paper
stringlengths 9
16
| proof
stringlengths 0
131k
|
|---|---|
math/0006158
|
Consider the category whose objects are pairs MATH where MATH is an algebraic MATH-group which is a negatively weighted extension of MATH and where MATH is a continuous, NAME dense representation which lifts MATH. A morphism MATH consists of a homomorphism MATH of MATH-groups that is compatible with the projections to MATH and satisfies MATH. Since the image of MATH is NAME dense, there is at most one morphism between any two objects. In addition, there is fibered product: MATH is defined to be MATH where MATH is the NAME closure of the image of MATH . It follows that this category is a projective system. The weighted completion of MATH with respect to MATH is then MATH, the projective limit of all objects of this category.
|
math/0006158
|
We need only prove the last statement. Since MATH has weights MATH, MATH has weights MATH, and its continuous dual MATH (a direct limit of finite dimensional MATH-modules) has weights MATH. It follows that the space MATH of degree MATH continuous cochains on MATH has weights MATH. Since the bracket is a morphism, the NAME complex MATH of continuous cochains on MATH is a complex of MATH-modules. Its cohomology is therefore a graded MATH-module and thus has a weight filtration induced from that of MATH. Since the MATH-cochains have weights MATH, it follows that the weights on MATH are also MATH.
|
math/0006158
|
The theorem above implies that MATH. Since MATH is pronilpotent, MATH is trivial.
|
math/0006158
|
The condition on MATH implies, by REF , that MATH. The last assertion in REF implies that MATH. The assumption about MATH and the last assertion of REF imply that MATH when MATH, so that MATH. Since MATH is an exact functor, it commutes with homology. So it follows that MATH vanishes. The result now follows from the following lemma.
|
math/0006158
|
By REF , there is a free graded NAME algebra MATH and a graded NAME algebra homomorphism MATH which is surjective and induces an isomorphism on MATH. Denote the kernel of this by MATH. Note that MATH is free if and only if MATH. Since MATH is an ideal in a negatively graded NAME algebra, MATH if and only if MATH. There is a spectral sequence MATH . Since MATH is free, MATH. It follows that MATH is an isomorphism. The result follows as MATH is the graded dual of MATH.
|
math/0006158
|
Since MATH is NAME dense, MATH is surjective. Denote the image of MATH in MATH by MATH. Since the diagram MATH commutes, the image of MATH in MATH is trivial. It follows that MATH is contained in the kernel of MATH and that the composite MATH is trivial. It remains to show that MATH contains the kernel of MATH. Note that MATH is the NAME closure of the image of MATH in MATH. Since MATH is normal in MATH, and since MATH is NAME dense in MATH, MATH is normal in MATH. The homomorphism MATH induces a homomorphism MATH. Because MATH is a negatively weighted extension of MATH, the universal mapping property of MATH gives a splitting MATH of MATH. Since the image of MATH is NAME dense in MATH, MATH has NAME dense image in MATH, which implies that this splitting is surjective, and therefore an isomorphism MATH.
|
math/0006158
|
REF implies that MATH. The result follows.
|
math/0006158
|
We will prove the sufficiency of the condition; necessity is left as an exercise. Since MATH is reductive and acts on the finite dimensional complex MATH of NAME algebra cochains, MATH . In particular, each MATH can be represented by a continuous MATH-invariant cocycle MATH. The NAME algebra MATH can be constructed as MATH with bracket MATH . The obvious action of MATH on MATH preserves the bracket as MATH is MATH invariant.
|
math/0006158
|
This follows directly from REF as the image of MATH is open.
|
math/0006158
|
This follows directly from REF , and REF .
|
math/0006158
|
Fix MATH. By replacing MATH by MATH for some MATH, we may assume that MATH is unipotent. By replacing MATH by MATH, we may assume that MATH is an inclusion. We will prove the result by induction on the dimension of MATH. First recall that if MATH is a unipotent group over a field MATH of characteristic zero, and MATH is a closed normal subgroup of MATH, also defined over MATH, then the group of MATH-rational points of MATH is isomorphic to MATH. We first consider the case where MATH is abelian. In this case, MATH is a compact subgroup of MATH. Since MATH is a PID, and since MATH is compact and torsion free, it is freely generated by MATH linearly independent elements of MATH. It follows that MATH is NAME dense in MATH if and only if MATH. This proves the result when MATH is abelian. Now suppose that MATH is not abelian. Note that the commutator subgroup MATH is NAME dense in MATH; it is clear that the NAME closure of MATH is contained in MATH. The reverse conclusion follows as the image of MATH in the MATH-rational points of the quotient of MATH by the closure of MATH is dense and abelian, which implies that MATH . If MATH is not abelian, then there is a least MATH for which MATH. Since the filtration MATH is central, MATH. Using the fact that MATH is NAME dense in MATH and also the fact that the result holds for MATH, it is not hard to see that MATH is NAME dense in MATH. By induction, the result holds for MATH from which the result for MATH follows as MATH .
|
math/0006158
|
We will use REF and its notation. Set MATH. Since MATH is a finitely generated pro-MATH group, MATH and MATH is a finite MATH-group. Also, MATH for all MATH if and only if MATH is torsion free for all MATH. Using universal mapping properties, one can show that the inclusion MATH is the MATH-adic unipotent completion of MATH. It follows that the natural homomorphism MATH descends to a natural homomorphism MATH which is injective. This induces a homomorphism MATH . The kernel of this morphism is finite. This is equivalent to the assertion that MATH has finite index in MATH, which follows from the compactness of MATH (MATH is finitely generated as a pro-MATH group) and the openness of MATH in MATH, a consequence of REF . Since MATH is a finite MATH-group, it follows that the kernel of both MATH are finite from which it follows that MATH is a finite index subgroup of MATH. Because the sequence MATH is exact, the kernel and cokernel of MATH are finite. Since MATH and MATH, MATH and MATH are abelian pro-MATH groups whenever MATH. Thus it follows that the kernel and cokernel of the mapping on each associated graded is a finite abelian MATH-group when MATH.
|
math/0006158
|
The first assertion follows from the assumption using the natural isomorphism MATH. Since the bracket MATH is surjective and MATH equivariant, it follows that MATH acts on MATH via the MATH-th power of the cyclotomic character. This implies the second and third assertions as it implies that every derivation of MATH that acts trivially on MATH has negative weight. The fourth follows from the universality of MATH. Because MATH is a module over MATH, it has a natural weight filtration. Since MATH acts on MATH via the MATH-th power of the cyclotomic character, next assertion follows. The remaining assertions follow as MATH and MATH are negatively weighted MATH-modules.
|
math/0006158
|
The surjection MATH restricts to a surjection MATH, which implies that MATH has NAME dense image. By strictness, it induces a surjection MATH. From REF , it follows that MATH is the inverse image of MATH under the natural mapping MATH. There is thus an injection MATH . REF imply that this induces isomorphisms MATH whenever MATH.
|
math/0006158
|
Since MATH is isomorphic to the free NAME algebra generated by MATH, it follows that MATH. By REF we have MATH . The non-degeneracy of the NAME bracket MATH implies that the mapping MATH, given by taking inner derivations, is an isomorphism. Thus MATH .
|
math/0006158
|
This follows directly from CITE - see REF .
|
math/0006158
|
Because of the previous lemma, we only need compute MATH. It is proved in CITE that there is a NAME type spectral sequence MATH . Since MATH is reductive and MATH, it follows that MATH . Since MATH is negatively weighted, the right hand side is zero when MATH. Since MATH is free and MATH is exact, MATH when MATH and MATH vanishes if MATH. When MATH, REF gives the result. For MATH, the assertion is obvious. The last assertion follows from REF and the fact that the regulator mappings MATH are isomorphisms for all MATH and MATH. This is due to CITE when MATH and CITE when MATH.
|
math/0006158
|
The first statement follows because MATH classifies extensions of MATH by MATH with property MATH. Since MATH is negative, each extension of MATH by MATH is a weighted MATH-module and hence a MATH-module. It therefore determines an element of MATH. The second assertion follows using the NAME Spectral Sequence CITE for the extension MATH and the fact that MATH is reductive, which implies that MATH vanishes whenever MATH. We also use the fact that MATH.
|
math/0006158
|
As in the proof of REF , the NAME spectral sequence implies that MATH. It thus suffices to prove that the natural mapping MATH is injective. Let MATH be a class in the left hand side corresponding to a two-step extension MATH . Since all MATH-modules are locally finite, we may assume MATH to be finite dimensional. By exactness of the weight filtration and the reductivity of MATH, we may assume that MATH, MATH, and that MATH, MATH. We assume that MATH is a trivial class as MATH-module. Then CITE implies the existence of a MATH-module MATH with MATH and MATH. Now the bootstrap property says that MATH has property MATH, and hence is a weighted MATH-module with property MATH. Thus this is a MATH module, which says that MATH is the trivial class as an extension of MATH-modules, too. Thus injectivity is proved.
|
math/0006158
|
By REF we have an injection. MATH . Since the right hand side vanishes, MATH is free by REF . The computation of MATH follows from REF and the facts (compare REF) that MATH when MATH and that there is an inclusion MATH corresponding to MATH via NAME characters. Thus, MATH where MATH is defined as in REF .
|
math/0006158
|
First recall the elementary fact that the set of torsion elements of a nilpotent group MATH forms a characteristic subgroup MATH and that MATH is torsion free. This implies that the set MATH of elements of MATH that are torsion modulo MATH is a characteristic subgroup, that MATH is a torsion free nilpotent group, and that MATH is a torsion group. It follows that the central series MATH is the most rapidly descending central series with each MATH torsion free. Let MATH. Since MATH is finite dimensional, each MATH is unipotent. Since unipotent groups are torsion free, and since MATH is a subgroup of MATH, it follows that MATH. It follows from NAME 's version CITE of NAME 's Theorem CITE that there is a unipotent MATH-group MATH that contains MATH as a NAME dense subgroup. The density implies that the length of MATH is MATH. The homomorphism MATH induces a homomorphism MATH which factors through MATH as MATH is unipotent of length MATH. It follows that MATH, and that MATH for all MATH, which implies the result.
|
math/0006158
|
This follows by a proof that is essentially the same as that of REF , taking the continuity of MATH into consideration in the current setting. The following lemma is a key ingredient.
|
math/0006158
|
The proof is similar to the proof of REF. Since MATH is finitely generated, there is a central filtration of MATH by closed normal subgroups, each of whose graded quotients is either MATH or MATH. We prove the lemma by induction on the length of this filtration. There is an exact sequence MATH where MATH is MATH or MATH. By induction, MATH is embedded in its MATH-adic unipotent completion MATH, which is algebraic by induction. If MATH, then MATH. The conjugacy action of MATH on MATH lifts by functoriality to a unipotent action on the NAME algebra of MATH, and thus extends to a unipotent action of MATH on MATH. It follows that MATH can be continuously embedded in the unipotent group MATH. If MATH, then choose MATH whose image generates MATH. Then MATH, and since MATH is uniquely divisible, there is a unique MATH with MATH. Then MATH can be imbedded in MATH by mapping MATH to MATH.
|
math/0006158
|
We may assume that MATH is the upper triangular unipotent subgroup of MATH for some MATH. We denote by MATH the group of matrices whose MATH-th entry lies in MATH when MATH, is MATH when MATH, and MATH when MATH. Since MATH is finitely generated, the image of MATH is contained inside MATH for some MATH. The filtration MATH is a basic set of neighbourhoods of the identity in MATH; each quotient is a finite group of MATH-power order. Since MATH, the inverse image of each MATH is a finite index subgroup of MATH of MATH-power order. The result follows.
|
math/0006158
|
It follows from the lemma that MATH is continuous and induces continuous homomorphisms MATH . By the universal mapping property of MATH-adic unipotent completion, these induce homomorphisms MATH . But the homomorphism MATH induces a homomorphism MATH whose composite with REF is the canonical isomorphism of REF . This completes the proof as the image of MATH is NAME dense.
|
math/0006158
|
Suppose that MATH is an automorphism of MATH. Since MATH, and since the bracket mapping MATH is surjective and commutes with MATH, we see that MATH acts trivially on MATH if and only if it acts trivially on MATH. It follows that the kernel MATH of MATH is a prounipotent group. The last statement follows from standard NAME theory as MATH.
|
math/0006158
|
This follows from REF as there is a natural action of MATH on MATH.
|
math/0006158
|
The short exact sequence MATH gives a long exact sequence CITE MATH . Since MATH is compact, the proof of CITE shows that MATH is torsion. So, after tensoring with MATH, we have an isomorphism MATH . When MATH, MATH is finite. This is obvious when MATH. When MATH it follows from class field theory. Indeed, reduce to the case where MATH has trivial NAME action by passing to a finite extension. Then MATH is MATH, which is finite. The corollary to CITE then implies that MATH . NAME with MATH completes the proof.
|
math/0006158
|
Let MATH and MATH. By CITE there is a long exact sequence MATH where MATH denotes étale cohomology with support on MATH. Thus it is enough to show that MATH vanishes for all MATH unless MATH or MATH. Let MATH denote the henselization of MATH at MATH. Then by CITE we have MATH so it suffices to establish the vanishing of the right hand side. Set MATH and note that this is just MATH. By a second application of the long exact sequence CITE applied to MATH, we see that if MATH and MATH both vanish when MATH, then MATH vanishes. Firstly, by CITE, we have MATH . As in REF , the right hand group is trivial unless MATH. Set MATH, where MATH is the maximal algebraic extension of MATH unramified at MATH. To prove the second vanishing we consider the NAME spectral sequence (compare CITE) MATH . According to CITE, MATH has cohomological dimension at most REF, so that MATH . Since the pro-MATH abelianization of the inertia group MATH is MATH as a NAME module, we have MATH . The required vanishing follows by plugging this, and the fact that MATH is MATH, into the spectral sequence.
|
math/0006159
|
The sketch of the proof is as follows: basically, the claim follows from REF , which implies that the denominator of any MATH in the standard basis of MATH is uniformly bounded, whence the period of the MATH-expansion of MATH is bounded as well.
|
math/0006159
|
It suffices to show that if MATH is weakly finitary, then MATH in question does exist. Let MATH be weakly finitary; then for any MATH there exists MATH such that MATH. Let MATH has the MATH-expansion MATH and MATH . Without loss of generality we may regard MATH to be greater than the preperiod MATH the period of the sequence MATH (as MATH is not necessarily the smallest period of MATH). Since MATH can be made arbitrarily small, we may fix it such that MATH . Put MATH. Let MATH. By REF the MATH-expansion of MATH is eventually periodic, and splitting it into the preperiodic and periodic parts, we have MATH. Let for simplicity of notation MATH (the whole picture is shift-invariant). It will suffice to check the condition for MATH. Put MATH. Then MATH . The expression in brackets in REF belongs to MATH and so does the second term. In view of REF and the definition of MATH the whole sum in REF belongs to MATH as well, because by our choice of MATH we have necessarily MATH. Since MATH is finite and the construction depends on MATH only, we are done.
|
math/0006159
|
Let MATH denote the distance to the closest integer, MATH be the MATH-coordinate of MATH and MATH denote the linear transformation of MATH defined by the matrix MATH. Let MATH be the mapping acting from MATH into MATH by the formula MATH . Then by REF , MATH where MATH. Therefore, it suffices to show that the diameters of the sets MATH are uniformly bounded for all MATH. We have (recall that MATH): MATH where MATH is the maximum of the absolute values of the conjugates of MATH that do not coincide with MATH. This proves the lemma.
|
math/0006159
|
By the well-known result, for any NAME MATH and MATH (where MATH denotes the trace of an element MATH of the extension MATH, that is, the sum of all its NAME conjugates) - see, for example, CITE. Since MATH is a unit, MATH implies MATH, whence MATH . Thus, if we regard MATH as a lattice over MATH, then by REF , MATH is by definition the dual lattice for MATH. Hence by the well known ramification theorem (see, for example, CITE) the equality REF follows with MATH, where MATH.
|
math/0006159
|
We have MATH, where MATH. As was mentioned above, the dimension of the unstable foliation MATH is REF, whence MATH, and since MATH, we have MATH, that is, MATH. Now the claim of the lemma follows from REF .
|
math/0006159
|
Suppose MATH is fundamental. Then the homoclinic point MATH whose MATH-coordinate is MATH can be represented as a finite linear integral combination of the powers MATH, that is, MATH whence MATH. Therefore, MATH is invertible in the ring MATH. Conversely, if MATH, then using the same method, we show that the claim of the lemma follows from the fact that the equation MATH always has the solution in MATH, namely, MATH.
|
math/0006159
|
By REF , MATH is finite and since it is shift-invariant, it must contain purely periodic sequences only. Let MATH. Then by REF , MATH as MATH, whence from REF , therefore, MATH, and MATH, because MATH.
|
math/0006159
|
Let MATH denote the set of all partial limits of the collection of sequences MATH, where MATH is the sequence MATH whose ``value" is MATH. It suffices to show that MATH. Let MATH; by definition, there exists a sequence of positive integers MATH such that MATH . Then MATH, and we are done.
|
math/0006159
|
Let the mapping MATH be given by REF and MATH. Let MATH and MATH . The NAME Separation Lemma CITE says that there exists a constant MATH such that if MATH and MATH are two sequences in MATH and MATH, then MATH. Hence MATH where MATH denotes the NAME measure on MATH. Since for any MATH is equivalent to MATH and the corresponding density is uniformly bounded away from MATH and MATH (see CITE), we have for some MATH, MATH whence by the fact that MATH is one-to-one except for a countable set of points, MATH and the claim of the lemma holds with MATH.
|
math/0006159
|
The claim follows from the definition of MATH (see Introduction) and the fact that the positive root MATH of the equation MATH is the smallest NAME number CITE. Indeed, MATH and MATH is a subshift of finite type, namely, MATH . Now the desired claim follows from CITE asserting that if MATH, then MATH.
|
math/0006159
|
We have MATH . Since MATH we have MATH . Now by REF , MATH being weakly finitary (see REF ) and the deifnition of MATH we have MATH for any MATH. Hence MATH and from REF we finally obtain the estimate MATH whence one can take MATH, and REF is proven.
|
math/0006159
|
We showed that MATH, whence MATH.
|
math/0006159
|
We already know that any bijective arithmetic coding is parametrised by a fundamental homoclinic point. Let MATH be such a point for MATH; then any other fundamental homoclinic point MATH satsifies MATH, where MATH and MATH are the corresponding MATH-coordinates and MATH - the proof is essentially the same as in REF . On the other hand, if MATH is a bijective arithmetic coding for MATH, then as easy to see, MATH is a toral automorphism commuting with MATH (this mapping is well defined almost everywhere on the torus, hence it can be defined everywhere by continuity). Finally, if MATH and MATH, then MATH belongs to MATH and commutes with MATH and vice versa.
|
math/0006159
|
Let MATH denote the bijective arithmetic coding for MATH parametrised by MATH and MATH. If MATH, then one can consider the mapping MATH; it will be well defined on the dense set and we may extend it to the whole torus. By the linearity of both maps, MATH is a toral endomorphism. Thus, we have MATH . Let MATH is given by the formula MATH. For the basis sequence MATH with the unity at the first coordinate we have MATH . Therefore, by the linearity and continuty, we have MATH. As MATH is REF-to-REF a.e., MATH will be MATH-to-REF a.e. with MATH. By definition, MATH is the determinant of the matrix of the multiplication operator MATH in the standard basis of MATH, whence MATH, because MATH is given by the companion matrix. Finally, MATH, as by the result from CITE, MATH whenever MATH is as in REF .
|
math/0006159
|
Let MATH be written column-wise as follows: MATH. Then by REF and the definition of MATH, MATH whence by the fact that MATH, we have MATH for MATH.
|
math/0006159
|
Let MATH, where MATH is a certain bijective arithmetic coding for MATH. Then MATH is a linear mapping from MATH onto itself defined a.e.; let the same letter denote the corresponding toral endomorphism. Then MATH. Therefore the matrix MATH of the endomorphism MATH satisfies REF , whence by REF , MATH for some MATH. Hence MATH, and we are done.
|
math/0006159
|
REF : see REF ; REF : see REF ; REF : also follows from REF ; REF : it is obvious that MATH satisfies this property (take MATH). Hence so does any MATH which is conjugate to MATH.
|
math/0006159
|
By REF , the definition of MATH and the NAME Theorem, MATH .
|
math/0006159
|
We have MATH. The case MATH thus leads to REF yields REF .
|
math/0006159
|
Since MATH and MATH are conjugate, they have one and the same characteristic polynomial. By the definition of MATH we have MATH which is equivalent to REF .
|
math/0006161
|
Consider a monad MATH in MATH. Let MATH be the (object part of) the coinserter of the pair MATH: MATH . Notice that REF-cell MATH sets up a lax cocone over MATH as follows: MATH . But this lax cocone need not satisfy the equations involving MATH and MATH, so we must enforce them. For the unit, pasting MATH and MATH, we obtain REF-cell MATH, which by (contravariant) local full and faithfullness of MATH, corresponds to a (unique) REF-cell in MATH, written MATH. Consider the coequifier of this REF-cell and the identity: MATH . We obtain thus a new cocone MATH with MATH. Again, we must impose on this data the condition for MATH. We have two REF-cells MATH . Using the following comma square MATH the above REF-cells determine, by adjoint transposition, REF-cells MATH which by (contravariant) local full and faithfullness of MATH, correspond to two parallel REF-cells MATH, which we coequify: MATH so that MATH is the NAME object, with universal lax cocone MATH and MATH. Universality and preservation of representability follow at once. MATH .
|
math/0006161
|
CASE: Recall that for a given object MATH we have the string of adjunctions MATH . For MATH, cartesianness implies that MATH is the comma object MATH. Thus since MATH is left adjoint to MATH, also MATH is left adjoint to the projection MATH; and so MATH is a split fibration. The same argument in MATH yields the split cofibration structure. CASE: Recall that MATH is the comma-object MATH and therefore the naturality square induces a canonical morphism MATH in MATH. Since MATH is a split cofibration, the canonical morphism MATH has a left adjoint MATH over MATH. Considering the following diagram MATH where both squares and the outer rectangle are pullbacks, we see that MATH is the pullback of MATH along MATH, and therefore has a left adjoint MATH. This adjunction is taken by the reflection sec:preliminaries. REF to an isomorphism MATH which corresponds to MATH. MATH .
|
math/0006161
|
CASE: MATH: Given a bimodule MATH define the invertible REF-cell MATH as the pasting composite MATH where the left isomorphism is that of REF and the right one is given by the homomorphism MATH applied to the naturality square for MATH. CASE: MATH: Use the same argument in MATH. MATH .
|
math/0006161
|
CASE: The square is a pullback and the lower arrow is a split monic (MATH), hence the top arrow is monic. CASE: Given `global generic' elements MATH, we must stablish a REF correspondence MATH . REF-cells into MATH are classified by REF into MATH and those into MATH are classified by REF into MATH. In the following diagram MATH both squares are pullbacks and MATH is monic, which yields the desired correspondence. MATH .
|
math/0006161
|
We must first verify that MATH is indeed a monad. This follows by a routine calculation using the unit and associativity axioms for a lax algebra REF . The normality condition is straightforward. It is furthermore clear that starting with a normal monad MATH we obtain the data for a normal lax algebra by taking adjoint transposes of MATH and MATH, and this correspondence is inverse to MATH. As for morphisms, to give MATH between monads is the same as to give a morphism of bimodules MATH (by change-of-base). But MATH, and so we have the following correspondence MATH which sets up the bijective correspondence between morphisms of lax algebras MATH and morphisms of monads MATH. As for REF-cells, given one in MATH, MATH for parallel morphisms MATH, we get one in MATH by precomposing with the adjoint transpose of MATH: MATH and in the opposite direction, given MATH the composite MATH where the isomorphism is that of pseudo-naturality of MATH, yields a REF-cell MATH and therefore a REF-cell MATH in MATH which is furthermore well-defined as a REF-cell in MATH. These correspondences of REF are readily verified to be mutually inverse. MATH .
|
math/0006161
|
CASE: MATH CASE: MATH . The monad axioms follow from those of MATH and pseudo-naturality of MATH. MATH .
|
math/0006161
|
CASE: Given MATH, define MATH as follows: MATH CASE: Given a MATH-algebra MATH, we have a lax cocone MATH where MATH is the adjoint transpose of MATH (associativity for MATH). Thus we have an induced morphism MATH from the NAME object MATH. Once again, associativity for MATH and universality imply that this induced morphism is a MATH-algebra morphism from MATH to MATH, which is the desired counit MATH. To show MATH fully faithful, first let us notice that the underlying morphism MATH in MATH is fully faithful as shown by the following diagram: MATH . Finally, full and faithfulness of MATH follows similarly, using that of MATH (. REF ). MATH .
|
math/0006161
|
Since the counit of the required adjunction is the identity, we must simply define the unit. Notice that MATH and MATH is the morphism of algebras MATH uniquely determined in MATH by the lax cocone MATH given by the adjoint transpose of MATH, while the underlying morphism of MATH is MATH. To give a REF-cell MATH amounts to give, by REF-dimensional universal property of the NAME object, a modification MATH between the respective lax cocones. To define such modification, notice that the adjoint transpose MATH in MATH corresponds to a REF-cell MATH. Define MATH as the composite MATH . The verification that REF-cell so defined is indeed a modification is quite delicate, so we outline the details. The equation MATH amounts to the equality (omitting objects to simplify the diagram) MATH the upper pasting can be simplified using the fact that MATH is a lax cocone for the monad MATH: MATH where MATH is the multiplication for the monad MATH, . CASE: So we will be done if we can show the following equality: MATH as pasting both sides with MATH gives the desired equality. The trick to prove this latter equality is to realise the morphisms of spans which induce these REF-cells. The left one is induced by MATH while the right one is induced by MATH where in turn MATH using cartesiannes of MATH. Finally, naturality shows MATH and thus the morphisms of bimodules induced from those of spans are equal, as desired. To conclude the proof, we must verify the adjunction equations for REF-cell MATH induced by the modification MATH, namely CASE: MATH, which is immediate by the definition of the lax cocone with REF-cell MATH which induces MATH. CASE: MATH, which is established using MATH (lax cocone condition for the unit of the monad MATH). Finally, universality of (representable) NAME objects guarantees that MATH is a well-defined REF-cell in MATH. MATH .
|
math/0006161
|
CASE: Let MATH have an adjoint-pseudo-algebra structure MATH, with MATH and MATH the unit and (invertible) counit of the adjunction MATH. Recall that the underlying morphism of MATH is MATH. Define MATH. We now intend to show that MATH represents MATH. Define MATH as the pasting MATH . We claim MATH is an isomorphism: its inverse MATH is given by the pasting MATH because the counit is a REF-cell in MATH and MATH by the adjunction equations for MATH (since MATH). We have thus shown that MATH is representable. To see that its structure REF-cell MATH is an isomorphism, we use the fact that MATH commutes which the structural REF-cells of these lax algebras. We have therefore MATH . Since MATH is an isomorphism, so is MATH, as well as MATH by adjointness, because MATH is an isomorphism and therefore MATH. From this adjointness we also conclude that the pasting of MATH and MATH is MATH. Hence, MATH . Thus, the equality of the pasting diagrams above imply that MATH is an isomorphism and so is MATH. Therefore, MATH is an isomorphism. Finally, MATH being an isomorphism allows us to conclude that MATH is one as well. CASE: Given MATH, the isomorphisms MATH and MATH endow MATH with a pseudo-MATH-algebra structure. We should therefore show that any such pseudo-algebra does endow MATH with a pseudo-MATH-algebra structure (which a fortriori would be an adjoint one, by REF ). Let MATH and MATH be the structural isomorphisms. The adjoint transpose of MATH produces a lax cocone MATH and we get an induced morphism MATH, which is equipped with structural isomorphisms CASE: MATH CASE: MATH uniquely determined by the given MATH and REF-dimensional universal property of the NAME objects involved. Furthermore, universality of NAME objects ensure that the induced canonical isomorphisms satisfy the pseudo-MATH-algebra axioms. MATH .
|
math/0006161
|
We must verify the compatibility with units and multiplications: CASE: MATH follows immediately from the definition of MATH that MATH (for MATH). CASE: MATH we have by definition of MATH that the right pasting instantiated at MATH is the upper morphism in the following commuting diagram, MATH while the bottom map is the corresponding instance of the left pasting. MATH .
|
math/0006161
|
The details are essentially contained in the proof of REF . Given a (pseudo-)MATH-algebra MATH (with structural isomorphisms MATH and MATH) we have MATH by definition of MATH. In the other direction, MATH is an isomorphism of (pseudo-)MATH-algebras. MATH .
|
math/0006161
|
Let MATH be an endomorphism in MATH with MATH a monad. Let MATH be the NAME object of MATH in MATH. Since MATH preserves it, there is a uniquely determined morphism MATH mediating between lax cocones in the following diagram: MATH . Uniqueness of mediating morphisms between lax cocones means that the algebra axioms for MATH follow from those of MATH. The above lax cocone in MATH is the NAME object of MATH. To verify its universality, consider another lax cocone MATH we have a mediating morphism MATH induced by universality of MATH and a REF-cell MATH induced by REF universal property of MATH, which furthermore guarantees the validity of the axioms making the above diagram a morphism in MATH. REF-dimensional property of the NAME object follows similarly. MATH .
|
math/0006161
|
Given a monad MATH in MATH, we transform it into a monad MATH in MATH. We claim that the resulting representable NAME object MATH in MATH yields one for MATH in MATH. Indeed, we obtain a lax cocone for MATH as follows MATH the latter correspondence by MATH being locally fully faithful. The universal property of such a representable NAME object in MATH restricts appropriately to MATH thanks to the preservation of representability, . CASE: Preservation by MATH is now evident, since the action of MATH on maps (representable bimodules) is simply that of MATH on the corresponding morphisms of MATH. MATH .
|
math/0006161
|
Notice first that for a given pseudo-MATH-algebra MATH, the (morphism part of the) monad MATH corresponds to a morphism in MATH. Now, our REF allows to apply REF . The corresponding NAME object yields the required left biadjoint by virtue of the analysis preceeding REF , while this latter guarantees that the unit of the biadjunction is an equivalence, as required. MATH .
|
math/0006161
|
For an internal category MATH, the corresponding `hom' bimodule is MATH and therefore MATH. MATH .
|
math/0006161
|
We define an explicit inverse MATH to the given REF-functor. Given a multicategory MATH we obtain a category MATH by pulling back along MATH: MATH . CITE. We then construe MATH itself as a bimodule MATH, whose action is obtained by `restriction' of the composition operation of MATH. Furthermore, the monad structure of MATH as a multicategory endows MATH with a monad structure in MATH, which is normal by the very definition of MATH. Given a morphism of multicategories MATH we obtain an internal functor MATH (since MATH is a REF-functorial construction) and a morphism of bimodules MATH commuting with the monads structures. Finally, a REF-cell MATH gives a REF-cell MATH, namely MATH, which being a morphism of MATH is also a morphism of MATH. MATH .
|
math/0006161
|
Given a monad MATH, we obtain via the lax functor MATH a monad MATH in MATH which is therefore an internal category. This category MATH is the (vertex of the lax cocone of the) NAME object of MATH. To define a lax cocone MATH REF-cell MATH yields an identity-on-objects functor MATH: MATH the description of MATH in the proof of REF . REF-cell MATH is given by MATH since MATH corresponds simply to the span MATH, because MATH is identity-on-objects (see the description of MATH above). For universality, given another lax cocone MATH we can take the mediating bimodule MATH to be MATH itself, as we have a split coequalizer: MATH . Clearly, if MATH is representable, so is the induced MATH and we can force this property to be REF-functorial. MATH .
|
math/0006165
|
The decomposition MATH determines a projection MATH such that MATH and MATH is also a projection, MATH. The same for MATH and MATH. The inclusion MATH gives MATH, that is, MATH; similarly, MATH. So, our projections commute with each other. Introducing MATH we have MATH; that is, MATH is also a projection, and MATH. It follows that MATH. So, the relation MATH holds in the topological sense (that is, when MATH is treated as a linear topological space). The following lemma completes the proof.
|
math/0006165
|
We introduce a Gaussian type space MATH and identify MATH with MATH. Subspaces MATH generate sub-MATH-fields MATH. Note that MATH generates MATH, the least MATH-field containing both MATH and MATH. The following two lemmas (and one definition) complete the proof.
|
math/0006165
|
`Only if': take a Gaussian measure MATH such that MATH makes MATH orthogonal, then MATH makes MATH independent. `NAME: some MATH makes MATH independent; however, MATH need not be Gaussian. We take any Gaussian measure MATH and introduce MATH-measurable densities MATH . The measure MATH still makes MATH independent, and MATH. So, with respect to MATH the spaces MATH are independent Gaussian spaces. Therefore their sum MATH is Gaussian with respect to MATH, that is, MATH.
|
math/0006165
|
We have MATH such that MATH and MATH are MATH-independent, while MATH and MATH are MATH-independent. Consider the MATH-measurable density MATH and the measure MATH, then MATH. The MATH-independence of MATH and MATH implies their MATH-independence. For every MATH, MATH, MATH .
|
math/0006165
|
We identify MATH with MATH of a Gaussian type space MATH and use the natural homeomorphism MATH between the space MATH of admissible norms and the space MATH of Gaussian measures. Subspaces MATH generate corresponding sub-MATH-fields MATH. The set MATH corresponds to MATH, where MATH consists of all measures making MATH independent. The following two lemmas complete the proof.
|
math/0006165
|
We have to prove that every MATH is measurable with respect to MATH. It suffices to prove that the set MATH belongs to MATH for every MATH such that the set MATH is negligible (indeed, such MATH are dense in MATH). We take MATH such that MATH in MATH, consider sets MATH and note that MATH and MATH.
|
math/0006165
|
Assume that MATH is nonempty; we have MATH such that MATH and MATH are MATH-independent. Let MATH be a finite set and MATH a MATH-measurable function satisfying MATH. Take MATH-measurable functions MATH such that MATH in MATH and MATH. The MATH-independent MATH-fields MATH are also MATH-independent, where MATH (since MATH). Thus MATH and MATH. However, such measures MATH (for all MATH and MATH) are dense in MATH. So, MATH, which is REF . Also, the MATH-independence of MATH implies MATH. However, MATH, and similarly MATH. So, MATH for all MATH of a dense set, therefore for all MATH, which is REF .
|
math/0006165
|
We choose an admissible norm on MATH, thus turning MATH into a NAME space. REF becomes MATH that is, MATH where the angle is defined by MATH. It is equivalent to MATH where MATH runs over finite-dimensional subspaces, and MATH. The following lemma completes the proof, provided that MATH tends to MATH slowly enough. Namely, in terms of MATH introduced there, it suffices that MATH.
|
math/0006165
|
We equip MATH with the norm MATH, thus turning MATH into a NAME space, and consider orthogonal projections MATH onto MATH respectively. Introduce subspaces MATH, MATH, MATH (here MATH is the orthogonal complement of MATH); the subspaces are orthogonal to each other, and invariant under both MATH and MATH. Therefore MATH where MATH is another subspace invariant under MATH (since these operators are Hermitian). Introduce MATH, MATH, then MATH for all MATH (since MATH commutes with MATH), and MATH. We may get rid of MATH by letting MATH . In other words, we'll construct MATH on MATH while preserving both the given norm on MATH and the orthogonality of MATH. Now we forget about MATH, assuming that MATH, MATH, MATH. So, we have MATH, MATH, MATH, MATH. The latter implies MATH. Similarly, MATH. Therefore MATH is finite-dimensional, MATH. Both MATH and MATH commute with the Hermitian operator MATH. The spectrum of MATH consists of some numbers MATH of multiplicity REF (though, some MATH may coincide), and MATH (the case MATH is excluded by MATH; the case MATH is excluded by MATH, MATH, MATH). Accordingly, MATH decomposes into the (orthogonal) direct sum of planes, MATH, MATH, invariant under MATH. Subspaces MATH, MATH are two lines on the plane MATH, and MATH; MATH; MATH. Clearly, MATH . We construct MATH on each MATH separately, while preserving their orthogonality. Elementary REF-dimensional geometry shows that the corresponding numbers MATH (two numbers for each plane) are, in the optimal case, MATH . The corresponding angle MATH between MATH and MATH is given by MATH therefore MATH here MATH, MATH.
|
math/0006165
|
We choose an admissible norm on MATH, thus turning MATH into a NAME space. Let MATH be a finite-dimensional subspace, MATH. For any given MATH consider the pair MATH. Its geometry may be described (similarly to the proof of REF ) via angles MATH, MATH. This time, zero angles are allowed, since MATH need not be MATH. It may happen that MATH, since MATH need not be MATH. However, MATH for large MATH we have MATH which implies MATH and MATH. We may send MATH into MATH rotating it by MATH. In other words, there is a rotation MATH such that MATH . We choose subspaces MATH such that MATH and MATH. Introduce MATH then MATH for each MATH. On the other hand, introduce MATH . Similarly to the proof of REF we have MATH for each MATH, therefore MATH if MATH tends to MATH slowly enough. However, we choose MATH, MATH so as to satisfy a stronger condition: MATH . We take MATH where rotations MATH satisfy MATH and MATH. Then MATH, and MATH due to REF , MATH is asymptotically orthogonal to MATH. We take admissible norms MATH such that MATH is orthogonal to MATH with respect to MATH. Consider the orthogonal complement MATH of MATH with respect to MATH; clearly, MATH is asymptotically orthogonal to MATH. We have MATH and MATH (in the NAME). On the other hand, MATH and MATH. REF states that the subspace MATH satisfies MATH and MATH (and also MATH).
|
math/0006165
|
Similarly to the proof of REF, we take MATH such that MATH are MATH-independent and MATH. We consider an arbitrary finite set MATH, an arbitrary MATH-measurable function MATH, and the corresponding vector MATH. We construct MATH-measurable functions MATH such that MATH in MATH, and corresponding vectors MATH. Independence of MATH with respect to MATH means that MATH, therefore MATH, where MATH. However, MATH, since MATH. Also MATH, therefore MATH. So, MATH, and MATH. It remains to note that such vectors MATH (for all MATH and MATH) are dense in MATH.
|
math/0006165
|
REF gives us MATH such that MATH and MATH (both in the FHS sense), and MATH, and MATH is asymptotically orthogonal to MATH. REF gives us a representation MATH for an arbitrary unit vector MATH; here MATH are unit vectors of MATH, MATH are unit vectors of MATH, and these MATH (unlike MATH) do not depend on MATH. We have MATH. However, the density matrix MATH on MATH that corresponds to the vector MATH is the same as the density matrix on MATH that corresponds to the vector MATH. The vector does not depend on MATH, therefore MATH does not depend on MATH. So, MATH.
|
math/0006165
|
Let MATH, MATH, MATH, then MATH (the notation being self-explanatory), which implies MATH. For every MATH however, MATH and MATH (since MATH); we have MATH. Note that MATH, since MATH for all MATH. We have MATH so, MATH . Denote MATH. We need only one ray of vectors MATH such that MATH. For every MATH there exists such MATH, satisfying MATH and MATH, which gives MATH; the supremum over MATH gives the first inequality. For the second inequality, the proof is quite similar. Only MATH and MATH change places, and MATH appears instead of MATH, which leads to MATH instead of MATH.
|
math/0006165
|
REF gives REF . Assume REF ; we have to prove REF . We choose a Gaussian measure MATH and apply REF to MATH: MATH which implies that MATH and MATH (when MATH) for all MATH, MATH (the dual to MATH). The latter, MATH, shows that MATH. The former, MATH, shows that MATH.
|
math/0006165
|
MATH, therefore MATH is a NAME subset of the standard NAME space MATH.
|
math/0006165
|
The projection is the limit (for MATH) of the orthogonal projection of MATH to MATH. Measurability of the latter implies that of the former.
|
math/0006165
|
The relation MATH means that, first, MATH are orthogonal in some admissible norm, and second, MATH is dense in MATH. The latter condition defines a measurable set of pairs MATH due to REF . The former condition may be expressed in terms of the infinite matrix MATH where MATH are as in REF. The relevant set of matrices is NAME measurable.
|
math/0006165
|
One may take MATH here MATH is equal to MATH if MATH and MATH otherwise.
|
math/0006165
|
CASE: The condition MATH may be expressed in terms of the measure MATH on MATH. (The choice of MATH does not matter.) That is the joint probability distribution of random variables MATH on MATH, and its marginal distributions are MATH. Clearly, MATH is a product measure if and only if MATH are independent with respect to MATH. Thus, the relevant condition on MATH says that MATH must be equivalent to a product measure. The set of all such measures on MATH is NAME measurable. The map MATH is the restriction to MATH of the map MATH measurable by REF . CASE: For such MATH the map MATH is an isomorphism REF between MATH and MATH, where MATH (and the MATH-field of MATH-measurable subsets of MATH is suppressed in the notation). Therefore, MATH where MATH, MATH are the marginals of MATH. I assume in addition that the MATH-fields MATH are nonatomic (atoms are left to the reader); then an additional transformation MATH turns MATH into NAME measure on MATH. The density of MATH (with respect to NAME measure MATH on MATH) is a NAME function of MATH (take an increasing (refining) sequence of finite partitions of MATH). The following lemma (or rather, its straightforward two-dimensional generalization) completes the proof.
|
math/0006165
|
CASE: Follows immediately from measurability of MATH in MATH. CASE: Let MATH be another probability measure on MATH, equivalent to MATH. Consider the distribution MATH we know that MATH is jointly measurable in MATH and MATH, and MATH is measurable in MATH, therefore MATH is jointly measurable in MATH and MATH. However, distributions MATH for all MATH determine MATH uniquely, and moreover, they generate the NAME MATH-field on MATH.
|
math/0006165
|
We choose some MATH and replace each MATH with the corresponding MATH according to their unitary correspondence determined by MATH, MATH . The disjoint union of MATH over all MATH is naturally a standard NAME space, since it is a NAME subset of the disjoint union of all subspaces of MATH. Thus, a NAME structure appears also on the disjoint union of MATH, and the NAME structure does not depend on the choice of MATH (since MATH is measurable in MATH, see REF ). REF is easily transferred from the disjoint union of all subspaces of MATH to the disjoint union of MATH. It remains to prove REF . REF gives us measurability of the set MATH, and measurability of MATH in MATH. It remains to verify measurability of MATH. We know (recall REF) that MATH that is, if MATH is a measure that makes MATH independent, then MATH the product in the right-hand side is just a pointwise product of two functions. Therefore MATH . However, the independence of MATH and MATH under MATH means that MATH so, MATH . By REF , MATH is NAME measurable in MATH, therefore MATH is measurable in MATH, which means that MATH is measurable in MATH.
|
math/0006165
|
Treating MATH as a subspace of MATH we choose measurable maps MATH such that every MATH is spanned by MATH. We have MATH where MATH, MATH. Applying the orthogonalization process we ensure that MATH form an orthogonal basis of MATH. Introducing NAME functions MATH for MATH we have MATH .
|
math/0006165
|
A sub-MATH-field MATH belongs to MATH if and only if MATH; here an element of MATH is identified with a pair MATH of elements of MATH. (The choice of a measure MATH does not matter.) For such MATH and MATH, MATH must be disjoint to the open set MATH.
|
math/0006165
|
Due to REF it suffices to prove measurability of the map MATH from MATH to MATH; of course, MATH means MATH. We know (see the proof of REF ) that MATH implies MATH. Therefore, for any open set MATH the set of all MATH such that MATH is closed.
|
math/0006165
|
We take a Gaussian type space MATH and consider MATH, so that MATH. Every subspace MATH generates a sub-MATH-field MATH. It is easy to see that MATH if and only if MATH. However, MATH is measurable in MATH by REF , and MATH is measurable by REF .
|
math/0006165
|
Consider some MATH treated as a sub-MATH-field of MATH. Given some MATH, we introduce on MATH a measure MATH for MATH; here, as before, an element of MATH is represented by a pair MATH where MATH. There is also a measure MATH on the MATH-field MATH such that MATH whenever MATH. The pair MATH belongs to the graph MATH that corresponds to MATH. Such pairs for all MATH (or for a countable dense subset) span the graph MATH. It remains to prove that, for a given MATH and arbitrary MATH, the pair MATH is measurable in MATH (you see, MATH depends implicitly on MATH). According to our definition of the NAME structure on MATH, we have to prove measurability in MATH of the density MATH. The latter is the restriction (to the second copy of MATH) of a density on the doubled space, MATH, namely, MATH, where measures MATH on MATH are defined (irrespective of MATH) by MATH, MATH. We apply REF to MATH on MATH. Though, MATH and MATH are not equivalent, but one can consider, say, MATH.
|
math/0006165
|
We know that every MATH generates MATH (see the proof of REF ), and the map MATH is measurable. The transition from MATH to the corresponding element of MATH is measurable by REF .
|
math/0006165
|
It is enough to consider the case MATH for an arbitrary MATH. The equality MATH, in combination with REF , reduces the problem to measurability of MATH in MATH, and MATH in MATH. However, a monotone function is always measurable.
|
math/0006165
|
By REF , MATH is measurable in MATH. According to REF there are MATH, measurable in MATH, such that every MATH is spanned by MATH. Therefore the graph of MATH is spanned by pairs MATH. Each pair is measurable in MATH and continuous in MATH (recall REF ), therefore, measurable in MATH (see CITE).
|
math/0006165
|
REF allows us to reformulate the conditions in terms of density matrices in product systems, thus making explicit their invariance under isomorphisms.
|
math/0006165
|
Choosing any MATH we have for large MATH where MATH, MATH. Taking into account that MATH and MATH, we get MATH . So, MATH which is the first claim of the lemma. In order to prove the second claim we note that the only properties of the function MATH used till now are the finite variation of MATH, and MATH for large MATH. Therefore the same argument may be applied to the function MATH with respect to MATH: MATH . Hence MATH .
|
math/0006165
|
First, the function MATH on MATH is of finite variation, thus, using REF , MATH for MATH. Second, MATH, since MATH is of finite variation on MATH. Hence MATH and MATH . Third, MATH and MATH, hence MATH . It is enough to prove that MATH for every function MATH as in REF . Taking into account that MATH we transform it into MATH it is enough to prove that MATH . We treat separately two cases, MATH and MATH. The first case, MATH, is simple; just using boundedness of MATH we have for REF as well, MATH . We turn to the other case, MATH. Now MATH. REF gives MATH, hence MATH summing over MATH gives MATH, a convergent series in MATH, which proves REF. It remains to prove the most delicate case, REF for MATH. Neglecting a finite number of terms, we get MATH large enough for using the asymptotic relation of REF : MATH . However, MATH and MATH, therefore MATH . It is enough to prove that MATH or, equivalently, MATH . It remains to note that MATH .
|
math/0006165
|
Here is an equivalent formulation: there is an operator MATH such that MATH and MATH is an NAME, in other words, an equivalence operator in the sense of CITE. It means that MATH is one-to-one onto, has a bounded inverse, and MATH (the NAME class of operators). The latter is equivalent to MATH, see CITE. Matrix elements of MATH are MATH; REF shows that MATH and, of course, MATH is bounded. It remains to prove that MATH has a bounded inverse. The range of MATH being evidently dense, we have to prove that MATH for some MATH, that is, MATH does not belong to the spectrum of MATH. The spectrum accumulates to MATH only (since MATH); we have to prove that MATH is not an eigenvalue, that is, MATH for all MATH. It is enough to prove that the following formula is a correct definition of (continuous) linear functionals MATH on MATH: MATH indeed, it will follow that MATH . The norm on MATH, defined in terms of MATH, uses MATH for MATH only. Therefore we may assume that MATH vanishes outside of some bounded interval (and still satisfies REF). For every MATH, MATH here MATH is the NAME transform (normalized as to be unitary) of MATH, and MATH - of MATH. The formula MATH defines a linear isometric embedding MATH on the dense subset MATH; we may extend MATH to the whole MATH by continuity. Every MATH gives a linear functional on MATH, namely, MATH. In order to get MATH, we take MATH such that MATH is the NAME transform of MATH; it remains to verify that such MATH belongs to MATH. The function MATH is of finite variation; its NAME transform is MATH; it remains to check that MATH . However, the continuous function MATH never vanishes, and MATH for MATH by REF .
|
math/0006165
|
Vectors MATH (defined by REF) are orthogonal with respect tosome admissible norm on the NAME MATH; the proof is quite similar to the proof of REF .
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.