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math-ph/0106002 | We denote by MATH the MATH-th differential of the basic complex and by MATH the corresponding differential of the reduced complex. MATH is a trivial MATH-module, hence MATH, and MATH. We remark that MATH if, and only if, for any MATH, MATH has no constant term as polynomial in MATH. Indeed, if MATH, MATH. Also, if MATH for any MATH, we can define MATH such that MATH and clearly MATH. Recall that MATH . Suppose MATH. Then MATH, that is, MATH. It follows that, for any MATH, MATH . On the other hand, MATH. Therefore, the polynomial MATH has no constant term for any MATH. But MATH, so this actually holds for any MATH. It follows that MATH, that is, MATH. |
math-ph/0106002 | This follows from REF , since MATH. |
math-ph/0106002 | The NAME algebra MATH acts completely reducibly on the space of reduced MATH-cocycles, hence there exists a MATH-invariant subspace MATH complementary to the space of trivial MATH-cocycles. Since MATH acts trivially on MATH, we conclude that MATH acts trivially on MATH. |
math-ph/0106002 | See REF. |
math-ph/0106002 | We remark that MATH as MATH-modules. By REF , we need to compute the cocycles corresponding to the MATH-invariants of MATH, which occur in the following cases MATH: MATH . Let MATH (respectively, MATH) be a NAME element for MATH (respectively, MATH), where MATH denotes the NAME operator. Let MATH (respectively, MATH) be the highest weight vector of MATH in MATH (respectively, MATH). Let MATH (respectively, MATH) be the highest weight vector of MATH in MATH (respectively, MATH), MATH. Then MATH . Let MATH (respectively, MATH) be the lowest weight vector of MATH, which we view as lowest weight module MATH in MATH (respectively, MATH), MATH. Then MATH . The action of MATH and MATH is given by the following formulas MATH: MATH . We remark that MATH for MATH and MATH for MATH. By REF , the corresponding cocycle is trivial. In all other cases, by REF we need to show that the cocycle pairing the highest weight vector and the lowest weight vector is trivial. The cocycle equation for the triples MATH, MATH and MATH, MATH shows that MATH. Finally, the cocycle equation for the triple MATH shows that MATH and MATH are trivial. Similarly, one can see that the cocycle equation for the triple MATH implies that MATH and MATH are also trivial. Therefore, MATH. |
math-ph/0106002 | Recall that MATH is a linearly compact NAME superalgebra and that it is a closed ideal of codimension MATH in the extended annihilation algebra MATH. By REF , due to the simplicity of MATH, MATH contains no non-central MATH-invariant ideals different from MATH. Let MATH be the center of MATH. Since the derived algebra of MATH is a non-central MATH-invariant ideal of MATH (otherwise MATH would be nilpotent), we conclude that MATH and therefore MATH is an irreducible central extension. Also, the center of MATH is zero, since otherwise its pre-image in MATH would be a proper non-central MATH-invariant nilpotent ideal. It follows that MATH contains no non-trivial MATH-invariant ideals and therefore MATH is a minimal ideal in MATH. Thus we may apply REF to the NAME theorem to obtain the isomorphism of linearly compact NAME superalgebras MATH, where MATH and MATH is a simple linearly compact NAME superalgebra. Next, we show that MATH. Let MATH be the ideal of MATH generated by MATH. We will show that MATH is a MATH-invariant ideal of MATH, which, of course, will imply that MATH. Indeed, due to REF , any continuous derivation MATH of MATH has the form MATH where MATH (respectively, MATH) is a continuous derivation of MATH (respectively, MATH) and MATH. Since MATH is even, MATH is an even derivation, hence MATH maps the ideal MATH into itself. It remains to show that MATH. If MATH leaves the ideal MATH of MATH invariant, then MATH is a non-trivial MATH-invariant ideal of MATH, which is impossible. Therefore there exists a continuous automorphism of MATH which transforms MATH to MATH (see for example, REF ). But in this case MATH, hence the ideal MATH of MATH is MATH-invariant. This implies that MATH. |
math-ph/0106002 | As we have just remarked, MATH, and by REF , MATH. |
math-ph/0106002 | By REF , MATH is an irreducible central extension of the NAME superalgebra MATH, where MATH and MATH is a simple linearly compact NAME superalgebra. It follows from REF that MATH is an algebra filtration of MATH such that MATH hence the growth of this filtration is at most MATH. This filtration induces an algebra filtration on MATH, whose growth is therefore at most MATH. We conclude (see REF ) that MATH and therefore, by REF , either MATH or MATH is one of the NAME superalgebras MATH, MATH, MATH, MATH. If MATH, then MATH, since MATH. In this case, due to REF , MATH is the universal central extension of MATH. Consequently, by the results of REF, MATH is one of the NAME superalgebras MATH, MATH, MATH, MATH, MATH. If MATH and MATH, then, by REF , MATH is an irreducible central extension of MATH. This gives us an embedding MATH. By REF , the universal central extension of MATH is MATH, where MATH is the universal central extension of MATH. Hence we have a surjective homomorphism MATH such that MATH is a central ideal. Also, since any derivation of a NAME superalgebra lifts uniquely to the universal central extension, we get an embedding (see REF ) MATH . Thus, the derivation MATH of MATH induces a derivation of MATH, which we denote by MATH. We have: MATH, where MATH, MATH. Define a filtration on MATH by letting MATH, MATH: MATH. Then MATH, otherwise MATH, being surjective on MATH, is surjective on MATH, hence on MATH, which is impossible by REF since MATH is not solvable. Hence we may assume that MATH, where MATH. Applying REF to MATH, MATH, MATH and MATH, we may find a continuous automorphism MATH of MATH that transforms MATH to MATH. But MATH is the extended annihilation algebra of the NAME conformal superalgebra MATH and the homomorphism MATH extends to a surjective homomorphism of extended annihilation algebras MATH satisfying the conditions of REF . Hence MATH is induced by a surjective homomorphism MATH. But MATH is simple, hence MATH induces an isomorphism MATH. It remains to consider the case MATH, where MATH and MATH is a central ideal. Recall that we may assume, from the proof of REF that MATH . Note that MATH is an algebra filtration of the extended annihilation algebra MATH. Denote again by MATH the induced filtration on MATH. Let MATH and consider the canonical filtration of MATH associated with the subalgebra MATH . Consider the following filtration of MATH . Since MATH is the canonical filtration of MATH associated to MATH and MATH has the form REF , it is easy to see that MATH is the canonical filtration of MATH associated to MATH. By NAME 's principle, MATH for some MATH, and since MATH is a canonical filtration, we conclude that MATH for all MATH. It follows that MATH (the last inequality follows from REF ). But MATH. Thus, the remaining case is impossible. |
math-ph/0106002 | We have: MATH, hence MATH is a MATH-invariant closed (hence linearly compact) subspace of a direct product of finite dimensional irreducible MATH-modules. The subspace MATH of MATH is MATH-invariant and closed too (by REF ). Hence there exists a closed MATH-invariant complementary subspace MATH. But MATH for any MATH, hence MATH if MATH. |
math-ph/0106002 | Let MATH be the even part of MATH, MATH, MATH. We have MATH, MATH, MATH respectively and the representation of MATH on MATH, MATH, MATH is the direct sum of the standard MATH and MATH- modules, the standard MATH- and MATH-module respectively. By REF , in all cases, MATH as MATH-module, and any MATH is a MATH-module homomorphism. We remark that MATH for MATH and MATH for MATH, MATH and MATH. It follows that any MATH preserves the standard gradation of MATH. By NAME Lemma MATH on MATH. It follows that the grading preserving derivation MATH is zero on MATH and it is a MATH-module homomorphism. Let MATH, and MATH. By induction on MATH we have MATH. Hence by transitivity we conclude that MATH, that is, MATH on MATH. Consequently MATH on MATH and MATH. On the other hand, MATH, hence MATH. Therefore MATH and every derivation of MATH is inner. A similar method can be used in the remaining cases. Notice that we may exclude the case MATH, which is isomorphic to MATH, so that MATH (respectively, MATH) is irreducible as a MATH-module (respectively, MATH-module), hence any derivation MATH acts on this subspace as a scalar matrix. Now we proceed as above using also that in both cases the MATH component is the bracket of the MATH component with itself. The result for MATH easily follows once one has established it for MATH. |
math-ph/0106002 | The NAME algebra MATH is the even part of MATH. By REF , we have MATH as MATH-modules and any MATH is a MATH-module homomorphism. Let us denote by MATH the irreducible MATH-module MATH whose eigenvalue with respect to the operator MATH is MATH. The irreducible MATH-modules which appear more than once in the standard gradation REF are: MATH, which occurs in MATH and MATH and MATH which occurs in MATH and MATH. Let MATH (respectively, MATH) be the highest weight vector of the module MATH in MATH (respectively, in MATH). Then MATH. Now, MATH is a MATH-module homomorphism, hence MATH and MATH maps MATH into itself. Also, MATH is sum of two non-isomorphic, irreducible MATH-modules, so by NAME Lemma we have that MATH. Consequently, MATH acts as MATH on MATH. Suppose MATH and MATH. By induction on MATH, we have MATH. Note that the homogeneous components of MATH have degree greater or equal than MATH. By transitivity, we conclude that MATH, hence MATH on MATH, and therefore MATH can be expressed as a linear combination of MATH, MATH and some inner derivation. Since MATH is an inner derivation, the lemma is proved. |
math-ph/0106002 | With respect to the action of MATH, MATH decomposes into eigenspaces relative to the eigenvalues MATH. The even part of MATH is contained in the zero eigenspace. Also, MATH (respectively, MATH) transforms the MATH (respectively, MATH) eigenspace into the MATH (respectively, MATH) eigenspace and kills the other eigenspaces. We will use the standard depth REF gradation of MATH, see REF . The even part of MATH is MATH. REF implies that MATH as MATH-modules and any MATH is a MATH-module homomorphism. The modules occurring more than once are MATH (in MATH and MATH), and MATH (in MATH and MATH). Let MATH (respectively, MATH) be the highest weight vector of MATH in MATH (respectively, MATH). Then MATH. The derivation MATH is a MATH-module homomorphism, so MATH, and MATH maps MATH into itself. By NAME Lemma, MATH. It follows that the derivation MATH acts as MATH on MATH. Let MATH. We have MATH so that MATH. For any MATH, MATH, hence MATH. Now, MATH and MATH occur only in MATH, hence MATH. The module MATH occurs in MATH and MATH, so MATH. Also, MATH. We may assume that MATH. Then MATH. MATH kills MATH and MATH. It follows that MATH. Also, MATH is still identically zero on MATH. Hence we have MATH, and MATH has homogeneous components of degree greater or equal to zero. Transitivity implies that MATH on MATH. Induction on MATH shows that MATH on MATH. Therefore, D can be written as a linear combination of MATH, MATH, MATH and some inner derivation. |
math-ph/0106002 | In the contrary case, we also have MATH and therefore MATH induces a surjective derivation of MATH. But for MATH, MATH, MATH and MATH we have MATH, MATH, MATH, MATH respectively. Hence by REF we reach a contradiction, unless MATH, MATH or MATH. The second case is also excluded since MATH has only inner derivations (this is immediate by REF ) and the third case is reduced to the first one since the even part of MATH is MATH. But if MATH, it is easy to see that one of the elements MATH or MATH does not lie in the image of MATH. |
math-ph/0106002 | Recall that MATH, where MATH is one of the linearly compact NAME superalgebras listed in REF and MATH is an even surjective derivation of MATH. If MATH is one of the NAME superalgebras listed in REF , consider the filtration MATH of MATH corresponding to the standard gradation MATH of MATH (compare REF). One checks directly that in all cases one has: MATH . Furthermore, due to REF we have MATH where MATH is non-zero, MATH and MATH, where MATH is one of the following subspaces of MATH: MATH . We may apply now REF to MATH (compare REF ) and the above MATH since REF holds due to REF also obviously holds. Hence by an inner automorphism of MATH we can bring MATH to the form: MATH . By rescaling we can make MATH and, using an inner automorphism of the NAME algebra MATH, MATH can be brought further, in all MATH cases, to the form MATH (if MATH is even) or MATH, MATH. The case of MATH has been treated in a similar fashion in the proof of REF . |
math-ph/0106002 | It follows from REF . |
math-ph/0106007 | Clearly a local trivialisation (compatible with the MATH action or not) defines a section, via MATH. Given a section, define MATH by MATH. This is clearly compatible with the MATH action. If we began with a trivialisation compatible wiht the MATH action, these constructions are mutual inverses, establishing the correspondence. |
math-ph/0106007 | It is clear that this map is linear. Additionally, it is surjective, because any MATH can be written as MATH for some MATH, and MATH. It is injective, since if MATH, then there is a MATH so that MATH, and so MATH, and finally MATH. |
math-ph/0106007 | This is an entirely standard argument. However, due to its importance, both in providing connections as technical tools, and underlying our interest in connections on spinor structures in REF, we give a proof in REF. |
math-ph/0106007 | Write MATH. Now MATH, since MATH. Further MATH, so MATH, proving the result. |
math-ph/0106007 | The method of construction is as described above - we simply add here that the integral curve giving the parallel transport can be extended so as to be defined over all of the interval MATH, following the argument of CITE, or of CITE. We omit these details here. The second part follows immediately from the fact that solutions of differential equations depend (at worst) continuously on a smoothly varying initial value CITE. In more detail, REF proves that MATH is smooth for each MATH, and since MATH is also smooth for each MATH, by the first part of this proposition, the map MATH is certainly continuous. It is a possible, but not necessary here, to prove a stronger result. |
math-ph/0106007 | See CITE. |
math-ph/0106007 | Define MATH by MATH. Thus MATH, and MATH. For an arbitrary MATH, choose a path MATH so MATH, and let MATH, where MATH. Then MATH . Here MATH, and MATH. Thus MATH and MATH agree on the second term of the expression above, and so MATH. Finally then MATH, and this is easily seen to imply the result. |
math-ph/0106007 | Let MATH be a curve in MATH, and let MATH, MATH. Suppose MATH, and parallel transport along MATH carries MATH to MATH. Thus parallel transport carries MATH to MATH, which is exactly MATH, since MATH in an invariant vector field. Thus parallel transport along any curve carries MATH to itself, and so, from the definition of the covariant derivative in terms of parallel transportation in REF, MATH . |
math-ph/0106007 | We construct this smooth structure as follows. Let MATH be an open covering of MATH by sets so that MATH maps MATH homeomorphically onto its image, and MATH is a chart for MATH, with coordinate map MATH for each MATH. Such an open covering certainly exists. Define MATH by MATH for each MATH. This map is a homeomorphism, because it is a composition of homeomorphisms. Further, the `transition maps' MATH are all diffeomorphisms, because MATH. Thus the collection MATH defines an atlas for MATH, and it is clear that MATH is a smooth map, and further a smooth covering map, with respect to this differentiable structure. Uniqueness is trivial, since for MATH to be a diffeomorphism, all of the charts described above must be in the atlas for MATH. The differentiable structure is uniquely determined by any atlas, establishing the result. |
math-ph/0106007 | See REF and the following discussion in REF, and REF of the same. Related results are given in CITE. |
math-ph/0106007 | This follows immediately from REF . Firstly it is surjective, because any path in a smooth manifold is homotopic to a smooth path. Secondly, it is injective, since if two smooth paths are continously homotopic, they are smoothly homotopic. |
math-ph/0106007 | A complete proof of this theorem, as stated, cannot be found in any one place. Furthermore, for later work we will need some of the details of the constructions involved. For this reason, we present here an outline of the proof, citing appropriate references for each intermediate result, and in places extending standard results to fit the particular circumstances of this theorem. The first part of the theorem, that the covering map induces an injective map of the fundamental groups, is very straightforward, using the lifting properties of covering maps. A proof is given in CITE, and CITE. Next, we consider the implications of the smoothness of MATH. Since MATH is a manifold, it is locally path connected and locally simply connected, on account of each point of MATH having a neighbourhood homeomorphic to an open ball in MATH. Further, connectedness implies that MATH is path connected. This is because local path connectedness means that the path connected components of MATH are open and closed, and so equal to connected components of MATH. See also CITE. The second part of the theorem, on existence of coverings, is proved in the continuous setting in CITE. It depends upon MATH being path connected, locally path connected, and locally (or semilocally) simply connected. As we have seen all these conditions are automatically true for smooth manifolds. To improve that result for this theorem, we need only show that this covering can be given a smooth structure so that the covering map becomes a smooth covering map, and this has already been achieved above, in REF . The statement about the fundamental groups remains true in the smooth setting, on account of REF . Finally, the last part, giving conditions for equivalence of covering spaces, is proved in the continuous case in CITE. To improve this for the current theorem, we need to show that if MATH and MATH are continuously equivalent covering maps, then they are smoothly equivalent covering maps, with respect to the differentiable structures defined above. This follows immediately from the definitions, and the fact that the continuous equivalence is given by a homeomorphism MATH such that MATH, which is then also a diffeomorphism. |
math-ph/0106007 | Take the trivial subgroup MATH in MATH, and form the associated covering space MATH. Since the covering map MATH induces an injective map MATH, MATH is itself trivial, and so MATH is simply connected. If MATH is any other covering map with MATH simply connected with base point MATH, then MATH has a trivial image, and so the covering map MATH is equivalent to the one we have constructed, MATH. |
math-ph/0106007 | This is proved in CITE. |
math-ph/0106007 | Consider a local cross section of MATH, defined on an open subset MATH, MATH. The composition MATH then defines a local cross section of MATH. We can use these cross sections to define local trivialisations of both bundles, by REF . MATH . Both of these maps are diffeomorphisms, and in fact principal morphisms. We can compose these maps with MATH, to obtain MATH . However this map acts very simply, as follows, MATH . Thus MATH, and as MATH and MATH are diffeomorphisms, we can write the covering map as MATH. This expresses MATH locally as a trivial map in the sense of covering spaces, and so MATH is a covering map. |
math-ph/0106007 | Since MATH is simply connected, every covering space is equivalent to MATH itself, and so MATH has no connected covering spaces larger than itself, and thus no spinor structure is possible. |
math-ph/0106007 | For the purposes of this proof, we will fix a connection on MATH. Such a connection always exists by the results in REF. Proving that MATH is surjective is relatively easy, so we first do that. Suppose MATH is any smooth loop in MATH based at MATH. Define MATH to the parallel transport of MATH along MATH. This curve will generally not be a loop. However, MATH, and since the fibres of MATH are path connected, we can find a path MATH so MATH and MATH. Now consider the path MATH, which is in fact a loop since MATH and MATH. Further MATH, and so MATH. Thus MATH is surjective. We now turn to the more technical problem of demonstrating that MATH is injective. The underlying result, however, has already been established, the idea here being to use a connection to `lift' a homotopy in MATH to a map into MATH, and then using the simply connectedness of fibres to modify this into the appropriate homotopy. Suppose MATH and MATH are elements of MATH, and MATH. Then there are smooth loops MATH and MATH in MATH, so MATH and MATH, and, further, there are smooth loops MATH and MATH in MATH so MATH, MATH and MATH, and MATH. Thus there is a smooth homotopy from MATH to MATH. Call this homotopy MATH, so MATH, and MATH. We will write MATH for the function MATH. According to the second part of REF , we can parallel transport MATH along MATH, to obtain a smooth curve MATH, so that MATH, and the map MATH is continuous. We will next modify MATH to form a homotopy between MATH and MATH. The particular properties of MATH that we require are MATH . Define MATH to be the boundary of MATH, that is MATH . Define MATH according to MATH for all MATH, and MATH, MATH. According to this definition, and the above properties of MATH, MATH on MATH, and so MATH, for some function MATH. Since MATH is simply connected, we can extend MATH to a continuous function MATH. Now define MATH by MATH. Thus on MATH, MATH and MATH agree, and so MATH is a continuous homotopy between MATH and MATH. Finally, this implies that there is a continuous homotopy between MATH and MATH, and so by REF , there is a smooth homotopy between MATH and MATH. This establishes the injectivity of MATH, and so proves that it is an isomorphism. |
math-ph/0106007 | Firstly, the map MATH is injective, according to the Covering Space Classification Theorem. We now consider the following commuting diagram, MATH and the restriction of MATH to MATH, MATH. Since MATH is injective and MATH is an isomorphism, by REF , MATH is injective. Further, MATH, so MATH must be surjective, and thus MATH is an isomorphism. |
math-ph/0106007 | See REF. |
math-ph/0106007 | Suppose MATH, and MATH. Suppose MATH is homotopically trivial. Then, according to the Path Lifting Lemma CITE we can lift MATH via the covering map MATH to a path MATH in MATH, and according to the Homotopy Lifting Lemma CITE, it is a loop homotopic to the constant loop. This loop lies within a single fibre, and, since the fibres are homeomorphic to MATH, they are simply connected, and so MATH is homotopic to the constant loop by a homotopy that stays within the fibre MATH. Applying MATH to this homotopy gives a homotopy of MATH to the constant loop by a homotopy that stays within the fibre MATH, and thus MATH is homotopic to the constant loop in MATH. Thus MATH, and so MATH is injective. |
math-ph/0106007 | The proof is in two steps. Say MATH, MATH, and MATH. Then, applying MATH to both sides, MATH since MATH has a representative lying within a single fibre. Now, since MATH restricted to MATH is an isomorphism, MATH also, and since MATH is injective by REF , MATH as well. Thus the two groups have a trivial intersection. Next, take any MATH. We define MATH as follows. Firstly let MATH. Then, as in the discussion of REF , let MATH be the parallel transport of MATH along MATH. Further, chose a path MATH so MATH and MATH is a loop in MATH. Now MATH. Define MATH. We see from this, and REF , that MATH, and moreover that MATH. Thus MATH for some loop MATH. Then MATH applying REF . Thus the internal direct product MATH generates all of MATH. |
math-ph/0106007 | The proof is in three parts. All are straightforward, but somewhat involved, especially the second. The action is free. Suppose MATH is such that MATH for some MATH. Take a path in MATH representing MATH, say MATH, so MATH. As in REF , MATH. Then MATH implies MATH and so MATH, by the hypothesis that MATH. Thus MATH is homotopically trivial, and so MATH. The action is transitive. Suppose we have two elements MATH of MATH within the same fibre, such that MATH and MATH for two paths MATH. Since MATH and MATH are in the same fibre, MATH for some MATH. Consider MATH for some MATH, and MATH so MATH and MATH. Such a MATH exists since MATH is path connected, and MATH acts transitively on the fibres of MATH. Further, MATH represents some MATH. We calculate MATH . Here MATH is any path in MATH so MATH and MATH, and MATH denotes the reversed path MATH, not the inverse path, and we have used REF in the last line. Note that by varying MATH, subject still to the conditions MATH and MATH, we can make MATH homotopic to any arbitrary loop MATH in MATH. This is achieved by setting MATH, so MATH. Thus the first two paths, MATH and MATH can be chosen to generate any element of MATH, since MATH. In particular, we can chose MATH and MATH so that MATH so MATH . With this choice, MATH . This proves that the MATH action is transitive on the fibres, as required. There are local trivialisations compatible with the MATH action. Since MATH is a principal fibre bundle, for any point MATH there is an open set MATH with MATH and a local section MATH, in accordance with REF . Find a simply connected open set MATH, and restrict MATH to MATH. By the monodromy principle CITE there is a lifting of MATH to a map MATH via the covering map MATH. This is then a local section of MATH, and applying REF a second time we find a local trivialisation. |
math-ph/0106007 | In the notation above MATH acts on MATH by taking MATH to MATH . Thus MATH . |
math-ph/0106007 | If MATH and MATH are equivalent as spinor structures then there is a principal bundle morphism MATH so MATH. This MATH is then a fortiori a diffeomorphism, and so MATH and MATH are immediately seen to be equivalent as covering maps. Conversely, suppose MATH and MATH are equivalent as covering spaces, so there is a diffeomorphism MATH such that MATH. Since MATH and MATH are spinor maps, we can easily calculate MATH . The equality between the first and last expressions then implies that MATH, for some MATH. Further, since MATH is a covering map, if we fix MATH, MATH depends continuously on MATH. Since MATH is discrete, MATH is constant, and since if MATH, MATH, we must have MATH for all MATH. That is, MATH is additionally a principal bundle morphism, and so MATH and MATH are equivalent as spinor structures. |
math-ph/0106007 | Suppose MATH and MATH are subgroups of MATH, each isomorphic to MATH via MATH, such that we can construct spinor structures in accordance with REF. Say these are MATH and MATH with spinor maps MATH and MATH. According to REF , these spinor structures will be equivalent if the spinor maps MATH and MATH are equivalent as covering maps. The classification of covering maps given in REF states that two such covering maps are equivalent if and only if the groups MATH and MATH are conjugate. Thus suppose MATH and MATH are conjugate, so there is a MATH so that MATH. Now MATH can be written as the product MATH, so MATH, for some MATH, and MATH in the image under MATH of MATH. Moreover, MATH lies in the centre of MATH, by REF . Thus MATH. This establishes the desired result. |
math-ph/0106007 | Suppose MATH is a homomorphism. Define MATH by MATH. Now take MATH and suppose MATH also. Now because MATH gives a unique decomposition, MATH, and so MATH. This establishes that MATH and MATH have a trivial intersection. Next, for any MATH, there is some MATH, MATH so MATH where MATH, and MATH. Thus the internal direct product of MATH and MATH is all of MATH, as required. Conversely, define an isomorphism MATH by MATH . Then we must have MATH for some maps (not necessarily, at this stage, homomorphisms) MATH, and MATH. Now MATH, so MATH, and since MATH restricted to MATH is an isomorphism, MATH for all MATH. Using this simplification, we write MATH in two ways. MATH . Thus by the uniqueness of the MATH decomposition and the injectivity of MATH, we conclude that MATH is a homomorphism. Finally, define MATH by MATH and note that now MATH, and so MATH is exactly the required homomorphism, relating MATH and MATH as in the first part of the proof. |
math-ph/0106007 | For any topological space MATH, the first homology group with integer coefficients is isomorphic to the commutative factor group of the fundamental group. That is, if MATH denotes the commutator subgroup of MATH, MATH . This called the NAME isomorphism and is a standard result from algebraic topology. See CITE for the proof. In particular, since MATH is commutative, by REF , for any homomorphism MATH, the commutator subgroup MATH is contained in the kernel, and so MATH descends to a map MATH. Clearly any such map extends to a homomorphism MATH. This has established that the homomorphisms MATH correspond naturally to the homomorphisms MATH. Finally, because MATH is commutative we can use the Universal Coefficient Theorem CITE, relating homology and cohomology, to show that MATH . Here MATH is precisely the group of homomorphisms MATH, and we do not define in detail MATH, pointing out that as MATH by CITE it is always trivial. Putting this together, we see that the spinor structures are classified by MATH . |
math-ph/0106007 | The maps MATH and MATH have been considered previously. It is obvious that the sequence is exact at MATH. Exactness at MATH states simply that MATH is onto. This is clear, since any path in MATH can be lifted arbitrarily to give a path in MATH, and the lift of a loop in MATH can be extended within the initial fibre to form a closed loop. This loop then maps down via MATH to give the original loop in MATH. Next we turn to the map MATH. The construction of this map in a similar context is mentioned in CITE. There are theorems proved in a very general setting giving exact sequences for homotopy groups of spaces with fibrations CITE, CITE. To use such a theorem here we would have to introduce relative homotopy groups CITE, which would take us rather far afield. However, in this particular situation, where we are content to assume that our spaces are smooth and paracompact, we can give a simple and geometric argument. Interestingly, the proof here will introduce a connection, but as it will turn out this particular choice will not affect the final construction. Providing our own argument here rather than the general one mentioned above simplifies the proof of REF below. We first give some notation for parallel transportation. For this purpose we will fix a particular connection on the principal fibre bundle. Given a path MATH in MATH, with initial point MATH, we can parallel transport MATH along MATH, to obtain a point in the bundle in the fibre of MATH. Denote this point by MATH, so that MATH becomes a map MATH. Parallel transportation along a loop in MATH is of interest because it returns MATH to the initial fibre. Thus MATH restricted to loops becomes a map MATH. For any MATH, there is a unique MATH such that MATH. This MATH is value of the translation function MATH. Define a new function MATH which, given a loop in MATH, produces this MATH. Thus MATH. Moreover, MATH acting on the constant loop gives the identity element of MATH, and so is base point preserving. In fact, MATH is actually a homomorphism, because of the reparametrisation properties of parallel transport, but we shall not need this fact. More importantly, MATH is continuous. Not having specified the topology for MATH, we cannot make this precise, but it is clear that REF ensures that MATH is relatively well behaved. Since MATH is a base point preserving map, it induces a map of the homotopy classes, MATH. The fundamental group of the loop space of MATH is just the second homotopy group of MATH, MATH, and so this MATH is of the form indicated in the statement of this Lemma. It is not too hard to prove that MATH is in fact independent of the particular choice of connection in the definition of MATH. However the argument is lengthy and unnecessary here. The remaining part of the series is MATH . Thus we want to prove that MATH. Suppose MATH is in MATH, so MATH for some MATH. Thus for each MATH, MATH. Define MATH by MATH so MATH. We now want to prove that MATH, that is, that MATH is homotopic to the constant map in MATH. For each MATH, define MATH as the path MATH traversing only the interval MATH. Thus MATH and in particular MATH, and MATH. We now define a homotopy MATH according to MATH . A calculation shows that this is a homotopy from MATH to the constant path at MATH. MATH . Also, MATH is a homotopy fixing endpoints, that is, MATH. This follows from the fact that MATH and MATH are both the constant path in MATH. Thus MATH, and so MATH. Next we want to prove that MATH, and so we suppose that MATH, and MATH. Now, MATH, and MATH. Therefore, from the hypothesis there must exist some homotopy MATH so that MATH . We will next use this homotopy to define an element MATH of MATH so that MATH. Let MATH. Thus, for each MATH, MATH is a path in MATH. Again, define MATH as the path MATH restricted to the interval MATH. We can use the connection to perform parallel transportations along these paths, resulting in a map MATH, given by MATH. This is a continuous function, by REF . However, there is no reason for MATH to be equal to MATH. On the other hand, it must be in the same fibre as MATH, since the parallel transport projects down to the original curve. Thus for each MATH, there is some MATH so MATH. Since MATH and MATH are continuous, MATH is a continuous function also. Now, MATH. Thus MATH in fact defines a homotopy between MATH and MATH. Moreover, this homotopy stays within the fibre of MATH, and so gives a homotopy of the loop MATH and MATH. This proves that MATH, and so MATH maps onto the kernel of MATH, completing the result. |
math-ph/0106007 | To prove that a diagram of this form commutes, we need only check that each square commutes. To prove the first square commutes we use the result of REF that the two parallel transports are the same, and so MATH for every MATH. Thus MATH. The second and third squares commute, using the identities MATH and MATH . The fourth square commutes trivially. Now that we have established that the diagram commutes, we can apply a powerful technique from homological algebra, the five lemma. (For a proof, see CITE.) The five lemma states that if we have two exact sequences, linked by four isomorphisms in a commuting diagram as above, then the central vertical map is also an isomorphism. Thus MATH is an isomorphism. |
math-ph/0106007 | We have seen in the proof of the Classification Theorem that the spinor structures are in one to one correspondence with subgroups of the fundamental group of the bundle with satisfy the hypotheses of the Existence Theorem. REF shows that such subgroups for MATH and MATH are in one to one correspondence via MATH. Thus, fix MATH and MATH, and form the associated spinor structures MATH and MATH. To construct the reduction map MATH, recall that in the construction of the spinor structures, each point of MATH is an equivalence class of paths in MATH, and each point of MATH is an equivalence class of paths in MATH. Thus the typical point of MATH is MATH where MATH, MATH, and MATH if and only if MATH and MATH. Similarly, the typical point of MATH is MATH, where MATH, MATH, and MATH if and only if MATH and MATH. Define MATH in the natural way, as MATH . This is well defined, since if MATH, MATH. Further, it is a reduction map. If MATH, then MATH is an equivalence class of homotopic paths in MATH. Say MATH is a representative, so MATH. Define MATH by MATH. This is well defined, since if MATH, then MATH, and the elements of MATH are homotopy classes of paths in MATH. Now, MATH as required. |
math-ph/0106007 | The proof is relatively straightforward, although requiring several technical calculations. It will be useful to define a partial inverse function to MATH for this proof. Since MATH is a covering map, there is a neighbourhood of the identity in MATH, say MATH, so that MATH is one to one. We will abbreviate MATH to simply MATH. Notice MATH. Firstly we need to check that vertical vectors are mapped appropriately into the NAME algebra. Firstly define functions MATH and MATH by MATH . Then, in accordance with REF , the condition on vertical vectors is that MATH . We will show that these conditions are equivalent. We easily see that for MATH, MATH if and only if MATH, since MATH. Moreover, every MATH such that MATH is of the form MATH, with MATH such that MATH. (That is, MATH maps MATH onto MATH.) Next since MATH, we can apply MATH to both sides and use the fact that MATH is a principal bundle morphism to obtain MATH, and so MATH. When MATH, we have MATH. Thus MATH, and MATH . Equivalently, MATH . Now suppose that MATH for all vertical vectors MATH. Then MATH applying REF . This holds for every vertical vector MATH. Conversely, suppose that MATH for all vertical vectors MATH. Then MATH . Applying REF , we obtain MATH . This completes this section of the proof. Secondly, to confirm that the `elevator properties', MATH are equivalent, we need to prove the following simple commutation relations. MATH . REF are obvious, because the adjoint map and MATH act on values in a NAME algebra. Similarly REF follows from the fact that MATH is a principal fibre bundle morphism. Finally, to establish REF , we calculate, for MATH, MATH . Thus for MATH, using MATH, we have MATH . We finish the proof as follows. Suppose firstly that the elevator property holds for MATH. Then for every MATH, MATH . Thus the elevator property holds for MATH. Conversely, suppose the elevator property holds for MATH. Then for every MATH, MATH . Next, expressing the MATH on the left hand side in terms of MATH, and applying MATH to both sides, we find MATH . Thus for every MATH, MATH . Now every MATH can be written as MATH for some MATH, and every MATH can be written as MATH for some MATH, and so we reach our desired result MATH . This completes the proof of the proposition. |
math-ph/0106007 | See CITE. |
math-ph/0106007 | We propose to define MATH by MATH. The problem with this is that MATH is not uniquely defined, so we must check that regardless of which inverse we use the same answer is reached. For this purpose, say MATH, and MATH is an inverse image, so MATH. Now all the inverse images are of the form MATH, where MATH. Suppose further that MATH, and the inverse of MATH taking MATH to MATH takes MATH to MATH. Then the inverse taking MATH to MATH acts as MATH. Now MATH . This has established that MATH is independent of the inverse used in the calculation, and so the proposed definition is well defined. Finally, it is clear that MATH, so REF applies. This proves that MATH is a connection. |
math-ph/0106007 | Clearly MATH. Further, MATH . Next, according to the definition of parallel transport as an integral curve of a horizontal vector field, in REF, MATH . We then calculate MATH . Finally, rearranging the result of REF shows that a vector MATH is determined by MATH and MATH, and so MATH . Thus the integral curves MATH and MATH are equal. |
math-ph/0106007 | We say that MATH is MATH-universal. The bijection is given by MATH, where MATH denotes the `pull-back' bundle, defined in CITE or CITE. A simple proof that this map is well defined, that is, that homotopic maps give isomorphic bundles, appears in CITE. The proposition itself is a deep result of the algebraic topology of bundles, and is discussed in CITE and CITE and proved in CITE. |
math-ph/0106007 | Since MATH is paracompact in particular it has a countable basis for its topology. With this fact, this Proposition is a slight weakening of a theorem proved in CITE. Another result which implies this theorem, but less obviously, is given in CITE. |
math-ph/0106007 | Say MATH is any spinor structure for MATH. Then MATH, as a bundle, is trivial, so MATH. The spinor map MATH is a principal fibre bundle morphism, and so gives a trivialisation of the bundle MATH. Since MATH is trivial, it has a global cross section, which is exactly the global orthonormal frame field. Conversely, if MATH is a trivial bundle, then the condition of the Existence Theorem is automatically satisfied, and so MATH has a spinor structure. |
math-ph/0106007 | To begin, we see from REF that MATH . Firstly suppose MATH is trivial as a bundle. Thus there is a bundle equivalence MATH. Define MATH. Since MATH, there is some function MATH so MATH . Now MATH and thus MATH . Define MATH by MATH. Therefore MATH. Now, as MATH is a diffeomorphism it induces an isomorphism of fundamental groups, and so the image of MATH is the same as the image of MATH. Thus as a spinor structure MATH is equivalent to MATH. We now need to prove that MATH. To see this, note that the general element of MATH is MATH where MATH and MATH is the constant path at MATH. The map MATH acts on this as MATH, and so MATH . Comparing this with REF demonstrates the MATH, completing this half of the proof. Conversely, suppose MATH is realisable as the map MATH. Define MATH as above, and, following the same argument, we still have MATH . Thus MATH is equivalent to MATH as a spinor structure, and thus the bundles MATH and MATH are equivalent as bundles, and so MATH is trivial. Finally, we briefly describe the freedom available in choosing MATH. Suppose MATH and MATH both induce the homomorphism MATH. It is easy to prove, using REF , that MATH, defined via the group product, induces the trivial homomorphism MATH. Such maps are exactly the smooth maps of the form MATH, where MATH CITE. That is, MATH and MATH differ by a map which lifts to a map into MATH. It is easily seen that, in the above argument, if MATH and MATH are derived from two bundle equivalence MATH, then these bundle equivalences differ by a map MATH, and MATH. |
math-ph/0106007 | Suppose MATH is a path, with MATH, and MATH and MATH is a path, with MATH, and MATH. Then MATH . In this last line here by MATH we really mean `the derivative of left multiplication by MATH acting on MATH'. Thus while MATH, MATH. Next, we calculate MATH . While MATH, we have MATH, as is appropriate. |
math-ph/0106007 | Let MATH be the vector space of self-adjoint MATH by MATH complex matrices. We let the group MATH act on MATH by MATH for MATH and MATH. This is a well defined map of MATH to itself, since MATH . That is, the action of MATH preserves the self-adjointness of MATH. This group action is a representation of MATH on MATH. We next make an identification of MATH with MATH, via the map MATH . This map is clearly a linear isomorphism. Further, the usual Lorentzian metric on MATH can be expressed as MATH . That is, MATH for all MATH. Additionally, the MATH component is recovered easily, as MATH . Using this identification, we define a map MATH from MATH to MATH by MATH . Now, MATH. Thus the image of MATH is contained in MATH, the group of isometries of MATH. Finally, since MATH is connected, and MATH is clearly continuous, the image of MATH must be connected, and so lies within MATH. That is, MATH. It is straightforward to see that MATH is a group homomorphism, since MATH . Using this, we can think of MATH as defining a representation of MATH on MATH. The map MATH is not just a linear isomorphism. It intertwines the representation of MATH on MATH and the representation via MATH on MATH. This follows trivially from the definition of MATH in REF , but it is nevertheless important. Next we prove that MATH is surjective, by exhibiting elements of MATH which are mapped to arbitrary rotations about the three coordinate axes, and elements of MATH which are mapped to arbitrary boosts in the MATH direction. Firstly, the rotations are given by MATH . Any rotation can be expressed as a product of rotations of these forms. Notice that each of these matrices actually lies within MATH. The boosts in the MATH direction are given by MATH . Further, a boost along any axis can be written as the product of a rotation taking that axis to the MATH axis, a boost along the MATH axis, and the inverse rotation. Since any element of MATH can be written as a product of rotations and boosts CITE, the map MATH is surjective. Finally, to establish that MATH is MATH to MATH, we find the kernel. Assume MATH. In particular MATH must preserve the MATH component of any vector in MATH, and so using REF we obtain MATH for any MATH. Write MATH . Substituting the following matrices MATH, MATH, MATH, MATH in MATH for MATH in REF we obtain the following conditions on MATH: MATH . This simply states that MATH, and so MATH is a unitary matrix, MATH. Thus there are MATH such that MATH, so that MATH . We next calculate MATH as MATH . If MATH, then MATH, and so either MATH or MATH is zero. If MATH, then MATH and so MATH, which is impossible. Thus MATH, and MATH. Next, we calculate MATH and the condition MATH implies that MATH. Both of these possibilities are clearly in the kernel of MATH, since MATH, and so the kernel is exactly MATH. Any such surjective homomorphism with a discrete kernel is always a covering map. In fact, strictly speaking, since both MATH and MATH are MATH dimensional, and connected, it is not necessary to explicitly demonstrate that MATH is surjective, given the other results. For the purpose of understanding the geometry, however, it is useful to have exhibited the matrices for the rotations and boost above. |
math-ph/0106007 | We need to check that the map is well defined. This requires two steps. Firstly, an element of the frame bundle MATH will have two inverse images under MATH. However these will differ by the generator of the kernel of the covering map MATH, so the two inverse images are of the form MATH and MATH. Now, MATH and so this ambiguity is removed. Secondly, if MATH, then MATH. We need to check that MATH. Choose MATH so that MATH, and apply REF . Then MATH as required. That the map is a linear isomorphism between the bundles follows immediately from the fact that MATH is a linear isomorphism between the underlying vector spaces. |
math-ph/0106007 | We simply calculate in components, using REF . MATH . Thus MATH. |
math-ph/0106007 | We choose adapted local trivialisations of MATH and MATH. Let MATH be a local trivialisation, and let MATH be defined by MATH. Fix MATH, and say MATH, and MATH. Let MATH be a vector field defined on MATH, and let MATH be the integral curve of MATH starting at MATH. We can form two parallel transports of the path MATH, via MATH and MATH, to obtain MATH and MATH. In the local trivialisation these parallel transports are MATH and MATH. Here MATH and MATH. Now, in accordance with REF , MATH, and so MATH . Thus MATH. Finally now we calculate the covariant derivative, using REF . MATH . |
math-ph/0106007 | The tensor MATH is an invariant tensor for MATH, according to REF , and so its associated tensor field is covariantly parallel, by REF . Thus MATH . |
math-ph/0106007 | Since we work in NAME coordinates, the covariant derivative is just a partial derivative. REF are equivalent to MATH and so MATH where we have used REF . Next, we fulfill the summation of MATH or MATH, and write these equations in matrix form. MATH . Using REF to rewrite the partial derivative operators with tensor indices, we obtain MATH . Finally, using MATH and MATH for MATH, we see that this agrees with the explicitly written NAME equation above. |
math-ph/0106007 | Consider the homotopy MATH . This proves MATH. A similar homotopy establishes the other half. |
math-ph/0106007 | This follows immediately from the argument given in the proof of REF . As seen there, elements of MATH are `trace preserving', and so fix the MATH component. Thus MATH maps MATH into MATH, and this restriction is clearly onto, because the pre-images of the rotations, as exhibited, all lie in MATH. Finally, the kernel of MATH lies in MATH, and so the restricted map is also MATH to MATH. |
math-ph/0106007 | Since MATH is topologically MATH it is simply connected. Thus it is the universal covering space for MATH. Covering space theory for NAME groups CITE (and see REF) states that the fundamental group of the base space is the kernel of the universal covering map. Thus MATH. |
math-ph/0106007 | See REF. Related results can be achieved directly by methods of linear algebra, as in CITE, or quite generally by means of the global NAME decomposition CITE. |
math-ph/0106007 | See CITE. |
math-ph/0106007 | Firstly, associate with each point MATH an open set MATH such that MATH is trivial over MATH. Next, choose for each point MATH an coordinate chart MATH, with MATH. Since MATH is an open set in MATH, there is a MATH so MATH, the open ball of radius MATH about MATH, is contained in MATH. Next, let MATH, and MATH. Since MATH is homeomorphic to MATH, it is relatively compact. Thus the collection MATH is a covering of MATH by relatively compact coordinate charts. Let MATH. The bundle is locally trivial over these sets, which are also relatively compact and coordinate charts. Any open subset of such a set also satisfies these properties. These sets form a covering of MATH, and so applying our assumption of paracompactness, we obtain a locally finite open covering MATH which is a refinement of the covering MATH, and so consists of relatively compact local trivialisations. |
math-ph/0106007 | The proof here follows that in CITE. A similar proof appears in CITE. Let MATH be an open covering of MATH as described in REF , and let MATH be a partition of unity subordinate to this open covering. We will define a connection on MATH using the NAME algebra valued form description, defining a connection on MATH for each MATH and patching these together using the partition of unity. We now define a connection form on the each of the sub-bundles MATH. Put simply, we choose the obvious flat connection relative to the local trivialisation MATH. Given MATH, MATH, say. If MATH, then MATH. This tangent space splits, since MATH. Thus we can always write MATH as MATH, where MATH, and MATH. Moreover, given this decomposition, MATH, and this represents a similar decomposition, since the action of MATH commutes with the trivialisation. We then define MATH. The map MATH is defined as before in REF, by MATH. The first property we require of a connection form, that it maps vertical vectors into the NAME algebra according to MATH, is satisfied since the vertical vectors MATH are those such that MATH, and so this definition gives MATH. Next, we calculate MATH and so MATH. Then MATH and so MATH is in fact a connection form on MATH. Finally, we obtain a connection form on the entire bundle simply by writing MATH. |
math-ph/0106007 | Firstly, MATH, by the elevator property for MATH, and so MATH. Next, since MATH, we have MATH. If MATH, then MATH and if MATH then MATH. Certainly MATH, and MATH, so MATH, as required. |
math-ph/0106007 | Suppose MATH is written in two ways, as MATH and MATH, so MATH and MATH for some MATH. Then MATH . Now MATH, and so MATH. Thus MATH . This holds also for any MATH. Applying this to REF , we obtain that MATH . Thus the value of MATH is independent of the particular presentation MATH chosen. The definition of MATH in REF guarantees that the elevator property is satisfied. Finally, we to check that vertical vectors are mapped into the NAME algebra according to REF . If MATH is vertical, so MATH, then MATH for every MATH. Thus MATH as required. |
math-ph/0106007 | We define two homotopies, MATH, as indicated in REF . Let MATH and MATH . It is easy to see that these piecewise definitions give continuous maps. Further, MATH, and MATH . Thus MATH establishes the first homotopy, and in the case MATH, MATH, MATH so MATH and MATH give the required homotopies for the second part. |
math-ph/0106009 | Given a ramified covering MATH with projections MATH of branch points on MATH, we construct the corresponding permutation representation as follows. Denote the projection of MATH on MATH by MATH. Generators MATH of permutation monodromy group are given by the following construction. Consider the pre-image MATH of the generator MATH. This pre-image is a union of MATH (not necessary closed) disjoint contours on MATH which start and end at some of the points MATH (by MATH we denote the point MATH-th sheet of MATH which has projection MATH on MATH). Denote by MATH the component of MATH which starts at the point MATH; the endpoint of this contour is MATH for some MATH. If MATH is not a branch point, then MATH, and contour MATH is closed; if MATH is a branch point, then MATH and contour MATH is non-closed. Then the monodromy matrix MATH has the following form: MATH and naturally corresponds to some element MATH of the permutation group MATH. On the other hand, starting from some permutation monodromy representation we obviously can glue the sheets of the NAME surface at the branch points MATH in such a way that it corresponds to the permutation monodromies REF . Moreover, this NAME surface is obviously compact. |
math-ph/0106009 | This theorem is valid for arbitrary multiplicities of the branch points. Here we check it under assumption that all branch points are simple and have different projections on MATH-plane. In this case the local parameter in the neighborhood of a branch point MATH is equal to MATH; this is the only local parameter which depends on the position of MATH. The NAME series REF looks as follows: MATH . Therefore, the NAME differential MATH where MATH is two-dimensional delta-function, describes the infinitesimal deformation of complex structure under variation of position of the branch point MATH CITE: MATH . Substitution of NAME differential REF into NAME variational REF gives REF and the following formula for variation of matrix of MATH-periods: MATH . In turn, this formula implies REF if we take into account the following REF . |
math-ph/0106009 | This lemma is also valid without any restrictions on ramification type at the points MATH; it is sufficient to check that MATH is a holomorphic differential on MATH. The only suspicious points are the branch points MATH. Regularity of this sum at point MATH follows from analysis of its NAME series in the neighborhood of MATH. For example, if the branch point MATH is simple, it is sufficient to observe that the local parameter MATH has different signs on the sheets glued at the branch point MATH. |
math-ph/0106009 | Choose in the Fay identity REF MATH and MATH. Then, taking into account the holonomy properties of the prime-form and asymptotics REF , we conclude that MATH which, being considered as function of MATH, does not vanish outside of the points MATH; thus MATH if MATH does not coincide with any of MATH. The normalization condition MATH is an immediate corollary of the asymptotic expansion of the prime form REF . REF for the monodromy matrices of function MATH follow from the simple consideration of the components of function MATH. Suppose for a moment that the function MATH, defined by REF , would be a single-valued function on MATH (as function of MATH). Then all monodromy matrices would be matrices of permutation: the analytical continuation of the matrix element MATH along contour MATH would simply give the matrix element MATH. However, since in fact the function MATH gains some non-trivial multipliers from crossing the basic cycles MATH, MATH and contour MATH, we get in REF an additional exponential factor. Its explicit form is a corollary of the definition of intersection indexes which enter this expression, and periodicity properties of the theta-function and the prime-form. |
math-ph/0106009 | REF can be rewritten in terms of the non-singular part of the NAME kernel REF , which immediately leads to REF taking into account the definition of projective connection MATH . |
math-ph/0106013 | Fix MATH and vary MATH. The product on the right hand side of REF defining MATH contains the MATH dependent sections MATH, MATH, MATH and MATH with the others constant as MATH varies. In CITE (and CITE for hyperbolic monopoles) it was shown using a bijection between nearby solutions that the assignment of MATH, etc, is continuously differentiable in MATH. Thus, the same is true of inner products involving the MATH dependent sections, such as MATH. |
math-ph/0106013 | We will prove only MATH since the proof of the limit of the MATH-point function is essentially the same. We define MATH for solutions of REF satisfying REF . If the connection is trivial and the NAME field is constant, MATH then MATH and MATH so MATH as required. As MATH, the connection and NAME field become more trivial and constant, respectively. More precisely, there exists a gauge in which MATH where MATH is constant and MATH as MATH. This follows from CITE. NAME 's theorem CITE uses a contraction mapping argument to show that solutions MATH on MATH and MATH on MATH of REF (using MATH and MATH respectively) are in one-to-one correspondence with solutions of REF . Moreover, the norm of the difference between corresponding solutions is controlled by the MATH norm of the perturbation term MATH. In other words, as MATH, the solutions MATH and MATH tend uniformly to the solutions of REF on MATH and MATH respectively, and in fact on any MATH and MATH. The inner product MATH can be calculated at any point MATH, in particular MATH so MATH uniformly. |
math-ph/0106013 | For MATH, where MATH, this follows simply from the definition. Taking limits and derivatives gives the result for general MATH. |
math-ph/0106013 | Since MATH it is sufficient to show along any geodesic that the solutions MATH of REF satisfy MATH, and in fact MATH so we will show that MATH. We have MATH where the last inequality uses the maximum principle MATH. Thus MATH . So the function MATH is strictly increasing, and by construction of MATH, MATH yielding the required inequality MATH . |
math-ph/0106013 | If MATH then MATH which contradicts REF . |
math-ph/0106013 | Consider REF-point function MATH where MATH is allowed to vary, MATH and MATH are fixed and different from MATH, and MATH, MATH are the solutions of REF along the geodesic running from MATH to MATH. We have MATH and this will be used to characterise MATH. By REF the NAME equation implies that MATH with MATH dependent coefficient, and MATH, since we are moving only one end of the geodesic. The limit MATH is independent of MATH so we can multiply MATH by a function depending on MATH and arrange that MATH, whilst preserving its normalisation at MATH. (We cannot do the same for MATH.) Thus MATH . If we differentiate REF then we get MATH . Let MATH. Then as shown in the proof of REF , MATH so MATH . Hence MATH for all MATH, so we get the relation MATH as required. |
math-ph/0106013 | CASE: We have MATH. This can be simplified to MATH by REF . CASE: In an open set of MATH choose solutions of REF normalised by REF so that MATH is independent of MATH and MATH is independent of MATH. (To achieve this choose a normalised solution of REF MATH for a fixed MATH and use MATH to define MATH for nearby MATH. Do the same for MATH around MATH.) Therefore, MATH and MATH, MATH for MATH independent of MATH and MATH independent of MATH, since we can calculate the coefficients in REF in the infinite limits. Then MATH is holomorphic away from it singular points: MATH . The meromorphicity of MATH at the singular points follows from REF where it is shown that MATH vanishes like MATH for a locally defined function MATH, anti-holomorphic in MATH and holomorphic in MATH, with the same zero set as MATH. Thus, ignoring the finite parts of MATH its singular part looks like MATH which has a pole in the variable MATH. CASE: For MATH defined in REF we can choose a local gauge in which MATH as follows. Choose MATH so that MATH (as in REF .) Now choose MATH so that MATH is real. This uniquely determines MATH up to a constant MATH gauge transformation given by the ambiguity in the phase of MATH. Then MATH . For MATH choose the solutions of REF normalised by REF along each family of geodesics, respectively MATH, MATH, MATH and MATH, so that MATH and MATH, and MATH and MATH (define MATH via MATH.) We can compare MATH and MATH by defining MATH so that MATH, then MATH . In particular the expression in REF is independent of MATH: MATH and real since it is the Laplacian of a real-valued function. |
math-ph/0106013 | This follows from the simple fact that MATH precisely when the solutions MATH of REF decay at both ends, which is the same condition for a geodesic to lie in the spectral curve. Notice that the invariance of MATH under the real structure MATH extends to REF-point function since MATH. |
math-ph/0106013 | Fix MATH and vary MATH. The NAME equation implies that the solution MATH of REF normalised by REF also satisfies MATH for MATH independent of MATH. In the limit, the section MATH gives a unitary gauge for the connection on the conformal boundary two-sphere, and hence MATH is the MATH component of MATH. Any other choice of MATH satisfying REF differs by MATH and hence MATH which is a change of the MATH gauge. In fact, without the normalisation REF , the MATH that arises gives the connection on the conformal boundary two-sphere which is Hermitian with respect to a Hermitian metric defined by MATH. As in the proof of REF we can choose MATH and MATH so that MATH and MATH is real. Then MATH so MATH gives the MATH part of MATH with respect to a well-defined MATH gauge (up to a constant gauge transformation) determined by the choice of MATH. Thus the first part of the proposition is proven. The curvature is given by MATH since MATH is real-valued, and MATH . This is independent of MATH, since it is a gauge invariant REF-form or we see it explicitly in REF . Thus we can take the limit MATH and since MATH the second term disappears to leave MATH . Since MATH then MATH thus MATH . |
math-ph/0106013 | In order to study the vanishing at MATH, we may ignore the normalisation REF of solutions MATH and MATH of REF since that simply involves multiplying the solutions by non-vanishing functions. Thus we may choose the solutions so that MATH and MATH. The inner product MATH is generically a transverse local section of the line bundle MATH so MATH vanishes like MATH and so too does MATH. |
math-ph/0106013 | The points MATH are determined by MATH and REF determines the behaviour of the singularities there. REF follow from REF . Any other REF-form with these properties must differ from MATH by MATH for a real-valued function MATH. By REF, MATH is a function defined outside the set of points MATH and by REF it is bounded and harmonic and hence constant. Thus MATH and the properties uniquely determine MATH. |
math-ph/0106013 | Suppose we have two monopoles MATH and MATH with respective algebras consisting of elements MATH and MATH. Fix MATH and vary MATH. The two monopoles have the same connection on the conformal boundary two-sphere precisely when MATH is harmonic in MATH, since the curvatures of the connections on the conformal boundary two-sphere must coincide. With respect to a local trivialisation of MATH in the neighbourhood of a point on MATH denote by MATH a section with zero set the spectral curve of MATH, and similarly MATH for MATH. (We are using the fact that a section of MATH over MATH extends to a section over MATH. This follows from considering the MATH-th formal neighbourhood of the diagonal MATH.) Then MATH since the left hand side of REF is well-defined everywhere, that is, we have canceled singularities, and for fixed MATH it is harmonic in MATH. Hence it is constant in MATH and when we evaluate at MATH we get the right hand side. Now fix MATH and take MATH of both sides of REF . The left hand side vanishes since REF is also harmonic in MATH by symmetry. Thus MATH is harmonic in MATH. If MATH is harmonic then it is the sum of a holomorphic and anti-holomorphic function since MATH is holomorphic for some (locally defined MATH) and MATH is anti-holomorphic. We can choose MATH to be real and positive on MATH so MATH and similarly for MATH. Thus MATH for MATH holomorphic and MATH anti-holomorphic. We conclude that MATH since the real analytic function MATH on MATH has a unique extension in a neighbourhood of MATH. But then MATH and MATH are both constant since MATH so the zero set of MATH cannot contain lines MATH or MATH. Thus, MATH is constant and hence MATH since they agree on MATH. |
math-ph/0106013 | We will use MATH to label a unit vector in the line MATH, and MATH its conjugate transpose, so MATH. Thus MATH is still ambiguous up to a phase, although MATH is well-defined. To show that MATH is smooth at MATH, choose a MATH so that MATH and choose a neighbourhood MATH of MATH so that MATH for MATH. Then fix a unit vector MATH and for each MATH choose a unit vector MATH so that MATH is real. Then by REF MATH is smooth in MATH so MATH is smooth in MATH. Thus the component MATH of MATH is smooth. This is true for almost all MATH so MATH is smooth on the linear span of the image of MATH. We may replace MATH by this linear span, since the representation annihilates the complement. Thus MATH is a smooth map. The holomorphicity of MATH is equivalent to the property MATH proven in REF . This can be seen by setting MATH. Then MATH and by acting on the left by any vector orthogonal to MATH we see that MATH for some function MATH, so MATH is holomorphic. (We use MATH and MATH to mean the same thing.) The degree of MATH is obtained by intersecting its image with a hyperplane. This corresponds to asking for the number of solutions MATH to MATH for a generic MATH, which is MATH, the charge of the monopole. Furthermore, the degree of MATH determines an upper bound for the dimension of the span of its image, thus MATH. The map MATH is one-to-one since the proof of REF shows not only that MATH in MATH but also that their images under the representation are unequal via MATH. |
math-ph/0106013 | This is simply a restatement of REF since the product of two projections is zero precisely when their images are orthogonal. The function MATH is quite different from the corresponding function MATH. In particular it is holomorphic, and hence can be represented by a polynomial. |
math-ph/0106013 | This is a simple result from linear algebra. For any two vectors MATH, MATH since MATH preserves REF for any MATH, so we may assume MATH, in which REF is easy. Put MATH for ease in reading the next set of formulae. MATH and the last expression defines the spectral curve by specialising the expression in CITE to the boundary value of the discrete NAME equations. |
math-ph/0106013 | The expressions MATH coincide since they both define holomorphic sections of MATH with the same zero set. Thus MATH for some MATH. |
math-ph/0106013 | In CITE it is proven that for each charge MATH monopole MATH there exists a holomorphic map MATH with two key properties. It determines and is determined by the spectral curve of MATH and satisfies the statement of REF , and it determines and is determined by the boundary value MATH of MATH. The curvature of MATH is obtained as the pull-back of the NAME form on MATH by MATH. As in the proof of REF , use MATH to label a unit vector in the line MATH, and MATH its conjugate transpose, so MATH is well-defined. We will prove that MATH is the image of MATH in a representation of MATH acting on MATH satisfying MATH. Since MATH for any MATH is obtained from derivatives and limits of such quantities, this is enough to show the representation satisfies REF . The functions MATH and MATH vanish to the same order on (an image under MATH of) the spectral curve of MATH and vanish nowhere else. Thus, MATH for a real valued nowhere vanishing function MATH. Fix MATH and choose MATH so that MATH for each MATH. Take the derivative of each side with respect to MATH so MATH since both MATH and MATH define MATH. Hence MATH so MATH is constant. It is identically MATH since MATH. Note that our assumption that MATH and MATH define the same gauge for MATH is unnecessary since if they differ by the gauge transformation MATH for a real-valued MATH, then we are left with MATH in which case MATH is harmonic and hence constant, thus MATH. The general case is proved analogously. Again since we know the vanishing behaviour of the respective functions, we have MATH for a nowhere vanishing MATH. Vary MATH and fix the other variables. Choose MATH so that MATH for each MATH. Then again MATH and MATH. Thus MATH is constant and it is REF on the diagonal MATH, so it is identically MATH. |
math-ph/0106023 | By REF, we have MATH so by REF MATH . Since MATH is positivity preserving and REF holds, MATH . So by REF again, MATH . Now MATH . Since MATH and MATH are uniformly in MATH and MATH bounded above and below, we see that REF is equivalent to REF. |
math-ph/0106023 | REF plus scaling implies that MATH which goes to zero if MATH since MATH. REF/REF imply MATH which implies MATH which goes to zero as MATH. |
math-ph/0106024 | To be a zero-energy state means that for every MATH, MATH . This necessitates MATH and MATH which in turn means MATH . Clearly, if there exist MATH and MATH such that MATH and MATH for all MATH, then MATH for all oriented bonds MATH, since MATH. Conversely if MATH for every oriented bond, then for any MATH not necessarily an oriented bond, consideration of a connecting sequence MATH leads us to the conclusion that MATH and MATH. Taking MATH and MATH for any MATH, we have have the same condition as before. It can easily be seen that MATH . So the lemma is proved once we observe that with MATH fixed, the map MATH is a strictly decreasing, continuous map from MATH onto MATH. |
math-ph/0106024 | We will first prove it for MATH. In this case, we are guided by the argument at the beginning of the section which proves that the ground states of the XXX model are symmetric tensors. The key is to realize the ground states of the XXZ model as symmetric tensors, as well, but with a different choice of action of the symmetric group. Let MATH be the group of permutations of MATH. The standard basis for MATH is the set of all MATH simple tensors of the form MATH where MATH is a sequence of MATH's and MATH's. Let MATH be the usual action defined by MATH . We define a weight MATH by MATH . We define a new basis MATH and a new action MATH by MATH . With respect to the original basis, MATH . Hence, if MATH and if MATH is the corresponding transpositon, then MATH where MATH acts on MATH by MATH with MATH and MATH. Note that, restricting attention to MATH and the representation of the two-element group MATH, the symmetric tensors are MATH and the antisymmetric tensor is MATH . Now we already know MATH from which it obviously follows that MATH . Thus a state minimizes the interaction MATH if and only if it is invariant under the action of MATH. Hence a state is frustration-free, that is, minimizes every interaction, if and only if it is invariant under every nearest-neighbor transposition. But it is well known that for a connected graph, MATH, the nearest-neighbor transpositions generate the entire symmetric group MATH. Hence, any state is frustration free iff it is invariant under the entire action of MATH of MATH. That is, the frustration free states exactly coincide with the tensors which are symmetric with repsect to MATH. We know a formula for the symmetric states using the basis MATH on which MATH has the standard action: MATH . This means that MATH is defined in terms of the usual basis by MATH . NAME by a constant MATH one obtains the formula MATH . This is the desired result for MATH. The result for higher spin systems follows by embedding any spin-MATH system with MATH sites into a spin-MATH spin system with MATH sites. Specifically, define MATH for be disjoint copies of MATH. Let MATH . For a single site MATH, there is a natural way to identify MATH with a subspace of MATH, namely as the symmetric tensors. We define an action of MATH on MATH, wherein each copy MATH acts by the usual permutation action on MATH. Let MATH be the projection onto the set of all vectors which are fixed by the entire action. Then by our previous consideration, this is precisely the projection onto MATH. Then we see that MATH . But also, the entire action of MATH commutes with MATH by the definitions of MATH. (All bonds are allowed between any MATH with MATH for all MATH and MATH. Therefore, permuting the indices MATH and MATH does not change the sum.) Hence, MATH also commutes with MATH, and we see that the subspace of ground states of MATH in MATH is exactly equal to MATH . But by its very definition as the symmetric tensors under the action MATH of MATH, one sees that the entire kernel of MATH is contained in the range of MATH. Hence the ground states of the spin-MATH model coincide with the ground states of MATH. Writing these in terms of the natural basis for the spin-MATH representation gives MATH and this last formula is exactly what we want. |
math-ph/0106024 | If MATH, then also MATH because MATH for any MATH. Since MATH, it is clear that there exist MATH, MATH with MATH. By REF , MATH . Using NAME, we estimate MATH and so MATH which proves the lemma. |
math-ph/0106024 | The proof is by contradiction. Thus we assume the existence of a sequence MATH satisfying the hypotheses of the proposition, and also such that MATH . We also assume, for convenience that each MATH is normalized. By taking an appropriate subsequence, we can replace the MATH in the formula above with a MATH, and we assume this is done. Our method of proof will be to show that under the hypotheses given, and if MATH in such a way that MATH, then we can construct a limit state MATH from the states MATH with the property that it is an infinite-volume ground state and also is orthogonal to every infinite-volume ground state, clearly a contradiction. But in order to prove this we must first show that MATH. If MATH is a sequence satisfying the hypotheses of the proposition, and also REF (with MATH replaced by MATH), then MATH. Let us suppose first that MATH independent of MATH. Then we can map MATH to a state MATH for every MATH. But MATH is a finite-dimensional NAME space, and the intersection of the range of MATH with MATH is exactly MATH by REF . So it is clear that MATH is strictly smaller than the identity operator on MATH. Thus the conditions of the proposition along with REF contradict REF . Now, in the general case, if MATH does not converge to MATH, then there is some MATH such that MATH infinitely often, and by taking the appropriate subsequence, we again have a contradiction. Therefore, it must be that MATH. By taking an appropriate subsequence, we may assume that MATH. We assume this is done. If MATH is a sequence satisfying the hypotheses of the proposition, and also REF (with MATH replaced by MATH), then MATH. We first assume MATH independent of MATH in order to prove a contradiction. The proof is similar to the previous lemma, and essentially follows from the fact that for a finite-dimensional vector space the spectral gap is always positive. But this time the finiteness comes from MATH, not MATH, and we need to demonstrate that the subspaces MATH actually converge to a single finite-dimensional space MATH. By the previous lemma, we know that MATH. By taking a unitary transformation, if necessary, we assume that MATH, with MATH. Then we can write MATH . We define two new vectors MATH where MATH is defined as the tensor product of MATH over MATH, and MATH . It is trivial to check that MATH converges in norm to MATH, as MATH. Indeed, from our definition MATH . From this it follows that MATH for each MATH such that MATH. Note that this is a finite set of multiindices MATH, specifically, the cardinality is MATH. Thus MATH which implies that MATH as MATH for every finite MATH (because the same is true for MATH by hypothesis). But also, MATH as MATH. So, since MATH, we have MATH . Since the subspace MATH is finite-dimensional, and all the MATH have norm REF, there is a limit point MATH of the sequence MATH. This vector satisfies MATH, MATH for every finite MATH, and MATH. But, since the intersection of MATH with MATH is MATH, the existence of such a vector MATH is impossible. Thus we have a contradicition. So, MATH does not equal any finite number infinitely often, and this implies MATH. With the hypotheses of the last lemma, MATH. Follows from the last lemma and simultaneous spin-flip/reflection symmetry of MATH. By choosing a subsequence, if necessary, we can assume MATH and MATH. Also, by taking an appropriate subsequence, we can assume that all MATH are equivalent modulo MATH, that is, that MATH for MATH and a number MATH independent of MATH. By taking a unitary transformation, we can assume MATH. Since MATH, it follows MATH; since MATH, it follows MATH. Let MATH be the vector MATH. Then every MATH can be written uniquely as MATH with MATH. (Recall that MATH is the sum of the parts of MATH, not the MATH-norm.) We observe that, for any fixed MATH with MATH, the sequence MATH is bounded-in-norm by REF. Thus we may choose a convergent subsequence MATH. By the NAME diagonal trick we can, in fact, choose a subsequence such that MATH converges for every MATH (since the set of MATH is countable). We assume this is done from the outset, so that MATH for all MATH. By NAME 's lemma MATH. But there is no guarantee at the outset that the opposite inequality holds, that is, that MATH. Next we will show that the coefficients MATH are small whenever any part of MATH is too large. This will allow the opposite inequality, and more. MATH . The proof of this fact is the most technical part of the paper. By hypothesis, MATH . For any finite MATH, and large enough MATH, MATH. In that case MATH . So for every finite MATH, MATH. Now, for any MATH, define MATH to be the sum of all those terms MATH for which MATH, and let MATH. We observe that MATH and MATH . But also MATH unless MATH, and if MATH, then MATH by NAME. Therefore, MATH where MATH . It is understood that we only take the supremum over those MATH for which MATH is defined. Thus MATH actually has an implicit dependence on MATH. But we claim that MATH has a bound, independent of MATH. MATH where MATH is a universal constant depending only on MATH (not on MATH, MATH, MATH or MATH). Suppose that MATH is a vector in MATH such that MATH is well-defined. Then for some MATH, MATH because MATH, and for some other MATH, MATH because MATH. We can estimate MATH upwards by MATH where MATH only symmetrizes in the two legs MATH, MATH. Now, obviously, MATH . So MATH . And MATH . The estimate of the right-hand-side of the last display is the type of calculation which may be carried out directly from REF . For details of these types of calculations see, for example, CITE. If MATH, then the first of the two factors above is bounded by a universal constant (depending only on MATH) times MATH, and the second factor is bounded by the same universal constant times MATH. Absorbing MATH, which is at most MATH, into the universal constant-squared, we have the result. Now, if we let MATH and if we let MATH and MATH, then by REF , we have MATH . Solving for MATH, we see that MATH lies between MATH, where MATH . If we let MATH stand for MATH, instead, then we can take MATH (because MATH). In this case, MATH, where MATH . Since MATH as MATH, we have MATH . But since MATH was arbitrary, we can take MATH, to obtain MATH . The reverse inequality is trivial, so the lemma is proved. Here is an important application of the previous lemma: MATH. For any MATH, we can find a MATH such that MATH . Since the set of all MATH in the sum is finite, we see that MATH as MATH. Thus MATH which implies MATH . Since MATH was arbitrary, the corollary follows. We now define a new sequence of vectors MATH as well as the vector MATH . Here, by MATH we mean the tensor product of all the MATH, MATH, where MATH is given by REF . It is trivial to check that for any fixed MATH, MATH as MATH. (It is a similar computation to that done in REF .) Thus, for any finite MATH, we have MATH where putting the superscript MATH means the same thing as before, namely truncating the terms to those involving only MATH with MATH. By the lemma MATH . Then following the argument in REF , MATH . By its definition, MATH is a ground state for MATH for every finite MATH, because each MATH is. Also, by hypothesis, for any MATH, MATH . Since MATH, it is true that MATH for any finite MATH. Then by REF , MATH is a ground state of MATH for every finite interval MATH. By REF , MATH is a ground state. Since it is pure it is MATH, MATH, some kink,MATH, or some antikink. The densely defined operator MATH distinguishes the different cases, and in particular all the infinite-volume kink states are eigenvectors for MATH, with eigenvalue equal to MATH. Similarly, each MATH is an eigenvector with eigenvalue equal to MATH. From this we see that MATH. But by hypothesis, MATH . Another easy calculation is the fact that MATH . So, this implies MATH, which is clearly a contradiction. Therefore the Proposition is proved. |
math-ph/0106024 | We will first prove that there is a nonzero gap above the infinite-volume kink ground states. The same will then hold for the infinite-volume antikink states by symmetry. The gap above the translation invariant states will be calculated exactly, using a different technique. Let MATH be the largest number such that MATH for all finite MATH, and all MATH. By REF , MATH. Then we claim MATH is a lower bound for the spectral gap above any of the kink states MATH. To prove this, it suffices to show that for any kink state MATH, and any local observable MATH, MATH . (This means that MATH is greater than MATH on its range in the GNS NAME space of all excitations of MATH, since the vectors MATH, MATH, are a core for MATH and all its powers.) We observe that since MATH is a limit of a sum of nearest-neighbor interactions, defining MATH, MATH. Thus, MATH and MATH . Now define MATH, and MATH. Then MATH restricted to MATH is a density matrix MATH, where each MATH and MATH. By the definition of MATH, MATH. So MATH . This is equivalent to REF . This proves the existence of a positive spectral gap above the kink and antikink states, although the value of MATH has not been calculated. For the translation-invariant ground states, the spectral gap can actually be calculated by standard techniques. First we obtain a lower bound. For this, suppose that MATH, where MATH is the all-up state in the NAME space MATH, and MATH. Then, (the boundary-field term is irrelevant since MATH is asymptotically MATH at MATH), MATH . Now MATH is diagonal in the basis MATH (where MATH with finite support), and the lowest eigenvalue, for any state other than MATH, is MATH, which occurs whenever MATH for some MATH. This shows that MATH. To obtain the reverse inequality, let MATH be the state MATH when MATH. We observe that MATH . Hence MATH . Let MATH. Then MATH . So, MATH which shows that MATH, as well. |
math-ph/0106024 | We prove the lemma for MATH, then the more general result holds by conjugating MATH by MATH (since MATH). Suppose that MATH, which is true for some MATH because MATH is local. Define MATH . Then MATH where MATH . What is most imporant is the value of MATH: MATH . Then MATH . Similarly, for MATH large enough MATH where MATH is a normalizing constant and MATH . Then defining MATH, MATH and MATH . The lemma will be proved if we show that MATH for each MATH. But MATH, is just the same as MATH, except that summands where some MATH is less than MATH are excluded. Then by MCT (or DCT since MATH is finite), the sums converge, and this proves the lemma. |
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