paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 co... |
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 RE... |
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/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. Simil... |
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... |
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 ... |
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... |
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-di... |
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/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 ... |
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. ... |
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... |
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... |
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... |
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 ... |
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 MA... |
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 ... |
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... |
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 i... |
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. CA... |
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 ... |
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... |
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 a... |
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... |
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 M... |
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 n... |
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 statem... |
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 th... |
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 ma... |
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 i... |
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 ... |
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 th... |
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... |
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 MA... |
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 ... |
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, sinc... |
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. H... |
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 compa... |
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 ... |
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 , f... |
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. Sim... |
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 ... |
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 w... |
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 MAT... |
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/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 ... |
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 fa... |
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-mod... |
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 tha... |
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,... |
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 t... |
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... |
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... |
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 corres... |
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 inject... |
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 ... |
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 argumen... |
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... |
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... |
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 MAT... |
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... |
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 r... |
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 ... |
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). Thi... |
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 ch... |
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 M... |
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... |
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 ... |
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/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. Th... |
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