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math/0105115 | MATH is generated by a unital copy of MATH and a subalgebra MATH, where MATH is a projection of trace MATH and where MATH and MATH are free in MATH. Let MATH be such that MATH. Recall that the canonical embedding MATH is the embedding MATH arising from the free product construction. Let MATH be projections such that MATH, let MATH and let MATH and MATH be systems of matrix units in MATH such that MATH . We may and do choose MATH and MATH so large that if MATH is equipped with the trace inherited from MATH then then MATH is a factor and MATH is a factor. Let MATH and MATH, where MATH is the tracial state on MATH. We have MATH where MATH . Thus MATH where MATH . We have MATH and, by REF , MATH and MATH are free over MATH with respect to the trace - preserving conditional expectation MATH. On the other hand, letting MATH, we have MATH and MATH where MATH. Therefore MATH. Furthermore, MATH while by REF , MATH and MATH are free over MATH with respect to the trace - preserving conditional expectation MATH. The NAME algebras MATH and MATH are isomorphic, and we can choose an isomorphism so that MATH and MATH. Using this isomorphism, sending MATH identically to itself and sending MATH and MATH, from REF we get an isomorphism MATH which is the identity on the embedded copies of MATH. By CITE, MATH. |
math/0105115 | Iterating REF , we see that the image of the canonical embedding MATH is freely complemented by an algebra isomorphic to MATH where MATH and MATH, By the results of CITE, MATH and MATH. |
math/0105115 | Use induction on MATH. When MATH then by construction, MATH . For MATH, MATH where MATH and MATH is as in REF. The isomorphisms above are from the nesting result REF , the induction hypothesis and, respectively, REF combined with CITE. |
math/0105115 | Suppose MATH. We will perform a variant of the construction in the proof of REF for MATH . We may rewrite MATH as MATH where MATH. Let MATH, MATH and MATH be the embeddings arising from the free product construction. We may choose, for each MATH, MATH so that MATH and MATH. Then MATH . But the family MATH is free with respect to the free product trace on MATH, while MATH so MATH where the third isomorphism above is because by CITE, every free product of infintely many MATH - factors is stable under taking the free product with MATH. This proves the isomorphism REF and that MATH is freely complementted in MATH by an algebra isomorphic to REF. Now suppose MATH, for some MATH. We may without loss of generality take MATH. Let MATH be generated by free subalgebras MATH and MATH, where MATH and MATH. Then using the nesting result REF , MATH . By REF , MATH via an isomorphism intertwining the canonical embedding MATH and the embedding MATH coming from the free product construction. Therefore, there is an isomorphism MATH intertwining the canonical embedding MATH and the composition of the embedding MATH coming from the free product construction and the canonical embedding MATH . Now applying the part of the theorem already proved shows the isomorphism REF and that MATH is freely complemented in MATH by an algebra isomorphic to MATH . |
math/0105115 | Let MATH be an infinite set of MATH such that MATH and such that MATH is also infinite. By the nesting result REF , MATH where MATH . If we can show MATH then, since MATH, by CITE MATH and the isomorphism REF will follow from REF . Hence we may without loss of generality assume MATH for all MATH. Let MATH with trace MATH be as in the proof of REF . Recall MATH where the projection MATH is arbitrary subject to its trace being MATH. Let us fix a projection MATH of trace MATH, and let us take MATH for all MATH. Let MATH and recall from REF that the family MATH is free over MATH with respect to the canonical trace - preserving conditional expectation MATH. Using partial isometries from MATH to bring everything under MATH, we see that MATH and that the family MATH is free over MATH with respect to MATH. Now MATH and, moreover, MATH and MATH are free by REF . It follows from REF that MATH is freely complemented in MATH by an algebra, let us call it MATH, isomorphic to MATH where MATH. Thus MATH and the family MATH is free with respect to MATH, yielding MATH with MATH freely complemented in MATH by an algebra isomorphic to MATH . Application of REF gives the isomorphism REF, and that MATH is freely subcomplemented in MATH by an algebra isomorphic to the one displayed at REF. |
math/0105115 | Write MATH as in REF with for every MATH, MATH for projections MATH satisfying either MATH or MATH. If MATH for all MATH then MATH and the conclusions of the lemma are clear. Assume MATH for all MATH and, for some MATH, MATH and either MATH or MATH. For every MATH, let MATH . Then MATH and MATH and MATH are free. By REF , MATH is freely complemented in MATH by an algebra isomorphic to MATH . Combining these embeddings, one obtains MATH where MATH, and that the algebra MATH is freely complemented in MATH by an algebra, call it MATH, isomorphic to MATH . Then MATH . Now the conclusions of the lemma are clear. |
math/0105115 | By REF , we have isomorphisms MATH that intertwine the corresponding canonical embeddings. These combined with the isomorphism MATH obtained from REF give the desired isomorphism REF. |
math/0105115 | In order to prove the isomorphism REF for all MATH, it will suffice to show it for all MATH. So assume MATH. If there is MATH such that MATH for infinitely many MATH, then the existence of the isomorphism REF with the required properties follows from REF . Hence we may assume either MATH for some MATH or MATH and MATH, (in which case we let MATH). Assume also MATH. If MATH then the conclusion of the theorem follows from REF . So assume there is MATH such that MATH and either MATH or MATH. Letting MATH by REF is freely complemented in MATH by an algebra isomorphic to MATH where MATH. By REF , MATH is freely complemented in MATH by an algebra isomorphic to MATH . If MATH then we are done. Otherwise, by REF , there is an isomorphism MATH intertwining the inclusion MATH and the canonical embedding MATH . Now REF shows that there is an isomorphism MATH intertwining the inclusion MATH and the canonical embedding MATH . This together with the fact that MATH is freely complemented in MATH by an algebra isomorphic to REF finishes the proof. |
math/0105115 | If MATH then this is immediate from the definition of free scaled products, REF . Let MATH denote the tracial state on MATH. Suppose first MATH, MATH. Then MATH and we may take MATH where MATH is a unital copy of MATH, MATH with MATH, MATH is a system of matrix units in MATH, the family MATH is free with respect to MATH, MATH with MATH a subalgebra of MATH isomorphic to MATH and the pair MATH is free with respect to MATH. Let MATH . Then MATH . We shall show that MATH and MATH are free with respect to MATH. Let MATH . Then we have MATH . Hence, for freenes of MATH and MATH, it will suffice to show MATH . After regrouping, any word MATH beloning to the LHS of REF is seen to be equal to MATH, for some MATH, where MATH . But freeness of the pair REF shows MATH and thus MATH. This shows the existence of the isomorphism REF in the case MATH. Now suppose MATH is not a reciprocal integer. Let MATH be such that MATH and let MATH, MATH and MATH . By the case just proved, regarding MATH as contained in MATH via the canonical embedding, MATH is freely complemented in MATH by a copy of MATH. Let MATH be a projection of trace MATH. By REF , MATH is freely complemented in MATH by a copy of MATH . On the other hand, by REF , there is an isomorphism MATH intertwining the inclusion MATH and the canonical embedding MATH. As MATH, we are done. |
math/0105115 | If MATH then this is just REF . Suppose MATH. Then by REF , since MATH, the image of MATH in MATH under the canonical embedding is freely complemented by an algebra isomorphic to MATH. On the other hand, by the definition of free scaled products REF , the image of MATH in MATH under the canonical embedding is freely complemented by an algebra isomorphic to MATH . From this, we can construct the isomorphism REF in the case MATH. Now suppose MATH. Denote by MATH the canonical embedding, let MATH and let MATH denote the canonical embedding. Let MATH be a projection of trace MATH. Then using REF , there is an isomorphism MATH intertwining MATH and the canonical embedding MATH . Since MATH, REF gives an isomorphism MATH intertwining MATH and the canonical embedding MATH . On the other hand, by REF , there is an isomorphism MATH intertwining MATH with the canonical embedding REF. The isomorphisms REF together with REF give the desired isomorphism REF. |
math/0105115 | We begin by proving REF and a special case of REF simultaneously. Suppose MATH for all MATH. Denote by MATH the tracial state on MATH. We may write MATH where MATH is a unital copy of MATH, MATH with MATH, MATH for a projection MATH of trace MATH, MATH is a subalgebra of MATH isomorphic to MATH if MATH and to MATH if MATH, MATH and MATH are free with respect to MATH and, finally, the family MATH if free over MATH with respect to the MATH - preserving conditional expectation MATH, (see REF ). Using REF , we get MATH where MATH with MATH, MATH and MATH are free, MATH is a subalgebra, for a projection MATH of trace MATH, MATH and, finally, MATH and MATH are free with respect to MATH. Thus MATH and we get an isomorphism REF, with MATH as in REF, intertwining the canonical embeddings. This proves REF in the case MATH for all MATH. For REF , if MATH, then MATH can be chosen making MATH. Then the isomorphism REF follows by REF . In order to prove the general case of REF , let MATH and let MATH . Using the nesting result REF and, twice in succession, the case of REF just proved, we get isomorphisms MATH where MATH, whose composition intertwines the canonical embeddings. |
math/0105115 | Suppose the free group factors are isomorphic. Then MATH where the second isomorphism is from REF , the third isomorphism is a consequence of isomorphism of free group factors by CITE and the last isomorphism is from REF . On the other hand, suppose REF holds in general. From REF we have MATH while the isomorphism REF gives MATH . |
math/0105117 | Since MATH is normal, MATH. We know that MATH, thus MATH for all MATH. Since MATH is bounded, MATH is the closure of MATH and MATH is the closure of MATH, it follows that MATH. Using the equality MATH, one proves in a similar way that MATH. |
math/0105117 | By REF we know that MATH is generated by the set MATH. Since MATH, it follows that the NAME algebra generated by MATH equals the NAME algebra generated by MATH. Thus, the NAME algebra generated by MATH equals MATH. Because MATH, it now follows that MATH is generated by MATH. Therefore the first statement holds, and as a consequence also the second one holds. Define the partial isometries MATH on MATH such that MATH and MATH for all MATH and MATH. Letting MATH denote the NAME algebra generated by MATH, the second statement implies that MATH. It is not so difficult to see that every rank one projector of the form MATH, where MATH, belongs to MATH, thus MATH. |
math/0105117 | We only prove the first equality, the other ones are even more straightforward to check. Take MATH such that MATH for all MATH and MATH such that MATH. Now MATH, whereas MATH. Let MATH and MATH. We consider REF different cases: CASE: If MATH and MATH, then MATH. Thus MATH. CASE: Now suppose that MATH and MATH. Then MATH. On the other hand, MATH. Since MATH, we have that MATH. Thus MATH. CASE: Suppose that MATH and MATH . Then MATH since MATH. Thus MATH. CASE: If MATH and MATH, then clearly MATH. So we see that MATH for this kind of function MATH. Thus MATH. |
math/0105117 | If MATH, this follows easily from REF . Now assume that MATH. By REF , the above equation is equivalent to MATH . This equality follows easily from REF . |
math/0105117 | Define the MATH-homomorphism MATH such that MATH for all MATH. Fix MATH. If MATH, MATH, then REF implies that MATH. It follows that MATH. As a consequence, MATH for all MATH. It follows that MATH for all MATH. Thus, MATH by REF . In a similar way, MATH. Hence MATH by the commutator theorem for tensor products. |
math/0105117 | Take MATH and MATH. REF implies that MATH for all MATH. If MATH and MATH, then REF implies that MATH . Thus, MATH . From this all we conclude that MATH. Notice that MATH. Thus MATH . Let MATH denote the flip map. If we let MATH denote the permutation map defined by MATH for all MATH and MATH, we get that MATH . |
math/0105117 | Define MATH. Set MATH and equip MATH with the strong-MATH topology. In MATH, we consider the set MATH that is the closure of MATH. Let MATH denote the set of all finite subsets of MATH and turn MATH into a directed set by inclusion. For MATH, we define the projection MATH. Thus, MATH converges strongly to REF. Fix MATH for the moment. By REF we see that MATH. Since MATH for all MATH, NAME 's density Theorem (see the proof of CITE), applied to the MATH-algebra MATH, implies that MATH. It is clear from REF that MATH and MATH. Thus, MATH. It follows that MATH, implying that MATH and the lemma follows. |
math/0105117 | Recall that MATH is also the modular conjugation of the trace MATH. So REF in the previous proof implies that MATH. Let MATH. By REF , MATH and MATH. So we get that MATH. Thus, since MATH, MATH and also the second claim follows. |
math/0105117 | The proof of this proposition is based on the properties of the functions MATH, in particular on REF . We now give details. Fix MATH and MATH. If MATH such that MATH, MATH and MATH, then it is clear that the element MATH is orthogonal to the element MATH. By REF we get moreover MATH . From this, it follows that we can define the element MATH as MATH . Note that MATH is just the right hand side of the formula we want to define MATH by. Now choose also MATH and MATH. Notice that REF together with the NAME equality implies that MATH is finite, which allows us to compute the next sum in any order we want: MATH . Thus, using the orthogonality relation in REF together with REF , we get that MATH . Because of the factor MATH in the above expression, REF implies that MATH . Now the lemma follows easily. |
math/0105117 | Choose MATH, MATH. Take also MATH and let us calculate the element MATH. Choose MATH and MATH. Then REF tell us that MATH by REF . So we see that MATH if MATH. If, on the other hand, MATH, then MATH . Now we have for MATH and MATH that MATH . So we see that MATH if MATH. If, on the other hand, MATH, then MATH and thus MATH . Applying MATH to this equation and using REF , we see that MATH . Or, again by REF , MATH so we conclude that if MATH, the element MATH belongs to MATH and MATH . Also recall that MATH if MATH. Notice that REF implies that MATH. So we get by REF that MATH . So we see that MATH if MATH. If, on the other hand, MATH, MATH . From REF , we conclude that MATH . We use the net of projections MATH introduced in the proof of REF . If MATH, we know by the previous equation that MATH and MATH . Therefore the MATH-strong-MATH closedness of MATH implies that MATH and MATH . Since the linear span of such linear functionals MATH is norm dense in MATH, the proposition now follows easily from REF and the MATH-strong-MATH closedness MATH. |
math/0105117 | Take MATH and MATH. Since MATH is in standard form with respect to MATH, there exists a vector MATH such that MATH. Take also an orthonormal basis MATH for MATH. It follows from the proof of CITE that MATH is an increasing net in MATH that converges strongly to MATH (as before, MATH denotes the set of finite subsets of MATH, directed by inclusion). Therefore the MATH-weak lower semi-continuity of MATH implies that MATH . By the previous proposition, this implies that MATH . So if we use first CITE and afterwards the fact that MATH is an isometry REF , we see that MATH and we have proven the proposition. |
math/0105117 | If MATH and MATH, then MATH. Hence, REF implies that MATH. This same lemma implies moreover for MATH and MATH: MATH . If MATH, the above result implies that MATH and MATH. By REF and the MATH-strong-MATH closedness of MATH this implies that for all MATH, the element MATH and MATH, thus MATH. Since MATH, it follows that MATH. Because MATH and MATH is left invariant, we get that MATH is right invariant. |
math/0105117 | We denote the modular element and scaling constant of MATH by MATH and MATH respectively, see the discussion after the proof of CITE. We know that MATH for all REF and MATH. Because MATH is right invariant, the uniqueness of right NAME weights implies the existence of MATH such that MATH. Thus, the uniqueness of the NAME derivative guarantees that MATH. But, since MATH, it follows that MATH. Therefore, REF implies that MATH. |
math/0105117 | Let MATH and MATH. By REF , we have for MATH satisfying MATH, MATH which by REF implies that MATH. It follows that the sum in the statement of this proposition is absolutely convergent. So, as a norm convergent limit of finite sums of vector functionals, we obtain MATH. Take MATH and MATH. Now MATH . By REF , we get immediately that MATH if MATH or MATH. Now suppose that MATH and MATH. Then REF imply that MATH . This sum is absolutely convergent so we can compute it in any order we deem useful. Therefore the orthogonality relations in REF imply that MATH where MATH is the NAME of MATH. If MATH, then MATH implying that REF also holds if MATH and MATH. Therefore the lemma follows. |
math/0105117 | Define MATH as in REF for MATH, MATH, MATH and MATH. Then REF implies that MATH. Thus CITE implies that MATH and MATH where, as is customary, MATH is given by MATH for all MATH. By definition, MATH . Thus, MATH where we used REF in the last equation. A simple change in summation variables MATH and MATH then reveals that MATH which upon close inspection shows that MATH is MATH times the functional in REF where MATH, MATH, MATH and MATH. By REF this implies that MATH . Therefore REF implies that MATH proving the result. |
math/0105117 | By REF and the discussion after the proof of CITE we know that there exists a strictly positive operator MATH on MATH such that MATH for all MATH and MATH. Note that MATH for all MATH. Choose MATH and MATH. Since MATH and MATH and MATH commute, we see that MATH. Thus, REF implies that MATH and MATH. Thus, REF and the NAME algebraic version of CITE guarantee that MATH and MATH. Therefore, REF tells us that MATH and MATH. Because such elements MATH form a core for MATH, we conclude that MATH. Taking the adjoint of this inclusion, we see that MATH. As a consequence, MATH. So if MATH and MATH, we see that MATH, implying that MATH. Note that MATH for all MATH, MATH and MATH. Now a straightforward calculation reveals that MATH for all MATH, MATH and MATH. Fix MATH and MATH for the moment. By REF. The previous paragraph implies that MATH. Therefore REF guarantees that MATH. Using the definition of MATH, one checks easily that also MATH and thus by the previous calculation, MATH. Since the NAME algebra MATH is generated by such elements MATH, we conclude that MATH for all MATH. |
math/0105117 | Let MATH and define MATH such that MATH for all MATH. The reason for introducing the function MATH stems from the fact that MATH for MATH and MATH (and which follows from REF ). Since MATH is the closed linear span of vector functionals, MATH is the closed linear span of elements of the form REF . Let MATH. Take MATH such that MATH. We consider two different cases: CASE: Note first of all that in this case, MATH for all MATH. REF implies the existence of a bounded function MATH such that MATH for all MATH. If MATH, we have that MATH, MATH and MATH. It follows that MATH as MATH. REF imply for MATH, MATH . If MATH, then MATH, where we used CITE. CASE: Let MATH and choose MATH such that MATH. Then REF implies that MATH . So if MATH (and thus MATH), MATH as MATH. If, on the other hand, MATH (and thus MATH), we see that MATH as MATH. If we combine this with REF and the convergence in REF , we conclude, since MATH, that CASE: MATH as MATH, CASE: if MATH, then MATH as MATH . But REF implies that MATH. By the convergence in REF , this implies that if MATH, also MATH as MATH. CASE: Note that in this case, MATH for all MATH. Completely similar as in the beginning of the first part of this proof, one shows that MATH as MATH. From the previous discussion we only have to remember that MATH as MATH. Provided MATH, we have that MATH as MATH. And MATH as MATH. CASE: Let MATH denote the norm closed linear span of MATH in MATH. Use the notation of REF . If MATH, then REF immediately guarantees that MATH. If, on the other hand, MATH, we must by REF necessarily have that MATH and MATH. Thus, REF implies in this case that MATH where we used REF in the last relation. Hence MATH, thus MATH. REF implies that MATH for all MATH and MATH. It follows that MATH for all MATH, MATH and MATH. Thus, MATH for all MATH and MATH. If MATH and MATH, it is easy to see that there exists MATH such that MATH. Hence, MATH in this case. Now fix MATH and MATH. Take MATH such that MATH. Because MATH, there exists MATH such that MATH uniformly as MATH. Consequently, MATH belongs to MATH by REF . As such, MATH. Moreover, REF guarantees that MATH. Hence MATH. It follows that MATH. Hence MATH, proving that MATH. |
math/0105117 | We only proof the inclusion for MATH. The other ones are dealt with in the same way. REF tells us that MATH and thus, MATH . By REF , MATH. Let MATH and MATH. There exists a sequence MATH in MATH such that MATH and MATH. Since MATH, MATH is bounded and MATH is closed, it follows that MATH. Therefore the net MATH converges to MATH. Hence, inclusion REF and the fact that the operator on the right hand side of this inclusion is closed imply that MATH and MATH . In other words, MATH . As a consequence, REF implies that MATH . |
math/0105117 | The previous proposition implies that MATH . CASE: By REF , we know that MATH for MATH and MATH. REF guarantees that MATH. As a consequence, MATH. In a similar way one shows that MATH. By REF , this guarantees that MATH. REF tells us that MATH and MATH. Taking the adjoints of this inclusions, we see that MATH and MATH are both restrictions of MATH. The previous discussion learned us that also the domains of MATH and MATH agree. Thus, MATH. Since MATH is the unitary in the polar decomposition of MATH and MATH, the uniqueness of the polar decomposition implies that REF and MATH. Since MATH, MATH, so we get that MATH. Applying functional calculus to this equation, we see that MATH for all MATH. Combining this with REF , it follows that MATH for all MATH (compare the proof of REF ). Because MATH is the isometry in the polar decomposition of MATH and MATH, the uniqueness of the polar decomposition implies that MATH. The lemma follows by REF . |
math/0105117 | Since MATH is generated by MATH, it only remains to show that MATH by the previous corollary. Define first of all the isometry MATH as MATH. Set MATH and use the uniform measure on MATH. Define MATH. Then MATH is obviously the orthogonal sum of MATH and MATH. Let MATH be a NAME space. Then we define a normal MATH-homomorphism MATH such that MATH and MATH. If MATH are two NAME spaces, MATH for all MATH, MATH. CASE: On MATH we define unitary operators MATH, MATH and MATH such that for MATH, MATH for almost all MATH. Now define the unitary element MATH in MATH as MATH. Since MATH is separable, we can extend the orthonormal basis for MATH given by the functions of REF to an orthonormal basis MATH for MATH. If MATH, we define MATH such that MATH for all MATH. Now define the unitary operator MATH such that MATH for all MATH and MATH. So the restriction of MATH to MATH equals MATH and MATH. Define the normal injective MATH-homomorphism MATH such that MATH for all MATH. Thus, if MATH, then MATH. By REF , this implies that MATH for all MATH. Take MATH and MATH. Choose MATH. One easily checks that MATH. Thus, MATH. Notice that MATH if MATH. Assume that MATH. Choose MATH, MATH and MATH. If MATH and MATH, then MATH which by REF guarantees that MATH . If, on the other hand, MATH or MATH, then MATH . Using notation REF , we define the function MATH such that for MATH, MATH, MATH, MATH equals MATH if MATH and MATH if MATH. Then it is clear from the above computations (and the proof of REF ) that MATH converges pointwise to MATH. CASE: Define MATH. By REF there exists a function MATH such that MATH for all MATH and MATH. There exists clearly a number MATH such that MATH for all MATH. Let MATH such that MATH. If MATH then MATH and since MATH, this implies that MATH. Hence, MATH belongs to MATH. So we can define an element MATH such that for MATH and MATH, MATH, we have that MATH if MATH and MATH if MATH. Then REF and the remarks thereafter imply that MATH for all MATH such that MATH. Therefore the convergence in REF and the dominated convergence theorem imply that MATH belongs to MATH and that the sequence MATH converges to MATH in MATH. Since the sequence MATH is a sequence of isometries and the linear span of such elements MATH is dense in MATH, we conclude that there exists an isometry MATH such that the sequence MATH converges strongly to MATH. CASE: Thus, MATH for all MATH and MATH. Using REF and the definition of MATH and MATH, one easily checks that MATH and MATH for all MATH and MATH (note that MATH for all MATH). As a consequence, MATH. CASE: Define MATH such that MATH and MATH. One sees that MATH for all MATH, MATH, implying that MATH is the restriction of MATH to MATH. So we get for every MATH that MATH, whereas MATH. It follows that MATH is the strong limit of the sequence of isometries MATH. As such, MATH belongs to MATH. We set MATH, which belongs to MATH because it is a multiplication operator. Multiplying REF from the right by MATH and using the fact that MATH, we see that MATH. Thus, MATH. CASE: Restricting the previous equation to MATH, we get MATH. Applying MATH to this equality, it follows that MATH. CASE: If we apply MATH to REF , we get MATH. So the restriction of this equation to MATH gives MATH. Multiplying this equation from the left by MATH, we see that MATH where in the last equality we used REF , the fact that MATH and REF . Now we calculate MATH. Let MATH. Since MATH belongs to MATH, REF guarantees that MATH which implies that MATH. Hence, MATH . If we let MATH tend to infinity in this equality, the expression we started with converges strongly to MATH whereas the expression we end up converges strongly to MATH. Thus, MATH. Inserting this in REF , we conclude that MATH. Comparing this with REF , we conclude that MATH . The initial projection of MATH is given by MATH where MATH is the orthogonal projection onto MATH. In other words, MATH. Thus, MATH, implying that MATH is a partial isometry with initial space MATH. Therefore REF guarantees that MATH. |
math/0105117 | The proof of this fact is pretty straightforward. For MATH and MATH, we define MATH as MATH . From NAME REF, we get immediately that the families MATH and MATH are both orthonormal bases of MATH. Moreover, REF imply that CASE: the space MATH is a core for the operator MATH, CASE: the space MATH is a core for the operator MATH, CASE: if MATH and MATH, then MATH . CASE: By REF , we get the following equalities (in which the infinite sums are MATH-convergent), MATH . Choose MATH and MATH. Look first at the case where MATH and MATH. Set MATH, MATH and MATH. Thus, MATH and MATH. Set MATH. The above equation, REF imply that MATH . If MATH or MATH, then REF implies immediately that MATH. So we see that MATH . REF implies that MATH . Define the anti-unitary operator MATH on MATH such that MATH where MATH are such that MATH for all MATH and MATH. Let MATH. If MATH, it follows from the considerations before REF that MATH and MATH. If on the other hand MATH, we have MATH and MATH. Thus, in both cases MATH. Now it is straightforward to check that MATH. Thus, MATH . Since MATH is supported on the set MATH and MATH for all MATH, we get that MATH . This implies for MATH, MATH and MATH, MATH . From these two equations and the properties listed in REF in the beginning of the proof, all the claims in the statement of this lemma easily follow. |
math/0105117 | Since the domain of MATH and MATH is MATH and clearly MATH, the statements about the adjointability of MATH follow easily. Using REF , one checks that MATH . So we get for MATH, MATH and MATH that MATH . The inverse MATH of MATH is such that MATH for MATH, MATH and MATH. Then the following rules are easily established for MATH, MATH and MATH. CASE: If MATH, then MATH. CASE: MATH. CASE: MATH. CASE: MATH. This implies that for MATH, MATH and MATH, MATH . |
math/0105117 | Define the sesquilinear form MATH such that MATH for all MATH and MATH. Under this pairing, MATH is anti-linearly isomorphic to the algebraic dual of MATH. So there exist an anti-linear anti-homomorphism MATH such that MATH for all MATH and MATH. If MATH and MATH then MATH equals the inner product in MATH. So if MATH, then MATH for all MATH (and the domain of MATH consists of all MATH such that MATH ). Moreover, MATH is adjointable if and only if MATH in which case MATH is the restriction of MATH to MATH. If MATH and MATH, we define MATH such that MATH for all MATH and MATH. We denote the restrictions of MATH to MATH by MATH respectively. Let MATH denote REF coordinate functions on MATH. Define also an auxiliary function MATH such that for MATH, MATH if MATH and MATH if MATH. It is not so difficult to check that MATH . Thus MATH . Let MATH denote the restriction of MATH to MATH. Clearly, MATH where MATH ranges over MATH, MATH, MATH, MATH and MATH belongs to MATH. Therefore MATH is adjointable. Moreover, MATH. Now an easy calculation reveals that MATH. Similarly, MATH. REF implies that MATH and MATH. By REF and the remarks before this lemma we get that MATH . Multiplying this inclusion from the right with MATH, we get that MATH. Fix a unit vector MATH. Define the bounded linear operator MATH such that MATH for all MATH. Then MATH. So if MATH, then MATH and MATH. Thus, MATH and MATH . So we have shown that MATH. Taking the adjoint of this inclusion we see that MATH. By the remarks in the beginning of the proof, this implies for MATH, MATH. Starting from the inclusion MATH, the statement concerning MATH is proven in the same way. |
math/0105117 | Fix MATH. Set MATH. On MATH we define the measure MATH such that MATH for all MATH. Define a unitary transformation MATH such that MATH for all MATH and MATH. Notice that MATH and MATH for all MATH and MATH such that MATH. Set MATH and MATH. Then, CASE: the space MATH is a core of MATH, CASE: the space MATH is a core of MATH. CASE: Choose MATH and MATH. By REF , we know that for MATH, MATH and MATH . Note the symmetry between the above formulas for MATH and MATH with respect to interchanging MATH and MATH. Define the auxiliary function MATH such that MATH for all MATH. We want to show that MATH. The difference of these inner products are sums over the whole area MATH but we will start by approximating these sums by sums over a finite subset of MATH obtained by cutting off MATH by REF rectangles (see REF later in the proof). Then we let these rectangles get bigger and bigger. In this way, these finite sums converge certainly to the sum over the whole of MATH. In the second part of the proof, we will then prove that these finite sums converge to REF, implying that MATH must be equal to MATH. So let us first calculate the contribution of one rectangle, lying in one of the quadrants, to the difference of the above inner product. Therefore let MATH such that MATH, MATH, MATH and MATH and set MATH where the statement MATH allows for MATH to be equal to MATH or MATH as well (similarly for MATH). Using REF and recalling discussion REF in Notations and conventions, we get that MATH . Define the set MATH. Define a new auxiliary function MATH such that MATH for all MATH. Now observe the following basic facts: MATH . Remembering that MATH and MATH are defined on MATH, these facts combined with REF imply that MATH . In the above sum, most of the terms of the sums in the left column are cancelled by the terms of the sums in the right column. What remains is MATH . Set MATH. If MATH we define MATH. Let us calculate the contribution of REF rectangles depicted in REF to MATH. Therefore set MATH. It is clear from REF and the dominated convergence theorem that MATH converges to MATH as MATH. CASE: If MATH, we set MATH and if MATH, we set MATH . Regardless of the sign of MATH, we set MATH and MATH . A bookkeeping exercise based on REF reveals that MATH . In order to show all the nitty gritty work involved, let us calculate MATH in the case of MATH (keep REF in mind). So in REF we have to set MATH, MATH, MATH and MATH. CASE: If MATH, then MATH since MATH. If MATH then also MATH since MATH. But note that MATH. So the first sum of REF equals MATH. CASE: For all MATH, clearly MATH. Thus the second sum in REF equals REF. CASE: For all MATH, we have that MATH and for all MATH, we have that MATH. Putting all these results into REF , we see that MATH . Similarly, one calculates MATH, MATH and MATH. Adding these four results together, one finds REF . The case MATH is treated similarly. Now we are going to make specific choices for our functions MATH and MATH. Therefore take MATH, MATH such that MATH and define MATH and MATH such that for MATH, MATH if MATH and MATH if MATH, and on the other hand, MATH if MATH and MATH if MATH. By REF , MATH,MATH are proportional to MATH,MATH respectively. In the next part we will show that for this choice of MATH and MATH, MATH as MATH. Notice that MATH is obtained from MATH by interchanging MATH and MATH and replacing the parameters MATH,MATH by MATH,MATH. There is also some symmetry with respect to interchanging MATH and MATH in the formulas defining MATH and MATH. Similarly, note the symmetry with respect to this variable interchange between MATH, MATH and MATH, MATH respectively. This is of course in correspondence with the symmetry observed in REF . We will make use of this symmetry to reduce our work. Let MATH. Then MATH for all MATH. Using this simple fact, a careful inspection learns that for MATH, MATH if MATH and MATH if MATH. On the other hand, MATH if MATH and MATH if MATH. Now we will show that the different summands of MATH as described in MATH converge to REF as MATH. CASE: First we quickly check that MATH as MATH. Let MATH be a sequence in MATH that converges to MATH. Then MATH and MATH for MATH big enough. Therefore REF imply, as in the proof of REF , that MATH where we used notation REF . We have also that MATH as MATH. From this all, we easily conclude that MATH as MATH. CASE: Let us now deal with MATH. First consider the case where MATH. Assume for the moment that MATH and MATH. Then MATH and MATH. Hence, MATH and MATH do not belong to MATH. As a consequence, MATH for all MATH. We conclude that MATH if MATH and MATH. Thus MATH as MATH. Now we look at the more challenging case where MATH. Referring to REF , we get the existence of a constant MATH, only depending on MATH, MATH and MATH such that MATH for all MATH. Therefore REF implies that MATH . Thus, MATH, which clearly implies that MATH as MATH. On the other hand, MATH since MATH belongs to MATH. So we also get that MATH as MATH. By the NAME inequality, we have MATH . Therefore the considerations above imply that MATH . Hence, the obvious symmetry between MATH and MATH guarantees that also MATH . Thus, we conclude that MATH as MATH. CASE: In the next step we prove that MATH as MATH. CASE: Let MATH such that MATH. Then MATH, thus MATH. Therefore REF imply for MATH, MATH if MATH and MATH if MATH (since MATH in this case). If MATH and MATH, then MATH, thus MATH, so MATH. So we get easily the existence of a constant MATH such that MATH for all MATH such that MATH and MATH. Moreover, REF implies the existence of CASE: a number MATH such that MATH for all MATH such that MATH and MATH such that MATH. CASE: a function MATH such that MATH for all MATH CASE: a function MATH such that MATH for all MATH and MATH such that MATH. Now define MATH as MATH. Then REF implies that MATH for MATH such that MATH and MATH. Define the function MATH such that for MATH, MATH if MATH and MATH if MATH (here we use the notation introduced in REF ). If we fix MATH, we see that MATH if MATH is small enough. By REF , we have moreover that MATH as MATH. Therefore the dominated convergence theorem implies that MATH and MATH as MATH. Thus, it follows that MATH as MATH. The aforementioned symmetry between MATH and MATH then guarantees that also MATH as MATH. CASE: In the last step we look at MATH. First assume that MATH. If MATH and MATH then MATH. Thus REF implies that MATH. So in this case, MATH as MATH. Now we look at the more challenging case where MATH. If MATH and MATH, then MATH and MATH clearly belong to MATH; thus REF imply that MATH . By REF , there exists a number MATH such that MATH for all MATH. Therefore the above equality and REF allow us to conclude that MATH also in this case. The symmetry between MATH and MATH thus guarantees that also the first term of MATH as MATH. Therefore MATH as MATH. Together with REF the convergence results proven in REF imply that MATH as MATH. By REF , this implies that MATH. So, by our choice of MATH and MATH, we find that MATH. Therefore the conditions stated in REF imply that MATH for all MATH and MATH. Consequently, REF . Since MATH and MATH are normal, this implies that MATH. But taking the adjoint of the inclusion REF , we arrive at the inclusion MATH. Hence, MATH. |
math/0105117 | If MATH, both sides of the above equation are easily seen to be equal to REF. Now we look at the case where MATH. Suppose that MATH. Then we have for MATH, MATH so if we let MATH tend to infinity, REF follows. If MATH, we apply REF , where we replace MATH by MATH and MATH by MATH. |
math/0105117 | By definition, we have that MATH and MATH. It follows that the MATH-th term of this series converges to the number MATH as MATH. Take MATH such that MATH. So there exists MATH such that MATH for all MATH. It is not so difficult to see that there exists MATH such that MATH for all MATH and MATH. Thus, since MATH, the dominated convergence theorem implies that the sequence MATH converges to MATH. |
math/0105117 | First suppose that MATH and MATH. Take a sequence MATH in MATH such that MATH and MATH for all MATH. Then NAME 's transformation formula CITE implies that MATH. If we let MATH tend to MATH in this equality, we obtain by the previous lemma the equality MATH. The general result follows by analytic continuation. |
math/0105117 | First assume that MATH and MATH do not belong to MATH. Take a sequence MATH in MATH such that MATH, MATH and MATH for all MATH. Let MATH and apply NAME 's transformation formula CITE for MATH, MATH, MATH and MATH. Thus, MATH. If we let MATH tend to MATH in this equality, REF implies that MATH. Thus, MATH, still under the assumption that MATH. The general formula now follows from analytic continuation. |
math/0105117 | By definition, we have MATH . On the other hand, MATH . |
math/0105117 | For MATH, we have MATH. Also notice that MATH for all MATH satisfying MATH. Hence, by the definition of MATH, we see that MATH where we used CITE in the last equality. Now the lemma follows. |
math/0105117 | Choose MATH such that MATH. Take MATH such that MATH and MATH. Let MATH such that MATH and MATH. Since MATH for all MATH, we get that MATH . This implies the existence of MATH such that MATH and MATH for all MATH such that MATH and MATH such that MATH. CASE: Define the function MATH such that for MATH, we have that MATH if MATH and MATH if MATH and MATH is such that MATH. Then it is clear that MATH. Take MATH such that MATH. Choose MATH such that MATH. Since MATH and MATH, we have that MATH, hence MATH. It is also clear that MATH, thus MATH. By estimate REF , we get for all MATH such that MATH, MATH . Also, REF guarantees that MATH . This estimate, together with the estimates in REF imply that MATH for all MATH and MATH such that MATH. Estimate REF also implies the existence of MATH such that MATH for all MATH such that MATH and MATH such that MATH. So if we set MATH, we find by the estimate in REF that MATH for all MATH and MATH such that MATH. |
math/0105124 | The map (in characteristic REF) MATH is injective, because its transpose MATH has connected kernel, by assumption. In particular, the map MATH is injective, for all prime numbers MATH. The map which MATH induces on tori MATH is then injective (in the appropriate sense), as one sees by lifting these tori to MATH. This injectivity corresponds to the surjectivity asserted by the lemma. |
math/0105124 | The cokernel in question is MATH . By the lemma, we have MATH, since MATH contains MATH. Therefore the cokernel is simply MATH . After some fooling with Ext groups, we see that this quotient map be identified with the torsion subgroup of MATH . However, the gcd in the formula is the order of the torsion subgroup of MATH. |
math/0105124 | Consider the map MATH gotten by composing MATH and the map MATH which is MATH. By the lemma, its cokernel is MATH. To obtain the formula, we notice that the image of MATH is generated by the elements MATH for MATH. For MATH and MATH in MATH, MATH maps the generator MATH of MATH to MATH, since MATH is deduced by restriction from the pairing MATH on MATH. It should be remarked here that we are using the compatibility between MATH and MATH which is summarized by the relation MATH for MATH and MATH. |
math/0105124 | Let MATH be our chosen generator of MATH. Then MATH so that MATH . The quantity MATH visibly coincides with the sum MATH. |
math/0105126 | As was observed in REF , the covering degree of MATH is MATH. Since MATH is unramified, from the NAME formula we have MATH, where MATH stands for the genus of the curve MATH. Since the points in MATH split completely and by the assumption on MATH, we have MATH . Hence we must have that the inequality above is actually an equality, as follows from NAME 's theorem and thus the curve MATH is MATH-maximal. Finally, the curve MATH is also MATH-maximal by NAME 's result CITE. |
math/0105126 | Let MATH be the maximal unramified abelian MATH-covering of MATH in which any point of MATH splits completely. The genus MATH of MATH satisfies MATH, so that we have MATH by the hypothesis on MATH. Therefore from the Theorem and the Lemma, the curve MATH is MATH-covered by the Hermitian curve and hence the curve MATH is also. |
math/0105127 | Perform MATH-twists on MATH, and change MATH to an unknot by using MATH. Then the resulting framed link is a NAME link whose coefficients are MATH and MATH. It is well known that this presents MATH. |
math/0105128 | Let us indicate the proof for a NAME measure. Let MATH and MATH be NAME exponents for MATH and MATH respectively. Take MATH. Then the map MATH commutes with MATH and so MATH. Tending MATH we get MATH-invariance of the repelling-expanding directions decomposition MATH. Applying this to all MATH different from NAME exponents and interchanging MATH and MATH we deduce for MATH from a full measure subset in MATH the common decomposition MATH which is: measurable, invariant and such that on every subspace MATH the NAME exponents of both MATH and MATH are constant. Therefore the NAME exponents of MATH are sums of the corresponding NAME exponents for MATH and MATH. The claim now follows from the NAME REF. The equality in REF is achieved if and only if we sum positive exponents for MATH and MATH in the above decomposition MATH almost everywhere with respect to MATH. |
math/0105128 | Actually MATH and using MATH we get the reverse inequality. |
math/0105128 | First note that any vector MATH generates left- and right-invariant vector fields MATH and MATH on MATH, which can be considered as functions on MATH. We also denote by MATH the invariant extensions of any function MATH, not necessarily linear. Any two MATH and MATH . NAME commute and the other NAME brackets are MATH, MATH (CITE, CITE), where the bracket MATH on MATH is induced by the usual commutator (this bracket is called NAME or NAME). So finding one left-invariant function MATH commuting with MATH will suffice, just take any MATH. We take the NAME function MATH for MATH, that is, such a function that MATH. This function always exists locally since MATH has odd dimension, but globally it can have singularities. In the case of MATH we have smooth NAME function MATH. For MATH. For the NAME group MATH. For the solvable group we consider MATH as a linear vector field on the plane MATH. We take MATH to be an integral of this vector field. For example for REF we take MATH. Note that these functions MATH have singularities along some axes in the solvable case. To overcome this difficulty we use the lemma. If REF-dim NAME group MATH is not semisimple, then for every left-invariant MATH there exist two commuting right-invariant integrals MATH. Actually due to NAME classification we can always find a two-dimensional commutative subalgebra MATH in non-semisimple case (MATH for solvable algebras). Now it's easy to check that in every case of non-holonomic metric we obtain REF functionally independent a.e. integrals. |
math/0105128 | MATH . But MATH because the matrix MATH is NAME and hence nondegenerate for MATH. |
math/0105128 | Actually one take a graph with vertices numerated by the components of MATH. This number is finite due to MATH. Applying sub-Riemannian NAME theorem CITE we get labeled directed edges of this graph with labels MATH realized as geodesic loops lifted to MATH. The rest of NAME 's proof CITE goes now unchanged. |
math/0105128 | Actually MATH, MATH. |
math/0105128 | Actually MATH. |
math/0105128 | Actually locally in NAME coordinates: MATH and MATH. Now it's obvious that MATH and MATH are equivalent to the condition that MATH does not depend on MATH [lifted from some MATH on MATH]. |
math/0105128 | Actually this follows from REF and commutativity. |
math/0105129 | Follows by CITE . |
math/0105129 | Let MATH be the blowup as in REF , where MATH is the proper transform of MATH. Let MATH be the exceptional divisor. Note that MATH has at worst isolated singularities (because the weights MATH in REF are prime to each other). Hence MATH. Clearly, MATH is the weighted projective plane (see for example, CITE) MATH, MATH and MATH in cases MATH, MATH and MATH, respectively. Put MATH. In all cases, MATH. Therefore, MATH, so MATH. In particular, MATH. By REF and the Inversion of Adjunction CITE the pair MATH is plt. CASE: In this case MATH is given by the equation MATH, MATH. It is easy to see that MATH is reduced, irreducible and contained into the smooth locus of MATH. Since MATH, the curve MATH have at most one singular point which is an ordinary double point or a simple cusp. In both cases MATH is lc. By the Inversion of Adjunction CITE, so is MATH. Using convexity of the lc property CITE one can show that the pair MATH is klt. Standard toric technique gives us MATH . Hence we can write MATH . Thus MATH is klt CITE. CASE: Here MATH is given by the equation MATH. We claim that MATH is lc. As above, MATH is contained into the smooth locus of MATH and MATH. If MATH is reduced and irreducible, we have that MATH is lc (because in this case the singularities of MATH are at worst ordinary double points or a simple cusp, compare REF ). Assume that MATH is reduced but not irreducible. Then MATH, where MATH and MATH are smooth rational curves and MATH. Hence MATH (see REF ). Finally, if MATH is reduced, then again MATH and MATH is smooth. Hence MATH is plt. This proves our claim. By Inversion of Adjunction CITE, MATH is lc. Similar to REF the pair MATH is klt. So, MATH is also klt. CASE: Here MATH is a cubic curve given by the equation MATH. Let MATH. We have MATH (see for example, CITE). Moreover, MATH whenever MATH is reduced (compare REF ) and MATH whenever MATH has no components of multiplicity MATH. Similar to REF the pair MATH is klt. Put MATH. Then MATH. Hence MATH . Since MATH both pairs MATH and MATH are klt. This proves the proposition. |
math/0105129 | By the Inversion of Adjunction CITE we have that MATH is lc. Assume that the pair MATH is nonexceptional. By REF , MATH. Hence, there are at least two divisors MATH and MATH such that MATH. Note that MATH . Since both pairs MATH and MATH are lc, we have MATH . Taking into account that the discrepancy MATH is a linear function in MATH, we obtain MATH . Clearly, MATH. Again by the Inversion of Adjunction, MATH is plt in codimension two. If MATH, then MATH. Hence MATH, a contradiction. Therefore, MATH is a point, say MATH. So, MATH. By CITE, MATH . On the other hand, MATH is plt. The contradiction proves our lemma. |
math/0105129 | Note that MATH is obtained from the minimal resolution MATH by contracting all the MATH-curves: MATH . Moreover, MATH and MATH are exceptional curves of the induced map MATH and the coefficient of the proper transform of MATH (respectively, MATH) in the fundamental cycle MATH is MATH (respectively, MATH). Write MATH where MATH and MATH are proper transforms of MATH and MATH, respectively, and the MATH for MATH are MATH curves on MATH. The rational constants can be computed from the system of linear equations: MATH . Since MATH, we have MATH. Hence, MATH is lc (respectively, klt) if and only if MATH (respectively, MATH). CASE: Now the assertion can be proved by elementary linear algebra computations. In our situation (when MATH and MATH has a multiple factor), the exceptional divisor is a tree of smooth rational curves. Therefore, all discrete invariants are uniquely defined by weighted dual graphs. Below the notation of CITE are used. Thus MATH denotes the vertex corresponding to exceptional curve MATH with MATH. The vertices MATH always have MATH while MATH can be MATH or MATH. We attach MATH to corresponding MATH-vertices with MATH and omit MATH everywhere. Note that each graph has at most one MATH-vertex. By CITE coefficients MATH in the fundamental cycle MATH satisfy MATH and MATH or MATH. Therefore, MATH is a MATH-vertex and MATH is MATH-vertex with MATH. Consider, for example, the case MATH. For weighted dual graph of MATH there are two possibilities: MATH or MATH . In REF , contracting MATH-curves, we get the surface MATH having two NAME points MATH and MATH of types MATH and MATH, respectively. Moreover, MATH and MATH. It is clear that MATH is plt at MATH. Hence MATH is klt at MATH. The dual graph of the resolution MATH over MATH is as follows: MATH . Thus system REF has the form MATH . The solution is MATH. Hence MATH is klt at MATH. Similarly in REF the surface MATH has three NAME points MATH, MATH, and MATH of types MATH, MATH, and MATH, respectively. Here MATH, MATH and MATH. Clearly, MATH is plt outside of MATH. Consider the dual graph of the resolution MATH over MATH: MATH . This gives us the following form of system REF: MATH . The solution is MATH. Hence MATH is klt at MATH. Computations in all other cases are similar. We omit them. |
math/0105130 | Actually this perturbation is MATH-small for MATH and large for MATH and hence MATH. To perform it one collects elliptic-hyperbolic points at pairs (after a small perturbation one can assume there's nothing more complicated than separatrix self-connection) and kill them one by one. The possibility of such a perturbation is an easy calculational argument similar to REF. |
math/0105130 | Let us change the almost complex structure MATH outside a small neighborhood MATH of MATH in such a manner that it be integrable. Moreover it can be done so that some bigger neighborhood is isomorphic to MATH near the boundary. Thus we can glue the neighborhood and obtain the manifold MATH with the almost complex structure MATH being integrable outside the fixed neighborhood of our NAME. (compare REF). Now we introduce a symplectic product-structure MATH on MATH and note that for sufficiently small neighborhood MATH symplectic structure MATH tames the structure MATH. If MATH the pseudoholomorphic spheres can lie only in the multiple of the homology class of the second factor in MATH. So we seek for them in the primitive class MATH. The NAME theory CITE implies that in the case of general position almost complex structure MATH the manifold MATH is foliated by NAME. Taking the intersection of this foliation with the neighborhood MATH we prove the statement. If the curve is rational MATH some more delicate arguments are required CITE. If the almost complex structure MATH is not of general position we apply the compactness theorem from CITE for a generic sequence MATH. |
math/0105130 | Recall that on almost complex manifold MATH there is always an almost complex connection MATH. Actually if MATH is any linear connection then MATH is also a linear connection, which preserves the structure MATH. Moreover we can assume that MATH is minimal, that is, the torsion of MATH equals to its antilinear by each argument part MATH CITE. MATH . There exists a minimal almost complex connection such that the curve MATH is a totally geodesic submanifold, that is .parallel transport of any vector MATH along a path in MATH belongs again to MATH. To see this let us make a gauge transformation MATH, where MATH is a REF-form with values in complex endomorphisms of the tangent bundle. If we require that this REF-form is symmetric MATH, then the new connection MATH is also almost complex and minimal. Let MATH be a vector field on MATH with nondegenerate critical points, MATH. Let MATH. We define MATH, MATH, MATH and for other vectors somehow preserving symmetry and MATH-linearity (near the critical points there's a more work, we need to choose MATH so that MATH has a zero of the second order at these critical points). Then MATH, MATH. Therefore by minimality MATH and also MATH. So MATH preserves MATH. Another way to prove this is to introduce trivial connection in MATH and then to check that procedures of making connection almost complex and then minimal do not destroy the property of MATH being totally geodesic. Let us denote this new connection again by the symbol MATH. We introduce a connection MATH to the bundle MATH by means of parallel transport as follows. Let MATH be the class of MATH and let MATH be a curve, MATH. Calculate parallel transport MATH of MATH along MATH. We define MATH to be the parallel transport of MATH along MATH. Since MATH is totally geodesic, the definition is correct (MATH-parallel transport of REF is REF). Moreover the connection MATH is MATH-linear. So as usual in the theory of generalized connections we conclude that MATH is a linear connection. Let MATH be a point in the normal bundle with projection MATH. Let MATH be the decomposition by horizontal and vertical subspaces induced by MATH. The first space MATH has a canonical complex structure MATH induced from MATH by MATH, while the second MATH inherits canonical complex structure MATH from MATH. So we introduce the structure on MATH by the rule MATH. Let us show that the constructed almost complex structure MATH does not depend on the choice of minimal connection MATH preserving MATH. Let us change the connection MATH, MATH. This affects in the change of decomposition MATH, where MATH and MATH. We state that MATH is a complex linear map. Actually MATH . So the complex structure MATH on MATH is canonically defined. |
math/0105130 | Actually: MATH . |
math/0105130 | Actually if the distribution MATH is nonintegrable in a neighborhood of the NAME MATH, then a neighborhood MATH of the curve is not isomorphic to a neighborhood of the zero section in the normal bundle MATH. |
math/0105130 | There are local coordinates MATH on MATH, MATH, MATH, MATH, such that MATH-lift of MATH is MATH, where the coefficients are linear functions of MATH and the vertical coordinates are complex linear coordinates but horizontal coordinates are complex only on the zero section. So the structure MATH is given by relations MATH and so MATH with constant coefficients MATH. The last claim follows because MATH and MATH is a linear subspace of MATH (of course invariant under MATH). Note that MATH can vary with MATH. |
math/0105130 | Let us define the structure by the formula MATH . Since MATH this structure MATH is linear on fibers. This proves the formula for MATH of the theorem. We first show that the structure MATH is almost complex. Note that by REF MATH for any MATH and MATH for MATH. So MATH . Now we show that this MATH is integrable. By NAME theorem CITE this is equivalent to the vanishing of the tensor MATH. Let us choose local coordinates MATH as in REF . In these coordinates MATH and we deduce from REF : MATH . Since the bivector MATH generates MATH the claim follows. |
math/0105130 | This statement is an analog of REF. Details of the construction can be found there. Here we present a short proof with the indication of differences. By the hypothesis the desired diffeomorphism MATH has REF-symbol MATH along the curve MATH, where MATH. Its REF-symbol MATH in local coordinates is given by MATH. Moreover these symbols are compatible with the condition MATH: MATH . The symbol satisfies the following condition: MATH where MATH is the differential with respect to some symmetric (and so not almost complex) curvature-free connection MATH. Equivalently this means that the tensor MATH satisfies the equation MATH . Now given MATH and MATH we have MATH and would like to find the corresponding MATH from REF . Form equality REF we deduce MATH which implies MATH for some MATH-tensor MATH. Now the identity MATH implies the possibility to satisfy REF by the choice MATH . Identity REF means exactly MATH. Note that the tensor MATH can be chosen to satisfy REF and all the constructions respect this condition (MATH is MATH-invariant). So we get MATH satisfying REF . Now the symbols MATH stand for the differential of the mapping MATH we seek for. Calculations above show the MATH-valued REF-form generated by MATH is closed. Then the condition MATH imply that it is exact. Hence the symbols MATH are integrated to the diffeomorphism MATH we sought for. |
math/0105130 | The values of the almost complex structure MATH and the almost complex MATH structure MATH on the right-hand side of the formula are equal at the points of the sphere MATH. Let us show that the same is true for their NAME tensors. Namely we show that the NAME tensor MATH calculated according to REF is equal to the prescribed NAME tensor MATH at all points MATH. The calculations are local and it is sufficient to consider the value of MATH on two complex independent vectors, say on MATH and MATH, where MATH and MATH are local coordinates defining MATH. Let us denote by MATH the equality MATH. We have: MATH . Thus MATH along MATH and the claim is proved. |
math/0105130 | We can assume the disks MATH have MATH as tangent planes at the points of MATH. We use the symbol MATH as in the proof of the corollary above. First note that REF implies MATH for vertical vectors (belonging to MATH). Thus we define MATH. Similarly to the proof of REF it is proved that MATH and that MATH. Thus MATH is a complex structure MATH CITE and we can take instead of it any complex structure MATH. |
math/0105130 | Let us take the universal covering MATH of MATH. The torus is covered by the entire line MATH. Changing the structure MATH at infinity in MATH and near the boundary to the integrable one we glue the manifold to the product MATH with the line MATH being glued to the first factor MATH. Then the introduction of the taming symplectic product-structure MATH yields a foliation of MATH by NAME MATH in the homology class of the first factor if we additionally demand MATH, that is, the homology class MATH of the first sphere-factor is symplectically simple. Here we use the fact that the dimension is REF: due to positivity of intersections CITE we actually have a foliation CITE. This foliation of MATH gives a family of big NAME on MATH parametrized by the radius MATH of disk in MATH out of which the almost complex structure is changed. To get estimates we use NAME reparametrization lemma as in CITE. The filling property is given by the choice of pertrubation of the structure in MATH: This is simple if we require the boundary of MATH to be MATH-pseudoconvex (CITE). |
math/0105130 | Actually take a covering of the neighborhood MATH which corresponds to one cycle of the torus. The torus is covered by the entire cylinder MATH. We can change the almost complex structure MATH at infinity so that it makes possible to "pinch" each end of the cylinder. This means we perturb the structure MATH so that it is standard integrable outside some MATH and the support is also a big cylinder MATH. Then we glue the ends to the disks. This operation gives us a sphere MATH instead of the cylinder MATH. We can also assume that neighborhoods of two cylinder ends are pinched REF . Thus we have a neighborhood MATH of the sphere MATH. It is foliated by NAME close to MATH. Actually, we can change the structure MATH near the boundary of this neighborhood, glue and get the manifold-product MATH. As before it is foliated by NAME. Thus MATH is foliated by NAME and in the preimage they give a NAME by cylinders. |
math/0105130 | The condition that MATH is attracting or repelling means that there are no self-intersections of the leaves MATH. If a leaf MATH transversally intersect another MATH with MATH then because of the convergence MATH there is an intersection of the spheres MATH from which cylinders are constructed. This is certainly impossible because our family of spheres is foliating. If the leaves MATH and MATH are tangent let us consider the first order of their jets which are different. This is the next number after the tangency order. Since the maps of finite cylinders MATH tends to the maps of MATH in MATH-topology the spheres in the manifold MATH do intersect. Every such intersection even nontransversal contributes positively to the intersection number CITE which contradicts MATH. It remains to note that if two pseudoholomorphic curves in an almost complex manifold have the infinite tangency they must coincide CITE. |
math/0105130 | A coordinate system MATH on a transversal disk MATH provides coordinates on all others MATH. These coordinates are multivalued because rotation along the second cycle MATH yields the monodromy mapping MATH. Now we apply NAME theorem CITE in complex dimension REF which states that the monodromy map is conjugate to the map MATH. |
math/0105130 | Actually the foliation provides a local bundle MATH. The hypothesys that all transports are complex is equivalent to the claim that the bundle MATH with fibers MATH is almost complex. Therefore the statement follows from REF . |
math/0105130 | First note that since MATH we have MATH. Thus MATH. Let us show REF has no nonzero solutions. Complex Laplacian of MATH equals MATH. Our torus neighborhood is the trivial cylinder MATH neighborhood glued by the rule MATH. Thus when MATH we have: MATH. So integrating over the cylinder gives: MATH . So using the calculation with the Laplacian we have: MATH . Therefore since MATH we get MATH. Thus there are no homologous to the zero section NAME MATH with MATH. If the homology class of MATH is a multiple of the zero section MATH a MATH-finite covering finishes the proof. |
math/0105130 | Actually as CITE noticed if the linearized equation MATH has a unique solution for any MATH then the torus is isolated and persistent. But the linearization we studied in REF . |
math/0105130 | First consider tori in the homology class of the zero section MATH. As was shown before REF the projection is a diffeomorphism. Since in linear almost complex bundles MATH the projection is an almost complex mapping, its restriction is a biholomorphism of the tori: MATH. The general case MATH follows from the case MATH by means of a MATH-covering. |
math/0105130 | Actually if there is a NAME by tori then the linearization of this foliation determines a holomorphic foliation of the normal bundle which is impossible by REF. |
math/0105132 | Let MATH. By REF we have MATH for every MATH, so that, passing to a subsequence, we may assume that MATH converges weakly in MATH to a function MATH such that MATH on MATH. By symmetry the traces MATH of MATH from MATH vanish on MATH. Let MATH be the characteristic function of MATH. From the definition of MATH it follows that MATH converges to MATH weakly in MATH, where MATH is the one-dimensional NAME measure. Since the trace operator is compact from MATH into MATH, the traces MATH of MATH from MATH converge to the corresponding traces MATH of MATH strongly in MATH. Therefore MATH converges to MATH weakly in MATH. As MATH on MATH (recall that MATH on MATH and MATH on MATH), we conclude that MATH on MATH, therefore MATH. By using the weak formulation REF we obtain MATH . Passing to the limit as MATH we obtain MATH . This implies that MATH coincides with the solution of REF . By uniqueness, the whole sequence MATH converges to MATH. |
math/0105132 | The case MATH and MATH is the Goa̧b theorem, for which we refer to CITE. For an independent proof for an arbitrary MATH see CITE. The case MATH is an easy consequence of the case MATH (see CITE). |
math/0105132 | If MATH, we have MATH where the first equality follows from our convention MATH a.e. in MATH, while the second one follows from REF . Equality REF means that MATH in MATH, hence MATH in MATH. As MATH is simply connected and has a NAME boundary, there exists MATH such that MATH a.e. in MATH. Since MATH on MATH, the tangential derivative of MATH (which is equal to the normal derivative of MATH) vanishes on MATH, and this implies that MATH is constant on each connected component of MATH. If MATH has a non-empty interior and a smooth boundary, then MATH is constant a.e. on the interior of MATH, since MATH a.e. on MATH. Therefore MATH is constant on MATH according to REF . The case of a general MATH can be obtained by approximating MATH by a decreasing sequence of compact sets with non-empty interior and a smooth boundary (we refer to CITE for the details). |
math/0105132 | Note that MATH is a minimum point of REF if and only if it satisfies REF ; similarly, MATH is a minimum point of REF if and only if it satisfies REF with MATH and MATH replaced by MATH and MATH. Taking MATH in the functional to be minimized, we obtain that the sequence MATH is bounded in MATH. Therefore, passing to a subsequence, MATH converges weakly in MATH to a function MATH. As MATH by REF , and hence MATH, is easy to see that there exists a function MATH, with MATH on MATH, such that MATH a.e. on MATH (we are assuming here that MATH, see CITE for the details). We will prove that MATH . As the limit does not depend on the subsequence, this implies that the whole sequence MATH converges to MATH weakly in MATH. Taking MATH and MATH as test functions in the equations satisfied by MATH and MATH, we obtain MATH . As MATH weakly in MATH and MATH strongly in MATH, from the previous equalities we obtain that MATH converges to MATH, which implies the strong convergence of the gradients in MATH. By the uniqueness of the gradients of the solutions, to prove REF it is enough to show that MATH is a solution of REF . This will be done by considering, for each MATH, its harmonic conjugate MATH given by REF . By adding a suitable constant, we may assume that MATH for every MATH. Since MATH a.e. on MATH, we deduce that MATH converges to MATH weakly in MATH, and by the NAME inequality MATH converges weakly in MATH to a function MATH which satisfies MATH a.e. on MATH. Let us prove that MATH is constant on MATH according to REF . This is trivial if MATH reduces to a point. If MATH has more than one point, then MATH; since the sets MATH are connected, we obtain also MATH, where the capacity MATH of MATH with respect to MATH is defined by MATH . As MATH on MATH for suitable constants MATH (see REF ), using the NAME inequality (see, for example, CITE) it follows that MATH is bounded in MATH, hence the sequence MATH is bounded, and therefore, passing to a subsequence, we may assume that MATH converges to a suitable constant MATH. Let us fix a constant MATH with MATH and MATH. Since MATH on MATH, by NAME 'sya's estimate (see CITE) there exist two constants MATH and MATH, independent of MATH, MATH, and MATH, such that for every MATH and every MATH where MATH. For MATH large we have MATH, so that for every MATH and every MATH we have MATH. As MATH is connected, there exists a constant MATH, independent of MATH, MATH, and MATH, such that MATH. Therefore REF yields MATH for every MATH and every MATH. Let us fix MATH and a sequence MATH converging to MATH. For every MATH we have MATH for MATH large enough, hence MATH for MATH large enough. Passing to the limit first as MATH and then as MATH we get MATH . As MATH we obtain that MATH is equal to MATH on MATH (see REF ). On the other hand, every MATH is constant on each connected component of MATH (see REF ). Since MATH weakly in MATH, we conclude that MATH is constant on each connected component of MATH. Therefore there exists a sequence MATH in MATH converging to MATH strongly in MATH, such that each MATH is constant in a neighbourhood of MATH and in a neighbourhood of each connected component of MATH. Let MATH with MATH on MATH., and let MATH with MATH on MATH and MATH on a neighbourhood of MATH. As MATH in MATH and MATH, we have MATH . Since MATH a.e. on MATH, passing to the limit as MATH we obtain MATH which shows that MATH is a solution of REF . |
math/0105132 | By REF. Assume by induction that MATH and let MATH be a constant such that MATH. Consider a minimizing sequence MATH of REF . We may assume that MATH for every MATH. By the Compactness REF , passing to a subsequence, we may assume that MATH converges in the NAME metric to some compact set MATH containing MATH. For every MATH let MATH be a solution of the minimum REF which defines MATH. By REF MATH converges strongly in MATH to MATH, where MATH is a solution of the minimum REF which defines MATH. By REF we have MATH and MATH, hence MATH. As MATH, we conclude that MATH. Since MATH is a minimizing sequence, this proves that MATH is a solution of the minimum REF . |
math/0105132 | For MATH, consider the functions MATH defined by MATH, with the convention that MATH. Then the functions MATH are NAME continuous with constant MATH for every MATH, and the functions MATH are non-increasing for every MATH. Let MATH be a countable dense subset of MATH. For every MATH there exists a countable set MATH such that MATH are continuous at every point of MATH. By REF we have MATH and MATH for every MATH and every MATH, MATH with MATH. This implies that MATH for every MATH and every MATH. Let MATH be the countable set defined by MATH, and let MATH. Then MATH for every MATH, and, by continuity, for every MATH, which yields MATH. This proves that MATH, hence MATH. |
math/0105132 | Let MATH be a countable dense subset of MATH. Using the Compactness REF and a diagonal argument, we find a subsequence, still denoted by MATH, and an increasing function MATH, such that MATH in the NAME metric for every MATH. Let MATH and MATH be the increasing functions defined by MATH where MATH denotes the closure. Let MATH be the set of points MATH such that MATH. As MATH and MATH satisfy REF , by REF the set MATH is at most countable. Since MATH for every MATH, we have MATH for every MATH. For every MATH we define MATH. To prove that MATH for a given MATH, by the Compactness REF we may assume that MATH converges in the NAME metric to a set MATH. For every MATH, with MATH, by monotonicity we have MATH. As MATH is closed, this implies MATH, therefore MATH by the definitions of MATH and MATH. Since MATH is at most countable, by a diagonal argument we find a further subsequence, still denoted by MATH, and a function MATH, such that MATH in the NAME metric for every MATH. Therefore MATH for every MATH, and this implies that MATH is increasing on MATH. |
math/0105132 | It is clear that MATH and MATH are increasing and satisfy REF . Therefore MATH is at most countable by REF . Let us fix MATH and a sequence MATH in MATH converging to MATH. By the Compactness REF we may assume that MATH converges in the NAME metric to a set MATH. For every MATH, with MATH, we have MATH for MATH large enough, hence MATH. As MATH is closed this implies MATH, therefore MATH by REF and by the definition of MATH. |
math/0105132 | Let us fix an integer MATH with MATH. From the absolute continuity of MATH we have MATH where the integral is a NAME integral for functions with values in MATH. This implies that MATH where the integral is a NAME integral for functions with values in MATH. As MATH and MATH on MATH, we have MATH . By the minimality of MATH and by REF we have MATH . From REF , and REF we obtain MATH where MATH by the absolute continuity of the integral. Iterating now this inequality for MATH we get REF with MATH. |
math/0105132 | As MATH is admissible for REF which defines MATH, by the minimality of MATH we have MATH, hence MATH for every MATH. As MATH is absolutely continuous with values in MATH, the function MATH is integrable on MATH and there exists a constant MATH such that MATH for every MATH. This implies the former inequality in REF . The latter inequality follows now from REF and from the inequality MATH, which is an obvious consequence of the minimality of MATH and of the fact that MATH. |
math/0105132 | By REF we have MATH for every MATH and every MATH. By REF this implies MATH for every MATH. |
math/0105132 | As MATH is a solution of the minimum REF which defines MATH, and MATH strongly in MATH, the conclusion follows from REF . |
math/0105132 | Let us fix MATH and MATH with MATH. Since MATH converges to MATH in the NAME metric as MATH, it is possible to construct a a sequence MATH in MATH, converging to MATH in the NAME metric, such that MATH and MATH as MATH. By REF this implies that MATH is bounded as MATH. The main difficulty in the construction of MATH is the constraint on the number of connected components. The proof of the details is quite long, but elementary, and is given in CITE. Let MATH and MATH be solutions of the minimum REF which define MATH and MATH, respectively. By REF MATH strongly in MATH. The minimality of MATH expressed by REF gives MATH, which implies MATH. Passing to the limit as MATH and using REF we get MATH. Adding MATH to both sides we obtain REF . |
math/0105132 | Let us fix MATH with MATH. Given MATH let MATH and MATH be the integers such that MATH and MATH. Let us define MATH and MATH. Applying REF we obtain MATH with MATH converging to zero as MATH. By REF for every MATH we have MATH strongly in MATH as MATH, and by REF we have MATH for every MATH. Given MATH, let MATH. As MATH for MATH small enough, we have MATH. Applying REF with MATH we get MATH . Passing to the limit as MATH we obtain MATH . Passing now to the limit in REF as MATH we obtain REF . |
math/0105132 | Let MATH. From the previous lemma we get MATH . On the other hand, by REF we have MATH. It is easy to see that the NAME differential MATH of MATH (with respect to MATH) is given by MATH where MATH is a solution of the minimum REF which defines MATH. Therefore we have MATH where MATH is a solution of the minimum REF which defines MATH. Together with the inequality MATH, this implies MATH . Since there exists a constant MATH such that MATH and MATH for MATH, from REF we obtain MATH which proves that the function MATH is absolutely continuous. As MATH strongly in MATH when MATH, if we divide REF by MATH, and take the limit as MATH we obtain REF . |
math/0105132 | Let MATH be the set defined in REF . By REF and by REF the differential MATH tends to MATH as MATH tends to MATH in MATH. Let us fix a point MATH in MATH such that the function MATH is differentiable at MATH and MATH is a NAME point of MATH. For every MATH we have MATH . Dividing by MATH and taking the limit as MATH we obtain MATH . The conclusion follows from REF . |
math/0105133 | From MATH follows that MATH, hence MATH must be of the form MATH, MATH. In turn, MATH can be obtained as follows: MATH where we have used the following property of MATH: MATH . In order to prove REF , we consider the map MATH, MATH where MATH is the central extension (see above). The pair MATH is a morphism between the bundles MATH and MATH. By the functoriality of the maps MATH and MATH, we have the following commutative diagram: MATH . The loop MATH in MATH corresponds to the coroot MATH in MATH. This loop is obviously mapped to the constant loop MATH by MATH, hence by MATH. So REF is true. |
math/0105133 | Put MATH. Look for MATH of the form MATH. We must have MATH hence: MATH where MATH are certain nonzero coefficients which arise in an obvious way and MATH for MATH (consequently, we have that MATH). If MATH is the sum of MATH with a strictly upper triangular matrix, then MATH equals MATH. The coefficients MATH are uniquely determined by the recursion formulae REF and the requirement MATH which determines the initial term MATH. If MATH is strictly upper triangular, then we consider the following relations: MATH . If MATH is minimal with the property that MATH, then MATH must be MATH and the coefficients MATH are uniquely determined by MATH. |
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