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math/9909030 | As tensoring commutes with direct limits, the necessary condition follows from the preceding Proposition. Conversely, if MATH is exact, then for any MATH one has the following exact sequence MATH . Because MATH and the object MATH is MATH-injective, it follows that MATH. |
math/9909030 | Since MATH, it follows that MATH. Similarly, we can define the functor MATH in the other direction. Both of the compositions MATH and MATH are exact functors that are equivalences on the respective categories of finitely generated projective objects. Therefore they are both natural equivalences. |
math/9909030 | For an arbitrary functor MATH by MATH denote a right MATH-module defined as follows. If MATH, then we put MATH. It is directly checked that MATH. From REF it follows that the functor MATH, MATH, defines an equivalence of categories MATH and MATH. Clearly, MATH. |
math/9909030 | By REF there is a pair of functors MATH, where MATH and MATH, defining MATH as a quotient category of MATH. In turn, by REF there is a pair of functors MATH, where MATH and MATH, defining MATH as a quotient category of MATH. It thus suffices to show that MATH is a fully faithful functor, the functor MATH is exact and left adjoint to MATH. Indeed, the composition MATH of fully faithful functors MATH and MATH is again a fully faithful functor and the composition MATH of exact functors MATH and MATH is an exact functor. The fact that MATH is right adjoint to MATH follows from the following isomorphisms: MATH . Hence MATH is equivalent to the quotient category of MATH with respect to the localizing subcategory MATH. |
math/9909030 | It follows from REF . |
math/9909030 | Equivalence MATH follows from REF . MATH. By assumption the ring MATH is right coherent. Therefore the category of right MATH-modules MATH is locally coherent. By REF MATH is closed in MATH. Our assertion would be proved if we showed that MATH is a closed subset of MATH. In view of REF this is equivalent to the localizing subcategory MATH of MATH to be of finite type. Thus we must show that MATH where MATH is a NAME subcategory of MATH. Clearly that MATH. Let us show the inverse inclusion. Let MATH; then there exists an exact sequence MATH where MATH, MATH, MATH is a monomorphism. By REF is right coherent, and so the exact sequence MATH is a direct limit of exact sequences MATH with MATH CITE. If MATH, then MATH. Consider a commutative diagram MATH with MATH and MATH the canonical homomorphisms. Because MATH, it follows that there exists MATH such that MATH. Similarly, given MATH, we can construct the commutative diagrams MATH . It is directly verified that the family MATH is direct. Now, taking a direct limit on the upper row in REF , one obtains MATH. Thus MATH. MATH. By REF the MATH-localization MATH of the MATH-coherent object MATH is MATH-coherent. MATH. Suppose MATH is a locally finitely generated category and MATH is a closed subset of MATH. By REF there is a NAME subcategory MATH of MATH such that MATH. From CITE it follows that MATH. Now our assertion follows from REF . |
math/9909030 | By REF the functor MATH, MATH, defines an equivalence of categories MATH and MATH. By the preceding Corollary MATH is right coherent if and only if MATH is of finite type (= of prefinite type). Because the family MATH is a family of generators of MATH, our assertion immediately follows from REF . |
math/9909030 | It follows from REF . The second part follows from REF . |
math/9909030 | Denote by MATH a localizing subcategory of MATH such that MATH and MATH (see REF ). Let MATH; then MATH is a NAME subcategory of MATH. Clearly, MATH. Obviously that MATH. Hence MATH. Let us show the inverse inclusion. Let MATH be a NAME topology corresponding to MATH. Our assertion would be proved, if we showed that MATH has a basis consisting of those ideals MATH of MATH such that MATH, where MATH is a coherent subobject of MATH such that MATH. So let MATH. Consider the following exact sequence MATH . Because MATH is a monomorphism, MATH. Let MATH; then MATH and MATH. Write MATH as a direct union of finitely generated subobjects MATH of MATH. Because each MATH is a subobject of the MATH-coherent object MATH, it follows that MATH is coherent. One has MATH . By REF the object MATH, and so there is a finite subset MATH of MATH such that MATH. Let MATH; then MATH and since MATH is a coherent objects, one has MATH. Hence MATH as was to be proved. |
math/9909030 | By REF there is a localizing subcategory MATH of MATH such that MATH is equivalent to MATH and by REF there is a localizing subcategory MATH of MATH such that MATH and MATH. Since both MATH and MATH are of prefinite type, it follows that MATH is of prefinite type. From the preceding Proposition it follows that MATH for some localizing subcategory of finite type MATH of MATH. Then MATH. |
math/9909030 | Adapt the proof for modules over a ring MATH CITE. |
math/9909030 | CASE: Let MATH be a MATH-absolutely pure object and MATH. We must show that MATH. Equivalently, any short exact sequence MATH of right MATH-modules splits. By REF MATH. By assumption, the morphism MATH splits, that is .there exists MATH such that MATH. Then MATH where MATH is a MATH-envelope of MATH. So MATH splits. Conversely, let MATH be an absolutely pure right MATH-module and let MATH be a MATH-exact sequence with MATH and MATH. By assumption, the short exact sequence MATH is pure-exact in MATH. Clearly that MATH. From REF it follows that MATH is a direct limit of split exact sequences MATH in MATH. Then MATH is a direct limit of split exact sequences MATH. Thus MATH is MATH-absolutely pure. CASE: Suppose MATH is a MATH-injective object in MATH and MATH is a monomorphism in MATH. Since MATH is of finite type, the morphism MATH is a monomorphism in MATH. Consider the commutative diagram MATH where vertical arrows are isomorphisms. Because MATH is an epimorphism, it follows that MATH is an epimorphism. NAME, suppose MATH is a monomorphism in MATH. Then there is a monomorphism MATH in MATH such that MATH. Indeed, we can embed MATH into the commutative diagram in MATH with exact rows: MATH . Because each MATH is MATH-closed and finitely generated projective in MATH, both MATH and MATH are finitely presented right MATH-modules. We put MATH and MATH. There is a unique morphism MATH. Since MATH and MATH, it follows that MATH. Consider the commutative diagram MATH where vertical arrows are isomorphisms. Because MATH is an epimorphism, it follows that MATH is an epimorphism. So MATH is MATH-injective in MATH. |
math/9909030 | It suffices to note that each MATH is MATH-closed (see REF ) and then apply the preceding Proposition. |
math/9909030 | CASE: Let MATH be absolutely pure. By the preceding Proposition it is an absolutely pure right MATH-module. Now let MATH be an injective envelope of MATH. Then MATH is a subobject of MATH. Because MATH is MATH-torsionfree, it follows that MATH is MATH-torsionfree. Conversely, since MATH, our assertion follows from REF . Let MATH be MATH-injective and MATH. Consider exact sequence REF MATH where MATH, MATH. Because MATH, it follows that the morphism MATH is a monomorphism in MATH. Consequently, the morphism MATH is an epimorphism, and so the morphism MATH is an epimorphism too. As MATH is a MATH-injective object, one has an exact sequence MATH . But MATH is an epimorphism, hence MATH. So MATH. Because MATH, the converse follows from the preceding Proposition and REF . |
math/9909030 | MATH: Let MATH be the family of flat right MATH-modules. By REF the module MATH is MATH-flat. Let MATH be a monomorphism in MATH. As MATH is left coherent, it follows that MATH is a direct limit of monomorphisms MATH in MATH CITE. Then the morphism MATH is a direct limit of monomorphisms MATH. So it is a monomorphism too. MATH is trivial. MATH: Let MATH be a finitely generated submodule of finitely presented module MATH. For each index set MATH we have a commutative diagram MATH where the horizontal arrows are monomorphisms. Since MATH is a monomorphism by REF , also MATH is a monomorphism. Thus MATH is finitely presented by REF . |
math/9909030 | MATH: It follows from CITE. The rest is proved similar to CITE. |
math/9909030 | Adapt the proof for modules over a ring (see CITE). |
math/9909030 | MATH: By assumption, given a finitely presented left MATH-module MATH, the right MATH-module MATH is finitely presented. From REF it follows that MATH is right coherent. Symmetrically, MATH is left coherent. Because the functor MATH where MATH is exact both on MATH and on MATH, it follows that MATH is an absolutely pure both left and right MATH-module. So MATH is an (two-sided) absolutely pure ring. MATH: Since MATH is a (two-sided) coherent ring, from REF it follows that MATH, and hence there is an equivalence of categories MATH and MATH. Similarly, there is an equivalence of categories MATH and MATH. In view of REF we have the following relations: MATH . Now our assertion follows from REF . MATH, MATH: Apply REF . MATH: It suffices to show that MATH is right coherent. To see this, consider a direct system of absolutely pure right MATH-modules MATH. Since each MATH, by assumption, is flat, it follows that the module MATH is flat, and so it is absolutely pure. Therefore MATH is right coherent by REF . MATH: By REF any absolutely pure right MATH-module is MATH-flat, and hence flat by REF . MATH: Since the ring MATH is right absolutely pure, the module MATH, where MATH and MATH is some set of indices, is also absolutely pure, and therefore it is flat. In view of REF MATH is left coherent. By REF any MATH-injective right MATH-module is absolutely pure, and hence flat. From REF it follows that MATH is left absolutely pure. |
math/9909031 | Let us first assume that MATH is satisfiable, with satisfying assignment MATH, MATH. Consider an edge MATH in the corresponding digraph. Since MATH, the presence of the edge MATH gives the logical implication MATH. A contradictory cycle MATH therefore gives the logical implication MATH, which is not compatible with any truth assignment for MATH. We prove the converse by induction on the number MATH of variables. For MATH there is nothing to prove. Turning to the induction step, suppose that the digraph MATH has no contradictory cycles. We claim that in this case MATH is satisfiable. To this end, we first recall the definition of strongly connected components for directed graphs. We say that two vertices MATH and MATH in a directed graph are strongly connected if MATH, that is, if the directed graph MATH has a cycle containing MATH and MATH. The strongly connected component of a vertex MATH is the induced subgraph of MATH containing the set of vertices MATH . Somewhat loosely, we call MATH the strong component of MATH. Clearly, the strong component partitions the vertex set MATH. We define a partial order MATH on the set of all strong components by taking MATH if MATH, and so MATH for all MATH and MATH. Let MATH be a minimal element in this partial order, that is, let MATH be a strong component such that MATH contains no edge MATH with MATH and MATH. For a set of literals MATH, let MATH . Since MATH has no contradictory cycle, MATH. Furthermore, since MATH is a minimal element in our partial order, MATH must be a maximal element. If we set all literals in MATH to FALSE, and so all literals in MATH to TRUE, then all clauses in MATH containing at least one literal from MATH are TRUE. This process removes all the variables corresponding to literals in MATH and MATH from MATH, and all clauses involving these variables from MATH, leading to a new REF-SAT formula MATH. Since the graph MATH is a subgraph of MATH, it contains no contradictory cycles either. Using the inductive hypothesis, we obtain a satisfying assignment for MATH, which completes the proof of the converse and hence of the theorem. |
math/9909031 | Suppose that MATH, and let MATH be a shortest directed path from MATH to MATH in MATH. Then no literal appears twice in the path, although a literal and its negation may well do so. Let MATH be the smallest positive integer such that MATH is not strictly distinct from all of MATH, and let MATH be such that MATH. Then MATH is a subformula of MATH that is satisfied by setting each of MATH to FALSE. On the other hand, MATH is UNSAT since in order to satisfy it, we would have to set MATH to TRUE, then MATH to TRUE, and so on, ending with the requirement that MATH be set TRUE. However, as MATH is TRUE, MATH is already set FALSE. This completes the proof that MATH implies that MATH. Conversely, suppose that MATH is SAT and MATH is UNSAT. Then MATH has a contradictory cycle MATH. Since MATH does not have a contradictory cycle, the cycle MATH of MATH contains the oriented edge MATH, say MATH. But then in MATH we have MATH, so MATH. Hence if MATH then MATH. |
math/9909031 | We start with the first equality in REF . If MATH, then MATH, so MATH is not strictly distinct. If MATH is not strictly distinct, then MATH for some literal MATH, and hence MATH and MATH. But MATH implies that MATH, which together with MATH implies MATH. To prove the second equality, we first note that the set of literals MATH for which MATH is not strictly distinct is obviously a subset of the set of literals MATH such that MATH is not strictly distinct. We are thus left with the proof that the statement that MATH is not strictly distinct implies the (apparently stronger) statement that MATH is not strictly distinct. So let us assume that MATH is not strictly distinct. By the first equality in REF , this implies MATH. Since the digraph of a REF-SAT formula does not contain any direct edges from MATH to MATH, we conclude that there must be a literal MATH strictly distinct from MATH such that MATH. The latter statement implies that both MATH and MATH, so that MATH is not strictly distinct. |
math/9909031 | By construction, REF are obvious. REF is not much more difficult. There are certain possible edges leading out of the vertex set MATH that were never tested, or that were tested and present, but then excluded from the trimmed out-graph MATH anyway. But each such edge either led to a literal already in MATH, or else led to a literal whose complement was in MATH. Thus if the literal set MATH were to contain more literals than MATH, then MATH would not be strictly distinct. On the other hand, if MATH and MATH are identical, then MATH is trivially strictly distinct by REF is similarly easy. First, for each literal MATH, we define MATH. By induction we shall prove that, at the beginning and end of each step of the search, the following properties hold: CASE: For every pair of literals MATH and MATH of the current graph, precisely two edges between MATH and MATH have been tested. CASE: For every literal MATH in the current graph but not in the frontier, and every literal MATH such that neither MATH nor MATH is in the current graph, precisely two edges between MATH and MATH have been tested, both results being ``no." CASE: For every literal MATH in the frontier, and any literal MATH such that neither MATH nor MATH is in the current graph, none of the edges between MATH and MATH have been tested. CASE: If none of MATH are in the current graph, then no edges between MATH and MATH have been tested. CASE: For any pair of strictly distinct literals MATH, either none or precisely one of the four edges between MATH and MATH appears in the current graph. The latter happens if and only if some test between MATH and MATH was positive. Indeed, assume that REF - REF hold at the beginning of a step. To prove that REF holds at the end of the step, we first note that no edge between MATH and MATH was tested in the current step if neither MATH nor MATH is new. If MATH is old and MATH is new, then either MATH was the selected vertex in the frontier, in which case the edges MATH and MATH have been tested in the current step, or MATH was not in the frontier, in which case precisely two edges between MATH and MATH were tested in a previous step (with answer ``no") by the inductive REF . If both MATH and MATH are new, then no edge between MATH and MATH was tested in a previous step by the inductive REF , and precisely two edges (the edges MATH and MATH) between MATH and MATH are tested in the current step. To prove REF , we note that if MATH is in the current graph but not in the frontier, it was in the frontier in some previous step, and got removed from the frontier after all edges from MATH to vertices MATH, with neither MATH nor MATH in the current graph at the time, were tested. This includes in particular the vertex MATH in question, and since we assume that neither MATH nor MATH is in the current graph, it follows that both tests must have given the result ``no" at the time. After that step, MATH is not in the frontier, so no edge containing MATH or MATH is ever tested again, implying REF follows from the observation that an edge between a vertex MATH in the current graph and a vertex MATH such that neither MATH nor MATH is in the current graph is only tested if MATH is the selected vertex in the current step, in which case it is not in the frontier after this step anymore. REF is obvious, since an edge MATH is only tested if either MATH is in the frontier (and hence in the current graph before the current step), or both MATH and MATH are new vertices, which means they are in the current graph after REF - REF . To prove REF , we consider three cases. In the first case, none of the vertices MATH, MATH, MATH and MATH is in the current graph, in which case no edge between MATH and MATH appears in the current graph by the inductive REF . The second case is the one in which exactly one of the four vertices MATH, MATH, MATH and MATH is in the current graph. Without loss of generality, let us assume that this is the vertex MATH. Then none of the edges between MATH and MATH appears in the current graph by REF . The third case is that precisely two of the four vertices MATH, MATH, MATH and MATH are in the current graph, say MATH and MATH. Then precisely two of the four edges between MATH and MATH have been tested by the inductive REF . Since the above search procedure always tests two of the four edges between MATH and MATH at a given time, and adds one (but not both) of them precisely when at least one of them tests positive, we get REF . We now use REF - REF above to prove REF. If we pick the unordered pairs of numbers between MATH and MATH in some arbitrary order, each time randomly saying ``present" (with probability MATH) or ``absent" (with probability MATH), then even if the order in which we pick the pairs depends on the previous random choices of present/absent, the result will be the random graph MATH. This is in effect what the trimmed local search does, except that it stops when the connected component containing MATH has been determined. Thus unoriented projection of MATH is just the connected component containing MATH in MATH. |
math/9909031 | For MATH, the probability in the integrand in REF is MATH by REF . Integrating, we find that MATH provided MATH. |
math/9909031 | By REF , the probability in the integrand in REF is MATH, provided MATH. In the middle region MATH we upper bound the probability in the integrand by MATH, and we bound it in the left region MATH by MATH as above. Integrating, we find that MATH provided MATH. |
math/9909031 | To get the bound on the left, we increase MATH from MATH to MATH, with MATH. For any pair of vertices, whether or not there was a directed edge between them before, afterwards the probability of finding such an edge is at least MATH. For each hourglass, for each pair of literals MATH and MATH in the out-portion of the hourglass, if the clause MATH appears, then we claim that the central vertex and the entire in-portion of the hourglass is afterwards part of the spine of the formula. Indeed, let MATH be such a vertex. Then MATH and MATH since MATH and MATH are in the out-portion of the hourglass. But the appearance of the clause MATH implies that MATH, so that we have MATH. Together with MATH, which is equivalent to MATH, we conclude that MATH. The probability of the event that the clause MATH appears is at least MATH. If furthermore a clause appears that contains two literals in the in-portion, then the formula is not satisfiable. These events are independent, so the probability that they both occur is at least MATH. But since with high probability there are MATH hourglasses, with probability at least MATH the formula becomes unsatisfiable. Setting MATH gives the desired upper bound on the left. |
math/9909031 | To get the bound on the right, we start with MATH (with MATH), where the probability that there is no giant hourglass is at most MATH, and then increase it to MATH. Any clauses of the form MATH where MATH and MATH are in the out-portion of the the giant hourglass beforehand will afterwards appear with probability at least MATH. This will cause the in-portion of the giant hourglass to become part of the spine of the formula, except with probability that can be bounded by MATH. Furthermore, any clauses of the form MATH where MATH and MATH are in the in-portion of the giant hourglass beforehand will afterwards appear with probability at least MATH. Therefore the formula will become unsatisfiable, except with probability that is again exponentially small in MATH. Setting MATH completes the proof. |
math/9909031 | We may assume that the vertex in question is vertex MATH. First, we shall inductively define a subtree MATH of MATH, rooted at MATH, with each edge oriented away from MATH. To this end, set MATH, MATH, and let MATH be the subtree of MATH with the single vertex MATH. Suppose that we have defined a pair MATH of subsets of MATH with MATH, and a subtree MATH with MATH. (We think of MATH as the set of vertices we have ``exposed", that is, tested for outgoing edges, and MATH as the set of vertices we have selected so far.) If MATH then MATH is our tree MATH. Otherwise, let MATH be the smallest element of MATH. Let MATH denote the set of vertices in MATH that can be reached by single edges of MATH that are oriented outward from MATH. Now set MATH, MATH, and take MATH to be obtained from MATH by adding to it the vertices in MATH, together with all the edges from MATH to MATH. The vertex set of the subtree MATH of MATH constructed in this way is clearly MATH; in particular, MATH iff MATH. Since the edges of MATH are oriented away from MATH, we may view MATH as an unoriented tree. Now, let us construct a subtree MATH of the random graph MATH rooted at MATH by precisely the same algorithm. The lemma will follow if we show that MATH for every tree MATH with vertex set MATH. Given MATH, we can define the MATH's as above. We have MATH if and only if the random digraph MATH is such that REF it contains all the edges of MATH (oriented away from vertex MATH), REF it contains no edge oriented from MATH to MATH. Similarly, MATH if and only if the random graph MATH is such that REF it contains all the edges of MATH, REF it contains no edge from MATH to MATH. Notice that the probability that MATH contains a given set MATH of oriented edges, and no edge of a second set MATH of oriented edges, is MATH provided that MATH. Moreover, this is equal to the probability that MATH contains a set MATH of unoriented edges, and no edge of a second set MATH of unoriented edges, provided that MATH, MATH and MATH. Thus relation REF holds, and we are done. |
math/9909031 | Let MATH be a set of MATH strictly distinct literals with MATH. For MATH, the dual of the implication MATH involves no literal in MATH. Therefore the probability that MATH is MATH, where MATH is the probability that every vertex of the random digraph MATH can be reached from MATH and MATH is the probability that the random REF-SAT formula MATH contains no implication from the set MATH. By REF , we have MATH, so we turn to the task of calculating the probability MATH. Note that there are MATH implications in the set MATH. However, a REF-SAT formula MATH contains none of the MATH implications MATH, MATH. Also, if MATH and MATH, then MATH and MATH are dual implications, that is, MATH contains MATH if and only if it contains MATH. In fact, both implications MATH and MATH belong to MATH if and only if MATH, MATH and MATH. Hence MATH contains MATH dual pairs, so that the probability that MATH contains no implication from MATH is MATH . Therefore, MATH . Since there are MATH choices for the set MATH, the lemma is proved. |
math/9909031 | By REF , MATH is strictly distinct if and only if MATH. As a consequence, MATH which proves REF. Rewriting REF in the form MATH and observing that MATH is a monotone increasing function of MATH, and hence of MATH (see REF ), we have MATH, obtaining an alternative proof of the bound REF . If MATH and MATH is bounded, then both MATH and MATH are bounded by a constant times MATH. By REF , we therefore have MATH and hence by REF MATH . As a consequence, MATH which implies the bound REF . In order to prove REF , we decompose MATH as MATH where MATH is the first and MATH is the second sum above. If MATH then MATH, so by REF we have MATH . By taking the ratio of successive terms in the series MATH we see that the summand is an increasing function of MATH for MATH . Since the MATH-st term is in the sum for MATH, we then have MATH . To bound MATH, we note that MATH so that MATH and therefore MATH . Combined with REF , this gives the desired bound REF . |
math/9909031 | With the assumption that MATH is bounded, we can use our bound REF on MATH together with the bounds MATH, MATH, and MATH to rewrite REF as MATH . Note that the ratio between the second and first terms in the exponential is MATH, so we can further rewrite REF as REF . The estimate REF for MATH follows immediately from the estimate for MATH and REF, and the estimate for MATH follows from that for MATH and REF. |
math/9909031 | We start with the proof of REF. The limit of the summation is MATH; we first sum the portion up to MATH separately. For this portion, the sum is MATH where here and throughout this proof, constants may depend upon MATH. Referring to our REF for MATH, we see that for the remainder of the terms in the sum, the additive error terms in the exponential tend to zero when MATH is small enough and MATH is large enough, and so they contribute only a multiplicative error of MATH. The multiplicative error terms in the exponential can also be made arbitrarily close to MATH by taking MATH large enough and MATH small enough. Therefore we have MATH where MATH and MATH where the MATH depends on MATH. Combining REF, we get REF. To prove REF, we use the well know fact that the cluster size distribution in MATH, with MATH, is stochastically dominated by a birth process with binominal offspring distribution MATH, which in turn is stochastically dominated by a NAME birth process with parameter MATH . Writing this parameter as MATH, we therefore get the estimate MATH . Bounding the sum on the right hand side by MATH, we obtain the estimate REF. |
math/9909031 | To prove the lemma, we again consider three regions of MATH: MATH, MATH, and MATH. In the first region, we use REF to approximate MATH by MATH. Combined with REF , this gives MATH . For MATH, we combine the second and the third region. Using the fact that MATH by REF, we then use REF to obtains the bound REF below threshold. Above threshold, the contribution from the second region is MATH . In the third region we use REF to write MATH . Consider the prefactor, that is, the summand ignoring the factor of MATH. Provided MATH, recalling the definition of MATH in REF , we can differentiate the logarithm of the prefactor to get MATH which will be nonpositive provided MATH which will hold if MATH is large enough, where ``large enough" depends upon MATH. Provided that these conditions are met, the prefactor takes on its largest value in the first term, MATH. Eventually, if MATH gets large enough, MATH may stop increasing, and take on the value MATH. If MATH then the prefactor will then increase up to MATH. Thus the maximum value of the prefactor is MATH . Since by REF, we have MATH, so the second term in the MATH is MATH. Thus we can write the maximum prefactor as MATH . Upon substituting the above expression into REF , we find MATH where we used REF to get the second line. The bounds REF imply REF above threshold, and REF imply the estimate REF. |
math/9909031 | Let MATH denote the event that MATH is strictly distinct, and MATH. So in particular MATH. Using the resolution of the identity MATH where MATH denotes the indicator function of the event MATH, we can decompose MATH in two different ways to obtain MATH . To estimate the probability on the right, consider the event MATH. With probability MATH either MATH or MATH is in MATH. If MATH then MATH. If MATH, the situation is more complicated, so we shall bound the probability below by MATH. If MATH contains neither MATH nor MATH, then any path from MATH to MATH avoids literals in MATH. In this case we may as well explore the out-graph MATH restricted to avoid the variables in MATH; with probability MATH the restricted out-graph will contain MATH. Thus we conclude MATH . Substituting this into REF proves the lemma. |
math/9909031 | Suppose MATH. Let MATH denote MATH, which then must be strictly distinct. We shall consider four cases depending on whether or not MATH and whether or not MATH: CASE: MATH and MATH. Then MATH. CASE: MATH and MATH. Then MATH. CASE: MATH and MATH. It is clear that if MATH, where MATH denotes MATH, then MATH. Suppose that conversely MATH. Since MATH, either MATH or MATH is in the path from MATH to MATH in MATH. The first such occurrence must be MATH, so we have MATH, and its contrapositive MATH. Since MATH within MATH, it must be that MATH occurs in the path from MATH to MATH. In particular MATH. We may assume without loss of generality that MATH and MATH occur only at the endpoints of this path. If a literal in MATH occurred in the path from MATH to MATH, consider the last such one. The next literal cannot be MATH by assumption, nor can it be MATH, since this occurs only at the beginning of the path. So the next literal would have to lie inside MATH, a contradiction. If a literal in MATH occured in the path from MATH to MATH, we could take the contrapositive and similarly derive a contradiction. Thus MATH using only literals in MATH, that is, MATH. Thus conditional upon MATH, we can write MATH, MATH, MATH iff MATH since the events on the right and the event MATH are determined by pairwise disjoint sets of variables. CASE: MATH and MATH. By symmetry, REF have the same probability. Putting these four cases together we see MATH . As a consequence, MATH . Hence by REF, we have MATH . In the above, all of the MATH terms are MATH. |
math/9909031 | We seek MATH and need only show that each summand is well-approximated by MATH. Define MATH by MATH . Since MATH, and MATH, it follows that MATH. Thus when we apply REF with MATH and MATH rather than MATH and MATH, we obtain an answer that differs by a factor of MATH, as desired. |
math/9909031 | We can now complete our estimate of the covariances. For strictly distinct literals MATH and MATH we have MATH . As a consequence, MATH where the upper entry applies to the subcritical regime, and the lower entry applies to the supercritical regime. In the above, the MATH terms are all MATH. This gives the variance bound above threshold. The bound on the second moment below threshold follows by combining the above variance bound with REF . |
math/9909031 | Proof of REF on the existence of many hourglasses Let MATH be a large positive number, but still small compared to MATH, and let MATH. We shall construct a process for growing hourglasses one by one. However, as we deplete variables, the distribution changes, so that the hourglasses so constructed are not identically distributed. We therefore use the following two variations of this naive procedure: In the first process, instead of drawing from MATH variables, we draw only from MATH variables, so that we have a ``buffer" of MATH, and replenish the variables as necessary. However, even this process can lead to trouble in the unlikely event that we have a very long run which uses up our entire reserve of variables. So we construct another process which aborts the growth of an hourglass when it would use up too many variables. To be explicit, consider the trimmed out-graphs and trimmed in-graphs of various literals, where the in- and out-graphs are restricted to sets of MATH variables. Recall from REF that the unoriented projection of the trimmed out-graph MATH of a literal MATH is identically distributed to the connected component MATH in MATH. We shall follow the convention that no matter how many trimmed out-graphs or trimmed in-graphs we have explored in the past, when exploring another in-graph or out-graph, we shall always restrict our attention to variables that are in none of the in-graphs or out-graphs found so far (except, possibly, for the root), and we shall add variables to ensure that there are MATH variables for the tree to grow within. (Recall that for the upper bounds we assume that we have a variable for each natural number.) In this way the sizes and structures of all the trees will be independent and identically distributed. As we alluded to above, later we shall consider a variation on this process. Since there are somewhat fewer than MATH variables in which we explore the out-graphs and in-graphs, this decreases the average out-degree of each literal, and has the effect of shifting the formula further into the subcritical regime. Specifically, if we define MATH by MATH we see that MATH. Pick a literal MATH, and look at its trimmed out-graph MATH within MATH unused variables. Recall the definition of MATH, the probability that MATH is a tree of size MATH. By using REF for MATH, we see that, for some MATH, there is a MATH chance that MATH is a tree of size between MATH and MATH. Here, as explained in the first sentence of the proof, we are assuming MATH, so by REF , in the remainder of this proof all MATH terms without subscripts are of the form MATH. By comparing with random graphs, if MATH is a tree, then it is uniformly distributed amongst the spanning trees on MATH vertices. Using the structural properties of random spanning trees (see for example . CITE), if we pick a random vertex MATH in MATH, with probability at least MATH the path connecting MATH to MATH has length at least MATH. Let MATH be the middle vertex in the path from MATH to MATH (in case of tie, we choose the vertex closer to MATH). Either the majority of the remaining vertices are connected to the first part of the path or to the second part of the path. Since the spanning tree is uniformly random, by symmetry, with probability at least MATH, at least half of the remaining vertices of the tree will be connected to MATH via the path from MATH to MATH - these vertices will be in the out-graph of MATH. In the event that the path from MATH to MATH has length at least MATH, and MATH has at least MATH descendents in the trimmed out-graph of MATH, say that vertex MATH is ``promising," and that the path from MATH to MATH is the tail. We thus have shown MATH . In the event that vertex MATH is promising, we proceed to explore the trimmed in-graphs of the first MATH vertices on MATH's tail. Again by REF , each individual in-graph will have size at least MATH with probability at least MATH. The probability that none of them is so large is at most MATH. If any of them is so large, we shall call MATH a ``central variable," and MATH together with its explored out-graph and in-graph both of size at least MATH an ``hourglass" (see REF ). Each time that we pick a literal and look for an hourglass as described above, we find one with probability at least MATH . Next let us compute how many variables we expect to use up while exploring the trimmed out-graph and trimmed in-graphs. At this point we recall from REF that if MATH is either the trimmed in-graph or trimmed out-graph of a vertex, MATH . We always explore one out-graph, and with probability MATH we explore MATH in-graphs. Thus the expected number of variables used up is MATH . If we look for an hourglass for MATH times, then the probability that we fail to find an hourglass is MATH, and the expected number of variables that we use up is MATH . The probability that we use up more than MATH times the expected number of variables is at most MATH. Therefore, with probability at least MATH (for large enough MATH and small enough MATH) we both find an hourglass, and do not use up more than MATH variables, where MATH. Now consider the following modification of the above procedure: as above we use the local search procedure in REF to explore the trimmed out-graphs and trimmed in-graphs, hoping to find an hourglass, but as soon as we use up MATH variables, we abort and stop looking for the hourglass. Then we can repeat this procedure MATH times, be guaranteed to use no variables other than the first MATH of them, and find a number of disjoint hourglasses that stochastically dominates the binomial distribution MATH. By NAME 's inequality CITE (see also CITE), the probability that we find fewer than half as many hourglasses as we expect will be no larger than MATH. |
math/9909031 | Let MATH be a large positive number, but still small compared to MATH. When MATH, as we have just seen, there will be MATH hourglasses with in- and out-portion of size at least MATH, except with probability MATH. We now increase MATH by a suitably large constant times MATH, say MATH. For any two hourglasses, the probability of an edge from the out-portion of the first hourglass to the in-portion of the second hourglass is therefore at least MATH. Then the central variable of the first hourglass implies the central variable of the second hourglass. Conceptually we can think of the directed graph whose nodes are the hourglasses, and place a directed edge from one node to another whenever the hourglasses connect up like this. The edges of this graph occur independently of one another, and the average out-degree of the graph is MATH. By choosing MATH large enough, we can make the average out-degree to be any convenient constant that we like. In particular, if the average out-degree is a constant larger than REF, then we might expect the connections to percolate, so that there is some node MATH that can reach a constant fraction of the other nodes through edges of this graph, and is reachable by a constant fraction of the other nodes. Provided this happens, each literal in the out-portions of the nodes reachable by MATH is implied by the central variable of node MATH, and each literal in the in-portions of the nodes that can reach MATH will imply the central variable of MATH, thereby giving the desired giant hourglass with in-portion and out-portion each of size MATH. It remains to be shown (in REF ) that we get the requisite percolation except with probability that is exponentially small in the number of nodes of the graph. |
math/9909031 | For convenience, let MATH, so that there are at least MATH vertices, and the probability of each directed edge is at least MATH. (We can throw out some of the edges to make the probability exactly MATH.) To search for a node MATH with a large in-graph and out-graph, we consider candidate vertices one at a time, and explore the in-graph and out-graph of the candidate, restricting the explorations of the in-graph and out-graph to disjoint sets of MATH vertices, none of which have yet been explored in the course of examining a previous failed candidate. In this way we ensure that the sizes of the in-graph and out-graph are independent, and both distributed in the same manner as the size of the component containing a particular vertex in the random graph MATH. We do the explorations in a parallel interleaved fashion, so that if either the in-graph or out-graph is found to be too small, then exploration of the other is immediately halted. We shall show that for large enough MATH, except with probability exponentially small in MATH, after looking at MATH candidates, the failed candidates have not wasted more than MATH variables, and we find a successful candidate with in-graph and out-graph each of size at least MATH. We first claim that for any real MATH and integers MATH and MATH such that MATH and MATH, there is a MATH between MATH and MATH so that MATH . Indeed, the probability that the component MATH containing the vertex MATH has size MATH can be bounded below by the probability that MATH is a tree of size MATH. To bound MATH from above, we note that the connectedness of MATH implies that MATH contains a tree of size MATH. Summing over all possibilities for this spanning tree, we therefore get MATH . This give a lower bound of MATH and an upper bound of MATH which establishes REF . Next define MATH by MATH . Using the bound REF on MATH we see that MATH . Let MATH be the number of variables lost on the MATH-th candidate if it is a failure; MATH if the MATH-th candidate is successful: MATH . Thus MATH, and hence MATH . Letting MATH be the total number of variables lost on failed candidates, we see MATH which is exponentially small in MATH for MATH. Next consider MATH. By REF we have MATH . Thus with probability MATH (for large enough MATH), both the in-graph and out-graph have size at least MATH. Therefore, if we try MATH candidates, the probability that we lose too many variables on failed candidates, or have enough variables but still fail to find a vertex with in-graph and out-graph of size at least MATH, is bounded by MATH which establishes the lemma. |
math/9909032 | We repeat the argument in CITE. We may assume that MATH is non-empty, and that MATH is contained in MATH. For every MATH and dyadic MATH, MATH, we let MATH denote the set MATH . In other words, MATH consists of those points MATH in MATH which lies in about MATH tubes from MATH, most of which make an angle of about MATH with MATH. From the pigeonhole principle we see that MATH . We now prove a technical lemma which allows us to uniformize MATH and MATH. This type of argument will also be used in the sequel. (For a more general formulation of this type of argument, see CITE). A somewhat similar lemma appears in CITE. Let the notation be as above. Then there exist quantities MATH, MATH and sets MATH and for each MATH, MATH there exists a set MATH such that MATH and MATH for some MATH satisfying MATH . The implicit constants may depend on MATH. The first stage shall be to construct sequences MATH and quantities MATH for all MATH and MATH, such that MATH and MATH for all MATH. To do this, suppose inductively that MATH is such that MATH and MATH have been constructed for all MATH. From REF we have MATH . By the pigeonhole principle, for every MATH one can thus find MATH, MATH such that MATH where MATH . By the pigeonhole principle again, there exists MATH, MATH independent of MATH such that the set MATH satisfies REF. It is clear that this construction gives the desired properties. By the pigeonhole principle, there must exist MATH and MATH, MATH such that MATH and MATH for MATH. The claim then follows by setting MATH and MATH. Let the notation be as in the above lemma. From REF we have MATH which we rewrite as MATH . From REF, the nesting MATH, and REF, the integrand is bounded by MATH. We thus see that MATH and MATH are naturally related by the estimate MATH . One can reverse the inequality in REF, but we shall not need to do so here. From REF, MATH is non-empty. Let MATH be an arbitrary element of MATH. Consider the ``hairbrush" MATH defined by MATH . From REF we see that MATH for all MATH. Integrating this using REF, we obtain MATH . From elementary geometry we see that MATH so we conclude that MATH . We will shortly combine REF with REF to prove the estimate MATH . Assuming this bound for the moment, let us complete the proof of REF. From REF we have MATH for all MATH and MATH. From the definition of MATH and the triangle inequality we thus see that MATH for all MATH in the set in REF. Integrating this and using REF, we thus obtain MATH . From REF we thus have MATH . However, from REF and the fact that MATH is constrained to a MATH box, we see from elementary geometry that MATH . Combining these two estimates we obtain (after some algebra) MATH and REF follows after some algebra from this and REF. It remains to prove REF. We first deal with a trivial case when MATH. In this case we simply use the bound MATH from REF and the fact from REF that MATH is non-empty, and REF follows since MATH and MATH. Now assume MATH. To prove REF we will in fact prove the stronger bound MATH where MATH and MATH . From REF, and elementary geometry we have MATH for all MATH. Summing this in MATH we obtain. MATH which we rewrite as MATH . We now use NAME 's argument. From NAME and the above we have MATH . From this and REF, it suffices to show that MATH since REF then follows from algebra. To prove REF, we expand the left-hand side as MATH which we break up further as MATH . From elementary geometry we have MATH . It thus suffices to show that MATH for each MATH, MATH. Fix MATH, MATH. The conditions MATH and MATH force MATH to lie in a MATH-neighbourhood of the MATH-plane spanned by MATH and (a slight translate of) MATH. Together with the condition MATH, this constrains MATH to live in one of MATH boxes, each of dimension MATH. The claim then follows from REF. |
math/9909032 | By squaring REF we have MATH . Now let MATH be a small number to be chosen later, and consider the quantity MATH . NAME MATH by finitely overlapping sets MATH where each MATH has diameter MATH, and such that for every MATH with MATH there exists a MATH such that MATH. We thus have MATH where MATH. Since MATH, we have the quasi-triangle inequality MATH (see for example, CITE), and so we may estimate REF by MATH . We now claim that MATH . To see this, first apply a mild affine map to make MATH centered at the origin, and apply the dilation MATH, and then apply REF to the result; compare CITE. Since our choice of MATH, MATH satisfy the scaling condition MATH, we may simplify REF using REF to MATH . Inserting this back into REF and using the elementary inequality MATH which follows since MATH, we obtain MATH . Comparing this with REF we see that MATH if we choose MATH to be a sufficiently small number depending only on MATH and MATH (so MATH). Now cover MATH by MATH balls of diameter MATH. By the pigeonhole principle and the above estimate we see that there must exist at least one pair MATH, MATH of such balls with MATH such that MATH . The claim follows. |
math/9909032 | We first prove REF. Let MATH, where MATH is a large constant to be chosen later. We trivially have MATH . We now claim that MATH the claim then follows by subtracting these two estimates from REF and choosing MATH suitably. To prove REF, we first observe that the left-hand side is bounded by MATH . By REF, this is bounded by MATH and REF follows. Now we prove REF, which is a dual of REF; the last two claims in the lemma then follow easily. As before we have MATH . It suffices to show that MATH for all MATH, where MATH by summing this for all dyadic MATH and using the exponential decay of the MATH we can obtain the analogue of REF. Fix MATH. By definition of MATH we have MATH . From NAME we thus have MATH . From REF we have MATH from REF we thus have MATH . Inserting this into REF and using REF we obtain MATH which simplifies to MATH and REF follows from the definition of MATH. |
math/9909032 | NAME MATH by MATH finitely overlapping intervals MATH of width MATH, and let MATH denote the slab MATH. For each MATH, we can then find a MATH such that MATH . It thus suffices to show that MATH where MATH . Partition MATH into about MATH refinements MATH, each of which is MATH-separated. We can split the left-hand side of REF as MATH where MATH . By NAME, we may estimate this by MATH . The sets MATH in the innermost sum can be rescaled to form a collection of MATH tubes which continue to satisfy REF. Also, the set of directions MATH satisfies the correct separation condition for the scale MATH. By a rescaled version of REF, we can therefore bound the norm in REF by MATH which can be estimated using REF and algebra by MATH . Inserting this back into REF, we may estimate the left-hand side of REF as MATH . Since we have MATH's, we can use NAME to obtain MATH . We can thus bound the left-hand side of REF as MATH . By NAME again, we bound this by MATH . Since MATH, we can bound this by MATH and REF follows. |
math/9909032 | The first step is to find MATH and MATH. Let MATH range over all dyadic integers from MATH to MATH. Let MATH denote the set MATH . Clearly we have MATH . Since the number of MATH and MATH is MATH, we can use the pigeonhole principle and conclude that there exist MATH, MATH for which REF holds with MATH and MATH. Fix this choice of MATH, MATH and MATH; this also fixes MATH. By construction we have MATH . Combining this with REF we have MATH for MATH. Combining this with REF we see that MATH . From the definition of MATH we thus have MATH. Since MATH, we see from REF that MATH. We now produce sets MATH and MATH with the properties that MATH for all MATH, MATH, MATH, MATH, and MATH . Clearly REF hold for MATH. Now suppose inductively that MATH is such that MATH have been constructed satisfying REF for this value of MATH. We perform a certain sequence of dance steps. From REF we have MATH which by REF implies MATH . By REF (noting that MATH; we shall need similar observations in the sequel), we thus have MATH where MATH is the set MATH . Now define the set MATH by MATH . From REF we have MATH by REF we thus have MATH . Combining this with REF we obtain (if MATH is sufficiently small) MATH . We may rewrite this using REF as MATH . By REF , we have MATH where MATH is the set MATH . In particular, from REF with MATH, we have MATH . By REF, we may rewrite this as MATH . By REF again, this implies MATH where MATH . Defining MATH we apply REF , and the preceding estimate as before to conclude MATH . By REF again, we rewrite this as MATH . By REF we have MATH where MATH . This completes the dance sequence. One can easily verify that REF, and REF are all satisfied for MATH and MATH. One now replaces MATH by MATH, and repeats the above dance. Of course, the implicit constants in the bounds will depend on MATH and hence on MATH. The quantities MATH are clearly monotone decreasing, and satisfy the trivial estimates MATH. By the pigeonhole principle one can then find MATH such that MATH . The lemma then follows by setting MATH, MATH, and MATH for MATH and MATH. |
math/9909032 | Firstly, from REF and elementary geometry we see that MATH must contain at least MATH parallel lines, which with REF gives MATH . It thus suffices to show MATH since REF follows by taking the geometric mean of these estimates and then using REF. To prove this estimate we invoke NAME 's argument as in the proof of REF. Summing REF over all MATH we obtain MATH which we rewrite as MATH . By the NAME inequality we thus have MATH . It thus suffices to show that MATH . Repeating the derivation of REF, we may estimate the left-hand side by MATH and the claim follows from the observation MATH which follows from REF and elementary geometry. |
math/9909040 | CASE: It is clear that the identity map MATH is continuous. Since MATH is compact it follows that the topologies are identical. CASE: Assume that MATH is an extreme point of the unit ball of the dual space MATH. By the NAME Theorem, the set of norm one extensions of MATH to MATH has an extreme point MATH. It is easy to check that MATH is also an extreme point of MATH . The argument is concluded by appealing to the very well-known fact that any extreme point of the last space is of the form MATH with MATH . CASE: Let MATH be a norm one element of MATH. By the NAME Theorem there is an extreme point MATH of the unit ball of MATH such that MATH. Thus REF follows from the previous part. |
math/9909040 | Follows from CITE II. REF. |
math/9909040 | To prove the first inclusion assume that MATH, and assume that MATH where MATH are norm one functionals on MATH . Let MATH be norm one extensions of MATH to functionals on MATH. By REF , there is a net MATH in MATH such that MATH and MATH uniformly on compact subsets of MATH. We have MATH hence MATH. Since MATH we get MATH. |
math/9909040 | Assume that MATH is a linear isometry. Set MATH in the last Theorem, and let MATH be as in that Theorem. We can define a module map MATH by MATH . That MATH is an isometry follows from the last Theorem and REF . |
math/9909040 | The only thing that is still not clear here is that (MATH). Note that if MATH is a MATH-isometric isomorphism, and if MATH, then MATH, MATH, and MATH on MATH. Since MATH for all MATH, it follows by a simple NAME point argument, that MATH on MATH. (Since if MATH with MATH, let MATH. By REF , there exists MATH and MATH on MATH. Hence MATH.) A similar argument shows that there exists MATH with MATH on MATH, and MATH on MATH. Hence MATH on MATH, and consequently on all of MATH. Thus MATH everywhere, so that MATH. |
math/9909040 | By REF , MATH. Now it is clear that REF gives (MATH). A similar argument proves the last statement. The rest of the numbered equivalences are clear. |
math/9909040 | The MATH direction is trivial. The MATH direction follows easily from the Theorem, but we will give a different proof, which will generalize later to the `almost isometric' case. Assuming that MATH-isometrically, as in the proof of REF , it follows that there exists a MATH such that MATH, and that the latter statement implies that MATH on MATH, and MATH on MATH. Moreover for MATH, we have MATH . |
math/9909040 | We will assume that MATH for simplicity, although the same argument works in general. Assume that MATH is a linear surjective isometry. Let MATH, for MATH, be the set of extreme points of unit ball in the dual space MATH. Let MATH be a restriction of the dual map MATH to MATH. Since MATH is a homeomorphism, in the weak* topology, as well as being an isometry, MATH is a homeomorphism of MATH onto MATH. Hence, for MATH, and MATH we have MATH . Since MATH is linear, the functions MATH and MATH depend only on MATH and we get the following representation of MATH: MATH where MATH is a surjective homeomorphism and MATH is a unimodular function. It is easy to see that this implies that MATH is continuous on MATH. (Since if MATH but MATH, then choose MATH with MATH. Then MATH. However MATH, but MATH.) For MATH we get that MATH. Thus the restriction of MATH to MATH is the restriction of a function in MATH. Also MATH . Hence it follows that MATH is also continuous on MATH. Since MATH is surjective there is MATH such that MATH, as functions on MATH . Hence MATH as functions on MATH, so MATH . Consequently we get MATH . Let MATH be such that MATH. We have MATH, so MATH on MATH. By REF the closure of MATH contains the NAME boundary of MATH so MATH on MATH, and consequently on MATH. This proves that MATH is an invertible element of MATH. Hence MATH. Note that the map MATH is linear, multiplicative, one-to-one, and indeed is isometric since MATH. As functions on MATH we have: MATH . Thus MATH may be viewed as an isometric automorphism MATH. In particular MATH is an invertible element of MATH and we have MATH as required. To finish the proof we need to show that MATH can be extended to a homeomorphism of MATH onto a subset of MATH, and that REF remains valid on the entire set MATH. Since MATH is an automorphism of MATH, it is given by a homeomorphism of the maximal ideal space of the algebra. That is, MATH can be extended to a homeomorphism of MATH onto itself mapping MATH onto MATH. For a given MATH in MATH, MATH and MATH are elements of MATH which, by REF , are identical on the NAME boundary of MATH. By dividing by MATH if necessary, we see that these elements are identical at any point of MATH and we get REF . |
math/9909040 | Let MATH be defined by MATH. We know from REF , that the restriction map from MATH to MATH is an isometry. By REF applied to MATH, there is a subset MATH of MATH, and a continuous function MATH from MATH onto MATH such that REF holds . Assume now that there is a MATH with MATH. At the beginning of the proof of REF, the set MATH is defined specifically as a subset of MATH, where we denote by MATH the norm of the ``evaluation at the point MATH" functional on MATH, and where MATH can be chosen to be any number satisfying MATH. Since MATH, there is a norm one element MATH of MATH such that MATH. Put MATH. We have MATH and MATH, while MATH . This contradicts REF and shows REF . To prove REF assume that there is a norm one element MATH of MATH such that MATH . Put MATH, let MATH be such that MATH, and let MATH be such that MATH. By REF we have that MATH which contradicts REF and shows REF . To finish the proof we need to show that REF implies MATH. To this end, choose a point MATH which is not in MATH. Without loss of generality we may assume that MATH. By REF there is a MATH such that MATH . By REF we have MATH . This contradiction proves that MATH. |
math/9909040 | That REF implies REF implies REF is left to the reader. Assume that MATH and MATH are almost isometric, and let MATH be such that MATH and MATH. Since MATH is surjective and MATH, there is a MATH such that MATH. By REF we get MATH so by REF we have MATH . From REF we obtain MATH where MATH. Let MATH be such that MATH. If MATH is sufficiently small, then REF gives MATH . Hence MATH for any MATH, and consequently for any MATH in the maximal ideal space of MATH. It follows that MATH and MATH are invertible in MATH. By REF the function MATH is approximately equal, on MATH, to MATH. Since MATH contains MATH, the function MATH is approximately equal to zero on the maximal ideal space of MATH. Thus, if MATH is sufficiently small, MATH is an invertible element of MATH. Consequently MATH is invertible. We can now define a MATH-module isomorphism MATH by MATH . Fix MATH and let MATH be such that MATH. By REF , MATH is close, on MATH, to MATH. It follows that MATH on MATH. Consequently by REF and the fact that MATH is an `almost isometry', we get MATH. Thus MATH is also an `almost isometry'. Now assume that MATH is equal to the NAME boundary. By REF we have that MATH on MATH . As we proved before, MATH is an invertible element of MATH. It follows that MATH . |
math/9909040 | This proceeds almost identically to the proof of REF. As in the proof of REF , it follows that for all MATH, there exists a MATH such that MATH and that the latter statement implies that MATH on MATH, and MATH on MATH. Thus MATH. Moreover for MATH, we have MATH . Since MATH is arbitrary, MATH. |
math/9909040 | We will only prove MATH. That MATH implies MATH follows from what we just did. The easy implications MATH are left to the reader. If MATH and MATH are module isomorphisms, let MATH and MATH. Then MATH, MATH, and MATH, MATH are invertible elements of MATH. If MATH then MATH and consequently MATH . Suppose that MATH. If MATH for some MATH, then by REF there is MATH such that MATH and MATH for MATH. We get MATH. This contradicts the assumption that MATH. Hence MATH so MATH . However MATH is invertible, so it attains minimum on MATH. Hence MATH . By REF it follows that MATH . Thus MATH; and since MATH is arbitrary we get MATH. |
math/9909040 | We use the injective tensor norm characterization of function modules CITE. Suppose that MATH is the MATH-isomorphism. Then for MATH, and MATH, we have: MATH where MATH is the injective tensor norm. Thus: MATH . By CITE, MATH is a function MATH-module. |
math/9909040 | Assume REF . As we said early in the introduction, the first two conditions in REF , together with a corollary in REF, shows that MATH-isometrically, where MATH, for some MATH on which MATH acts as a function algebra. Now REF follows from REF . By REF , MATH and MATH, showing REF . Given REF , it follows by REF that MATH is a function MATH-module, and now REF is clear. Clearly REF implies REF . |
math/9909040 | Let MATH let MATH be an open neighborhood of MATH and let MATH be such that MATH . Assume that MATH, and let MATH be such that MATH . Multiplying MATH by appropriate numbers of absolute value MATH we may assume that MATH. Then MATH . Put MATH . We have MATH . By the NAME MATH criterion REF page REF and remark on p. CASE: MATH is a p-point of MATH. That is MATH. |
math/9909040 | Suppose that MATH are MATH-tuples with MATH converging uniformly to MATH. Given MATH, we have MATH for all MATH and sufficiently large MATH. For such MATH, and for any complex Euclidean unit vector MATH we have MATH . Thus MATH. Letting MATH gives MATH. Since MATH was arbitrary, we get the result. |
math/9909040 | By the Lemma and REF (respectively, REF), MATH (respectively, in MATH), and MATH. Now use REF (respectively, REF). |
math/9909040 | REF is proved as for NAME and logmodular algebras, using NAME lemma and the fact that for a MATH-tuple MATH, the function MATH achieves its maximum modulus on the NAME boundary. (The latter fact may be seen by considering MATH for MATH). Similarly, REF follows the classical line of proof: Suppose that MATH are representing measures for MATH, and that MATH are MATH-tuples with MATH for all MATH. By NAME and NAME, we have MATH . The remainder of the proof follows REF Finally, REF follows from REF by REF. |
math/9909040 | CASE: Clearly MATH is a unitary subequivalence bimodule if and only if MATH and MATH both contain inner functions (functions of constant modulus REF on MATH), which is clearly equivalent to MATH. CASE: Suppose that MATH is given, and that MATH are as in REF . Then MATH as one allows MATH. Hence MATH. Similarly MATH. For the converse, notice that the argument given is reversible. CASE: If MATH uniformly as MATH, where MATH, then we may write MATH, and clearly MATH and MATH. This obviously implies the condition in REF with MATH. Also, this argument is reversible. CASE: This is obvious. |
math/9909040 | First suppose that MATH is antisymmetric on MATH. By hypothesis, we have MATH and MATH, for MATH-tuples MATH with MATH. By the converse to NAME, MATH, where MATH is the complex conjugate. Thus MATH. By antisymmetry, MATH a nonnegative constant. Therefore, if MATH then MATH. We have MATH, so that MATH. If MATH is a disjoint union of a finite number of antisymmetric pieces MATH, then each MATH is open and compact, and by the first part we have MATH, for an invertible MATH. Put MATH if MATH, then MATH. It follows from REF say, that MATH and MATH. Thus MATH. |
math/9909040 | Assume that MATH. We will show that MATH are in different NAME parts of MATH. In the definition in the introduction of MATH, take MATH for a natural number MATH, and choose the corresponding MATH with MATH. By that definition we may assume that MATH , and MATH, for all MATH. Here MATH is a sequence of real numbers decreasing to REF. Multiplying the functions MATH by constants with absolute value one we may also assume that MATH . Write MATH, and MATH. So MATH and MATH. By the NAME Inequality we get MATH . Hence MATH . Obviously, the same formula holds with MATH replaced by MATH. Expanding out the following square as an inner product and using REF we get MATH . Hence, by the Pythagorean Identity MATH . Put MATH. Since MATH and MATH we get MATH . On the other hand, for any Euclidean vector MATH of norm REF we have MATH where MATH is the norm of MATH considered as a functional on MATH. From the above it follows that MATH. Thus MATH . Since MATH, we get MATH. Thus MATH lie in different NAME parts of MATH. To prove the other implication assume now that MATH are in different NAME parts of MATH. Choose a sequence of functions MATH analytic in the disc MATH with MATH and MATH for all MATH. Such functions can be found by taking a conformal equivalence of MATH with a `smile shaped region' inside the annulus MATH, with the two tips of the smile at MATH and MATH. Since MATH are in different NAME parts of MATH, for any MATH large enough, there is a MATH in MATH such that MATH, MATH, MATH. Also the norm, on MATH of the ``evaluation at MATH" functional is equal to one. Thus there is a MATH such that MATH and MATH. Put MATH. Notice that MATH so MATH. Moreover the spectrum of MATH is contained inside the annulus MATH. Hence MATH is an invertible element of MATH, and consequently MATH is an invertible elements of MATH. Thus MATH. By REF this proves that MATH is a strong NAME equivalence bimodule. |
math/9909040 | Suppose that MATH. As in the proof of the previous theorem, take MATH for a natural number MATH, and choose MATH as in that proof. Write MATH, and MATH. So MATH and MATH. Again by NAME we get MATH for all MATH; and the same formula holds for MATH. By expanding out the following square as an inner product, we see again that MATH uniformly as MATH. This gives the one direction of the proposition. However, the argument is reversible until at the end we obtain that MATH uniformly in MATH. Thus MATH is invertible in MATH, and MATH. |
math/9909040 | This follows essentially as in pure algebra REF. Suppose that MATH is any strong NAME equivalence MATH-bimodule. Then by the basic NAME theory, any right MATH-module map MATH is simply left multiplication by a fixed element of MATH. Indeed, via this identification, we have MATH isometrically and as algebras (CITE and REF). For fixed MATH, the operator MATH on MATH, is a right MATH-module map, with completely bounded norm MATH. Therefore by the above fact, there is a unique MATH such that MATH for all MATH. The map MATH is then seen to be an isometric unital automorphism MATH of MATH. In this way we have defined a surjective group homomorphism MATH. This homomorphism has a REF-sided inverse MATH, namely MATH. In this way, we see the `semidirect product' statement. Note that MATH, and MATH is symmetric. |
math/9909040 | We use the notation and facts established in the second paragraph below the proof of REF . If MATH is singly generated over MATH, then so is MATH over MATH, and again it follows that MATH completely MATH-isometrically. Here is one way to see this (which adapts to the noncommutative case): if MATH is the generator, then MATH is a strictly positive element in MATH (see for example REF ). Thus MATH is invertible, so that MATH is invertible in MATH. Thus MATH, and MATH. The `converse' assertion follows from the earlier correspondences between the classes MATH and MATH, and strong NAME equivalence and subequivalence bimodules. |
math/9909040 | We just saw that MATH. As we saw in REF is equivalent to REF . REF shows that REF implies REF . |
math/9909040 | By the `harmonicity' associated with the MATH-class, we can assume that MATH. Then REF follows immediately from REF . However here is another argument, which adapts immediately to give REF too. Suppose that MATH-isometrically, where MATH. Then by REF we have MATH. This implies that MATH by the last theorem. Thus MATH. This also gives REF , in view of REF. A similar argument to REF also proves REF . |
math/9909040 | Let MATH be a c.a.i. for MATH, and define MATH to be multiplication by MATH. These satisfy the requirements for a rigged module. |
math/9909040 | As observed in REF, MATH is a (singly generated) left MATH-module over MATH. Also MATH may be regarded as a MATH-submodule of MATH (in the obvious way). Let MATH be the single generator of MATH and MATH. Suppose that MATH is the ideal in MATH generated by the range of the inner product. Now MATH is a full MATH-module over MATH, and so MATH is dense in MATH, since MATH-modules are automatically nondegenerate. Since MATH is dense in MATH, MATH, and MATH is dense in MATH, we see that MATH is dense in MATH. Hence MATH is singly generated by MATH over MATH. The obvious map MATH is adjointable and MATH have dense range, so by the basic theory of MATH-modules (see for example, REF ), MATH, MATH-isometrically. Hence MATH , MATH-isometrically. Since MATH is singly generated by MATH, MATH is a strictly positive element in MATH, by the argument before REF above. Of course MATH is not strictly positive on MATH, unless MATH. Clearly MATH is MATH-isometric to the closure of the submodule MATH of MATH. |
math/9909040 | MATH: For MATH, we have MATH . Since MATH achieves its maximum modulus on MATH, we have proved this direction. MATH: If MATH, then this is well known, following by the usual NAME boundary point argument (such as we've seen, for example, in REF). For general MATH, fix MATH. Then we have MATH, for all MATH. Thus by the MATH case, MATH for all MATH, which gives what we need. |
math/9909040 | Suppose that MATH is MATH-Hilbertian. We may suppose, by REF, that MATH in MATH. We will take MATH to be the NAME boundary of MATH. We use the notation of the beginning of the section, but assume, as we may, that the norms of MATH are strictly less than REF. The map MATH may be written as MATH for some MATH, and without loss of generality, we can assume that MATH, with MATH. Thus MATH may be associated with a MATH-tuple MATH, and without loss of generality, MATH for all MATH. See REF . Also, MATH is completely determined by its action on MATH, so we can associate MATH with a unique MATH-tuple MATH. Now it is easily seen that without loss of generality, MATH may be regarded as (multiplying by) the element MATH of MATH. We have MATH uniformly. Next we note that by REF , the identity MATH for all MATH, implies that MATH for all MATH. Thus it follows that for MATH, we have MATH for a constant MATH that does not depend on MATH. Set MATH. These give new factorization nets, but now we have that MATH as MATH. This says that MATH is a rigged module. The converse is trivial. |
math/9909040 | If one studies the proof of REF , one sees that the ideas there yield that if MATH then MATH is a rigged module. If MATH, then MATH vanishes on MATH. By REF , the functions MATH form a c.a.i. for MATH, and also for MATH. We deduce from REF that MATH is a p-set. Since MATH is strictly positive on MATH, it follows that MATH is a MATH, from which we deduce, using CITE II. REF, that MATH is a peak set. This gives REF . If MATH is a rigged module then the ideas of the proof of REF show that REF holds for MATH. Set MATH, then MATH vanishes on the set MATH. We saw in that Corollary that MATH uniformly on MATH. By the NAME theorem MATH is dense in MATH. Thus the functions MATH, which are in MATH, form a c.a.i. for MATH. Thus by NAME 's lemma, the restriction of MATH to MATH is in MATH. |
math/9909040 | If MATH is a compact subset of MATH, and MATH is given, then MATH uniformly for MATH and MATH sufficiently large. From this one easily sees that MATH uniformly on MATH, since MATH is bounded away from zero on MATH. The second statement is similar. Finally, for MATH, we see that MATH . |
math/9909040 | REF is obvious. CASE: Let MATH be MATH-tuples as in REF . For any unit vector MATH in the complex Euclidean space of the appropriate dimension, and any MATH, we have MATH . Thus MATH. Letting MATH, gives MATH. To get the other inequality, we may assume that MATH is bounded below. Thus MATH. As above, we obtain MATH. Hence MATH. Now let MATH. REF : A function is harmonic on a domain in MATH if and only if it satisfies the Mean Value Principle. Let MATH be as in REF , and fix MATH. Since MATH is analytic for MATH, we have that MATH is subharmonic on MATH. Thus, for any ball MATH center MATH in MATH, we have MATH . Since this is true for all such MATH we have MATH. Taking the limit as MATH gives MATH. A similar argument using MATH gives the other direction of the Mean Value Property. |
math/9909040 | Let MATH be vectors in complex Euclidean space. Then we have MATH for all MATH, and consequently for all MATH. |
math/9909040 | As we saw above, REF is equivalent to saying that the functions MATH, which are in MATH, form a c.a.i. for MATH. However, MATH, so that MATH is a c.a.i. for MATH. Hence by CITE II. REF and II. REF, taking the function MATH there to be a continuous strictly positive function which is MATH on MATH and MATH on a compact subset MATH of MATH, we see that MATH uniformly on MATH. Now by REF , we have MATH for all MATH. This then implies that MATH uniformly on any compact subset MATH of MATH. Since MATH and MATH are uniformly bounded above on MATH, we deduce that they are also uniformly bounded away from zero on MATH. By REF , we have MATH . Thus we have MATH uniformly on MATH, for MATH large enough. Thus the sequence MATH, and by symmetry the sequence MATH, are uniformly NAME on MATH. Let MATH be the uniform limit of MATH on MATH. Varying over all compact MATH gives a well defined continuous MATH on MATH. Clearly MATH extends MATH. We next show that MATH. Let MATH be given, and let MATH. For MATH to be determined, choose (by CITE II. REF again) MATH with MATH on MATH, MATH, and MATH on MATH. For MATH and any Euclidean vector MATH of norm REF, we have MATH , if MATH is smaller than a certain constant which depends only on MATH and MATH. Hence MATH, for all MATH. Hence MATH. Letting MATH we see that MATH. In particular, for MATH, we have MATH. Thus indeed MATH. Define MATH to be zero on MATH. Clearly MATH for all MATH. For MATH we obtain from REF , that MATH, and also MATH. Thus MATH. Finally, for the uniqueness, we suppose that MATH with MATH, and MATH with MATH, both fulfill the definition of MATH. If MATH for all MATH, then MATH on MATH, and hence, by REF , on MATH. Hence MATH for any MATH. This obviously implies what we want, by symmetry. |
math/9909040 | The equivalence of REF is classical (CITE p. REF). Suppose that MATH is a singly generated rigged module over MATH. By REF and the fact mentioned in the previous paragraph, we have MATH-isometrically, where MATH is a subset of MATH of NAME measure MATH, and MATH. With the earlier notation, we have MATH, and MATH, for all MATH. This implies that there is a nonzero function MATH, such that MATH on MATH. Then MATH, a.e. on MATH. Since MATH is integrable on MATH, so is MATH . Conversely, let MATH, with MATH continuous on MATH. Let MATH be the subset of MATH on which MATH vanishes, which is a a closed subset of measure MATH. Let MATH, then MATH is integrable. We choose a function MATH on MATH such that MATH, and such that MATH is continuously differentiable wherever it is finite. Indeed, one may assume that MATH lies in a thin strip about MATH. We define MATH . Then MATH, and MATH on MATH. Next, choose an open subset MATH of MATH containing MATH, such that MATH, and such that MATH on MATH. We may suppose that MATH is a finite collection of disjoint open intervals, of total combined length MATH. Choose MATH outside MATH . On MATH we define MATH so that MATH lies between MATH and MATH on MATH, and so that MATH is finite and continuously differentiable on all of MATH. It is not hard to see that this is possible. Then we have MATH. We define MATH by the formula defining MATH above, but with MATH replaced by MATH. Then MATH is nonvanishing on MATH, and MATH on MATH. Setting MATH, we may write, for fixed MATH: MATH . The first of the three terms on the right equals MATH. Supposing that MATH, we have MATH for MATH, whence MATH . We now check that REF holds. By an easy compactness argument, we may assume that the compact subset MATH there, is a finite closed interval. Pick MATH so small in relation to MATH, that for any MATH close enough to MATH, we have MATH for any MATH with MATH. For the MATH associated with this MATH, we have, for MATH close to MATH, that MATH . Thus MATH for all MATH close enough to MATH . Thus MATH, on MATH. From this it is clear that the conditions of REF are met, so that MATH is a rigged module over MATH. |
math/9909040 | Clearly the multiplication map MATH is completely contractive, and has dense range. Conversely, choose MATH as in REF , and let MATH as before. For MATH, define MATH. The MATH notation here, is commonly used with reference to the NAME tensor product. Namely, for two finite tuples MATH , the expression MATH means MATH. If MATH, then MATH. From the definition of the NAME tensor product CITE we see from this that MATH is well defined and completely contractive. It is easy to see that MATH , for all MATH. Hence MATH is a complete isometry. We leave it to the reader to check that if MATH, then MATH. Thus MATH is a unital semigroup. The last assertion is also an easy exercise. |
math/9909040 | MATH: This follows from the definition of MATH; given MATH there exist MATH-tuples MATH with MATH, such that the norms of MATH and MATH are close to REF. These may be associated with maps MATH and MATH as in the proof of REF . MATH: Follows from the definition of MATH-Hilbertian, and the fact that such MATH is MATH-rigged if and only if it is MATH-Hilbertian. MATH: The maps MATH in the definition of MATH-Hilbertian, may be associated as in the proof of REF , with certain MATH-tuples MATH. We have MATH. Hence MATH, so that by the common NAME series trick, MATH is an invertible element MATH of MATH, and MATH. Replace MATH by MATH. Correspondingly we get an adjusted MATH, which now satisfies REF . MATH we state as the next result. |
math/9909040 | The MATH direction is REF of the previous theorem. Supposing REF, we proceed as in the proof of REF , to associate with MATH and MATH, MATH-tuples MATH and MATH. We have that MATH, and as in that proof we get MATH on MATH, and MATH on MATH. Applying NAME to MATH, and using these inequalities, shows that MATH on MATH, and MATH on MATH. Thus MATH. That MATH follows as in the proof of REF. |
math/9909040 | MATH: Follows since MATH. MATH : Assuming REF , then by REF , MATH-isometrically, where MATH. Since every strong NAME equivalence bimodule is a rigged module, REF now gives REF . MATH: This is true for any algebraic NAME equivalence MATH-bimodule CITE. MATH: It follows by REF, and REF below, that MATH is unital. Thus MATH is continuous, so that MATH is closed and open. If MATH is connected, it follows that MATH is the empty set. MATH: Every singly generated MATH-rigged module is of the form MATH, which is a faithful function MATH-module. Now appeal to REF . |
math/9909040 | The proof requires some technical knowledge of rigged modules CITE. Here is one way to see REF . If MATH is a left MATH-rigged module, then from REF, MATH is a left MATH-rigged module. But MATH, MATH-isometrically. To get REF , suppose that MATH is a singly generated MATH-rigged module. Then by the previous results, there is a peak set MATH and a function MATH such that MATH-isometrically. Since MATH is a strictly positive element of the ideal MATH in REF , we see that MATH. Suppose that MATH is the dual rigged module of MATH (see CITE for details), where the reader may also find the definition of MATH, which we shall need shortly. Since MATH may be taken to be MATH, it follows that the linking MATH-algebra for MATH is MATH. Thus we can make the following deductions from REF. Firstly, MATH may be identified with a closed subalgebra MATH of MATH, and MATH has a contractive approximate identity which is a contractive approximate identity for MATH. Also, MATH may be regarded as a subspace of MATH, and the canonical pairings MATH, and MATH may be regarded as the commutative multiplication in MATH . Thus it follows that MATH is the closure of the span of the range of the canonical pairing MATH . Thus MATH is also a subset, indeed a closed ideal, of MATH. By the commutativity of the multiplication in MATH, MATH is a strong NAME equivalence MATH-bimodule. We also see that MATH is dense in MATH. Since MATH has a contractive approximate identity, we may appeal to REF to see that MATH for a p-set MATH. Since MATH contains a c.a.i. for MATH, it is clear that MATH. On the other hand, if MATH, then there is a peak set MATH containing MATH, with MATH. Choose MATH with MATH on MATH and MATH. Then MATH, so MATH. This is impossible, so that MATH. Note that MATH is dense in MATH, MATH is dense in MATH and MATH. Hence MATH is dense in MATH. Thus MATH. |
math/9909041 | Suppose that MATH is an ideal in MATH whose intersection with MATH is zero. Let MATH. Since MATH is REF on MATH, it is completely isometric on MATH. Let MATH be the restriction of MATH to MATH. By REF, MATH is a complete order injection. Extend the map from MATH which is the inverse of MATH, to a map MATH. Since MATH, it follows by rigidity that MATH. Thus MATH. |
math/9909041 | Clearly MATH, where MATH is as in REF . Thus MATH is an essential ideal in MATH, by that Proposition. To see the second assertion, notice that if MATH and MATH, then MATH, which implies that MATH for all MATH. It follows immediately that MATH. Hence by the first assertion, MATH. Thus MATH is an essential ideal in MATH. That MATH follows from the universal property of the multiplier algebra, namely that MATH contains a copy of any MATH-algebra containing MATH as an essential ideal (REF, say). |
math/9909041 | Without loss of generality, we may assume that MATH. By replacing MATH by MATH, we may also suppose that MATH, where this last `REF' is the identity of MATH. Let MATH. Define MATH, for MATH. Since MATH for MATH, we obtain MATH, by rigidity. Thus MATH, which (we showed in the proof of REF ) implies that MATH. |
math/9909041 | By REF , we have a completely order isomorphic copy MATH of MATH inside MATH. By REF , there exists a surjective *-homomorphism from the MATH-subalgebra MATH of MATH generated by MATH, onto the MATH-envelope MATH, which fixes the copies of MATH. Let MATH be a completely positive map extending the *-homomorphism. Since MATH fixes the diagonal scalars MATH, it follows by REF (this is a common argument), that MATH decomposes as a MATH matrix of maps, each corner map defined on the corresponding `corner' of MATH. In particular, we have MATH for a map MATH, MATH, and for all MATH. By the previous lemma, MATH is a `MATH-bimodule map'. Hence for MATH we have: MATH . Thus MATH. The uniqueness of MATH follows from REF . Similarly for MATH. For REF , note that for MATH and MATH we have MATH. Now use REF to conclude that MATH is a homomorphism. The last assertion follows from the fact that a contractive representation of a MATH-algebra is a *-homomorphism. |
math/9909041 | CASE: Represent MATH nondegenerately on a NAME space MATH. We obtain a multiplication situation on MATH given by the actions of MATH and MATH on MATH. By the multiplication theorem, we get a completely contractive homomomorphism MATH such that MATH, for all MATH and MATH. Hence MATH. Thus by REF , we have MATH. By NAME 's theorem CITE, the latter MATH-algebra may be identified with MATH. Taking adjoints gives MATH. As noted in REF below implies that MATH. Finally, note that we have the following MATH-subalgebras: MATH. The last inclusion is a fact from CITE which is reproved at the end of our paper. By the injectivity of MATH there exists a completely contractive projection MATH from MATH onto MATH; since MATH fixes MATH we see by rigidity that MATH and MATH. CASE: We apply REF but with MATH replaced by the right MATH-module sum MATH. Clearly the latter is a full MATH-module over MATH, and it is injective since it is a corner of MATH. Using REF if necessary, we obtain that MATH, from which it is easy to see that MATH is a self-dual MATH-module. However MATH by NAME 's factorization theorem. CASE: It is well known that for any full right MATH-module MATH over a MATH-algebra MATH say, we have that MATH is a copy of the linking MATH-algebra of MATH. Also MATH. Thus we have MATH. This of course splits into four corners. REF corner is MATH by REF . REF corner is MATH by REF . REF corner is MATH by REF . This gives the first result. The second follows just as MATH in REF . |
math/9909041 | CASE: By REF, any MATH may be viewed as a bounded, and hence completely bounded, module map on MATH. We obtain a canonical sequence of completely contractive homomorphisms MATH given by restriction of domain. By REF , these homomorphisms are REF. On the other hand, we have a multiplication situation given by the action of MATH on MATH. Hence, by the multiplication theorem, there exists a completely contractive homomorphism MATH such that MATH for all MATH. Thus MATH. Thus MATH is onto, and since MATH is REF we obtain that MATH completely isometrically and as operator algebras. CASE: The first statement follows from a result which may be found in REF or REF , which asserts that any contractive homomorphism from a MATH-algebra into a NAME algebra, is a MATH-homomorphism onto its range, which is a MATH-algebra (with the norm and algebra structure inherited from the NAME algebra). Hence the canonical homomorphism MATH (or MATH) is a MATH-homomorphism onto a MATH-algebra. Since the homomorphism is REF it is therefore isometric. The second statement follows by considering the following canonical isometric inclusions MATH the last inclusion following from REF. Thus by restriction of domain, we get a contractive unital homomorphism MATH. Now we can apply REF to deduce this last homomorphism is an isometry. A similar argument works for MATH. |
math/9909041 | As in CITE the difficult part is to prove the first statement, the existence of MATH. (The uniqueness follows from REF.) Indeed it suffices to find NAME spaces MATH, a complete isometry MATH, and a linear complete contraction MATH, such that MATH, and such that MATH, for all MATH. For in that case MATH would form a multiplication situation on MATH. Now use the multiplication theorem above, together with the fact that MATH, to obtain the existence of MATH. The existence of such MATH etc., follows easily from NAME 's proof of the `BRS' theorem (CITE). Namely, first suppose that MATH. By the `multilinear NAME theorem of CITE, we may write MATH, for completely contractive maps MATH and MATH. Similarly, MATH, where now MATH say. Inductively we obtain, MATH, where MATH, say. Since MATH, we see that each MATH is a complete isometry. Let MATH, and define MATH by MATH. It is easy to check that MATH, for MATH, as in CITE. Hence (by the parallelogram law if necessary) it is clear that MATH defines a seminorm which gives rise to a NAME space norm (on the quotient of MATH by the nullspace of the seminorm). Write this resulting NAME space as MATH. There is an obvious map MATH, given by MATH. It is easy to see that this is completely contractive as in CITE. The map MATH given by MATH is clearly a complete contraction, too. On the other hand, for any MATH, we have MATH . Thus MATH, showing that MATH is an isometry. A similar argument shows that MATH is a complete isometry. |
math/9909041 | That REF is equivalent to REF is a restatement of the NAME representation theorem. MATH: Obvious. MATH: MATH form a multiplication situation on MATH. The result then follows by the multiplication theorem. |
math/9909041 | This follows immediately from the proof of REF , and of the implication MATH of REF. The MATH is as in the proof of REF , which is easily seen to be a homomorphism on MATH. |
math/9909041 | Clearly MATH. Any minimal MATH-projection CITE is the identity on MATH, and is therefore the identity map. The rest is clear. |
math/9909041 | One direction of REF is obvious. Namely, suppose that MATH is MATH-injective. By CES, MATH may be realized as a MATH-submodule of some MATH, where MATH is a NAME MATH-module and MATH is a NAME MATH-module. Indeed MATH is an operator MATH-bimodule. By the MATH-injectivity of MATH, there is a completely contractive projection from MATH onto MATH. Since MATH is injective as an operator space, so is MATH. The other direction of REF is harder. In a previous version of this paper we had a proof which used almost all the results established until now. Instead we shall only use a few results from REF , and REF above. We will also use the fact, which is a simple consequence of NAME 's original result, or NAME 's modification of this result CITE, that if MATH is a NAME MATH-module, then for any NAME space MATH, we have that MATH is MATH-injective. Indeed their result gives that MATH is MATH-injective if MATH isometrically. However in the contrary case, one may use the following kind of trick: Pick a NAME space MATH in which MATH is faithfully represented, then one obtains a faithful representation of MATH on MATH. Then one may apply the NAME or NAME result to conclude that MATH is MATH-injective, from which it is easy to see by compression that MATH is MATH-injective. See REF for another proof of this simple consequence. Suppose that MATH is injective. Represent the MATH-algebra MATH faithfully and nondegenerately on a NAME space; then the two diagonal projections determine a splitting of the NAME space as MATH, say. So MATH is a unital *-subalgebra of B REF , and so on. Now by injectivity there is a completely positive projection MATH from MATH onto MATH. As in the proof of REF, the NAME Lemma implies that this projection is an I REF -module map, and that MATH decomposes as a MATH matrix of maps. Let MATH be the `REF corner' of MATH. Thus MATH is a completely contractive projection onto MATH, and its easy to see, as in REF, that MATH is a left MATH-module map. However if MATH is an operator MATH-bimodule, then by REF above there is a unital *-homomorphism MATH implementing the left module action. Hence MATH is a NAME MATH-module, via MATH. Since MATH maps into MATH, we see that the projection MATH is a left MATH-module map onto MATH. Similarly, MATH is a right MATH-module map onto MATH. Since MATH is MATH-injective, so is MATH. REF is obvious, given REF and the fact, observed earlier, that the MATH-action on MATH extends to make MATH an operator MATH-bimodule. |
math/9909041 | We may suppose without loss of generality that MATH is a full MATH-module over MATH. Thus we may regard MATH as a MATH-imprimitivity bimodule. Let MATH equipped with its MATH-imprimitivity bimodule structure. Here MATH as usual. From CITE, we know that MATH, as imprimitivity bimodules. Thus MATH, giving REF . Similarly we get REF . It is well known, and fairly obvious, that REF implies REF . |
math/9909041 | Represent MATH nondegenerately and faithfully on a NAME space MATH. If MATH is an increasing contractive approximate identity for MATH, then MATH in the SOT. We also have MATH. Let MATH be a minimal MATH-projection. So MATH is a completely contractive idempotent map whose range contains MATH. If MATH we would be done, since in that case MATH is completely positive, and then we can deduce that MATH is a unital MATH-algebra as in the introduction (that is, by the NAME result quoted there). In order to see that MATH, we choose MATH with MATH. Let MATH, for MATH. Then MATH and MATH. It is no doubt well known and easy to see that this implies that MATH. One way to do this is to write MATH, where MATH is a unital *-representation of MATH on a NAME space MATH, and where MATH. Set MATH, and let MATH be the projection onto MATH. Then it is easy to see that the net MATH has MATH as a WOT limit point. For if MATH is a WOT limit point of MATH, then for MATH we have MATH . Thus MATH. By the converse to NAME, MATH. Thus MATH . |
math/9909042 | Given any choice of defining function MATH, let MATH and set MATH, so MATH and MATH. Thus MATH so the condition MATH is equivalent to MATH . This is a non-characteristic first order PDE for MATH, so there is a solution near MATH with MATH arbitrarily prescribed. |
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