paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9906174 | Since MATH. So the right hand side of the above identity is MATH . By REF , MATH is non-zero if and only if MATH and MATH. Since we chose the measure MATH so that the volume of this set is one, MATH is the characteristic function of MATH. Therefore, REF is MATH which is the definition of MATH. |
math/9906174 | We have MATH, which implies that MATH since MATH or MATH. Since MATH, MATH. In particular, MATH and so MATH. If MATH then MATH and we reach a contradiction. Hence MATH. The order of MATH is either MATH (if MATH) or MATH (if MATH). If MATH this forces MATH and if MATH it forces first MATH and then MATH. |
math/9906174 | The conditions imply that MATH and MATH are integral. Since MATH, MATH. Thus MATH, which implies that MATH. If MATH then MATH and so MATH is a negative, even integer. This is a contradiction and so MATH. The reduction of the polynomial MATH has no roots in MATH and thus MATH for all MATH. It follows that MATH. This gives MATH and MATH, as required. |
math/9906174 | If MATH then MATH and MATH since MATH. The statement follows in this case. We now assume that MATH. Let MATH and consider the polynomial MATH. Since MATH, MATH and so MATH. The discriminant of MATH is equal to the discriminant of MATH and so if MATH is a root of MATH then MATH. It follows that MATH and hence that MATH. But the standard orbital representative was chosen so that MATH and the statement follows in this case also. |
math/9906174 | If MATH then MATH and so MATH by REF . But MATH and it follows that MATH. This establishes the first part of REF . We now have to show that if MATH then MATH. The orbital representatives have already been fixed in REF - REF and the notation introduced there will be used without comment below. We begin with the cases (sp), (in) and (rm). Let MATH; we have to show that MATH. By REF , MATH is left MATH-invariant and so we may assume that MATH. Since MATH we have MATH in case (sp) and MATH in cases (in) and (rm). The assumption that MATH implies that MATH in case (sp) and that MATH in cases (in) and (rm). In case (sp) we have MATH and in cases (in) and (rm) we have MATH . Let MATH or MATH. Then, by assumption, MATH or MATH. Consider case (sp). Let MATH for MATH, and MATH. Then MATH if and only if MATH are integral. So MATH. We assume MATH and deduce a contradiction. Suppose MATH is not a unit. Then MATH . Then MATH is a unit. This implies MATH, which is a contradiction. So MATH is a unit and similarly MATH are units also. Then the order of MATH is MATH, which is a contradiction. This implies MATH. Then MATH for MATH. Cases (in) and (rm) are similar using MATH in the places of MATH above. The only difference is that we consider elements in MATH. Next we treat the case (sp rm). Suppose MATH. Then MATH for MATH. We may assume that MATH are lower triangular. Note that MATH. So MATH is integral. Since MATH, MATH by REF . In this case, we can regard MATH as MATH. Instead of the third factor, we can use the first and the second factors to make equivariant maps similar to MATH. Then because of the symmetry of our element MATH, MATH by REF again. This concludes the verification in this case. Now we consider the case (sp ur). Let MATH and MATH. In this case there are four possibilities as follows: CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. For these cases, MATH respectively. The argument in REF is similar to that used in case (sp rm). In REF . Since MATH is integral, this corresponds to the case MATH in REF . Therefore this cannot happen. REF are similar to REF because of the symmetry (considering an equivariant map using the second MATH factor in REF ). Now we consider the case (in ur). Suppose that MATH. This implies that MATH. We have MATH and, since MATH is a unit by assumption, MATH by REF . Since MATH is left MATH-invariant we may assume that MATH and that MATH is lower triangular, say MATH. Note that MATH and this is a primitive integral vector. Computation gives MATH where MATH with MATH and both these matrices must be integral. Let MATH. Then MATH and MATH and so MATH and MATH are integral if and only if MATH are integral. Since MATH, it follows that MATH. Also MATH and it remains to show that MATH and MATH are units. From the definition of MATH we know that MATH. Let MATH; we assume that MATH and deduce a contradiction. We have MATH and, from REF , we conclude that MATH. Thus we may write MATH where MATH and MATH. Then MATH and, since MATH, it follows that MATH . Regarding this as a linear system for MATH, the determinant of the coefficient matrix is MATH. This is a unit by the choice of MATH and so REF implies that MATH. This contradicts MATH and so MATH. This completes the case (in ur). Next we must deal with the case (in rm). Suppose that MATH. By similar arguments to those of the previous case, using REF in place of REF , we see that MATH. Hence we may assume that MATH and that MATH is lower triangular. Then MATH where MATH and MATH are given by REF . Since MATH, MATH and MATH are integral if and only if MATH are integral. Let MATH; we shall again assume that MATH and derive a contradiction. We have MATH, so that MATH. Thus MATH and, since MATH is an NAME polynomial, it follows that MATH. Also, MATH and, using our conclusion about MATH together with the fact that MATH is an NAME polynomial, we deduce that MATH. But MATH and MATH is always even, so this last congruence is impossible. This contradiction completes the case (in rm). Finally we must deal with the case (rm ur). Suppose that MATH. There are apparently two possibilities: either MATH or MATH and MATH. However, REF shows that the second possibility cannot occur and, moreover, that MATH. Thus we may assume, as usual, that MATH and MATH. The matrices MATH and MATH given by REF must be integral and, since MATH is a unit, this happens if and only if the quantities enumerated in REF are all integral. Again assume that MATH and that MATH. Then MATH. If MATH is dyadic then MATH and so this congruence forces MATH to be a unit. However, since MATH is ramified, MATH and so MATH, which implies that MATH is not a unit. This contradiction completes that proof in the dyadic case. Now assume that MATH is not dyadic. Then MATH and so MATH is not a unit. We can write MATH with MATH and a suitable choice of uniformizer MATH. Since MATH, we conclude that MATH is not a unit and hence that MATH is not a unit. However, MATH and MATH is a unit. This contradiction completes the proof in the non-dyadic case. |
math/9906174 | Let MATH and MATH. From REF and the choices made above we have MATH where MATH is the representative we have chosen here to represent the orbit of MATH. Let MATH. From REF we obtain MATH . However, from REF in the definition of an omega set, MATH if MATH. Thus the sum really only extends from MATH to MATH and MATH for MATH. This makes it clear that MATH for all MATH. Since MATH is the square of a rational character, MATH if MATH is odd and this gives the last statement. Finally, again by REF, MATH and so MATH. |
math/9906174 | We begin with the case MATH. Define MATH . Suppose that MATH where MATH (recall that all the standard orbital representatives have this form). We claim that MATH and that MATH . The elements of the group MATH have the form described in REF . This makes it clear that MATH. Since the last entry in elements of MATH is unrestricted, we need only show that the equation MATH is always solvable for MATH, MATH and MATH and MATH satisfying MATH provided that MATH and the matrix MATH is non-singular. If REF holds, we must take MATH and MATH and then the equation is equivalent to MATH which is solvable for MATH and MATH since the coefficient matrix is non-singular by hypothesis. If MATH then we have MATH and MATH. Multiplying the first equation by MATH, the second by MATH and subtracting, we obtain MATH, contrary to hypothesis. This proves the second claim. Let MATH. Then MATH is a left NAME measure on the (non-unimodular) group MATH. From what we have just shown, it follows that MATH always has measure zero. Thus we have MATH for all MATH. Now if MATH is left invariant under MATH then the NAME decomposition implies that MATH where MATH and MATH denotes the right NAME measure on the group MATH. It is easy to check that MATH and applying this in REF we obtain the conclusion. Finally, almost identical arguments apply in the case when MATH is not split and we shall not repeat them. |
math/9906174 | Our work will be simplified if we compute with the element MATH with MATH or MATH instead of with the element MATH. By REF , MATH and so this is permissible. Suppose that MATH. Then, by REF , elements of MATH have the form MATH where MATH is determined by the other two entries. Note that the conjugation by MATH does not change the first two components. Let MATH . From REF and the NAME decomposition it follows that MATH and MATH. Since MATH is MATH-invariant, MATH . Computation gives MATH . Introducing the variables MATH we have MATH. So MATH . But we know from REF that the constant term in MATH is MATH and so MATH has the stated value. When MATH the calculation is a simple variation on the above and we shall not reproduce it here. |
math/9906174 | The standard orbital representative is MATH for some quadratic polynomial MATH which is irreducible over MATH (we may assume that MATH since MATH). Let MATH, MATH, MATH and MATH be as in REF . By REF , MATH for some constant MATH. Since MATH, MATH which implies that MATH for all MATH. (Recall that MATH if MATH is odd by REF .) Computing, we find that MATH where MATH with MATH . If we make MATH units and MATH integers then MATH and the volume of the set MATH under MATH is MATH and so It follows from this, REF that MATH . Therefore, from REF again, MATH for all MATH. We introduce new variables defined by MATH . Then MATH . Note that MATH are monomials of MATH. So they correspond to a lattice in MATH. Since the correspondence between MATH and MATH is bijective, this lattice must be unimodular. This implies that MATH . Suppose that MATH. Then MATH. Since MATH is the maximum of MATH for MATH, MATH, which implies that MATH. The conditions that the MATH entry in MATH and the MATH and MATH entries in MATH are integers may be expressed as MATH where MATH . This matrix factorizes as MATH where we have set MATH, MATH and MATH . Let MATH . Then the three conditions are equivalent to MATH, which in turn is equivalent to the conditions MATH . By computation, MATH and MATH . The remaining condition for MATH is that MATH. Expressing MATH in terms of the coordinates MATH we find that MATH where MATH. Since MATH and MATH, MATH if and only if MATH . We claim that at least one of MATH, MATH and MATH must be greater than or equal to MATH. Suppose to the contrary that MATH. Then MATH also, by REF , and so MATH. Furthermore, since MATH, MATH and it follows from REF that MATH and so MATH. However, by the choice of the standard orbital representatives, MATH and we have a contradiction. This establishes our claim. Next we claim that MATH. Suppose to the contrary that one of these quantities is less than MATH. In light of the symmetry between the roles of the pairs MATH, MATH and MATH we may suppose without loss of generality that MATH is the greatest of MATH, MATH and MATH and that MATH. By the previous paragraph, MATH. Dividing REF through by MATH we obtain MATH . We have MATH and so we may drop the terms involving MATH to obtain MATH . Now MATH and hence MATH. This implies that MATH and so MATH. By REF , MATH . These last two facts allow us to apply NAME 's lemma to conclude that MATH, which contradicts the assumption that MATH. Thus MATH, as claimed. Changing variables to MATH in REF and using REF , we obtain MATH where, on the domain of integration, MATH and MATH for MATH. Note that MATH on MATH. Carrying out the integration with respect to MATH, MATH and MATH we get MATH . Note that the volume of the set MATH is MATH and MATH. Put MATH. Using the formul MATH valid for MATH, we obtain MATH valid for MATH, where MATH is given in the statement of the proposition. This completes the proof. |
math/9906174 | The structure of this proof will be very similar to that of the proof of REF and so we shall abbreviate somewhat. We have MATH for some irreducible quadratic polynomial MATH. Let MATH, MATH, MATH and MATH be as in REF . Arguing as in the previous proposition we obtain the inequality MATH for all MATH. Calculation gives MATH where MATH with MATH . We introduce new variables defined by MATH . Then MATH . Since we are dealing with coordinates in two different fields, MATH and MATH, a small digression is required to calculate the relationship between MATH and MATH. Let us fix an element MATH which satisfies MATH. For MATH, we define MATH and MATH. Both MATH and MATH lie in MATH and since MATH, MATH and MATH serve as MATH coordinates for MATH. We use this notation replacing MATH by other letters. The measure corresponding to MATH is invariant under addition and hence there is a constant MATH, depending only on MATH, MATH and MATH, such that MATH . We also have MATH and a calculation gives MATH so that MATH. Multiplying both sides by MATH we obtain MATH . Suppose that MATH. Then MATH and MATH. Also MATH by REF . If we set MATH then the MATH entry in MATH is MATH and the MATH entry in MATH is MATH and it follows that MATH . We have MATH and so MATH . Hence MATH which implies that MATH . The remaining condition for MATH is that MATH. In the coordinates MATH we have MATH where MATH . Thus MATH if and only if MATH . Note that for any MATH we have MATH. We claim that either MATH or MATH. Suppose to the contrary that MATH and MATH, so that MATH. Then MATH and so MATH. Also MATH . So, from REF , MATH. Thus MATH, which is a contradiction. The claim follows. Next we claim that MATH. Suppose to the contrary that MATH. Then, from the previous paragraph, MATH. Dividing REF by MATH we obtain MATH . Since MATH this containment implies that MATH . Now MATH . Hence MATH . This implies that MATH and so MATH. By REF , MATH . Thus MATH and so MATH. We may now apply NAME 's lemma to conclude that MATH, which contradicts the assumption that MATH. Thus MATH. Changing variables to MATH in REF and using REF , we obtain MATH where, on the domain of integration, MATH, MATH, MATH and MATH. Carrying out the integration with respect to MATH and MATH we get MATH since MATH is unramified. Set MATH. Using REF , we obtain MATH valid for MATH, where MATH is given in the statement of the proposition. This completes the proof. |
math/9906174 | Suppose we have two series MATH with MATH for all MATH and MATH. Then MATH with MATH and so if we set MATH then MATH, MATH and MATH for all MATH. We have shown that if MATH then MATH if MATH and MATH if MATH (the right hand side comes from writing MATH in REF over a common denominator). Multiplying these two gives the value of MATH recorded in REF . The case MATH is similar. From their construction, the series for MATH in REF have non-negative coefficients and constant term MATH. It follows by inspection that these series converge when MATH. The discussion in the first paragraph shows that MATH for all MATH and MATH. Finally, it is well-known that the series MATH is absolutely and locally uniformly convergent in the region MATH. The usual convergence test for products now shows that MATH has the stated convergence properties. |
math/9906174 | The conditions on MATH, MATH and MATH which are proposed in the statement simply say that MATH lies in the standard maximal compact subgroup in that factor for MATH (respectively MATH) and so they are certainly necessary. To see that they are also sufficient we must show that they imply that the last entry in MATH also lies in the relevant maximal compact subgroup. But this follows immediately from REF . |
math/9906174 | We have to show that in every case listed above MATH. Of the triple (sp), (in), (rm) we deal with the first, since the other two are very similar. For this orbit MATH, we have MATH and the map MATH is MATH. Note that in this case we may choose MATH in REF . Then computing MATH in REF explicitly for MATH, we get the formula for MATH. If MATH then MATH where MATH and, since MATH, it follows from REF that MATH. This leaves the cases (sp ur) and (sp rm). Recall that MATH was chosen so that MATH. We have MATH and the map MATH is MATH . If MATH then MATH and so we may find MATH so that MATH and MATH. If we set MATH then MATH. Thus MATH and MATH. |
math/9906174 | For these cases MATH and MATH is MATH. Note that we get this formula by explicitly computing MATH for MATH in REF . Given MATH we wish to determine whether there are integers MATH such that MATH, because MATH if and only if this is possible. Elementary linear algebra shows that MATH . Thus MATH if and only if MATH divides MATH and MATH. If MATH is unramified over MATH then MATH is a unit and this is no condition on MATH and MATH. Thus MATH in this case and the first claim follows. If MATH is ramified over MATH then MATH and so we must have MATH. With this condition MATH and so the second condition also holds. Thus MATH where the isomorphism is by the measure preserving map MATH. Thus MATH under the normalized multiplicative NAME measure on MATH. The second claim follows. |
math/9906174 | The map MATH is MATH as pointed out in the proof of REF . If MATH and MATH then MATH and so MATH. Moreover, MATH and since MATH and MATH is stable under MATH, it follows that MATH is a unit in the ring MATH. Thus MATH. Suppose that MATH. Then MATH and MATH and it follows from REF that MATH. This establishes the reverse inclusion. |
math/9906174 | If the type is (rm ur) then MATH either has the form MATH with MATH a non-square unit, if MATH, or is an NAME polynomial, if MATH. By hypothesis, MATH is irreducible when regarded as an element of MATH. Thinking of MATH as the ground field, these facts imply that MATH and so MATH, by REF . If the type is (in rm) then MATH is an NAME polynomial and, since MATH is unramified, MATH is still an NAME polynomial when regarded as an element of MATH. Thus MATH and, again by the Lemma, MATH. |
math/9906174 | We may choose MATH as the NAME polynomial associated to the extension MATH. Then MATH and so MATH by REF . Since MATH, we have MATH . On the other hand, MATH and so MATH is MATH . From this it is clear that MATH and so the volume has the indicated value. |
math/9906174 | We briefly sketch the proof here. The conditions in REF determine the structure of MATH as a scheme over MATH regarding MATH as variables. Note that the inequalities MATH can be regarded as equations MATH after adding variables MATH. Then MATH is by definition the set of MATH-valued points of this scheme and so REF will be regarded as an equation over MATH and the above inequalities, after reduction modulo MATH, can be regarded as the condition that MATH are units. Since MATH is defined over MATH, this proves the lemma. |
math/9906174 | We first consider the case MATH. Suppose MATH. Let MATH be as in REF where the entries are elements of MATH. By computation, MATH . Since MATH, MATH . If MATH is a unit then MATH. Thus MATH are units. So MATH. This implies MATH is a unit and so MATH. If MATH is not a unit, MATH must be a unit. By a similar argument, we can conclude that MATH and MATH are units. Multiplying by an element of MATH if necessary, we may assume that MATH. Then it is easy to see that MATH. It follows that MATH. The case MATH is similar. We now assume MATH is quadratic and unramified. Suppose MATH. Let MATH. We choose representatives of MATH in MATH and use the same notation. Since MATH is unramified, if MATH, MATH is a unit if and only if either MATH or MATH is a unit. The MATH-entry of MATH is MATH and the MATH-entry is MATH. Since either MATH or MATH is a unit, we may choose MATH and MATH to make the MATH-entry of MATH zero. Replacing MATH by an element of the form MATH (with MATH determined by MATH), we may assume that MATH . If MATH and MATH, MATH if and only if MATH and MATH. By computation, MATH where MATH . Therefore MATH, and MATH . If MATH is not dyadic, MATH. So MATH. Then MATH are both the identity matrix or both MATH. If MATH is dyadic, MATH is an NAME polynomial. This means MATH and MATH is a unit. By the first condition of REF , MATH. Substituting in the second condition and simplifying, MATH. Since MATH is a unit, MATH. If MATH is a unit, MATH, and if MATH is not a unit, MATH. Note that this argument is valid in MATH (even though it is not a field). If MATH then MATH and MATH are both MATH. If MATH then MATH and MATH. Then since MATH, MATH. This proves the proposition when MATH. Suppose now that MATH. Let MATH be as in REF where entries of MATH are in MATH and entries of MATH are in MATH or MATH, according as MATH or MATH. We first show that the right MATH-coset of MATH contains an element of the form MATH . If MATH, roots of MATH still generates the unique unramified quadratic extension of MATH and so an element MATH with MATH is a unit if and only if MATH or MATH is a unit. We choose MATH. Then either MATH or MATH is a unit. So there exist MATH such that MATH. Multiplying this element by MATH, we may assume that MATH. If MATH then MATH is unramified over MATH. Note that MATH is irreducible modulo MATH by assumption. Therefore, MATH. Since MATH is a perfect field, one can use the same argument as in REF, p. REF to determine the stabilizer MATH. So the proposition is true if MATH. By the previous step, we can multiply MATH by an element of MATH to assume that MATH. Let MATH. Then MATH and MATH. So MATH. Therefore, there exist MATH such that MATH. Multiplying MATH by this element, we may assume MATH. In both cases, MATH and so MATH. Multiplying by an element of the form MATH, we may further assume that MATH. This done, MATH is given by REF with MATH replaced by MATH. Therefore, by the same argument, MATH, and REF holds also. After this, the argument is the same as in the case MATH. |
math/9906174 | By the choice of orbital representatives, MATH is a unit when MATH is not ramified (including the case MATH) and MATH when MATH is ramified. In REF , we have MATH and MATH is a unit. This congruence may be rewritten as MATH. By NAME 's lemma, a unit is a square if and only if it is a square modulo MATH. Note that this is true whether or not MATH is dyadic. So MATH is a square. This implies that MATH in all these cases. In REF , the condition implies that MATH and hence that we have MATH, which again implies that MATH, since MATH is not dyadic. The same argument works for REF because MATH in this case. In REF , the assumption implies that MATH and MATH. So MATH. Since MATH is not dyadic, MATH. Therefore, MATH. |
math/9906174 | We first deal with REF - REF . For every one of these orbit types, we showed in REF that there exists an omega set for MATH. Suppose that MATH. Then MATH and so MATH. Since MATH is also integral, it follows from the first property of omega sets that MATH. That is, MATH for some MATH. We have seen above that the congruence conditions on MATH also imply that MATH and so MATH. This proves the claim for REF - REF . Consider REF . Let MATH . Then MATH fixes the set MATH. Consider the usual coordinates MATH as in REF . Since MATH, MATH. Applying this element, we may assume that MATH. Since MATH, MATH. Applying this element, we may assume that MATH. Since MATH, MATH. Applying this element, we may assume that MATH. Since we are assuming MATH, we chose MATH so that MATH. So MATH. So by a similar argument, we may assume MATH. Also MATH and MATH. We may assume that MATH. Let MATH where MATH. Then MATH. Note that MATH. So applying this element, we may assume that MATH. Then MATH. So MATH with MATH by NAME 's lemma. Applying MATH, we get MATH. |
math/9906174 | Let MATH in all cases. Then MATH. By REF , MATH. Consider REF . In this case, MATH. So its order is MATH. Therefore, MATH . Consider REF . Let MATH be the prime ideal. In both cases, MATH, and MATH. So its order is MATH. Therefore, in REF . In REF . Consider REF . In this case, MATH and MATH. So its order is MATH. Therefore, MATH . Consider REF . Let MATH be the prime ideal. In this case, MATH and MATH. So its order is MATH. Therefore, MATH . |
math/9906174 | We prove that MATH. Let MATH be the NAME polynomial corresponding to MATH. Since MATH is not dyadic, we assume MATH. Consider REF . Elements of the form MATH are in MATH if and only if MATH (see REF ). So the order of MATH is MATH. Since MATH, this is equivalent to MATH. Suppose MATH. Since MATH reduces to MATH modulo MATH, MATH. Let MATH. Then MATH is a unit scalar multiple of MATH modulo MATH. By computation, MATH . So MATH are units. By a similar argument as in the proof of REF , MATH. This implies that MATH are units. Since MATH for MATH, the right coset MATH contains an element MATH where MATH are in the form REF . Moreover, it is easy to see that MATH are determined by the coset MATH. We use REF to determine the possibilities for MATH. By computation, MATH . The condition MATH is equivalent to the condition that MATH, and MATH. Since MATH is not dyadic, MATH and MATH. Therefore, MATH. So there are MATH possibilities for MATH modulo MATH. This proves that MATH. In REF , by a similar argument as in REF , we can assume that MATH is in the form REF with MATH unit and MATH. Note that the order of MATH is MATH in this case. By the same consideration, we get MATH, and the rest of the conditions turn out to be the same. Since MATH, there are MATH possibilities for MATH. So MATH in this case also. Since the volume of MATH in REF is MATH, if MATH, MATH and if MATH, MATH . |
math/9906174 | Suppose MATH. It is easy to see that if MATH satisfies all the conditions in REF except for the last condition, then MATH and MATH. Suppose MATH where MATH is a unit or zero for all MATH (see I. REF, p. REF). Let MATH where MATH. Then by REF , MATH satisfies the condition in REF if and only if MATH has an expansion of the form MATH such that MATH if MATH (and no condition on MATH if MATH). If MATH, the volume of the set of such MATH is MATH if MATH and MATH if MATH. The volumes of the sets of MATH are MATH, the volumes of the sets of MATH are MATH, and the volume of the set of MATH is MATH. Therefore, MATH . Simplifying, we get REF . If MATH, similar considerations apply to MATH as to MATH in the case MATH. The volumes of the sets of MATH are MATH and MATH, respectively, the volumes of the sets of MATH are MATH and MATH, respectively, and the volume of the set of MATH is MATH. Therefore, MATH . Simplifying, we get REF . |
math/9906174 | A simple, direct calculation shows that elements of MATH and MATH preserve all the conditions for membership in MATH, with the possible exception of the last. Consider REF and suppose that MATH with MATH and MATH. Then MATH and so MATH is equal to MATH . We must investigate the order of this number. Since MATH and MATH are units, they may be ignored. Both MATH and MATH are units and, since MATH, MATH is also a unit. Thus the order of MATH equals MATH if MATH and is greater than or equal to MATH if MATH. Also, MATH and MATH, in the first case because MATH and in the second because MATH. It follows that the order of MATH is MATH if MATH and is greater than or equal to MATH if MATH. Thus MATH, as required. Similar arguments apply in REF . |
math/9906174 | Consider REF . We first show that MATH, where MATH runs through the standard orbital representatives for the orbits of type (sp rm) whose corresponding fields have a fixed discriminant. As in the statement, the discriminant is related to MATH by MATH. By construction, if MATH then MATH and so if MATH is the standard representative for the orbit corresponding to MATH then MATH. Since MATH, the theory of omega sets implies that there exists MATH such that MATH. But then MATH and so MATH. That is, MATH. Each one of the standard orbital representatives itself lies in MATH and so the MATH-translates of MATH cover MATH. In order to find the volume of MATH it remains to determine the number of distinct MATH-translates of MATH. Suppose that MATH and MATH; say MATH and MATH. We shall show that MATH. Since MATH both reduce to unit scalar multiples of MATH modulo MATH, the assumption implies that MATH. Since MATH, we may assume MATH. Let MATH and MATH be the reduction of MATH modulo MATH. We define MATH, MATH similarly. Then MATH . Since MATH or MATH, MATH. This implies that MATH and therefore MATH. Thus MATH, as claimed. It follows from this and REF that two MATH-translates of MATH are either disjoint or equal and that the number of them is MATH by REF . Using REF , the result follows. Similar arguments apply to prove the formula in REF . |
math/9906174 | Consider MATH in REF . Obviously MATH. In both cases, MATH consists of elements of the form MATH where MATH. So MATH. We show that MATH. Suppose MATH. We consider the coordinates REF again. Since MATH is a unit scalar multiple of MATH modulo MATH, MATH. Then since MATH is a unit scalar multiple of MATH, MATH. So MATH are units. This implies that the right MATH-coset of MATH contains an element of the form MATH. We use REF again. By computation, MATH . So MATH, and MATH. Since MATH, MATH, and MATH. Since MATH, there are MATH possibilities for MATH. Therefore, in both cases, MATH . |
math/9906174 | As noted above, it suffices to compute MATH where MATH, the element introduced in REF . Let MATH be the product of the coordinate measures on each of the three factors of MATH so that MATH. Let MATH denote the set of MATH such that MATH and MATH lie in the big NAME cell. Then MATH is dense in MATH and it is on this set that we shall carry out the comparison of measures. Any element MATH of MATH may be written as MATH with MATH and when MATH is written in this form, MATH, MATH and MATH for MATH may be regarded as coordinates on MATH. Note that the map MATH is injective. With respect to these coordinates, the Jacobian determinant of the map MATH is found to be MATH. Note that this map is a double cover because MATH. Since MATH and MATH, it follows that the pullback of the measure MATH to MATH is MATH, where MATH denotes the coordinate measure on the third factor. (The measure has been divided by MATH because the map MATH is a double cover and, in REF , the measure MATH was defined via an integral over MATH.) In REF , MATH may be regarded as coordinates on MATH and as such MATH. It follows from the definition and the remarks above that MATH and so all that remains is for us to determine the relationship between the coordinate measure and the measure MATH on the big cell inside MATH when it is coordinatized as MATH. A simple Jacobian calculation shows that in fact these two measures are equal. This gives MATH or equivalently MATH. Using the remarks before the proof, this gives the stated values. |
math/9906174 | We shall again use MATH as the orbital representative. Let MATH be the complex place of MATH which divides MATH. Let MATH be the product of the coordinate measures on the two factors of MATH so that MATH. We let MATH denote the set of MATH such that MATH lies in the big NAME cell. An element of MATH may be written as MATH with MATH and when MATH is written in this form, MATH, MATH and MATH, MATH may be regarded as coordinates on MATH. With respect to the real coordinates MATH, MATH, MATH, MATH, MATH, the Jacobian determinant of the map MATH is found to be MATH. Since the map is a double cover, MATH and the canonical measures MATH and MATH are MATH and MATH, it follows that the pullback of the measure MATH to MATH is MATH, where MATH denotes the coordinate measure on the second factor. In REF , MATH may be regarded as coordinates on MATH and as such MATH. It follows from the definition and the remarks above that MATH and since the coordinate measure restricted to the big cell inside MATH is MATH, we have MATH. |
math/9906174 | We shall use MATH, where MATH, as the orbital representative for this orbit. The roots of MATH are MATH. Let MATH be as in REF . As we discussed in REF , any MATH has the form MATH where MATH . The isomorphism MATH from MATH to MATH may be taken as MATH . Recalling that the canonical measure on MATH is MATH where MATH is twice NAME measure, we see that MATH . For any MATH and any MATH we may find MATH and MATH such that MATH and it follows from this and the NAME decomposition that any MATH may be expressed as MATH. Thus any MATH may be expressed as MATH where MATH. Then MATH, MATH and MATH, MATH may be regarded as coordinates on MATH. With respect to these coordinates, the Jacobian determinant of the map MATH is found to be MATH. Since the map is a double cover, MATH and MATH this shows that the pullback of the measure MATH to MATH is MATH where MATH is the coordinate measure on the third factor. An easy Jacobian calculation shows that if MATH then the coordinate measure is MATH and so, with MATH denoting the product of the coordinate measures on the three factors, MATH. Since MATH, it follows that MATH. |
math/9906174 | We again use MATH, where MATH, as the orbital representative. We let MATH be the complex place of MATH dividing MATH. With MATH as in REF and MATH any element of MATH has the form MATH where MATH. The isomorphism MATH from MATH to MATH may be taken as MATH . From this it follows that MATH . It was shown during the proof of REF that any matrix MATH in MATH may be written in the form MATH provided that MATH. Since the complement of the set of matrices satisfying this condition has measure zero it suffices to make the comparison of measures on the set of elements of MATH whose first entry satisfies this condition. Any MATH in this set may be expressed as MATH where MATH. We may use MATH, MATH, MATH, MATH, MATH and MATH as real coordinates on (a set of comeasure zero in) MATH. With respect to these coordinates the Jacobian determinant of the map MATH is MATH. This shows, as usual, that the pullback of the measure MATH to MATH is MATH where MATH is the coordinate measure on the second factor. (Note that MATH and MATH again.) An easy Jacobian calculation shows that if MATH then the coordinate measure is MATH and so, with MATH denoting the product of the coordinate measures on the two factors, MATH. Since MATH, it follows that MATH. |
math/9906175 | We have MATH and so if MATH is the minimal polynomial of MATH over MATH then MATH and MATH. If MATH then MATH and hence MATH. If MATH then MATH and so MATH, which gives MATH. |
math/9906175 | The different of MATH is MATH and so, from the definition of the different, MATH implies that MATH. Multiplying by MATH, we find that MATH implies that MATH. Let MATH. Then MATH and so MATH. Thus MATH and so MATH. |
math/9906175 | If MATH and MATH are uniformizers of MATH then MATH with MATH and MATH. If MATH is the NAME polynomial associated to MATH then the NAME polynomial, MATH, associated to MATH is MATH. Say MATH satisfies the congruences MATH and MATH. Then it is easy to check that MATH satisfies the congruences MATH and MATH. Since MATH, the map MATH induces a well-defined map on MATH with inverse induced by MATH. This establishes a one-to-one correspondence between the two sets and REF follows. Fix a uniformizer MATH of MATH and let MATH be the corresponding NAME polynomial. Consider MATH. We may assume that MATH, since we have evaluated the numbers MATH, MATH and MATH in REF and they satisfy the second claim. With this assumption, every element of MATH is (the class of) a uniformizer in MATH. Suppose MATH. Fix MATH. We will use the corresponding NAME polynomial MATH to evaluate MATH. Every other element MATH of MATH has the form MATH with MATH and MATH. Moreover, the conditions on MATH imply that MATH and MATH satisfy the congruences MATH . We define MATH. Then, using the facts that MATH and MATH, it is easy to check that MATH and so MATH. Suppose MATH and MATH. If we write MATH with MATH then MATH and MATH. By computation, MATH . It is easy to check that this element belongs to MATH and so the map MATH induces a well-defined map from MATH to MATH. Reversing the roles of MATH and MATH we obtain a similar map from MATH to MATH induced by the map sending MATH to MATH. It is easy to check that these maps are inverse to one another and so MATH and MATH have the same cardinality. |
math/9906175 | There are two cases to consider. If two of the fields are generated by adjoining the square-root of a uniformizer then they have equal discriminants and the third field has a smaller discriminant (since it is obtained by adjoining the square root of a unit). Therefore, we have the statement of this lemma in this case. Otherwise, all the fields are obtained by adjoining the square root of a unit. Let MATH, MATH and MATH be the units whose square roots generate MATH and MATH respectively. We may assume that MATH where MATH and MATH for MATH. We may also assume MATH. Then MATH . If MATH then MATH and so MATH. The case MATH is similar. If MATH then MATH. If MATH then MATH and the inequality holds true. Otherwise, the largest number, MATH, such that MATH has the form MATH with MATH. Then MATH and again the inequality is true. |
math/9906175 | Let MATH so that MATH is either a unit of a uniformizer of MATH. Since MATH for MATH and MATH, we conclude that MATH is an integer. Thus the ideal generated by MATH in MATH is contained in MATH. But MATH and so the ideal generated by MATH is MATH. The inequality follows. |
math/9906175 | Let MATH. We choose NAME polynomials MATH so that MATH. Then MATH and our assumptions imply that this lies in MATH. Using the previous lemma we obtain MATH and the inequality follows. |
math/9906175 | Consider REF . Suppose, without loss of generality, that MATH. Then, according to REF , we must have MATH and so the inequality in REF becomes MATH. Since MATH, MATH. REF is obvious from REF because MATH. |
math/9906175 | Suppose MATH and MATH. We first assume MATH is unramified. Then MATH is not totally ramified. Therefore, by REF , p. REF, MATH contains an unramified quadratic extension of MATH. Since MATH are ramified, the remaining quadratic subfield MATH must be unramified over MATH. Let MATH, so that MATH. Multiplying MATH and hence MATH by a square, if necessary, we may assume that MATH. Then MATH and MATH have the same order in MATH and, multiplying them both by the same square, we may assume that they are either both units or both uniformizers without altering MATH. If MATH are both uniformizers then MATH. By the assumption on MATH, MATH for some MATH. Let MATH for MATH. Then MATH are uniformizers of MATH respectively and MATH . This implies that MATH. Suppose MATH are both units. Then MATH with MATH. Let MATH and MATH with MATH. Then MATH . Let MATH . Then MATH, MATH and MATH. Therefore, MATH and so MATH. Let MATH and MATH. We put MATH for MATH. Then MATH is a uniformizer of MATH satisfying the NAME equation MATH for MATH where MATH. Thus MATH and since MATH, we have MATH. Conversely, suppose MATH and MATH. Let MATH be as before. Then by REF , MATH. This implies that MATH is unramified. Since MATH is generated by roots of an NAME equation and they also generate the field extension MATH, this extension is unramified also. |
math/9906175 | If MATH, we choose MATH. Then MATH. Since any element of MATH is a square modulo MATH, we can choose MATH so that MATH. We now assume MATH. Note that if MATH then MATH or MATH and so MATH. This condition is obviously satisfied in the second case. Suppose MATH is the NAME equation satisfied by MATH. Let MATH. Then MATH . Note that since MATH in both cases MATH. Let MATH with MATH and MATH. Then MATH . The last congruence is satisfied because of the condition MATH. So MATH . Note that the orders of MATH are odd and even respectively and they can be any odd or even integer greater than or equal to two. So we can choose suitable MATH so that MATH . |
math/9906175 | We put MATH. Let MATH be any uniformizer. Using REF we can arrange that MATH and MATH if MATH. By REF , MATH and so MATH. Thus MATH and MATH if MATH. When MATH, the equality MATH then follows from REF . |
math/9906175 | Note that MATH are both uniformizers. So we may assume that MATH with MATH. Then MATH. Therefore, MATH. This implies MATH. |
math/9906175 | We first consider the case MATH. For any uniformizer MATH, MATH. So REF follows from the fact that any unit in MATH is a square modulo MATH. Consider REF . By REF , there exist MATH such that MATH. Let MATH with MATH. Then MATH . So MATH also. This proves the proposition when MATH. Suppose MATH and MATH. Let MATH, where MATH for MATH. For REF , we look for an element of the form MATH with MATH. If MATH satisfies the condition of REF , MATH is of the above form by REF . Therefore, in both cases we consider MATH of the above form. Then MATH . So MATH . Consider the case MATH. We have MATH . Since the orders of MATH, MATH can be any even or odd integer greater than or equal to two, we can choose MATH so that MATH. By the first congruence, we still have MATH. This proves REF . If MATH, MATH by the first congruence also. So this proves REF . Consider the case MATH. We have MATH . As long as MATH, MATH. By the same consideration as the previous case, we can choose MATH so that MATH. This proves REF . We now turn to REF . By assumption, MATH. So MATH. Since the orders of MATH are even and odd, MATH. Since MATH, MATH. So MATH. This implies that MATH which proves REF . We now assume MATH. Since MATH by assumption, MATH. We first consider the case MATH. Then MATH if and only if there exists MATH such that MATH. Then by REF , MATH . Let MATH. Consider REF . We choose MATH so that MATH and MATH. This is possible because MATH. Then MATH and MATH. Note that MATH and it can be any even integer between MATH and MATH. Also MATH and it can be any odd integer between MATH and MATH. Since MATH, MATH. So we can choose MATH so that MATH. Since MATH, the condition MATH is still satisfied. This proves REF . Consider REF . If MATH then MATH because MATH, by assumption. So MATH. Therefore, MATH. So we assume that MATH. If MATH then MATH and so MATH. If MATH then MATH. Since the orders of these elements are even and odd, MATH. In both cases, MATH. Since MATH, MATH. Therefore, MATH. This proves REF . We now assume MATH and so MATH. Then by REF , MATH if and only if there exists MATH such that MATH. Then MATH . Let MATH and MATH. Then it is easy to see that MATH. Suppose MATH. Consider REF . We choose MATH. Then MATH and MATH can be any odd integer between MATH and MATH. Since MATH, MATH. So MATH can be any even integer greater than or equal to MATH. Therefore, we can choose MATH so that MATH. This proves REF . Consider REF . By assumption, MATH . If MATH then MATH . So MATH. If MATH then MATH and the orders of MATH are odd and even respectively. This implies that MATH also. In both cases, MATH and so MATH. Therefore, MATH. This proves REF . |
math/9906175 | Consider the first claim in REF . If MATH then MATH and MATH are unramified, by REF , and so REF holds. We may now assume that MATH. We choose MATH which satisfies the condition of REF . Let MATH be the NAME equation with roots MATH, which generate the field MATH. Let MATH. Then MATH is equivalent to the following equation MATH . Since MATH, the order of the third term in REF is one. Therefore, REF is an NAME equation whose roots generate MATH. If MATH then MATH and MATH. So MATH. If MATH then MATH and MATH. So MATH also. Since MATH and MATH, MATH. Consider the second claim in REF . By REF , we choose an element MATH. We may assume that MATH is one of the roots of MATH. Let MATH be the polynomial REF . Then the roots of MATH generate the field MATH. We evaluate the order of the element MATH in REF , which is the same as that in REF . By REF, MATH, MATH and MATH. Therefore, MATH. Now MATH . Note that MATH and MATH. If MATH then MATH since MATH. This implies that the order of MATH is MATH. If MATH then MATH. Since MATH, the order of MATH is MATH. Therefore, in both cases, REF is an NAME polynomial with the order of the coefficient of the middle term MATH. Therefore, MATH. Consider REF . By REF , MATH. Let MATH. Then MATH by the second statement of REF . Therefore, using the first statement of REF , MATH (see REF , p. REF which is a local version of REF , p. REF). This implies that MATH and MATH. Thus REF . |
math/9906175 | We shall show first that MATH in general. Let MATH and choose MATH with minimal polynomial MATH such that MATH and MATH. Consider the element MATH of MATH; we claim that it is an integer. In fact, by REF , MATH by hypothesis. Thus MATH and it follows that MATH, as claimed. We know that MATH and so MATH and MATH if MATH. We now assume that MATH to complete the proof. There are MATH such that MATH and one of MATH is a unit (for otherwise MATH would not be the least integer with its defining property). Taking norms from MATH to MATH we find that MATH. Since MATH, MATH. It follows that MATH. Furthermore, MATH and MATH from which it follows that MATH. Let us set MATH. Then MATH is a uniformizer of MATH and MATH. Let MATH be REF minimal polynomial of MATH over MATH. Then MATH . Since MATH has odd order in MATH and MATH has even order, it follows that MATH and MATH which implies that MATH. Thus MATH. This proves the proposition. |
math/9906175 | The ring MATH is a MATH-order in MATH and so if MATH satisfies MATH then there is some MATH such that MATH . From the previous proposition we see that MATH. Then MATH . The normalized additive NAME measure on MATH is MATH and so the normalized multiplicative NAME measure on MATH is MATH, where MATH is the module of MATH. Since MATH, MATH . In case the index is (rm rm ur), MATH and MATH and we have MATH. In case the index is (rm rm rm), MATH and we have MATH. |
math/9906175 | REF is obvious. Consider REF - REF . We put MATH for REF - REF , respectively. Then the group in question is MATH in all parts. If MATH then MATH and MATH. Thus it is enough to verify the claims for MATH and MATH separately. We begin with MATH. For MATH in the form REF , let MATH and consider similar coordinates for MATH. Then MATH for MATH. Suppose MATH or MATH. Note that MATH in all cases. So MATH, MATH and MATH. Therefore, MATH by REF . We also have MATH since MATH and MATH. Since MATH and MATH, we further have MATH. Note that MATH in all cases. By REF , MATH . Therefore, MATH for MATH. Also MATH. By REF and so MATH. In REF and so MATH. In all cases MATH and so MATH. We have MATH and MATH. Thus MATH. All the conditions for MATH to lie in MATH, MATH or MATH have now been verified and so MATH or MATH. We now assume that MATH. It is easy to verify that MATH preserves all the conditions in REF - REF , with the possible exception of the last. The necessary calculation to show that this condition is also preserved by MATH has already been carried out in REF and will not be repeated here (note that MATH stands in for MATH in the proof of REF ). This proves the proposition. |
math/9906175 | We first consider the first implication in each of REF - REF . Let MATH, MATH, or MATH. Applying the element MATH to MATH, which is permissible by REF , we may assume that MATH; note that this doesn't change MATH. Further, applying MATH we may also assume that MATH. This implies that MATH and MATH . In REF and so MATH. Note that MATH for REF - REF , respectively, and so MATH in these cases also. Therefore, MATH . By this and the last conditions in REF - REF , MATH is an NAME polynomial such that the corresponding MATH is MATH for REF - REF , respectively. In REF and so MATH or MATH, by REF . The first implication in REF is now clear. Consider REF . By assumption, MATH . So, by REF , MATH. This proves the first implication of REF . Consider REF . We have MATH . Therefore MATH. So the only possible types are (rm rm ur), (rm rm)*, by REF . By similar considerations, MATH in REF . So the only possible type is (rm rm)*, by the remark after REF . We now consider the second implication of REF - REF . Suppose the roots of an NAME equation MATH generate MATH. In REF , there exists MATH such that MATH and MATH, by REF . In REF , by REF , there exists MATH such that MATH and MATH. In REF , by REF , there exists MATH such that MATH. Let MATH in all cases. Then MATH and so MATH satisfies REF or REF . This proves the second implication of REF - REF . |
math/9906175 | Let MATH and MATH be roots of MATH, respectively. Since MATH are both uniformizers of the same field, there exist MATH such that MATH. Let MATH. Then MATH and MATH. If MATH then, by REF , MATH and MATH since MATH. If MATH then MATH . Note that we are regarding MATH and MATH as elements of MATH here. Since MATH, MATH is an element of MATH regarded as embedded in MATH. Therefore MATH in this case also. |
math/9906175 | As shown in the proof of REF , we may assume that MATH and MATH. Let MATH. Then, by REF , MATH . Consider REF . By REF there exists MATH such that MATH. By REF , MATH and so MATH. REF are similar. |
math/9906175 | By REF , MATH. If MATH then MATH and, since MATH, REF follows. In REF , it is clear that MATH. If MATH then REF gives MATH and if MATH then it gives MATH. This completes the verification of REF . |
math/9906175 | REF is obvious. Consider REF . Suppose MATH and MATH satisfies REF except possibly for the last condition. Then MATH by REF . Since MATH, MATH, by REF , and MATH. Thus the last condition of REF is automatically satisfied. The volumes of the sets of MATH satisfying REF are MATH, respectively. Therefore MATH . Now suppose that MATH and again assume that MATH satisfies the conditions of REF except possibly for the last. We have MATH and so, by REF , MATH. Since MATH, MATH and so MATH . If MATH then it follows from this and REF that MATH satisfies the last condition of REF if and only if MATH but the corresponding congruence with MATH in place of MATH is false. With the other variables fixed, the volume of the set of MATH satisfying REF is MATH and hence the volume of the set of allowable MATH is MATH. This gives MATH in this case. If MATH the reasoning is the same, except that REF is the only condition on MATH. We thus obtain a similar formula for MATH with one fewer factors of MATH. This proves REF . Consider REF . Suppose MATH and MATH satisfies the conditions of REF for MATH except possibly for the last condition. Since MATH, MATH by REF . The order of the first term is MATH, by REF , and the order of MATH is MATH. So, when MATH are fixed, for MATH to be of order MATH, MATH has to be a unit which is not congruent to MATH modulo MATH. Therefore MATH . Consider REF . Suppose that MATH satisfies all the conditions of REF except possibly for the last. We shall show that the last condition follows automatically. First suppose that MATH and MATH. Then MATH, by REF , and MATH. Thus, by REF , MATH, as claimed. Now suppose that MATH and MATH. Then MATH, by REF , and MATH. Thus, by REF , MATH and again the last condition holds. This implies that MATH . Consider REF . Suppose MATH satisfies REF except possibly for the last condition. Then MATH. Since MATH if MATH and MATH if MATH, so the last condition of REF is always satisfied. Therefore, MATH . This finishes all the cases. |
math/9906175 | Suppose MATH is as in REF . Since both MATH and MATH are congruent to unit scalar multiples of MATH modulo MATH, MATH. Since MATH for every MATH, we may assume that MATH. Since MATH, MATH and MATH, we have MATH. This implies that MATH. By REF , MATH . Consider REF . Since MATH, if MATH then MATH by REF . Since MATH, this is a contradiction. Since MATH, REF are similar. |
math/9906175 | Consider REF . Let MATH. By REF , MATH is a disjoint union of translates of MATH and the number of translates is MATH. So, by REF , MATH . REF - REF are similar using REF - REF . |
math/9906175 | Since MATH and MATH, MATH and the similarly defined set MATH are conjugate within MATH and so it suffices to find the order of MATH. Let MATH . It was proved in REF that MATH consists of elements of the form MATH where MATH, MATH, MATH and MATH and MATH are related to MATH and MATH by the equation MATH . Note that MATH if and only if MATH. The expression for the order follows immediately. |
math/9906175 | Let MATH denote the reduction of MATH modulo MATH and MATH the stabilizer of MATH in MATH. Our first step will be to show that every right MATH coset in MATH has a representative of the particular form given in REF below. For MATH, let MATH be the MATH-module spanned by MATH and MATH inside MATH. As was stated in REF , if MATH then there exists MATH such that MATH if and only if MATH. Suppose that MATH. Since MATH reduces to MATH modulo MATH, MATH. Using this fact and examining the second component of MATH modulo MATH, we find that MATH. This implies that MATH and MATH lie in MATH. If we put MATH, MATH and MATH then MATH, the MATH-entry of MATH is MATH and the MATH-entry is MATH. Since MATH, we may further multiply on the left by MATH to find a representative for the right MATH coset of MATH having the form MATH with MATH and MATH. It is easy to check that each coset has exactly one representative in this form. Since MATH and MATH, it easily follows that every right MATH coset in MATH also has a unique representative in the form REF . Our second step will be to determine when such an element actually lies in MATH. Suppose that MATH is in the form REF . Then MATH if and only if MATH. Computation gives MATH where MATH . Note that MATH is in MATH if and only if MATH and MATH. Thus our element lies in the stabilizer of MATH if and only if MATH . Since MATH must be a unit, the first equation is equivalent to MATH, MATH. Using this, the second two equations become MATH and MATH. Making use of the second of these, the first is equivalent to MATH. Thus REF is equivalent to the system MATH . In the analysis of this system it will be convenient to adopt the usual abuse of notation by which classes in MATH and their representatives in MATH are denoted by the same symbol. Since MATH, the third equation in REF is equivalent to the condition that either MATH or MATH. These two possibilities are mutually exclusive and it is easy to check that the bijection MATH carries the set of solutions to REF satisfying the first inequality onto the set of solutions satisfying the second. Thus we may assume henceforth that MATH provided we then double the number of solutions found. Since MATH, MATH, by REF , and so MATH. Thus the second equation in REF is a consequence of the third and may be deleted from the system. Now suppose that MATH, so that MATH. Since MATH, we may write MATH with MATH. The fourth equation in REF is then equivalent to MATH. Thus MATH with MATH. It follows that MATH. But MATH and so MATH . Thus the first equation of REF is a consequence of the third and fourth. There are thus MATH choices for MATH and, for each choice of MATH, MATH choices for MATH. This gives MATH solutions to REF with MATH. Thus there are MATH solutions in all in this case. Finally, we must consider the case where MATH. We may assume that the uniformizer, MATH, has been chosen so that MATH. Since MATH, we may write MATH with MATH. Again MATH and so MATH with MATH. Let us write MATH and MATH where MATH and MATH. This is possible since MATH is ramified. A simple calculation gives MATH . Now MATH and MATH and so the last two terms lie in MATH. Thus MATH and so MATH if and only if MATH. Since MATH are determined modulo MATH, we can regard MATH as elements of MATH. There are MATH pairs MATH satisfying the congruences relating MATH and MATH and these lead to MATH pairs MATH. Thus there are MATH solutions in all. |
math/9906175 | In light of the previous two lemmas and REF , we have MATH . |
math/9906182 | This is proven in the same manner as in CITE. We review the key elements. One takes a sequence of compact polyhedra MATH converging to the ideal maximal MATH, and smears MATH uniformly about MATH to get a measure chain on MATH, where the copies of MATH with opposite orientation to M have negative sign. This measure chain is a cycle, since any face MATH of MATH will be matched with MATH as a face of the polyhedron obtained by reflecting MATH through the geodesic plane containing MATH. One can approximate this measure cycle by choosing a NAME domain and a basepoint. Then send each vertex of a polyhedron in a measure cycle to the nearest basepoint vertex. Each polyhedron then is weighted by the measure of the set of polyhedra sent to it under this map. One checks that for large enough MATH we still have a fundamental MATH cycle of the same norm. Then this chain has norm MATH, and taking the limit as MATH gives the desired result. The non-compact case works similarly to REF in NAME 's notes CITE. |
math/9906182 | First, make MATH transverse to S. Consider the universal cover MATH. Then MATH is a disjoint union of properly embedded planes. MATH has preimage a locally finite cycle MATH in MATH which is equivariant under the MATH action. For each pair of vertices in MATH, choose an edge connecting these points which meets MATH in as few points as possible, such that the choice is equivariant. Construct a new cycle MATH inductively. Map the vertices of each MATH to the corresponding vertices of MATH. Then extend MATH to MATH by the choices of efficient edges. For a given lift MATH of MATH, and a component MATH of MATH, MATH meets a cutset of edges in MATH in at most one point per edge. We can extend these to normal curves in MATH, and then to normal disks in MATH REF . To see this, look at a face MATH of MATH. If one edge of MATH meets MATH, then exactly one of the other two edges of MATH meets MATH. So we may connect the two points of MATH by a normal arc in MATH. We may always choose these arcs disjoint, for different components of MATH. If not, then there would be components MATH and MATH such that MATH and MATH are linked in MATH. But MATH divides MATH into two pieces, which lie on different sides of MATH. So only once piece could intersect MATH, which means the points are not linked. Now, extend MATH to these curves, and then to the disks equivariantly so that the normal disks map to MATH. This can be done since MATH is just a union of planes, and the action of MATH is free. We may now extend to MATH so that the complement of the normal curves misses MATH, and then we can fill in the balls in P complementary to the normal disks equivariantly, and missing MATH, since the complementary pieces of MATH are contractible. This chain MATH is equivariantly homotopic to MATH, so the chain MATH is the desired normalized chain. |
math/9906182 | First, let's outline the steps involved: CASE: Normalize the fundamental cycle. CASE: Cut the manifold and cycle along MATH. CASE: Lift the cut up cycle to coverings corresponding to MATH. CASE: Put a hyperbolic metric on MATH and compactify the cusps to points. CASE: Straighten the cycle. CASE: Collapse to a new cycle. CASE: Show that each triangle contributes only once to MATH . CASE: Estimate the contribution of each cell in the original cycle to the boundary cycle. CASE: Find a lower bound in terms of MATH. CASE: Since MATH has finite volume, MATH may have cusps. In this case, replace MATH and MATH by their doubles along the cusps, to get a closed manifold and surface. Denote MATH. MATH has components MATH and MATH. As an example of these definitions, consider the link shown in REF , which is the double of the tangle on one side of the NAME sphere. Double along the link to obtain MATH. MATH is the double of the NAME sphere. Cutting along MATH gives an atoroidal manifold which is the double of the tangle inside the sphere. Then MATH is two copies of the double of the tangle, the characteristic submanifold is a regular neighborhood of the tangle, and MATH is boundary of a neighborhood of the tangle. MATH is four copies of the tangle complement. Take a MATH-cycle MATH, such that MATH, and MATH . By the normal cycles lemma, we may assume MATH is normal. CASE: For each singular cell MATH, we can cut it along MATH, to obtain cells MATH, where MATH(see REF ). Then we get a cell cycle MATH . Since CITE= REF , MATH is locally degree one. So MATH is also locally degree one. We choose a decomposition of each face of MATH into triangles, without introducing any new vertices. CASE: Choose a component MATH of MATH. Then MATH injects into MATH. Take the cover MATH of MATH corresponding to MATH. Then there is a lift MATH, which is a homotopy equivalence since it is an isomorphism on fundamental group (we will call the image of the lift MATH too). There are simply connected complementary pieces for each component of MATH. So there is a retraction MATH crushing each component of MATH to its corresponding component of MATH. The cell chain MATH has preimage a locally finite chain MATH in MATH. Since MATH is compact, we can get a finite chain MATH by taking only the cells of MATH which intersect the lift of MATH. Then we project MATH to MATH by the retraction MATH. CASE: By NAME 's geometrization theorem, MATH has a geometric structure of finite volume with totally geodesic boundary. Crush each cusp of MATH to a point. This is equivalent to adding parabolic limit points to MATH. NAME call these parabolic points MATH, and the new manifold MATH with cycle MATH. Remark: The subsequent argument may be made without choosing a metric, but it makes some later choices canonical, and gives a better intuition for the argument, in our opinion. CASE: Now, we will straighten MATH inductively. This will be done in a certain order. First, we straighten interior edges. This will be done in a manner which keeps each end point of each edge on the boundary component on which it started (or keeps interior endpoints fixed). If the endpoint of an edge is mapped to a point MATH, then it will remain fixed at MATH, unless it was originally mapped to one of the two boundary components incident with MATH, in which case it can move around on that component. Each edge will be homotoped to a unique arc or cusp point in this manner: each pair of boundary components in MATH has a unique geodesic connecting them, or a unique tangent point in MATH. By the normalization, each interior edge has end points on distinct boundary components, when lifted to the universal cover, so no arc will be homotopic into a component of MATH. Next, homotope the edges in the boundary to be straight, keeping the endpoints fixed. Do this in such a way that edges which homotope to the same geodesic have the same parametrization. Finally, homotope the triangles in each face in the boundary (chosen at the end of REF ) to be straight. We may choose these so that parametrizations of edges are canonical, for example, by projecting down from the Lorentzian model of MATH. Call the resulting cell cycle of MATH. We need to show that MATH is degree one on MATH. The chain MATH is locally degree one on the interior of MATH, since MATH and MATH are locally degree one. Then MATH is a locally degree REF chain on MATH. MATH has support in MATH. When we retract MATH to MATH by the map MATH, then all the edges of MATH will lie inside MATH. All edges connecting points of MATH will lie in MATH, since we chose MATH to include every cell which intersects MATH. So when we crush MATH to MATH, and straighten, MATH will remain in MATH, so MATH will remain degree one in MATH. CASE: Many triangles in MATH will have collapsed to edges or points under this straightening process. The set of all collapsed triangles forms a degree REF subcycle of MATH, so we may eliminate it and retain a degree REF cycle in MATH. We will call this new cycle MATH. CASE: We need to show that each triangle of MATH contributes at most once to MATH. That is, we took the preimage of the cycle MATH in MATH to get MATH in REF . A particular triangle will have many preimages in MATH, for each MATH, and we need to show that at most one preimage contributes non-trivially to MATH. To see this, first notice that the position of each triangle after straightening REF is determined by the interior edges of the cycle MATH (the original cut up cycle from REF ) to which it was attached. Take a particular triangle MATH in MATH which has edges MATH attached to it in the particular cell of MATH in which it lies. Now, consider the lift of MATH to MATH, the universal cover of MATH. Suppose that MATH contributes twice to MATH. Let MATH be the preimage of MATH in MATH. Then a lift of MATH to MATH is homotopic into two different components MATH and MATH of MATH. There is some infinite strip or plane component MATH of the preimage of MATH (the pared locus of MATH) in MATH which separates MATH and MATH. If MATH is a plane, then MATH can contribute only to the MATH which lies on the same side of MATH as MATH does. So MATH is an infinite strip, covering an annulus in MATH. MATH will be incident with two components of MATH, MATH and MATH. We may assume that the endpoints of MATH all lie on distinct components of MATH, otherwise MATH will always collapse. Let's say that MATH lies to the left of MATH, and MATH lies to the right of MATH. If MATH lies on a component of MATH to the left of MATH, then it will be crushed to a cusp in MATH when we straighten. Symmetrically for the right, so MATH must lie on MATH (or MATH ) if it is to contribute to both MATH and MATH. We will assume MATH is on MATH. There are some cases to consider now: CASE: MATH ends on MATH, MATH and MATH end on other components of MATH to the left of MATH. In this case, all the edges will be homotopic to MATH in MATH, so MATH will not contribute to MATH. CASE: MATH ends on MATH, MATH ends on a boundary component on the left of MATH and MATH ends on a boundary component on the right of MATH. In this case, MATH and MATH will be collapsed to MATH in MATH, so MATH will not contribute to MATH (or to MATH). CASE: MATH and MATH end on boundary components to the left of MATH and MATH ends on a boundary component on the right of MATH. In this case, the ends of MATH and MATH will be sent to MATH in MATH, and the edge of MATH connecting MATH and MATH will collapse in MATH, so MATH does not contribute (see REF ). CASE: MATH and MATH end on boundary components to the left of MATH. In this case, all the ends will be mapped to MATH in MATH, and MATH will not contribute to MATH. So we see that each triangle can contribute in only one way to the cycle MATH. CASE: Here is a way to compute the number of triangles from a polyhedron MATH which contribute to the collapsed cycle MATH. MATH will have faces which came from MATH, called MATH, and new faces which came from MATH, called MATH. Collapse each quadrilateral and triangle in MATH to an edge by collapsing the edges adjacent to MATH - this collapsing is compatible with the straightening of edges which we did in REF (see REF ). Now, MATH will consist of truncated triangle faces after the collapsing (some MATH may collapse to line segments, which we throw out). In REF , each face of MATH was divided up into triangles, and after collapsing MATH, some of the boundary of these triangles will collapse, so we get rid of them, since they will contribute to the collapsed cycle MATH. Now, each triangle of MATH which contributes to the cycle MATH (from REF ) must come from one of these triangles left over. So we need to estimate how many of these triangles there are. Collapse the faces of MATH to points. We then get a set of triangulations of boundaries of balls, one for each MATH (except for the segments which we got rid of). Each new vertex corresponds to a cell which is divided up into triangles of the collapsed cycle MATH constructed in REF . If the vertex has degree MATH, then it contributes at most MATH triangles to MATH. The union of triangulations has the same number of faces as the original triangulation of MATH, but there are more components than the original. We may as well assume every vertex contributes, since this is the maximal possible case. Let there be MATH vertices, where vertex MATH has degree MATH, MATH edges, and MATH components to the triangulation. Then MATH, by euler characteristic. So there are at most MATH triangles contributing to MATH from the polyhedron MATH. CASE: The cycle MATH is locally degree one. In the metric on MATH, each triangle has area at most MATH. We have MATH . Letting MATH go to zero, then MATH . |
math/9906182 | Since MATH is a handlebody, it is atoroidal. So the only NAME pieces in the characteristic sub-pair of MATH must be solid tori. Suppose the characteristic MATH-bundle of MATH has MATH. Then there is a sub-bundle which is MATH, where MATH is a pair of pants. First, notice that the disks MATH must be MATH-incompressible, that is there is no MATH-compression with one boundary arc on MATH and the other boundary arc on MATH. If not, then the MATH-compression would represent a separating vertex of MATH. But the graph MATH is REF-connected, since it is REF-connected, and has at least REF vertices, a contradiction. This is REF. Look at the intersection of MATH with the product pair of pants MATH. Make MATH minimal. We need to show that MATH consists of product rectangles. The intersection will have no closed curves, because MATH is incompressible, and MATH will consist of product lines, by the MATH-incompressibility of MATH. Suppose that some component MATH of MATH is not a product rectangle in MATH. Then there are two arcs of MATH running over MATH. Taking a path connecting these arcs on MATH, we see that this path is homotopic rel endpoints into MATH, so MATH has a boundary compression by a well known argument, which we outline. Make the homotopy boundary compression transverse to MATH, with no fake branching. Then an innermost arc of intersection on the boundary compression gives a disk with interior disjoint from MATH. The loop theorem allows us to replace it with an embedded MATH-compressing disk. So there is a MATH-compression of MATH inside MATH. But since MATH is MATH-incompressible in the complement of MATH, when we surger along the MATH-compression, we get something isotopic to MATH, with smaller intersection with MATH, a contradiction. So MATH must be a product rectangle. If we cut MATH along MATH, we must get a set of disks MATH, otherwise there would be a simple closed curve in the sphere which compresses MATH, a contradiction. Call one of these disks MATH. MATH intersects MATH in a collection of at least three rectangles. Each of these rectangles intersects two vertices of MATH and cuts the graph into two pieces. Since MATH is REF-connected, the part of MATH lying to one side of each rectangle must contain no vertices of MATH. So we must see a picture like REF . Thus, MATH must meet every vertex in MATH. This picture is impossible, since it is not REF-connected. |
math/9906182 | Let MATH. If MATH, then MATH is a union of MATH. The MATH-bundle part of the characteristic sub-pair of MATH must have planar base surface, since MATH is a REF-sphere. Also, it must be a product bundle, since otherwise there would be a properly embedded MATH-bius band in MATH, contradicting that MATH is a MATH-homology ball. The other components of the characteristic sub-pair are solid tori, such that the meridian meets MATH at least REF times. We will denote these pieces by MATH, where MATH. Fill in the components of MATH to get a book of MATH-bundles. Notice that MATH is a connected surface, since its complement in MATH is a union of disks. Take a path MATH in MATH which connects points MATH and MATH on opposite sides of a product piece of MATH, with the property that this path hits the annuli in MATH a minimal number of times. This number must be MATH, since the piece is a product. Consider the first annular piece of MATH which the path crosses. It will first cross the boundary of an annulus MATH, then an annulus MATH, and then proceeds back into the product part. The path must hit these components of MATH and MATH at most once, otherwise we could find a path hitting MATH fewer times. Now, suppose the path crosses the other component of MATH at some point. Then we could find a subpath which connects opposite points of MATH and which intersects MATH fewer times (see REF ). So we may assume that MATH hits MATH at most once. Make a closed path MATH by connecting MATH and MATH by a path in the product part (see REF ). Then we have a closed path in MATH which hits the annulus MATH once, contradicting that MATH is a MATH-homology ball. Thus, MATH. |
math/9906182 | By the lemma, MATH, so MATH. |
math/9906182 | Put a partial order on subsurfaces of MATH, such that if MATH are subsurfaces of MATH, then MATH if and only if MATH is isotopic to a subsurface of MATH. Let MATH be the union of components of MATH with MATH. Let MATH. For a collection of curves MATH in MATH, let MATH represent the set of homotopy classes of the curves. MATH. By a result of NAME, MATH is NAME minimizing in MATH CITE. If MATH, then MATH, where MATH, and MATH is a union of annuli. Then MATH is a surface homologous to MATH with smaller NAME norm, a contradiction. Similarly, MATH. (Note: NAME 's result simplifies this argument, but is not mandatory.) MATH is a decreasing sequence, and for each MATH, either MATH, or MATH. MATH can be isotoped so that MATH is incompressible, and divides each component of MATH into product pieces (see NAME, IX. REF). The union of first pieces of each component is a product from MATH to MATH. By the characteristic submanifold theory, this product part is isotopic into MATH, so since each component has MATH, it must be isotopic into MATH (they couldn't be isotopic into a solid torus component of MATH) REF . Thus MATH. By an isotopy, we may assume that MATH. Suppose that MATH. Then MATH is a disjoint union of annuli. For each curve in MATH, the other boundary curve in the annulus in MATH must be a curve in MATH, otherwise MATH would have an annulus component, contradicting that each component has MATH. So we have MATH. Now, assume that MATH and MATH. Then MATH is isotopic to MATH. We can isotope MATH so that MATH dissects MATH into product pieces. Let MATH be the union of these pieces which have a boundary component on MATH. Then MATH is a product from MATH to MATH. MATH can be isotoped into MATH REF . So MATH will be a subsurface of MATH. MATH is homeomorphic to MATH. Then as with MATH, MATH is isotopic to MATH. Thus, MATH. Since MATH, we may take each component of MATH to a parallel component of MATH. Then MATH gives a product from MATH to MATH. So we may map each component of MATH to a component of MATH which cobounds an annulus with it, coming from the boundary of the product. Then this component of MATH corresponds to a component of MATH, since MATH. So we have a map from MATH. Some iterate of this map has a fixed curve. The sequence of annuli connecting the iterates of this curve give a torus in MATH which is non-trivial, contradicting that MATH is hyperbolic. Thus we see that if MATH, then MATH, which is equivalent to the last part of the claim. The sequence MATH can have length at most MATH. We can assume that we have nested sequence of surfaces MATH in MATH, such that for each MATH, either MATH, MATH. Let MATH. Then MATH. Each curve of MATH can be in MATH at most once, since once a curve of MATH disappears from MATH, it never appears later in MATH, MATH. Also, MATH can increase at most MATH times. Thus, MATH. So MATH. |
math/9906188 | The vector representation of MATH on the second component of MATH defines an action of MATH on MATH that preserves the class of properly supported operators and the products of such operators. It also normalizes the group MATH of translations, and hence it maps MATH-invariant operators to MATH-invariant operators. This property extends right away to the action of MATH on families of operators on MATH, and hence MATH maps MATH isomorphically to itself. From the isomorphism REF, we see that MATH also maps MATH to itself. This gives an action by automorphisms of MATH on MATH, which is the sum of MATH and MATH. |
math/9906188 | Let MATH be the space of sections of the bundle of algebras MATH. If MATH is a flat MATH-space, then the isomorphism MATH follows, for example, from the results of CITE. In general, this local isomorphism gives MATH. Our assumptions imply that MATH is infinite dimensional, and hence its group of automorphisms is contractable, see CITE. Consequently, there is no obstruction to trivialize the bundle of algebras MATH, and hence MATH. We obtain MATH . The last part of lemma follows from the above results and from the natural isomorphism MATH, valid for any MATH-algebra MATH. |
math/9906188 | It is clear from definition that if we can find MATH with the desired properties, then MATH. Suppose now that MATH is elliptic and has vanishing analytic index. Using the notation MATH, we see that MATH is invertible between the indicated NAME spaces if, and only if, MATH is invertible as a bounded operator on MATH. We can hence assume that MATH and MATH. Because MATH satisfies MATH, by REF , we can use some general techniques to prove that the vanishing of MATH implies that MATH has a perturbation by invariant, regularizing operators in MATH that is invertible on each fiber. We fix an isomorphism MATH. We now review this general technique using a generalization of an argument from CITE. Let MATH be the algebra introduced in REF . We denote by MATH the unit of the matrix algebra MATH. Also, denote by MATH the closure of MATH in norm. Choose a sequence of projections MATH, MATH, such that MATH in the strong topology. Because MATH is invertible modulo MATH, we can find a large MATH and MATH such that MATH is surjective and MATH, for all MATH. Then MATH is not in the spectrum of MATH, and we can consider MATH, which by construction satisfies MATH. Consequently, MATH is a projection in MATH. Because MATH is invertible modulo MATH, MATH . Let MATH, which is also a projection, by construction. Moreover, MATH and hence both MATH and MATH are projection in MATH (for any algebra MATH, we denote by MATH the algebra with adjoint unit). REF gives that, by definition, MATH defines a MATH-theory class in MATH. From definition, we get then MATH . Now, if MATH, then we can find MATH such that MATH is REF equivalent to MATH. By replacing our original choice for MATH with MATH, we may assume that MATH and MATH are equivalent, and hence that we can find MATH with the following properties: there exists a large MATH and MATH such that, if we denote MATH, then MATH and MATH. Then MATH is in MATH (more precisely MATH) and is invertible. Consequently MATH is also invertible. But MATH is a perturbation of MATH, and hence also of MATH, by an element in MATH. Since MATH is dense in MATH, this gives the result. |
math/9906188 | Assume first that there exists a NAME group bundle morphism MATH. (In other words, there exists a smooth map MATH that is a morphism on each fiber.) Let MATH denote the kernel of this morphism and let MATH be obtained from MATH by the same deformation construction by which MATH was obtained from MATH. Then we obtain a smooth map MATH that is a group morphism on each fiber, and hence MATH . Moreover, all above isomorphisms are natural, and hence compatible with the morphisms MATH. Assuming now that the result was proved for all NAME group bundles of smaller dimension, we obtain the desired result for MATH using NAME 's NAME isomorphism in MATH-theory CITE, which in this particular case gives: MATH . This will allow us to complete the result in the following way. Let MATH be the open subset of MATH consisting of those MATH such that MATH has dimension MATH. From the Five Lemma and the six term exact sequences in MATH-theory associated to the ideal MATH of MATH, for each MATH, we see that it is enough to prove our result for MATH for all MATH. Thus, by replacing MATH with MATH, we may assume that the rank of the abelianization of MATH is independent of MATH. Consequently, the abelianizations of MATH form a vector bundle MATH on MATH. A similar argument, using the NAME exact sequence in MATH-theory and the compatibility of MATH-theory with inductive limits CITE, shows that we may also assume the vector bundle MATH of abelianizations to be trivial. Then the argument at the beginning of the proof applies, and the result is proved. |
math/9906188 | For continuous families MATH that are multiplication operators on the boundary MATH, the degree is a local quantity - it depends only on the principal symbol - so we can follow word for word CITE to prove that MATH . When MATH, using MATH, we get MATH . This completes the proof. |
math/9906188 | We cannot use REF directly because our family MATH does not consist of multiplication operators on the boundary. Nevertheless, we can deform MATH to a family of operators that are multiplication operators at MATH, for suitable MATH. We now construct this deformation. Let MATH be a vector bundle over MATH. We consider classical symbols MATH whose support projects onto a compact subset of MATH. Let MATH . First we need to define a nice quantization map MATH. To this end, we proceed as usual, using local coordinates, local quantization maps, and partitions of unity, but being careful to keep into account the extra structure afforded by our settings: the fibration over MATH and the action of MATH. Here are the details of how this is done. Fix a cross section for MATH and, using it, identify MATH with MATH as MATH-spaces. Denote by MATH the natural projection. We cover MATH with open sets MATH that are diffeomorphic to open balls in a Euclidean space such that MATH and MATH. We also cover MATH with open domains of coordinate charts MATH. Then we let MATH, with MATH. The natural coordinate maps on MATH then give rise to a quantization map MATH where we identify MATH . Denote by MATH, MATH, and MATH the corresponding coordinate maps. Then MATH acts on MATH as MATH . Choose now a partition of unity MATH subordinated to MATH and let MATH . The main properties of MATH are the following: CASE: if MATH has order MATH, then MATH, modulo symbols of lower order; CASE: there exist maps MATH such that MATH if MATH and MATH; CASE: the maps MATH define a quantization map MATH where we regard MATH as defining a bundle of algebras, MATH, on MATH, first, and then on MATH, by pull-back. Similarly, we regard MATH as defining a bundle over MATH, which we then pull back to a bundle on MATH. (See also the discussion related to REF .) The deformation of our family is obtained as follows. Let MATH and MATH be the norms MATH and MATH. Define then MATH which are chosen to satisfy MATH. For any symbol MATH, MATH, we let MATH . We can define then MATH, MATH, which is the same as saying that MATH, and these operators will satisfy the following properties: CASE: For each fixed MATH, the operators MATH define a section of MATH over MATH and these sections depend smoothly on MATH (in other words, MATH depends smoothly on both MATH and MATH, in any trivialization); CASE: MATH, for all MATH; CASE: For each nonzero MATH and MATH, the limit MATH exists and is a multiplication operator; CASE: If MATH are such that MATH, MATH is homogeneous of order zero outside the unit ball, and if we define MATH and MATH, then there exists a constant MATH such that MATH and similarly MATH . For all MATH and MATH; CASE: All these estimates extend in an obvious way to matrix valued symbols. These properties are proved as follows. We first recall that, for any vector bundle MATH, we can identify the space of classical symbols MATH with MATH, the space of compactly supported functions on MATH, the unit ball of MATH, by MATH. For any quantization map, the norm of the resulting operator will depend on finitely many derivatives. Because we can extend MATH to a smooth function on the radial compactification of MATH, the first property follows. The second property is obvious. The third property is obtained using the same argument and observing that, for MATH, we can further extend our function to the radial compactification in MATH also. By investigating what this limit is along various rays, we obtain the third property. The fourth property is obtained using the following observation: there exist a constant MATH and seminorms MATH on MATH and MATH on MATH such that, for any symbols MATH, MATH being all derivatives in the symbolic directions (whose coordinates are denoted by MATH). Finally, the fifth property is obvious. We now turn to the proof of the formula for the degree of MATH stated in our theorem. We prove it by a sequence of successive reductions, using the facts established above. First, it is easy to see that MATH depends only on its principal symbol, and hence we can assume that MATH has order zero and MATH, where MATH. The above deformation can be used to prove our theorem as follows. Fix MATH large enough, and restrict the families MATH to the closed ball of radius MATH in MATH. For MATH large enough, all operators MATH, MATH are invertible, so the degree MATH of these families is defined and does not depend on MATH or MATH, provided that MATH is large enough. Since MATH, by definition, it is enough to compute MATH, for any given MATH. Choose then MATH arbitrary, and let MATH. Then the family MATH extends to a continuous family on the radial compactification of MATH, which consists of multiplication operators on the boundary. Moreover, the symbol class of MATH is nothing by the extension of MATH to the radial compactification in MATH and MATH (which is a manifold with corners of codimension two). We can use then REF to conclude that MATH . But MATH is homotopic to MATH through symbols that are invertible outside a fixed compact set, so MATH. We get MATH . To obtain the second form of our formula for MATH, and thus finish the proof, we proceed as at the end of the proof of REF , using MATH. |
math/9906188 | Let MATH be as above. The algebra MATH of MATH-families of pseudodifferential operators on MATH acting on fibers of MATH, contains as an ideal MATH, the space of families of smoothing operators that vanish to infinite order at the boundary of MATH. If MATH is an elliptic family, as in the statement of the lemma, and if MATH is large enough, then MATH, the indicial family of MATH, defines by restriction an element of MATH that is invertible modulo MATH. Recall that the boundary map MATH in algebraic MATH-theory associated to the ideal MATH of the algebra MATH gives MATH by definition. Also, the boundary map MATH in algebraic MATH-theory associated to the ideal MATH of the algebra MATH gives MATH. We want to prove that MATH. The desired equality will follow by a deformation argument, which involves constructing an algebra smoothly connecting the ideals MATH and MATH. Consider inside MATH the subalgebra of families MATH such that MATH for MATH. (In other words, MATH, if MATH, and MATH is arbitrary.) Denote this subalgebra by MATH. Also, let MATH be the set of families MATH, MATH, MATH, if MATH, MATH arbitrary such that the families MATH and MATH are in MATH, for all families MATH. It follows that MATH is a two-sided ideal in MATH and that the natural restrictions of operators to MATH and, respectively, to MATH, give rise to morphisms of pairs MATH . Moreover, the indicial family of the operator MATH gives rise, by restriction to larger and larger balls MATH, to an invertible element in MATH, also denoted by MATH. Let MATH be the boundary map in algebraic MATH-theory associated to the pair MATH. Then MATH and MATH. Since MATH and MATH are natural isomorphisms, our result follows. |
math/9906188 | Note first that we can deform the bundle of NAME groups MATH to the bundle of commutative NAME groups MATH as before, using MATH. Moreover, we can keep the principal symbol of MATH constant along this deformation. This shows that we may assume MATH to consist of commutative NAME groups, that is, that MATH is a vector bundle. The result then follows from REF . |
math/9906188 | Consider the map MATH . We need to show that this is the only invariant trace functional. In terms of indicial families, the infinitesimal generators of the MATH - action correspond to the multiplication operators with the functions MATH. Let MATH be the first homology group of MATH. It remains to show that the subspace of MATH spanned by MATH has codimension MATH. Indeed, the kernel MATH of the evaluation at MATH is the span of MATH. This proves the lemma. |
math/9906188 | This follows from the fact that the right hand side of REF is holomorphic in MATH and MATH for MATH. The same formula guarantees at most simple poles at MATH. Since MATH, in our case, the result follows. |
math/9906188 | The proof for MATH or MATH is the same as that for MATH and MATH, so we shall assume that we are in the latter situation. Also, by replacing MATH by MATH, we can assume that MATH. We first prove the lemma for MATH, that is for MATH. Denote by MATH the algebra of compact operators acting on MATH and by MATH the normed ideal of trace class operators. For any MATH, the product MATH is in MATH (here MATH denotes the completed projective tensor product). Also, because MATH is invertible and positive, the function MATH is differentiable, with bounded derivatives, and holomorphic in MATH, for MATH. Since MATH is continuous, it follows that the function MATH is differentiable, with bounded derivatives, and holomorphic in MATH for MATH. Since MATH is arbitrary, this proves REF for MATH. The last statement is an immediate consequence of REF , because MATH is holomorphic. Using now the fact that the lemma is true for MATH in the residual ideal, we may assume, using a partition of unity, that MATH and that the NAME convolution kernels of MATH are contained in a fixed compact set. Let MATH be the constant coefficient NAME on MATH and MATH, respectively. We define MATH. To prove the lemma for MATH, MATH, we shall first assume that MATH. Clearly, MATH. If MATH, for a symbol MATH, then MATH . This gives, by the standard calculus, that MATH for MATH . From the above relation, we obtain MATH . Using the asymptotic expansion of MATH in homogeneous functions in MATH and the substitution MATH, and the asymptotic expansion of MATH in powers of MATH at MATH, we obtain REF for this particular choice of MATH. We obtain REF directly from REF using REF . The case MATH arbitrary follows by writing MATH and observing that MATH is an entire function. |
math/9906188 | Assume MATH. The proof for arbitrary MATH is completely similar. The function MATH is a holomorphic extension of the function MATH, which is convergent for MATH large. The result is obtained then by integration in polar coordinates and by combining REF with REF of the above proposition. |
math/9906188 | It is clear that the restriction of MATH to a line MATH is in MATH, whenever MATH is in MATH. Moreover, we obtain isomorphisms MATH . These isomorphisms allow us to assume, using a partition of unity argument, that MATH. Using the fact that a symbol MATH restricts to a symbol in MATH, MATH, and the relation MATH if MATH, we see that the restriction of MATH to MATH is the indicial family of an operator in MATH, denoted MATH. Since MATH the isomorphism MATH follows. |
math/9906188 | The map MATH is obviously well defined and MATH-invariant, in view of the above lemma. In order to check the tracial property, we use the definition. Fix MATH of length one, arbitrarily, then MATH . Since MATH is a morphism, MATH is central, and MATH is a trace CITE, the tracial property of MATH follows. To complete the proof, we need only prove that MATH extends MATH, and this follows by integration in polar coordinates. |
math/9906188 | The function MATH is defined for all MATH, and it is seen to be holomorphic on MATH using the definition of MATH and REF , which also gives the existence of the desired holomorphic extensions. Finally, using again REF , we see that MATH and all the other functions in the stated equation coincide for MATH. Because they are holomorphic on a connected open set containing both MATH and MATH (MATH), they must coincide at MATH also. |
math/9906188 | The function MATH is holomorphic on the indicated domain by REF. Its poles are simple by REF and by the definition of MATH in terms of MATH, see REF . For the rest of the proof, it is enough to assume that MATH is independent of MATH. We set then MATH. For example, the proof of the fact that MATH is a trace and that it is independent of the choice of MATH is obtained from a standard reasoning, as follows. We first write MATH and observe that MATH is a holomorphic function vanishing at MATH. This shows that MATH vanishes on commutators. The independence of MATH on MATH is a consequence of MATH using that MATH is a holomorphic function vanishing at MATH. |
math/9906188 | By the definition of MATH using integration with respect to the orthogonal group, it is enough to prove the result for MATH. By definition, MATH is completely determined by MATH. This reduces our analysis to a lemma about integrals of functions in MATH. Fix MATH, and define MATH to be the space of functions MATH, MATH, MATH, with the following properties: CASE: MATH is smooth in MATH and holomorphic in MATH, for each fixed MATH; CASE: For any MATH there exist MATH, MATH, and complex valued functions MATH, and MATH satisfying MATH for MATH, and MATH for MATH, MATH, and MATH. CASE: MATH, MATH are holomorphic in MATH, MATH is holomorphic in MATH, for each fixed MATH, and MATH, for each fixed MATH in the strip MATH. Of course, the choice of MATH is not important in the above definition. A simple but crucial observation is that MATH if MATH. Moreover, for any MATH, with the coefficients of its canonical asymptotic expansion denoted MATH, MATH, and MATH, as in the above definition, we have MATH which gives MATH . The main idea is to prove that the familiar function MATH is in MATH, for MATH large. Then MATH. This is, of course, a refinement of REF . Now, by definition, MATH, for any MATH. For any such MATH, MATH is holomorphic in MATH, for MATH, and hence MATH (and MATH, but this will play no role in our reasoning), by REF . Consequently, MATH. Let MATH be the corresponding coefficients in the canonical asymptotic expansion of MATH. We know, by classical results, that the functions MATH and MATH have holomorphic extensions in MATH. Moreover, for MATH, the difference MATH is a polynomial in MATH, which shows that the coefficients of MATH in the asymptotic expansions of these two functions are the same. Consequently, MATH, for MATH. But by the definition of MATH, MATH is holomorphic in a neighborhood of MATH, and hence MATH has a holomorphic extension to a neighborhood of MATH, as claimed. Finally, using MATH, for MATH in a small neighborhood of MATH, and REF , for MATH, we obtain MATH . This completes the proof. |
math/9906188 | Since all the statements of the above theorem are statements about the local behavior in MATH of certain functions, we may assume that MATH is a flat MATH-space. This means, we recall, that MATH and MATH. Then we just repeat the proof of REF including an extra parameter MATH, with respect to which all functions involved are smooth. |
math/9906188 | If MATH is a flat MATH space, then this result follows right away from REF . In general, we can choose trivializations of MATH such that the transition functions preserve the metric on MATH, and hence the transition functions are in MATH. Because the isomorphisms of REF commute with the action of the orthogonal group, the result follows. |
math/9906188 | Everything in this proposition follows from the case when MATH is reduced to a point, except the independence of isomorphism MATH. For this we also use REF . |
math/9906188 | This follows by applying the above constructions to MATH, MATH elliptic, using REF . |
math/9906188 | The map MATH is onto because if MATH is invertible in MATH with inverse MATH, then its image MATH in MATH is also invertible with inverse MATH . (The sum is actually finite for our algebras.) The relations MATH are easily checked. From the naturality of the boundary map in algebraic MATH - theory, we obtain that MATH and hence MATH . This simple relation will play an important role in what follows because it reduces the computation of MATH to the computation of MATH. Lift MATH to an element MATH and its inverse MATH to an element MATH, as in the statement of the lemma. This gives for MATH (the inverse of the image MATH of MATH in MATH with respect to the MATH product) the explicit lift MATH . We now proceed by direct computation (as in CITE, for example), using the explicit formula MATH with MATH and MATH . (all products and powers are with respect to the MATH-product). Then, MATH . Then we notice that MATH, because MATH. This relation and its analogue obtained by switching MATH with MATH then give that MATH, and the lemma follows. |
math/9906188 | We have that MATH . Now we observe that MATH . Consequently, MATH for some holomorphic function MATH such that MATH. The result then is an immediate consequence of REF . |
math/9906188 | We shall denote MATH as before. Moreover, MATH and MATH will have the meaning they had before. We shall use REF . Let MATH be as in that lemma and evaluate the commutator MATH (with respect to the MATH product). We obtain MATH the sums being of course finite. We next observe that MATH and MATH because MATH is a graded trace on MATH (with the usual product). Using this we obtain from REF that MATH . This proves the first part of our formula. We now prove the second part of our formula. The commutator MATH maps to MATH in MATH, and hence MATH is in MATH. Consequently, MATH . Next we observe that MATH and hence MATH (defined in the canonical representation) is actually in MATH, in spite of the fact that MATH is not in the algebra MATH. Moreover, MATH is holomorphic at MATH. Using that MATH for all MATH such that MATH is not an integer, we finally obtain MATH where MATH. Since MATH is a holomorphic function in a neighborhood of MATH with MATH it further follows from REF that MATH because MATH is a closed graded trace. Putting toghether the above formulae, we obtain the desired result. |
math/9906188 | The proof is word for word the same, if we replace MATH with MATH, MATH with MATH, MATH with MATH, and so on. |
math/9906198 | For a fixed MATH consider the system of REF . By NAME 's theorem (see section four for a convenient form of this result), there is a nonempty NAME open set MATH of MATH where MATH and; MATH such that, if not empty, the zero set MATH of the above system of equations on MATH minus the set of common zeroes of MATH is smooth and of dimension MATH. Left multiplying this MATH vector of equations with an invertible MATH matrix MATH of the form MATH where MATH is a MATH matrix and MATH is a MATH matrix, results in the equivalent system of REF . Thus we can assume that the set MATH is invariant under this action by the matrices MATH. Thus we have a nonempty NAME open set MATH of MATH such that for each MATH the equivalent system is of the form MATH and has smooth nonsingular zeroes when MATH. Thus we have the first assertion of the Lemma. To see the second assertion of the Lemma, note that the solutions of MATH with MATH are naturally identified with the solutions of the system MATH . The second assertion follows now from the Algorithm in CITE. We prove the third assertion of the lemma by induction on MATH. If MATH, then it is a tautology that the solutions of MATH with MATH are the same as the solutions of MATH with MATH. So we can assume that the result is true for MATH where MATH. Note that a solution of MATH with MATH but MATH is a solution of the system MATH . Since the solutions of MATH with MATH are isolated and nonsingular, a generic choice of MATH will not be zero on any of the solutions. But this means that the solutions of MATH for generic MATH will have no solutions with MATH. |
math/9906198 | Let MATH denote the dimension of an analytic set MATH at the point MATH. Let MATH denote the zero set of MATH on MATH. Since there are MATH-functions, the dimension of each irreducible component of MATH is at least MATH, and in particular MATH . Since MATH is isolated we have that MATH. Since MATH is smooth, we have MATH . Thus we conclude that MATH . Thus the union MATH of the components of MATH passing through MATH has pure dimension MATH. It follows, for example, use CITE, that there are neighborhoods MATH of MATH and MATH of MATH such that the projection MATH of MATH to MATH is proper and finite. By shrinking MATH and MATH we can assume that MATH and that MATH. Now MATH is a local complete intersection, and thus since the map MATH is proper with finite fibers, we conclude by the discussion before the discussion of flatness, that this map is flat. Thus the direct image MATH of the structure sheaf of MATH is locally free, for example, CITE. On MATH, MATH is defined by the functions MATH, for MATH. By definition, at a point MATH we have MATH is MATH where MATH is the local ring of convergent power series on MATH centered at MATH and MATH denotes the ideal in MATH generated by MATH. The statement that MATH is locally free, is equivalent to ranks of MATH at different points of MATH being equal. The rank of MATH at a point MATH is by REF where MATH is the maximal ideal generated MATH consisting of convergent power series vanishing at MATH. Thus comparing to the definition of the multiplicity of the solution MATH of MATH being MATH, we see that the rank of MATH is MATH. Unwinding the definition of MATH, MATH for any fixed MATH is the direct sum of the modules MATH with index set the set of distinct points MATH with MATH. This proves the assertion of the Lemma. |
math/9906199 | It follows immediately from REF . |
math/9906199 | Let MATH be a maximal linearly independent subset of MATH. Fix MATH, and assume there exist non-zero constants MATH so that MATH . Let MATH be arbitrary. Applying the operator MATH in REF , it follows that MATH . Since MATH is linearly independent and MATH are non-zero, by REF we have MATH . Since MATH is arbitrary, MATH . Hence, MATH, and REF follows. |
math/9906199 | Given MATH, the map MATH is a homeomorphism. So MATH . Hence, we may assume that MATH. Reducing MATH if necessary, we may also assume that for some MATH, MATH . In particular, MATH, for MATH. It will then suffice to show the following: Claim: For every MATH and MATH, MATH. To prove the claim, let MATH and suppose the claim is true for MATH. Then for each MATH we have MATH since MATH. So given MATH, MATH . Thus, MATH in MATH, and MATH. So the claim holds. |
math/9906199 | Consider the function MATH defined by MATH where each MATH is the MATH-homogeneous polynomial defined by MATH . Now, since MATH is of exponential type, there exists MATH so that MATH . Given MATH with MATH and MATH, MATH and so MATH . Thus, MATH is entire (moreover, it is of bounded type), and non-constant, since MATH is non-constant. So the sets MATH are both open, non-empty. Hence, according to REF , MATH are both dense subspaces of MATH. Next, notice that if MATH, given MATH . By REF , MATH . Also, by REF there exists a (possibly discontinuous) linear map MATH determined by MATH which by REF satisfies MATH . By REF , MATH is hypercyclic. |
math/9906204 | This is very well known. See for instance REF . See also REF . |
math/9906204 | First observe that if MATH is contained in a line then MATH and all its subsets are complete intersections. The resolution of a complete intersection is well known. We let the reader check the theorem in this case. Thus we may assume that MATH is not contained in a line. Now we show that it is enough to prove the theorem for MATH. To do this, it is enough to show the following. Let MATH be a subset consisting of MATH points, such that MATH has truncated NAME function and MATH has the predicted rank, for any MATH. Assume that there is a subset MATH consisting of MATH points such that MATH has truncated NAME function, and such that for all MATH, the rank of MATH is what is predicted in the theorem if we take MATH and MATH. Then we have to show that this is the same rank that is predicted by the theorem if we had taken MATH and MATH. The fact that MATH has the predicted rank says that for all MATH, either MATH is surjective or MATH. By our assumption on MATH, we get that for all MATH either MATH is surjective or MATH. For those MATH with MATH surjective we are done. For those MATH with MATH we are done. This leaves only those MATH for which MATH but MATH. But if MATH then MATH is surjective. Furthermore, the NAME function of MATH in degree MATH is different from that of MATH in degree MATH. Since MATH has truncated NAME function, this says that MATH imposes MATH independent conditions on forms of degree MATH, and MATH imposes MATH independent conditions on forms of degree MATH. Hence MATH is surjective. Thus it is enough to prove the theorem for the case MATH. Part of the proof will be by induction on MATH. The reader can easily get this induction argument started by checking directly that the theorem is true for small values of MATH. Let MATH be the smallest positive integer such that MATH imposes MATH conditions on forms of degree MATH. Then MATH is generated in degrees less than or equal to MATH, REF . Let MATH be any subset with MATH points. Then MATH is also generated in degrees less than or equal to MATH. Thus for MATH the multiplication map MATH is surjective and thus satisfies the conclusion of the proposition. Next consider MATH. From now on, unless specified otherwise, we assume that our subset MATH has truncated NAME function. Then MATH imposes at most MATH conditions on forms of degree MATH. By REF , Z imposes the smaller of MATH and the number of conditions imposed by MATH on forms of degree MATH. Thus, MATH. This says that MATH and MATH are the same map. Certainly the conclusion of the proposition follows in this case. We are only left to consider the case MATH. We have to show that among subsets with cardinality MATH and with truncated NAME function, we can find one with the right number of minimal generators in degree MATH. Consider REF with MATH. Regardless of whether or not MATH has truncated NAME function, MATH has codimension one in MATH, and similarly for MATH. Let MATH be a basis for MATH and let MATH be a form of degree MATH in MATH, so that MATH, MATH is a basis for MATH. The proof breaks down into four cases according to the codimension of the image of MATH in MATH, in other words, the number of generators MATH needs in degree MATH. Note that if MATH is zero dimensional then MATH is one dimensional, so MATH is injective. We may assume that MATH has positive dimension. CASE: MATH is generated in degrees MATH. This says that MATH is surjective. We wish to show that MATH is also surjective, for any MATH (hence in particular one with truncated NAME function). Since MATH has codimension one in MATH we simply need to find a single form in the image of MATH not in the image of MATH. Let MATH be a linear form not vanishing on the single point of MATH. Note that MATH also does not vanish on the single point of MATH. MATH is certainly in the image of MATH, but not in the image of MATH because MATH does not vanish on all of MATH. CASE: The image of MATH has codimension one in MATH. We need to show that there is at least one subset MATH, with cardinality MATH and truncated NAME function, so that MATH is surjective. For this case we consider all subsets of MATH of cardinality MATH. Set MATH, MATH. Let MATH be a form of degree MATH in MATH. Note that MATH is well defined up to elements of MATH. One can see that MATH form a basis for MATH. Indeed, since MATH has codimension MATH in MATH there are the right number of them to be a basis, and any linear relation MATH would need to have MATH, MATH, since MATH but all the other MATH's and MATH's vanish at MATH. This would give a linear relation among the MATH's which is impossible because they form a basis for MATH. We only need to find one MATH such that MATH is surjective, and such that MATH has truncated NAME function. We will first argue that in this situation, if MATH is surjective then MATH must have truncated NAME function. Let MATH be a general linear form and let MATH be the corresponding ideals in MATH. Note that MATH is not a zero divisor on MATH or MATH. By slight abuse of notation, we will call the rings MATH and MATH the NAME reductions of MATH and MATH, respectively. MATH and MATH (respectively, MATH and MATH) have the same number of minimal generators, occurring in the same degrees. This is standard. See for instance CITE, p. CASE: MATH is equal to MATH in degrees greater than or equal to MATH if and only if MATH does not have truncated NAME function. Notice that MATH for all MATH, and notice that MATH in degrees MATH, so it is enough to prove that MATH in degree MATH if and only if MATH does not have truncated NAME function. Consider the NAME functions of MATH and of MATH: MATH . Since MATH we have MATH for all MATH. We also have MATH and MATH. It follows that for one value of MATH, say MATH, we have MATH, and for all other MATH we have MATH. Since MATH has truncated NAME function if and only if MATH, this completes the proof of the claim. It follows from REF that if MATH does not have truncated NAME function then it is impossible that MATH has a minimal generator in degree MATH but MATH does not have a minimal generator in degree MATH. So if we prove the existence of a MATH with no minimal generator in degree MATH then the truncated NAME function will follow automatically. Suppose that MATH is never surjective. Since MATH, MATH, and by REF we see that for every MATH the kernel of MATH must have dimension at least two larger than the dimension of the kernel of MATH. That is, there must be two degree one relations of the form MATH . These relations must be linearly independent of each other and no linear combination of the two of them can involve only MATH's and not MATH. From this one can see that all MATH of these relations (as you vary MATH) are linearly independent elements of the kernel of the multiplication map MATH which remain independent modulo the kernel of MATH. Using our assumption on the codimension of the image of MATH in MATH we conclude that this image has codimension MATH in MATH. Comparing MATH with the multiplication map MATH we see that MATH. However, from the previous paragraph we know that the dimension of the kernel of MATH is at least MATH larger than the dimension of the kernel of MATH. Counting dimensions we get that MATH is not surjective. But, it is a well known triviality that MATH is surjective. This contradiction finishes REF . CASE: The image of MATH has codimension two in MATH. This will involve quite a bit more work and will be done in REF. CASE: The image of MATH has codimension MATH in MATH. Let MATH, MATH, MATH, MATH and MATH be as in REF . The codimension of the image of MATH in MATH is MATH. We want to show that there is a MATH with truncated NAME function, such that MATH has MATH minimal generators in degree MATH. By REF , if MATH does not have truncated NAME function then MATH in degrees MATH. Hence MATH and MATH have the same number of minimal generators in degree MATH, and by REF , the same is true of MATH and MATH. So just as in REF , it is enough to prove the existence of a MATH with the right number of minimal generators, and it will automatically have truncated NAME function. The proof will be by induction on MATH. Hence we can assume that the theorem is true for all the MATH, but suppose that it fails for MATH. In this case we conclude that for each MATH we have at least one degree one relation of the form MATH. If there were always two or more such relations we could arrive at a contradiction as in REF , so assume for MATH there is only one such relation. As indicated above, we may assume that the theorem holds for MATH. Thus we can find MATH, MATH such that MATH satisfies the conclusion of the theorem with respect to MATH. A basis for MATH consists of MATH and a basis for MATH consists of MATH. The relations MATH for MATH say that the codimension of the image of MATH in MATH is exactly MATH (because we assumed only one such relation) and the codimension of the image of MATH in MATH is at least MATH. But the assumption that MATH satisfies the theorem says that the codimension of the image of MATH in MATH is MATH. This contradiction finishes REF . |
math/9906204 | By REF , there is at least one subset MATH whose NAME function is the truncation of that of MATH. We will find our desired MATH from among these subsets, so from now on we will assume that this is the NAME function of MATH. Then we have that the ideal of MATH agrees with that of MATH in degrees MATH and MATH and MATH both impose independent conditions on curves of degree MATH. It follows that MATH . We are assuming, furthermore, that MATH has precisely two minimal generators in degree MATH. We need to show that MATH can be chosen with no minimal generator in degree MATH. The assumption about the dimension of the zero locus means that MATH has a GCD, MATH. Let MATH be the degree of MATH. By abuse of notation we will use MATH both for the curve in MATH and for the polynomial. MATH. We will use ideas from REF (closely related to work of CITE). Let MATH be the subset of MATH lying on MATH and let MATH be the subset not lying on MATH. We have MATH and MATH, which is already saturated. For MATH we have MATH . Notice that MATH. From this we deduce two things. First, MATH imposes independent conditions on forms of degree MATH since MATH does on forms of degree MATH. Second, MATH. Let MATH be the ideal generated by MATH. We have just seen that MATH. In degree MATH we have the inequality MATH, where the failure to be an equality is measured by the number of minimal generators of MATH in degree MATH. Let MATH be the corresponding NAME function and consider MATH. From the above considerations, one can check that MATH . On the other hand, since MATH has two minimal generators in degree MATH, we have MATH. This gives MATH and this proves the claim. If MATH then MATH consists of exactly MATH points on MATH. If MATH then MATH consists of exactly MATH points on MATH. Let us collect the following facts. CASE: The initial degree of MATH is MATH. CASE: MATH since MATH imposes independent conditions on forms of degree MATH (MATH). CASE: MATH. CASE: MATH. This follows because we are assuming that MATH has two minimal generators in degree MATH. It can be seen, for example, by applying REF , since in our situation certainly MATH is contained in a complete intersection of type MATH with MATH. CASE: MATH. CASE: MATH. CASE: MATH. If one now considers the NAME function of the NAME reduction of MATH (that is, the function given by MATH) and applies REF , in the case MATH (respectively, MATH) one gets from the above facts that MATH (respectively, MATH) as claimed. We consider the cases MATH and MATH separately. Suppose that MATH consists of MATH points on either a smooth conic or else a union of two lines. In the latter case, either one point lies at the intersection of the two lines or else there are MATH points on one line and MATH points on the other. (Otherwise MATH fails to impose independent conditions on forms of degree MATH.) In any of these cases, the removal of a suitable point MATH leaves MATH points which form the complete intersection of MATH and some curve MATH of degree MATH. Let MATH be the subset of MATH obtained by removing this point. We know that there exists an element of MATH which is not in MATH. We first claim that such an element must meet MATH in finitely many points. Certainly the base locus of the linear system MATH cannot contain all of MATH since then it contains the deleted point MATH, contradicting the fact that MATH imposes independent conditions on forms of degree MATH. Hence the assertion is clear if MATH is irreducible. Suppose that MATH is reducible and suppose (without loss of generality) that MATH is in the base locus of MATH. Then any element of this linear system consists of the product of MATH with a homogeneous polynomial of degree MATH containing the remaining points of MATH. By construction, the remaining points include MATH points on MATH, so in fact all of MATH is in the base locus, a contradiction. Hence without loss of generality we may assume that the subset of MATH lying on MATH is the complete intersection of MATH and a form MATH (that is, MATH contains all of MATH). Now, if MATH form a basis for MATH and MATH form a basis for MATH, then any linear relation MATH implies MATH since no factor of MATH is a factor of MATH, and MATH. The conclusion follows from this fact. We now turn to the case MATH. We have that MATH is the union MATH, where MATH consists of MATH points on the line MATH. Since MATH has a zero-dimensional base locus, a general element, MATH, of this linear system does not contain MATH as a component. Hence in particular MATH is the complete intersection of MATH and MATH. Also, in particular we have that MATH. We now note that MATH is a liaison addition (compare CITE, CITE)! Hence its ideal is of the form MATH . Since MATH has two minimal generators of degree MATH, clearly MATH must be one of these and MATH must have exactly one minimal generator in degree MATH (which is the maximum possible degree). Let MATH be the subset of MATH obtained by removing a point, MATH, of MATH. Let MATH be the subset of MATH obtained by removing MATH. Exactly as above, MATH is a liaison addition, since the linear system MATH does not have all of MATH in its base locus. Its ideal is of the form MATH where MATH, MATH and MATH form a complete intersection. We want to show that MATH can be removed in such a way that MATH has no minimal generator in degree MATH. The following are equivalent: CASE: MATH has a minimal generator in degree MATH. CASE: MATH has a minimal generator in degree MATH other than MATH. CASE: MATH is not a minimal generator for MATH. The equivalence of REF is clear since we have already observed that MATH has exactly one minimal generator in degree MATH. The fact that REF implies REF follows from the equation MATH. Now assume REF , and let MATH be the minimal generator in MATH other than MATH. We want to show that MATH is a minimal generator for MATH. Suppose not, and let MATH be a basis for MATH. Then we have MATH or equivalently MATH . Since MATH does not divide MATH, we get that MATH up to scalar multiples, contradicting the assumption that MATH is a minimal generator for MATH other than MATH. As a result of REF we have in particular that MATH has a minimal generator in degree MATH if and only if MATH is not a minimal generator of MATH. We want to show that we can find the subset MATH with no minimal generator in degree MATH. We thus have to show that among the MATH collinear points of MATH, there is at least one point MATH whose removal leads to a MATH as above which is not in the image of MATH. We will do this by contradiction. Let MATH be the points of MATH. For each MATH, MATH, let MATH such that MATH does not contain MATH as a factor, as was done with MATH above. Note that MATH are linearly independent in MATH, as was done in REF in the previous section. We want to show that it is impossible for MATH to all be in the image of MATH. Suppose otherwise. Note that MATH is a non zero-divisor of MATH, and let MATH . We may view MATH as the NAME reduction of MATH, and in particular MATH has a minimal generator in degree MATH since we saw that MATH must have a minimal generator in degree MATH. On the other hand, if we let MATH be the image of MATH in MATH, the same argument as above gives that the MATH are linearly independent in MATH. Note that MATH, so the MATH form a basis. If the MATH are all in the image of MATH then none of the MATH is a minimal generator of MATH, so MATH has no minimal generator in degree MATH, a contradiction. This concludes the proof of REF . |
math/9906204 | We use the NAME Theorem REF . The first difference function for the NAME function of the complete intersection MATH is: MATH with MATH repeated MATH times. The first difference function for the NAME function of MATH does not reach MATH until degree MATH. Since MATH, when we subtract these and read backwards to get the first difference function for MATH we see its maximum is less than MATH. |
math/9906205 | This follows immediately from the explicit construction of the completion in REF. Alternatively, it can be deduced from the universal property as follows. Consider the subspace MATH of MATH spanned by the sets of the form MATH. Endow it with the bornology MATH of all subsets of sets of the form MATH. Verify that MATH is a complete bornological vector space and show that the universal property of MATH implies that MATH has the universal property as well. |
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