paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010066 | For MATH, we already know this from REF , so fix some MATH. Let MATH be a partition of MATH into positive integers. We can compute REF for each such partition. The maximum value is found to be achieved for MATH and MATH for MATH and for MATH; for MATH and MATH for MATH; and for either MATH and MATH or MATH and MATH for MATH and MATH. Applying REF completes the proof. |
math/0010066 | Define MATH, MATH, etc. analogously to the definitions of MATH, MATH, etc. in the proof of REF . Then MATH for MATH except that MATH, MATH, and, for MATH, MATH (compare CITE); therefore, MATH and MATH . As in REF , MATH . Both numerator and denominator are analytic except for a pole at MATH, so using singularity analysis CITE there we get MATH pointwise in MATH as MATH, and by CITE, the generating function of the limiting probabilities is MATH. |
math/0010067 | Given an irredundant primary decomposition of MATH, the ideal MATH is the intersection of the minimal primary components. (See REF; compare REF.) The ideal MATH is the intersection of the primary components contracting to zero. Thus MATH is the intersection of the minimal primary components and those contracting to zero. |
math/0010067 | We may assume that MATH is affine. Thus MATH corresponds to a homomorphism MATH from a discrete valuation ring to an equidimensional ring. Let MATH be the fiber ring, and let MATH be a uniformizing parameter. If MATH is not flat, then MATH is a zero divisor and thus is an element of some associated prime MATH of MATH. Since MATH does not contract to zero, it must be minimal. Thus MATH. |
math/0010067 | Assume REF , but that MATH is not flat over MATH. We will first show that MATH does not specialize to MATH, and subsequently that the NAME class fails to specialize (that is, that REF does not hold). Since MATH is equidimensional, so is MATH. And since MATH is internally flat, so is the morphism MATH. Thus by REF , MATH and thus the dimension of MATH exceeds that of MATH. This tells us that MATH is properly contained in MATH. Now we turn to NAME classes. In view of REF, the NAME class of the normal cone may be calculated using MATH where MATH denotes projection. Because MATH is a proper subscheme of MATH, the blowup MATH is likewise a proper subscheme of MATH. Both of these schemes are equidimensional and have the same dimension, namely, MATH. Hence the difference of fundamental classes MATH must be a positive cycle of dimension MATH. Moreover, since the two cycles comprising MATH can only disagree over the center of the blowup, MATH must be supported on MATH. The exceptional REF divisor MATH on MATH restricts to MATH on MATH. Thus MATH . (Note that the first cap product takes place on the blowup, and the second in MATH.) Thus, by REF , the difference between the NAME classes of MATH and MATH is given by MATH where MATH is projection. To show that this cycle class is nonzero, we consider an arbitrary component MATH of MATH. Set MATH. Then MATH if MATH, while MATH for some positive integer MATH. (Note that the dimension of MATH is MATH and that MATH, so that MATH.) Now suppose MATH, MATH, , MATH are those components of MATH for which MATH achieves its minimum value MATH. Then MATH and MATH will differ in dimension MATH by a positive linear combination of MATH, MATH, , MATH. Since MATH is projective, this linear combination cannot be the zero class. |
math/0010067 | Let MATH be an elliptic curve. Then the morphism MATH (projection onto the first factor followed by MATH) is a flat projective morphism from an equidimensional scheme. The subscheme MATH, embedded in MATH via MATH, is nowhere dense in MATH, and its normal cone MATH is internally flat over MATH since MATH is. The NAME class of MATH is obtained from that of MATH by pullback via the projection of MATH. Similarly, MATH is obtained from MATH by pullback via the morphism MATH. Therefore, the NAME class MATH specializes to MATH. By REF , we see that MATH is flat over MATH and MATH. Now MATH is naturally isomorphic to the product of the projective completion of MATH with MATH. Hence MATH is flat over MATH and MATH. Thus MATH is flat over MATH and MATH. |
math/0010068 | We apply the NAME decomposition to split the expression inside the projection MATH in REF as MATH . As it turns out, all the terms except for REF will be of the form MATH. We first consider the high-frequency contributions REF. We can rewrite REF as MATH . To show that the contribution of this term is MATH, it thus suffices by REF and the triangle inequality to show that MATH . We use NAME to split MATH into two MATH norms and apply REF to estimate the left-hand side by MATH which by REF is bounded by MATH . But this is acceptable from our choice of MATH and the fact that MATH. Now we consider REF. By symmetry it suffices to consider the contribution when MATH. In this case we may assume that MATH since the contribution to REF vanishes otherwise. By the triangle inequality it thus suffices to show that MATH . We use NAME, splitting MATH, MATH, MATH, and use REF and decomposition into projections MATH to estimate the left-hand side by MATH which is acceptable by the same calculation used to treat REF. Now consider REF. We may assume that MATH since the contribution to REF vanishes otherwise. We can thus simplify REF as MATH . By NAME it thus suffices to show that MATH . But this is immediate from REF, and (a trivial modification of) CASE: The contribution of REF to REF is always zero, so we turn to REF. In light of REF and NAME it suffices to show that MATH . But this is immediate from REF, and a breakdown into projections MATH. The terms REF are equal. We have thus shown that MATH . Since MATH, it only remains to show the commutator estimate MATH where MATH is the matrix MATH and MATH is the function MATH. From REF and summing over NAME pieces we have MATH . A similar argument using REF gives MATH . Combining these together with REF we obtain MATH while from REF we have MATH . Thus to finish the proof of this Proposition it suffices to use the standard commutator estimate (with MATH, MATH, and MATH) We have MATH for all smooth functions MATH, MATH on MATH and all MATH such that MATH. For a previous application of this type of lemma to wave maps, see CITE. We begin with the identity MATH and use the Fundamental theorem of Calculus to rewrite this as MATH . Since MATH is integrable, the claim then follows from NAME and NAME. |
math/0010068 | We first observe that REF is easy to show if the derivative is moved to the low frequency term: MATH . Indeed, we simply place MATH in MATH using REF and MATH in MATH using REF. From this and the product rule it thus suffices to show that MATH . Consider the expression MATH. Heuristically, this quantity is approximately MATH, by the same type of reasoning used to obtain the linearization REF. On the other hand, since MATH lies on the sphere, MATH. Accordingly, we shall rewrite the above estimate as MATH . We now split MATH as MATH plus other terms which vanish when MATH is applied. The first term of the above expression can be refined to MATH since all the other components vanish after applying MATH. The contribution of this term to REF can thus be estimated using the product rule and NAME by MATH . Applying REF, this can be bounded by MATH which is acceptable by REF and our choice of MATH. The other two terms are equal to each other. It thus remains to show the commutator estimate MATH where MATH. Since the expression inside the MATH has NAME support on MATH, we may discard the derivative MATH. By REF we may estimate the left-hand side of this by MATH and this is acceptable by the argument used to treat REF (as the commutator estimate has effectively moved the derivative from the high-frequency term to the low-frequency term; compare the ``NAME trick in CITE, CITE, CITE). |
math/0010068 | As noted in the introduction, we have the identity MATH whence MATH for all MATH. We now show inductively that MATH for all MATH. This is clearly true for MATH. Now suppose that MATH and the claim has been proven for all smaller MATH. Then from REF and NAME we have MATH for all MATH. By REF, and REF we thus have MATH and the induction REF can thus be closed by REF, if MATH is sufficiently small. From the above analysis we see that the first part of REF obtains, as does the first part of REF. In particular MATH, and thus MATH, are invertible, and this gives the second part of REF. We now show REF; the second part of REF will then follow from REF, and the product rule. We shall again use induction, showing that MATH and MATH for all MATH and some sufficiently large absolute constant MATH. The claim is trivial when MATH. Now suppose that MATH and the claim has been proven for all smaller MATH. By differentiating REF and using NAME, we obtain MATH and MATH . By applying REF, and the induction hypothesis we thus see that MATH and MATH . If MATH is sufficiently small and MATH is sufficiently large depending on MATH then one can close the induction hypothesis. This gives REF. Next, we prove REF. We can write MATH as MATH where MATH . We thus have the telescoping identity MATH . To estimate MATH, we use REF to obtain MATH . This term is thus acceptable by NAME in time if MATH is sufficiently large depending on MATH, MATH, MATH. By REF and the triangle inequality it thus suffices to show that MATH for all MATH. We expand out MATH and MATH . In both expressions, the dangerous terms REF occur when the derivative falls on a high frequency term MATH instead of a low frequency term such as MATH, MATH, MATH. (Indeed REF is the only reason why MATH fails to be in MATH, and is the only reason why we need a renormalization by MATH in the first place). Fortunately, we have chosen MATH so that the dangerous terms REF cancel each other. From the triangle inequality it thus suffices to show the bounds MATH . The first term in REF is acceptable by REF for the first two factors (dyadically decomposing the latter factor and using REF) and REF for the last. The second term is acceptable by REF for the first factor, REF for the second, and REF for the last. The third term is treatable by the same argument as the first term. Finally, the fourth term is acceptable by REF for the first term, REF for the second term, and the estimate MATH which can be proven from REF and estimating MATH using REF, MATH using REF, and MATH using REF. The only remaining estimate to prove is REF. In principle this is the same type of estimate as REF, but there is a minor complication arising from the double time derivative in MATH, which are not directly treatable by the MATH norms. To get around this we will have to use REF . More precisely, we shall need For all MATH, we have MATH . Morally speaking this estimate obtains from REF if we treat time derivatives like spatial ones. We could have modified our NAME operators to project in time as well as space in order to make this heuristic rigorous, but this creates other difficulties having to do with time localization which we wished to avoid. We shall show REF for MATH to simplify the exposition; the reader may verify that the argument below is scale invariant and thus extends to all MATH. Applying REF, we see it suffices to show that MATH . Let us first consider the contribution of MATH . In this case it suffices by NAME 's inequality (or NAME 's inequality) to obtain MATH estimates. From REF and the definition of the MATH norm we have MATH . From REF and the assumptions on MATH we thus have MATH and this contribution is thus acceptable by REF. Now consider the contribution of MATH . In this case we can replace the first factor by MATH, since the error in doing so vanishes after applying MATH. We now modify the above argument, the only difference being that we now place the first term in MATH and the second in MATH. In fact, the summation is much better because the derivative is now on the low frequency term. A similar argument deals with MATH and so we are left with MATH . We can split this into MATH and MATH plus some other terms which vanish when MATH is applied. For each of these three terms we place the high frequency REF factor in MATH and the other two factors in MATH. Regardless of the position of the derivatives, the high frequency factor has a norm of MATH by REF. Of the other two factors, both are bounded by MATH by REF, and at least one contains a derivative and therefore has a norm of MATH by REF. The claim then follows (if MATH is sufficiently small depending on MATH). We now prove REF. As before, we shall use induction and in fact prove MATH for all MATH and some large MATH. The claim is trivial for MATH. Now suppose that MATH and the claim has been proven for all smaller MATH. We apply MATH to REF and take absolute values (ignoring any possibility of cancellation) to obtain MATH . We will show that all six terms on the right-hand side have a MATH norm of MATH so that we can close the induction if MATH is sufficiently small and MATH sufficiently large. For the first term we use REF for the first factor, REF for the second factor, and the induction hypothesis for the third factor. For the second and third terms we use REF for the MATH term, and place the other two terms in MATH using REF. For the remaining three terms we place the first factor in MATH using REF and the other two factors in MATH using REF for the second factor and REF for the third. |
math/0010069 | Let MATH, MATH, MATH, MATH, MATH, MATH, MATH be as above. Define a vertical line segment to be a pair MATH such that MATH. (The notation comes from depicting MATH, MATH as one-dimensional sets, so that MATH becomes a two-dimensional set and MATH is the projection to the first co-ordinate). Let MATH denote the space of all vertical line segments. From an easy NAME argument we have MATH . Now define a trapezoid to be a pair MATH of vertical line segments such that MATH and MATH, and let MATH denote the space of all trapezoids. From NAME we have a lower bound for the cardinality of MATH: MATH . On the other hand, we can obtain an upper bound for the cardinality of MATH as follows. Consider the map MATH defined by MATH . We claim that this map is one-to-one. At first glance, this seems unlikely, as MATH has four degrees of freedom with respect to MATH (MATH has REF degrees of freedom, hence MATH has MATH, hence MATH has MATH) and MATH only specifies three of these four degrees. However, we can use REF to recover the fourth degree of freedom. Specifically, we take advantage of the identity MATH for all MATH, to conclude that the quantity MATH determines MATH, which then determines MATH by REF. From some straightforward linear algebra one can then check that MATH determines all of MATH, or in other words that MATH is one-to-one. Hence MATH . Combining this with REF we obtain REF with MATH as claimed. |
math/0010069 | (Informal) Let MATH be the union of all the tubes MATH as before; this set can be thought of as the union of about MATH-balls. Since each tube occupies about MATH of these balls, we can expect two random MATH, MATH to have a probability of about MATH of intersecting. Define a quadrilateral to be four tubes MATH, MATH, MATH, MATH in MATH such that MATH intersects MATH in a MATH-ball for MATH (with MATH). The total number of REF of tubes is MATH, but there are four constraints, and so by the above heuristic we expect the total number of quadrilaterals is bounded below by MATH. (This can be proven by two applications of NAME as before). Let MATH be the center of the ball where MATH and MATH intersect. A quadrilateral is mostly determined by the three points (which are usually in MATH): MATH . Given these three points, one can determine MATH as a linear combination. This determines the tube MATH, because the tubes point in different directions. Thus MATH (for instance) has only one degree of freedom, and once this degree is specified one can reconstruct the entire quadrilateral. Thus the number of quadrilaterals is bounded by MATH. Combining these bounds gives the same bound MATH as previously. |
math/0010070 | Recall that a NAME space MATH is universally negligible if there is no NAME probability measure on MATH that vanishes at singletons. By CITE, there is a universally negligible set MATH of cardinality MATH (see also CITE). Pick a non-null set MATH of size MATH and fix a bijection MATH. If MATH is such that MATH is continuous, then we may transport NAME measures on MATH to MATH, and therefore MATH is universally negligible and thus NAME negligible. (See also CITE.) |
math/0010070 | MATH . Follows from REF MATH (take MATH). MATH . If MATH, then any MATH (for MATH) work. So assume MATH and let MATH. Then MATH. For MATH put MATH . We are going to show that these MATH's are as required. To this end suppose that MATH is such that MATH (for all MATH). Then also MATH (for MATH) and by REF MATH we get MATH . |
math/0010070 | Assume toward contradiction that MATH for all MATH. Choose inductively fronts MATH of MATH such that CASE: MATH, CASE: MATH, CASE: MATH for all MATH. Note that then (by REF) for each MATH we have MATH . Since the right hand side of the inequality above approaches MATH (as MATH), we get an immediate contradiction with the demand MATH. |
math/0010070 | CASE: Let MATH; clearly we may assume that MATH for all MATH. Also we may assume that MATH (remember REF) and MATH. Fix MATH such that MATH for a moment. Let MATH. For each MATH pick a front MATH of MATH such that CASE: if MATH, then MATH, CASE: if MATH, then MATH. Let MATH, MATH, MATH, and MATH . Apply REF MATH for MATH (note that MATH) to pick MATH and MATH such that MATH and CASE: if MATH, MATH, then MATH, MATH, and MATH . Note that, if MATH, then MATH, and thus MATH . Also, letting MATH, MATH . Together CASE: MATH. Since MATH is strongly finitary, considering MATH (and using REF MATH), we find MATH such that MATH and MATH for all MATH, and MATH . Note that also, as MATH, MATH so CASE: MATH. Now, starting with MATH, build a tree MATH and a system MATH such that MATH. It should be clear that in this way we will get a condition in MATH (stronger than MATH). Why is MATH in MATH? Let MATH, MATH and MATH. Using MATH, we may show by downward induction that for every MATH we have MATH . Now we may easily conclude that MATH is special. CASE: Let MATH be a front of MATH, MATH normal (so, in particular, MATH for MATH). Fix MATH for a moment. For each MATH pick a front MATH of MATH such that MATH. Let MATH and MATH for MATH. By downward induction one can show that for all MATH we have MATH. Then, in particular, we have MATH and hence (letting MATH) MATH. The reverse inequality is even easier (remember REF). |
math/0010070 | Let MATH (for MATH); note that MATH. Fix MATH for a while. Let MATH be a front of MATH such that MATH . By downward induction, for each MATH, we define MATH and MATH such that CASE: If MATH, then MATH, MATH. CASE: If MATH, then MATH, MATH. CASE: If MATH, MATH, then: if MATH, then MATH, else MATH. Clauses MATH define MATH for MATH; MATH are not defined then (or are arbitrary). Suppose MATH, MATH. If MATH, then we let MATH (and MATH are not defined). So assume now that MATH . Then also (as MATH) MATH and we may apply REF MATH to pick MATH and MATH such that CASE: MATH, CASE: if MATH, MATH, then MATH, MATH, and MATH. Suppose now that MATH, MATH, and MATH have been defined for all MATH (and they satisfy clause MATH). If MATH then we let MATH (and MATH are not defined). So assume MATH . Then for MATH we let MATH and we note that MATH . Applying REF MATH choose MATH and MATH such that MATH and CASE: if MATH, then MATH, MATH and MATH, CASE: if MATH, then MATH, MATH and MATH. Now look at the definition of MATH. If MATH, then MATH, so MATH. Therefore MATH . Hence we may apply REF MATH again and get MATH and MATH such that MATH and if MATH, MATH, then MATH, MATH and MATH. Note that (as MATH) MATH . Therefore, MATH . Hence also (by MATH) MATH . Now, if MATH, MATH, then we build inductively a finite tree MATH as follows. We declare that MATH, MATH, and MATH. If we have decided that MATH, MATH (and MATH), then we also declare MATH, MATH (note MATH for MATH). Then, if MATH is defined, MATH, and, if MATH is defined, MATH. Also, if we ``extend" MATH using MATH (for MATH), then we get a condition MATH such that MATH. Likewise, if we ``extend" MATH using MATH (for MATH), then we get a condition MATH such that MATH. If for some MATH we have MATH, then we use the respective condition MATH to witness the demand REF. So assume that for each MATH we have MATH, and thus MATH . Apply the NAME Lemma to find an infinite set MATH such that for all MATH, MATH, we have MATH . Then MATH, MATH (for sufficiently large MATH) determine a condition MATH witnessing the first assertion of the lemma. |
math/0010070 | Let MATH consist of all MATH such that CASE: MATH and there is a normal condition MATH stronger than MATH and such that MATH, MATH, MATH, and for some front MATH of MATH, for every MATH: CASE: MATH and the condition MATH decides the value of MATH, and CASE: no initial segment of MATH has the property stated in MATH above. Note that MATH is an antichain of MATH, and MATH for every condition MATH such that MATH (by REF). For each MATH fix a condition MATH witnessing clause MATH (for MATH). Now apply REF: REF there is not possible by what we stated above, so we get a condition MATH as described in REF . It should be clear that it is as required here. |
math/0010070 | We may assume that MATH is normal, MATH, MATH, and MATH for MATH. We build inductively a sequence MATH such that CASE: MATH is a normal condition, MATH, MATH, MATH, CASE: MATH is a front of MATH, MATH, CASE: if MATH, then MATH, and for each MATH such that MATH we have MATH, CASE: if MATH, then MATH, CASE: for each MATH, the condition MATH decides the value of MATH, CASE: MATH. The construction can be carried out by REF (MATH are obtained by applying REF to MATH and MATH; if MATH have been defined, then we apply REF to MATH and MATH for MATH; remember REF). Next define MATH so that MATH, each MATH is a front of MATH, and if MATH then MATH. It is straightforward to check that MATH is as required in REF. |
math/0010070 | Should be clear at the moment; compare the proof of CITE. |
math/0010070 | Let MATH, MATH. We try to choose inductively partial functions MATH from MATH to MATH such that CASE: MATH, MATH, CASE: MATH. Note that in REF MATH, the left hand side expression is not more than MATH, so if the inequality holds, then (as MATH) CASE: MATH. Consequently, in the procedure described above, we are stuck at some MATH satisfying MATH. Let MATH . So this defines MATH, but we have to check that MATH. For this note that MATH . (So indeed MATH, and plainly MATH.) Also note that MATH . Now, suppose that MATH, MATH. Let MATH. We cannot use MATH as MATH, so REF MATH fails for it. Therefore MATH . For each MATH, we have MATH . Assume that MATH is such that MATH. We know that MATH for each MATH, so MATH . Hence MATH and so MATH, a contradiction. Consequently, we get that MATH so MATH satisfies the demand MATH. But we also know that for each partial function MATH from MATH to MATH, if MATH and MATH, then MATH and thus MATH satisfies the demand MATH as well. |
math/0010070 | Clauses REF should be obvious, so let us check REF MATH only. Let MATH, MATH, MATH be as in the assumptions of REF MATH. So in particular MATH . For MATH let MATH. First, we consider the case when both MATH and MATH are not smaller than MATH. Then we may apply REF and get MATH such that MATH, MATH and MATH . Then MATH and we are done. So suppose now that MATH. Then MATH and using REF we find MATH such that MATH, MATH, and MATH . |
math/0010070 | Let MATH, MATH, MATH. First note that MATH . Let MATH. For each MATH we may use REF to pick MATH such that MATH and MATH . Hence, MATH . Consequently, MATH and hence MATH . |
math/0010070 | Assume that MATH is a set of outer REF measure MATH. We have to show that, in MATH, it is still an outer measure one set. Let MATH be a MATH - name for a tree such that MATH and the NAME measure MATH of the set MATH of MATH - branches through MATH is positive, and suppose that some condition MATH forces MATH. Take a condition MATH such that CASE: MATH is special (remember REF) and MATH, and MATH for all MATH, and MATH, CASE: for some MATH, MATH, the condition MATH forces that MATH, CASE: for some MATH, letting MATH, we have that for each MATH, the condition MATH decides the value of MATH (remember REF). Fix MATH for a moment, and let MATH. For MATH and MATH we let MATH . If MATH, MATH, then MATH . [If MATH, then we stipulate MATH.] We show it by downward induction on MATH. If MATH, then MATH decides MATH, and if MATH forces that MATH, then MATH. Hence, by MATH, we have MATH. Let us assume now that MATH. Apply REF to MATH, MATH, MATH, MATH (for MATH), and MATH. Note that, as MATH is special, MATH, so MATH. Also note that CASE: MATH defined as in REF is MATH. [Why? First suppose that MATH. By the definition of MATH we may find MATH such that MATH, MATH for MATH, and MATH, and MATH. Note that MATH for all MATH, and thus MATH . By the definition of MATH, the last expression is MATH, a contradiction. Now suppose MATH. Take MATH such that MATH and MATH; clearly we may request that MATH for MATH. Let MATH (for MATH) be positive numbers such that if MATH for MATH, then MATH (compare REF). Pick conditions MATH such that MATH, MATH as in definition of MATH, and let MATH be such that MATH, MATH, and MATH for MATH. Then MATH giving an easy contradiction.] Thus we get MATH as required. Now suppose MATH, and we have proved the assertion of the claim for all MATH. We again apply REF, this time to MATH, and MATH, MATH, MATH (for MATH) and MATH. We note that MATH so the assumptions of REF are satisfied. Therefore we may conclude that MATH as needed. Applying REF to MATH we get MATH and hence MATH. Then necessarily MATH (remember MATH). Let MATH. Look at the set MATH - it is a NAME set of positive REF measure, and therefore we may pick MATH such that MATH. For each MATH such that MATH choose a condition MATH such that CASE: MATH, MATH, MATH, and CASE: MATH, and CASE: MATH. By NAME Lemma (remember MATH is strongly finitary) we find an infinite set MATH such that for each MATH from MATH we have MATH . Let MATH be such that MATH, MATH and if MATH, then MATH and MATH for sufficiently large MATH. It should be clear that MATH is a condition stronger than MATH, and it forces that MATH, a contradiction. |
math/0010070 | Let MATH, MATH, MATH. Note that MATH . Let MATH consist of all triples MATH such that MATH and MATH and fix MATH for a moment. Let MATH be such that MATH and MATH. Apply REF (to MATH and MATH for MATH, MATH) to pick MATH such that MATH, MATH and MATH . Next note that MATH, so MATH . Therefore, MATH . Hence, MATH and therefore, as MATH and MATH, MATH . |
math/0010070 | Assume toward contradiction that MATH is a MATH - name for a tree included in MATH, and MATH is a condition such that MATH . (Here, MATH stands for the product measure on MATH.) Passing to a stronger condition and shrinking the tree MATH (if necessary) we may assume that CASE: MATH is special and MATH, and MATH for all MATH, and MATH, CASE: for some MATH, MATH, the condition MATH forces that MATH, CASE: for some MATH, letting MATH, we have that for each MATH, the condition MATH decides the value of MATH. Fix MATH for a moment, and let MATH. Let MATH, and for MATH and MATH let MATH . So we are at the situation from the proof of REF (with MATH there replaced by MATH), and we may use REF to conclude that CASE: MATH. Now, for each MATH we define MATH by MATH . (If MATH, so MATH, then MATH and MATH.) If MATH, MATH, then MATH where MATH. The proof, by downward induction on MATH, is similar to that of REF, but this time we use REF. First note that if MATH, then our assertion is exactly what is stated in MATH. So suppose that MATH, MATH, and that we have proved our claim for all MATH. We are going to apply REF to MATH, MATH, MATH (and MATH being interpreted as MATH), and MATH, and MATH (for MATH, MATH), so we have to check the assumptions there. Note that (as MATH is special) MATH (so the demand in REF is satisfied). Also, by the inductive hypothesis, MATH (so REF holds). Finally note that if MATH, MATH, and MATH is such that MATH, MATH, then MATH defined by REF is MATH. So, by REF, we may conclude that MATH as needed. In particular, it follows from REF that MATH and hence MATH . Let MATH be such that MATH. Now we define: MATH . Note that MATH, and therefore MATH . Now we may finish like in REF: the set MATH is a NAME set of positive REF measure, so we may choose MATH such that for infinitely many MATH we have MATH. For each such MATH pick a condition MATH such that CASE: MATH, MATH, and CASE: MATH, and CASE: MATH. By NAME Lemma, we may find a condition MATH stronger than MATH, and an infinite set MATH such that CASE: if MATH are from MATH, then MATH and MATH . Then clearly MATH, a contradiction. |
math/0010070 | The equivalences MATH are well known (see CITE; also compare with the proof of CITE). MATH: Assume MATH, and suppose that MATH is a non-measurable set. Then we may find a closed set MATH such that CASE: for each MATH, the set MATH is either empty or is a perfect set of positive NAME measure, CASE: for every NAME set MATH of positive measure, both MATH and MATH (that is, both MATH and MATH are of full outer measure in MATH). Pick a NAME isomorphism MATH such that CASE: if MATH, then MATH, CASE: if MATH is NAME, then MATH has measure REF if and only if its image MATH has measure zero. Now note that the set MATH has outer measure REF and inner measure REF (in MATH), so we may apply MATH to it, and we get a suitable function MATH. Let MATH, and let MATH be defined by MATH for some (equivalently: all) MATH such that MATH. Easily MATH is a NAME function. Take any NAME extension MATH of MATH - it is as required in MATH for MATH. MATH: NAME easier. (Note that, since all MATH's are powers of MATH, we have a very nice measure preserving homeomorphism MATH.) |
math/0010070 | Start with universe MATH satisfying CH. Let MATH be countable support iteration such that each iterand MATH is (forced to be) the forcing notion MATH (defined in the second section; of course it is taken in the respective universe MATH). It follows from REF (and CITE) that the limit MATH is proper and MATH - bounding. Also it satisfies MATH - cc, and consequently the forcing with MATH does not collapse cardinals nor changes cofinalities (and MATH). We are going to prove that MATH . By REF, it is enough to show that MATH. To this end suppose that MATH is a MATH - name for a subset of MATH such that both MATH and its complement are of outer measure one. By a standard argument using MATH - cc of MATH (and the fact that each MATH for MATH has a dense subset of size MATH), we may find MATH of cofinality MATH, and a MATH - name MATH such that MATH . Let MATH be the MATH - name for the continuous function from MATH to MATH added at stage MATH by MATH (as defined right before REF). Then, by REF (applied to MATH and to its complement), in MATH the set MATH has outer measure REF and inner measure REF. By REF + REF, in MATH we may use CITE (and CITE) to conclude that MATH preserves non-nullity of sets from MATH. Consequently, MATH finishing the proof. |
math/0010070 | Assume MATH. Let MATH be a NAME isomorphism preserving null sets (see, for example, CITE), and let MATH. Let MATH be given by REF for MATH. Put MATH. We know that MATH is a non-null set (and consequently it is non-measurable), so applying MATH we may pick a NAME function MATH such that the set MATH has positive outer measure, and so does MATH. Let MATH be defined by MATH . Clearly MATH is NAME and for each MATH we have MATH. Finally, using NAME 's theorem (see, for example, CITE) we may pick a continuous function MATH such that the set MATH is not null (just take MATH so that it agrees with MATH on a set of large enough measure). |
math/0010070 | It is a standard application of fusion arguments somewhat similar to the proof of CITE; compare also the proof of preservation of ``proper+MATH - bounding" in CITE or CITE. Of course, we use here REF . |
math/0010070 | For a set MATH, MATH, and MATH we let MATH . We may assume that MATH (otherwise the lemma is easier and actually included in this case). Fix an enumeration MATH such that MATH, and let MATH . We are going to define inductively a decreasing sequence MATH of closed (non-empty) subsets of MATH such that MATH and CASE: for each MATH and MATH we have MATH . (Note that MATH implies MATH.) Suppose that MATH has been defined already, MATH. Let MATH enumerate the set MATH in the increasing order. By downward induction on MATH we chose open sets MATH. So, the set MATH is such that (remember MATH): CASE: MATH, CASE: MATH. Now suppose that MATH have been chosen already so that MATH for each MATH. Let MATH . Note that (by our assumptions) MATH . Let MATH and MATH. Note that MATH and hence MATH . Pick an open set MATH such that MATH and MATH. Finally we let MATH. It is easy to check that MATH is as required. After the sets MATH are all constructed we put MATH. It follows from MATH that the demand MATH is satisfied for each MATH. |
math/0010070 | For MATH let MATH be defined by MATH . This defines a function MATH on MATH. Plainly, MATH is a nice measured tree creating pair (we are going to use it to simplify notation only). Let MATH be a tree such that MATH and MATH. For MATH let MATH be such that MATH . Let MATH. It should be clear that (as MATH has positive NAME measure) MATH is a condition in MATH (note: MATH, not MATH!). Moreover, possibly shrinking MATH and MATH, we may request that CASE: MATH for all MATH, CASE: MATH, and MATH for each MATH (remember REF, or actually its proof). Let MATH. Fix an integer MATH for a moment. Let MATH (so it is a front of MATH). For each MATH, by downward induction, we define MATH and a real MATH such that CASE: MATH. If MATH, then we let MATH (and MATH is not defined). Suppose that MATH has been defined for all MATH so that MATH holds. Then MATH (remember our requests on MATH). Consequently we may apply REF (for MATH, MATH and MATH) to pick MATH such that CASE: MATH, and CASE: MATH. This completes the choice of MATH's and MATH's. Now we build a system MATH such that MATH is a finite tree, MATH, MATH and MATH for MATH. Next, applying NAME Lemma, we pick an infinite set MATH and a system MATH such that MATH and MATH . It follows from our construction that necessarily MATH, and it is a condition stronger than MATH, and MATH. |
math/0010070 | For MATH let MATH be a MATH - name for the generic real added at stage MATH (so it is a member of MATH if MATH, and a member of MATH if MATH). Suppose that MATH is a MATH - name for a function from MATH to MATH, and MATH. For each MATH pick a template with a name MATH, an enumeration MATH of MATH, and a condition MATH such that CASE: MATH, MATH behaves well for MATH (see REF), CASE: MATH and MATH obeys MATH, CASE: MATH and MATH. Using NAME Lemma (and REF) we find a template with a name MATH, ordinals MATH and MATH, an enumeration MATH of MATH, and a stationary set MATH such that for each MATH we have CASE: MATH is isomorphic to MATH by an isomorphism mapping MATH to MATH, and MATH, MATH, MATH, and CASE: MATH, MATH, and MATH and CASE: MATH. Let MATH be the MATH - name for the set MATH and let MATH be the canonical homeomorphism (induced by a bijective mapping from MATH onto MATH). Now, in MATH, we define a mapping MATH by: MATH (MATH is as defined before REF). Let MATH for some (equivalently: all) MATH. MATH . Assume not. Then there are an ordinal MATH, a condition MATH, and a MATH - name MATH such that CASE: MATH, MATH, and MATH, CASE: MATH is a MATH - name for a REF null subset of MATH, and CASE: MATH`` MATH ". (Note that above we use the fact that the forcing used at MATH is the random real forcing, so the conditions are closed sets of positive measure. This allows us to replace MATH by MATH.) Fix any MATH larger than MATH and let MATH be the increasing enumeration of MATH and let MATH, and MATH. Note that the conditions MATH and MATH are compatible. Also, as MATH is (a name for) a random real over MATH, we have MATH . Using REF , we may pick (a MATH - name for) a closed set MATH such that the condition MATH forces (in MATH): CASE: MATH, CASE: MATH, and CASE: the condition MATH of REF holds true for every MATH. (For the first demand remember that MATH is well behaving, so the set on the right-hand side has positive NAME measure.) But now, using REF, we may inductively build a condition MATH stronger than both MATH and MATH (and with the support included in MATH) such that MATH getting an immediate contradiction. Pick any MATH and let MATH be as defined in the proof of REF above. Let MATH be a MATH - name for the set MATH . So MATH is (a name for) a closed subset of MATH. Let MATH be a name for a NAME function from MATH to MATH such that if MATH, and MATH, then for each MATH (remember REF ). It should be clear that MATH is (a name for) a continuous function and MATH finishing the proof. |
math/0010073 | Let MATH be a simple polytope. Choose a linear function MATH which is generic, that is, it takes different values at all vertices of MATH. For this MATH there is a vector MATH in MATH such that MATH. Note that MATH is parallel to no edge of MATH. Now we view MATH as a height function on MATH. Using MATH, we make REF-skeleton of MATH a directed graph by orienting each edge in such a way that MATH increases along it (this can be done since MATH is generic). For each vertex MATH of MATH define its index, MATH, as the number of incident edges that point towards MATH. Denote the number of vertices of index MATH by MATH. We claim that MATH. Indeed, each face of MATH has a unique top vertex (a maximum of the height function MATH restricted to the face) and a unique bottom vertex (the minimum of MATH). Let MATH be a MATH-face of MATH, and MATH its top vertex. Since MATH is simple, there exactly MATH edges of MATH meet at MATH, whence MATH. On the other hand, each vertex of index MATH is the top vertex for exactly MATH faces of dimension MATH. It follows that MATH (the number of MATH-faces) can be calculated as MATH . Now, the second identity from REF shows that MATH, as claimed. In particular, the number MATH does not depend on MATH. On the other hand, since MATH for any vertex MATH, one has MATH . |
math/0010073 | In CITE the statement was proved directly, using MATH-vectors. Instead, we use MATH-vectors, which somewhat simplifies the proof. It is sufficient to prove that the affine hull of the MATH-vectors MATH of simple MATH-polytopes is a MATH-dimensional plane. Set MATH, MATH. Since MATH, REF gives MATH . It follows that MATH, MATH. Therefore, the vectors MATH, MATH, are affinely independent. |
math/0010073 | For any subset MATH denote by MATH the square-free monomial MATH. Let MATH be a square-free monomial ideal. Set MATH . Then one easily checks that MATH is a simplicial complex and MATH. |
math/0010073 | Let MATH be a regular sequence of degree-two elements of MATH. Then MATH is a graded algebra generated by degree-two elements, and MATH. Now the result follows from REF . |
math/0010073 | Since MATH is a MATH-vector, there exists a graded algebra MATH generated by degree-two elements such that MATH REF . In particular, MATH. Since MATH is generated by MATH, the number MATH can not exceed the total number of monomials of degree MATH in MATH variables. The latter is exactly MATH. |
math/0010073 | Using the NAME resolution MATH for the definition of MATH we calculate MATH . |
math/0010073 | Set MATH, MATH, MATH, MATH. Then MATH is a free MATH-module and MATH. Therefore, REF , gives a spectral sequence MATH . Since MATH is a regular sequence, MATH is a free MATH-module. Therefore, MATH . It follows that MATH for MATH. Thus, the spectral sequence collapses at the MATH term, and we have MATH which concludes the proof. |
math/0010073 | We apply REF to a minimal resolution of MATH. It follows from REF that the numerators of the summands in the right hand side of REF are exactly MATH, MATH. Hence, MATH . Using REF we get MATH . |
math/0010073 | This follows from REF . |
math/0010073 | This follows from the fact that the MATH-faces of MATH are in one-to-one correspondence with pairs MATH of faces of MATH such that MATH. |
math/0010073 | Subdividing each cube of MATH as described in Construction REF we obtain a simplicial complex, say MATH. Then applying Construction REF to MATH we get a cubical complex that subdivides MATH and embeds into some MATH (see REF ) as the subcomplex MATH. |
math/0010073 | Realise the simple polytope as a lattice polytope MATH. Let MATH be the corresponding toric variety. REF is already proved REF . It follows from REF that the multiplication by MATH is a monomorphism MATH for MATH. This together with REF gives MATH, MATH, thus proving REF . To prove REF , define the graded commutative MATH-algebra MATH. Then MATH, MATH for MATH, and MATH is generated by degree-two elements (since so is MATH). It follows from REF that the numbers MATH, MATH, are the components of a MATH-vector, thus proving REF and the whole theorem. |
math/0010073 | The MATH-th NAME number equals the number of MATH-dimensional cells in the cellular decomposition constructed above. This number equals the number of vertices of index MATH, which is MATH by the argument from the proof of REF . |
math/0010073 | At the beginning we choose signs of the edge vectors at MATH arbitrary, and express MATH as in REF . Then MATH is the edge vector corresponding to the edge MATH opposite to MATH, MATH. It follows that MATH and MATH for MATH. Hence, MATH (see REF ). Since MATH is a primitive vector, it follows from REF that MATH. Changing the sign of MATH if necessary, we obtain MATH which together with REF gives MATH, as needed. |
math/0010073 | The orientation of MATH is determined by the canonical orientation of the almost complex manifold MATH and the orientation of the torus REF . Since the almost complex structure on MATH is MATH-invariant, it induces almost complex structures on the MATH-fixed submanifolds MATH. It follows that for any vertex REF the vectors MATH constitute a positively oriented basis of MATH. |
math/0010073 | Since the one-dimensional subtori MATH, MATH, generate MATH, we may define MATH as any automorphism of MATH that maps MATH to MATH, MATH. |
math/0010073 | The characteristic map for MATH has the form MATH where MATH, MATH, is the diagonal subgroup in MATH. Let MATH be a quasitoric manifold over MATH with characteristic map MATH. We may assume that MATH, MATH, by REF . Then it easily follows from REF that MATH where MATH, MATH. Now define the automorphism MATH by MATH . It can be readily seen that MATH, which together with REF completes the proof. |
math/0010073 | In order to construct a required embedding we provide a smooth function MATH such that the equation MATH defines a hypersurface diffeomorphic to MATH. For MATH we take MATH; then we proceed by induction on MATH. Suppose we have a function MATH such that MATH defines a smooth embedding MATH and MATH for any MATH satisfying MATH. Then set MATH . The hypersurface MATH is easily seen to be diffeomorphic to MATH. |
math/0010073 | Construction REF provides the atlas MATH for MATH as a manifold with corners. The set MATH is based on the vertex MATH and is diffeomorphic to MATH. Then MATH. We claim that MATH can be realised as an open set in MATH, thus providing a chart for MATH. To see this we embed MATH into MATH as a closed hypersurface MATH REF . Since the normal bundle is trivial, the small neighbourhood of MATH is homeomorphic to MATH. Taking the cartesian product with MATH we obtain an open set in MATH homeomorphic to MATH. |
math/0010073 | It follows from REF that MATH meets every isotropy subgroup only at the unit. This implies that the action of MATH on MATH is free. By definitions of MATH and MATH, the projection MATH descends to the projection MATH which displays MATH as a principal MATH-bundle over MATH. |
math/0010073 | Recall that the cubical decomposition of MATH consists of the cubes MATH based on the vertices MATH. Note that MATH is contained in the open set MATH (see Construction REF). The inclusion MATH is covered by an equivariant inclusion MATH, where MATH is a closed subset homeomorphic to MATH. Since MATH and MATH is stable under the MATH-action, the resulting embedding MATH is equivariant. |
math/0010073 | In this proof we identify the polyhedrons MATH and MATH with their images MATH and MATH under the map MATH, see REF . For each vertex MATH denote by MATH the union of MATH-cubes of MATH that contain MATH. Alternatively, MATH is MATH. These MATH are analogues of facets of a simple polytope. Moreover, if MATH for some MATH, then MATH is the image of a facet of MATH under the map MATH (Construction REF). As in the case of simple polytopes, we define ``faces" of MATH as non-empty intersections of ``facets" MATH. Then the ``vertices" (that is, non-empty intersections of MATH ``facets") are the barycentres of MATH-simplices of MATH. For every such barycentre MATH denote by MATH the open subset of MATH obtained by deleting all ``faces" not containing MATH. Then MATH is identified with MATH, while MATH is homeomorphic to MATH. This defines a structure of manifold with corners on the MATH-ball MATH, with atlas MATH. At the same time we see that MATH is a manifold, with atlas MATH. |
math/0010073 | Indeed, MATH is a union of ``moment-angle" blocks REF , and each MATH is the closure of the cell MATH. |
math/0010073 | Since MATH is the closure of the cell MATH, the statement follows from REF . |
math/0010073 | To prove that MATH is a cellular subcomplex of MATH we just mention that it is the closure of the MATH-cell MATH. So, it remains to prove that MATH is contractible within MATH. To do this we show that the embedding MATH is homotopic to the map to the point MATH. On the first step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, our MATH can be contracted to MATH within MATH. On the second step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, MATH can be contracted to MATH within MATH, and so on. On the MATH-th step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, MATH can be contracted to MATH within MATH. We end up at the point MATH to which the whole torus MATH can be contracted. |
math/0010073 | Indeed, REF-skeleton of our cellular decomposition of MATH is contained in the torus MATH. |
math/0010073 | Consider the decomposition REF . Since each MATH is MATH-stable, the NAME construction MATH is patched from the NAME constructions MATH, MATH. Suppose MATH; then MATH (see REF ). By definition of NAME construction, MATH. The space MATH is the total space of a MATH-bundle over MATH. It follows that there is a deformation retraction MATH, which defines a homotopy equivalence between the restriction of MATH to MATH and the cellular inclusion MATH. These homotopy equivalences corresponding to different simplices MATH fit together to yield a required homotopy equivalence between MATH and MATH. |
math/0010073 | Note that MATH and REF-skeleton of MATH coincides with that of MATH. If MATH is MATH-neighbourly, then it follows from REF that the MATH-skeleton of MATH coincides with that of MATH. Now, both statements follow easily from the exact homotopy sequence of the map MATH with homotopy fibre MATH REF . |
math/0010073 | Since MATH acts freely on MATH, we have MATH . |
math/0010073 | The differentials of the spectral sequence are all trivial by dimensional reasons, since both MATH and MATH have cells only in even dimensions (see REF ). |
math/0010073 | The monomorphism MATH takes MATH to MATH, MATH. Since MATH is a free MATH-module (note that this follows from REF , so we do not need to use REF ), MATH is a regular sequence. It follows that the kernel of MATH is exactly MATH. |
math/0010073 | Let us consider the NAME - NAME spectral sequence of the commutative square MATH where the left vertical arrow is the pullback along MATH. REF shows that MATH is homotopy equivalent to MATH. By REF , the map MATH induces the quotient epimorphism MATH, where MATH denotes the cellular cochain algebra. Since MATH is contractible, there is a chain equivalence MATH. Therefore, there is an isomorphism MATH . The NAME - NAME spectral sequence of commutative square REF has MATH and converges to MATH REF . Since MATH it follows from REF that the spectral sequence collapses at the MATH term, that is, MATH. REF shows that the module MATH is an algebra isomorphic to MATH, which concludes the proof. |
math/0010073 | The spectral sequence under consideration converges to MATH and has MATH . The differential in the MATH term acts as in REF . Hence, MATH (the last identity follows from REF ). |
math/0010073 | A routine check shows that the cochain homotopy operator MATH for the NAME resolution (see the proof of REF) establishes a cochain homotopy equivalence between the maps MATH and MATH of the algebra MATH to itself. That is, MATH . We just illustrate the above identity on few simple examples. MATH . |
math/0010073 | It follows directly from the definitions of MATH and MATH that the map is an isomorphism of graded algebras. Thus, it remains to prove that it commutes with differentials. Let MATH, MATH and MATH denote the differentials in MATH, MATH and MATH respectively. Since MATH, MATH, we need to show that MATH, MATH. We have MATH, MATH. Any REF-cell of MATH is either MATH or MATH, MATH. Then MATH where MATH if MATH and MATH otherwise. Hence, MATH. Further, any REF-cell of MATH is either MATH or MATH. Then MATH . Hence, MATH. |
math/0010073 | We make calculations with the cochain complex MATH. The module MATH has the basis consisting of monomials MATH with MATH and MATH. Since MATH, MATH, the bigraded component MATH is generated by monomials MATH with MATH and MATH. In particular, MATH if MATH or MATH, whence REF follows. To prove REF we observe that MATH is generated by REF, while any MATH, MATH, is a coboundary, whence MATH, MATH. Now we prove REF . Any MATH has MATH, while any simplex of MATH is at most MATH-dimensional. It follows that MATH for MATH. By REF , MATH for MATH, so it remains to prove that MATH for MATH. The module MATH is generated by monomials MATH, MATH. Since MATH, it follows easily that there are no non-zero cocycles in MATH. Hence, MATH. REF follows from REF . It also follows from REF that MATH. The basis for MATH consists of monomials MATH, MATH. We have MATH and MATH. It follows that MATH is a cocycle if and only if MATH is not a REF-simplex in MATH; in this case two cocycles MATH and MATH represent the same cohomology class. REF now follows easily. The remaining REF follows from the fact that the monomial MATH of maximal total degree MATH necessarily has MATH, MATH, MATH. |
math/0010073 | It follows from REF that MATH . Then MATH . Denote MATH. Then it follows from REF that MATH . Substituting MATH for MATH above, we finally obtain from REF MATH which is equivalent to REF . |
math/0010073 | We have MATH . Now the statement follows from REF . |
math/0010073 | Since MATH and MATH, we have MATH . Combining REF we get MATH . Hence, MATH where we used REF . |
math/0010073 | This follows from REF . |
math/0010073 | It follows from REF that MATH. By definition, the module MATH is spanned by monomials MATH such that MATH, MATH, MATH. Any such monomial is a cocycle. Suppose that MATH are two MATH-simplices of MATH that share a common MATH-face. We claim that the corresponding cocycles MATH, MATH, where MATH, MATH, represent the same cohomology class (up to a sign). Indeed, let MATH, MATH. Since any MATH-face of MATH is contained in exactly two MATH-faces, the identity MATH holds in MATH. Hence, MATH (as cohomology classes). Since MATH is a simplicial sphere, any two MATH-simplices can be connected by a chain of simplices such that any two successive simplices share a common MATH-face. Thus, all monomials MATH in MATH represent the same cohomology class (up to a sign). This class is the generator of MATH, that is, the fundamental cohomology class of MATH. |
math/0010073 | Every subgroup of MATH of dimension MATH intersects non-trivially with any MATH-dimensional isotropy subgroup, and therefore can not act freely on MATH. |
math/0010073 | Since any isotropy subgroup for MATH is coordinate (see REF ), it intersects with MATH only at the unit. |
math/0010073 | The map MATH defines the epimorphism of tori MATH. It is easy to see that if MATH is a regular colouring, then MATH acts freely on MATH. |
math/0010073 | It follows from REF that the orbits of the MATH-action on MATH corresponding to the vertices of MATH have maximal (rank MATH) isotropy subgroups. The isotropy subgroup corresponding to the vertex MATH is the coordinate subtorus MATH. The subgroup REF acts freely on MATH if and only if it intersects each isotropy subgroup only at the unit. This is equivalent to the condition that the map MATH is injective for any MATH. The latter map is injective whenever the image of the corresponding map MATH is a direct summand of MATH. The matrix of the map MATH is obtained by adding MATH columns MATH (REF stands at the place MATH, MATH) to MATH. This matrix defines a direct summand of MATH exactly when the same is true for each MATH. |
math/0010073 | REF shows that if MATH admits a characteristic map MATH, then the MATH-dimensional subgroup MATH acts freely on MATH, whence MATH. Now suppose MATH, that is, there exists a subgroup REF of rank MATH that acts freely on MATH. The corresponding MATH-matrix MATH defines a monomorphism MATH whose image is a direct summand. It follows that there is a MATH-matrix MATH such that the sequence MATH is exact. Since MATH satisfies the condition of REF , the matrix MATH satisfies REF , thus defining a characteristic map for MATH. |
math/0010073 | REF shows that MATH . The quotient MATH is identified with MATH. |
math/0010073 | Since MATH and MATH, we have MATH . By REF , MATH . Combining the above two identities with REF we get MATH, which concludes the proof. |
math/0010073 | The inclusion of the subgroup MATH defines the map of classifying spaces MATH. Let us consider the commutative square MATH where the left vertical arrow is the pullback along MATH. It can be easily seen that MATH is homotopy equivalent to the quotient MATH. The NAME - NAME spectral sequence of the above square converges to the cohomology of MATH and has MATH where the MATH-module structure on MATH is defined by the matrix MATH, that is, by the map MATH. It can be shown in the same way as in the proof of REF (using cellular decompositions) that the spectral sequence collapses at the MATH term and the following isomorphism of algebras holds: MATH . Now put MATH, MATH, MATH, and MATH in REF . Since MATH here is a free MATH-module and MATH, the spectral sequence MATH arises. Its MATH term is MATH and it converges to MATH. Since MATH is a free MATH-module, we have MATH . Thus, the spectral sequence collapses at the MATH term, and the following isomorphism of algebras holds: MATH which together with REF proves the theorem. |
math/0010073 | The map MATH takes the cohomology ring MATH to the subring MATH of MATH. This subring is isomorphic to the quotient MATH. Now the assertion follows from the fact that a polytope MATH is MATH-neighbourly if and only if the ideal MATH does not contain monomials of degree MATH. |
math/0010073 | To prove the first assertion we construct homotopy equivalence MATH as it is shown on REF . This map is covered by an equivariant homotopy equivalence MATH. The second assertion follows easily from the definition of MATH. |
math/0010073 | We need only to check that the differential MATH adds MATH to bidegree. This follows from REF and the following formulae: MATH . |
math/0010073 | The definition of MATH (see REF ) shows that MATH (see REF) is a cell of MATH if and only if the following two conditions are satisfied: CASE: The set MATH is a simplex of MATH. CASE: MATH. Let MATH denote the number of cells MATH with MATH, MATH, MATH, MATH, MATH, MATH. It follows that MATH where MATH is the MATH-vector of MATH (we also assume MATH and MATH for MATH or MATH). By REF , MATH . Now we calculate MATH using REF : MATH . Substituting MATH above we obtain MATH . Since MATH we get MATH . The second sum in the above formula is exactly MATH (see REF ). To calculate the first sum we observe that MATH (this follows from calculating the coefficient of MATH in both sides of the identity MATH). Hence, MATH since MATH (remember that MATH). Finally, using REF we calculate MATH . |
math/0010073 | By REF , MATH and MATH. Moreover, it can be seen in the same way as in REF that relative NAME duality isomorphisms REF regard the bigraded structures in the (co)homology of MATH and MATH. It follows that MATH . Using REF we calculate MATH . Substituting the formula for MATH from REF and the above expression into REF we obtain MATH . Calculating the coefficient of MATH in both sides after dividing the above identity by MATH, we obtain MATH, as needed. |
math/0010073 | Since MATH, MATH, we have MATH (the coefficient MATH can be dropped since for odd MATH the left hand side is zero). |
math/0010073 | Suppose MATH is a coordinate subspace arrangement in MATH. Define MATH . Obviously, MATH is a simplicial complex. By the construction, MATH depends only on MATH (that is, on MATH) and MATH, whence the proposition follows. |
math/0010073 | Take MATH. Let MATH be the maximal subset of MATH such that MATH (that is, MATH). Then it follows from the definition of MATH (see REF ) that MATH is a simplex of MATH. Hence, MATH and MATH. Thus, the first statement is proved. Since MATH is the quotient of MATH, the second assertion follows from the first one. |
math/0010073 | First, we construct a deformation retraction MATH. This is done inductively. We start from the boundary complex of a MATH-simplex and remove simplices of positive dimensions until we obtain MATH. On each step we construct a deformation retraction, and the composite map would be a required retraction MATH. If MATH is the boundary complex of a MATH-simplex, then MATH. In this case the retraction MATH is shown on REF . Now suppose that MATH is obtained by removing one MATH-dimensional simplex MATH from simplicial complex MATH, that is MATH. By the inductive hypothesis, the there is a deformation retraction MATH. Let MATH be the point with coordinates MATH and MATH for MATH. Since MATH is not a simplex of MATH, we have MATH. At the same time, MATH (see REF ). Hence, we may apply the retraction from REF on the face MATH, with centre at MATH. Denote this retraction by MATH. Then MATH is the required deformation retraction. The deformation retraction MATH is covered by an equivariant deformation retraction MATH. |
math/0010073 | If MATH then vectors MATH (see REF ) are linearly dependent, which is impossible. |
math/0010073 | A point from MATH has the non-trivial isotropy subgroup with respect to the action of MATH on MATH if and only if at least one of its coordinates vanishes. It follows from REF that if a point MATH has some zero coordinates, then the corresponding facets of MATH have at least one common vertex MATH. Let MATH. The isotropy subgroup of the point MATH with respect to the action of the subgroup MATH is non-trivial only if some linear combination of columns of MATH lies in the coordinate subspace spanned by MATH. But this implies that MATH, which contradicts REF . Thus, MATH acts freely on MATH. To prove the second part of the theorem it is sufficient to show that each orbit of the action of MATH on MATH intersects the image MATH at a single point. Since the embedding MATH is equivariant with respect to the MATH-actions, the latter statement is equivalent to that each orbit of the action of MATH on MATH intersects the image MATH (see REF ) at a single point. Let MATH. Then MATH lies in some MATH-face MATH of the cube MATH, see REF . We need to show that the MATH-dimensional subspace spanned by the vectors MATH is in general position with the MATH-face MATH of MATH. This follows directly from REF . |
math/0010073 | This follows from REF . |
math/0010073 | Let us consider the NAME - NAME spectral sequence of the NAME fibration MATH with fibre MATH, where MATH is the NAME - Reisner space REF and MATH is the path space over MATH. By REF , MATH and the spectral sequence converges to MATH. Since MATH is contractible, there is a cochain equivalence MATH. We have MATH. Therefore, MATH which together with REF shows that the spectral sequence collapses ate the MATH term. Hence, MATH. Finally, REF shows that MATH, which concludes the proof. |
math/0010073 | Consider the bundle MATH with fibre MATH. It is not hard to prove that the corresponding loop bundle MATH with fibre MATH is trivial (note that MATH). To finish the proof it remains to mention that MATH. |
math/0010075 | We prove the theorem using the atomic decomposition of MATH. See CITE and Latter CITE. Since finite sums of atoms are dense in MATH we will work with such sums and we will obtain estimates independent of the number of terms in each sum. The general case will follow by a simple density argument. Write each MATH, MATH as a finite sum of MATH-atoms MATH, where MATH are MATH-atoms. This means that the MATH's are functions supported in cubes MATH and satisfy the properties MATH for all MATH . By the theory of MATH spaces, see CITE page REF, we can take the atoms MATH to have vanishing moments up to any large fixed specified integer. In this article we will assume that all the MATH's satisfy REF for all MATH. For a cube MATH, let MATH denote the cube with the same center and MATH its side length, that is, MATH. Using multilinearity we write MATH . We now fix MATH and MATH and we consider the following cases: CASE: MATH . CASE: MATH lies in the complement of at least one of the cubes MATH. Let us begin with REF . We fix MATH and without loss of generality (by permuting the indices) we assume that MATH and that MATH. Assume without loss of generality that the side length of the cube MATH is the smallest among the side lengths of the cubes MATH. Let MATH be the center of the cube MATH. Since MATH has zero vanishing moments up to order MATH, we can subtract the NAME polynomial MATH of the function MATH at the point MATH to obtain MATH for some MATH on the line segment joining MATH to MATH by NAME 's theorem. We have MATH, since MATH. Similarly we obtain MATH for MATH. Set MATH and note that MATH. The estimates for the kernel MATH and the size estimates for the atoms give the following pointwise bound for the expression above: MATH . Integrating the above over MATH, MATH, MATH we obtain that the expression above is bounded by a constant multiple of MATH . But the integral above is easily seen to be controlled by a constant multiple of MATH. Since the cube MATH was picked to have the smallest size among the MATH, the expression above is bounded by a constant multiple of MATH since MATH. Summing over all possible MATH and all possible combinations of subsets of MATH of size MATH we obtain the pointwise estimate MATH for all MATH which belong to the complement of at least one MATH REF . Now using REF we obtain MATH where MATH and MATH correspond to cases MATH and MATH respectively and are given by MATH . Applying NAME 's inequality with exponents MATH and MATH we obtain the estimate MATH where we used the MATH-subadditivity of the MATH quasi-norm and the easy fact that the functions MATH have MATH norms bounded by constants. We now turn our attention to REF . Here we will show that MATH . To prove REF we will need the following lemma whose proof we postpone until the next section. Let MATH . Then there is a constant MATH such that for all finite collections of cubes MATH in MATH and all nonnegative integrable functions MATH with MATH we have MATH . We momentarily assume REF and we prove REF . Using the assumption that MATH maps MATH, it was proved in CITE that MATH maps all possible combinations of products MATH into MATH with norm at most a multiple of MATH. MATH denotes here the space of all MATH functions with compact support. Now fix atoms MATH, MATH, MATH supported in cubes MATH, MATH, MATH respectively. Assume that MATH, otherwise there is nothing to prove. Since MATH, we can pick a cube MATH such that MATH and MATH. Without loss of generality assume that MATH has the smallest size among all these cubes. Since MATH maps MATH into MATH we obtain that MATH since MATH and MATH. It follows from REF that MATH which combined with MATH gives MATH . We now have the easy estimate MATH and using REF , estimate REF , and the last inclusion in REF we obtain MATH . This proves REF which combined with REF completes the proof of the theorem. |
math/0010075 | Let us set MATH for all MATH. Given MATH, we define MATH to be the unique MATH such that MATH. We also set MATH . We now observe that if MATH and MATH, then MATH is constant on MATH. Therefore for MATH the sets below are well-defined MATH . For MATH and any MATH, we let MATH . Note that MATH is a constant since both functions MATH and MATH are constant on MATH. We claim that for all MATH we have the identity MATH . To prove REF observe that if MATH, then MATH and the conclusion easily follows. Otherwise, there is a smallest MATH such that MATH . Then all the cubes that contain MATH from the collection MATH belong to MATH and the cube that contains MATH from MATH belongs to MATH. It follows that MATH since the last sum has only one term. This proves REF . Next we set MATH . Then using REF we obtain MATH . Now MATH is supported on MATH. NAME 's inequality and REF give MATH . Summing the above over all MATH and using the fact that MATH we obtain the required estimate MATH where the last inequality follows by summation by parts. |
math/0010075 | Given the cubes MATH we can find a finite collection of dyadic cubes MATH with MATH and MATH where MATH. We apply REF to the functions MATH. (We collapse terms when the same dyadic cube is used twice). We obtain MATH where MATH. Inserting this estimate in REF gives MATH and the required conclusion follows from the last inclusion in REF . |
math/0010075 | We discuss the multilinear interpolation needed to prove this theorem for all indices MATH. REF is valid when all the MATH's satisfy MATH as proved in CITE. In REF we considered the case when all MATH. We now fix indices MATH so that some of them are bigger than MATH and some of them are less than or equal to MATH. We pick MATH and MATH so that MATH . Using that MATH is bounded from MATH into MATH with norm at most MATH, it follows from CITE that MATH with norm at most a constant multiple of MATH. Now define MATH be setting MATH . Then it is easy to see that MATH for all MATH and by REF we have MATH with norm at most a constant multiple of MATH, where MATH. Here we need REF with MATH. Identity REF gives MATH where MATH. Interpolating between REF we obtain that MATH where MATH. But MATH if MATH or MATH if MATH and the required conclusion follows (see for example, CITE). |
math/0010075 | The proof is similar to that for MATH. First we consider the case where all the MATH's are less than or equal to one. It follows from CITE that MATH is bounded on the same range as MATH with bound at most a multiple of MATH. Thus the estimates in REF follow as before. Next observe that the kernels MATH satisfy REF uniformly in MATH. Hence the estimates in REF for MATH equally apply to MATH uniformly in MATH and the same conclusion follows. The remainder of the argument is then similar. One treats the multilinear maps MATH as maps MATH for any finite set MATH and uses complex interpolation as before. |
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