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math/0005274
By REF MATH and MATH are singular vectors of MATH in the case MATH. Assume first that MATH. Let MATH and MATH be the MATH-submodules generated by MATH and MATH, respectively. We form the MATH-submodule MATH and consider MATH. The set MATH is a set of MATH-generators for MATH, since MATH has MATH-weight MATH. We have MATH . Similarly, the following is a set of MATH-generators for MATH. MATH . Therefore MATH is a set of MATH-generators for MATH, which implies that MATH is a MATH-basis for MATH in the case when MATH. In the case when MATH the set MATH generate MATH. But MATH, and hence MATH is also a MATH-basis for MATH in this case as well. In the case when MATH we note that MATH and MATH generates MATH over MATH. From the formulas above one sees that a set of MATH-generators for MATH is given by the set MATH above, but with MATH removed. Hence the quotient module is again generated freely over MATH by MATH. Hence in the case when MATH the quotient module MATH is generated freely over MATH by MATH. Now MATH has MATH-weight MATH, MATH and MATH both have MATH-weight MATH, and MATH has MATH-weight MATH. Therefore MATH has rank MATH over MATH. So it remains to show that MATH is irreducible. We again study MATH as a MATH-module. It is easy to check that MATH, MATH, annihilates MATH and hence MATH is a direct sum of the following four irreducible MATH-modules: MATH, MATH, MATH and MATH, where MATH is the MATH-submodule generated by the vector MATH. Again MATH as MATH-modules. Now we compute MATH . Therefore MATH is irreducible. Now consider the case of MATH. By REF MATH, MATH and MATH are singular vectors inside MATH. Let MATH, MATH and MATH be the MATH-submodules generated by MATH, MATH and MATH, respectively, and put MATH. We note that MATH, MATH and MATH have MATH-weight MATH, hence MATH is generated over MATH by MATH and similarly for MATH and MATH. We first compute a set of MATH-generators for MATH. MATH . A set of MATH-generators for MATH is given as follows. MATH . Finally we have the following set of MATH-generators for MATH. MATH . From this it follows that MATH generate MATH over MATH. But MATH, and thus MATH is generated over MATH by the vectors MATH, MATH, MATH and MATH, which takes us back to the case when MATH, except that here MATH is not irreducible. It contains a unique irreducible submodule isomorphic to MATH generated by MATH. But then the above calculation plus the fact that MATH show that MATH is irreducible of rank MATH.
math/0005274
As a module over MATH we have MATH is a direct sum of MATH copies of MATH, generated by the highest weight vectors MATH, where MATH. Since the MATH-weights of the MATH's are all distinct for distinct MATH's it follows that these modules as MATH-modules are all non-isomorphic. Therefore if MATH is irreducible over MATH, then MATH is irreducible over MATH. From this and REF we thus conclude that in the case when MATH and MATH the MATH-module MATH is irreducible. By symmetry we conclude that if MATH and MATH, then MATH is irreducible over MATH as well. Therefore MATH is possibly reducible only if both MATH and MATH satisfy one of the two linear equations MATH and MATH. But the case MATH and MATH is not possible, since both MATH and MATH are non-negative integers. By the same token MATH and MATH is not possible, either. Hence either we have MATH and MATH or else MATH and MATH. In either case we must have MATH.
math/0005274
Since as a MATH-module MATH is a direct sum of MATH copies of MATH we obtain a description of the vector space spanned by all proper MATH-singular vectors by virtue of REF. But as a MATH-module MATH is also a direct sum of MATH copies of MATH, from which we obtain similarly a description of the vector space spanned by all proper MATH-singular vectors (see REF). The intersection of these two spaces is the space of proper singular vectors. In the case when MATH it follows from REF that the space of proper MATH-singular vectors is spanned by MATH, MATH and MATH, for MATH. On the other hand the space of proper MATH-singular vectors is spanned by MATH, MATH and MATH, for MATH. It is not hard to see that the intersection of these two spaces is the one-dimensional space spanned by MATH, which is MATH. Other cases are analogous and so we omit the details.
math/0005274
By REF MATH is a singular vector in MATH. Consider MATH, the MATH-submodule generated by MATH. Then we have MATH, where MATH is the irreducible MATH-submodule generated by MATH. Let us compute the space MATH, the space of MATH-invariants inside MATH. Since the MATH-weight of MATH is MATH, we know that MATH is a free MATH-module generated over MATH by MATH. We have MATH . It follows that MATH is generated over MATH by the set MATH . In the case when MATH it follows from the description of MATH that MATH is a MATH-basis for the MATH-invariants of the quotient space MATH. (The choice of MATH instead of just MATH will be explained later.) The MATH-weights of MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH are MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, respectively. Hence MATH is a free MATH-module of rank MATH. So we need to show that MATH is irreducible. Now MATH, MATH, together with MATH and MATH generate a copy of MATH, which thus allow us to study the MATH-module structure of MATH. We can easily check that MATH, for MATH, annihilates the vectors MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. (We want to point out that MATH is not annihilated by MATH, for MATH, hence the choice of MATH.) Thus MATH as a MATH-module is a direct sum of the following eight irreducible modules: MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, where MATH is the irreducible MATH-module generated by MATH, for MATH, and MATH is generated by MATH, and finally MATH denotes the irreducible MATH-module of highest weight MATH. Note that as MATH-modules they are all non-isomorphic and thus to show that MATH is irreducible, it suffices to show that one may send a MATH-highest weight vector in any irreducible MATH-component to the irreducible component containing the MATH-highest weight vectors. This follows from the following computation. MATH . Now if MATH the vectors MATH. Therefore MATH is MATH. But then the above calculation also shows that MATH is irreducible. The rank of MATH is then MATH, which is equal to MATH in the case when MATH. Finally when MATH, the vectors MATH and MATH reduces to MATH. Hence MATH is the trivial module and so has rank MATH.
math/0005274
By REF MATH is a singular vector in MATH. Let MATH be the MATH-submodule generated by MATH so that MATH, where MATH is the irreducible MATH-submodule generated by MATH. Consider MATH, the subspace in MATH of MATH-invariants. Now the MATH-weight of MATH is MATH and so MATH is a free MATH-module generated over MATH by MATH. We have MATH . It follows that in the case MATH that MATH is generated over MATH by the set MATH . Hence in this case MATH is a MATH-basis for the MATH-invariants of MATH. The MATH-weights of MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH are MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, respectively. Hence MATH is a free MATH-module of rank MATH. So we need to show that MATH is irreducible. Again we will study the MATH-module structure of MATH. We can check directly that MATH, for MATH, annihilates the vectors MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. Thus MATH as a MATH-module is a direct sum of the following eight irreducible modules: MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, where MATH is the irreducible MATH-module generated by MATH, for MATH, and MATH is generated by MATH, and MATH is the irreducible MATH-module of highest weight MATH. Note these modules are all irreducible. Note further that they are all non-isomorphic. So as before to show that MATH is irreducible, it suffices to show that one may send a MATH-highest weight vector in any irreducible MATH-component to the irreducible component containing the MATH-highest weight vectors. For this purpose we compute MATH . This settles the case when MATH. In the case when MATH is generated over MATH by MATH . Therefore MATH contains a MATH-invariant (and hence MATH-invariant) vector MATH. Since in this case the vectors MATH, MATH as a MATH-module is isomorphic to MATH. Every component is irreducible except for MATH, which contains a unique (irreducible) MATH-submodule isomorphic to MATH generated by the highest weight vector MATH. But then the above calculation plus the fact that MATH also shows that MATH is irreducible.
math/0005278
Let MATH be MATH-narrow. If MATH has a zero, we have to show that, given MATH, there is some MATH such that both MATH and MATH. Now, if MATH, which is a proper subset of MATH, and MATH such that MATH, then MATH as well. Conversely, if a closed proper subset MATH is given, pick some MATH such that MATH on MATH, MATH off a neighbourhood MATH of MATH. If MATH, MATH and MATH, then in particular MATH on MATH. Hence it is possible to replace MATH by a function MATH such that MATH, which proves that MATH is MATH-narrow.
math/0005278
Let us denote MATH and MATH. If MATH is a left semi-Fredholm operator, then, since MATH is complemented by a finite-codimensional subspace MATH, MATH is bounded from below, because MATH acts as an isomorphism from MATH onto MATH. On the other hand, if MATH is bounded from below on some finite-codimensional subspace MATH, then MATH and, consequently, MATH must be closed, and MATH is finite-dimensional, since otherwise MATH. This shows that MATH is not a left semi-Fredholm operator if and only if CASE: MATH is not bounded from below on any finite-codimensional subspace MATH. Now, if MATH satisfies REF , MATH and MATH, then MATH, that is, MATH. Conversely, if MATH, MATH is finite-codimensional and MATH is the quotient map, then, since MATH, MATH satisfies REF . Thus, we have shown the announced characterisation of MATH and, moreover, we have shown that REF provides another characterisation of MATH. It follows from REF that MATH if and only if CASE: for every MATH there exists an infinite-dimensional subspace MATH such that MATH; see CITE. From REF it is clear that every strictly singular operator belongs to MATH. On the other hand, if MATH is not strictly singular and is bounded from below on some infinite-dimensional subspace MATH, then we have for the quotient map MATH that MATH is bounded from below. Since MATH obviously satisfies REF , this shows that MATH.
math/0005278
REF is clear from the definition. For REF we note first that MATH . Indeed, by definition of MATH we have MATH whence MATH . On the other hand, an application of REF with MATH replaced by MATH gives MATH in REF . Now, by elementary arithmetic involving MATH and MATH we have, writing MATH for short, MATH . Because MATH is a semigroup, one can easily deduce that MATH; indeed, MATH . Therefore MATH completing the proof that MATH is a semigroup. Finally, REF is the special case MATH of REF .
math/0005278
Let MATH be a maximal subsemigroup of MATH. Put MATH. We have proved above in REF that MATH is a subsemigroup, so MATH is a subsemigroup, too. By definition of MATH we have MATH. So the maximality of MATH implies that MATH. This proves the inclusion MATH. Let us now prove the inverse inclusion. Let MATH. Then there is some MATH such that MATH does not belong to MATH. Consider the maximal subsemigroup MATH of MATH which contains MATH. Then MATH cannot contain MATH, so MATH cannot contain MATH either.
math/0005278
Let MATH. Then for every MATH we have MATH. This means that for every MATH and MATH there is an element MATH such that MATH. This in turn implies that MATH and MATH. So MATH. Now let MATH. Then for every MATH, every MATH and MATH there is an element MATH such that MATH. But by the definition of tubes, MATH. So MATH and MATH.
math/0005278
We assume without loss of generality that MATH. Given MATH pick MATH such that MATH. If MATH and MATH is chosen according to REF , then MATH hence MATH which proves the lemma.
math/0005278
MATH if and only if for every pair MATH and MATH there is an element MATH such that MATH. This in turn is equivalent to the following condition: for every pair MATH and MATH there is an element MATH such that MATH, MATH and MATH belongs to the tube MATH (just put MATH). Evidently, the last equation coincides with the strong NAME property of the operator MATH.
math/0005278
The fact that every MATH-narrow operator is a strong NAME operator has been proved in a slightly different form in CITE. Consider the converse implication. Let MATH. Fix a closed subset MATH and MATH. According to the definition it is sufficient to prove that there is a function MATH for which the restriction to MATH is less than MATH and MATH (compare REF ). Let us fix a neighbourhood MATH of MATH and an open set MATH, MATH. Select inductively functions MATH and MATH as follows. All the MATH are supported on MATH, and the MATH are non-negative functions supported on MATH. Given MATH and MATH pick MATH with MATH, and let MATH. Then choose MATH subject to the above support condition such that MATH coincides on MATH with MATH, and let MATH be a non-negative continuous function supported on the subset of MATH where MATH attains its supremum on MATH up to MATH, that is, on the set MATH, etc. (There is no initial restriction in the choice of MATH and MATH apart from the support and positivity conditions.) We first claim that MATH . This is certainly true for MATH since MATH. Now induction yields for MATH . (We have tacitly assumed that MATH since the induction step is clear otherwise, because MATH.) Next, we have that MATH . Indeed, the functions MATH and MATH are disjointly supported; hence by the definition of MATH there is a point in the support of MATH at which MATH is bigger than MATH. Thus, the above inequality holds for MATH. To perform the induction step we use the same argument to find a point MATH in the support of MATH at which MATH exceeds MATH. At this point MATH the function MATH attains its supremum on MATH up to MATH. So MATH . Therefore MATH, and on the other hand we have MATH. So for MATH big enough the function MATH will satisfy the desired conditions.
math/0005278
Since a narrow operator is a strong NAME operator and a strong NAME operator is MATH-narrow REF , it is left to prove that a MATH-narrow operator MATH on MATH is narrow if MATH is perfect. Let MATH be a functional, represented by a regular NAME measure MATH; we have to show that MATH is a strong NAME operator. Thus, let MATH and MATH. Let MATH, and consider the open set MATH. Pick an open non-empty subset MATH with the property that MATH is almost constant on MATH in that for some number MATH and MATH; the latter is possible since MATH has no isolated points. Since MATH is MATH-narrow, there is some MATH vanishing off MATH such that MATH; in fact, MATH can (and will) be chosen positive CITE. Let MATH. Then MATH, and MATH furthermore MATH so that MATH which proves that MATH is a strong NAME operator.
math/0005278
Fix some MATH and find an element MATH in norm-interior of MATH such that MATH. By REF , for every MATH there is an element MATH such that MATH. If MATH is small enough, then MATH. So, if in turn MATH is small enough, then MATH satisfies our requirements.
math/0005278
Select MATH. There are two symmetric cases: MATH or MATH. Consider for example the first of them. If we assume that our statement is not true, then we obtain MATH a contradiction.
math/0005278
CASE: First of all let us fix elements MATH such that MATH. Applying repeatedly REF with sufficiently small MATH to MATH, MATH and MATH we may select elements MATH with MATH, MATH, in such a way that for every MATH (to get the last inequality, we need to apply the previous lemma at each step). The element MATH will be as required. CASE: This follows from REF since given MATH there is a convex combination MATH of slices such that MATH; see CITE or CITE.
math/0005278
It only remains to show that the above condition is sufficient for MATH to be narrow. We first note that an operator satisfying that condition will also satisfy the conclusion of REF ; see the proof of that lemma. Now, if MATH, MATH and MATH, consider the relatively weakly open set MATH . By REF there exists some MATH such that MATH and MATH; note that MATH. By definition this means that MATH is a strong NAME operator; that is, MATH is narrow.
math/0005278
The convexity is evident. To prove the density we need to show, by the NAME theorem, that for every MATH and every MATH there is an element MATH such that MATH (in other words, MATH). Let us fix an element MATH with MATH and consider the operator MATH. Consider further MATH and a MATH-neighbourhood MATH of MATH in MATH. By the NAME property of the set MATH there is a convex combination MATH of slices of MATH in MATH. The preimages in MATH of these slices of MATH are slices in MATH. The corresponding convex combination MATH of these slices in MATH lies in the preimage of MATH in MATH, so this convex combination is contained in MATH. Fix an element MATH. By our construction MATH. On the other hand, MATH so MATH.
math/0005278
Let us fix MATH, MATH and MATH such that MATH. According to the definition of MATH there exists a convex combination MATH of slices of the unit ball such that MATH and MATH. By REF there is an element MATH such that MATH and MATH. But the inclusion MATH means that MATH. So MATH . Because MATH is arbitrarily small, the last inequality shows that MATH satisfies the definition of a strong NAME operator. Now let MATH and consider MATH. This is a strong NAME operator, too. So MATH is a strong NAME operator by what we have just proved; by definition, this says that MATH is narrow.
math/0005278
REF follows from the previous theorem, because every finite-rank operator is a strong NAME operator. For REF use REF and note that MATH . REF is a restatement of REF .
math/0005278
Consider an auxiliary space MATH and an auxiliary matrix MATH, MATH. Since MATH contains no copies of MATH either and since MATH is a MATH-limit point of MATH, there is, according to REF , a sequence of the form MATH which converges to MATH in MATH. This means in particular that MATH in MATH and MATH. So MATH and MATH both tend to MATH, which, after passing to a subsequence, provides the desired sequence.
math/0005278
Using inductively REF we can select a doubly indexed sequence MATH in MATH with the following properties: CASE: for every MATH, MATH is a MATH-limit point of every column MATH; CASE: for every MATH, if MATH, then MATH. Applying REF and passing to a subsequence if necessary, we obtain strictly increasing sequences MATH, MATH such that MATH and MATH converges to MATH in MATH. To finish the proof put MATH.
math/0005278
Assume there exist MATH and MATH such that for every finite-dimensional subspace MATH there is a slice MATH containing MATH with MATH for all MATH. Take a weak-MATH cluster point MATH of the net MATH and let MATH. We have MATH since MATH and therefore MATH. Now if MATH, then MATH and therefore MATH for some MATH that contains MATH. So by REF, which contradicts REF when applied to the slice MATH.
math/0005278
Suppose MATH has the NAME property. Let MATH be a dense sequence in MATH. We select a sequence MATH of finite-dimensional subspaces of MATH by the following inductive procedure. Put MATH. Suppose MATH has already been constructed. Fix a MATH-net MATH, MATH, in MATH provided with the sum norm, select by REF finite-dimensional subspaces MATH, MATH, for MATH and define MATH. If MATH is defined to be the closure of the union of all the MATH, then MATH and MATH has the NAME property by REF . Conversely, let MATH, MATH and let MATH be a slice. Fix a point MATH. If MATH is a separable subspace with the NAME property containing MATH and MATH, then by REF there exists some MATH such that MATH. Again by REF this shows that MATH has the NAME property.
math/0005278
First, let MATH be a perfect compact metric space. It follows from REF and CITE that the set of all narrow operators on MATH is stable under the operation MATH, that is, it is a semigroup. (In fact, CITE only deals with MATH, but the arguments work as well for a metric MATH.) We shall now reduce the general case to the metric one. Let now MATH be a perfect compact NAME space, and let MATH and MATH be two narrow operators on MATH; we shall verify that MATH is narrow, using REF above. Thus, let MATH be a separable subspace of MATH. We shall first argue that there is a separable space MATH containing MATH such that MATH and MATH are strong NAME operators. Let MATH be a countable dense subset of MATH. For every pair MATH in MATH and every MATH there is some MATH (respectively, MATH) according to the definition of the strong NAME property of MATH (respectively, MATH). The countable collection of these MATH's and MATH span a closed separable subspace MATH. Repeating this procedure starting from MATH yields some closed separable subspace MATH, etc. The closed linear span MATH of MATH then has the desired property. Now by CITE there is a separable space MATH isometric to some space MATH for a perfect compact metric space MATH. By the same token as above, we can extend MATH to a separable space MATH so that MATH and MATH are strong NAME operators on MATH, and we can extend MATH to a separable space MATH isometric to some space MATH for a perfect compact metric space MATH, etc. Let MATH be the closed linear span of MATH. Then MATH and MATH are strong NAME operators, and MATH is isometric to some space MATH for a perfect compact metric space MATH. By what we already know, MATH is a narrow operator on MATH; recall that the classes of narrow and strong NAME operators coincide on MATH. Finally, REF implies that MATH is narrow on MATH, which proves the theorem.
math/0005278
We shall argue by induction on MATH. First of all consider MATH. Every MATH-neighbourhood of MATH can be represented as MATH, where MATH. Since MATH is a strong NAME operator by definition of the central part, there is an element MATH such that MATH and MATH. The last inequality means, in particular, that MATH and MATH. Now suppose our assertion is true for MATH, let us prove it for MATH. Let MATH, and let us assume that an element MATH such that MATH and MATH, MATH, has already been selected. Then there is a weak neighbourhood MATH of MATH such that the inequalities MATH, MATH, hold for every MATH. The intersection MATH is a MATH-neighbourhood of MATH, so according to our inductive assumption for MATH, there is an element MATH such that MATH and MATH. This element MATH satisfies all the requirements.
math/0005278
REF implies that every operator which does not fix a copy of MATH can be factored through a space without MATH-subspaces. So every operator which does not fix a copy of MATH can be majorized by an operator which maps into a space without MATH-subspaces. Since the class of narrow operators is an order ideal, it is enough to prove our theorem for MATH, where MATH has no MATH-subspaces. Also, by REF we may assume that MATH and MATH are separable. Let us fix a narrow operator MATH, MATH and MATH. Let us introduce a directed set MATH as follows: the elements of MATH are finite sequences in MATH of the form MATH, MATH, with MATH. The (strict) ordering is defined by MATH and of course MATH if MATH or MATH. Now define a bounded function MATH by MATH where MATH . Due to REF , for every weak neighbourhood MATH of MATH in MATH, every MATH and every finite collection MATH there is some MATH for which MATH and MATH. This means that MATH is a weak limit point of the function MATH. So, by REF there is a strictly MATH-increasing sequence MATH for which MATH tends weakly to MATH, MATH tends to REF and MATH tends to MATH. Passing to a subsequence we can select points MATH in such a way that the sequence MATH is MATH-equivalent to the canonical basis of MATH. According to NAME 's theorem, there is a sequence MATH such that MATH. Evidently MATH and MATH, which means that MATH and thus proves the theorem by REF .
math/0005278
In each MATH there is a separable subset whose closed convex hull contains MATH. So, passing to the linear span of these separable subsets we may assume that MATH is separable. Introduce a directed set MATH as follows: the elements of MATH are of the form MATH where MATH, MATH, MATH, MATH, MATH. Define MATH as follows: let MATH, MATH; then MATH if MATH. Define MATH by the formula MATH. Now, MATH is a weak limit point of MATH; see the proof of CITE. So, by REF there is a sequence of elements MATH such that MATH and MATH tends weakly to zero. To finish the proof one just needs to apply NAME 's theorem.
math/0005278
Consider elements MATH, MATH, a slice MATH and MATH. According to our assumption the quotient map MATH is a narrow operator. So there is an element MATH such that MATH and MATH. The last condition means that the distance from MATH to MATH is smaller than MATH, so there is an element MATH with MATH. The norm of MATH is close to MATH, viz. MATH. Put MATH. For this MATH we have MATH, so MATH and MATH.
math/0005278
REF follows from REF from REF , and REF follows from REF .
math/0005278
Due to REF every finite-codimensional subspace of a space with the NAME property has the NAME property itself (see also CITE); this is the reason for the equivalence of REF . The implication MATH follows immediately from the definition of a wealthy subspace; REF are consequences of REF .
math/0005278
First of all let us fix a slice MATH from the definition of a MATH-fine pair and fix a MATH such that the set MATH, MATH still has diameter less than MATH. Now let us find a finite-codimensional subspace MATH such that CASE: MATH on MATH, CASE: if MATH and MATH, then MATH; the last condition can be satisfied by a variant of the NAME argument leading to the basic sequence selection principle; see CITE. According to our assumptions MATH has the NAME property. So there is an element MATH such that MATH. Let us represent MATH in the form MATH, where MATH, MATH. By choice of MATH this means that MATH and MATH. Thus, MATH and MATH. Finally we have that the element MATH belongs to MATH, which concludes the proof.
math/0005278
Let MATH be the quotient map and let MATH; further let MATH and let MATH be the corresponding quotient map. Then MATH or MATH is MATH-codimensional in MATH. Now, in either case we have MATH for all MATH. Since MATH is a strong NAME operator by assumption, so is MATH, and MATH is narrow.
math/0005278
If MATH intersects all the elements of MATH, then the quotient map MATH is unbounded from below on every element of MATH. So the quotient map belongs to MATH which coincides with the class of strong NAME operators by REF . Now consider the converse statement. If MATH is almost rich, then for every MATH the map MATH is unbounded from below on every set of the form MATH. This means that there is an element MATH for which MATH. In this case MATH belongs to MATH, so the intersection of this set with MATH is non-empty.
math/0005278
According to REF we need to prove that for every positive MATH and every pair MATH the subspace MATH intersects MATH. To do this, according to REF , it is enough to show that for every MATH and every pair MATH there is a MATH-fine pair MATH which approximates MATH well; that is, MATH. Let us fix a positive MATH and select an element MATH in such a way that for every MATH and for every MATH (we use REF ). Put MATH, MATH. To show that MATH is a MATH-fine pair it is sufficient to demonstrate that, for every MATH with MATH, MATH. To do this let us argue ad absurdum. Take some MATH with MATH and assume that MATH. Then MATH . So MATH. But in this case MATH and MATH which provides a contradiction.
math/0005278
It is clear that REF , see the remark following REF . Now suppose REF . Every MATH-codimensional subspace of MATH is wealthy by REF and is hence almost rich by REF . An appeal to REF completes the proof.
math/0005278
Let MATH, MATH, and MATH. Consider a slice in MATH of the form MATH . Applying REF to this slice, the elements MATH, MATH and MATH we get a function MATH such that MATH . Denote by MATH the set MATH. The condition MATH implies that MATH, so MATH . Next, introduce MATH. By the last inequality MATH and MATH to see this observe that MATH . Put MATH with MATH so that MATH. Since MATH we have from MATH that MATH. By REF we conclude that MATH and MATH and if MATH is small enough, by REF MATH. This proves the inclusion MATH. To prove the opposite inclusion we use REF . Let us fix MATH. Let MATH, MATH and MATH be such that MATH. Without loss of generality we may assume that there is a partition MATH of MATH such that the restrictions of MATH, MATH and MATH on MATH are constants, say MATH, MATH and MATH respectively. By our assumption MATH is unbounded from below on each of the MATH for every MATH, MATH. Let us fix functions MATH such that MATH, MATH, and put MATH . By definition of balanced MATH-peaks MATH, MATH, and MATH and MATH become arbitrarily small when MATH is small enough. Thus MATH can be chosen so that MATH fulfills the conditions MATH and MATH.
math/0005278
Denote MATH . Since MATH and MATH coincide off MATH, we clearly have MATH . We also have that MATH . Indeed, we can write MATH as a countable union of disjoint (half-open) intervals; denote by MATH any one of these. Then MATH is constant on MATH, and MATH. Hence MATH . Summing up over all MATH gives the result. Next, we claim that MATH . To see this, we label the intervals MATH from the previous paragraph as follows. For every MATH write MATH and MATH. Each MATH can be written as MATH where MATH is some subset of MATH with cardinality MATH. Let us write MATH. We then have the estimates MATH and MATH . Summing up over all MATH gives us MATH . On the other hand, by the triangle inequality MATH hence the claim follows. The lemma now results from REF - REF
math/0005278
Let us fix MATH and MATH. Without loss of generality we may assume that MATH for a big enough MATH to be chosen later. Put MATH. Then MATH with MATH. So MATH and MATH. Since MATH we can pick MATH big enough to satisfy MATH. This shows that MATH is a strong NAME operator. To show that MATH fails the NAME property if MATH, take MATH and MATH. Since MATH, we get MATH . Thus, MATH. We show that there is no MATH in this slice such that MATH. Suppose, on the contrary, that there is such a MATH. Without loss of generality we can assume that MATH where MATH is as in REF . It follows from our conditions that MATH . Hence, MATH and since MATH, we get MATH . By REF , MATH thus REF yield MATH . But now REF imply MATH which yields MATH, that is, MATH, which is false for MATH.
math/0005280
We need to show that MATH is stably congruent to a unidiagonal matrix. In fact MATH is congruent to MATH, which is congruent to MATH and which, in turn, since MATH is almost even, is congruent to MATH, where MATH is a diagonal matrix all of whose entries are MATH or MATH. But this is congruent to some unidiagonal matrix.
math/0005280
MATH represents an element in MATH and so its bordism class is determined by the degree of MATH. It follows that MATH is bordant to the identity map of MATH. This gives us the manifold MATH and map MATH without the desired handlebody structure. We must eliminate the handles of index not equal to MATH. First of all we can eliminate handles of index MATH and MATH in the usual way. Now a handle of index MATH represents a boundary connect sum with MATH. We can replace this with MATH, thereby changing MATH. The only problem is how to replace the map MATH on this altered piece. Since MATH is a retract of MATH, MATH represents an element of MATH. By sliding one foot of this MATH - handle around a representative of this element in MATH we can arrange that MATH is null-homotopic. Thus, after replacing this MATH - handle with a MATH - handle, we can also replace MATH. To get rid of the MATH - handles we regard them as MATH - handles on MATH and apply the same argument since MATH is onto.
math/0005280
This follows from CITE. The isomorphism MATH in REF is determined, since the maps MATH are isomorphisms. The commutativity of MATH corresponds to the diffeomorphism equivalence of the pairs MATH.
math/0005280
Suppose that MATH is bordant to the identity on MATH by MATH, where MATH consists of MATH - handles adjoined to MATH. Let MATH be an associated matrix. Suppose MATH is bordant to MATH by a MATH - homology bordism MATH. By pasting MATH together we create a bordism MATH from the identity map on MATH to itself. The intersection pairing on MATH is represented by MATH. Now suppose MATH is bordant, rel boundary, to the projection MATH. Then the standard argument shows that the intersection pairing on MATH is metabolic. Thus by REF the proposition is proved. The obstruction to this bordism is an element of the bordism group MATH. Now MATH and MATH is generated by MATH and so this bordism will exist after we connect sum, say, MATH with a number of copies of MATH. But this can be achieved by adding to the framed link defining the handlebody decomposition of MATH a number of trivial components with MATH - framing. The effect of this is to block sum MATH with a unidiagonal matrix.
math/0005280
Since each component MATH of MATH is null-homotopic in MATH, it is homotopic in MATH to a product of meridians of MATH. Thus we can connect sum several meridians of MATH to MATH to get a new knot MATH which is null-homotopic in MATH and is clearly isotopic to MATH in the complement of the other components of MATH. To see that MATH, when MATH is null-homotopic, we only need note that any lift MATH to the universal cover MATH of MATH is null-homotopic in MATH, for any lift of MATH. If MATH is algebraically split then the linking elements MATH, where MATH, differ from the entries of a unidiagonal matrix by members of the two-sided ideal MATH of MATH generated by elements of the form MATH, where MATH. REF shows how to modify MATH to change MATH by an element: CASE: MATH if MATH, CASE: MATH if MATH for MATH, without changing any other linking element MATH except when MATH. The dotted curves connecting MATH to the basepoint are those used to specify the lifts - these are needed to define the linking elements. Note that we can represent MATH by the boundary of a disk in MATH which is disjoint from MATH and the arcs used in the modification, since MATH is a product of meridians of MATH. Thus the modified MATH is isotopic to MATH in MATH. Since elements of the form REF generate MATH, we only need show, by REF , that self-conjugate almost even elements of MATH are linear combinations of elements of the form REF , which we will call norm-like, to conclude that MATH can be chosen to be algebraically split. Choose a subset MATH so that, for every MATH, exactly one of MATH belongs to MATH. For each MATH choose MATH so that MATH; choose MATH. Now suppose MATH is a self-conjugate even element of MATH. We can write MATH uniquely in the form MATH where MATH, the augmentation ideal of MATH. Clearly the terms of the first summation in REF are norm-like, so we consider each term MATH of the second summation. Let us write MATH and so MATH where MATH are distinct elements of MATH, MATH and MATH. Since MATH is self-conjugate we have MATH and so, for each MATH there is some MATH so that MATH. If MATH, then replace MATH in REF by MATH. If MATH then MATH is of order MATH and so MATH is even. In this case rewrite MATH in REF , as MATH. Now REF will look like MATH where still MATH. If we now subtract MATH from REF we get MATH which is a sum of norm-like terms.
math/0005280
Suppose MATH is an algebraically split link determining MATH in MATH and MATH determines MATH in MATH. We can apply REF to MATH and the meridians MATH of MATH in MATH to allow us to assume that the components of MATH are null-homotopic in MATH. Now MATH is algebraically split in MATH and determines MATH in MATH. Since, by REF , MATH, the linking matrix of MATH, which represents MATH is the block sum of the linking matrix of MATH, which represents MATH and the image under MATH of the linking matrix of MATH, which represents MATH.
math/0005280
Suppose that MATH is an algebraically split link which defines MATH. Now let MATH be any algebraically split link - NAME can assume that MATH is disjoint from the meridians of MATH and so lies in MATH. In fact, by REF applied to MATH and the meridians of MATH, we can assume that the components of MATH are null-homotopic in MATH and that MATH is algebraically split in MATH. It is clear that MATH defines the element MATH. Suppose that MATH has MATH components and so MATH. Then we have: MATH . Thus we see that MATH, which shows that MATH, and that MATH, which shows that MATH is onto.
math/0005280
Pick a maximal forest MATH for MATH, that is, a maximal tree for each connected component of MATH. If MATH, then we can assume that MATH, a base point of MATH, that is, that it factors though a map MATH. This map is determined by the induced one on the level of MATH. A different choice of a maximal forest or a different choice of a base point of MATH results in maps on MATH that differ in each connected component of MATH by independent inner automorphisms of MATH.
math/0005280
It suffices to consider a vertex-oriented, edge-oriented connected graph MATH. To a map MATH, we will associate a map MATH and vice versa. Given a map MATH, (which in view of relation MATH we may assume that it is a decoration of the edges of MATH by elements of MATH) we define a map MATH as follows. For a closed path of oriented edges MATH, we set MATH. It is easy to see that this defines a group homomorphism MATH, compatible with the relations MATH and MATH. Conversely, given a map MATH, choose a maximal tree MATH and define MATH. Since MATH can be identified with the free group on MATH, MATH will then determine MATH on these edges.
math/0005280
Choose a base point MATH of MATH and a basing MATH of MATH, that is, a choice of disjoint paths MATH in MATH from MATH to points MATH, one for each component of MATH. Choose a framing of MATH and a lift MATH of MATH. Then, there is a unique lift MATH in MATH of MATH that contains MATH and a well-defined linking matrix MATH of MATH. Choose a regular homotopy MATH from MATH to MATH, which we can assume is stationary on every MATH. We may assume that MATH is a link except for finitely many times MATH where MATH is an immersion with a single transverse double point. The linking matrix MATH and MATH of MATH and MATH are related as follows: if the double point MATH involves the components MATH and MATH of MATH, with MATH, construct a loop MATH by starting at MATH, going along MATH to MATH, then over to MATH and back to MATH via MATH - see the figure below. It is easy to see that MATH where MATH is the local orientation sign of MATH and MATH is the matrix with all zeros except in the MATH place where it equals MATH. Thus, if MATH is the linking matrix of MATH and MATH is the linking matrix of an unlink with the same framing and number of components as MATH, we have MATH, where the sum is over all double points of the homotopy. Since MATH is a MATH - AS link, it follows that for every pair of components MATH and MATH of MATH there is a pairing of the double points of MATH and MATH into classes MATH such that MATH and MATH. In other words, one can undo MATH by a sequence of double crossing changes shown below in REF , where the loop MATH is nullhomotopic. These double crossing changes can be achieved by surgery on MATH - graphs whose leaves are nullhomotopic, see CITE and also REF below. So far, each of the MATH - graphs have two leaves that bound a disk that intersects MATH at most once and a nullhomotopic leaf. Observe that every nullhomotopic leaf in MATH bounds a disk with clasp intersections as shown below. Using repeatedly Move MATH of CITE (the so-called, move of Cutting a Leaf), as follows MATH we may assume that every leaf of each MATH - graph bounds a disk that either intersects MATH geometrically once, or none. In all cases, the MATH - link MATH that consists of all these MATH - graphs is lacing the unlink MATH. It is easy to verify that the rest of the statements of the proposition.
math/0005280
MATH .
math/0005280
Since MATH, we need only show the opposite inclusion. REF implies that for every MATH - AS link MATH in MATH, there exists a trivial MATH - link-MATH that ties a trivial unimodular link MATH such that MATH. Let MATH denote the image of MATH under surgery on MATH. MATH is a MATH - link (with nullhomotopic leaves) and MATH. The result follows.
math/0005280
This follows by elementary properties of NAME 's calculus applied to the unit-framed knots MATH shown in the pictures above.
math/0005280
First of all we prove the invariance of the MATH under surgery equivalence. If MATH is chosen so that MATH is MATH - AS, then we can, by tubing, arrange that the surfaces MATH used to define MATH are disjoint from the lifts of MATH and so pass unchanged into the MATH - covering of the surgered link. In particular the intersections which define MATH are unchanged. Now suppose that MATH and MATH are two MATH - AS links such that MATH. By REF we know that MATH can be transformed into MATH by surgery on a set of MATH - links whose leaves are meridians of MATH. Since surgery on such a MATH - link is the same as a sequence of disjoint MATH - moves in the terminology of CITE - see REF - it is easy to see the effect of such a surgery on the MATH. Suppose a lift of the MATH - link MATH into MATH has its meridians on three components MATH, where MATH. If any two of these components are the same then there is no change in any of the MATH. If the three components are distinct then MATH is changed by MATH and every other MATH, where MATH, is unchanged. Thus our assumption about MATH says that the transformation from MATH to MATH is accomplished by a sequence of surgeries of two types: CASE: surgery on pairs of MATH - links MATH, where MATH and MATH can be lifted to MATH - links in MATH with oppositely oriented trivalent vertices and which have leaves on the same three distinct components, and CASE: surgeries on individual MATH - links MATH with at least two leaves on the same component. In REF it is easy to see that surgery on MATH does not change the surgery equivalence class since we can undo the Borromean part of the MATH - link by crossing changes, using the second part of REF , on the two rings attached to the same component of MATH. Thus it remains to show that the effect of surgery on a pair of MATH - links MATH with leaves on the same three distinct components of MATH does not change the surgery equivalence class of MATH. First of all we can consider the case where MATH is an inverse of MATH in the sense of CITE. In this REF surgery on MATH and MATH does not change MATH at all. For any other MATH we can assume that there is a homotopy in MATH from MATH to an inverse of MATH which is stationary on the leaves of MATH. Such a homotopy is a sequence of isotopies in MATH together with REF crossings of an edge of MATH and a component of MATH and REF crossings of an edge of MATH with a leaf of MATH. It suffices to show that these two types of crossings do not change the surgery equivalence class of the MATH surgery on MATH. For REF the effect of this crossing on surgery of MATH is pictured in REF . The surgery equivalence is given by REF and the double crossing change in REF . For REF we invoke the following: Suppose MATH is the union of two MATH - links MATH in the complement of a link MATH, whose leaves are meridians of MATH, and MATH is obtained from MATH by a single crossing change of a leaf of MATH with a leaf of MATH. Then the link produced by surgery on MATH using MATH is surgery equivalent to the link produced by surgery on MATH. By CITE and MATH moves of CITE, surgery on MATH is the same as surgery on MATH together with surgery on a clover of degree MATH with the shape of MATH. Thus we need to see that surgery on such clovers does not change the surgery equivalence class of MATH. The effect of surgery on such a clover is shown in REF . A double crossing change which will undo this surgery is illustrated in REF . This completes the proof of REF .
math/0005280
Concordance, just as in the classical case, is generated by the following ribbon move MATH. Given a MATH - AS link MATH, consider also a finite number of disks MATH in MATH, disjoint from each other and MATH. For each MATH choose a band MATH connecting MATH to a component, which we denote MATH, of MATH. The band cannot intersect MATH or any MATH except at its ends. Then MATH is defined to be the band-sum of MATH with MATH. Now choose some place where a band MATH penetrates a disk MATH. Choose a path MATH from MATH to nearby the penetration so that the closed path consisting of MATH followed by the path from MATH along MATH and back along MATH to the starting point of MATH is null-homotopic. Now thicken MATH to a band (or finger) and apply REF as follows: This removes the penetration. Eventually we can remove all the penetrations and the resulting link will be isotopic to MATH.
math/0005281
The metric introduced in REF is equivalent to the metric described in CITE. The induced topologies are therefore the same. The result follows therefore from CITE.
math/0005281
Let MATH be a minimal basic encoder of MATH and let MATH be the first column of MATH. Note that MATH and that there is at least one entry of MATH which does not contain the factor MATH. Let MATH and consider the sequence of code words MATH. For each MATH one has that MATH. However MATH is in MATH. This shows that MATH is not a closed set inside MATH. The closure MATH is obtained by extending the input space MATH to all of MATH. The image of MATH under the encoding map REF is closed by REF , hence the closure is a code in the sense of REF .
math/0005281
CITE.
math/0005281
The proof of the completeness part of the Theorem is analogous to the proof of REF . In order to show controllability, let MATH be an encoding matrix for a code MATH and consider two code words MATH and MATH. The codeword MATH required by REF can be constructed in the form MATH .
math/0005281
Let MATH. If MATH is not controllable, then MATH is not left prime and one has a factorization MATH, where MATH is left prime and describes the controllable sub-behavior MATH. Since MATH is an autonomous behavior it follows that MATH . It follows (compare with REF ) that the completion MATH.
math/0005287
For simplicity, assume MATH, the general case being quite similar. The diagonal MATH is obviously an invariant subset for the group MATH, where MATH runs over the set of all MATH-preserving transformations of the space MATH. Thus it suffices to show that if MATH is concentrated on the set MATH, then MATH, and if MATH is concentrated on MATH, then MATH. In the first case let MATH be an arbitrary partition of the space MATH into MATH sets of equal MATH-measure MATH. Denote by MATH the corresponding partition of the space MATH, that is, MATH, where MATH. The group MATH acts transitively on the set of non-diagonal elements of MATH. Thus all non-diagonal elements have equal MATH-measure. Denote MATH and MATH. Since MATH is concentrated on MATH, we have MATH as MATH. Considering finer partitions and using the above argument, we obtain that for each MATH, if a rectangle MATH and MATH, then MATH. But then the restriction of MATH on the set MATH equals MATH. Letting MATH, we obtain MATH. In the second case, identifying the diagonal MATH with MATH, we obtain that MATH is a measure on MATH which is invariant under all MATH-preserving transformations, hence obviously MATH.
math/0005287
Let MATH be a MATH-preserving transformation of MATH. This transformation acts on the space MATH by substituting coordinates, that is, MATH, and it is clear that the law MATH of a homogeneous NAME process on MATH is invariant under MATH. Denote by MATH the conditional measure of MATH given the conic part equal to MATH. The transformation MATH acts ``fibre-wise", that is, it does not change the conic part, hence MATH preserves almost all conditional measures MATH. In particular, if we denote by MATH the conditional distribution of the first MATH points MATH on the space MATH, then the transformation MATH preserves MATH. Now it follows from REF that for almost all MATH for all MATH, that is, MATH, and REF follows.
math/0005287
Given fixed MATH and a probability vector MATH, consider a partition MATH of the space MATH such that MATH, MATH. Let MATH be the sequence of i.i.d. variables with common distribution MATH and assume MATH, if MATH. Then the random variables MATH form a sequence of i.i.d. variables, and MATH. Consider a random process MATH where the sequence MATH is independent of MATH and obeys the law MATH. Let MATH. It is easy to see that for arbitrary MATH where MATH is a step function such that MATH, if MATH. Now let MATH be the conic part of the law MATH of some NAME process. Then, by REF , the process MATH defined by REF obeys MATH, and it follows from the NAME transform REF that the right-hand side of REF equals MATH . Since the left-hand side of REF is the NAME transform of the common distribution of the variables MATH, we obtain that MATH are independent, and MATH obeys the law MATH, that is, MATH is of product type. Conversely, let MATH be a measure of product type corresponding to an infinitely divisible law MATH. Define a random process MATH on an arbitrary measurable space MATH satisfying the conditions of REF by REF . The above argument shows that MATH satisfies REF with MATH equal to the NAME measure of MATH for all positive step functions MATH, and one can easily extend this to all bounded positive NAME functions by continuity. Thus MATH obeys MATH, and MATH is the conic part of MATH.
math/0005287
Fix MATH and let MATH. Consider an arbitrary function MATH. Then MATH. Thus, in view of REF , the NAME transform MATH equals MATH . Using REF once more, we may consider the last factor as the NAME transform of MATH calculated on the function MATH. Denote MATH. Then we have MATH and REF follows.
math/0005287
Let MATH be a MATH-measurable functional on MATH which is invariant under all MATH that is, MATH a.e. with respect to MATH. Consider an arbitrary NAME function MATH. Then for each MATH where MATH denotes the expectation with respect to MATH. But in view of REF the last factor equals MATH hence we have MATH . Thus MATH is independent of every functional MATH, and REF follows.
math/0005287
The representation is correctly defined and its unitarity follows from the invariance property of MATH. The irreducibility follows from the ergodicity of the action of the group MATH of multiplicators.
math/0005287
In case of a MATH-stable process, we have MATH, and REF follows immediately from REF .
math/0005287
It follows from REF that the NAME transform of the measure MATH equals MATH . But MATH as MATH, hence MATH and REF follows.
math/0005287
Easy calculation.
math/0005287
Using REF and the NAME theorem we obtain that the right-hand side of REF equals MATH and Theorem follows.
math/0005287
CASE: Denote the left-hand side of the desired identity by MATH and the right-hand side by MATH. Using the identity MATH we obtain MATH . By the NAME transform REF , the expectation equals precisely MATH, thus MATH and REF follows by changing variables. CASE: Follows from REF by letting MATH.
math/0005292
Let MATH be a smooth path of representations starting at the inclusion MATH corresponding to MATH. MATH where the sign equals MATH. Applying REF to the last expression gives MATH . Thus MATH has the same sign as MATH as claimed. To prove REF, apply REF and the chain rule to obtain: MATH . Since MATH REF follows from REF.
math/0005292
Here are two constructions for MATH, the first based on the Riemannian geometry of MATH with the NAME metric and the second based on NAME 's earthquake flows. Let MATH be the NAME geodesic satisfying REF. By REF, the geodesic length function MATH is strictly convex along MATH and the directional derivative MATH, for any MATH. Therefore MATH is monotonically increasing for MATH. However, in general the NAME metric is geodesically incomplete (CITE, CITE), so that MATH is only defined for MATH where MATH. We show this is impossible under our assumptions on MATH. By NAME 's compactness theorem (CITE, CITEEF or CITEEF), the subspace of moduli space consisting of hyperbolic surfaces whose injectivity radius is larger than any positive constant is compact. An incomplete geodesic on a Riemannian manifold must leave every compact set. Therefore, if the NAME geodesic MATH cannot be extended to MATH, then MATH contradicting monotonicity of MATH. Hence MATH is defined for all MATH. As above, convexity implies REF. Alternatively, take MATH to be the earthquake path introduced by NAME (see CITE and CITE). For the given tangent vector MATH, there exists a unique measured geodesic lamination MATH such that the corresponding earthquake path MATH satisfies REF REF . By CITE (see also CITE), each length function MATH is convex along the earthquake path MATH, implying REF. Indeed, MATH is strictly convex along MATH since the lamination MATH fills up MATH - that is, every nonperipheral simple closed curve MATH intersects MATH. For otherwise MATH would be constant along MATH, contradicting MATH .
math/0005301
First we will show homotopy equivalence and remark on MATH-homotopy equivalence later. We work with MATH, the barycentric subdivision. Notice that the associated poset of MATH contains the poset MATH of nontrivial abelian subgroups of MATH as a subposet, they are merely the commuting sets whose elements actually form an abelian subgroup (minus identity). Let MATH denote this inclusion. We now define a poset map MATH as follows: If MATH is a set of nontrivial, pairwise commuting elements of MATH, then MATH, the subgroup generated by MATH will be a nontrivial abelian subgroup of MATH, thus we can set MATH. It is obvious that MATH is indeed a poset map, and that MATH and so MATH is homotopic to the identity map of MATH by REF . Furthermore it is clear that MATH and so MATH is a deformation retraction of MATH onto MATH. Thus MATH is homotopy equivalent to MATH. To see this is a MATH-homotopy equivalence, we just need to note that MATH is indeed a MATH-map, and maps a commuting set invariant under conjugation by a subgroup MATH into a subgroup invariant under conjugation by MATH and thus induces a homotopy equivalence between MATH and MATH for any subgroup MATH. Thus MATH is indeed a MATH-homotopy equivalence by REF . The MATH-local version follows exactly in the same manner, once one notes that the subgroup generated by a commuting set of elements of order MATH is an elementary abelian MATH-subgroup.
math/0005301
If MATH has a nontrivial center MATH, then MATH is conically contractible via MATH for any abelian subgroup MATH of MATH. Thus MATH is also contractible as it is homotopy equivalent to MATH. Let MATH be the natural inclusion of posets. Take MATH and let us look at MATH. However since MATH is nilpotent, it has a nontrivial center MATH and hence MATH is conically contractible. Thus by REF the result follows.
math/0005301
First we show that MATH is path-connected. Take any two vertices in MATH, call them MATH and MATH, then these are two noncentral elements of MATH, thus their centralizer groups MATH and MATH are proper subgroups of MATH. It is easy to check that no group is the union of two proper subgroups for suppose MATH where MATH are proper subgroups of MATH. Then we can find MATH (it follows MATH) and MATH (hence MATH). Then MATH is not in MATH as MATH and MATH. Similarly MATH so MATH which is an obvious contradiction. Thus no group is the union of two proper subgroups. Thus we conclude that MATH and so we can find an element MATH which does not commute with either MATH or MATH and so the vertices MATH and MATH are joined by an edge path MATH. (The MATH stands for concatenation.) Thus we see MATH is path-connected. In fact, any two vertices of MATH can be connected by an edge path involving at most two edges of MATH. To show it is simply-connected, we argue by contradiction. If it was not simply connected, then there would be a simple edge loop which did not contract, that is, did not bound a suitable union of REF. (a simple edge loop is formed by edges of the simplex and is of the form MATH where all the MATH are distinct except MATH.) Take such a loop MATH with minimal size MATH. (Notice MATH is just the number of edges involved in the loop.) Since we are in a simplicial complex, certainly MATH. Suppose MATH, then MATH can be connected to MATH by an edge path MATH involving at most two edges by our previous comments. This edge path MATH breaks our simple edge loop into two edge loops of smaller size which hence must contract since our loop was minimal. However, then it is clear that our loop contracts which is a contradiction so MATH. So we see MATH so we have three cases to consider: CASE: MATH: Here MATH with MATH. But then it is easy to see MATH is a set of pairwise non-commuting elements and so gives us a REF-simplex MATH in MATH which bounds the loop which gives a contradiction. CASE: MATH: Here MATH with MATH. Thus MATH forms a square. Notice by the simplicity of MATH, the diagonally opposite vertices in the square must not be joined by an edge in MATH, that is, they must commute, thus MATH commutes with MATH and MATH commutes with MATH. Since MATH and MATH do not commute, MATH is a set of mutually non-commuting elements and so forms a REF-simplex of MATH. Since MATH commutes with MATH but not with MATH, it does not commute with MATH and thus MATH also is a REF-simplex in MATH. Similar arguments show that MATH and MATH form REF-simplices in MATH. The union of the four MATH-simplices mentioned in this paragraph, bound our loop giving us our contradiction. Thus we are reduced to the final case: CASE: MATH: Here MATH with MATH. Thus MATH forms a pentagon, and again by simplicity of MATH, nonadjacent vertices in the pentagon cannot be joined by an edge in MATH, thus they must commute. Similar arguments as for the MATH case yield that MATH, MATH and MATH are MATH-simplices in MATH and the union of these three simplices contract our loop MATH into one of length four namely MATH which by our previous cases, must contract thus yielding the final contradiction.
math/0005301
The remarks about NAME characteristics follow from the fact that if a finite group MATH acts freely on a space where the NAME characteristic is defined, then MATH must divide the NAME characteristic. So we will concentrate mainly on finding such actions. First for the action of MATH. MATH acts by taking a simplex MATH to a simplex MATH. Notice this is well-defined since MATH is noncentral if MATH is noncentral and since MATH commutes with MATH if and only if MATH commutes with MATH. Furthermore if MATH is not the identity element, this does not fix any simplex MATH since if the set MATH equals the set MATH then MATH is one of the MATH's but MATH commutes with MATH so it would have to be MATH but MATH gives MATH, a contradiction. Thus this action of nonidentity central MATH does not fix any simplex and so we get a free action of MATH on MATH. Now for the action of MATH, first note that MATH is well-defined since if MATH is a set of mutually non-commuting elements of MATH, so is MATH. Clearly MATH. Furthermore, if the two sets above are equal, then MATH would have to be one of the MATH. But since MATH commutes with MATH it would have to be MATH, that is, MATH would have to have order MATH. Similarly, all the MATH would have to have order REF. Thus in a group of odd order, MATH would not fix any simplex of MATH or MATH, and hence would not have any fixed points. The final comment is to recall that if MATH were MATH-acyclic, its NAME characteristic would be REF and hence the center of MATH would be trivial and MATH would have to have even order by the facts we have shown above.
math/0005301
Follows from the proof of REF , once we note that left multiplication by a central element of order MATH takes the subcomplex MATH of MATH to itself and that the map MATH maps MATH into itself, as the inverse of an element has the same order as the element. For odd primes MATH, MATH is fixed point free on MATH always as no elements of order REF are involved in MATH.
math/0005301
Since MATH does not commute with any nontrivial element, the center of MATH is trivial and MATH. Furthermore, it is clear that MATH is a cone with MATH as vertex and hence is contractible.
math/0005301
We first recall that the map MATH from REF has order REF as a map of MATH. However it might have fixed points, in fact from the proof of that proposition, we see that MATH fixes a simplex MATH of MATH if and only if each element MATH has order MATH and it fixes the simplex pointwise. Thus the fixed point set of MATH on MATH is nothing other than MATH the MATH-local non-commuting complex for MATH. Since MATH has order REF, we can apply NAME Theory to finish the proof of the first statement of the proposition. The identity on the NAME characteristics follows once we note that under the action of MATH, the cells of MATH break up into free orbits and cells which are fixed by MATH and the fixed cells exactly form MATH.
math/0005301
First, from the condition that MATH for MATH we see that no nonidentity element of MATH commutes with anything outside of MATH. Thus conjugating the picture, no nonidentity element of any MATH commutes with anything outside MATH. Thus if MATH is a complete list of the conjugates of MATH in MATH, we see easily that MATH is partitioned into the sets MATH and two elements picked from different sets in this partition will not commute. Thus it follows that the non-commuting complex based on the elements of MATH decomposes as a join of the non-commuting complexes based on each set in the partition. To complete the picture one notices that each MATH is isomorphic to MATH and so contributes the same non-commuting complex as MATH and furthermore since the conjugates of MATH make up MATH, a simple count gives that MATH. The sentence about MATH follows from the fact that if we define MATH for a simplicial complex MATH, then MATH. Thus since MATH this proves the stated formula concering MATH. Finally when MATH and MATH are abelian, MATH and MATH are just a set of points namely the nonidentity elements in each group. The short exact sequence of the join together with the fact that MATH is simply-connected finishes the proof.
math/0005301
The order of MATH is MATH of course. It is easy to check that there are five NAME REF-subgroups MATH which are elementary abelian of rank REF and self-centralizing, that is, MATH, and are ``disjoint", that is, any two NAME subgroups intersect only at the identity element. Thus the picture for the vertices of MATH is as REF sets MATH of size REF. (Since each NAME REF-group gives REF involutions.) Now since the NAME REF-groups are self-centralizing, this means that two involutions in two different NAME REF-subgroups, do not commute and thus are joined by an edge in MATH. Thus we see easily that MATH is the join MATH. Using the short exact sequence for the join (see page REF), one calculates easily that MATH has the homology of a bouquet of MATH-spheres. Since the join of two path connected spaces MATH and MATH is simply-connected, it follows that MATH is homotopy equivalent to a bouquet of MATH-spheres, and since it is obviously MATH-dimensional, this completes all but the last sentence of the claim. The final sentence follows from REF which says that MATH has the same NAME characteristic as MATH mod REF.
math/0005301
Let MATH be the non-commuting complex associated to the core of MATH. Let us define a facet to be a face of a simplicial complex which is not contained in any bigger faces. Thus the facets of MATH consist exactly of maximal non-commuting sets of noncentral elements in MATH. The first thing to notice is to every facet MATH of MATH, there corresponds a facet MATH of MATH, and further more this correspondence preserves the dimension of the facet (or equivalently the number of vertices in the facet). Since the link of a facet is always empty (a MATH-sphere), in the wedge decomposition of REF , we get the suspension of this empty link as a contribution. If the facet MATH has MATH vertices in it, then we suspend MATH times to get a MATH-sphere. Thus to every maximal non-commuting set of size MATH, we get a MATH sphere contribution from the corresponding facet in MATH. In fact we get MATH-many such spheres from the facet MATH. However above each facet MATH in MATH, there correspond MATH many facets in MATH. Thus in the sum over the facets of MATH stated in the theorem, we divide MATH by MATH in order to count the contribution from the facet MATH in MATH the correct number of times. Notice that MATH if all the centralizer classes have size bigger than one. So if MATH has the property that the size of the centralizer classes of noncentral elements is always strictly bigger than one, for example if MATH is odd or if MATH has a nontrivial center, then MATH for all MATH. In particular MATH whenever MATH has a maximal non-commuting set of size MATH. Also observe that MATH whenever MATH has maximal non-commuting set of size MATH, that is, a singleton consisting of a nontrivial central element (except for the trivial case when MATH has order REF.)
math/0005301
This is because it is easily seen that MATH is a cone on the vertex MATH as everything outside MATH does not commute with MATH as MATH.
math/0005301
The cycle MATH in MATH has MATH by a simple calculation. Thus MATH and so the result follows from REF .
math/0005301
If MATH is an equivalence relation, it is easy to see that the centralizer classes are exactly the MATH equivalence classes. Thus one sees that the non-commuting complex for the core, MATH, is a simplex. Thus in REF , all terms drop out except that corresponding to the maximum face in MATH where the link is empty. This link is suspended to give a sphere of dimension equal to the number of equivalence classes minus one. The number of these spheres appearing in the wedge decomposition is the product MATH where MATH is the size of equivalence class MATH and the product is over all equivalence classes. (Thus this can be zero if one of the equivalence classes has size one, in which case the complex is contractible). In the case of MATH, for MATH a MATH-group, one just has to note that MATH is the centralizer class MATH for any noncentral element MATH.
math/0005301
Follows from a direct interpretation of the inequalities in CITE, page REF . One warning about the notation in that paper is that MATH means the number of vertices in MATH and the empty face is considered a simplex in any complex.
math/0005301
By CITE, if MATH are the components of MATH, then under the MATH-action, MATH acts transitively on the components with isotropy group MATH under suitable choice of labelling. Thus the components are all simplicially equivalent and there are MATH many of them. However we have seen that MATH is MATH-homotopy equivalent to MATH and so we have the same picture for that complex. Thus MATH is the disjoint union of MATH copies of some simplicial complex MATH. Thus by REF , MATH is the MATH-fold join of the dual of MATH with itself. To finish the proof one just has to note that a MATH-fold join of nonempty spaces is always MATH-connected.
quant-ph/0005013
Write the four-qubit state MATH as MATH . The tensor MATH can be regarded as a MATH matrix in three different ways: MATH and the requirement of maximal entanglement is that MATH and MATH should all be unitary matrices. We show first that by local unitary transformations we can arrange that the coordinates of MATH satisfy MATH . To do this, we find normalised states MATH which maximise MATH. Such states certainly exist, since MATH is a continuous function on the compact space MATH. Change basis in each of the single-qubit spaces so that MATH becomes the basis state MATH; then MATH. But if MATH were non-zero we could change basis for states of the first qubit so as to increase MATH; since we have maximised it, we must have MATH. Similarly MATH. The matrices MATH which have to be unitary are now MATH . Hence the following three conditions must hold: CASE: MATH or MATH; CASE: MATH or MATH; CASE: MATH or MATH. Following the branching consequences of these, and requiring that the matrices are unitary, leads to a conclusion of the form that three MATH matrices MATH must all be unitary. In this situation all the matrix elements must be non-zero: if, for example, MATH, then MATH and MATH would all have unit modulus, which is impossible if the second matrix is to be unitary. Now unitarity gives MATH so MATH . But MATH so MATH, which is impossible.
quant-ph/0005018
The convergence of MATH to MATH is understood to use some kind of norm for the density matrices that is continuous in the matrix entries MATH. (The operator norm MATH, for example.) The entropy MATH is a continuous function of the finite set of eigenvalues of MATH. These eigenvalues are also continuous in the entries of MATH.
quant-ph/0005018
This follows from the fact that the fidelity between two mixed states MATH and MATH equals the maximum `pure state fidelity' MATH, where MATH and MATH are `purifications' of MATH and MATH. (See CITE for more details on this.)
quant-ph/0005018
This follows directly from the definition MATH.
quant-ph/0005018
The proof follows from the existence of a universal quantum NAME machine, as proven by CITE. Let MATH be this UTM as mentioned above. The constant MATH represents the size of the finite description that MATH requires to calculate the transition amplitudes of the machine MATH. Let MATH be the state that witness that MATH, and hence MATH for every MATH. With the description corresponding to MATH, MATH can simulate with arbitrary accuracy the behavior of MATH. Specifically, MATH can simulate machine MATH on input MATH with a fidelity of MATH. Therefore, by REF , MATH.
quant-ph/0005018
Consider the quantum NAME machine MATH that moves its input to the output tape, yielding MATH. The proposition follows by invariance.
quant-ph/0005018
This is clear: the universal quantum computer can also simulate any classical NAME machine.
quant-ph/0005018
Take MATH and their minimal programs MATH (and hence MATH). Let MATH be the completely positive, trace preserving map corresponding to the universal QTM MATH with fidelity parameter MATH. With this, we define the following three uniform ensembles: CASE: the ensemble MATH of the original strings, CASE: MATH the ensemble of programs MATH, and CASE: the ensemble of the MATH-approximations MATH, with MATH. By the monotonicity of REF we know that for every MATH, MATH. The chi factor of the ensemble MATH is upper bounded by the maximum size of its strings: MATH. Thus the only thing that remains to be proven is that MATH, for sufficiently big MATH, is `close' to MATH. This will be done by using the insensitivity of the NAME entropy. By definition, for all MATH, MATH, and hence MATH. Because the ensembles MATH and MATH have only a finite number REF of states, we can use REF , to obtain MATH. This shows that for any MATH, there exists a MATH such that MATH. With the above inequalities we can therefore conclude that MATH holds for arbitrary small MATH, and hence that MATH.
quant-ph/0005018
Apply REF to the set of classical strings of MATH bits: MATH for all MATH. All MATH are pure states with zero NAME entropy, hence the lower bound on MATH reads MATH. The average state MATH is the total mixture MATH with entropy MATH, hence indeed MATH.
quant-ph/0005018
All the pure states have zero entropy MATH, hence by REF : MATH. Because all MATH-s are mutually orthogonal, this NAME entropy MATH of the average state MATH equals MATH.
quant-ph/0005018
First we sketch the proof, omitting the effect of the approximation. Consider any qubit string MATH whose minimal-length program is MATH. To produce MATH copies of MATH, it suffices to produce MATH copies of MATH and make MATH runs of MATH. Let MATH be the length of MATH; we call MATH the MATH-dimensional NAME space. Consider MATH, the MATH-fold tensor product of MATH. The symmetric subspace MATH is MATH-dimensional, where MATH. The state MATH sits in this symmetric subspace, and can therefore be encoded exactly using MATH qubits, where the MATH and MATH terms are used to describe the rotation onto MATH. Hence, the quantum NAME complexity of MATH is bounded from above by MATH qubits. For the full proof, we will need to take into account the effect of the imperfect fidelities of the different computations. To achieve a fidelity of MATH, we will compute MATH copies of the minimal program MATH to a fidelity of MATH. On each copy, we simulate the program with fidelity of MATH, and thus obtain the strings MATH (MATH), each of which has (according to REF ) fidelity MATH with the target string MATH. By REF we get a total fidelity of at least MATH. We now proceed to the details of the proof. First we introduce some notation. Assume that for some QTM MATH, MATH, where MATH-computes MATH (with fidelity MATH for any MATH.) Let MATH be the rotation that takes qubit strings MATH to qubit strings of length MATH. More precisely, MATH is the rotation that takes the MATH-th basis state of MATH to the MATH-th classical basis state of the NAME space of dimension MATH. For any fidelity parameter MATH, MATH can be computed efficiently and to arbitrary precision. By that we mean that for any MATH, there is a transformation MATH for which the following holds: Let MATH for some MATH. If MATH, then for each MATH, the mixed state MATH obtained from MATH by tracing out all components that do not correspond to the MATH-th copy of MATH, is such that MATH. We now define the program that witnesses the upper bound on MATH claimed in the theorem. Let MATH be the quantum NAME machine that does the following on input MATH. CASE: Computes MATH. (When MATH is a MATH-proper input, which we specify below, then MATH for some MATH.) CASE: On each `copy' MATH of MATH, runs the QTM MATH. (That is, MATH is the result of tracing out all but the positions of MATH that correspond to the MATH-th block of MATH qubits.) The input MATH is an `MATH-proper input' if for some MATH, MATH. (Note that MATH is exactly MATH, not an approximation up to some fidelity.) If we run the above QTM MATH on input MATH then the output of this MATH is MATH. (Recall that MATH is the length of MATH.) It remains to show the following claims. MATH. The length of the program above for MATH is MATH, where MATH. REF follows immediately from the fact that the total length of the inputs MATH is MATH. We prove REF . Since we chose a precision MATH in REF , MATH, MATH. Furthermore, since the computation at REF introduces at most an error of MATH, MATH, MATH (by REF .) Therefore by REF , MATH. This completes the proof of REF together give us that MATH, where MATH is the length of MATH and an upper bound on its complexity. By invariance, we can conclude that MATH, which proves the theorem.
quant-ph/0005018
Fix MATH and MATH and let MATH be the MATH-dimensional NAME space. Consider the (continuous) ensemble of all MATH-fold tensor product states MATH: MATH, where MATH is the appropriate normalization factor. The corresponding average state is calculated by the integral MATH. This mixture is the totally mixed state in the symmetric subspace MATH (see REF), and hence has entropy MATH. Because all MATH are pure states, we can use REF to prove the existence of a MATH for which MATH.