH . This completes the induction.
|
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 function MATH such that MATH is finite for all MATH . Define real-valued functions MATH and MATH on MATH as follows. If MATH let MATH . If MATH for some MATH and MATH let MATH and MATH . Notice that MATH where MATH . It follows from REF that MATH for all MATH . Since MATH and MATH for all MATH . Thus MATH for all MATH . Hence MATH . We now turn to the calculation of MATH and MATH . First notice that the sets MATH, MATH, form a partition of MATH into relatively open sets for any MATH and that MATH is constant on each set MATH . Hence the restriction of MATH to MATH is a continuous function for each MATH . It follows readily by induction that for any MATH for all MATH . But MATH on MATH . Thus MATH . Therefore MATH . Finally, consider the function MATH. Let MATH be given. The set MATH is finite. List the elements of MATH in a finite sequence MATH in lexicographical order. Then MATH for all MATH . Hence MATH . Note that MATH on MATH . Thus MATH for all MATH such that MATH . Let MATH be such that MATH . Then MATH . Therefore, MATH . Repeating the argument, we see that MATH where MATH . Since MATH for all MATH . As MATH increases to MATH . Hence MATH for any MATH . It follows that MATH . Summarizing, we have functions MATH and MATH such that MATH, MATH and MATH . Since MATH is arbitrary, the theorem is proved.
|
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 MATH is necessarily unique. If MATH for MATH and MATH, then the canonical rational section MATH of MATH is a rational function on MATH without zeroes nor poles on MATH. NAME 's lemma (see REF below) implies that MATH is constant, so that MATH. We remark that any effective cycle MATH on MATH is rationally equivalent to a cycle that does not meet MATH. Indeed, if MATH is the parameter of a subgroup of MATH isomorphic to MATH, we can consider the specialization of the cycles MATH when MATH.
|
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 such that the line MATH is fixed under this action. As MATH has no nontrivial characters, MATH itself is fixed. The divisor MATH is MATH-invariant; therefore, MATH does not meet MATH and is necessarily a sum MATH such that MATH. Because of REF , every other such section will be proportional to 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 MATH with basis MATH) such that MATH. Moreover, we can write uniquely MATH for some functions MATH. Let MATH be the irreducible component of MATH containing MATH. Necessarily, MATH stabilizes MATH so that the function MATH is identically MATH in a NAME neighborhood of MATH in MATH and a fortiori, MATH vanishes on MATH. By REF hence MATH is a multiple of MATH: there exists a unique MATH such that MATH and MATH . The lemma is proved.
|
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 MATH. Necessarily, MATH, hence we have to prove that these integers are positive. Fix any MATH in the smooth part of MATH, so that there exists a unique MATH such that MATH. Pick local coordinates MATH in a neighborhood MATH of MATH in such a way that in MATH, MATH is defined by the equation MATH. Write MATH, with MATH. We have seen in REF that for any MATH, MATH. Hence, MATH.
|
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 function on MATH given by MATH is the norm of the canonical basis MATH and therefore extends to a locally constant function on MATH. Its restriction to the compact subset MATH is uniformly continuous. Since it is locally constant and equal to MATH on MATH, there exists a neighborhood of MATH on which it equals MATH. Such a neighborhood contains a neighborhood of the form MATH, where MATH is a compact open subgroup of MATH. This proves the lemma in the effective case. In the general case, we write MATH for two effective line bundles (each having a nonzero global section). We can endow MATH with any locally constant MATH-adic metric and MATH with the unique (necessarily locally constant) MATH-adic metric such that the isomorphism MATH is an isometry. By the previous case, there exist two compact open subgroups MATH and MATH contained in MATH which act isometrically on MATH and MATH, respectively. Their intersection acts isometrically on MATH.
|
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 such that for any MATH, MATH . We may now apply NAME 's theorem and obtain the desired finiteness.
|
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 converges for MATH big enough, say MATH. In the proof of REF we may choose some MATH which works for any MATH, so that for any MATH . Since MATH is bounded from below on MATH it follows that for any MATH . Therefore, the height zeta function converges absolutely and uniformly on the tube domain MATH.
|
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 bounded continuous function of MATH. (Only at a finite number of places do they differ, and at these places MATH, a comparison is provided by the compactness of MATH.) As MATH is compact, one may take for MATH any compact set containing a fundamental domain. This implies the uniform convergence of the series once MATH belongs to the tube domain of REF . The integrability follows at once : for such MATH, MATH .
|
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 boundary divisor MATH has strict normal crossings. The map MATH induces a map from the set MATH of irreducible components of MATH to MATH. Moreover, MATH has a canonical section obtained by associating to a component of MATH its strict transform in MATH. This allows to view MATH as a subset of MATH, the complementary subset MATH consisting of exceptional divisors. Via the identifications of MATH with MATH and MATH with MATH, the map MATH on line bundles induces a linear map MATH which is a section of the first projection MATH. (We also extend MATH by linearity to the vector spaces of complexified line bundles.) Concerning metrizations, it is possible to endow line bundles on MATH with adelic metrics which extend the metrics coming from MATH via MATH. Remark that with such a choice, the height function on MATH and its NAME transform on MATH extend those on MATH. For instance, for any MATH and any MATH, one has MATH . As MATH maps MATH into MATH, it follows in particular that MATH is integrable for MATH and that for any MATH and any MATH, one has MATH . Unless MATH is an isomorphism, MATH is not smooth and MATH has a nonzero coefficient at each exceptional divisor. For any effective MATH, this implies that MATH lies on the boundary of MATH. (Recall that the classes of MATH and MATH in MATH and MATH are denoted MATH and MATH.) It follows that MATH and that MATH can be identified with MATH. Consequently, any ``exceptional" factor MATH for MATH extends to a holomorphic bounded function on MATH . This shows that REF still stands true without the hypothesis that MATH has strict normal crossings.
|
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 local measures glue to define a global measure MATH on MATH. Recall that NAME 's MATH-function of MATH is equal to MATH. Using NAME 's theorem on NAME 's conjectures (but it will also follow from our calculations), NAME has shown that the NAME product MATH converges absolutely and he defined the NAME number of MATH as MATH . (Note however that NAME apparently doesn't use the selfdual measure for his definition in CITE, but inserts the appropriate correcting factor MATH.) On the open subset MATH, we can define the local measure with the differential form MATH induced by the canonical coordinates of MATH. Let MATH be such that MATH. On MATH, we thus have MATH so that, the boundary subset being of measure MATH, MATH . Moreover, it follows from these results and the absolute convergence of the NAME products of NAME zeta functions for MATH that for any MATH, MATH is integrable on MATH and that one has MATH . The estimates of REF below allow us to interchange the limit MATH and the infinite product MATH, so that MATH because of the product formula.
|
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 compactification whose boundary is a divisor with strict normal crossings obtained by intersecting the components of the boundary of MATH with MATH. The arguments of REF show that we do not lose any generality by assuming this on MATH itself.
|
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 and and MATH. Therefore, with notations from REF , MATH . The proposition is proved.
|
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 which we have a set coordinates MATH so that in MATH the divisor MATH is defined by the equations MATH for some MATH. In MATH there exist continuous bounded functions MATH (for MATH) such that MATH . The integral over MATH is comparable to an integral of the same type with functions MATH replaced by MATH. The lemma is now a consequence of the following well-known lemma.
|
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 that MATH. If MATH is a local equation of MATH in a neighborhood MATH of some MATH-point of MATH, then there is a MATH function MATH on MATH such that for MATH we have MATH . It then follows from REF that for each MATH, MATH extends to a MATH function on MATH. From the equality MATH we deduce by induction the existence of an isobaric polynomial MATH of degree MATH (with the convention that each MATH has weight MATH) and such that MATH . This implies that there exists a constant MATH such that MATH . Choosing MATH such that MATH is maximal gives MATH, hence an upper bound MATH where MATH is the trivial character and MATH some positive constant. To conclude the proof it suffices to remark that for any MATH, MATH is bounded on the set MATH (but the bound depends on MATH). The case MATH is treated using a similar integration by parts. (The exponent MATH on the numerator comes from the fact that for a complex place MATH, MATH is the square of a norm.)
|
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 MATH denotes the fixed measure on MATH, one has the equality of measures on MATH: MATH . Consequently, MATH where the last equality follows from MATH . Summing these equalities for MATH gives the desired formula.
|
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 's estimate CITE to the geometrically irreducible smooth MATH-scheme MATH.
|
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 MATH and from the fact that NAME zeta functions have a polynomial growth in such vertical strips.
|
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, there is a trivial estimate MATH which implies that for some constant MATH, MATH . Using REF , we have for all MATH, MATH . Replacing MATH by MATH and MATH by MATH concludes the proof of the proposition.
|
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, this is true since MATH. If MATH, one has MATH since we assumed MATH.
|
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, add MATH-rational local étale coordinates corresponding to a basis of the subspace in MATH which is complementary to the span of MATH for MATH.
|
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 . Then the function MATH is in MATH. We next prove the following lemma. Let MATH. Assume that the real valued function MATH satisfies REF and that there exists constants MATH such that MATH . Then: CASE: For all MATH, the function MATH is in MATH. CASE: For MATH the function MATH is in MATH. Assume first that MATH. Let MATH, then by REF, MATH . By equivalence of norms in MATH there is a constant MATH such that MATH . The last inequality is an application of the following inequality (with MATH): (see CITE) NAME Inequality: Let MATH and MATH with MATH. Let MATH and MATH. Then there exists a sharp constant MATH, independent of MATH and MATH, such that MATH . This proves REF when MATH. Assume now that MATH. We assume without loss that MATH. Then, with MATH, MATH . For MATH, using REF and equivalence of norms in MATH, we get MATH . As for MATH, we apply the following inequality; for the convenience of the reader, we give the proof (borrowed from CITE). For MATH: MATH . Proof of REF. By NAME 's inequality we have, for MATH, MATH, MATH . Substituting MATH for MATH, with MATH, this gives MATH . So, for MATH, using MATH and the triangle inequality in MATH, we have MATH . In this way, by REF and equivalence of norms in MATH, MATH since MATH. This finishes the proof of REF . The proof of MATH is similar to that of MATH so we omit the details. The proof of the following fact is straightforward: There exist constants MATH and MATH such that MATH for all MATH. Using this and REF , we shall prove the following lemma on the regularity of the functions MATH and MATH from REF. Let MATH and MATH be as in REF. Then CASE: MATH for all MATH. CASE: MATH for all MATH. Herefrom follow the regularity properties of the function MATH stated in REF . Proof of REF . Firstly, by REF , MATH which gives REF for MATH. Next we verify that for MATH, REF is fulfilled. Then REF can be applied with MATH, and the NAME of MATH and MATH follows. Given MATH, and MATH. Using that (see e. CASE: CITE) (here, MATH, MATH) MATH and that MATH for all MATH we get, with REF, MATH so REF follows for MATH. This proves REF . To prove REF , we write MATH as in the proof of REF : MATH, with MATH and MATH as in REF. Then MATH . Using the idea from REF twice (on MATH and MATH, respectively), the estimates REF, and REF, we have, with MATH, and MATH: MATH . By REF , with MATH, this implies that MATH. Using the same ingredients, writing MATH, gives REF with MATH and so by REF , MATH. The remaining terms are those involving the function MATH, namely MATH, MATH and MATH. Note that MATH and so MATH . In this way, MATH . Note that MATH. Using the ideas above, REF implies that the following functions from REF (with the mentioned choices of MATH satisfying REF) are all in MATH: MATH . Likewise, REF implies (with the mentioned choices of MATH satisfying REF) that the following functions from REF are all in MATH: MATH . From the decomposition of MATH in REF we are left with MATH . Note that MATH . The function MATH is smooth for MATH and therefore in MATH. The function MATH satisfies REF (by the same ideas as above), so REF implies that the function MATH is in MATH. The functions in REF are therefore in MATH. As for the functions in REF, these are all in MATH, which can be seen by applying the previous ideas, in particular REF . This proves that MATH. As for MATH (see REF), with MATH, MATH . The terms in the first sum with MATH are in MATH, due to REF , with MATH satisfying REF. (To see this, use the previous ideas; to apply the idea from REF to MATH we use that MATH is smooth). The terms in the second sum in REF are all in MATH, due to REF , applied with MATH. The term with MATH is in MATH. This can be seen by following the ideas in the proof of the regularity properties of the function in REF, now using REF with MATH. The statements and proofs are similar for MATH . That is, the functions in the first sum in REF with MATH and those in the second sum are all in MATH, whereas the function in the first sum with MATH is only in MATH. To prove this we use the inequality (with MATH): MATH . This inequality follows from REF (remember that MATH). This proves that MATH, and so finishes the proof that MATH. (See REF). This proves REF and therefore REF . That MATH is a consequence of the foregoing and of the following proposition: Assume MATH, MATH. Then MATH, where MATH . Let MATH. For all MATH, choose MATH and MATH such that MATH . This is possible, since MATH. Then MATH . Using compactness of MATH, choose MATH such that MATH . Choose MATH such that MATH . Then, for all MATH and all MATH there exists MATH such that MATH and therefore by REF, MATH . So with MATH, MATH . This implies that MATH . This proves that MATH. To prove that MATH, we apply the following: Assume MATH. Then MATH, where MATH . The function MATH is continuous in MATH, since it is in MATH. Let MATH. Then MATH . Using the supremum of MATH on a sufficiently large compact set in MATH as a dominant, NAME 's Dominated Convergence Theorem gives us that MATH . Therefore MATH. Recall the proof of the fact that MATH. In fact, the only terms in the decomposition of MATH (see REF, and REF) that are only in MATH and not in MATH are the functions MATH . Comparing REF, it can be seen that all the terms in REF stem from the function MATH, namely from MATH, MATH, and MATH. All other terms in the decomposition of MATH are in MATH. When integrating them over MATH, we get something continuous in MATH, according to REF above. For the terms in REF we note that they are all of the form MATH . In each case, we have MATH . To see this, apply REF to each of the coordinate functions MATH, MATH. The integrands are easily seen to satisfy REF in each case, by the previous ideas. To get continuity in MATH of the functions in REF we use REF, and the following lemma: Assume MATH. Then MATH, where MATH . The same as for REF , noting that for all MATH and fixed MATH: MATH . This holds even in the case MATH, for which MATH . This proves that the functions in REF are in MATH. Therefore MATH, which finishes the proof of REF .
|
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 curve or simple and modular (by REF). CASE: Immediate, by REF .
|
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 of codimension at least MATH, and MATH is a holomorphic embedding into MATH. As MATH is a connected commutative NAME group its universal cover is MATH. As MATH has co-dimension at least two in MATH, MATH is simply connected, and so MATH lifts to a holomorphic map MATH (see CITE). Let MATH. Then MATH is a holomorphic map from MATH into MATH which agrees with MATH off the thin analytic subset MATH. But then by NAME 's removable singularity theorem CITE MATH, contradicting the fact that MATH. REF is proved. CASE: Let MATH be the graph of MATH. We will first show that for all MATH in some dense NAME open subset MATH of MATH, MATH is finite. If not, then for a NAME open subset MATH of MATH, the above set of MATH's has positive dimension. It follows from REF that MATH, contradicting irreducibility of MATH. It now follows by the implicit function theorem that REF MATH is holomorphic on MATH. For MATH let MATH be MATH, a meromorphic mapping from MATH to MATH. Note that if MATH is defined (single valued) at MATH and MATH is defined at MATH then MATH is defined at MATH and equals MATH. It follows that if MATH and MATH, then MATH. As MATH is NAME in MATH it follows that REF MATH is a meromorphic mapping into MATH, yielding REF . CASE: The same argument as above shows that for any MATH, MATH is a meromorphic mapping from MATH to MATH. Let MATH be the irreducible components of MATH. Note that the image of the meromorphic mapping MATH from MATH to MATH (that is, projection of its graph on second component) is all of MATH. But for MATH, MATH. Thus the image of the meromorphic mapping MATH is all of MATH. We work model-theoretically. Fix MATH. Let MATH be a generic point of MATH over MATH. So MATH for some MATH. It follows that MATH is a meromorphic mapping from MATH into MATH. CASE: Thus MATH must be a permutation of MATH. If for some MATH, and MATH as above, MATH is not a generic point of MATH over MATH, then there is a proper analytic subset MATH of MATH such that MATH has image contained in MATH. By REF , we contradict the fact that MATH has image all of MATH. Thus for MATH generic, MATH is generic in MATH over MATH. It follows that MATH gives a definable action of MATH on MATH. As MATH is connected this has to be trivial. This gives REF .
|
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 MATH let MATH be a coordinate neighborhood of MATH in MATH such that the closures MATH of the MATH in MATH are disjoint. Note that MATH is contained in MATH, so in MATH, but is not compact, so not closed in MATH. Let MATH be the boundary of MATH in MATH, namely MATH. Let MATH be the quotient map which collapses each MATH to a point MATH. Then MATH is compact, MATH is holomorphic (in fact is a modification), and is biholomorphic outside the union of the MATH's. Let MATH be defined by MATH for MATH and MATH. Then MATH is a definable, holomorphic embedding. CASE: MATH is a NAME of MATH. If MATH there is nothing to do. Otherwise, (as MATH is commutative) REF applies. Let MATH be as there. We will show that MATH is holomorphic on MATH, and in fact acts as the identity. Note that the generic type of MATH is orthogonal to any set of dimension less than that of MATH (MATH being strongly minimal). In particular MATH is orthogonal to MATH. Fix a component MATH of MATH. REF gives us a generic action of MATH on MATH. Let MATH be generic independent elements of MATH and let MATH be generic in MATH over MATH. Then by the orthogonality mentioned above, each of MATH and MATH is independent from MATH. It follows that MATH, and thus MATH. But MATH is generic in MATH and independent from MATH. It follows that MATH acts generically trivially on MATH. So the holomorphic map from MATH to MATH taking MATH to MATH agrees generically with the meromorphic mapping MATH. By REF , these mappings agree. This shows that MATH is a NAME of MATH.
|
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, contradicting REF Thus MATH is the semidirect product of MATH and some MATH, contradicting REF .
|
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 hypothesis, MATH is contained in finitely many translates of MATH as desired. So we may assume that MATH is infinite. By REF , and the fact that MATH is simple, MATH must contain MATH. Note that MATH. But it is well-known that the NAME rank of the stabilizer of a NAME degree MATH set MATH is at most the NAME rank of MATH, with equality if and only if the stabilizer is connected and MATH is, up to a set of smaller NAME rank, a translate of this stabilizer. Thus MATH, MATH and up to a set of smaller NAME rank, MATH is a translate of MATH, so we finish again by induction.
|
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 MATH which is a dense NAME subset of a compact complex manifold MATH. Let MATH be the canonical surjective homomorphism. Then MATH takes MATH onto a cofinite subset MATH of MATH. CASE: We may assume that for any MATH, MATH (a translate of MATH). MATH is a definable subset of MATH of NAME rank MATH. By REF , MATH is contained in finitely many translates of MATH, namely finitely many fibers of MATH. Remove these fibers from MATH. Let MATH denote MATH. MATH extends to a meromorphic function MATH from MATH to MATH. Further restricting MATH we may assume: CASE: For all MATH, MATH. Let MATH be the finite set MATH. Then we can find MATH such that MATH. Let MATH be a preimage of MATH. Let MATH be multiplication by MATH. MATH is not defined everywhere but is holomorphic on the open set where it is defined and so extends to a meromorphic map MATH. By REF (see REF), there is a modification MATH and a holomorphic map MATH such that MATH. In particular, for MATH such that MATH is defined at MATH, MATH. CASE: MATH. This holds generically, so holds everywhere. We will now construct the required compactification MATH of MATH as a holomorphic image of MATH. Let MATH. So MATH or MATH. As a set MATH will be the disjoint union of MATH with MATH. The manifold structure of MATH is as follows: MATH is given its canonical manifold structure. Now let MATH. Let MATH. Choose an open neighborhood MATH of MATH in MATH such that MATH. Then MATH is an open neighborhood of MATH. The transition maps are clearly holomorphic, yielding a structure of a complex compact manifold on MATH containing MATH as an open (dense) subset. Now we define a holomorphic surjective map MATH from MATH to MATH. Let MATH. If MATH define MATH (so MATH). On the other hand, if MATH, define MATH. Note that in this latter case MATH and so MATH and MATH. It is easy to check, given our assumptions, that MATH is holomorphic and surjective. So MATH is a compact complex manifold, containing MATH as a dense NAME open subset (the embedding of MATH in MATH is definable and holomorphic). This completes REF: The action of MATH on itself extends to a trivial action on the boundary MATH. Let MATH be the irreducible components of MATH. Note that for each MATH, MATH. CASE: For each MATH there is a surjective holomorphic map from MATH to MATH (so finite-to-one outside a proper NAME closed subset MATH of MATH). By REF we have a surjective holomorphic map MATH such that MATH and MATH is precisely the canonical surjective homomorphism from MATH to MATH. So MATH is MATH where MATH. Consider the map MATH from MATH to MATH. This is definable and holomorphic so extends to a meromorphic map from MATH to MATH, which we also call MATH. By REF this map is holomorphic. In particular, for any MATH and MATH, MATH is a holomorphic map from MATH into MATH. We must show that for suitable MATH, this is surjective. For MATH, let MATH be the meromorphic map from MATH to MATH whose restriction to MATH is multiplication by MATH (so MATH is MATH in previous notation). Let MATH. The map taking MATH to MATH is the constant map with value MATH. It follows that whenever MATH and MATH is single valued at MATH then MATH. Choose MATH generic in MATH. Then for a dense open set MATH of MATH in MATH, MATH is defined and in MATH. So for each MATH, MATH. Thus MATH is generically surjective, so surjective. In any case this map is finite. By REF , for each MATH, MATH induces a generic holomorphic action of MATH on MATH. Let MATH be the subgroup of MATH consisting of those MATH such that for all MATH in some NAME open subset of MATH, MATH. For dimension reasons MATH is a definable infinite subgroup of MATH, so contains MATH. Moreover we have an induced generic action of MATH on MATH. Let MATH be as in REF . Let MATH be such that for any MATH in some dense NAME open subset of MATH, MATH. This gives a meromorphic map from MATH to MATH whose image contains infinitely many points outside MATH. Composing with the holomorphic map from MATH to MATH given by REF yields a meromorphic nonconstant map from MATH into MATH which is impossible. Thus MATH and the generic action of MATH on MATH is trivial. This holds for each MATH. So the generic action of MATH on MATH is trivial. That is, if MATH, then the meromorphic mapping MATH agrees with the identity map on a dense NAME open set. This implies that MATH is the identity map. So MATH witnesses MATH being NAME meromorphic. The proof of REF is complete as well as the proofs of REF
|
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 combination of cosets of definable subgroups. Simplicity implies that MATH is strongly minimal. By REF, MATH is a complex torus.
|
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. MATH has an an infinite center. By the hypotheses, MATH contains an infinite definable linear algebraic group and thus, by the structure of commutative linear algebraic groups, MATH contains a definable MATH-dimensional connected linear algebraic group MATH. MATH is normal in MATH, so by induction hypothesis MATH is (meromorphically isomorphic to) a linear algebraic group. It makes sense to talk about the algebraic dimension MATH of MATH. Note that MATH so the map MATH witnesses that MATH. If MATH, then MATH is already isomorphic to an algebraic group, so a linear algebraic group. Otherwise MATH, and it is well-known (see CITE) that the general fiber of the algebraic reduction MATH is an elliptic curve MATH. But the map MATH must meromorphically factor through MATH. The general fiber of the first map is MATH and thus we see that MATH is an image of MATH under a meromorphic (that is, rational) map, a contradiction. Thus MATH is linear algebraic. CASE: MATH is solvable. We may assume, by Case I, that MATH is finite. But then MATH is centerless and easily MATH is linear algebraic iff MATH is. So we may assume that MATH is centerless. As is shown in REF or REF, the commutator subgroup MATH of MATH is connected, and nilpotent, so MATH is infinite and contains a minimal definable connected MATH-normal subgroup MATH. MATH defines an infinite group of automorphisms of MATH. By again results in CITE or CITE, MATH is the additive group of a definable field MATH. As MATH is by hypothesis linear algebraic, the field MATH has to be (definably isomorphic to) MATH and MATH. MATH is by induction hypothesis algebraic, and as in REF we deduce that MATH is (definably) algebraic. CASE: MATH is nonsolvable. Note that MATH is among other things a connected complex NAME group, and as such we have the NAME decomposition MATH where MATH is the maximal normal solvable connected subgroup of MATH, MATH is a semisimple (complex) NAME group (unique up to conjugacy in MATH), and MATH is discrete. Finite NAME rank considerations (see REF) show that MATH is definable, so linear algebraic by Case II (or induction). It is probably then well-known that MATH must be isomorphic as a complex NAME group to a linear algebraic group. However we want MATH to be definably isomorphic to such a group. So we must do a little more work, although maybe there is a more direct way. We will show Claim. MATH is a definable subgroup of MATH. We may assume that MATH is a proper, nontrivial subgroup of MATH, definably isomorphic to a linear algebraic group. We will first reduce to the case where MATH is commutative and unipotent. Let MATH be the connected component of the center of the commutator subgroup of MATH. MATH is then a nontrivial commutative connected linear algebraic group, normal in MATH. So MATH is the direct product MATH of a commutative unipotent group MATH and an algebraic torus MATH. Note that both MATH and MATH are definable connected normal subgroups of MATH. MATH has no infinite definable group of automorphisms, so is central in MATH. By REF we may assume MATH to be trivial. Thus MATH is unipotent. By induction hypothesis, MATH is linear algebraic. Clearly MATH is the solvable radical of MATH. Thus MATH is an almost direct product of MATH with a semisimple algebraic group MATH (where MATH is a definable connected subgroup of MATH containing MATH). As MATH is unique up to conjugacy we may assume that MATH. Note that the homomorphism MATH is an isomorphism on MATH. Note that MATH is linear algebraic, by induction hypothesis among other things. Now MATH (being semisimple) is isomorphic (uniquely) as a complex NAME group to a linear algebraic group, so it makes sense to talk about an element of MATH being unipotent. MATH is an almost direct product of almost simple groups MATH. Fix a nontrivial unipotent element MATH in some MATH. Work now inside the definable group MATH. Let MATH. MATH is then unipotent, and let MATH be a MATH-dimensional definable unipotent subgroup of MATH containing MATH. Let MATH. Then MATH is an extension of a unipotent linear algebraic group REF by a linear algebraic unipotent group MATH so is (by induction) linear algebraic unipotent. MATH. We can find a definable commutative connected subgroup MATH of MATH containing MATH. MATH is definably a vector space over MATH. The MATH-dimensional subspace MATH generated by MATH is a definable subgroup of MATH contained in MATH. We have found an infinite connected subgroup MATH of MATH which is definable in MATH. The group generated by all the MATH, where MATH is definable and must be equal to MATH. So MATH is definable. As MATH was arbitrary, MATH is definable. The claim is proved. We want MATH to be definably isomorphic to a linear algebraic group. Recall that MATH as a complex NAME group is the almost direct product of almost simple (discrete centre) groups MATH. As the centre of MATH is definable, each MATH has finite centre. By considering centralizers, each MATH is definable. MATH being almost simple is almost strongly minimal, hence by the validity of the NAME trichotomy in MATH, modular or nonorthogonal to MATH. But MATH is nonabelian. So it must be nonorthogonal to MATH, so algebraic. Thus MATH is definably an algebraic group, so by semismplicity, linear algebraic. Finally MATH, being the almost semidirect product of linear algebraic MATH with linear algebraic MATH, must be linear algebraic. Case III is complete, as well as the proof of REF .
|
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 each of MATH and MATH. As both MATH are strongly minimal, both projections MATH and MATH are finite-to-one, and MATH is strongly minimal with MATH. Replacing MATH by its normalization we may assume MATH is normal. MATH has no proper infinite analytic subsets, in particular no co-dimension MATH analytic subsets. By the purity of branch theorem, MATH is an unramified covering. Thus MATH is a complex torus. As MATH is also an unramified covering MATH is a complex torus.
|
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 finite set of realizations of MATH. Then by the modularity of MATH, MATH is a generic point of a translate of a (strongly minimal) subtorus MATH of MATH. After translating we may assume that MATH is a generic point of MATH. By elimination of imaginaries the finite set MATH is coded by some MATH. As MATH, we obtain a surjective meromorphic map MATH from MATH to a compact complex manifold MATH where MATH is a generic point of MATH. But the map taking MATH to MATH extends to a meromorphic map from MATH to MATH. Modularity of MATH implies that this induces a bimeromorphic map between MATH and MATH for some finite group MATH of automorphisms of MATH. Thus MATH is NAME. So if REF hold, the only possibility left for MATH is to be trivial. CASE: If REF fails then MATH is nonorthogonal to MATH so is nontrivial. If REF fails and there is a surjective meromorphic map to the NAME manifold MATH then clearly MATH is also simple and its generic type is nontrivial. So MATH is nontrivial. The ``in particular" clause follows from the observations that any NAME manifold MATH is in MATH as well as any compact complex manifold which maps meromorphically and generically finite-to-one on MATH.
|
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, once we have proved that MATH and MATH are isomorphic as MATH-modules. Now MATH acts on MATH by a character MATH, hence on MATH, as well. With MATH being dense in MATH, it follows that MATH for all MATH. It is now obvious that MATH decomposes as stated. Thus we have REF. For the last statement, it is enough, of course, to consider the case where MATH is a line having MATH-weight say MATH. By REF, there is a MATH-subspace MATH of MATH, on which MATH acts by the character MATH restricted to MATH, so that MATH contains MATH. Let MATH be a regular one parameter subgroup of MATH so that MATH. Thus, there is an induced surjective morphism MATH. Let MATH be a MATH-module, into which MATH embeds equivariantly with MATH. Then MATH is a MATH-stable subvariety of MATH, which certainly contains MATH, the latter being a vector bundle over MATH. This means in particular that MATH is a trivial bundle. Moreover, it is easy to see, that over MATH, the map MATH is a closed MATH-equivariant immersion, MATH acting trivially on the first factor MATH. Summarizing, there is a global section MATH of MATH with MATH and with MATH, which, over MATH has the form MATH for suitable MATH. Let MATH be the decomposition into MATH-weightspaces, so that MATH with MATH. Now compare weights in the expansion MATH . Since MATH extends to zero and MATH, the term on the right hand side of REF of degree zero occurs when MATH. Furthermore, for every nonzero term on the right hand side, MATH. Since MATH is regular, it follows that MATH, where MATH. This says MATH, and we are done.
|
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 subset of the unique irreducible component of MATH through MATH. Since MATH is smooth somewhere it follows that MATH, so MATH is the unique component of MATH through MATH. It follows that MATH is an attractive fixed point of MATH, and therefore MATH is nonsingular. Passing to the normalization MATH of MATH, we obtain an equivariant finite map MATH, which is étale in codimension one, because the natural map MATH is clearly an isomorphism over MATH. Thus, by REF, MATH is étale everywhere. Hence for some point MATH which maps to MATH we have MATH via MATH. This implies that MATH is attractive, forcing MATH to be an isomorphism. Thus MATH is birational. But being finite, MATH is also an isomorphism, so we are through.
|
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 locus of MATH. According to REF , we are done if MATH. By assumption, if MATH, then MATH. It follows that MATH if and only if MATH has a nontrivial kernel MATH for some MATH. But then MATH. With MATH being the projection to MATH, the latter is trivial, so MATH and hence MATH both are equal to MATH. We conclude that MATH is empty. Thus, by REF , MATH forcing MATH thanks again to REF . This ends the proof.
|
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 that the functions corresponding to MATH and MATH have weights MATH and MATH respectively (thus, the minus sign for MATH). Except for the case where MATH, MATH and MATH are contained in a copy of MATH this actually implies that MATH for suitable nonnegative integers MATH. Using the multiplicity freeness of the representation of MATH on MATH, one is done in these cases. In the remaining case, it turns out that MATH is nonsingular at MATH unless MATH are orthogonal long roots in MATH CITE. Let MATH. Then MATH isomorphic to MATH where MATH correspond to MATH-lines in MATH of weights MATH and MATH.
|
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 when MATH is short, the last statement is obvious.
|
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 span MATH of the MATH-curves MATH. This means that the differentials MATH are independent along MATH. As MATH is nonsingular, we are done if MATH is the ideal MATH of MATH. We know that MATH at every stalk MATH for MATH, since the MATH are independent. As a subset of MATH, MATH is equal to the support of the NAME subscheme MATH of MATH. Under the natural restriction, a function MATH on MATH, which vanishes on MATH, defines a global section of MATH with support contained in MATH. It is well known that on a NAME scheme, the only such section is zero. So MATH restricts to zero, and we are done.
|
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-module, it follows immediately that MATH. Since MATH is long, REF implies MATH . Moreover, since MATH, it also follows that MATH, so MATH constitute a complete MATH-string occuring in MATH. Since MATH is nonsingular at MATH and MATH, we get the inequality MATH. Thus MATH is nonsingular at MATH. Letting MATH be the good MATH-curve in MATH such that MATH, we have MATH, so the string MATH also has to occur in MATH. In particular, MATH, and hence MATH. But this means MATH so MATH. This is a contradiction, so MATH. We next prove REF . Recall MATH and suppose to the contrary that MATH. If MATH, we can argue exactly as above, so we are reduced to assuming MATH. If MATH, then MATH due to the fact that MATH. Thus MATH, whence MATH and MATH since MATH. Moreover, MATH since MATH. Hence, we have MATH . Now MATH and MATH form a MATH-string in MATH, and since MATH, it follows as above that MATH. Therefore, MATH. But MATH, so we have a contradiction. Hence, MATH. To prove REF , note that, as usual, MATH, so MATH by the MATH-module property. To obtain REF , note that if MATH, then MATH, so MATH. As MATH, this implies that MATH, so in fact MATH. The proof of REF is clear, so the proof is now finished.
|
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 corresponding to MATH, and let MATH be those corresponding to the MATH-curves MATH through MATH. Consider the unique linear projections MATH which restrict respectively to MATH. Since the (restriction of the) projection MATH has a finite fibre over MATH, MATH is a finite MATH-module. In particular MATH is integral over MATH. We obtain a relation MATH where MATH is a suitable integer and the MATH. Without loss of generality we may assume that every summand on the right hand side is a MATH-eigenvector with weight MATH. Let MATH be polynomials restricting to MATH, having the same weight MATH as MATH. Then every monomial MATH of MATH has this weight too. If for all MATH every such monomial MATH has degree MATH, then MATH is an element of MATH, where MATH is the maximal ideal of MATH in MATH. This means that MATH vanishes on the tangent cone of MATH, so MATH, which is a contradiction. Thus, there is a MATH and a monomial MATH of MATH, such that MATH. Let MATH, with integers MATH and a nonzero MATH. So MATH. Let MATH be the weight of MATH. Then we have MATH . After choosing a new index, if necessary, we may assume that MATH for all MATH. Let MATH be a nondegenerate bilinear form on MATH which induces the length function on MATH. We have to consider two cases. First suppose that MATH is a long root, with length say MATH. Then MATH with equality if and only if MATH. Thus, MATH and so there is a MATH with MATH and we are done, since this implies MATH. Note that, although we are considering all roots short in case MATH is simply laced, this contains actually the case that all roots have the same length. Now suppose MATH is short, having length MATH. In this case MATH. Since MATH and since MATH, it follows that all MATH satisfy MATH. If there is a MATH such that MATH, then, as above, we are done. Otherwise for each MATH, MATH is long, and MATH and MATH are contained in a copy MATH of MATH. There is a long root MATH with MATH. We have to show that there are MATH and MATH so that MATH. Fix MATH and let MATH, MATH. Then MATH. This gives us the result: MATH. Now if all MATH are less or equal MATH, this last equation cannot hold, since MATH. We conclude that there is a MATH so that MATH, hence MATH, and we are through. The statement for MATH simply laced follows from this, since in this case all roots have the same length, hence are long.
|
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 that MATH. REF follows from the existence of a local MATH-equivariant section of MATH, the smoothness of MATH and REF .
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.