paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9906041 | Apply MATH to MATH and MATH using REF , then both MATH and MATH have the common factor MATH. The parts of the calculation involving MATH and MATH or MATH proceed just like those with MATH and MATH . |
math/9906041 | The non-zero cases are all MATH summations. For the first case, MATH thus MATH . This uses the NAME sum MATH and MATH for arbitrary MATH and MATH. The other formulas are proved in the same way. |
math/9906041 | Substituting MATH in the generating functions yields MATH . The latter term expands to MATH . Now multiply top and bottom by MATH replace MATH by MATH (also a simple calculation to multiply the resulting series by MATH), and in the denominator expand MATH . |
math/9906043 | The proof is given in REF. |
math/9906043 | The proof is given in REF. |
math/9906043 | The proof is provided in REF . |
math/9906043 | The proof is provided in REF . |
math/9906043 | The proof is provided in REF. |
math/9906043 | Let us recall MATH. MATH is in the spectrum of MATH and, because REF , MATH is in the spectrum of MATH. But MATH . By appliying a well-known perturbation formula: MATH . Let us define MATH . If MATH and MATH are close enough, it is possible to neglect the higher order terms MATH. Then, it is clear that the algorithm converges REF if and only if MATH. On the other hand, from REF : MATH . Therefore MATH . But, premultiplying REF and posmultiplying REF by MATH: MATH . So MATH . On the other hand, MATH . So MATH and MATH . |
math/9906043 | The invariance with respect to transformations MATH shall be proven, being the other case esentially equal. Firstly, note that MATH and MATH are invariant under the transformation MATH which proves the MATH invariance. As MATH, MATH is also invariant. Let us consider now: MATH . It has been used that MATH to go from REF to REF , that MATH to go from REF to REF and that that MATH to go from REF to REF . On the other hand under the considered transformation MATH becomes MATH . It has been used the MATH REF to write REF , and REF to simplify it. |
math/9906044 | We use the following simple lemma from linear algebra without proof. Let MATH be a non-degenerate linear pairing of finite dimensional vector spaces and MATH a subspace of MATH. Then the induced pairing MATH with MATH is also non-degenerate. Applying this lemma to the non-degenerate pairing MATH, CITE, and MATH we have MATH if and only if MATH for MATH. Hence MATH if and only if MATH. Consequently MATH, where MATH denotes the pre-image of MATH under MATH. Since the induced pairing is also non-degenerate and MATH by definition, the assertion of the lemma is proved. |
math/9906044 | Since MATH, MATH. For MATH, MATH, and MATH we thus get MATH . Since the pairing is non-degenerate the first assertion follows. Since MATH and MATH are morphisms, the corresponding subspaces are invariant. Let MATH, i. CASE: MATH, and suppose MATH for all MATH. Then MATH. Since the pairing is non-degenerate, MATH. Similarly one shows that MATH separates the elements of MATH. |
math/9906044 | CASE: Since MATH and MATH are MATH-invariant, and since MATH and MATH are MATH-invariant, the mappings MATH and MATH are well-defined on both quotients MATH and MATH respectively, MATH and MATH. It follows from CITE that for MATH, MATH, and MATH . Thus REF applies to our situation. |
math/9906044 | CASE: It is well known that MATH, and MATH, and MATH form a linear basis of MATH. Using REF it is easy to see that the mapping MATH defines a linear isomorphism MATH. Since MATH is bi-invariant if and only if MATH, REF is proved. CASE: The elements MATH, MATH, and MATH form a basis of MATH. Using the graphical calculus it is not difficult to check that MATH acts as the identity on MATH: CASE: MATH acts as the identity on MATH, MATH, and MATH. Consequently, MATH and MATH. By REF MATH. CASE: By CITE, MATH. Further by CITE, MATH. Let MATH . Using the above identities, REF twice, and finally REF one gets MATH . Using MATH, the above calculation, and again REF it follows that MATH . Consequently, MATH and MATH. By REF applied to the projection MATH onto the space MATH, the pairing MATH is non-degenerate. Since MATH, MATH by REF , and MATH by the result of REF we get MATH. This completes the proof. |
math/9906044 | CASE: Since all four mappings appearing in REF are morphisms of corepresentations one easily checks that MATH. Hence the space is closed under the right coaction. Now we compute the right adjoint action. Set MATH. By REF MATH . The last but one equation becomes evident by taking a look at the graphical presentation of these equations. CASE: MATH is closed under the right adjoint action. Consequently MATH is closed under the right adjoint action. Hence MATH is a bicovariant subbimodule. CASE: The first part follows from REF and the fact that MATH is the identity of MATH. In addition MATH and MATH as well as MATH and MATH are pairwise orthogonal idempotents, respectively. Hence the sum is direct. Let us turn to the second part. Let MATH denote the eigenvalue of MATH with respect to the idempotent MATH, MATH, namely MATH, MATH, and MATH. Note that MATH as well. Put MATH. Then by REF MATH . Since MATH is not a root of unity, MATH for MATH. Hence MATH if and only if MATH. We thus get MATH as linear spaces. By REF each space on the right hand side generates a bicovariant subbimodule. This completes the proof. |
math/9906044 | By REF the canonical left-invariant basis of MATH is MATH . The proof is in two steps. First we compute MATH and obtain elements MATH . The graphical presentation of MATH and MATH is as follows. CASE: The elements MATH and MATH. First we will show that MATH . By REF one has MATH . CASE: The right adjoint action of MATH on MATH. In particular for MATH we compute REF : The proof of REF. In the first step we replaced the crossing in the dash box using REF. In the second step we did the same with the MATH-matrix in the first dash box. This gives the first three terms in the next line. Moreover the dash box in the second summand is multiplied by MATH and gives MATH (no crossing). Similarly, a second crossing in the same term gives another MATH. With the third summand we proceed in the same way; in addition the curl gives the factor MATH. Since MATH, REF follows immediately. Note that for MATH, MATH is obvious from the first line in REF since MATH and no crossing appears there. Moreover MATH and MATH and REF are valid. Since MATH is one-dimensional there is nothing to prove. Now we fix MATH. We shall eliminate MATH from REF. Multiplying REF by MATH and using MATH, MATH gives REF. Since MATH is not a root of unity, MATH is invertible with inverse MATH . Set MATH and multiply REF by MATH. Then we obtain MATH. Consequently, MATH for MATH and MATH. In the second step we again compute the right adjoint action of MATH but on elements MATH. We obtain elements MATH . Obviously MATH. Graphically they are represented by REF : The elements MATH and MATH. By REF one has REF : The right adjoint action of MATH on MATH. Replacing one crossing MATH by MATH similarly to the graphical calculations in the first part of the proof one can show that MATH . Since MATH is not a root of unity, MATH is invertible with inverse MATH. Therefore MATH belongs to MATH. Finally we have MATH which completes the proof. |
math/9906049 | CASE: Consider the algebraic group MATH. In view of REF , we have the exact sequence MATH . (If MATH, then MATH.) Since MATH is reductive and Abelian, both MATH and MATH lie in MATH. Hence MATH induces a surjective homomorphism MATH. By a standard property of diagonalizable groups, this means that there exists a REF-dimensional torus MATH such that MATH and MATH is onto. On the NAME algebra level, this yields MATH and MATH. The restriction of the Killing form to either of the reductive subalgebras MATH and MATH is non-degenerate CITE. Hence MATH can be chosen so that MATH. As MATH, we obtain MATH. Restricting the differential of MATH to MATH yields an isomorphism MATH. Define MATH by MATH, MATH. Then MATH have rational eigenvalues and satisfy REF . CASE: Suppose MATH is another characteristic of MATH. Since MATH and MATH, we obtain MATH. Thus MATH and MATH are two maximal diagonalizable subalgebras of the solvable algebra MATH. By the standard conjugacy theorem (see CITE), there exists MATH such that MATH. It then follows from REF that MATH. CASE: Because the characteristic constructed in the first part of the proof had rational eigenvalues, we conclude by the second part. |
math/9906049 | Let MATH be the MATH-stable decomposition. Then MATH and therefore MATH is orthogonal to MATH. That is, MATH is a characteristic of MATH. From the proof of REF , it follows that MATH contains a reductive NAME subalgebra of MATH. Thus, MATH. |
math/9906049 | CASE: Arguing by induction on MATH, we first note that the claim is true for MATH. CASE: Assuming that MATH has fractional eigenvalues, one may replace MATH by the smaller semisimple subalgebra MATH, where MATH. Indeed, MATH is reductive and MATH. Then, being nilpotent, MATH belong to MATH. Since MATH are orthogonal to MATH and the latter contains the centre of MATH (if any), we have MATH. Thus, MATH is a characteristic of MATH relative to MATH and, by the inductive assumption, MATH contains finitely many MATH-orbits of characteristics. Clearly these orbits generate finitely many MATH-orbits in MATH. CASE: Assume now that MATH, that is, the eigenvalues of MATH are integral. Fix a NAME subalgebra MATH of MATH such that MATH. Then MATH. Choose a set of simple roots MATH with respect to MATH so that MATH is dominant for all sufficiently small positive MATH. For all MATH, we then have MATH and if MATH, then MATH. CASE: Assume that MATH for some MATH. Then we may just throw it away! That is, consider MATH and the corresponding NAME subalgebra MATH. The constraint on MATH implies that MATH for MATH. Hence MATH and, as in REF , we even have MATH. Thus, we may apply the inductive assumption to MATH. CASE: Certainly, the constraint that MATH for all MATH leaves no much room and one gets only finitely many possibilities for MATH. By symmetry, the same argument applies to MATH. It follows that, up to conjugacy, there are finitely many possibilities for MATH. This completes the proof. |
math/9906049 | CASE: Inductive steps used in the proof of REF provide us with regular subalgebras in MATH. Hence, under our hypothesis on MATH and with the same choice of MATH and MATH, we already see that the eigenvalues of MATH must be integral, MATH, and MATH, if MATH. CASE: If either MATH or MATH for some MATH, one can again, as in the proof of REF , throw away this MATH and get a regular semisimple subalgebra containing MATH. Thus, this cannot occur. CASE: It remains to obtain the lower bound for MATH, if MATH. Notice that MATH induces in MATH the MATH-grading MATH and that MATH is a set of simple roots for MATH. We also know that MATH for MATH. Look what is happening on the next level. Suppose MATH is such that MATH, and let MATH be a nonzero root vector. Then MATH is a lowest weight vector of an irreducible MATH-module. Denote this module by MATH. We have MATH. Let MATH be the root of MATH corresponding to the highest weight of MATH. Then, more precisely, MATH . Let us say that MATH is the height of MATH-grading on MATH. If MATH, then MATH and, as above, we could drop the simple root MATH. Under our assumption, this is however impossible and we must have MATH. It is thus enough to give an upper bound on MATH for any such MATH. The problem can be stated as follows: The algebra MATH and a MATH-module MATH are compatibly MATH-graded; the grade of each simple root in MATH is either REF or REF. Give an upper bound on the height of the MATH-grading on MATH. The answer was essentially given by NAME, who considered the height of representation in case, where all simple roots are of grade REF. (Actually, NAME considered not gradings but the associated partitions of the weights of a representation into `layers', the height being the number of layers-MATH.) If some of the simple roots have grade REF, then the height can only decrease. Hence the bound given by NAME applies in our situation as well. Being adapted to our setting, it reads MATH, where MATH is the sum of all positive coroots of MATH and MATH is a NAME group invariant inner product on MATH. Whence MATH, and we may take MATH. |
math/9906049 | Since REF implies REF , it suffices to demonstrate that REF . Suppose MATH. Then replacing MATH by its projection to the MATH-eigenspace of MATH, we obtain the MATH-triple MATH. Further, MATH. Hence MATH lies in the MATH-eigenspace of MATH in MATH. As the latter is positively graded, this forces MATH. Similarly, MATH. Thus MATH belong to the reductive subalgebra MATH. Since the restriction of the Killing form to MATH is non-degenerate, it follows from the condition MATH that MATH is orthogonal to MATH and hence MATH. Thus, MATH and MATH can be included into the MATH-triple inside of MATH. |
math/9906049 | Relations REF show that different summands in definition of all limits belong to different eigenspaces (relative to MATH and MATH respectively). |
math/9906049 | CASE: The `only if' part is obvious. Suppose MATH is nilpotent in MATH and MATH satisfies REF . Let MATH be a MATH-invariant decomposition and MATH, MATH. Then MATH also satisfy REF . This shows that MATH contains sufficiently many elements to ensure that REF holds for MATH in place of MATH. CASE: Let MATH be wonderful in MATH. It follows from REF that there exists a characteristic MATH lying in MATH. The following is obvious. |
math/9906049 | By symmetry, it is enough to prove one equality in each item. CASE: Using REF we obtain MATH. CASE: Applying the formula in REF with MATH gives MATH for all MATH. Then summation over MATH yields the first formula. |
math/9906049 | CASE: The first equality follows from the relations MATH and MATH. The hypothesis on MATH also implies that the kernel of the map MATH is of dimension MATH. Thus MATH is onto. CASE: Using an invariant nondegenerate bilinear form on MATH and surjectivity of MATH, one obtains MATH is injective. Hence MATH is concentrated in nonnegative degrees. The second claim is just a reformulation of the fact that MATH is onto. |
math/9906049 | CASE: In view of REF , the previous Lemma applies to reductive NAME algebras MATH and MATH. CASE: In view of REF , the previous Lemma applies to NAME algebras MATH and MATH. |
math/9906049 | We have MATH and MATH. By REF with MATH, MATH. This means that MATH (and also MATH) lies in the dense MATH-orbit in MATH. Hence we may assume that MATH. Let MATH denote the stabilizer of MATH in MATH. Then MATH is NAME algebra of MATH. By REF with MATH, MATH. This means that MATH (and hence MATH) lies in the dense MATH-orbit in MATH. Thus we can make MATH. |
math/9906049 | If MATH is a fractional eigenvalue of MATH in MATH, then, applying nilpotent endomorphisms MATH and MATH to MATH, we eventually obtain an eigenvalue in MATH of the form MATH with MATH. |
math/9906049 | The argument used in the proof of REF applies here verbatim. For convenience of the reader, we reproduce it. Assume MATH is nonzero for MATH and MATH. It follows from the invariance of the Killing form on MATH that MATH if and only if MATH. By definition, put MATH. For each MATH, consider the finite set MATH, with the lexicographic ordering. This means MATH or MATH and MATH. Denote by MATH the unique maximal element in MATH. Let MATH be an element such that MATH for all MATH. Then MATH is a nonzero element in MATH. By REF , there is MATH such that MATH. Then MATH. Since MATH, we have MATH is nonzero and belongs to MATH. However, MATH. Therefore MATH, which contradicts the choice of MATH. Thus, the case MATH is impossible. |
math/9906049 | ` REF This is REF . ` REF Making use of the Killing form on MATH, one can translate these three conditions in ones about the positive quadrant. Namely, REFst: MATH for MATH; REFnd: MATH for MATH; REFrd: MATH for MATH. These three together show that MATH, MATH, and MATH generate MATH. In particular, applying MATH and MATH to MATH, we obtain the whole space MATH. |
math/9906049 | Let MATH and MATH be commuting MATH-triples and let MATH be the grading determined by MATH. Then MATH and MATH. Set MATH. Recall that MATH in the rectangular case. It is easily seen that MATH and MATH are isomorphic MATH-modules. (In the notation of REF, MATH.) As MATH, we obtain MATH . Since MATH, we see that MATH is even if and only if MATH (that is, MATH is even) and MATH is even in MATH. Then either by symmetry or by a direct argument one concludes that MATH is actually even in MATH. |
math/9906049 | Left to the reader. |
math/9906049 | CASE: MATH - We always have MATH. MATH. As the centralizer of MATH entirely lies in the positive quadrant, we have MATH is injective for all MATH and MATH. (Otherwise, applying MATH to a nonzero element in the kernel we would eventually arrived at an element in MATH with MATH.) Similarly, MATH is injective for all MATH and MATH. We thus have more than enough to apply REF and conclude that MATH. CASE: This follows from the first part and from REF . |
math/9906049 | We first prove that MATH has prescribed properties as nilpotent pair in MATH and then go down to MATH. CASE: Take a characteristic MATH of MATH and consider the corresponding bi-grading MATH. Set MATH. Each nonempty coset MATH determines a subspace in MATH containing a MATH-fixed vector (compare REF). Hence MATH lies in a unique such coset. For the same reason, MATH for all MATH, and MATH has a unique `northeast' corner, that is, MATH. Let MATH be this corner. Since MATH is self-dual, MATH is centrally-symmetric. Whence MATH is the unique `southwest' corner of it. (Note that, although MATH are not necessarily integral, MATH implies MATH.) It follows that MATH lies inside of the rectangle having opposite vertices MATH and MATH. It is however easy to see that the conditions MATH and MATH force that MATH is ``equal" to this rectangle, that is, MATH. CASE: Since the eigenvalues of MATH in MATH form a rectangle, MATH is a rectangular principal nilpotent pair in MATH. Indeed, it is not hard to write a formula for the nilpotent operators MATH such that MATH and MATH (MATH). Hence MATH is rectangular. Set MATH. As MATH and MATH is an irreducible MATH-module, an explicit computation shows that MATH (compare CITE). Hence MATH is principal in MATH. CASE: Now, one has the following: MATH is semisimple, MATH is a MATH-triple in MATH, and MATH. This easily implies that MATH and MATH is rectangular in MATH. CASE: It follows from the general theory of principal nilpotent pairs CITE and is also easily seen in our situation that MATH is a NAME subalgebra (the set of diagonal traceless matrices). Whence MATH is a NAME subalgebra in MATH. Because the eigenvalues of MATH in MATH are even (MATH), all irreducible MATH-submodules in MATH, and hence in MATH, have zero weight. Thus MATH and MATH is principal in MATH. |
math/9906049 | Since MATH is MATH-stable, the first equality follows for dimension reason. CASE: Suppose MATH is integral. That the case MATH is impossible follows from REF . This already means that MATH is wonderful. Consequently, results of REF apply. By REF , neither MATH nor MATH can occur. We are thus left with the case MATH. CASE: Suppose MATH is fractional. Let MATH be the sum of all fractional eigenspaces. Then MATH is an orthogonal MATH-module, MATH, and MATH. Applying REF , we conclude that MATH, MATH is rectangular principal in MATH, and MATH is a NAME subalgebra of MATH. Moreover, it follows from the orthogonality that both MATH must be fractional (compare REF ). Finally, MATH, that is, MATH is almost principal in MATH. |
math/9906049 | The rows of the inverse of the NAME matrix yield the expressions of the fundamental weights through the simple roots. |
math/9906049 | As in REF , consider the decomposition MATH and set MATH. Then MATH and MATH is a MATH-module. Now we conclude by the previous lemma. |
math/9906049 | (Compare Remark after REF) Assume that MATH is contained in a proper regular semisimple subalgebra and let MATH be a maximal one among them. It follows from CITE and NAME 's description of periodic automorphisms of MATH (see for example, CITE) that MATH is a fixed-point subalgebra of some inner automorphism of MATH of finite order. Since MATH, this means MATH contains non-trivial semisimple elements. But MATH is connected unipotent for such MATH (see CITE and CITE). |
math/9906049 | Using REF , and the fact that MATH is NAME (hence the case MATH is impossible), one sees that we have to only prove that MATH. By CITE and CITE, MATH is regular nilpotent in MATH. It then follows from REF that the MATH-grading in MATH defined by MATH is nothing but the standard grading associated with the function MATH on the set of roots of MATH (that is, MATH is the linear span of the root spaces such that the height of the corresponding root of MATH is MATH). Therefore MATH. If MATH is NAME in MATH, then MATH. Hence MATH, which is exactly what we need. |
math/9906049 | CASE: First, assume that MATH is either principal or almost principal integral. Since MATH is wonderful and integral in both cases, the formulas in REF become simpler. In particular, MATH. Using REF , we obtain MATH. Because MATH, this implies that the eigenvalues of MATH in MATH and MATH are the same. Therefore MATH are just the eigenvalues of MATH in MATH. Since MATH for MATH (see REF), the partition dual to MATH consists of the MATH-eigenvalues in MATH. But, since the MATH-grading of MATH is the standard grading associated with the height of roots (see the proof of REF), the dual partition consists also of the exponents of MATH. This is a classical result of CITE, see also CITE. This argument is completely symmetric with respect to MATH and MATH, because we do not need the assumption (in the almost principal case) that MATH is NAME. CASE: Assume that MATH is almost principal non-integral. Then MATH is principal in MATH (see CITE or REF ) and we conclude by the first part of the proof. |
math/9906066 | Let the vertices of MATH be MATH. Let MATH, where MATH. Then for any MATH we have MATH, hence MATH, that readily implies MATH. Similarly one shows MATH and MATH, that implies the statement of the lemma. |
math/9906066 | By REF the centre of an inscribed similar copy MATH of MATH is at the centre of MATH, and its edges of length MATH are parallel to some MATH axes of MATH. This gives a partition of the MATH axes of MATH to classes of sizes MATH; the number of such partitions is MATH. Fixing one such partition, MATH is determined, up to magnification from its centre. Its size is determined by the requirement that a given vertex of it should belong to MATH (and then, by symmetry, all of its vertices belong to MATH). |
math/9906066 | We have MATH, and MATH is the set of skew symmetric MATH matrices. Let MATH. Then MATH. We have to show that for MATH we have that MATH does not lie on the diagonal of MATH (thus, in particular, is not MATH). We will argue indirectly. For MATH we have MATH, taking in account the skew symmetry of MATH. Here by REF for MATH. If MATH, then the quadratic form MATH assumes equal values for each MATH. Like in REF , this implies MATH for each MATH, hence MATH as asserted. |
math/9906066 | The proof of this proposition follows the proof of REF , with the exception that we need the following consequence of REF : there are three ways of inscribing a non-cubical square based box with given ratio of edges into an ellipsoid in MATH with axes of different lengths. Now the proof of the proposition is complete noting that MATH is odd and therefore MATH does not vanish. |
math/9906066 | As in the case of the cube we can define a map MATH by setting MATH. We have to show that the image of MATH intersects the diagonal MATH in MATH. Notice that the rotation group of a square based box contains MATH (and in general equals MATH). Now an identical argument as in the case of the cube in REF yields the statement of this theorem, using REF instead of REF . |
math/9906066 | Let us consider a square based box, with the given ratio of the height to the basic edge, that is inscribed to MATH, and has vertex set MATH. Like at the reduction of REF to REF , we suppose that the centre of MATH is the origin, and we let MATH be the NAME norm in MATH, associated to MATH. We apply REF for MATH and the MATH chosen above, obtaining MATH such that MATH is a point MATH on the diagonal MATH of MATH. Then the square based box with vertex set MATH is inscribed in MATH, has the given ratio of the height to the basic edge, and is centred at the centre of MATH. |
math/9906066 | We will show that the MATH-bundle MATH defined at the beginning of REF is actually trivial. Then the MATH-bundle MATH will have a section, that implies by construction the existence of MATH-equivariant maps, that are also MATH-equivariant for any MATH. Think of MATH as an orthonormal basis. Let MATH be the forgetful map associating to MATH its MATH'th basis vector. These maps are MATH-equivariant as MATH is the rotation group of the regular tetrahedron. In this way we get three linearly independent sections MATH and MATH of the vector bundle MATH. Thus MATH is trivial indeed. (Another way of seeing this is the following. If MATH is a discrete subgroup of MATH, then the bundle MATH is the tangent bundle of MATH (left cosets), and this is trivial. Namely the tangent bundle of the quotient of a NAME group with respect to a discrete subgroup is trivial.) |
math/9906066 | Recall that MATH is the action of the subgroup MATH of the symmetric group MATH of the regular tetrahedron, the four letters corresponding to the four vertices of the regular tetrahedron. In other words, for MATH we have that MATH is that element of MATH, whose restriction to the vertex set of the regular tetrahedron inscribed to MATH equals the permutation MATH of these vertices. For MATH any MATH, equidistant to the MATH'th and MATH'th vertices, is a fixed point for MATH. For MATH any vector MATH, orthogonal to the lines connecting the MATH'th and MATH'th vertices, and the MATH'th and MATH'th vertices of the tetrahedron, respectively, is a fixed point of MATH. For MATH being the subgroup of permutations fixing some MATH, the MATH'th vertex is a fixed point of MATH. Then a MATH-equivariant map can be given in all three cases as a constant map MATH, having as value a fixed point MATH of MATH. |
math/9906066 | We show that there is a section of the MATH bundle MATH over the orbit space MATH. This is equivalent to the statement of the theorem. Obstruction theory (compare CITE, pp. REF) tells us that the existence of such a section is equivalent to the vanishing of the obstruction class MATH sitting in a cohomology space with twisted coefficients. We show the vanishing of MATH. We use REF (compare CITE) which claims that there is a homomorphism MATH which sends the second obstruction class MATH to MATH. Therefore it is sufficient to show that the second obstruction class vanishes. On the other hand the second obstruction class MATH coincides with the second NAME class of MATH, that is, MATH (compare CITE, p. REF). Thus the proof of MATH gives the theorem. The vector bundle MATH decomposes as the direct sum of a rank MATH and a rank MATH vector bundle, where MATH is the non-trivial representation of MATH on MATH and MATH is an effective representation of MATH on MATH (the generator is sent to the rotation through the angle MATH), and MATH. Thus we get that MATH . This formula shows that the vanishing of MATH and MATH yields the vanishing of MATH. The first NAME class MATH vanishes since the representation MATH is orientation preserving and hence MATH is oriented. The second NAME class MATH vanishes since, as we are going to show, MATH with some MATH principal bundle MATH (here MATH is considered as a MATH principal bundle). Hence the NAME class MATH is even, yielding the desired vanishing of MATH. Thus we are left with constructing a MATH principal bundle MATH with MATH. The universal covering of MATH is the composite covering MATH. Moreover, we claim that this covering is given by the MATH action MATH on MATH, sending the generator MATH to the transformation MATH of MATH, the unit sphere of the quaternionic algebra MATH. To see this, divide out first by the MATH action defined by MATH. The quotient is MATH which can be identified with MATH, by associating to any element MATH a special orthogonal transformation of MATH, namely the conjugation with MATH. Now we see that the MATH action on MATH inherited from the MATH action MATH of MATH corresponds to the MATH action MATH of MATH with respect to the axis MATH (that is, to the rotation of the space Im-MATH through MATH about the axis MATH), via the above identification. Hence indeed MATH is a lens space. Let MATH be the principal MATH bundle on MATH given by the representation MATH of MATH on MATH, sending the generator to the rotation through the angle MATH. Now the relation MATH is immediate. The proof is complete. |
math/9906066 | Let us consider a box similar to the given box, that is inscribed to MATH, and one of whose faces has vertex set MATH. Like at REF , we suppose that the centre of MATH is the origin, and we let MATH be the NAME norm in MATH, associated to MATH. We let MATH. Then Theorem D of CITE (quoted in REF) implies for MATH and the set MATH of the vertices of a planar rectangle inscribed to MATH, that there exists MATH such that MATH is a point MATH on the diagonal MATH of MATH. Then the box with vertex set MATH is inscribed in MATH, is similar to the given box, and is centred at the centre of MATH. |
math/9906066 | We proceed identically as above, with the only difference that we apply REF not for a cube, but use its consequence mentioned in the proof of REF , again observing that MATH is odd. |
math/9906066 | Similarly as above MATH induces a map MATH given by coordinates MATH. Let MATH be the standard basis vectors of MATH. By abuse of notation MATH will stand for MATH if MATH. Let MATH, the action of MATH on MATH be given as right multiplication by the rotation group of MATH, the rhombic dodecahedron, which is easily seen to be isomorphic to the rotation group of the cube. Moreover define MATH to be the MATH action on MATH given on the standard basis of MATH by the following rule: if MATH is a permutation of the letters MATH then MATH . The construction of the map MATH and the oddness of MATH imply that MATH has to be a MATH equivariant map from MATH to MATH. As we have seen above, to prove the theorem we need MATH such that MATH lies in the MATH-subspace MATH of MATH. We prove more, namely that for any MATH equivariant map MATH there exists a MATH such that MATH. The MATH-space MATH is spanned by any three of the four vectors MATH, as we get these when MATH respectively. Note that MATH is an invariant subspace of MATH. Let REF-subspace MATH of MATH be spanned by any three of the four vectors MATH . Now one checks that MATH is the orthogonal complement of MATH with respect to the standard Euclidean scalar product of MATH. As the action MATH preserves this standard scalar product, it follows that MATH, being the orthogonal complement of an invariant subspace MATH, is itself invariant under the action MATH. Let MATH denote the MATH action MATH on MATH. Now finding MATH with MATH is equivalent to showing that the MATH equivariant map MATH vanishes somewhere. By REF this is the case as MATH is isomorphic to MATH, which is the action where MATH acts as the symmetry group of the regular tetrahedron. This last statement can be seen by checking that MATH faithfully permutes the four vectors MATH, which form a regular tetrahedron in MATH. The result follows. |
math/9906073 | We have CITE MATH. For the reverse inequality, suppose MATH and let MATH. We will prove that MATH is bilateral reducible in MATH, that is there exist MATH, MATH in MATH such that MATH . Let MATH be the closed two-sided ideal in MATH generated by MATH. It was proved in CITE that MATH is bilateral reducible in MATH if MATH is dense in MATH. We prove now, using several techniques borrowed from CITE, that MATH holds for all two-sided closed ideals of MATH. Let MATH be a selfadjoint element of A. Let MATH be a real number. Let MATH be the positive integer such that MATH. Set MATH. Let MATH . Then MATH and MATH for MATH. Using CITE we get MATH where MATH . For MATH (and so for all MATH such that MATH is sufficiently large) we have (compare CITE) : MATH where MATH . This implies that MATH . It follows CITE from NAME 's CITE condition of regularity that MATH is locally normal in MATH. This means CITE that there is a commutative NAME MATH-subalgebra MATH in MATH such that MATH and MATH belong to MATH and such that MATH is a dense normal subalgebra of MATH. It was proved in CITE that if MATH is a closed two-sided ideal in MATH, then MATH is dense in MATH. This completes the proof. |
math/9906073 | By CITE, the map MATH, MATH, REF is a NAME fibration and, with the same proof, the map MATH, MATH, is also a NAME fibration. Here MATH denotes the space of last columns of invertible MATH matrices with entries in MATH : MATH . The stability subgroup of MATH induced by the action of MATH under MATH and MATH consists of matrices whose last column is MATH, that is matrices of the form MATH where MATH and MATH is an arbitrary row in MATH of length MATH. Viewing MATH as a subset of MATH via the embedding MATH we obtain a deformation retract of the stability subgroup of MATH onto MATH by carrying the off-diagonal entry MATH linearly to zero. Since MATH is a NAME fibration, one has the homotopy exact sequence CITE MATH . This long exact sequence ends with CITE MATH viewed as pointed sets. The base points in the groups are taken to be their identity elements, while the base points in MATH and MATH are taken to be MATH. For MATH one has CITE MATH and MATH. Here MATH, the last columns connected stable rank is, by definition CITE, the least integer MATH such that for all MATH the set MATH is connected. Since CITE MATH we have MATH and MATH is trivial. From the proof of CITE it follows that there is a bijection MATH . Since MATH, we get that MATH is also trivial. From the long exact sequence above we obtain the surjectivity of MATH . It follows again from the homotopy exact sequence that the map MATH is injective if and only if the map MATH is surjective. Suppose that MATH and thus the map MATH is surjective. Let MATH be a map preserving the base point MATH, and so representing an element of MATH. Then MATH can be identified with an element of MATH such that MATH. By the surjectivity, there is MATH such that MATH, MATH. In particular, we have MATH and thus MATH. Then the map MATH is such that MATH and MATH. Since MATH, we obtain that the map MATH is surjective and thus MATH is injective. To obtain the first consequence (which is the result of NAME and NAME), note that MATH and CITE MATH. For the second consequence, we use a result due to CITE : if MATH is commutative, then MATH . This completes the proof. |
math/9906073 | The first statement was proved by CITE for the so-called NAME. As was proved by CITE and CITE, if MATH is a symmetric NAME MATH-algebra, then the matrix algebra MATH is also symmetric. It follows that symmetric NAME MATH-algebras are NAME. The fact that MATH follows from the density theorem in MATH-theory. Indeed, since MATH is symmetric, MATH is dense and spectral invariant subalgebra of MATH. Indeed, MATH has the same spectrum in MATH or in MATH. |
math/9906073 | Recall MATH is a dense and spectral invariant subalgebra of MATH. Then we have (compare for instance CITE) MATH . According to CITE and CITE, we have MATH and MATH . The result now follows from the above NAME and NAME periodicity theorem. |
math/9906080 | This is essentially REF, together with some general observations in CITE (see REF there). |
math/9906080 | To show MATH has dense range, we suppose the contrary, and let MATH be the nonzero quotient map MATH, where MATH. Then MATH, so that there exists MATH with MATH as a map on MATH, where MATH is the natural transformation MATH. Hence MATH, and thus MATH for some MATH. By REF , MATH for some MATH, so that MATH for MATH. Hence MATH, which is a contradiction. To show MATH is contractive it is sufficient to show that if MATH and MATH, then MATH. Let us rewrite the last expression. Let MATH be regarded as a map in MATH via right multiplication MATH; then clearly MATH. By REF , MATH, so that we may regard MATH as an element MATH of MATH of norm MATH. We claim that MATH. This follows because MATH, where MATH is the inclusion of MATH as the MATH-th entry in MATH, so that MATH . Thus MATH. The complete contraction is similar. |
math/9906080 | This is also almost identical to the analoguous result in CITE. One first establishes, for example, that for MATH, we have MATH, and this gives the REFnd commutant assertions as in CITE. We shall simply give a few steps in the calculation showing that MATH is a complete isometry; the missing steps may be found by comparison with CITE: MATH where we used the last part of REF in the last line. Thus MATH is a complete isometry. |
math/9906080 | We will use the facts stated in the first part of the proof of the previous lemma. By REF we know that MATH. Hence, using the second equation in (MATH), we see that MATH . If MATH is such that MATH, then MATH. Thus MATH, so that MATH. Since MATH we have MATH. Thus MATH for MATH . Since MATH has dense range in MATH, we see the multiplier assertion. |
math/9906080 | We first observe that as in CITE the natural transformations MATH and MATH imply the following equation: MATH for all MATH. Replacing MATH by MATH for MATH we have, as in CITE, that MATH A similar argument shows that for MATH, we have MATH . As in CITE this implies that MATH is a lowersemicontinuous element in MATH , for each MATH, and that MATH, as an element in MATH, corresponds to a lowersemicontinuous element in MATH (which we recall, is MATH-isomorphic to MATH). The remainder of the proof in CITE is the same, merely replacing the MATH which appears in the last few paragraphs there, by MATH, and replacing the element MATH there by MATH, where MATH is a c.a.i. for MATH. We obtain MATH. Similarly MATH for MATH. |
math/9906080 | We will use some elementary theory or notation from MATH-modules as may be found in CITE for example. It follows by the polarization identity, and the previous theorem, that MATH is a RIGHT MATH-module over MATH with inner product MATH . The induced norm on MATH from the inner product coincides with the usual norm. Similarly MATH (or equivalently MATH) is a right MATH-module over MATH. Also, MATH is a LEFT MATH-module over MATH, indeed it is clear that MATH, the so-called imprimitivity MATH-algebra of the right MATH-module MATH. The inner product is obviously MATH. We will show that MATH. NAME statements hold for MATH and MATH, and we will assume below, without writing it down explicitly, that whenever a property is established for MATH, the symmetric matching assertions for MATH. Let MATH be the linking MATH-algebra for the right MATH-module MATH , viewed as a subalgebra of MATH. We let MATH. It is easily seen, using equation REF , that MATH is an operator algebra containing MATH, and that the c.a.i. of MATH is a c.a.i. for MATH. We let MATH, and we define MATH to be the following subset of MATH: MATH . This is a subalgebra by REF . It is also easy to check that MATH and MATH. Therefore from REF we conclude that MATH. Comparing corners of these algebras yields MATH and MATH. Thus we see that MATH, from which it follows that MATH, since MATH is the MATH-algebra generated by MATH in MATH. Thus we have finally seen that MATH is a left MATH-module, and that MATH is a left MATH-module map. By symmetry, MATH is a left MATH-module and MATH is a MATH-module map, so that MATH . Also MATH and so MATH. Thus MATH. By symmetry note that MATH, and that MATH. Thus the conclusions of the theorem all hold. |
math/9906080 | We keep to the notation used until now. We will begin by showing that MATH is the maximal dilation of MATH, and MATH is the maximal dilation of MATH. We saw above that the set which we called MATH, equals MATH, so that MATH generates MATH as a left operator MATH-module. We have the following sequence of fairly obvious maps, using REF above: MATH . It is easily checked that MATH corresponds under the last two maps in the sequence to the map MATH, which lies in MATH since MATH is a left MATH-module map. Thus if MATH is the composition of all the maps in this sequence, then the range of MATH is contained in MATH. Moreover, MATH is an inverse to the restriction map MATH. Thus MATH. Hence by REF, MATH is the maximal dilation of MATH. A similar argument works for MATH. Let MATH. By REF above, and REF, we have MATH as left MATH-operator modules, where the last MATH is from REF . Now the latter space is a left MATH-operator module, hence MATH is a MATH-operator module, and (by a comment in REF, which is REF) the identity MATH above, is also valid as MATH-operator modules. One may easily check that this last identity is a natural isomorphism. But MATH is clearly a MATH-module functor. Hence MATH is MATH-restrictable. |
math/9906081 | Only the uniqueness requires proof. We shall require some facts about NAME algebras which may be found in CITE. Suppose that MATH and MATH are two contractive homomorphisms MATH, extending MATH. By REF , the ranges MATH and MATH of MATH and MATH are each MATH-algebras, but with possibly different involutions. We will write these involutions as MATH and MATH respectively. With respect to these involutions MATH and MATH are `*-homomorphisms'. Choose a c.a.i. MATH for MATH, and let MATH, and MATH. Then MATH is a NAME algebra with c.a.i. MATH. If MATH is a NAME algebra with c.a.i. we define MATH if MATH is unital, otherwise we let it be the unitization of MATH, with its `multiplier norm'. In the nonunital case, it is easy to see that MATH may be defined equivalently to be the subalgebra of MATH generated by MATH and a weak*-limit point of the c.a.i. In any case, the `unitized' MATH-algebras MATH and MATH may be viewed as subalgebras of MATH, with the same unit. Let MATH, and let MATH be a state on MATH (or equivalently on MATH). Then for MATH, MATH restricted to MATH is a state on MATH. Thus MATH. Thus MATH is a Hermitian element in MATH (or MATH) with numerical radius MATH, and consequently MATH. Therefore MATH on MATH. |
math/9906081 | If MATH is as above, then by the functoriality of the NAME tensor product, MATH is completely bounded. Composing this with the module action MATH gives the required map MATH. Its easy to see that MATH has the right properties. The uniqueness assertion is obvious. |
math/9906081 | Note that by definition of MATH, REF implies REF (and in fact one may replace MATH by any operator space). Put MATH in REF , and observe that MATH. Dualizing MATH now yields REF . Since, therefore, REF implies REF , we see that REF implies that MATH is a row contraction, so that MATH is also. |
math/9906081 | Only the last part still requires proof. Every nondegenerate NAME MATH-module MATH is a complemented submodule of a direct sum of MATH copies of the universal representation, where MATH is some cardinal. Thus the last assertion of the theorem reduces to proving that: if the restriction map gives a complete isometry MATH, then also MATH completely isometrically. One way to see this is to first check that MATH (see CITE), and similarly MATH. Here MATH, for an operator space MATH, is the collection of `columns' of length MATH with entries in MATH, whose truncated finite subcolumns are uniformly bounded. |
math/9906081 | If MATH, then the bounded net MATH has a weak*-convergent subnet, which easily converges weak* to MATH. That proves the first assertion. Next notice that if MATH satisfy REF , and if MATH is the canonical map MATH above, then MATH is MATH-essential if and only if MATH is MATH-essential. From this it is easy to see the MATH direction. Conversely, given a complete contraction MATH, then MATH lifts to a complete contraction MATH in MATH. A weak*-accumulation point of the MATH will be the desired extension of MATH. We leave it to the reader to fill in the remaining details. |
math/9906081 | Suppose that MATH is the module action on MATH. We have the following sequence of canonical complete contractive MATH-module maps: MATH . These maps compose to MATH, which yields the assertion. |
math/9906081 | By adjoining MATH to MATH, NAME 's result fairly obviously extends to the case when MATH is a nonunital MATH-subalgebra of MATH acting nondegenerately on MATH. Hence if MATH is the universal representation of MATH, and if MATH is a direct sum of copies of MATH, then MATH is MATH-injective. However, every nondegenerate NAME MATH-module MATH is a MATH-complemented submodule of such a MATH, and if MATH is the MATH-module projection onto MATH, then MATH as MATH-operator modules. Thus MATH is MATH-injective. If MATH is not nondegenerate, we let MATH be the essential part of MATH. To show that MATH is injective, is sufficient to show that MATH is MATH-injective, since any MATH-module map MATH into MATH has range inside MATH. We may assume MATH is nontrivial, otherwise the result is clear. However, by a routine NAME space cardinality argument MATH may be regarded as a MATH-complemented submodule of MATH where MATH is a large enough direct sum of copies of MATH. Finally, the MATH case is clear from the above, whereas the right injectivity of MATH follows from the left injectivity of MATH by noting that MATH is the `conjugate operator module' of MATH . |
math/9906081 | Note that just as in the Remarks after REF, it suffices to take MATH in REF to be MATH, where MATH is a NAME space, and MATH is an NAME MATH-module. By an argument similar to that given in those same Remarks (the main difference being that the map MATH there is a complete quotient map), this is equivalent to REF (in fact, one may replace the word `complete' in REF with `row'). To see that REF implies REF , we first observe that as in REF, we may assume MATH is nondegenerate. Using the functoriality of MATH, and the fact that every nondegenerate NAME MATH-module is a complemented submodule of a direct sum of copies of the universal representation, the result reduces to proving that if REF holds for MATH, then it also holds for MATH for some cardinal MATH. However this is easily seen from the injectivity of the NAME tensor product CITE, together with the operator space identification MATH, where MATH is the row NAME space of dimension MATH. That REF is equivalent to REF follows easily from the universal properties of MATH-injectivity, and REF. For the MATH direction take a completely contractive MATH-module map MATH. By REF we get a completely contractive MATH-module map MATH. Hence, by our hypothesis and MATH-injectivity of MATH, there is a completely contractive MATH-module map extension MATH. Then MATH restricted to MATH is a completely contractive MATH-module extension of MATH to MATH. The other direction follows easily by showing that the `closure of MATH' in MATH has the correct universal property (in the remark after REF). |
math/9906081 | By REF implies REF . Clearly REF implies REF , since MATH is naturally a complemented MATH-submodule of MATH. That REF implies REF is in CITE, since as we said, an injective NAME module is orthogonally injective, but in any case the proof is immediate by extending the inclusion MATH to a completely contractive MATH-module map MATH. Clearly MATH is the projection onto MATH, and since it is a MATH-module map we obtain REF . A similar idea shows that REF implies REF ; if MATH is as in REF , and if MATH is MATH restricted to MATH, let MATH, which is a representation of MATH on MATH. If MATH is as above, then MATH commutes with MATH. If REF holds, MATH commutes with MATH, which easily gives REF . Thus REF are all equivalent. Clearly REF implies REF . If we have REF , and if MATH is any MATH-operator module, then MATH . Taking MATH to be a NAME MATH-module shows REF . Assuming REF , namely that MATH is a NAME space, write MATH for the canonical map MATH mentioned in REF, and let MATH be the identity map. By REF, there is a completely contractive MATH-module map MATH such that MATH. By REF , MATH is a MATH-module map. Hence MATH is onto, which shows that MATH. Hence, by the universal REF , given a MATH-submodule MATH of any NAME MATH-module MATH as in REF , the inclusion map MATH is a MATH-module map; and so we see that REF holds. Thus REF are all equivalent. Clearly REF implies REF , whereas REF implies REF by REF. |
math/9906081 | Suppose that MATH is an equivalence functor. For MATH, we have MATH is a subalgebra of MATH. By an obvious argument (see for example REF), the map MATH from MATH to MATH is a isometric homomorphism. Hence its restriction to the MATH-algebra MATH is a *-homomorphism, and consequently maps into MATH. From this we see that if MATH, then MATH is a MATH-algebra for all NAME MATH-modules. By the implication ` MATH ' in REF , we see that MATH is a MATH-algebra. Now suppose that MATH and MATH. Let MATH, and let MATH and MATH be, respectively, the inclusions and projections between the MATH and MATH. Thus MATH, so that MATH. From REF it follows that MATH and MATH are MATH-module maps. By the first part, MATH is a MATH-module map. Thus MATH is a MATH-module map. We have shown that MATH restricts to a functor from MATH to MATH. Similarly for MATH, and now the category equivalence is clear. Note that the natural transformation maps are unitary and commute with the action of the operator algebra, and hence also commute with the action of the generated MATH-algebra. Thus we have REF . |
math/9906081 | It is easy to see that REF follow from REF . We shall simply show that MATH generates MATH as a left MATH-operator module. Since the pairing MATH has dense range, we can pick a c.a.i. for MATH which is a sum of terms of the form MATH, for MATH. This c.a.i. is also one for MATH, and hence sums of terms of the form MATH, for MATH are dense in MATH. However, MATH (where MATH is the other pairing). So MATH generates MATH as a left MATH-operator module. |
math/9906081 | By the previous result, MATH and MATH generate MATH and MATH respectively as left operator modules. Thus we have a complete contraction MATH with dense range. On the other hand MATH . However, the pairing MATH determines a complete contraction MATH, and so we obtain a complete contraction MATH. One easily checks that the composition of these maps MATH is the identity, from which it follows that MATH. Similarly MATH is the dilation of MATH. |
math/9906081 | If MATH and MATH are as usual the maximal MATH-algebras of MATH and MATH respectively, and if MATH is a left dilatable subcontext of MATH then, using REF , we have MATH completely isometrically. On the other hand, we have the canonical complete contraction MATH coming from the restricted pairings in REF . It is easy to check that the composition of the maps in these two sequences agree. Hence the canonical map MATH is a completely isometric isomorphism. Similarly, MATH completely isometrically. Thus by the remark before REF (or see REF and the `sketch' beneath it), MATH and MATH are strongly NAME equivalent operator algebras. The last statement is proved in CITE. |
math/9906081 | Write MATH and MATH for the pairings discussed in REF . Notice firstly, that there is a natural map MATH coming from these pairings. Hence we get a sequence MATH where the second last map comes from REF. However, the composition of maps in this sequence agrees with the inclusion of MATH in MATH. Hence the map MATH is a complete isometry. That this map is onto follows by the argument of REF . A similar proof shows that MATH as operator bimodules, and that MATH and MATH (completely isometrically) as operator algebras. Define MATH and MATH, we will show that MATH and MATH are are completely contractive equivalence functors between the category operator MATH-submodules of MATH-operator modules, and the category of MATH-submodules of MATH-operator modules, which compose (up to natural completely isometric isomorphism) to the identity functor. If MATH is a MATH-operator module, then by REF , we have MATH . Moreover, this, together with the corresponding result for MATH, shows that MATH. For a general MATH-operator module MATH, there is a canonical complete contraction MATH given by MATH, for MATH. Suppose that MATH is a MATH-operator module, and that MATH is a MATH-operator module containing MATH. Then we get the following sequence of complete contractions MATH . The first map here is MATH. The composition of maps in this sequence is the inclusion map, and so MATH is a complete isometry. To show that MATH is onto in the unital case is a simple exercise in algebra. In the nonunital case, to show that MATH is onto, one may use an argument similar to those in the proof of REF showing that the maps there are onto. That NAME modules in these categories are taken by this equivalence to NAME modules follows easily from the observations above. |
math/9906081 | Only the last statement still needs a word of proof, and this is similar to the proof that REF implies REF. |
math/9906082 | In the MATH case, note MATH, and it has norm REF (as may be seen by noting that it corresponds to the identity map after employing the canonical completely isometric identification MATH). Applying MATH, we find MATH has norm REF. However, via the canonical completely isometric isomorphism of MATH with MATH, MATH corresponds to the canonical morphism MATH. So this latter morphism is a complete contraction. Similarly the canonical morphism MATH is a complete contraction. Applying MATH to this morphism, gives a complete contraction MATH, which yields a complete contraction MATH. This proves the lemma for MATH. The MATH case is similar. |
math/9906082 | REF direction is easy and is omitted CITE. A simple proof of the other direction may be found in CITE (REF , setting MATH). For completeness, we sketch a slight simplification of their argument. We use canonical operator space identifications, which may be found in CITE, and the notation of CITE. By the complete injectivity of the minimal and NAME tensor product, (see CITE for example), and the fact that column NAME space is determined by its finite dimensional subspaces being column space, it follows that MATH, for any NAME space MATH. Choose MATH so that MATH. The last ``MATH" may be rewritten as MATH, where MATH is the operator space projective tensor product. Next, recall that the functors MATH and MATH are the same. Applying this functor to the identity MATH yields the identity MATH, where MATH is the operator space predual of MATH. Taking the operator space dual yields MATH. Thus the inclusion map MATH factors through NAME column space. Hence MATH is NAME column space. |
math/9906082 | Using REF , and the fact that MATH, we have MATH completely isometrically. Sorting through these identifications shows that an element MATH corresponds to the map MATH in MATH. A similar proof works for MATH. |
math/9906082 | We shall simply prove the assertions for MATH; those for MATH are similar. The module map assertions are fairly clear, for instance MATH . Next we show the MATH assertion. By REF , we have a MATH-isomorphism MATH. Note MATH for MATH, and so also MATH. Together these imply that MATH. The matching assertion for MATH has the additional complication of the maps MATH, however since they are unitary as remarked above, they disappear from the calculation. Finally, we turn to the complete isometry. The equalities in the following calculation follow from, in turn, REF above, the definition of MATH, REF , the definition of MATH, REF again, and REF : MATH . Thus MATH is a complete isometry. |
math/9906082 | If MATH for all MATH, then by the polarization identity, and the previous lemma, MATH is a RIGHT MATH-module over MATH with i.p. MATH . We can also deduce that MATH is a LEFT MATH-module over MATH by setting MATH. This last quantity may be seen to lie in MATH by using the polarization identity and the following argument: Since the range of MATH is dense in MATH, we can find a c.a.i. MATH for MATH, with terms of the form MATH (using REF ). Here MATH depend on MATH. Then MATH is also a c.a.i. for MATH. Since MATH, it follows that MATH is a norm limit of finite sums of terms of the form MATH, where MATH. Thus MATH. A similar argument shows that MATH (or equivalently MATH) is both a left and right MATH-module. At this point we can therefore say that the right module actions on MATH and MATH are nondegenerate. Notice also, that if we choose a contractive approximate identity for MATH of form MATH as above, then MATH is also a c.a.i. for MATH. However MATH where MATH. Since MATH is a positive matrix, it has a square root MATH, say, with entries MATH. Thus MATH where MATH. From this one can easily deduce that the MATH-valued innerproduct on MATH has dense range, that is, MATH is a full right MATH-module over MATH. Similar arguments show that MATH is a full left MATH-module over MATH, and that MATH is also full on both sides. Thus MATH and MATH are strong NAME equivalence bimodules, giving the strong NAME equivalence of MATH and MATH. Observe that by the basic theory of strong NAME equivalence (see for example, CITE) MATH. Thus if MATH is a c.a.i. for MATH, then MATH is a c.a.i. for MATH, where MATH. Observe too, by REF , that MATH in norm MATH completely isometrically, where MATH is an approximate identity for MATH. Applying the functor MATH and REF , we see the last set is completely isometrically isomorphic to MATH in norm MATH, which is completely isometrically isomorphic to MATH in norm MATH, which in turn equals MATH, since MATH. Thus we have shown that MATH completely isometrically, and it is an easy algebra check that this is also as left MATH-modules. Setting MATH gives MATH, so that MATH. It is easily checked that this last relation is as bimodules too. In REF, we showed that MATH completely isometrically. Thus MATH completely isometrically and as MATH-modules, for all MATH. Its an easy algebra check now that MATH as functors. By symmetry, we get the matching statement for MATH. The last statement of REF , about the mapping of subcategories, follows because MATH coincides with the interior tensor product on the subcategories concerned. |
math/9906083 | Suppose that MATH, as a nondegenerate MATH-subalgebra. Let MATH be a minimal MATH-projection on MATH, and let MATH. As noted earlier, MATH is completely positive and *-linear. By REF REF , MATH is a unital MATH-algebra with respect to the old linear and involutive structure, but with product MATH. Also, MATH is, by the note above the statement of the theorem, a copy of the injective envelope of MATH. Let MATH be the MATH-subalgebra of MATH generated by MATH, with respect to the new product. By the universality of the injective envelope, it is clear that as a MATH-algebra generated by a copy of MATH, MATH only really depends on MATH and its identity element. With respect to the usual product on MATH, the MATH-subalgebra of MATH generated by MATH contains MATH, the MATH-subalgebra of MATH generated by MATH. A key part of the NAME theorem is the relation MATH, for MATH. Hence by induction it follows that MATH, for MATH. Hence MATH is a *-homomorphism MATH, with respect to the new product on MATH. Since MATH extends the identity map on MATH it clearly also maps into MATH. Since MATH has dense range, it is necessarily surjective. |
math/9906083 | Here is one proof using the construction above. Since MATH acts nondegenerately on MATH, we may view MATH. Then if MATH corresponds to such a left multiplier, and if MATH corresponds to an element in MATH, we may write (by NAME 's factorization theorem) MATH for an operator MATH (respectively, MATH) corresponding to an element in MATH (respectively, MATH). Hence MATH. Thus it is clear that MATH is a left operator MATH-module. |
math/9906083 | Suppose MATH has generators MATH. The map MATH which takes MATH, is onto. By the open mapping theorem there is a constant MATH such that for any MATH, there exists MATH with MATH, such that MATH. Given any MATH and MATH, choose MATH as above. We have MATH . Here MATH is as in the remark after REF . Thus we see that MATH. Thus MATH is e.n.v. |
math/9906083 | Most of these follow simply from the fact that MATH for MATH. For example, if MATH is MATH-faithful then MATH, so that MATH is bicontinuous. However since norm equal spectral radius on function algebras, MATH is isometric. We leave the remaining assertions as exercises. |
math/9906083 | CASE: If MATH is any single generator of MATH, then MATH is nonvanishing (since MATH for all MATH). If MATH then MATH, so that MATH for some MATH, and if MATH then MATH. Conversely, if MATH then MATH. The last part is clear. From REF , the open mapping theorem, and REF , we get REF . That MATH is isometric in this case is because norm equals spectral radius on function algebras. |
math/9906083 | Clearly REF implies REF . By REF implies REF . If REF holds then by the previous theorem we obtain a unital *-homomorphism MATH, whose range is a MATH-subalgebra which separates points. Thus MATH is onto, so again by the previous theorem MATH, giving REF . |
math/9906083 | The REF direction is easy. The REF direction follows from the proof of the theorem as follows. If MATH is a.s.g. then MATH. Since MATH is bounded away from MATH, it follows that MATH is uniformly closed. If in addition MATH is faithful then it follows from REF and the open mapping theorem, that MATH is isometric. The rest is clear. |
math/9906083 | Suppose that MATH is left e.n.v. The image of MATH in the corner of MATH, together with the the identity of the MATH corner of MATH, generates a unital MATH-algebra MATH inside MATH. We do not assert yet that the identity of MATH is the identity of MATH. Inside MATH, the image of MATH and MATH, and the two idempotents on the diagonal of MATH corresponding to MATH and to the identity of the MATH corner of MATH, form an operator system MATH. By NAME 's lemma, the obvious map MATH is a complete order isomorphism. By the NAME theorem (REF above), MATH extends to a surjective *-homomorphism MATH. If MATH is the embedding into REF-corner, and if MATH is the embedding of REF-corner of MATH inside MATH, then it is easy to see that MATH for MATH. Hence the restriction of MATH to MATH is the identity map on MATH. If MATH is the idempotent in REF corner of MATH then MATH. So MATH. |
math/9906083 | CASE: That MATH and MATH follows from REF, or by REF below. Hence MATH, and so within MATH, we have MATH. If MATH is a unital operator algebra then we remarked earlier that one can take MATH to be a unital homomorphism, so that MATH, implying that MATH. If MATH is a system, then MATH implies that MATH (since MATH). This yields the last assertion. CASE: See REF for example. |
math/9906083 | It is easy to check that any MATH is linear, and bounded by the closed graph theorem. We leave it to the reader to check that MATH is a MATH-algebra with the MATH norm, and involution MATH, where MATH is as in the definition of MATH. For example, for MATH and with MATH we have: MATH . CASE: It is an easy exercise to check that MATH as *-algebras. Hence the *-isomorphism must be isometric. CASE: By the above, MATH isometrically. On the other hand, if MATH, with MATH, then MATH is a finite convex combination of unitaries in MATH CITE. For such a unitary MATH, MATH, for MATH. Hence MATH, so that MATH . Hence MATH. Thus we have proved REF also. CASE: This follows from REF and the above, with MATH replaced by MATH. CASE: Let MATH, and let MATH be such that the row MATH has norm MATH. Then from REF, we have that MATH. |
math/9906083 | We need only prove the assertions about MATH. The map MATH above clearly maps MATH into MATH, since any MATH is adjointable in the usual sense on MATH. This map MATH is clearly a REF *-homomorphism. If MATH is a unitary in MATH, then MATH for all MATH. Since MATH is dense, there is one possible extension of MATH to a MATH, and its easy to see that MATH is well-defined and isometric (compare proof of REF). Clearly MATH. Since the unitaries span a MATH-algebra, MATH is onto. |
math/9906083 | This follows from the relation MATH (see REF) and the facts MATH for any right NAME MATH-module MATH CITE. Putting MATH , and appealing to the definition of MATH gives the result. |
math/9906083 | CASE: Any MATH satisfying REF , is clearly in MATH, and moreover we have MATH. Thus MATH over all MATH as in REF . CASE: There are many ways to see this. For example, we saw at the end of the introduction that there is a complete isometric injection MATH of MATH into MATH, such that for every MATH there is a MATH, such that MATH for every MATH. Let MATH be the MATH matrix in MATH with MATH in REF corner and zero elsewhere. This gives REF . Since MATH equals the MATH-norm of MATH, the latter dominates the infimum of MATH possible in REF . REF : if MATH, we define MATH by MATH, for MATH and MATH. We see that MATH is well defined and bounded, since if MATH then MATH . Clearly MATH, and MATH . We have also, by the way, established the final assertion of our theorem. The least value of MATH possible in REF is achieved by the MATH above. The REF direction in REF , and the fact that MATH in REF , may be proved almost identically to the REF direction in REF ; after using the fact that MATH (established in REF). Next we prove REF . Notice the hypothesis MATH defines a function MATH given by MATH. Also we have MATH for MATH. It is very clear that MATH in REF . To check that MATH, note that in REF we may as well assume that MATH is the embedding of MATH into its injective envelope, or triple envelope, by NAME 's universal REF . Then if MATH is selfadjoint, we have that MATH so that by polarization, we see that MATH is left adjointable on MATH, with MATH. If MATH we may obtain MATH exactly as in CITEEF. And if MATH for all MATH, then one can say that since MATH, we have MATH both positive and negative in MATH. Consequently MATH, and since MATH we have that MATH is a (orthogonal) projection in MATH. |
math/9906083 | This can be proved directly from the definition of MATH, and the universal property of MATH. Or it follows immediately from REF above. |
math/9906083 | Note that MATH is a NAME MATH-extension of MATH, since by REF , any c.a.i. for MATH is also one for MATH. Thus by the universal property of MATH (see REF), we see that MATH is the quotient of MATH by a closed ideal MATH say. Therefore MATH is a MATH-algebra generated by MATH. Therefore MATH is a MATH-algebra generated by MATH, and by the NAME theorem, there exists a *-homomorphism MATH, which restricts to a *-homomorphism MATH, extending MATH. Thus it is clear that MATH, so that MATH. It is easy to see from the fact mentioned above that any c.a.i. for MATH is also one for MATH, that MATH. Similarly, MATH, since we may represent MATH nondegenerately on the same NAME space. REF-sided multiplier algebra of an operator space which we defined earlier (above REF), is thus equal to the usual multiplier algebra of MATH CITE. Now MATH, since any minimal MATH-projection on MATH is a MATH-projection, and is consequently the identity. Thus MATH, since for any MATH-algebra MATH we have MATH (see for example the end of CITE). Since MATH, we see that any minimal MATH-projection on MATH is a MATH-projection and is consequently the identity. Thus MATH. |
math/9906083 | We will write MATH for the identity in MATH. The MATH-algebra MATH is injective, and has as sub-systems MATH where we have identified the diagonal idempotents in MATH and MATH with MATH. As we said in the introduction, NAME 's results imply that there is a minimal MATH-projection MATH on MATH. As in CITE, we may write MATH . Now MATH fixes MATH , so by rigidity of MATH, MATH. Thus MATH is a MATH-projection. Now one sees that MATH fixes MATH, which is a unital MATH-algebra. Hence by NAME 's multiplicative domain lemma (see for example REF ) MATH is a MATH-bimodule map. Thus MATH. Hence MATH, which proves REF . REF is straightforward to check. |
math/9906083 | It is not hard to see by the ordinary NAME theorem, that MATH is equivalent to the first condition in MATH. It is also not hard that MATH. We will therefore be done if we can show that the first condition in REF implies the second. To that end, note that if MATH is a closed MATH-submodule containing MATH, then MATH. The latter set is a *-subalgebra, and is therefore dense in MATH by the first condition in REF . Choose (by basic C*-module theory), an approximate identity MATH for MATH, such that each MATH is of the form MATH, for MATH. For any such MATH and MATH, we have MATH, since MATH is a module. Thus MATH. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.