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math/0103039 | The elements MATH form a NAME MATH-family in MATH, so there is a homomorphism MATH with MATH and MATH. We trivially have MATH for MATH. To see that MATH is injective, we use the universal property of MATH to build an action MATH such that MATH note that MATH converts the gauge action on MATH to MATH, and apply the gauge-invariant uniqueness theorem CITE. It follows from CITE that MATH converges strictly to a projection MATH such that MATH . Thus MATH is spanned by the elements MATH with MATH, and by applying the NAME relations we may assume MATH also, so that the range of MATH is the corner MATH. To see that this corner is full, suppose MATH is an ideal containing MATH. Then CITE implies that MATH is a saturated hereditary subset of MATH; since MATH certainly contains MATH, we deduce that MATH. But then MATH by CITE. Finally, to see that the diagram commutes, we just need to check that MATH and MATH agree on generators. |
math/0103039 | Let MATH. Then MATH is a NAME MATH-family which generates MATH. The universal property of MATH gives a homomorphism MATH such that MATH and MATH, which is an isomorphism by the gauge-invariant uniqueness theorem CITE. It is easy to check on generators that MATH is a MATH-isomorphism. When MATH is a tree with one sink MATH, there is precisely one path MATH in MATH from MATH to MATH, and hence all the new paths from a vertex MATH to MATH have the form MATH. Thus if MATH, MATH . On the other hand, if MATH, then MATH . Together these calculations give REF . |
math/0103039 | Suppose that there exists MATH such that MATH. Then since MATH . (see CITE), we have MATH, and MATH is not essential. Conversely, suppose that MATH. To show that MATH is an essential ideal it suffices to prove that if MATH is a representation with MATH, then MATH is faithful. So suppose MATH. In particular, we have MATH. For every MATH there is a path MATH in MATH such that MATH and MATH. Then MATH, and hence MATH. Since every loop in a REF-sink extension MATH must lie entirely in MATH, every loop in MATH has an exit in MATH; thus we can apply CITE to deduce that MATH is faithful, as required. |
math/0103039 | Since the statement is symmetric in MATH and MATH, it suffices to prove this for MATH. Choose a path MATH in MATH such that MATH and MATH. Since MATH is finite, going along paths from MATH to MATH and then to and fro between MATH and MATH must eventually give either REF a loop MATH which visits both MATH and MATH, and a path MATH with MATH and MATH, or REF a vertex MATH and a path MATH with MATH and MATH. We deal with REF first. Since there are boundary edges MATH and MATH with MATH on MATH, we can perform outsplittings along MATH to get new tree extensions MATH, where MATH is the loop MATH relabelled so that it ends at MATH. Because MATH and MATH have the same vertices as MATH in a different order, REF gives MATH so we have MATH. Since MATH, and in forming both MATH we have performed an outsplitting at MATH, MATH is a boundary vertex in both MATH; say MATH has MATH. Write MATH, MATH for the path obtained by going MATH times around MATH, and define MATH . We now compute the NAME vectors using REF : for example, MATH . The formula for MATH is the same except for the last term, so MATH as required. In REF , we can dispense with the first step in the preceding argument: we choose boundary edges MATH with MATH, and then MATH have the required properties. |
math/0103039 | As we indicated earlier, it suffices to prove the theorem when MATH and MATH are tree extensions. It also suffices to prove that we can perform boundary outsplittings on MATH and MATH to achieve extensions MATH and MATH with the same NAME vector; REF then imply that we can take for MATH the common simplification of MATH and MATH. We can write MATH for some finite set MATH. We shall prove by induction on MATH that we can perform the required outsplittings. If MATH, then MATH, and there is nothing to prove. So we suppose that we can perform the outsplittings whenever MATH has the form MATH, and that MATH. Let MATH be the subgraph of MATH with vertices MATH and edges MATH. Since MATH is a finite graph it contains either a sink or a loop. If MATH contains a sink, then by relabelling we can assume the sink is MATH. Since MATH for all MATH, we have MATH . Thus either MATH or MATH has at least MATH boundary edges leaving MATH: we may as well assume that MATH, so that MATH. We can then perform MATH boundary outsplittings on MATH at MATH to get a new extension MATH. From REF , we have MATH, and therefore MATH . Since MATH is formed by performing boundary outsplittings to the essential tree extension MATH, it is also an essential tree extension, and the inductive hypothesis implies that we can perform boundary outsplittings on MATH and MATH to arrive at extensions with the same NAME vector. If MATH does not have a sink, it must contain a loop MATH. If necessary, we can shrink MATH so that its vertices are distinct, and by relabelling, we may assume that MATH lies on MATH. Because the extensions are essential, we have MATH and MATH, so we can apply REF . Thus there are REF-sink tree extensions MATH and MATH formed by performing boundary outsplittings to MATH and MATH, and for which MATH . But because MATH this implies that MATH where MATH if MATH lies on MATH, and MATH otherwise. We can now invoke the inductive hypothesis to see that we can perform boundary outsplittings to MATH and MATH to arrive at extensions with the same NAME vector. This completes the proof of the inductive step, and the result follows. |
math/0103039 | We first recall from CITE that MATH is MATH if and only if MATH has no loops. Now we proceed as in the proof of REF . Everything goes the same until we come to consider the finite subgraph MATH associated to the support of the vector MATH. Since there are no loops in MATH, MATH must have a sink, and the argument in the second paragraph of the proof of REF suffices; this does not use essentiality. |
math/0103039 | For any REF-sink extension MATH, the map MATH is a homeomorphism of MATH onto the closed subset MATH. If MATH, then MATH, and hence MATH is always a homeomorphism of the closed set MATH in MATH onto the corresponding subset of MATH. So the only issue is whether the closures of the sets MATH and MATH match up. But MATH . Since other sets MATH associated to sinks are never MATH, the ideals on the right-hand side are those associated to the maximal tails lying in MATH, and the result follows. |
math/0103039 | Suppose MATH is a maximal tail such that MATH in MATH and MATH; we want to prove MATH. We know MATH for some MATH. If MATH, there is no problem. If MATH, then MATH for some MATH with MATH, so MATH. Now suppose MATH in MATH and MATH; we want to prove that MATH. We know that there is a path MATH with MATH and MATH. If MATH, we have MATH. If MATH and MATH, we have MATH. The one remaining possibility is that MATH and there is no path of length at least REF from MATH to MATH. Because MATH is a tail, there exists MATH such that MATH and MATH. Now we use MATH to get a path MATH with MATH and MATH, and we are back in the first case with MATH. |
math/0103039 | Since the closures MATH and MATH are unaffected by simplification and boundary outsplitting, we can run the argument of REF . In doing so, we never have to leave the common closure MATH: by hypothesis, MATH for some MATH, so all the vertices on the subgraph MATH used in the inductive step lie in MATH. When MATH has a sink, the argument goes over verbatim. When MATH has a loop MATH, all the vertices on MATH lie in MATH, and the hypothesis MATH implies that MATH, so we can still apply REF . The rest of the argument carries over. |
math/0103039 | Since the maximal tails comprising MATH are backwards hereditary, there are no paths from MATH to MATH. Thus MATH decomposes with respect to the decomposition MATH as MATH, and MATH. Writing MATH as MATH and noting that MATH has support in MATH shows that MATH, which by the hypothesis on MATH implies MATH. But this says exactly what we want. |
math/0103039 | The only edges which are affected in forming MATH are MATH and the edges MATH with MATH. Since none of the new edges in MATH have range in MATH, they are not affected by the outsplitting. Simplifying collapses paths which end at one of the sinks MATH, and forming MATH adds only paths of length REF ending at MATH, so there is nothing extra to collapse in simplifying MATH. |
math/0103039 | As in REF-sink case, it suffices by REF to prove the result when MATH and MATH are simple. So we assume this. Our proof is by induction on MATH, but we have to be careful to get the right inductive hypothesis. So we shall prove that by performing MATH standard constructions on both MATH and MATH, we can arrive at simple MATH-sink extensions of MATH with all their NAME vectors equal; these graphs are then isomorphic, and we can take MATH to be either of them. REF says that this is true for MATH (see REF ). So we suppose that our inductive hypothesis holds for all simple MATH-sink extensions satisfying the hypotheses of REF . Then MATH and MATH are simple MATH-sink extensions of MATH with NAME vectors MATH and MATH for MATH. So the NAME vectors of MATH and MATH satisfy REF . Since MATH in MATH, and we have not deleted any edges except those ending at MATH and MATH, we still have MATH in MATH for MATH, and similarly MATH in MATH. By the inductive hypothesis, therefore, we can perform MATH standard constructions on each of MATH and MATH to arrive at the same simple MATH-sink extension MATH of MATH. By REF , MATH and MATH are obtained from MATH and MATH by MATH standard constructions. We now view MATH and MATH as two simple REF-sink extensions of the graph MATH. Since the standard constructions have not affected the path structure of MATH inside MATH, and we assumed MATH in MATH, we still have MATH in MATH, and similarly MATH in MATH. Because any sink in MATH has to be a sink in MATH, the hypothesis MATH in MATH implies that MATH has no sinks; thus every vertex in MATH lies on an infinite path MATH, and hence in the maximal tail MATH. Thus MATH says precisely that MATH is the closure of MATH in MATH. Of course the same is true of MATH in MATH. We can therefore apply REF to deduce that we can by one more standard construction on each of MATH and MATH reach the same REF-sink extension MATH of MATH; since all the boundary vertices of MATH lie in MATH, this standard construction for extensions of MATH is a also standard for extensions of MATH, and hence MATH can also be obtained by performing MATH standard constructions to each of MATH and MATH. This completes the proof of the inductive hypothesis, and hence of the theorem. |
math/0103039 | We first suppose that MATH is simple. Then MATH is the group MATH considered in CITE, and it suffices to show that MATH is the homomorphism MATH considered there. To do this, we need to check that the map MATH of MATH onto MATH in CITE satisfies MATH. The map MATH is built up from the homomorphisms induced by the embedding of MATH in the dual crossed product MATH, the NAME duality isomorphism MATH, and the map MATH of MATH into MATH determined by a rank-one projection MATH. The formulas at the start of the proof of CITE show that, because MATH is fixed under MATH, the duality isomorphism carries MATH into MATH, where MATH is the projection onto the subspace spanned by the basis element MATH. Thus MATH has the required property, and the result for simple extensions now follows from CITE. If MATH is an arbitrary REF-sink extension, we consider its simplification MATH and the embedding MATH of MATH in MATH provided by REF , which by CITE induces an isomorphism MATH in MATH-theory. But now it is easy to check that MATH, and the result follows. |
math/0103039 | The first equation is a translation of the condition MATH. For the second, let MATH be the homomorphism such that MATH, which induces the usual isomorphism of MATH onto MATH. If MATH is given by MATH, then we have MATH, and similarly for MATH. Thus MATH . Now fix MATH. Since MATH is surjective, there exists MATH such that MATH. From REF we have MATH and hence there exists MATH such that MATH. Now because MATH is constant on the image of MATH, we have MATH and MATH will do. |
math/0103039 | By REF , for each MATH there exists MATH such that MATH. We define MATH. A calculation shows that for any MATH we have MATH . Now let MATH. On one hand, we have from REF that MATH . On the other hand, since MATH, its composition with MATH is also MATH. Thus the class REF must vanish in MATH, and there exists MATH such that MATH . Comparing REF shows that MATH and MATH . Since we are supposing MATH, we deduce that MATH. We have now proved that MATH which implies MATH, as required. |
math/0103039 | The forward direction follows from the previous corollary. For the converse, we seek to apply REF . To see that MATH has support in the common closure MATH, recall that MATH decomposes as MATH with respect to MATH. Thus MATH is an eigenvalue for the MATH corner MATH of MATH if and only if it is an eigenvalue for MATH, and hence if and only if it is an eigenvalue for MATH. So REF applies, MATH lies in MATH, and the result follows from REF . |
math/0103046 | Observe that MATH where MATH and MATH divides MATH for MATH. |
math/0103046 | We have MATH . But MATH is divisible by MATH for MATH, so MATH . Thus, MATH. From REF , MATH . But MATH, so the minimum of the right-hand side is less than MATH, so the minimum of the left-hand side is also less than MATH. |
math/0103046 | First, we will translate the periodic orbit so that it passes through REF; this simplifies the algebra in our proof. Let MATH. Then the function MATH also has a periodic orbit of length MATH, namely MATH. But MATH, so MATH, and similarly for higher derivatives; likewise, any iterate MATH, so MATH. So computations assuming that MATH will also hold for arbitrary MATH. Now, for MATH we have MATH and MATH, where MATH denotes a polynomial in MATH in which every term has degree at least MATH. Since MATH is the order of MATH, each of MATH is coprime to MATH. Thus, for any MATH, we can project to MATH and apply the following lemma, which implies that each of MATH is divisible by MATH, completing the proof of the Proposition. |
math/0103046 | Write MATH. Then the compositions MATH and MATH . Since MATH, the coefficients of MATH are equal, so MATH, so MATH. Since MATH is a root of unity, it is not a zero-divisor. Therefore if MATH, then MATH would be a zero-divisor, contradicting the hypothesis. |
math/0103048 | For a root system MATH of type MATH, the vertices of alcoves belong to the coweight lattice MATH. Moreover, if MATH, the orbit MATH is exactly the set of elements MATH such that MATH, as one sees using the relation MATH, for any MATH. Since MATH admits MATH as MATH-basis, any element MATH can be written uniquely in the form MATH with MATH. Thus, it is sufficient to prove the inclusion MATH for any simple root MATH. Let MATH be the integer MATH. Obviously, MATH and we are done. |
math/0103048 | Suppose the gallery MATH is not minimal. This means there exist two integers MATH such that MATH. Let MATH denote this hyperplane and suppose MATH for any integer MATH in between MATH. By hypothesis the alcove MATH lies in MATH. By induction, we know that MATH also lies in MATH for any MATH since the only hyperplane separating MATH and MATH is MATH. Thus MATH, which contradicts the hypothesis MATH. In an arbitrary minimal gallery MATH, for every MATH, the alcoves MATH and MATH lie in the same connected component of MATH, and the alcoves MATH and MATH lie in the other connected component. If the gallery MATH is supposed to be in the MATH-direction, MATH and MATH and consequently we have MATH and MATH. This means for any hyperplane MATH separating MATH and MATH, MATH and MATH. The same argument now proves that any other minimal gallery going from MATH to MATH is also in the MATH-direction. |
math/0103048 | If MATH then for every root MATH, the inequality MATH holds. Hence, for any hyperplane MATH separating MATH and MATH, MATH lies in the MATH-negative half-space MATH and MATH lies in the MATH-positive half-space MATH. As shown in the proof of the above lemma, any minimal gallery going from MATH to MATH is in the MATH-direction. |
math/0103048 | The union MATH obviously covers the whole set MATH for any MATH. |
math/0103048 | If MATH is dominant and lies in MATH, MATH belongs to MATH for all MATH. If now MATH is dominant but not necessarily in MATH, let us choose a point MATH in MATH fixed by MATH, for instance we can take the barycenter of MATH. Then MATH belongs to MATH a fortiori to MATH. |
math/0103048 | This is only a matter of rephrasing the proof of REF . In fact, we have seen there that MATH if and only if for any wall MATH separating MATH and MATH, we have MATH and MATH. In other words, MATH . |
math/0103048 | According to CITE (see also CITE) every NAME group satisfies the following property : let MATH and MATH be a simple reflection, then MATH whenever MATH and MATH. It follows easily by induction on MATH that MATH is equivalent to MATH whenever MATH for MATH. Therefore what we really have to prove is that the lengths add MATH . Of course, we only need to prove the first of these equalities. Let us choose a minimal gallery MATH going from MATH to MATH, with MATH, and MATH a minimal gallery going from MATH to MATH, with MATH. According to REF , any minimal galleries going from MATH to MATH or going from MATH to MATH are automatically in the MATH-direction. Obviously, the concatenation of two galleries in the MATH-direction is still in the MATH-direction. By REF again, the concatenated gallery MATH is minimal. Let us prove that the concatenated gallery MATH is minimal only if the equality MATH holds. Let MATH be the set of reflections by walls of MATH; obviously MATH is also a NAME system. Now, by viewing MATH as base alcove, the gallery MATH gives rise to a reduced expression MATH with MATH in MATH. Thus the length of MATH in the NAME system MATH is equal to MATH. Consequently, the length of MATH in the NAME system MATH is equal to MATH. |
math/0103048 | Let us denote MATH and MATH. We can easily reduce to the case MATH, where MATH is some reflection MATH and MATH. In other words MATH and MATH are symmetric with respect to the hyperplane MATH ; MATH lies in the MATH-negative half-space MATH ; MATH lies in the MATH-positive half-space MATH. We can also suppose MATH, thus MATH and MATH lie in the interiors of the corresponding half-spaces determined by MATH. Let MATH be an alcove in the MATH-negative half-space MATH. The whole acute cone MATH belongs then to MATH. Let MATH be an alcove lying in MATH, a fortiori in the MATH-negative half-space MATH. Just as in the proof of the last lemma, it will be convenient to consider MATH as base alcove. Let MATH denote the set of reflections by walls of MATH. Let MATH and let MATH denote the subset of reflections by the walls of MATH containing the vertex MATH. Since the conjugation by MATH makes the NAME MATH isomorphic to the NAME system MATH, the inequality we want to prove in the former system MATH is equivalent to the inequality we are going to prove MATH in the latter system. In the remainder of the proof, all inequalities are understood to be relative to the NAME system MATH. Obviously, MATH and MATH. The alcove MATH, respectively, MATH, is minimal among the alcoves sharing the vertex MATH, respectively, MATH. Like MATH, the alcove MATH lies in the MATH-negative half-space MATH. Like MATH, the alcove MATH lies in the MATH-positive half-space MATH. Let MATH be a minimal gallery going from MATH to MATH. Since MATH and MATH, there exists an integer MATH such that MATH and MATH share the wall MATH. The minimal gallery MATH corresponds to a reduced expression MATH with MATH. By removing the reflection MATH from the reduced expression, we get MATH . Therefore MATH. But the alcove MATH contains the vertex MATH. Thus by minimality, we know MATH, therefore MATH. |
math/0103048 | The idea of the proof is the remark, due to NAME and NAME, that nearby the extreme elements MATH, the picture looks like the NAME order in the finite NAME group MATH. (This is proved in a precise form in REF below.) Let MATH be a regular dominant coweight. For simplicity, suppose MATH. Let MATH be chosen as in REF . Let us consider the element MATH . We prove MATH is MATH-permissible but not MATH-admissible, if MATH is sufficiently regular. Since MATH, we know MATH for any MATH. Therefore to prove MATH is not MATH-admissible, it is enough to show MATH. Since MATH is regular dominant, by REF the maximal length element in the coset MATH is MATH. Moreover we have MATH so MATH is equal to MATH. Let us prove that MATH is MATH-permissible. Let MATH be a vertex of MATH. By construction of the elements MATH and MATH, the difference MATH lies in MATH, since MATH is dominant. For MATH sufficiently regular dominant, we have MATH . (This relation can be used as the definition of ``sufficiently regular".) According to the following well-known statement, MATH is MATH-permissible. |
math/0103048 | By REF CITE, MATH for MATH. The lemma is a consequence of these equalities, as explained already in the proof of REF . |
math/0103048 | We need to show that if MATH is a sum of positive coroots for all dominant coweights MATH, then MATH, the reverse implication being a general fact for all root systems which is easily proved. Recall that for root systems of type MATH every vertex MATH of the base alcove is either a dominant fundamental coweight or zero. Let MATH denote the longest element of MATH. The assumption on MATH and MATH implies that, for every vertex MATH, MATH and thus MATH by REF . By REF below, MATH and MATH both belong to MATH. Therefore by REF , MATH, as desired. |
math/0103048 | Since the concatenation of two galleries in the MATH-direction is still in the MATH-direction, it suffices to prove that MATH and MATH. The first statement results from REF , and since the translation of a gallery in the MATH-direction is still in the MATH-direction, the second is equivalent to MATH. Choose a reduced expression MATH, where for each MATH, MATH is the reflection corresponding to a simple root MATH. We claim that the gallery MATH is in the MATH-direction. Fix MATH and let MATH denote the hyperplane separating MATH and MATH. We need to show that MATH and MATH . But since MATH, we have MATH and MATH yielding the desired inclusions. |
math/0103048 | In light of CITE MATH, to prove MATH is a based root system it remains only to show that, REF every element of MATH is an integral linear combination of elements of MATH, and REF MATH. For REF , consider the following property of a MATH-orbit MATH: MATH . If MATH holds for every MATH, then each MATH is a highest root and thus MATH. Therefore REF follows from the analogous property of the root system MATH. By examining automorphisms of simple NAME diagrams, we see that MATH holds for every MATH unless MATH is of type MATH (which we assume realized in MATH), MATH, and MATH; therefore, MATH. In this case one can verify by direct calculation that REF holds (and in fact MATH is of type MATH). We note that REF is equivalent to the statement that MATH for every MATH. Next, one can show, following CITE MATH, that MATH is the group of affine transformations of MATH generated by the reflections through the hyperplanes MATH, for MATH and MATH. It follows that MATH is the affine NAME group for the root system MATH in MATH, and therefore the fixed-point group MATH is precisely the corresponding coroot lattice. But then MATH, and REF is proved. Finally, if MATH is the highest root of MATH, then MATH is also highest and thus MATH is the unique highest root of MATH. Therefore MATH is irreducible. |
math/0103048 | First note that if MATH, then MATH is either MATH or MATH (the latter occurring only in the case MATH is of type MATH and MATH; compare the proof of REF ). Thus it is enough to prove the following statement. MATH . Given MATH, there exist MATH and MATH such that MATH. For any MATH, we have MATH, where MATH for every MATH-orbit MATH. We will prove MATH by induction on the quantity MATH. Fix a positive root MATH; we may assume that MATH is not supported in any MATH. Since the pairing MATH is positive definite, we have MATH and thus there is a MATH-orbit MATH such that MATH. By CITE MATH, MATH permutes the positive roots not supported on MATH, and thus MATH is positive. On the other hand MATH and thus MATH. By induction MATH, whence the first statement follows. The converse statement, as well as the one concerning positivity, is obvious. |
math/0103048 | If MATH are in the same alcove MATH, they belong to the same connected component of MATH; therefore MATH for some alcove MATH in MATH. If MATH belong to two different alcoves in MATH, there exists MATH such that MATH and MATH. Let MATH be a corresponding affine root in MATH (that is, MATH and MATH). We have MATH and MATH, thus MATH and MATH belong to different alcoves in MATH. An element MATH satisfies the inequalities MATH for all positive roots MATH of MATH, and the inequality MATH for the highest root MATH of MATH. Since MATH these inequalities are equivalent with MATH for all positive roots MATH of MATH, and MATH. Therefore MATH. |
math/0103048 | This follows easily from the definitions, noting that MATH if and only if MATH under the assumption MATH. |
math/0103048 | This follows easily from REF and the description of acute cones of alcoves as intersections of half-spaces as in REF . |
math/0103048 | According to REF, the statement is already valid if MATH is replaced by MATH and MATH is replaced by MATH. Since the NAME order on MATH (respectively, MATH) is extended in the obvious way to MATH (respectively, MATH) as explained in REF, the proposition follows. |
math/0103048 | Under the assumption that MATH is of type MATH, we have MATH . If MATH, we have MATH according to REF , thus MATH. Suppose MATH. Let MATH and MATH; the alcoves MATH in MATH and MATH in MATH satisfy MATH. By REF , there is an element MATH such that MATH. By REF , MATH. Since MATH, we have MATH for every vertex MATH. We can now apply REF . to MATH and MATH and we obtain MATH relative to the NAME order on MATH. By REF . the same inequality holds for the NAME order on MATH. |
math/0103054 | CASE: Since MATH is not in a cycle, all the nodes in the (infinite) ``pruned" MATH tree MATH with root node MATH have distinct values. We can find an infinite path MATH from the root node MATH in MATH which contains infinitely many nodes satisfying REF . View MATH as a REF-adic integer, encoded as MATH in which MATH and MATH for all sufficiently large MATH. Then MATH has a prefix in the certificate MATH, of length MATH, say. By REF of the certificate, there exists a path in the associated tree MATH from leaf node MATH to MATH with MATH and MATH where MATH is the depth of MATH. Now MATH is not in a cycle and MATH, so we can repeat the same construction; finding a prefix of MATH of length MATH, say, with tree MATH of depth MATH having a path from leaf node MATH to MATH with MATH and MATH . Thus MATH and MATH . Continuing inductively, we find an infinite chain MATH for which REF hold. CASE: We use a similar argument. Suppose first that MATH is not in a cycle. Then at the first round we find two different paths in MATH with MATH, MATH satisfying REF . At the MATH-th stage we produce MATH distinct elements MATH satisfying REF . Furthermore, if the largest depth tree MATH for MATH is MATH then, all the elements MATH are at depth at most MATH from the root node MATH, hence each such element satisifes MATH. Let MATH count the number of elements MATH which satisfy REF . It follows that for MATH we have MATH with MATH. Now suppose that MATH is in a cycle, that is, is periodic. Then in the first tree MATH at least one of the two preimages MATH and MATH of MATH that satisfies REF cannot be in the periodic cycle containing MATH. Let it be MATH, and we may then apply the argument above to the tree with root MATH for which MATH . But any such element MATH by adjoining the path from MATH to MATH, and REF gives MATH . Thus MATH since MATH. |
math/0103054 | The certificate was found by computer search, of size indicated in REF . The computer search required examining trees to depth MATH, with some paths having MATH ones, although the path with the largest ones ratio occured in a tree of depth MATH, having MATH ones. It is an interesting feature of this certificate that the ``worst" tree is not one of maximal depth in the search. |
math/0103054 | The proof is similar to that for REF . The bound for the exponent MATH comes from REF . |
math/0103055 | Since MATH is essential, MATH is a countable set of mutually orthogonal nonzero projections and we may use REF to lift them to a collection MATH of mutually orthogonal nonzero projections in MATH. Now each MATH is infinite-dimensional, and for each MATH we define MATH. Then each MATH is infinite-dimensional and for each MATH we can let MATH be a partial isometry with initial space MATH and final space MATH. Also for each MATH we shall let MATH be the projection onto MATH. Then MATH is a NAME MATH-family. By the universal property of MATH there exists a homomorphism MATH such that MATH and MATH. Let MATH. Then MATH is a degenerate extension and MATH. Furthermore, for all MATH we have that MATH so MATH is essential. |
math/0103055 | Since MATH and MATH are NAME, there exists a partial isometry MATH such that MATH and MATH. Now notice that MATH commutes with MATH. Thus for any MATH we have MATH and MATH . Hence MATH. Now let MATH be the vector given by MATH. Then for any MATH we have MATH so MATH and MATH. |
math/0103055 | Since MATH and MATH are equal in MATH it follows that they are NAME. By interchanging MATH and MATH if necessary, we may use REF to choose an isometry MATH in MATH for which MATH and MATH. For each MATH define MATH and MATH. By REF there exists a degenerate essential extension MATH with the property that MATH for all MATH. Then MATH will be a representation of MATH (MATH is multiplicative since MATH is an isometry), and thus MATH will be a degenerate extension with the property that MATH. Now since MATH is essential we have that MATH . Therefore MATH for all MATH and it follows from REF that MATH, and thus MATH is essential. Now recall that MATH and MATH. Since MATH is an isometry, we see that MATH is a partial isometry with source projection MATH and range projection MATH. Therefore by REF it follows that MATH and MATH equals MATH in MATH. |
math/0103055 | Let MATH and MATH be elements of MATH and choose the representatives MATH and MATH such that MATH. Let MATH and MATH be degenerate essential extensions such that MATH and MATH. If we let MATH, then it is straightforward to see that MATH. Also since MATH is weakly stably equivalent to MATH, REF implies that we have MATH in MATH. Putting this all together gives MATH in MATH. Thus MATH is additive. |
math/0103055 | Let MATH be an essential extension of MATH and suppose that MATH equals MATH in MATH. Use REF to choose a degenerate essential extension MATH of MATH such that MATH for all MATH. Also let MATH. By hypothesis, there exists MATH such that MATH. Since MATH is essential, for all MATH we must have that MATH. Since MATH is a projection, this implies that MATH. Therefore for each MATH we may choose isometries or coisometries MATH in MATH such that MATH. Extend each MATH to all of MATH by defining it to be zero on MATH. Let MATH. It follows that this sum converges in the strong operator topology. Notice that for all MATH we have MATH . Since MATH commutes with MATH for all MATH, we see that MATH is a unitary in MATH. Hence we may consider MATH . Using the above identity we see that for each MATH we have MATH . Now since MATH is a partial isometry with source projection MATH and range projection MATH, we may use REF to conclude that MATH . This combined with REF implies that MATH . Combining REF with REF gives MATH . Thus by REF there exists an operator MATH such that the restriction of MATH to MATH is a unitary operator and MATH. Let MATH. Then MATH is a partial isometry that satisfies MATH and MATH. One can then check that MATH is a NAME MATH-family in MATH. Thus by the universal property of MATH there exists a homomorphism MATH such that MATH and MATH. Let MATH. Then MATH is a degenerate extension of MATH. Furthermore, since MATH we see that MATH for all MATH. Since MATH satisfies REF , it follows from REF that MATH and MATH is a degenerate essential extension. In addition, we see that for each MATH . Thus MATH for all MATH, and since the MATH's generate MATH, it follows that MATH for all MATH and hence MATH. In addition, since the MATH's are either isometries or coisometries on MATH with finite NAME index, it follows that MATH. Therefore, for any MATH we have that MATH . Again, since the MATH's generate MATH, it follows that MATH for all MATH. Similarly, MATH for all MATH. Thus MATH for all MATH and MATH. Now because the MATH's are all isometries or coisometries on orthogonal spaces, it follows that MATH, and hence MATH, is a partial isometry. Therefore, MATH in MATH and since MATH is a degenerate essential extension it follows that MATH in MATH. This implies that MATH is injective. |
math/0103055 | Suppose that MATH. Then MATH for some MATH. Then MATH and MATH does in fact map MATH into MATH. Thus MATH induces a homomorphism MATH of MATH into MATH. In the same way, MATH induces a homomorphism MATH from MATH into MATH, which we claim is an inverse for MATH. We see that MATH and similarly MATH is the identity on MATH. |
math/0103055 | Since MATH, and MATH is an isomorphism by REF , the result follows from REF . |
math/0103055 | Let MATH be the set of all paths in MATH whose range is MATH. Since MATH is an essential REF-sink extension of MATH, it follows that MATH. Thus for every MATH there exists a path from MATH to MATH. If MATH is infinite, this implies that MATH is also infinite. If MATH is finite, then because MATH it follows that MATH is a finite graph with no sinks, and hence contains a loop. If MATH is any vertex on this loop, then there is a path from MATH to MATH and hence MATH is infinite. Now because MATH is infinite it follows from CITE that MATH. |
math/0103055 | For MATH let MATH. Since MATH is a REF-sink extension of MATH we know that MATH is finite. We shall prove the claim by induction on MATH. If MATH, then MATH and MATH. Assume that the claim holds for all MATH with MATH. Then let MATH with MATH. Since MATH is a REF-sink extension of MATH there are no loops based at MATH. Thus MATH for all MATH with MATH. By the induction hypothesis MATH for all MATH with MATH. Since the projections MATH are mutually orthogonal we have MATH . |
math/0103055 | Let MATH be the canonical NAME MATH-family in MATH, and let MATH be the canonical NAME MATH-family in MATH. Choose an isomorphism MATH, and let MATH and MATH be the homomorphisms that make the diagram MATH commute. Then MATH is the Busby invariant of the extension associated to MATH, and since MATH is an essential REF-sink extension, it follows that MATH and MATH are injective. For all MATH and MATH define MATH . Note that MATH implies that MATH. Also since MATH is a rank REF projection, and since the above diagram commutes, it follows that MATH is a rank REF projection. Thus MATH is REF-dimensional. Furthermore, by REF we see that MATH and MATH for all MATH and MATH. In addition, since MATH for any MATH and because the MATH's are mutually orthogonal projections, it follows that the MATH's are mutually orthogonal subspaces for all MATH. For all MATH define MATH . Then for every MATH, we have MATH since MATH is injective. Therefore, the rank of MATH is infinite and hence MATH and MATH. Now for each MATH and MATH let MATH be the projection onto MATH and MATH be a partial isometry with initial space MATH and final space MATH. One can then check that MATH is a NAME MATH-family in MATH. Therefore, by the universal property of MATH there exists a homomorphism MATH with the property that MATH and MATH. Define MATH. Then for all MATH we have that MATH . Thus MATH for all MATH. By REF it follows that MATH and MATH is an essential extension of MATH. Now since MATH is a projection onto a subspace of MATH with finite codimension, it follows that MATH. Thus MATH has the property that for all MATH . By the definition of the NAME map MATH it follows that the image of the extension associated to MATH will be the class of the vector MATH in MATH, where MATH. Now MATH is equal to the NAME index of MATH in MATH. Since MATH is a partial isometry with initial space MATH and final space MATH, and since MATH is a partial isometry with initial space MATH it follows that MATH in MATH. Furthermore, MATH is a partial isometry with initial space MATH and final space MATH and MATH is a partial isometry with initial space MATH. Therefore, since MATH for all MATH we have that MATH . Thus MATH for all MATH. |
math/0103055 | From REF we have that MATH in MATH, where MATH is the vector given by MATH for MATH. By the definition of MATH we have that MATH in MATH. Thus MATH equals the class of the vector MATH given by MATH . Hence for all MATH we have MATH-so MATH. Thus MATH and MATH in MATH. |
math/0103055 | Let MATH be those vertices of MATH that feed into a loop; that is, MATH . Now consider the set MATH. Because MATH has no sinks, and because MATH and MATH implies that MATH, it follows that MATH cannot have a finite number of elements. Thus MATH is either empty or countably infinite. If MATH then list the elements of MATH as MATH. Now let MATH. Choose an edge MATH with the property that MATH and define MATH. Continue in this fashion: given MATH choose an edge MATH with MATH and define MATH. Then MATH are the vertices of an infinite path which are all elements of MATH. Since these vertices do not feed into a loop it follows that they are distinct; that is, MATH when MATH. Now if every element MATH has the property that MATH for some MATH, then we shall stop. If not, choose the smallest MATH for which MATH for all MATH. Then define MATH and choose an edge MATH with MATH. Define MATH. Continue in this fashion: given MATH choose an edge MATH with MATH and define MATH. Then we produce a set of distinct vertices MATH that lie on the infinite path MATH. Moreover, since MATH for all MATH we must have that the MATH's are also distinct from the MATH's. Continue in this manner. Having produced an infinite path MATH with distinct vertices MATH we stop if every element MATH has the property that MATH for some MATH. Otherwise, we choose the smallest MATH such that MATH for all MATH. We define MATH. Given MATH we choose an edge MATH with MATH. We then define MATH. Thus we produce an infinite path MATH with distinct vertices MATH. Moreover, since MATH for all MATH, it follows that the MATH's are distinct from the MATH's for MATH. By continuing this process we are able to produce the following. For some MATH there is a set of distinct vertices MATH given by MATH with the property that MATH, and for any MATH there exists an edge MATH for which MATH and MATH. Now define MATH-and let MATH. We shall now show that MATH has the appropriate properties. We shall first show that MATH. Let MATH and consider four cases. (Throughout the following remember that the entries of MATH are nonnegative integers.) REF MATH. Then MATH REF MATH. Since MATH and MATH feeds into a loop, there must exist an edge MATH with MATH and MATH. Thus MATH . CASE: MATH. Then there exists an edge MATH with MATH and MATH. Thus MATH . CASE: MATH. Then MATH . Therefore MATH. We shall now show that for all MATH there exists MATH such that MATH and MATH. If MATH, then MATH and MATH for some MATH. But then there is an edge MATH with MATH and MATH. Thus we have that MATH . On the other hand, if MATH, then MATH feeds into a loop. Since MATH satisfies REF this loop must have an exit. Therefore, there exists MATH such that MATH and MATH is the source of two distinct edges MATH, where one of the edges, say MATH, is the edge of a loop and hence has the property that MATH. Now consider the following three cases. CASE: MATH. Then MATH and hence MATH for some MATH. But then MATH and MATH as above. CASE: MATH and MATH. Then MATH . CASE: MATH and MATH. Then MATH . |
math/0103055 | By REF we see that there exists MATH with the property that MATH and for all MATH there exists MATH for which MATH and MATH. Since MATH we may let MATH be a REF-sink extension of MATH with MATH vector MATH. Let MATH be the sink of MATH. We shall show that MATH is essential. Let MATH. Then there exists MATH for which MATH and MATH. But then MATH and MATH is a boundary vertex of MATH. Hence MATH and we have shown that MATH. Thus MATH is essential, and furthermore MATH in MATH. |
math/0103055 | If MATH is any vector in MATH, then by REF there exists an essential REF-sink extensions MATH for which MATH. If MATH is the Busby invariant of the extension associated to MATH, then by REF we have that MATH. Thus MATH for all MATH. Now because MATH is separable and nuclear, it follows from CITE that MATH is a group. Because MATH is the positive cone of MATH, and hence generates MATH, the fact that MATH for all MATH implies that MATH. |
math/0103055 | This follows from the fact that MATH, and MATH is an isomorphism. |
math/0103055 | If MATH then we can let MATH and the claim holds. Therefore, we shall suppose that MATH. Since MATH has no sinks and satisfies REF , there must exist an edge MATH with MATH and MATH. Then since MATH has no sinks we may find an edge MATH with MATH , and an edge MATH with MATH. Continuing in this fashion we will produce an infinite path MATH with MATH. Since MATH is finite, the vertices MATH of this path must eventually repeat. Let MATH be the smallest natural number for which MATH for some MATH. Note that because MATH we must have MATH. Now MATH will be a loop, and since MATH satisfies REF , there exists an exit for this loop. Thus for some MATH there exists MATH such that MATH and MATH. For each MATH define MATH . Note that MATH may be empty. This will occur if and only if MATH. Now let MATH. To see that MATH, note that MATH, and consider four cases. CASE: MATH and MATH. Since MATH we have that MATH . CASE: MATH and MATH. Then MATH . CASE: MATH, MATH, and MATH. Then MATH . CASE: MATH. Then MATH and MATH . To see that MATH let MATH and consider the following three cases. CASE: MATH and MATH. Then MATH and we have MATH . CASE: MATH and MATH. Then MATH . CASE: MATH. Then either MATH or MATH. In either case there exists an edge MATH with MATH and MATH. Thus MATH and MATH. |
math/0103055 | For each MATH we may use REF to obtain a vector MATH such that MATH and MATH. Now write MATH in the form MATH. Let MATH. Then by linearity, MATH and MATH. Let MATH be a REF-sink extension of MATH with sink MATH and MATH vector equal to MATH. Then MATH in MATH. Furthermore, since MATH for all MATH it follows that MATH and MATH is an essential REF-sink extension. |
math/0103055 | For each MATH let MATH be the graph formed by adding a sink to MATH at MATH, and let MATH be the graph formed by adding a sink to each vertex in MATH. In each case we shall let MATH denote the sink that is added at MATH. As examples we draw MATH and MATH: MATH . We shall now assume that MATH is semiprojective and arrive at a contradiction. Let MATH and for each MATH let MATH. Also let MATH. Set MATH. Then MATH is an increasing sequence of ideals and MATH. Now MATH and for each MATH, MATH by CITE. Thus if we identify MATH and MATH, then by semiprojectivity there exists a homomorphism MATH for some MATH such that MATH. Note that the projection MATH is just the projection MATH. Now if we let MATH be the canonical NAME MATH-family in MATH and let MATH be the canonical NAME MATH-family in MATH, then by the universal property of MATH there exists a homomorphism MATH such that MATH . Since MATH is a REF-sink extension of MATH, we have the usual projection MATH. One can then check that the diagram MATH commutes simply by checking that MATH and MATH agree on generators. This, combined with the fact that MATH on MATH, implies that MATH. Hence the short exact sequence MATH is split exact. Therefore this extension is degenerate. Since MATH by CITE we have that this extension is in the zero class in MATH. However, the MATH vector of MATH is MATH. Since MATH we see that every vector in the image of MATH has entries that are multiples of MATH. Thus MATH, and MATH is not zero in MATH. But then REF imply that the extension associated to MATH is not equal to zero in MATH. This provides the contradiction, and hence MATH cannot be semiprojective. |
math/0103056 | Straightforward. |
math/0103056 | Since the extension associated to MATH is zero, one of two things must occur. If MATH, then MATH is the Busby invariant of MATH. If MATH, then since MATH is unital it follows that MATH and the Busby invariant of MATH must be the zero map. In either case, the Busby invariant of the extension MATH is zero. From CITE it follows that MATH via the map MATH, where MATH denotes the (unique) map for which MATH is the identity. The fact that MATH then follows by checking each on generators of MATH. |
math/0103056 | Let MATH and MATH be the extensions associated to MATH and MATH, respectively. It follows from REF that MATH and MATH are in the same equivalence class in MATH if and only if one of the MATH's may be MATH-embedded into the other. Since MATH and MATH are essential REF-sink extensions of MATH, the graph MATH contains no sinks. By CITE the MATH map MATH is an isomorphism, and by CITE the value of the MATH map on MATH is the class MATH in MATH. |
math/0103056 | Let MATH be a MATH-embedding. Also let MATH be the projection which determines the full corner MATH. Now for MATH we have that MATH is homeomorphic to MATH via the map MATH, where MATH by CITE. Furthermore, since MATH embeds MATH onto a full corner of MATH it follows that MATH is NAME equivalent to MATH and the NAME correspondence is a homeomorphism between MATH and MATH, which in this case is given by MATH CITE. Composing the homeomorphisms which we have described, we obtain a homeomorphism from MATH, where MATH is the unique element of MATH for which MATH. We shall now show that this homeomorphism MATH is equal to the map MATH described in CITE; that is MATH restricts to the identity on MATH. Let MATH. We begin by showing that MATH. Let MATH be the canonical NAME MATH-family, and let MATH be the canonical NAME MATH-family. Since MATH it follows that MATH. Therefore MATH and MATH. Consequently, MATH, and since MATH and MATH it follows that MATH. Thus MATH and MATH. It follows that MATH, and hence MATH. We shall now proceed to show that MATH. Since MATH it follows that MATH. Thus MATH. Now let MATH. If we let MATH be the canonical NAME MATH-family, then since MATH we have that MATH. It then follows that MATH . Thus MATH. We shall now show that MATH. To do this we suppose that MATH and arrive at a contradiction. If MATH, then we would have that MATH. Thus MATH and MATH. Now MATH and MATH. Hence MATH and we have that MATH. Since MATH is injective this implies that MATH and MATH and MATH which is a contradiction. Therefore we must have that MATH and MATH and MATH. Hence MATH. To show inclusion in the other direction let MATH. Then MATH and MATH. As above, it is the case that MATH. Therefore, MATH and since MATH it follows that MATH or MATH. Thus MATH and MATH and MATH. Hence MATH. Thus MATH for any MATH, and the map MATH restricts to the identity on MATH. Since this map is a bijection it must therefore take MATH to MATH. Therefore MATH is precisely the map described in CITE, and it follows from CITE that MATH. |
math/0103056 | Let MATH and MATH with MATH. If MATH, then MATH and hence MATH for some MATH with the property that MATH. Since maximal tails are backwards hereditary this implies that MATH. Hence MATH and MATH which is a contradiction. Thus we must have MATH and MATH is hereditary. Suppose that MATH. Then MATH and MATH for some MATH with the property that MATH. Since maximal tails contain no sinks there exists an edge MATH with MATH and MATH. Thus MATH and MATH. Hence MATH is saturated. |
math/0103056 | Since MATH is an ideal which contains MATH and MATH, it follows that MATH. Conversely, if MATH then MATH and MATH. Thus MATH for some MATH. Hence MATH and MATH. Thus MATH. In addition, the commutativity of the above diagram implies that MATH. Since MATH is surjective it follows that MATH and from the previous paragraph MATH. |
math/0103056 | Since MATH satisfies REF and MATH is a REF-sink extension of MATH, it follows that MATH also satisfies REF . Thus MATH for some saturated hereditary subset MATH. Let MATH be the canonical NAME MATH-family in MATH. Now because MATH is a rank REF projection, it follows that MATH and thus MATH. Since MATH is hereditary this implies that for any MATH we must have MATH. Hence MATH. Now let MATH; that is, MATH is the graph given by MATH and MATH. Then by CITE we see that MATH. Thus we may factor MATH as MATH to get the commutative diagram MATH where MATH is the standard projection and MATH is the monomorphism induced by MATH. From the commutativity of this diagram it follows that MATH is injective. Let MATH be the canonical NAME MATH-family in MATH. Also let MATH be the ideal in MATH generated by MATH, and let MATH be the ideal in MATH generated by MATH. Using CITE and the fact that any path in MATH with range MATH is also a path in MATH, we have that MATH . From the commutativity of the above diagram it follows that MATH is the (unique) map which makes the diagram MATH commute. Since MATH is injective, it follows from CITE that MATH is an essential ideal in MATH. Now suppose that there exists MATH with MATH in MATH. Then for every MATH with MATH we must have that MATH. Hence MATH. Since MATH it follows that MATH. Since MATH this would imply that MATH is not an essential ideal. Hence we must have that MATH for all MATH. Furthermore, if MATH is a path with MATH and MATH, then MATH. So if MATH in MATH, then we must have that MATH in MATH. Consequently, if MATH, then MATH in MATH, and we cannot have MATH because there is a path in MATH from every element of MATH to MATH, and hence a path in MATH from every element of MATH to MATH. Thus MATH, and MATH. Hence MATH. |
math/0103056 | We shall first show that MATH is a saturated hereditary subset of MATH. To see that it is hereditary, let MATH. If MATH with MATH, then one of two things must occur. If MATH, then MATH must be in MATH and hence MATH. Since we know from REF that MATH is a saturated hereditary subset of MATH, it follows that MATH. On the other hand, if MATH, then MATH, and hence MATH. Thus MATH is hereditary. To see that MATH is saturated, let MATH. Then MATH and MATH for some MATH with the property that MATH. Since maximal tails contain no sinks, there exists MATH with MATH and MATH. But then MATH and MATH. Since MATH this implies that MATH. Thus MATH is saturated. Now since MATH we see that MATH is a saturated hereditary subset which contains MATH. Thus MATH. Conversely, suppose that MATH. If MATH is any saturated hereditary subset of MATH which contains MATH, then for every vertex MATH we know that MATH cannot be a sink, because if it were we would have MATH. Thus we may find an edge MATH with MATH and MATH. Furthermore, since MATH we must also have that MATH. Thus if MATH, we may produce an infinite path MATH in MATH with MATH and MATH for all MATH. If we let MATH, then MATH and MATH. Hence MATH and MATH which is a contradiction. Thus we must have MATH for all saturated hereditary subsets MATH containing MATH. Hence MATH and MATH. |
math/0103056 | Consider the following two cases. CASE: The set MATH contains finitely many elements. Then from the definition of the extension associated to MATH, we have that MATH. However, if MATH has only finitely many elements then MATH for all MATH. Hence MATH and MATH. CASE: The set MATH contains infinitely many elements. Then MATH, and from REF we have that MATH. Also REF implies that MATH. Since MATH, we see that from REF that MATH. Now if we let MATH be the canonical NAME MATH-family in MATH and MATH be the canonical NAME MATH-family in MATH, then MATH . |
math/0103056 | Suppose that there were infinitely many such paths. Then since MATH is row-finite there must exist an edge MATH with MATH and with the property that there exist infinitely many MATH for which MATH and MATH. Likewise, there exists an edge MATH with MATH and with the property that there are infinitely many MATH for which MATH and MATH. Continuing in this fashion we produce an infinite path MATH with the property that MATH for all MATH. If we let MATH, then MATH and MATH. Since MATH, it follows that MATH and MATH, which is a contradiction. |
math/0103056 | If there was such an edge MATH, then MATH would be a maximal tail and since MATH it would follow that MATH. Since MATH this implies that MATH which contradicts the fact that MATH. |
math/0103056 | If MATH, then MATH and by REF we have MATH. Since MATH is a partial isometry MATH. Therefore we need only consider the case when MATH. Let MATH denote the boundary edges of MATH. Also let MATH. By REF we see that MATH is finite. We shall prove the claim by induction on MATH. CASE: MATH. Let MATH be the boundary edges of MATH which have source MATH. Then it follows from CITE that for MATH where MATH is the set of paths from MATH to MATH. Also if MATH is an edge with MATH, then because MATH . REF implies that MATH. Furthermore, since MATH we must have that MATH. Therefore, just as before we must have MATH. Now since the projections MATH are mutually orthogonal, we see that MATH . Inductive Step: Assume that the claim holds for all edges MATH with MATH. We shall now show that the claim holds for edges MATH with MATH. Let MATH be the exits of MATH with source MATH. As above we have that MATH for all MATH. Now if MATH is any edge with MATH, then REF implies that MATH. Thus we must have that MATH, and by the induction hypothesis MATH. Furthermore, since the projections MATH are mutually orthogonal, we see that MATH . |
math/0103056 | Notice that because of the way MATH was defined, MATH will have no sinks. Also note that the diagram MATH commutes. Let MATH be the canonical NAME MATH-family in MATH. For each MATH let MATH and for each MATH let MATH . Also for each MATH define MATH to be the projection onto MATH and for each MATH define MATH to be the partial isometry with initial space MATH and final space MATH. Then MATH is a NAME MATH-family in MATH. If we let MATH be the canonical NAME MATH-family in MATH, then by the universal property of MATH there exists a homomorphism MATH such that MATH and MATH. Let MATH. Since MATH satisfies REF , it follows that the quotient MATH also satisfies REF . Because MATH for all MATH this implies that MATH and MATH is an essential extension of MATH. Because MATH is a lift of MATH for all MATH we see that MATH equals the NAME index of MATH in MATH. Since MATH is a partial isometry with initial space MATH and final space MATH, and since MATH is a partial isometry with initial space MATH and final space MATH, it follows that MATH . Also, MATH is a partial isometry with initial space MATH and final space MATH, and MATH is a partial isometry with initial space MATH and final space MATH. Because MATH we see that MATH . Thus MATH . Now if MATH is any boundary edge of MATH, then by CITE we have that MATH where MATH is the set of paths in MATH from MATH to MATH. Also if MATH is an edge with MATH, then MATH by REF . Therefore, MATH . Thus MATH . |
math/0103056 | By REF. From REF we see that MATH, where MATH for all MATH. Therefore, MATH is equal to the class MATH in MATH where MATH is the vector given by MATH. Hence for all MATH we have that MATH and thus for all MATH we have that MATH . Hence MATH, and MATH in MATH. |
math/0103056 | It follows from REF that if one of the MATH's is MATH-embeddable in the other, then MATH. Thus we may let MATH and form the graph MATH as discussed in REF . If we let MATH and MATH be the Busby invariants of the extensions associated to MATH and MATH, then it follows from REF that MATH. Thus for each MATH, we may factor MATH as MATH where MATH is the standard projection and MATH is the monomorphism induced by MATH. It then follows from REF that one of the MATH's may be MATH-embedded into the other if and only if MATH and MATH are NAME. Since MATH we see that MATH and MATH are NAME if and only if MATH and MATH are NAME. Furthermore, since MATH and MATH are essential extensions we see from REF that MATH and MATH are NAME if and only if MATH and MATH are equal in MATH. If MATH is the MATH map, then this will occur if and only if MATH, and by REF we see that this happens if and only if MATH in MATH. |
math/0103056 | Using the notation established in REF and the proof of REF , we see that if MATH and MATH are totally inessential, then MATH. Hence MATH and MATH. Thus MATH and MATH are trivially NAME. Hence MATH and MATH are NAME and it follows from REF that one of the MATH's can be MATH-embedded into the other. Alternatively, we see that if MATH and MATH are totally inessential, then MATH, and provided that we interpret the condition that MATH in MATH as being vacuously satisfied, the previous theorem implies that one of the MATH's can be MATH-embedded into the other. |
math/0103058 | The proof of existence will be given later. Here we check the statement involving the scalar product, assuming existence. The following holds for MATH and MATH: MATH . Taking the limit when MATH gives MATH . Taking the complex conjugate: MATH which completes the proof (assuming existence.) |
math/0103058 | Everything is either stated explicitely in REF , or follows from their proofs. We will give more details for MATH as this is not treated in CITE. |
math/0103058 | A simple recurrence. |
math/0103058 | The formula defining MATH is equivalent to (after integration by parts and the change of variable MATH): MATH . This proves the existence of MATH, its analytic character in MATH, and the uniform MATH bound on MATH. The formula can be rewritten as: MATH . When MATH is in the critical strip the integral MATH is absolutely convergent and its value is MATH from well-known integral formulae, so that: MATH . The first term is MATH and the second term can be explicitely evaluated using the series expansion of MATH with the final result MATH which shows MATH, on MATH, uniformly for MATH. |
math/0103058 | Clearly a corollary to REF. |
math/0103058 | To establish this we first consider, for MATH: MATH . If MATH an integration by parts shows that it is MATH. On the other hand when MATH its exact value is MATH. With this information the theorem follows directly from REF as (for example) the leading divergent contribution as MATH to MATH is MATH which gives MATH. The rescaling MATH is chosen so that a finite limit for MATH is obtained. As the scalar products involving distinct zeros have a smaller divergency, the rescaling let them converge to MATH. |
math/0103058 | We have MATH, and MATH so MATH. From CITE we know that MATH is MATH so MATH is the function MATH. It is thus MATH as MATH, and from REF we then deduce that the scalar products MATH admit finite limits as MATH. This settles the case MATH. For MATH, one uses the uniformity with respect to MATH in REF to get MATH which gives MATH. |
math/0103058 | Let MATH be a non-empty finite set of zeros. We showed that MATH is the NAME space distance from MATH to MATH, and that the vectors MATH for MATH are perpendicular to MATH. So MATH is bounded below by the norm of the orthogonal projection of MATH to the finite-dimensional vector space MATH spanned by the vectors MATH, MATH, MATH. This is given by a well-known formula involving the inverse of the NAME matrix of the MATH's as well as the scalar products MATH. From REF the NAME matrix converges to diagonal blocks, one for each zero, given by NAME matrices of sizes MATH. From NAME we know that the top-left element of the inverse matrix is MATH. Combining this with the scalar products evaluated in REF we get that the squared norm of the orthogonal projection of MATH to MATH is asymptotically equivalent as MATH to MATH. The proof is complete. |
math/0103060 | Let MATH denote the center of MATH, that is, the MATH such that MATH for all MATH. (The argument applies equally well to the supercenter defined by MATH.) One first checks that symmetric polynomials in MATH are central following the argument in the proof of CITE. Conversely, take MATH where each MATH. Let MATH be maximal with respect to the NAME order such that MATH. Assume for a contradiction that MATH. Then, there exists some MATH with MATH. Consider MATH. By REF , this looks like MATH plus a linear combination of terms of the form MATH for MATH and MATH with MATH in the NAME order. So in view of REF , MATH is not central, a contradiction. Hence, we must have that MATH. Considering the form of the center of MATH one easily shows that MATH in fact lies in MATH. To see that MATH is actually a symmetric polynomial, write MATH where the coefficients MATH lie in MATH. Applying REF to MATH now gives that MATH for each MATH, hence MATH is symmetric in MATH and MATH. Similar argument shows that MATH is symmetric in MATH and MATH for all MATH to complete the proof. |
math/0103060 | Since MATH, REF implies at once that the given MATH span MATH. But they are linearly independent too by REF , proving the first statement. The second follows immediately using REF once more. |
math/0103060 | The isomorphism MATH is length preserving. Equivalently, MATH for each MATH with MATH. Using this, it is straightforward to check that the map MATH defined as above respects the defining relations on generators. |
math/0103060 | We define a bilinear map MATH by MATH. For MATH, MATH which is all that is required to check that the map is MATH-balanced. Hence there is an induced MATH-bimodule map MATH. Finally, to prove that MATH is bijective, note that MATH is a basis of the induced module MATH as a vector space. In view of REF , the image of these elements under MATH is a basis of MATH. |
math/0103060 | This follows from REF and the isomorphism MATH which is easy to check. |
math/0103060 | CASE: Since MATH, MATH is isomorphic as a MATH-bimodule to MATH, the isomorphism being simply the map MATH for MATH, compare REF . Now REF follows because MATH is simply this isomorphism composed with the projection from MATH to MATH along the bimodule decomposition MATH. CASE: We show by downward induction on MATH that MATH whenever we are given MATH and MATH with each MATH and MATH. Since MATH is the longest element of MATH, the induction starts with MATH: in this case, the conclusion is clear as MATH. So now suppose MATH and that the claim has been proved for all higher MATH. Pick a basic transposition MATH such that MATH and MATH. Then, MATH for MATH with MATH. But now the induction hypothesis shows that MATH. CASE: We remind the reader that MATH denotes the map with MATH (the sign being MATH always in this case). Given this and REF , it is straightforward to check that MATH is a homomorphism of MATH-bimodules. To see that it is an isomorphism, it suffices by dimension to show that it is injective. Suppose MATH lies in the kernel. Then, MATH for all MATH. Hence MATH by REF . |
math/0103060 | CASE: According to a special case of REF , the top factor MATH in the bimodule filtration of MATH defined in REF is isomorphic to MATH as a MATH-bimodule. The map MATH is simply the composite of this isomorphism with the quotient map MATH. CASE: Recall that MATH forms a basis for MATH as a free left MATH-module, and MATH is isomorphic to MATH as a left MATH-module. It follows that the maps MATH form a basis for MATH as a free right MATH-module, where MATH is the unique left MATH-module homomorphism with MATH for all MATH. The analogous maps MATH defined by MATH for MATH form a basis for MATH as a free right MATH-module. So in view of REF , we can find a basis MATH for MATH viewed as a right MATH-module such that MATH for each MATH, that is, MATH for every MATH. But MATH, so the elements MATH also form a basis for MATH as a right MATH-module, and MATH since MATH on MATH. Thus MATH maps a basis of MATH to a basis of MATH (as free right MATH-modules), hence MATH is an isomorphism of MATH-bimodules. |
math/0103060 | Let MATH be the bimodule isomorphism constructed in REF . Then, there are natural isomorphisms MATH the second isomorphism depending on the fact that MATH is a free left MATH-module, see for example, CITE. |
math/0103060 | One routinely checks using REF that the functor MATH (from the category of finite dimensional MATH-modules to finite dimensional MATH-modules) is right adjoint to MATH. Hence, it is isomorphic to MATH by uniqueness of adjoint functors. Now combine this natural isomorphism with the previous corollary (with MATH and MATH swapped and MATH replaced by MATH). |
math/0103060 | Define a total order MATH on MATH so that MATH . We have a corresponding reverse lexicographic ordering on MATH: MATH if and only if MATH for some MATH. It is important that MATH has a smallest element with respect to this total order. Define a function MATH by MATH where MATH . We claim that for MATH, MATH where MATH is some unit in MATH. In other words, MATH(lower terms). Since MATH is a bijection and we already know that the MATH form a basis for MATH viewed as a right MATH-module by REF , the claim immediately implies the lemma. To prove the claim, proceed by induction on MATH. If MATH, the statement is quite obvious. Now assume MATH and the statement has been proved for MATH. We need to consider MATH for MATH. If MATH, then the conclusion follows without difficulty from the induction hypothesis, so assume MATH. Let MATH and MATH. Then by induction MATH where MATH is some unit in MATH. Multiplying on the left by MATH and on the right by MATH, using REF , one deduces that MATH where both MATH and MATH are units in MATH. Noting that MATH this is of the desired form. |
math/0103060 | CASE: Proceed by induction on MATH, the case MATH being immediate from REF . For MATH, MATH applying REF , the relations in MATH especially REF , and the induction hypothesis. CASE: Proceed by induction on MATH, the conclusion being immediate in case MATH. If MATH, an application of REF , combined with REF to commute MATH past MATH, gives that MATH . Now apply the induction hypothesis. CASE: By considering the antiautomorphism MATH of MATH, one sees that MATH is generated by the elements MATH for MATH and MATH for MATH, and the latter satisfy the braid relations. We show that each of these generators of MATH leave MATH invariant. First, consider MATH for MATH. Expanding the definition of MATH and commuting MATH past the leading terms, it equals MATH . By the relations, MATH. Now the conclusion in this case follows immediately since MATH. Next, consider MATH for MATH. Expanding and commuting again gives that it equals MATH . Applying the braid relation gives MATH which again equals MATH as required. |
math/0103060 | Let MATH denote the subalgebra of MATH generated by MATH, so MATH. We first claim that MATH . To prove this, it suffices to show that MATH lies in the right hand side for each MATH. Proceed by induction on the NAME order on MATH, the case MATH being trivial. For the induction step, we can find MATH and MATH such that MATH where MATH. Now an argument using REF gives that MATH modulo lower terms of the form MATH with MATH. The claim follows. Now let MATH. Clearly MATH. So it suffices for the lemma to show that MATH. Noting that MATH and using the result in the previous paragraph, we get that MATH . Now, each MATH. Also each MATH is contained in MATH thanks to REF . |
math/0103060 | We proceed by induction on MATH, the case MATH being almost obvious. So now suppose that MATH. Clearly we can assume that MATH. It will be convenient to write MATH for the two-sided ideal of MATH generated by MATH, so MATH by the induction hypothesis. Let MATH . Obviously MATH. So in view of REF , it suffices to show that MATH for each MATH and each MATH. Consider first MATH. Write MATH for MATH. Expanding MATH in terms of the basis of MATH from REF , we see that MATH . This is contained in MATH thanks to REF . Finally, consider MATH with MATH. Write MATH for MATH. By the induction hypothesis, MATH . Now we need to consider the cases MATH and MATH separately. The argument is entirely similar in each case, so suppose in fact that MATH. Then we show by induction on MATH that MATH for each MATH. This is immediate if MATH, so take MATH and consider the induction step. Expanding MATH using REF , the set MATH looks like the desired MATH plus a sum of terms lying in MATH with MATH. It now suffices to show that each such MATH. But by REF , MATH and each such term lies in MATH by induction, since MATH. |
math/0103060 | By REF , the elements MATH form a basis for MATH viewed as a right MATH-module. Hence REF implies that the elements MATH form a basis for a complement to MATH in MATH viewed as a right MATH-module. The theorem follows at once. |
math/0103060 | Rearrange REF . |
math/0103060 | For REF , by REF and dimension considerations, we just need to check that MATH is generated as a right MATH-module by the given elements. For this, it suffices to show that all elements of the form MATH lie in the right MATH-module generated by MATH . This involves considering terms of the form MATH for MATH and MATH, for which REF is useful. For REF , define a map MATH. This is MATH-balanced, so induces a well-defined epimorphism MATH of MATH-bimodules. We know from REF that MATH is a free right MATH-module on basis MATH for MATH. But MATH maps these elements to MATH which, again using REF , form a basis for MATH as a free right MATH-module. This shows that MATH . Now the remaining parts of REF are obvious consequences of REF . |
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