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math/0011183 | We start by proving the result in the case of MATH being a continuous piecewise affine map, that is, letting MATH be the support of MATH, there is a finite number of domains MATH for which MATH and MATH (the gradient of MATH) is constant on each MATH. We define MATH and MATH the horizontal MATH-neighborhood of the graph of MATH. That is, MATH is equal to the set of points MATH for which there is MATH with MATH and MATH such that MATH and MATH. We observe that MATH. Indeed, given MATH, and since MATH, by the continuity of MATH there is MATH such that MATH. Taking MATH we have MATH. Hence MATH . For each MATH we define MATH and MATH the horizontal MATH-neighborhood of MATH. We have MATH . Letting MATH denote the gradient (constant) vector of MATH, we have that MATH is orthogonal to MATH. Taking MATH we have MATH where MATH is a constant depending only on the volume of the unit ball in MATH. We have MATH and MATH . Altogether this yields MATH . Taking into account that in this case MATH we obtain the result for any continuous piecewise affine map. The next step is to deduce the result for any MATH map MATH. In this case, we may take a sequence MATH of continuous piecewise affine maps such that MATH (here we are take derivatives only in the interior points of the elements of the partition associated to each piecewise affine map). This implies that MATH and MATH . Taking into account the case we have seen before, this implies the result also for the case of MATH being a MATH map. For the general case, we know by REF that given MATH there is a sequence MATH of MATH maps for which MATH . We have MATH . Since MATH where MATH, the result for general MATH follows from REF and the previous case. |
math/0011183 | Take some small MATH and define MATH the MATH-neighborhood of the critical set of MATH in MATH. We divide MATH into a finite number of domains of injectivity of MATH whose collection we call MATH. We observe that if MATH is close enough to MATH, then MATH also contains the critical set of MATH, and so we may define an analogous MATH for MATH in such a way that for each MATH there is one (and only one) MATH for which the NAME measure of MATH is small. For each MATH let MATH be the element in MATH naturally associated to MATH, and define MATH . We have MATH . Let us estimate the expressions on the right hand side of the inequality above. For the first one we have MATH . Since MATH (recall REF ), it follows from NAME 's inequality that MATH and MATH . Let MATH be some small constant (to be determined later in terms of MATH). Using REF and taking MATH and MATH sufficiently small we can make MATH . By a change of variables induced by MATH we deduce for each MATH and MATH associated to MATH and this last expression is bounded from above by MATH . Choosing MATH sufficiently small, then MATH implies that MATH on MATH (which contains MATH). Hence MATH . Thus, applying REF and choosing MATH sufficiently small we obtain MATH . Let us finally estimate the terms involved in REF (the same method can be applied to obtain a similar estimate for REF ). MATH . Using REF and taking MATH is sufficiently small, then MATH . Putting estimates REF above together we obtain MATH . So we only have to take MATH in such a way that MATH. |
math/0011183 | Fixing some small MATH, we are going to prove that there is some MATH for which MATH . We have MATH . By REF we know that there is an integer MATH and MATH for which MATH . Now we take MATH sufficiently large in such a way that MATH. We split each one of the sums in REF into two sums and write MATH where MATH and MATH . We have MATH and MATH which together with REF yield MATH . On the other hand, we have for MATH which is bounded from above by MATH . Here we also consider the transfer operator for the iterated maps MATH and MATH defined analogously as for MATH in REF . We have MATH and MATH . Taking into account REF we can make MATH if MATH is sufficiently large. Using REF we can also make MATH as long as we take MATH large enough. This completes the proof of the proposition. |
math/0011183 | It follows from REF that MATH converges to MATH in the weak* topology. Hence, given any MATH continuous we have MATH . On the other hand, since MATH is MATH-invariant we have MATH . It suffices to prove that MATH . We have MATH . Since MATH is uniformly close to zero when MATH is large (at least in the compact set MATH that contains the supports of the measures), we have that the first term in the sum above is close to zero for MATH sufficiently large. On the other hand, since MATH converges to MATH in the weak* topology we also have that the second term in the sum above is close to zero if MATH is taken large enough. |
math/0011183 | See CITE and CITE. |
math/0011183 | See CITE. |
math/0011183 | Fix some MATH with MATH and let MATH. We have MATH and MATH for some MATH and MATH depending only on the diameter of MATH. Now, since MATH the result follows from CITE. |
math/0011183 | Take MATH and let MATH. The set of points MATH for which there is some MATH that is a return for MATH is contained in MATH. Hence, defining MATH we have MATH and so MATH . Taking into account estimates REF above, it suffices to study the decay of MATH with MATH. We define for each MATH and MATH . Using REF we obtain MATH . Fixing some MATH and MATH we deduce from REF MATH . Similarly we prove that MATH . Hence MATH . It follows from the definition of MATH that the iterates of points in MATH do not hit the critical region MATH from time MATH to MATH. From REF we deduce that there is some constant MATH for which MATH . On the other hand, it follows from CITE that there is some absolute constant MATH for which MATH . From REF we obtain MATH which together with REF gives MATH . |
math/0011183 | The same proof of CITE with the improved estimate REF in the place of REF . |
math/0011183 | The proof of this proposition will be made in four steps. In the first one we prove that the image of MATH by MATH has height bounded from below by a constant of order MATH. In the second step we prove that a vertical segment of order MATH becomes, after a finite number of iterates, an interval with height bounded from below by a constant of order MATH. In the third step we show that iterating vertical segments of order MATH they become segments with length bounded from below by a constant not depending on MATH. In the final step we make use of the properties of the quadratic map MATH to obtain the result. CASE: There is a constant MATH such that for every MATH, MATH and MATH, MATH . This follows from CITE. It is easy to check that the estimate in REF above in the place of the estimate of CITE is enough to yield this dependence on MATH. CASE: There is a constant MATH and an integer MATH such that if MATH is an interval with MATH, then for every MATH . We start by remarking that we may assume that MATH intersects MATH. (If this is not the case, we take MATH the first integer for which MATH intersects MATH. It follows from REF that MATH for some MATH, and so we may start with MATH). Take MATH a subinterval of MATH such that MATH . It follows from REF that for MATH . Let MATH be the first integer for which MATH intersects MATH. It follows from REF that MATH . Now we proceed inductively and prove that for each MATH there is an interval MATH and a sequence of integers MATH for which MATH . We stop when MATH, and take MATH and MATH (note that MATH only depends on MATH which does not depend on MATH). CASE: There is a constant MATH and an integer MATH such that if MATH is an interval with MATH, then for every MATH . Arguing as in REF we can prove an analog to REF MATH . Letting MATH be the first integer for which MATH intersects MATH we have MATH . In both cases we have that the MATH-component of MATH contains some interval MATH not intersecting MATH with an end point at MATH or MATH and whose length is at least MATH. From now on we use MATH to denote any large constant depending only on the map MATH. Take MATH the smallest integer for which MATH is a periodic point for MATH and let MATH be its period. Denote MATH and note that by CITE we must have MATH. Fix MATH with MATH and MATH, and take MATH small enough in order to obtain MATH (and MATH sufficiently small). Since REF is pre-periodic for MATH, there exists some constant MATH such that MATH for every MATH. From this we deduce MATH as long as MATH is sufficiently small. By REF we may write MATH with MATH at every point MATH. This, together with REF , gives for every MATH and so we have for some MATH . For MATH and MATH we denote MATH . Take MATH and MATH sufficiently small in such a way that MATH . If MATH and MATH are such that MATH and MATH, then MATH and so, inductively, MATH . In particular for the points MATH we have MATH. Now we take MATH the smallest integer for which MATH. This choice of MATH implies MATH . Now we consider the following two possible cases: CASE: MATH for every MATH . This implies that MATH (recall that MATH). CASE: MATH for some MATH . Since MATH, it follows that MATH . In both cases we have some integer MATH for which MATH . Thus, taking MATH we have MATH . CASE: There is an integer MATH such that if MATH is an interval with MATH, then for every MATH . Since we are taking MATH a NAME parameter, it follows that the pre-orbit of the repelling fixed point MATH of MATH is dense. So, there is some integer MATH such that for every interval MATH with MATH we have that MATH covers a neighbourhood of MATH with a definite size (depending only on MATH). By a finite number of iterates MATH we transform this neighbourhood in the whole interval MATH. Hence, taking MATH we have by continuity MATH for sufficiently small MATH. Now it suffices to take MATH and we complete the proof of REF . |
math/0011183 | Since we have MATH, it follows that MATH . Thus, if MATH, then MATH, and so we have proved the first item. Now let MATH. Since MATH, there must be some MATH for which MATH. On the other hand, since MATH we have MATH, and so MATH belongs to MATH. Hence MATH. |
math/0011183 | Take an arbitrary MATH. Since MATH is a regular measure, there are compact sets MATH with MATH for MATH . Let MATH and take MATH such that MATH . For MATH we divide MATH into MATH groups, whose unions we call MATH, by putting MATH if MATH intersects MATH. Note that each MATH intersects at most one MATH. If it does not intersect any MATH, then we include it arbitrarily in some MATH. We have MATH . Since MATH is arbitrary and MATH is fixed, we have proved the result. |
math/0011183 | Assume by contradiction that there are MATH and a sequence of integers MATH going to MATH such that MATH . We know from REF that for every MATH there is a partition MATH of MATH such that MATH, MATH are unions of atoms of MATH, and MATH . Take MATH and MATH sufficiently large in order to MATH . Letting MATH, we know that there is some MATH for which MATH . From REF we have in particular MATH for every MATH, and so summing over all MATH we have MATH . Finally, from REF we get MATH which gives a contradiction (recall our choice of MATH). |
math/0011183 | We observe that the constant in CITE that bounds the distortion does not depend on the integer MATH that we have used for starting the construction of the partition (it essentially depends on the expansion rates of the maps MATH and MATH). This means that the same proof of REF applies to this situation with the constant MATH not depending on MATH. |
math/0011184 | For the first assertion, consider an invertible adjointable bounded operator MATH, MATH for MATH. Note, that the MATH-linear spans of the sequences MATH and MATH are non-trivial orthogonal to each other subsets of MATH. Since MATH for all MATH the linear span of the set MATH cannot be dense in MATH. To show the second assertion, fix an index MATH. By supposition MATH . Therefore, MATH by the property of the frames to be normalized tight, and MATH for every MATH. This implies the identity MATH. |
math/0011184 | By REF the frame transforms of similar standard frames of a fixed NAME C*-module have always the same range. So we can restrict our proof to the consideration of standard normalized tight frames. Note that the property to be standard transfers from the summands of an inner sum to the inner sum itself. We gave already the arguments for the first and the second statements in this special situation. To demonstrate REF let MATH be the standard orthonormal basis of MATH. If MATH and the operator MATH has a closed range, then MATH by the positivity of MATH, and the canonical norm MATH on MATH is equivalent to the standard norm on MATH. By CITE this equivalence of norms forces the equivalence MATH for some fixed constants MATH and any MATH, MATH. We obtain the following equality: MATH . Together with the equivalence REF above we get the frame properties of the inner sum MATH for the NAME MATH-module MATH. In case MATH the optimal constants MATH in REF are equal to one and the frame MATH is normalized tight. We leave the easy demonstration of the converse implications in REF to the reader. Suppose MATH and MATH are weakly disjoint. If there exists a non-zero element MATH then MATH for some non-zero MATH, MATH. The equality MATH shows the existence of a non-trivial orthogonal complement to the MATH-linear span of the sequence MATH, a contradiction to the assumptions on it. Conversely, assume MATH and the existence of an element MATH that is orthogonal to the MATH-linear span of the sequence MATH. Then MATH and MATH. Consequently, MATH, and by the injectivity of frame transforms we obtain MATH and MATH. So the frames MATH and MATH are weakly disjoint. |
math/0011184 | Let MATH be invertible adjointable bounded operators such that the sequences MATH and MATH are orthogonal bases of the NAME MATH-modules MATH and MATH, respectively. For the standard normalized tight frames MATH and MATH we have the identity MATH for MATH. By CITE the operator MATH has to be unitary. Consequently, the sequence MATH is an orthogonal NAME basis that is unitarily equivalent to the orthogonal NAME basis MATH. By CITE there exists a unitary operator MATH such that MATH holds for every MATH. Finally, we obtain MATH for MATH, the desired result. |
math/0011184 | By REF the ranges of the corresponding frame transforms MATH and MATH are orthogonal to each other subsets if and only if REF holds. Since any standard frame is similar to its canonical dual frame and strong disjointness is invariant under summand-wise similarity we obtain the equivalence of REF . To show the implication MATH observe that the frame transform MATH of the inner sum MATH equals MATH since the ranges of both the frame transforms are orthogonal to each other by REF . Consequently, MATH, and MATH. The frames MATH and MATH are always standard normalized tight frames of MATH and MATH, respectively. If they are strongly disjoint then the inner sum of them is a standard normalized tight frame of MATH since MATH. By REF and its proof the equalities MATH hold in MATH and MATH, respectively, for any MATH, MATH. Since the operators MATH and MATH are invertible we arrive at REF omitting them at the right end. We derive REF if we take into account that MATH for every MATH and MATH, where MATH runs over the entire set MATH if MATH does so. Analogously we derive REF . Conversely, if one of the formulae REF holds then the ranges of the frame transforms of the participating two frames are orthogonal in MATH and they are strongly disjoint by REF . Again, summand-wise similarity and invariance of strong disjointness under this equivalence relation gives REF , and therewith the other conditions. Starting with REF and MATH, MATH a short counting shows MATH and, therefore, MATH. We get REF that were already shown to be equivalent to REF . |
math/0011184 | Let MATH be a standard orthonormal basis of MATH, let MATH, MATH be the frame transforms of the frames MATH and MATH, respectively. The operators MATH are isometries with orthogonal ranges in MATH by assumption. Moreover, MATH and MATH, MATH for any MATH. Consider the new operator MATH and observe that MATH . So MATH is an isometry, and MATH is a standard normalized tight frame of MATH. |
math/0011184 | If MATH for any MATH then define a new operator MATH by MATH for every MATH. Obviously, the equality MATH holds for every MATH and, hence, MATH is unitary. Also, MATH for every MATH. So the st.n.t. frames MATH are strongly complementary. Conversely, if the st.n.t. frames MATH are strongly complementary then MATH for every MATH and, consequently, MATH, where the operators MATH possess orthogonal ranges by the choice of the frames. |
math/0011184 | Fix MATH. The operator MATH is adjointable and invertible since MATH . Because of the invertibility of MATH in the C*-algebra MATH the operator possesses an polar decomposition MATH where MATH is a unitary operator. Since MATH we obtain MATH and MATH where MATH is unitary. Consequently, MATH is the sought decomposition by REF . If MATH is a adjointable bounded MATH-linear map of MATH onto itself that has a polar decomposition MATH inside MATH, then MATH is a partial isometry with range MATH and MATH. Replacing MATH by MATH in REF we get MATH with partial isometries MATH. |
math/0011184 | Consider the adjoint MATH of the frame operator MATH induced by the frame MATH of MATH. It possesses a polar decomposition and is surjective. Consequently, MATH is a linear combination of two partial isometries by the previous lemma. Since MATH maps the elements of the standard orthonormal basis precisely to the frame elements, and since partial isometries applied to orthonormal bases yield standard normalized tight frames of the image module by REF we get the desired result. |
math/0011184 | We can replace the open maps by the surjective ones in CITE without changes in the proof, a fact first observed by NAME in CITE. Hence, we know that the open (respectively, surjective) bounded linear mappings on MATH form an open subset of the set of all bounded linear mappings on MATH. Intersecting the complement of this open set with the closed subset of all [adjointable] bounded MATH-linear operators on MATH and taking again the complement we obtain that the open (respectively, surjective) [adjointable] bounded MATH-linear operators on MATH form an open subset in MATH. Since the subset of all invertible [adjointable] bounded MATH-linear operators is an open subset of MATH by the spectral theory of NAME algebras we conclude that the invertible [adjointable] bounded MATH-linear operators form an open subset of the subset of all surjective [adjointable] bounded MATH-linear mappings on MATH. Again, by CITE the set of all invertible bounded linear mappings on MATH becomes a closed subset of the subset of all surjective bounded linear operators on MATH. Intersecting with the closed in MATH subset MATH the analogous statement for [adjointable] bounded MATH-linear maps yields. |
math/0011184 | If MATH is adjointable and invertible then MATH by the second part of REF , where MATH are unitary operators. Conversely, suppose MATH for complex numbers MATH and unitary operators MATH. Then MATH is naturally adjointable. If either MATH or MATH are zero, then MATH is invertible. So, without loss of generality we start the investigation with MATH for MATH and unitary operators MATH dividing our original decomposition by that complex number of MATH with the smaller absolute value. Set MATH for real MATH. Since MATH for every MATH and since MATH for MATH we obtain the injectivity of MATH for every MATH. On the other hand, MATH and MATH was supposed to be surjective. Applying REF we obtain for MATH close enough to MATH that the operator MATH has to be surjective since the sequence MATH converges to MATH in norm for MATH. Furthermore, MATH has to be invertible. |
math/0011184 | Consider the frame transform MATH of this frame. Then by the previous result its adjoint MATH is invertible if and only if it is a linear combination of two unitary operators. However, every orthonormal basis of MATH is connected to the fixed standard orthonormal basis of MATH by a unitary operator. What is more, MATH for every MATH and the standard orthonormal basis MATH of MATH. So the claimed decomposition property of the frame MATH is equivalent to the invertibility of MATH, and MATH has to be invertible, too. This in turn shows the property of the frame MATH to be a basis. |
math/0011184 | Let MATH be an orthonormal basis of MATH. The operator MATH defined by the formula MATH is the adjoint of the frame transform MATH, MATH is surjective and MATH equals the upper frame bound MATH of the frame MATH. It can be represented as MATH for three unitary operators MATH on MATH that depend on the choice of MATH, see REF . Setting MATH, MATH, MATH we are done. |
math/0011184 | Denote by MATH the (existing) projection of MATH onto the range of MATH. Note, that MATH for MATH and for the fixed orthonormal basis MATH of MATH used for the definition of the frame transform. Then MATH and MATH are both orthonormal bases of MATH, and MATH for every MATH. |
math/0011184 | The proof is given in the same manner as for REF , with slight modifications. If MATH is the adjoint operator of the frame transform arising from the given standard frame MATH of MATH then define an operator MATH by the formula MATH for any fixed MATH. Again, MATH and MATH forces MATH to be invertible and to possess a representation MATH for a unitary operator MATH on MATH by REF . (Note that MATH replaces MATH of the proof of REF .) Resolving the equation for MATH with respect to MATH we obtain MATH . Since MATH is unitary the sequence MATH forms an orthonormal basis of MATH, where MATH denotes the standard orthonormal basis of MATH used to build the frame transform MATH. The second summand MATH is injective and adjointable. Since MATH the operator MATH multiplied by MATH has to be invertible, that is, it is also surjective. Consequently, the sequence MATH forms a standard NAME basis of MATH. In total we obtain MATH as desired. |
math/0011187 | Let MATH and MATH, and let MATH. Then MATH, defined by letting MATH, is a function in MATH so MATH. |
math/0011187 | We prove REF by induction on MATH, working in MATH. The first part follows immediately. We fix the notation that MATH is the generic filter for MATH, for MATH. The case MATH is given by REF . For the case where MATH, there are two subcases. If MATH, then we may assume by the induction hypothesis that MATH and that there is a MATH-semi-generic MATH such that MATH and MATH. Then since MATH forces that MATH will be elementary in MATH, where MATH is the generic filter for MATH, we can replace MATH with a MATH with the additional property that MATH forces that MATH will be equal to MATH for some MATH. By REF , then, there is a MATH such that MATH, MATH and MATH is MATH-semi-generic. Such a MATH suffices. For the subcase MATH, by the induction hypothesis there is a MATH-semi-generic MATH such that MATH and MATH. Then since MATH forces that MATH will be elementary in MATH, where MATH is the generic filter for MATH, we can replace MATH with a MATH with the additional property that MATH forces that MATH will be equal to MATH for some MATH, and we can let MATH. Then as in the previous paragraph, by REF there is a MATH as desired. The case where MATH is a limit and MATH is similar, now fixing MATH and inducting on MATH. By the induction hypothesis for MATH we may assume that MATH and by the induction hypothesis for MATH we may assume that there is a MATH-semi-generic MATH such that MATH and MATH. Since MATH is forced to be MATH-semi-proper, by REF (or REF ) there is a MATH-semi-generic MATH below MATH such that MATH. For the case where MATH and MATH are both limits there are two subcases (in each of which we may assume that MATH). If MATH has cofinality MATH or MATH, or if MATH for each MATH, then we fix an increasing sequence MATH cofinal in MATH, with MATH. If the cofinality of MATH is countable, fix an increasing sequence of ordinals MATH cofinal in MATH, with MATH. Otherwise, let MATH be any ordinal in MATH greater than MATH, again with MATH. Let MATH and let MATH. Alternately choose conditions MATH, such that CASE: each MATH is a condition in MATH, CASE: each MATH is a condition in MATH, CASE: each MATH, CASE: each MATH, CASE: for all MATH, MATH, CASE: each MATH is MATH-semi-generic, CASE: each MATH is MATH-semi-generic, CASE: each MATH forces that for some condition MATH, MATH. For the case where no condition in any MATH makes MATH, we modify REF as follows: CASE: each MATH forces that for some condition MATH such that for some MATH, MATH. That such conditions exist is immediate by the induction hypothesis (REFa follows from REF ). Then the limit of the MATH's (call it MATH) will be the desired MATH-semi-generic, as long as it is below each MATH. This fact follows from the fact that MATH forces that MATH will be cofinal in MATH. This is clear if MATH, and if MATH it follows from the fact that each MATH will be cofinal in MATH since MATH is MATH-semi-generic. For the remaining case it follows from REF . If MATH for some MATH, then we may assume that MATH and so MATH in the MATH-extension. Then we may apply the previous argument in the MATH-extension, along with REF . |
math/0011187 | By induction on MATH. If MATH has a last model, then we are done by the extension assumption on countable elementary submodels of MATH. For the limit case, applying the induction hypothesis we may assume that every condition already holds, except REF . Let MATH . Applying the end-extension assumption, let MATH be a countable subset of MATH such that MATH and such that, letting MATH, we have CASE: MATH, CASE: MATH, CASE: MATH. Let MATH be a bijection, and fix an increasing sequence MATH cofinal in MATH with MATH. Now let each MATH where MATH is the largest MATH such that MATH. Since these are all finite extensions, each initial sequence of the new sequence is an element of the later models. By the choice of MATH, each MATH, and so MATH is as desired. |
math/0011187 | By induction on MATH, using REF and working in the extension by MATH. Fix MATH and MATH with MATH, and let MATH be as in REF , with respect to MATH. If MATH has a final model (that is, if MATH is a successor), then we can choose a MATH-semi-generic MATH extending MATH by the induction hypothesis (see REF ). Then since MATH any MATH-generic for MATH extending MATH will suffice as the desired condition. For the case where MATH is a limit, we let MATH and MATH be as in the proof of REF , and fix an increasing sequence MATH of ordinals cofinal in MATH. Then letting MATH be MATH we can successively pick conditions MATH such that each MATH is a MATH-semi-generic in MATH extending MATH, and such that the last member of each MATH contains MATH. By REF from REF , the limit of the MATH's (adjoined by their union) will be a condition in MATH, and it will be MATH-semi-generic since it extends a MATH-semi-generic for each MATH. |
math/0011187 | Given MATH-generic for MATH, one can easily build a corresponding potential-MATH-generic MATH, letting MATH be the empty condition when MATH . The point then is just to find a MATH-generic filter MATH with MATH, such that MATH for all MATH. Inducting primarily on MATH, and secondarily on MATH, we show that if MATH and MATH is a MATH-generic filter for MATH such that MATH and MATH for all MATH, then there exists a MATH-generic MATH such that MATH, MATH and MATH for all MATH. Note that for each MATH, since MATH (and MATH), the MATH-genericity of MATH can be verified in MATH (MATH itself is not in any member of MATH, so this is not automatic). The base case MATH is trivial. For the successor step, we may assume that MATH . Then since for the successor case there is no restriction on the last coordinate of the MATH-generic filter (other than MATH-genericity), MATH can be extended in any fashion to a MATH-generic for MATH. For the limit, fix an increasing sequence MATH cofinal in MATH, and let MATH. Now in MATH many steps alternately pick CASE: MATH-generic MATH in MATH (applying the induction hypothesis plus the elementarity of MATH plus the fact that MATH, as in REF ) with MATH such that CASE: MATH for all MATH, CASE: MATH for all MATH, CASE: MATH in MATH meeting the MATH-th dense set in MATH for MATH such that MATH (such a MATH exists because MATH is MATH-generic for MATH). Since each MATH, its initial segments are automatically in the corresponding MATH's. Then MATH generates a MATH-generic filter MATH for MATH, and for each MATH, MATH. Then by the induction hypothesis, MATH for all MATH, since for all MATH with MATH, MATH. |
math/0011187 | Towards a contradiction, fix the least MATH for which the lemma fails, and let MATH be a condition in MATH forcing that MATH adds a new MATH-sequence of ordinals. Let MATH be a MATH-system with MATH and MATH a potential-MATH-generic, as given by the hypothesis of the lemma. Then there exists a MATH-name MATH in MATH such that MATH forces that MATH will be a new MATH-sequence of ordinals. Let MATH be a MATH-generic filter witnessing that MATH is a potential-MATH-generic. We wish to see that MATH is below each member of MATH. Then we will be done, as for each integer MATH there is a member of MATH intersecting the antichain in MATH determining the MATH-th member of MATH. So we will show by induction on MATH that MATH for all MATH (simultaneously). The cases where MATH or MATH is a limit and MATH is cofinal in MATH are clear. For the successor step from MATH to MATH, MATH is a MATH-name in MATH for a countable ordinal, so by REF , MATH forces that MATH . Then MATH forces that MATH. By REF , MATH, so MATH forces that MATH extends MATH. Now fix MATH. Since MATH forces that MATH extends MATH, we have that MATH. Lastly, for the case where MATH is a limit and MATH is not cofinal in MATH, note that MATH has uncountable cofinality, and also that we have shown that each MATH, MATH is MATH-distributive. By REF then, densely many conditions in MATH (and therefore MATH) are conditions in some MATH, MATH, and so densely many conditions in MATH are in some MATH with MATH. |
math/0011187 | Since MATH is regular, each MATH has bounded range below MATH. If MATH is in MATH then this bound exists in MATH. |
math/0011187 | Let MATH be countable with MATH. For some MATH we build MATH, MATH and MATH satisfying the following conditions. CASE: Each MATH is a countable elementary submodel of MATH. CASE: Each MATH is a countable elementary submodel of MATH. CASE: Each MATH is the MATH-th ordinal in MATH (MATH being the MATH-th ordinal). CASE: MATH. CASE: For all MATH, MATH. CASE: For all MATH, MATH. CASE: For all MATH and for all MATH, MATH. CASE: For each MATH, MATH is a minimal MATH-extension of MATH, CASE: If MATH is a limit ordinal, then MATH. CASE: If there exists a MATH such that MATH, then MATH is the least such MATH; otherwise MATH. Given MATH, by applying REF in the statement of REF repeatedly, once for each MATH with MATH, we can choose each MATH to meet REF . While the model MATH resulting from this repeated application may not be a minimal MATH-extension of MATH, MATH will be, as MATH, so we can take this MATH as our MATH. The rest of the construction is straightforward. We claim that for each MATH the sequence MATH lists the first MATH ordinals of MATH in increasing order. This follows by induction on MATH. It is clear when MATH, and when MATH it follows from REF and the fact that MATH (so in this case MATH). If MATH is a limit ordinal, we have from the fact that MATH and the induction hypothesis that MATH lists the first MATH ordinals of MATH in increasing order. Then the definition of MATH finishes the proof of the claim. Next we claim that MATH for some MATH, and so MATH. Given this we are done, letting MATH and MATH. All the conditions of REF are satisfied trivially, aside from REF is satisfied since for each MATH, there is some MATH such that MATH (and so MATH). Then MATH and MATH were chosen to satisfy REF by REF of the construction, and this relationship was preserved for all later MATH by REF . Assume to the contrary that MATH. If MATH, then since every ordinal in MATH is equal to some MATH, there is a limit ordinal MATH such that MATH . But then MATH, contradicting MATH. So we may assume that MATH. We will show that the cofinality of MATH is countable, which is a contradiction since MATH is increasing and cofinal in it. If MATH for some MATH with MATH, then MATH, which is countable. If not, MATH. If MATH is singular, let MATH be the cofinality of MATH. Then MATH for some MATH with MATH, and there exists a cofinal map MATH in MATH. Since MATH, MATH is a countable set cofinal in MATH. The last remaining case is that MATH is a regular limit cardinal. Let MATH be least with MATH, and fix a cofinal sequence in MATH. By REF of the construction, this sequence is cofinal in MATH. |
math/0011187 | We first note that the lemma follows from the restricted version of the lemma where MATH. To see this, fix MATH as given by the hypothesis of the unrestricted version, and note that the restricted version then gives a MATH such that MATH and MATH is a MATH-system. Now, if MATH is MATH-generic with MATH, then MATH (since MATH is a potential-MATH-generic and MATH) and the following hold. CASE: MATH, CASE: MATH . Furthermore, in MATH, CASE: MATH CASE: MATH, MATH is MATH-semi-generic, CASE: MATH. Applying the elementarity of MATH in MATH, we see that there exists a condition MATH satisfying these conditions. By REF , then, there is a condition MATH such that MATH and MATH forces that MATH is a MATH condition extending such a MATH. This MATH suffices. We have REF by the properites listed above for MATH, REF by our extension, and the others by the assumptions of the lemma. The restricted version of the lemma follows by induction on MATH. Note that our induction hypothesis entitles us (once we have fixed MATH and MATH) to assume that the unrestricted version holds whenever MATH. Now, when MATH there is nothing to show. The argument for increasing MATH by one follows from the case where MATH, and this case is also trivial (see REF ). The only remaining case then is when MATH and MATH is a limit ordinal. Fix an increasing sequence MATH cofinal in MATH with MATH, and succesively apply the induction hypothesis (in MATH, as opposed to some submodel). That is, let MATH, and given MATH, let MATH be such that MATH and MATH is a MATH-system. Then each MATH is MATH-semi-generic, and the limit of the MATH's is the desired MATH. REF is then easily satisfied (the second half being vacuous), and REF , being local properties, are satisfied by the induction hypothesis. |
math/0011187 | It suffices to prove the theorem for the case MATH, since then we may repeat the construction any countable ordinal number of times. To see that MATH is preserved, note that if after repeating the construction some countable number of times we have the MATH-end-extension MATH of MATH and we have added MATH (possibly among other ordinals) to MATH and MATH is in MATH, then MATH. Further, if MATH, then by REF , MATH. Now, to prove the theorem for the case MATH, let MATH be the MATH-least normal measure on MATH, and fix MATH as in the statement of the theorem. Let MATH be any member of MATH, and let MATH . The key point is that if MATH is a regressive function in any elementary submodel of MATH, then MATH is constant on a set MATH, and since MATH is MATH-least, MATH and therefore MATH and the constant value are also in the model. Along with the fact that MATH for any MATH with MATH, this shows that MATH satisfies MATH. To see that MATH is a proper end-extension of MATH, note that MATH, and since any ordinal MATH is of the form MATH for some MATH in MATH, if MATH then any corresponding MATH is regressive on a set in MATH, and so MATH, by the key point. |
math/0011187 | Let MATH be a continuous increasing sequence of cardinals with supremum MATH strongly inaccessible such that each MATH is a measurable cardinal. Fix a regular cardinal MATH, and let MATH be a wellordering of MATH. Let MATH be such that each MATH is the set of countable subsets of MATH of ordertype greater than MATH. Then by REF are satisfied (REF by the fact that small forcing preserves measurable cardinals), so any forcing MATH as in the statement of REF is semi-proper and MATH-distributive. We may further require that for every function MATH added by some initial stage of the iteration there is a MATH such that MATH. Each MATH forces that MATH and that any canonical function for MATH dominates MATH on a club, so MATH forces Bounding. Further, since MATH in MATH and MATH forces MATH, MATH forces CH. |
math/0011187 | For the forward direction, we adapt the proof that MATH sets are projections of trees on MATH (see CITE). Let MATH be a tree on MATH, let MATH and let MATH. Recall that for all MATH, MATH is wellfounded. For each MATH, let MATH be the least ordinal (necessarily countable) such that there exists a function MATH with the property that if MATH are such that MATH is a proper initial segment of MATH, then MATH. Fix MATH, and for each MATH let MATH. Now fix MATH and a bijection MATH such that for every MATH in a fixed club MATH, MATH. Let MATH be a bijection such that MATH for all MATH. Let MATH be the tree on MATH consisting of all pairs MATH such that if MATH and MATH is a proper initial segment of MATH with MATH both in MATH, then MATH. Then MATH, and, for each MATH, if MATH is the least (again, necessarily countable) ordinal MATH such that MATH and MATH is an order preserving embedding of MATH into MATH, then MATH is a path through MATH. Since MATH for all MATH, this shows that MATH satisfies the definition of NAME Bounding for MATH. For the other direction, let MATH be the set of functions coding wellordings of MATH under some fixed coding with the property that for each MATH and MATH, MATH codes how MATH compares with each MATH. Each function from MATH to MATH induces a corresponding function from MATH into MATH. Let MATH be such a function, and let MATH be the tree given by NAME Bounding, with MATH the witnessing club set. We have a partial, not necessarily transitive, order MATH on the sequences in MATH, where MATH if one extends the other, and, MATH being the longer one, the first cordinate of the last element of MATH codes that MATH in the corresponding ordering. Let MATH be the least transitive ordering containing MATH. Then MATH is a wellfounded partial order. Seeing this requries checking that the resulting order is antireflexive and wellfounded. To see antireflexivity, assume that MATH is the shortest possible counterexample. First note that MATH and MATH cannot be comparable in MATH (or identical) since then there would be a shorter counterexample, removing the MATH's and placing either MATH and the beginning of the sequence or MATH at the end if they are unequal. Therefore MATH and MATH must be incompatible extensions of MATH, and MATH must be comparable with MATH and distinct from MATH. But then depending on whether MATH or MATH, there is a shorter counterexample, removing either MATH or MATH from the original sequence. To see wellfoundedness, let MATH REF be a descending sequence in MATH. First note that if some MATH has infinitely many extensions in the sequence, then all but finitely many of them must be extensions of a fixed immediate successor of MATH, since otherwise initial segments of MATH are visited infinitely often by the sequence, which is impossible (using antireflexivity of MATH), there being only finitely many of them. But the empty sequence has infinitely many extension in the sequence, which means that we can build an infinite chain though MATH all of whose members have this property. We claim that infinitely many members of this chain must be in the sequence, which gives a contradiction since the chain codes a wellordering. To see the claim, fix a member MATH of the chain, and an arbitrary integer MATH. Then there is some MATH, MATH, extending MATH, and a member MATH of the chain with length greater than MATH. Let MATH be an extension of MATH, for some MATH. Then if MATH is not on the chain, there must be some MATH in the interval MATH such that MATH is an initial segment of MATH. Since MATH was arbitrary, the claim follows. Now extend MATH to a wellordering MATH of MATH, and let MATH be the length of MATH. Let MATH be a bijection, and define MATH by letting MATH be the ordertype of MATH restricted to MATH. Then MATH is a canonical function for MATH. Furthermore, for a club MATH of MATH, MATH. For these MATH, MATH is greater than the ordertype of every wellordering in the projection of MATH, and thus greater than MATH. |
math/0011187 | This follows from the following standard facts about measures, where MATH is an extension by a forcing MATH such that MATH. CASE: For every MATH-complete measure MATH on MATH in MATH, MATH. (Otherwise, densely often in MATH there is a set in MATH whose membership in the measure is undecided; by genericity then there will be a subset of the measure of size MATH with empty intersection.) CASE: For every MATH-complete measure on MATH in MATH, every positive set contains a positive set in MATH. (For each MATH-name for a positive set and each condition in MATH, consider the set of sequences MATH forces into the positive set.) CASE: Every MATH-complete measure on MATH in MATH extends to one in MATH. (All sets containing positive sets from the ground model.) For the forward direction, let MATH be a MATH-homogeneous tree on MATH in MATH such that MATH, for some MATH. Let MATH witness that MATH is MATH-homogeneous. Each MATH extends a measure in MATH on MATH, and since MATH is MATH-distributive, the corresponding function MATH taking each MATH to the restriction of MATH to MATH exists in MATH. For each MATH, let MATH be a witness to the fact that MATH is not countably complete. For each MATH, let MATH . Since every positive set for each MATH contains one from MATH, and since MATH is MATH-distributive, we can assume by shrinking if necessary that MATH is in MATH. Now let MATH be the set of pairs MATH such that MATH. Since the measures are all MATH-complete, each MATH is positive for MATH. Then MATH is MATH-homogeneous (with MATH as a witness) with the same projection as MATH. For the other direction, assume that MATH (on MATH) and MATH in MATH witness that MATH is MATH-homogeneously NAME. Extend the MATH's to MATH-measures, inducing a function MATH. Since each positive set in MATH contains one in MATH, and since no MATH-sequences of ordinals have been added by MATH, for each MATH the countable completeness of the corresponding tower is not changed by MATH. Since no countable sets of ordinals have been added the projection of MATH is the same, so MATH (along with MATH) witnesses in MATH that MATH is MATH-homogeneously NAME. |
math/0011187 | Fix MATH and MATH. By REF , it suffices to show that we can deal with MATH, as we can just repeat the process MATH times. For each MATH, let MATH. Then by the definition of MATH-homogeneity (that is, countable completeness) there exists MATH such that for all MATH, MATH and MATH. Now MATH is as desired, since by the MATH-completeness of each MATH, if MATH and MATH then MATH is constant on a set MATH, and so this constant value must be an element of any elementary submodel of MATH with MATH and MATH as members. Further, if MATH is in MATH, then MATH, so MATH is the corresponding constant value. |
math/0011190 | The sequence REF splits if and only if the induced map MATH is zero. We have the commutative diagram MATH and we can naturally identify MATH with MATH. Therefore, the map REF factors through the NAME map MATH, the restriction MATH and the surjection MATH . In short, we have MATH . The last map MATH is actually an isomorphism by the following argument. By the standard exact sequence MATH we have the exact sequence MATH . Notice that MATH classifies the embedded deformations of MATH and MATH classifies the versal deformations of MATH. To show that MATH maps nontrivially to MATH, it suffices to show that as MATH varies in the pencil MATH, the corresponding NAME map to the tangent space of the versal deformation space of MATH at the origin is nontrivial, or equivalently, the map to the versal deformation space of MATH is unramified over the origin. To see this has to be true, we only need to localize the problem at the node MATH of MATH: if the map to the versal deformation space is ramified over the origin, then MATH is locally given by MATH at MATH for some MATH; however, this is impossible since MATH is smooth at MATH. Therefore, the map MATH is nonzero and hence must be an isomorphism. Thus we conclude that MATH is an isomorphism. We have the exact sequence MATH . Combining REF, we are left to show that the image of the map MATH does not contain MATH as in REF. We claim that that the image of MATH is contained in the subspace MATH of MATH perpendicular to MATH, that is, MATH . By NAME duality, we have the following commutative diagram: MATH . So we may identify the map MATH with MATH, which is the same as MATH on the NAME cohomologies. For any MATH, we have MATH . So REF follows. On the other hand, we have REF. It is trivial that MATH and MATH are linearly independent in MATH. So MATH and a general NAME class MATH does not lie in MATH and hence MATH. Therefore, MATH maps nontrivially to MATH. Consequently, the map REF is not zero and the sequence REF does not split. |
math/0011190 | We start with MATH which is smooth at MATH. Choose local coordinates such that MATH is cut out by MATH at MATH. Blow up MATH along MATH and we obtain that MATH where MATH is the affine coordinate of MATH such that MATH is given by MATH and MATH is given by MATH. We see from REF that MATH has a rational double point at MATH. At a point MATH and MATH, that is, for MATH, MATH is analytically equivalent to REF for MATH. At MATH, that is, at MATH, MATH is given by MATH where MATH; this is equivalent to REF for MATH. Notice that MATH is cut out by MATH. Blow up MATH along MATH and we obtain that MATH where MATH is the affine coordinate of MATH such that MATH is given by MATH and MATH is given by MATH. Obviously, MATH is given by REF at MATH. At a point MATH and MATH, that is, for MATH, MATH is analytically equivalent to REF for MATH. At MATH, that is, at MATH, MATH is given by MATH where MATH; this is equivalent to REF for MATH. Apply this argument inductively for MATH and we are done. |
math/0011190 | See CITE. |
math/0011190 | See CITE. |
math/0011190 | Let MATH be a maximal NAME and NAME . Since MATH is maximal, there does not exists MATH such that MATH and there is no curve MATH. So MATH has to pass through MATH. Similarly, there is no point MATH such that MATH and hence MATH must pass through MATH. Applying REF to the point MATH, we see that the branch of MATH at MATH is joined to either the branch of MATH at MATH or a component MATH dominating MATH over MATH. If it is the former case that the branch of MATH at MATH is joined to the branch of MATH at MATH over MATH, it contradicts the fact that the dual graph of MATH is a tree. Otherwise, if MATH is joined to MATH over MATH, we continue to apply REF to the point MATH and see that MATH is joined to either MATH or a component MATH dominating MATH over MATH. If it is the former case, we again get a circuit in the dual graph of MATH. We may continue this argument and obtain that MATH is joined to MATH over MATH, MATH is joined to MATH over MATH and so on; finally, we have MATH is joined to MATH over MATH, where MATH is a component dominating MATH. As mentioned before, there is no curve MATH. So MATH is joined to MATH over MATH. Once again, we obtain a circuit in the dual graph of MATH. Contradiction. REF illustrates our argument. Here MATH passes through MATH and MATH. Then there will be a loop between MATH and MATH on MATH and consequently, MATH. This is a contradiction. |
math/0011190 | By REF, we have MATH for MATH. So REF is equivalent to the statement that MATH meets MATH only at the points MATH and meets MATH only at the points MATH. Obviously, MATH must pass through MATH since there is no curve MATH. For the same reason, MATH. Suppose that MATH for some MATH and MATH is the largest number for this to hold. Applying REF to MATH, we see that MATH is joined to a component MATH dominating MATH over MATH; continuing applying REF, we see that MATH is joined to a component MATH dominating MATH over MATH, MATH is joined to MATH dominating MATH over MATH and so on. Finally, we have MATH dominating MATH is joined to MATH over MATH and we obtain a circuit in the dual graph of MATH. Contradiction. Therefore, MATH. Similarly, MATH. If MATH is not totally separated at MATH, we have three cases CASE: a component MATH dominating MATH is joined to a component MATH dominating MATH over MATH; CASE: a component MATH dominating MATH is joined to MATH over MATH; CASE: a component MATH dominating MATH is joined to MATH over MATH. In either of these cases, we can argue in the same way as before to show that there is a circuit in the dual graph of MATH. Therefore, MATH is totally separated at MATH. As a consequence, by REF MATH can be neither tangent to MATH at MATH nor tangent to MATH at MATH. So if MATH meets MATH and MATH at MATH and MATH, it must meet MATH and MATH transversely at these points. Combining this with the fact that MATH, we obtain REF. Finally for REF , if MATH is not totally separated at MATH, then MATH will be joined to a component MATH dominating MATH over MATH. Again, we may use the same argument as before to show that there is a circuit in the dual graph of MATH. |
math/0011190 | Since MATH, all the equalities in REF must hold. Then REF follow immediately. As for REF , we notice that the equality in REF has to hold. So we must have MATH, MATH, MATH and so on, where MATH are defined by REF. It follows immediately that REF holds for MATH. Similarly, REF holds for MATH. And by REF, we see that MATH meets MATH and MATH transversely everywhere. Obviously, REF holds for MATH and MATH since MATH. Suppose that REF fails for some MATH with MATH and MATH is the largest number with this property. Then there exists a component MATH dominating MATH with a map of degree at least MATH. We claim that REF MATH is joined to at least two different components MATH over the point MATH, where MATH or MATH dominates MATH for MATH. If the map MATH is not totally ramified over MATH, there are at least two distinct points MATH such that MATH for MATH where MATH is the map from MATH to MATH. Then by REF, the branch of MATH at MATH is joined to a component MATH over the point MATH for MATH, where MATH or MATH. This justifies our claim REF in the case that MATH is not totally ramified over MATH. If MATH is totally ramified over MATH, MATH meets MATH at MATH with multiplicity at least MATH. Again by REF (see also REF), MATH is joined to a union of components MATH over MATH such that MATH and MATH meets MATH at MATH with multiplicity at least MATH. Our assumption on MATH implies that REF holds for MATH, that is, every component of MATH dominating MATH maps birationally to MATH. And since MATH meets MATH transversely at MATH if MATH, we see that MATH contains at least two different components dominating either MATH or MATH and hence REF follows. Starting with (MATH), we may argue as before to show that each MATH is joined by a chain of components over MATH to MATH for MATH. And hence there is a circuit in the dual graph of MATH. Contradiction. So REF holds for each MATH with MATH. A similar argument shows that REF holds for each MATH with MATH. |
math/0011190 | It follows from the correspondence MATH . |
math/0011191 | As observed in the introduction, MATH. It is known that the NAME Conjecture with coefficients in an arbitrary MATH-algebra is true for the group MATH (and much more generally: CITE CITE CITE). This implies that MATH coincides with its ``MATH-part", and MATH may be computed as the limit of a NAME spectral sequence CITE. The initial terms of the spectral sequence are MATH, the MATH homology of the group MATH with coefficients in the module MATH. (See CITE for an explanation of spectral sequences and their convergence.) Noting that MATH for MATH odd (since the algebra MATH is MATH), it follows that MATH for MATH odd. Also MATH for MATH, and the differential MATH is zero. Thus MATH . To clarify notation, write MATH. We have a free resolution MATH of MATH over MATH given by MATH . It follows CITE that MATH is the homology of the complex REF . Therefore MATH . Convergence of the spectral sequence to MATH (see CITE) means that MATH and that there is a short exact sequence MATH . The group MATH is free abelian. Therefore the exact sequence REF splits. This proves the result . |
math/0011191 | Note that MATH. Now MATH . If MATH then MATH is a minimal projection in MATH, and so its class in MATH equals MATH. Therefore MATH . Consequently MATH . This proves the result. |
math/0011191 | REF follow from the definitions. To prove REF note that MATH. |
math/0011191 | Denote by MATH the equivalence classes in the relevant homology groups. REF-homology: If MATH, then MATH, the zero element in REF-homology group. REF-homology: Let MATH with MATH. Then MATH in REF-homology group. Likewise for MATH. REF-homology: Let MATH with MATH. Then MATH in REF-homology group. |
math/0011191 | The direct limit of maps makes sense because the diagram MATH commutes. The diagram MATH commutes by REF . Since MATH the result follows from the uniqueness assertion in the universal property of direct limits CITE. |
math/0011191 | We have MATH. Hence, by REF (and REF ), MATH . Also MATH . Since MATH and MATH, it follows that MATH . Finally MATH where the last equality follows from the NAME normal form for integer matrices. |
math/0011191 | Define MATH by MATH for MATH and MATH. Then MATH is the type preserving isometry of MATH given by reflection in the edge MATH. (See REF .) Now define a permutation MATH by MATH. If MATH then it is clear that MATH. The situation is illustrated, not too cryptically we hope, in REF , where the tiles are located in the building MATH and, for example, the tile labeled MATH is the range of a suitable isometry MATH with MATH. Let MATH be the permutation matrix corresponding to MATH. Then MATH. Clearly MATH. Therefore MATH. A similar argument proves the other equality. |
math/0011191 | The NAME of a rank one NAME algebra MATH can be characterized as follows (see CITE): MATH . By REF , we have MATH for the algebra MATH. Since stably isomorphic algebras have the same NAME, it follows that if MATH is stably isomorphic to a rank one NAME algebra then MATH is torsion free. On the other hand, suppose that MATH is torsion free. Let MATH be the class in MATH of the identity element of MATH. By a result of CITE, there exists a simple rank one NAME algebra MATH such that MATH with the class of the identity in MATH being MATH. Since MATH is torsion free we necessarily have MATH and by REF we also have MATH. Thus MATH and the identity elements of the two algebras have the same image in MATH. Since the algebras MATH and MATH are purely infinite, simple, nuclear and satisfy the Universal Coefficient Theorem, it now follows from the Classification Theorem of CITE that they are isomorphic. |
math/0011191 | CASE: Let MATH be the standard set of generators for the free abelian group MATH and MATH that of MATH. Define the map MATH by MATH. Observe that the diagram MATH commutes. Therefore so does the diagram MATH . Hence there is a well defined map of cokernels, which is surjective because MATH is. The kernel of MATH is generated by MATH. Now if MATH then according to the hypothesis of the lemma, MATH and so MATH. Hence the kernel of MATH is contained in the image of the map MATH. It follows by diagram chasing that the map on cokernels is injective. CASE: The argument in this case is a little harder but similar. Note that the vertical maps in the diagrams go up rather than down. |
math/0011191 | Suppose that MATH has been chosen. Refer to REF . In the link of the vertex MATH, let the vertices of type MATH correspond to points in MATH and the vertices of type MATH correspond to lines in MATH. There are then MATH choices for a line incident with the point MATH; therefore there are MATH choices for MATH. After choosing MATH there are MATH choices for the point MATH. That choice determines MATH. There are therefore MATH choices for MATH. This proves that for each MATH, there are MATH choices for MATH such that MATH. That is, each column of the matrix MATH has precisely MATH nonzero entries. A similar argument applies to rows. |
math/0011191 | If MATH with MATH then the covariance condition for the action of MATH on MATH implies that MATH. The result now follows because equivalent idempotents belong to the same class in MATH. |
math/0011191 | Referring to REF , we have for each MATH, MATH, where the sum is over all MATH such that MATH lies as shown in REF ; that is the sum is over all MATH such that MATH. Now by REF , MATH and so MATH . It follows that MATH . By REF , there are MATH nonzero terms in this double sum and each term MATH occurs exactly MATH times. Thus MATH, which proves the result. |
math/0011191 | By CITE, the algebra MATH is isomorphic to the algebra MATH, which is in turn stably isomorphic to the algebra MATH CITE. We refer to REF for notation and terminology. Recall that MATH where MATH. The isomorphism MATH has the effect MATH and the isomorphism MATH sends MATH to a minimal projection MATH. As an abelian group in terms of generators and relations, we have MATH . By REF MATH is isomorphic to MATH. Under this identification, MATH maps to the coset of MATH. By REF that coset maps to MATH under the injection of REF . Thus the order of MATH is equal to the order of MATH in MATH. Each of the relations in REF expresses a generator MATH as the sum of exactly MATH generators. It follows that there exists a homomorphism MATH from MATH to MATH which sends each generator to MATH. As MATH has MATH terms, MATH . Consequently, the order of MATH is MATH . The result follows since the order of MATH is necessarily a multiple of the order of MATH. |
math/0011192 | Let MATH be a nonempty simplex of MATH and define MATH . Then MATH is a compact subgroup of the locally compact group MATH. (See, for example, the proof of REF.) Since MATH is a discrete subgroup of the automorphism group of MATH, it follows that MATH is a finite subgroup of MATH. Since MATH is torsion free, MATH. |
math/0011192 | Fix MATH. By REF , MATH, a set containing precisely MATH elements. Since MATH acts freely without inversion REF , the set MATH also contains precisely MATH elements, each of which maps to MATH under MATH. Now suppose that MATH for some MATH. Then MATH for some MATH. Thus MATH and so MATH . This proves that MATH is MATH-MATH. |
math/0011192 | If MATH with MATH then the covariance condition for the action of MATH on MATH implies that MATH. The result now follows because equivalent idempotents belong to the same class in MATH. |
math/0011192 | REF implies that MATH, where MATH is the greatest common divisor of MATH and MATH. There are different outcomes, depending on the value of MATH. CASE: If MATH, then MATH is a multiple of MATH, since MATH is an integer. Using the fact that MATH gives MATH . CASE: If MATH then MATH and so MATH. This does not improve REF . CASE: If MATH then again the fact that MATH implies that MATH . |
math/0011207 | Note that for each MATH - module MATH, MATH . |
math/0011207 | Consider the commutative diagram MATH where MATH is the adjunction isomorphism, MATH is the mapping analogous to MATH with respect to the pairing MATH and MATH is the homomorphism induced by MATH. This last morphism is monic and MATH is monic. Therefore, MATH is a monomorphism. |
math/0011207 | Assume MATH is MATH - pure and let MATH. Assume MATH for all MATH. By REF there are MATH and MATH such that MATH . As MATH we have MATH and by purity MATH as an element in MATH. By REF MATH is injective. Assume now MATH is injective. The diagram MATH is commutative where MATH denotes inclusion and MATH. Since MATH is injective, we get MATH is injective and MATH is MATH - pure. |
math/0011207 | The proofs of these facts are formally the same that CITE, using REF instead of CITE . |
math/0011207 | Define MATH and consider the MATH - submodule MATH generated by the elements MATH. Since MATH is noetherian, MATH is finitely generated, and there are MATH for which MATH. For any MATH we have the set MATH . For each MATH, choose MATH. This gives a map MATH. Now. it is clear that there are maps MATH such that MATH. Finally, MATH. But this is an equality in MATH, whence, for each MATH we obtain MATH, and hence MATH. |
math/0011207 | CASE: By CITE MATH is a coalgebra. The rest of the first statement is similar to the argument in CITE due to the fact that over noetherian rings, submodules of finitely generated modules are finitely generated. CASE: Let MATH and assume MATH to be a cofinite left ideal contained in MATH. Since MATH is noetherian, it is easy to check that MATH is a left cofinite ideal contained in MATH. A diagram chase shows that MATH is a MATH - coalgebra map and the second statement is proved. REF is clearly satisfied. By CITE MATH is pure in MATH, so REF give injectivity of MATH for each MATH - module MATH. CASE: Since MATH is injective MATH can be viewed as a MATH - submodule of MATH. Let us see that MATH. If MATH and MATH, then MATH by REF . It follows that MATH is finitely generated by MATH as MATH - module for every MATH. Moreover, REF easily implies that the comultiplication and the counit on MATH are induced from MATH. |
math/0011207 | Assume MATH for all MATH. Let MATH and consider MATH for all MATH. Then MATH and so MATH for all MATH, which implies MATH, since the pairing MATH is a rational pairing by REF . |
math/0011207 | Following CITE, we consider MATH, as the argument for MATH is similar. Straightforward computations show that MATH is an algebra morphism. To see the injectivity we define MATH and MATH as follows: MATH where MATH is the composition inverse of MATH. We can see that MATH as in CITE. Moreover, MATH is a rational pairing by REF , so MATH is injective. |
math/0011207 | The computations in CITE remain valid here once we have proved REF . |
math/0011207 | Let MATH and let MATH. Assume MATH for every MATH. Notice that for each MATH, MATH for some cofinite ideal MATH of MATH. Put MATH. Then MATH is cofinite. Since MATH there exists some ideal MATH such that MATH is finitely generated and projective as MATH - module (and so MATH). Let MATH be a finite dual basis for MATH. Since MATH for all MATH, we get MATH . Hence MATH is a rational system by REF . |
math/0011207 | It follows directly from REF . |
math/0011207 | Let MATH be a cofinite left ideal. Then MATH is a cofinite left ideal because MATH is noetherian. Since MATH belongs to MATH, there exists MATH such that MATH is finitely generated and projective in MATH. By the natural isomorphism MATH is finitely generated and projective in MATH. Since MATH and MATH we get MATH is in MATH. |
math/0011207 | By purity MATH and MATH are MATH - submodules of MATH. Since the diagram MATH is a pushout diagram, the result follows. |
math/0011207 | Consider the canonical maps MATH and MATH . Put MATH and MATH. Since MATH is noetherian, MATH and MATH are cofinite left ideals of MATH and MATH, respectively. Since MATH there exist MATH and MATH such that MATH and MATH are finitely generated and projective in MATH. Let MATH. Since MATH and MATH are pure submodules we have MATH as desired. By REF MATH hence MATH is finitely generated and projective and MATH is in MATH. |
math/0011207 | Since MATH are in MATH, MATH is pure in MATH and MATH is pure in MATH by REF . So MATH. Let MATH be the morphism: MATH . By CITE this map is injective, so the statement will be clear once we have seen MATH. So let MATH and let MATH and MATH be left ideals contained in MATH and MATH respectively and such that MATH, MATH are finitely generated and projective (they exist because MATH belong to MATH). Since MATH and MATH are pure, by REF MATH is a cofinite left ideal, which is contained in MATH. As MATH is bilinear it is clear that MATH. Let MATH, and assume MATH to be a cofinite left ideal contained in MATH. By REF there exist left ideals MATH and MATH such that MATH and MATH are finitely generated and projective in MATH and so that MATH. By REF there is an epimorphism MATH which induces a monomorphism MATH . So there exist elements MATH and MATH such that MATH. This completes the proof. |
math/0011207 | Let MATH be a reversible polynomial (that is, MATH in a unit in MATH). Notice MATH . Put MATH and consider the MATH - linear map MATH . Clearly MATH is a MATH - algebra homomorphism and MATH. Moreover, MATH. Clearly MATH is surjective if and only if MATH. Notice MATH . So MATH . Hence MATH and we conclude that MATH is surjective. |
math/0011207 | CASE: Let MATH be a reversible ideal. Then MATH contains a reversible polynomial MATH. By REF , MATH which implies, by CITE, that MATH is finitely generated as a MATH - module. Therefore, MATH is finitely generated as a MATH - module. CASE: Since MATH embeds in the finitely generated MATH - module MATH, we get that MATH is finitely generated as a MATH - module. By CITE, there exists a monic polynomial MATH. We know MATH is a NAME MATH - algebra with antipode MATH . Since MATH is bijective, MATH as MATH - modules. So there exists a monic MATH. Hence we have that MATH. An easy computation shows that MATH and MATH contains the reversible polynomial MATH. By REF MATH and so is finitely generated and projective (in fact free) as a MATH - module. |
math/0011207 | By REF it is easy to see that MATH is in MATH, so the first statement follows from REF . Since MATH is a group algebra, the last assertion follows from REF . |
math/0011211 | By symmetry it is enough to prove the existence of MATH. We claim that it is possible to choose MATH such that for all MATH one has MATH. It follows that MATH. Hence there exists an integer MATH such that MATH and this proves the lemma. It remains to show the claim. If MATH for all MATH, then we may choose MATH arbitrary because MATH. Otherwise there exists an ideal MATH with MATH. In this case we may choose MATH such that MATH since MATH. Let MATH be arbitrary. Then MATH because MATH. We also have that MATH and this implies that MATH by the choice of MATH. This gives the claim. |
math/0011211 | By symmetry it is enough to show this theorem only for MATH. Let MATH for MATH and MATH for MATH. Then MATH because MATH. We claim that: CASE: For MATH one has MATH. CASE: For MATH one has MATH . This yields the theorem. We show REF by induction on MATH. For MATH we have the following exact sequence MATH which proves this case. Let MATH. Since MATH we get MATH. If MATH, then MATH. Assume that MATH. There exists an integer MATH such that MATH. Then by MATH we have MATH because MATH. This gives also MATH. On the other hand let MATH. If MATH, then by MATH we get MATH because MATH. Similarly MATH. Therefore we obtain that MATH by the induction hypothesis. We prove REF also by induction on MATH. The case MATH was shown in REF , so let MATH. Assume that MATH. For MATH one has MATH . Then we get MATH because MATH. Similarly MATH and therefore MATH. If MATH, then MATH. Assume that MATH. There exists an integer MATH such that MATH. Consider MATH . If MATH, then MATH. This is a contradiction by NAME lemma because MATH is a finitely generated MATH-module. Hence MATH and thus MATH. |
math/0011211 | By symmetry we only have to prove REF . Without loss of generality MATH is an almost regular sequence for MATH with respect to the MATH-degree because a generic minimal system of generators of MATH-forms for MATH has this property. By REF one has MATH if and only if MATH. By definition of MATH this is equivalent to the fact that, for all MATH and all MATH, we have MATH . Equivalently, for all MATH we obtain MATH . This concludes the proof. |
math/0011211 | In CITE and CITE it was shown that CASE: MATH can be generated by a MATH-sequence (with respect to the reverse lexicographic order) if and only if MATH can be generated by a MATH-sequence. CASE: MATH can be generated by a MATH-sequence if and only if MATH can be generated by a MATH-sequence. Together with REF these facts conclude the proof. |
math/0011211 | This follows easily from the fact that MATH is bistable. |
math/0011211 | Set MATH, choose MATH generic for MATH and let MATH such that MATH. We may assume that the sequence MATH is almost regular for MATH with respect to the MATH-degree. Furthermore by REF the sequence MATH is almost regular for MATH with respect to the MATH-degree. We have MATH . It follows from CITE that MATH . By REF we get the desired result. |
math/0011211 | If MATH, then MATH for MATH. This means that for all MATH one has MATH. For fixed MATH with MATH we have MATH . As a MATH-vector space MATH because MATH is stable. Thus MATH which is REF . To prove REF we replace MATH by MATH and may assume that MATH is generated in MATH-degree MATH. Then MATH where MATH for all MATH. Let MATH be maximal with MATH and define MATH. We show that MATH for MATH and this gives REF . By a similar argument as in REF we have MATH for all MATH. If MATH, then MATH. Assume that MATH. We claim that MATH . Assume this is not the case, then either MATH for some MATH and MATH which contradicts to MATH. Or MATH for MATH. It follows that MATH. Let MATH be the largest integer MATH such that MATH. Then MATH because MATH is bistable, and this is again a contradiction. Therefore MATH is true and we get MATH for MATH. This concludes the proof. |
math/0011211 | We choose an almost regular sequence MATH for MATH over MATH with respect to the MATH-degree. We have that for all MATH the sequence MATH is almost regular for MATH over MATH in the sense of CITE because MATH as graded MATH-modules and MATH . Define MATH for MATH as in REF. Since MATH it follows that MATH . By a characterization of the regularity of graded modules in CITE we have MATH . Hence the assertion follows from REF. |
math/0011211 | This follows from REF. |
math/0011211 | We prove the case MATH. For MATH one has the exact sequence MATH . In CITE it was shown that MATH and this concludes the proof. |
math/0011211 | We know that MATH where MATH is the minimal bigraded free resolution of MATH over MATH. Let MATH . Then by the definition of the MATH-regularity we have MATH for all MATH. Thus MATH for MATH. The assertion follows. |
math/0011211 | This statement follows from REF and the fact that MATH or MATH respectively. |
math/0011211 | We prove the corollary for MATH. By REF and by standard arguments (see REF for the bigraded case) we get that for MATH . Since MATH, it follows from REF that MATH . Hence MATH. |
math/0011211 | The NAME MATH complex with respect to MATH provides a minimal bigraded free resolution of MATH because these elements form a regular sequence. Observe that MATH is an exact functor on complexes of bigraded modules. Note that MATH is a complex of free MATH-modules because MATH and MATH . Furthermore MATH is minimal by the additional assumption MATH. We have for MATH and this is independent of MATH. If in addition MATH for all MATH, then we obtain MATH . |
math/0011211 | Since MATH is of linear type, we have MATH with the ideal MATH. One knows that REF MATH. Since MATH is defined by MATH equations, we conclude that MATH is a complete intersection. Now apply REF. |
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