paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0105202 | Observe that for the homotopy classes MATH from REF , we have that MATH . This implies that the formal sum MATH . Thus, the proof follows from associativity as before. |
math/0105202 | Combine the refinements from REF with those of REF . |
math/0105202 | Let MATH be the intersection point MATH. It follows as in the proof of REF that there are no triangles MATH with MATH, MATH and MATH, and MATH. Hence, counting holomorphic triangles whose MATH-vertex is MATH, we obtain a map MATH. On the chain level, this map has the form MATH, where MATH where MATH is the intersection point on MATH closest to MATH, MATH is the unique small triangle (supported in the neighborhood of MATH and the support of the isotopies between MATH and MATH with non-negative multiplicities) and lower order is taken with respect to the energy filtration on MATH. Moreover, there is a relative MATH-grading on both complexes, given by the NAME index (where we take an ``in" domain for MATH). The map preserves this grading. Moreover, there are only finitely many generators in each degree. It follows then that the induced map is an isomorphism. |
math/0105202 | This follows from the fact that on the chain level, MATH has the form MATH . REF , together with the fact that MATH preserves the relative MATH grading. |
math/0105202 | Fix an equivalence class of intersection points between MATH, all of which are MATH-induced. According to REF, if we wind sufficiently many times along MATH and move the basepoint MATH sufficiently close to MATH, then MATH can be made arbitrarily large. By moving the basepoint to change the MATH structure, we have that the complexes MATH and MATH (respectively, MATH and MATH) are identical. Moreover, if MATH and MATH are sufficiently positive, then the map MATH is independent of the MATH structure. Choose a degree MATH sufficiently large that MATH and MATH, and note under these identifications, the map induced on homology MATH agrees with MATH. For fixed MATH and sufficiently large MATH, MATH on MATH vanishes. Since MATH is zero for all sufficiently large MATH, it follows from the long exact sequence induced from MATH that MATH is an isomorphism. It follows that the kernel of MATH in degree MATH is trivial. From this, it follows in turn that the kernel of MATH is trivial in all larger degrees. Since MATH decreases degree more than MATH, it is easy to see that the cokerenel of MATH in dimension MATH is trivial, as well. The lemma then follows. |
math/0105202 | Recall that the NAME homologies of a rational homology three-sphere have an absolute MATH grading, specified by MATH . From the exact sequence of REF , we have that MATH . The hypothesis in the sign guarantees that the degree shift occurs at MATH (using the absolute MATH grading on each group). It follows that MATH vanishes in all odd degrees. Indeed, since this is true when we take coefficients in MATH for all MATH; hence, MATH has no torsion in even degrees. Since MATH for all rational homology three-spheres, the result follows. |
math/0105202 | We adapt the proof of REF . We start with MATH representing the knot complement MATH, and then choose MATH to represent MATH and MATH to represent MATH: that is, MATH represents MATH and MATH represents MATH. There is an added feature now, since the divisibility MATH of MATH could be greater than one. It is still the case that for sufficiently large winding, all the intersection points are represented from MATH or MATH, and, as in REF , all homotopy classes of maps MATH with MATH admitting holomorphic representatives (connecting any two intersection points) satisfy that the property that MATH uses the central point MATH either once or zero times. Recall MATH is the map defined using those homotopy classes which meet MATH once. Now, there is a difference map MATH which is defined by MATH . There are corresponding splittings MATH labeled so that MATH if MATH and MATH, and MATH. and MATH. The proof of REF gives us that MATH (for MATH). Also, analogues of REF still hold: both MATH and MATH are isomorphisms. Now, the proposition easily follows as before. |
math/0105202 | The proof is obtained by modifying the above proof of REF , with minor modifications, which we outline presently. For the case where MATH, we adapt the proof of REF , thinking of MATH as MATH. In this case, REF is replaced by an isomorphism MATH (with the same proof). Next, we observe that rather than having MATH and MATH cancel, as in the proof of REF , we have that MATH. In fact, for some choice of orientation convention, we can arrange for MATH. The result then follows easily from the long exact sequence connecting MATH, MATH, and MATH observing that the map MATH injective, with cokernel MATH (with trivial action by MATH). The same modifications work to prove the general case (arbitrary MATH) as well. We now turn to the uniqueness assertion on the orientation system. For the various equivalence classes of orientation systems, it is always true that MATH as a MATH module. In fact, we saw (compare REF ) that as a MATH module, the isomorphism class of the chain complex MATH is independent of the choice of orientation system. Moreover, from REF , it is clear that the MATH different equivalence classes of coherent orientation system give rise to all MATH different MATH-module structures on MATH which correspond naturally to MATH, with a distinguished module for which the action by MATH is trivial. |
math/0105202 | First observe that the sign comparing MATH and MATH in REF is universal, depending on the relative sign between MATH and MATH. Checking these signs for MATH, the Proposition follows. |
math/0105202 | The short exact sequence MATH induced a long exact sequence in homology MATH which shows that MATH . Moreover, REF implies that MATH where MATH is the divisibility of MATH in MATH. The result now follows from the ``wall-crossing formula": MATH for NAME 's torsion (see CITE). |
math/0105202 | This follows in the same manner as the previous corollary, except that now MATH is a non-trivial vector space, so its exterior algebra has NAME characteristic zero: thus, MATH. |
math/0105202 | As before, we have a short exact sequence MATH and hence a long exact sequence: MATH . Note that we are using a relative MATH grading here, which we can do since MATH is torsion. When MATH is sufficiently large, the coboundary map MATH is zero, since on MATH, the map MATH is an injection. It follows that for all sufficiently large MATH, MATH . On the other hand, we still have a short exact sequence: MATH inducing MATH . Note that with the earlier grading conventions, MATH must decrease the grading by one. Of course, MATH is a finite-dimensional graded vector space, so the above gives the following relation for all sufficiently large MATH: MATH . But from REF applies in the present case, to identify MATH. Note that the proof of the that proposition does not really require that MATH be negative; it suffices to consider the case where MATH, MATH and MATH are negative, and MATH is torsion. Combining this result, REF , and REF , we obtain that: MATH . Suppose that MATH. Then, (according to REF ) for all sufficiently large MATH, MATH if MATH is even and MATH when MATH is odd. Similarly, when MATH, we have MATH . |
math/0105202 | The proof proceeds exactly as in the proof of REF (with the sign pinned down in REF , and REF in the case where the MATH structure is torsion), together with the observation that now MATH multiplies by MATH. |
math/0105202 | We still have the long exact sequence MATH . We place a reference point MATH at the intersection of MATH (the perturbing curve) with MATH. It is clear that MATH. Moreover, the coboundary splits as MATH, where MATH is is a primitive MATH root of unity, and MATH and MATH are the maps obtained from the MATH and MATH using MATH coefficients, by a base-change to MATH. In particular, both MATH and MATH are isomorphisms REF . Thus, in view of REF (indeed, we're using here the special cases from Subsection REF), we have exactness for MATH . |
math/0105204 | Because MATH is a MATH-deformation of MATH, we have MATH . Therefore all the equalities in REF should hold, and MATH . Now the corollary follows. |
math/0105204 | First note that when MATH goes to MATH, MATH goes to a maximal vector MATH of MATH in the classical case (see CITE). Therefore we know that MATH is a nonzero vector. Next observe that the weight of MATH is same as the weight of MATH, which is MATH by construction, where MATH. Now let's prove that MATH is annihilated by the action of MATH for MATH. Note MATH . CASE: First we consider the case MATH. In this case MATH maps MATH for MATH or MATH, because MATH. We fix a special standard tableau MATH corresponding to partition MATH which is the shape of MATH. We give an example below and do not bother to give the precise definition of MATH. If MATH then MATH . Note that the entries in MATH increase across rows from left to right for the first MATH rows, then entries increase down the columns. From REF with MATH, we have that MATH for some MATH. Note we know MATH is nonzero because the left side of REF is nonzero by specializing MATH. Combining REF we have MATH for some nonzero MATH. Now from REF MATH for some MATH. We will show that MATH is a linear sum of simple tensors which are killed by the actions of MATH for MATH. Write MATH and MATH. Then we may write the simple tensor MATH as MATH . Recall that MATH. Let's fix a MATH. Then MATH is a linear sum of simple tensors MATH where MATH are the same as MATH up to scalar multiplications except the orders of the entries MATH in MATH are changed. This is because if MATH moves entries in the MATH-th row of MATH, MATH, then the action of MATH on MATH does not create any new simple tensors because MATH maps MATH to MATH. And if MATH moves entries in the MATH-th row of MATH, MATH, then MATH is a linear sum of MATH where MATH are simple tensors same as MATH except the orders of entries in MATH are changed because MATH where the MATH's are basis vectors of MATH. Thus MATH maps REF to a linear sum of simple tensors MATH . Now note that MATH is the permutation which transforms MATH to MATH, and MATH maps the word MATH to the word MATH by place permutation. Moreover, the permutation MATH does not move entries from MATH. Thus the action of the NAME element MATH on REF produce a linear sum of simple tensors MATH where MATH are the same as MATH except only the orders of entries in MATH are changed, which are in fact same as MATH except only the orders of entries in MATH are changed. Thus we may also write MATH just like REF. MATH . Our next goal is to show MATH maps REF to a scalar multiple of MATH, where MATH is a simple tensor which is the same as MATH except the order on entries MATH's is different. Write the simple tensor MATH as MATH . Now note that MATH maps the word MATH to the word MATH by place permutation. We decompose MATH, so that MATH as in the followings ways: First we define a sequence MATH of standard tableux such that MATH is a standard tableau of shape MATH whose entries increase by one down the first MATH columns, and then other entries increase by one just like they are in the MATH for the rest of columns, that is, entries increase by one across rows for first MATH rows then entries increase by on down the columns. Then we define MATH. We give an example to explain our idea. If MATH then MATH, MATH, MATH, MATH, MATH, MATH, MATH, and MATH are MATH . And our decomposition MATH is illustrated in REF . Note the MATH's entries of our MATH and MATH are not nicely ordered as in NAME REF, but we don't need to consider those orders. Now we notice that the situation explained in REF does not happen in decomposition MATH, because once an entry moves to the left, then it is fixed by the following series of transformations MATH so that it goes straight down, and it does not produce any further crossings of edges. This is clear from REF . Thus MATH and so, MATH . Next we decompose each MATH into a product of transpositions. For example a decomposition for MATH in REF is explained in REF . If each MATH is expressed as a product MATH of transpositions as shown in REF , then the expression is reduced for the same reason as decomposition in REF is reduced. So NAME element MATH is also a product of MATH's which correspond to the transpositions in the decomposition. Note also that we only exchange MATH and MATH or MATH and MATH during the process of applying the above place permutations. Because MATH maps MATH to MATH for MATH and MATH to MATH, we have MATH . Thus the actions of MATH coming from the decomposition of MATH on the simple tensor MATH are the same as the place permutations except for scalar multiples. Thus we have MATH maps REF to MATH except for a scalar multiple of a power of MATH. Hence MATH is a linear sum of MATH . Now we show that MATH is killed by the action of MATH for MATH. The action of MATH on MATH commute with the actions of the NAME algebra, so that MATH. Note MATH is a linear sum of simple tensors MATH such that each MATH has a tensor factor where MATH has been changed to MATH or MATH has been changed to MATH. If the first case happens, then there is a MATH such that MATH (note MATH). So MATH . Hence we obtain MATH as expected. If the second case happens, then for some MATH, where MATH are in the same column of MATH, the vectors in the MATH'th and MATH'th tensor slots are both MATH. Note the vectors between the MATH'th and MATH'th tensor slots are from MATH. Consider NAME element MATH. Without loss of generality we assume MATH. Then MATH which explains that MATH . Note MATH and MATH if MATH, so that MATH . Therefore we have MATH again this case. Hence MATH. And we have shown here MATH for MATH. CASE: Second we consider the case MATH. This case is somewhat easier than the other case. First note MATH is a linear sum of simple tensors MATH such that one of MATH in tensor slots of MATH is changed to MATH. Then for some MATH, MATH because MATH. Thus MATH . Thus we have MATH and we have MATH for MATH as expected. |
math/0105205 | Let us first see how REF follows from REF . If MATH is noncompact, or if MATH is nonempty, then MATH is a free group, and therefore bi-orderable. Thus we are reduced to considering closed surfaces. According to the standard classification, such surfaces are either a connected sum of tori, or of projective planes in the nonorientable case. We shall first consider non-orientable surfaces. Of course MATH is certainly not left-orderable. For MATH, the NAME bottle, MATH is a well-known example of a group which is left-orderable (being an extension of MATH by MATH), but not bi-orderable, as the defining relation would lead to a contradiction. By REF , the surface MATH has bi-orderable fundamental group. We shall picture it as a torus with a small disk removed, and replaced by sewing in a NAME band. Consider a MATH-fold cover of the torus by itself, and modify the covering by replacing one disk downstairs, and MATH disks upstairs, by NAME bands. This gives a covering of MATH by the connected sum of a torus with MATH copies of MATH. Thus the fundamental group of MATH injects in that of MATH, and is therefore bi-ordered. We now turn to orientable closed surfaces. The cases of genus zero or one being easy, we consider a closed surface of genus MATH. This surface is the oriented double cover of MATH. Therefore its fundamental group is a subgroup of a bi-orderable group. This completes the proof of REF , assuming REF . |
math/0105205 | We shall proceed by induction on MATH. For MATH, we have that MATH is left-orderable by REF . Moreover, we have a short exact sequence MATH where MATH is induced by forgetting one of the strands. Now MATH is free and hence left-orderable, and MATH is left-orderable by induction; an application of REF completes the induction step. |
math/0105205 | Let MATH. Our strategy for constructing a bi-invariant order on MATH is to apply REF , where the normal subgroup MATH of MATH will be chosen so that MATH. To define the subgroup MATH, we note that MATH has presentation MATH where MATH and MATH represent free generators of the punctured torus in MATH and MATH the generator corresponding to the central curve of the NAME band in MATH. We define MATH, the normal subgroup generated by MATH. Note that a word in MATH belongs to MATH if and only its exponent sums in MATH and MATH are both zero. The covering MATH of MATH with MATH is very easy to imagine: consider the universal covering MATH, and modify MATH by taking a family of small disks (say MATH) centered at the integral points MATH. Remove each of these and replace by a NAME band MATH. This defines a covering MATH. The group of covering translations of MATH is just MATH, with MATH acting by translation MATH on the MATH part of MATH, and taking each MATH to MATH. Therefore we have (as required) an exact sequence MATH . We now turn to the task of proving that the orderability hypotheses of REF are satisfied. The group MATH can be bi-ordered, say lexicographically. (In fact, there are uncountably many different bi-orders on MATH; for example, there are already two for each line of irrational slope in MATH through MATH.) All that remains to be proven is that MATH has a bi-ordering which is invariant under conjugation by elements in MATH. We note that MATH is an infinitely-generated free group. There is a free basis for MATH consisting of the generators MATH represented by a loop that goes around the central curve of the NAME band MATH, connected by a tail to the basepoint in some (non-canonical) way; for definiteness we may take MATH as a free generating set for MATH. Now MATH acts upon MATH by conjugation, which may be described in terms of the generators as follows. Suppose MATH has exponent sums MATH and MATH in MATH and MATH, respectively. Then MATH where MATH. We just take MATH; by calculating the exponent sums of MATH and MATH in MATH we can verify that indeed MATH. For the following, MATH denotes the abelianization of MATH, which is an infinitely generated free abelian group, with generators, say MATH; the abelianization map MATH is just MATH. Any automorphism MATH of MATH induces a unique automorphism MATH of MATH. For example, in the above lemma, under abelianization the conjugation map is just the shift MATH. Now REF completes the proof of REF . |
math/0105205 | We use the NAME expansion CITE, sending MATH into the ring of formal power series in the infinitely many noncommuting variables MATH. Since there are infinitely many generators, some care must be taken in defining the ring MATH, which we take to be the ring consisting of formal power series in the MATH, but we consider only such series which involve just finitely many different variables. The NAME map MATH is given by MATH . Clearly the image of MATH lies in the group of units of the form MATH inside MATH, and it is an embedding of groups, by the same proof as in CITE. Elements of MATH may be written in standard form, arranged in ascending degree, and within a degree terms are arranged lexicographically by their subscripts (which in turn are ordered lexicographically). Then two series are compared according to the coefficient of the first term at which they differ (here is where the finiteness assumption is necessary). The proof that this defines a (multiplicative) bi-invariant ordering of MATH, is routine - compare CITE. Via the injection MATH, we may regard MATH as a subgroup, and hence it is also biordered. Finally, we argue that this ordering has the desired invariance property; equivalently, that MATH preserves the positive cone of MATH. Consider an automorphism MATH such that MATH is a shift MATH . This means that MATH, where MATH is in the commutator subgroup MATH . Since MATH maps into MATH under the NAME embedding, the effect of MATH is reflected in MATH by the substitution MATH . Therefore, if the NAME expansion of MATH is MATH, where MATH is the sum of all degree MATH terms, then MATH terms of higher degree. Therefore, the lowest degree nonzero terms of the NAME expansions of MATH and MATH are identical, except that the subscripts are shifted. Thus the ``first" nonconstant terms of both MATH and MATH have the same coefficient, and we conclude that MATH preserves the positive cone of MATH in the ordering we described. |
math/0105205 | The proof is virtually the same as for REF . We order the variables MATH by defining MATH to ``come before" MATH if MATH in the MATH-invariant bi-order of MATH. Then in order to define which of two words in MATH is the larger we compare their images under the NAME map MATH. Finally the invariance property is proved precisely as in REF . |
math/0105207 | It suffices to consider the case MATH and to prove the statement in a local chart. Take a function MATH such that MATH. Then MATH. Now let us change coordinates to MATH, where MATH are the coordinates along MATH from the regularity condition and MATH are arbitrary coordinates along MATH. We have MATH and MATH. Write the function MATH in the form MATH where MATH. Then MATH, hence MATH, and MATH, so that MATH. Thus, MATH. |
math/0105207 | Suppose that MATH is the compatibility operator for MATH. Linearizing the equality MATH, we get MATH. Hence, MATH. Conversely, consider an operator MATH satisfying the condition MATH. Since MATH, we see that MATH. Define the operator MATH by the equality MATH. We have MATH, so that MATH. This completes the proof. |
math/0105207 | Since MATH is a vector field and MATH, it is sufficient to evaluate MATH on functions linear along the fibers of the natural projection MATH. Such functions can be naturally identified with MATH-differential operators belonging to MATH, where MATH is a graded MATH-module. Define an odd MATH-differential operator MATH by the formula MATH where MATH. Trivially, MATH and MATH. It is easy to see that if MATH then MATH, so that MATH. |
math/0105207 | See CITE. |
math/0105207 | Pick up an element MATH. It can be thought of as a nonlinear operator MATH. Obviously, MATH, where MATH is the parity of MATH. Hence, MATH. |
math/0105207 | We prove the statement by identifying in a natural way elements MATH with graded symmetric multilinear MATH-differential operators MATH . It is not hard to check that MATH if and only if the corresponding operator MATH is selfadjoint: MATH (since MATH is symmetric, self-adjointness in one argument implies self-adjointness in the other arguments). Consider the projector MATH given by MATH. Obviously, MATH and MATH. Now, if a cycle MATH belongs to MATH then by REF we have MATH. This gives MATH, which is the desired conclusion. |
math/0105209 | What follows is the argument for the case when MATH separates MATH. The argument for the other case is analogous and is left to the reader. To start, remark that the given MATH structures MATH can be patched together over MATH with the specification of an isomorphism over MATH between the corresponding lifts, MATH. In this regard, note that the choice of a Riemannian metric on MATH which is a product flat metric on a tubular neighborhood, MATH, of MATH determines principal MATH reductions of the general linear frame bundles of MATH which are consistent with the inclusions of MATH in MATH. Having digested the preceding, note next that the space of isomorphisms between MATH and itself which cover the identity on MATH has a canonical identification with the space of maps from MATH to the circle; thus the space of isomorphisms MATH which cover the identity on MATH has a non-canonical identification with MATH. This implies that the set of homotopy classes of such maps is a principal bundle over a point for the the group MATH. In this regard, note that a pair of isomorphisms between MATH and MATH yield the same MATH structures over MATH if and only if they differ by a map to MATH which extends over either MATH or MATH. In any event, a choice of isomorphism from MATH to MATH covering MATH is canonically equivalent to a choice of isomorphism between the restrictions to MATH of the associated MATH line bundles MATH. Meanwhile, as MATH, the data MATH mapping to MATH canonically determines a homotopy class of isomorphisms from MATH to MATH. Likewise, MATH determines a homotopy class of isomorphisms from MATH to MATH. With the preceding understood, use the composition of an isomorphism MATH in the MATH determined class with the inverse of one between MATH to MATH from the MATH determined class to construct the required isomorphism between MATH and MATH. |
math/0105209 | The proof sticks closely to a well worn trail initially blazed by NAME in CITE and translated to the MATH - manifold context at the start of REF. In fact, the argument follows almost verbatim the discussion in the proof of CITE's REF . Here is a brief synopsis of the argument: First of all, in the case where MATH, use the NAME - NAME formula MATH on MATH, and the two NAME - NAME equations to conclude that MATH. Here, MATH denotes the covariant derivative on sections of MATH which is defined by MATH and the NAME - NAME connection on the MATH frame bundle. As MATH is covariantly constant, either MATH or MATH, or both. In particular, the first point in REF is only consistent with both vanishing. In the case where MATH, NAME multiplication on MATH by MATH defines a skew Hermitian, covariantly constant endomorphism of MATH. Decompose MATH into the eigenbundles for this endomorphism and write MATH with respect to this decomposition as MATH. In so doing, follow the steps in CITE or at the beginning of REF. Here, it may be useful to consider lifting the story to MATH with MATH viewed as MATH to make the connection with the MATH - dimensional framework in these references. In any event, with MATH written as MATH, the second equation in REF becomes a coupled system of equations for the pair MATH. Continuing the analysis in either CITE or CITE then leads directly to the conclusion that MATH and that when MATH is small, then MATH is constant and MATH. |
math/0105209 | This is a straightforward computation since the operators involved have constant coefficients. The details are left to the reader. |
math/0105209 | Write MATH in REF as MATH and then take MATH to infinity in the resulting equation to obtain bounds on MATH at points MATH. The latter are identical to those in REF after shifting MATH in REF to MATH. |
math/0105209 | The MATH bound on MATH by any MATH immediately yields a MATH bound on MATH since MATH with MATH a universal constant. Thus, since MATH is a constant, the first point in REF implies that there is a constant MATH and a bound of the form MATH with MATH depending only on MATH. The next step obtains bounds on the MATH norm of MATH, and the NAME - NAME formula for the NAME operator is the principle tool for doing so. Without assumptions on the Riemannian metric, the connection MATH and the section MATH, this formula reads MATH . Here, MATH denotes the metric's scalar curvature while MATH and MATH denote the formal, MATH adjoints of the NAME operator MATH and the covariant derivative MATH. Also, MATH denotes the NAME multiplication induced homomorphism from MATH into MATH. As indicated first by NAME in CITE, this formula in conjunction with the NAME - NAME equations can be used with great effect to analyze the behavior of solutions to REF. In particular, the left hand side of REF is zero when the second line in REF holds, while the first line can be used to control the term with MATH. In any event, for the purposes at hand, take the inner product of both sides of REF with MATH and use the equations in REF to rewrite the result as MATH . Here, MATH denotes the formal MATH adjoint of the exterior derivative MATH. REF implies that the square of the MATH norm of MATH over MATH is bounded by MATH, with MATH only dependent on MATH. Indeed, to obtain such a bound, first replace MATH by MATH in the first term on the left side. Then, multiply both sides of the resulting equation by a smooth, non-negative function which equals MATH on MATH and vanishes near the boundary of MATH. Next, integrate the result over MATH and then integrate by parts to remove the derivatives in MATH from MATH. Finally, an appeal to REF and a suitable application of the inequality MATH produces the asserted bound. With the MATH norms of MATH, MATH and MATH bounded in terms of MATH over the interior cylinder MATH, the next step proves that there is, given MATH, a value for MATH which implies that the MATH has MATH distance MATH or less from a pair MATH in the gauge orbit on MATH of MATH. This step is a straightforward argument by contradiction which invokes fairly standard elliptic techniques. In this regard, the only novelty is that the action of MATH must be used to make REF an elliptic system. Indeed, ellipticity can be achieved by using the MATH action to write MATH, where MATH are constrained so that the expression in the top line of REF holds. The details of all of this are left to the reader. Finally, given that MATH is MATH close to a point in the gauge orbit on MATH of MATH, the final step proves that MATH is MATH close to MATH. This last part of the proof is also left to the reader as it constitutes a direct application of standard procedures in elliptic regularity theory. |
math/0105209 | The lemma can be proved by invoking, with only minor changes, the argument which CITE uses to prove its REF . However, a slightly more direct argument can made by filling out the sketch that follows. The sketch starts with the remark that MATH provides a positive upper bound for the MATH norm over MATH of MATH that has the following significance: When the MATH norm of MATH is less than this bound, then MATH is gauge equivalent to MATH, where the MATH norms of MATH are bounded by MATH on the sub-cylinder MATH; and where the NAME - NAME equations in terms of MATH have the form MATH . Here, MATH is the tangent vector field to the line segment factor in MATH, and MATH is a linear, symmetric, first order differential operator. Meanwhile, MATH in REF is a `remainder' term which is formally second order in MATH. To be precise here, the MATH norm of MATH on each constant MATH slice of MATH is bounded by the product of MATH and the MATH norm of MATH on the same constant MATH slice. What follows are some important points about MATH. First, MATH is determined solely by the metric on MATH, MATH and MATH, and thus only by the metric and MATH. Second, the MATH spectrum of MATH is discrete, real, and lacks accumulation points. Third, MATH is not in the spectrum of MATH. This last conclusion is essentially the statement of REF and plays the starring role in the subsequent part of the argument. With MATH introduced, let MATH denote the projections of MATH onto the respective eigenspaces of MATH with positive REF and negative REF eigenvalues. Next, introduce functions MATH on the interval MATH whose values at a point MATH are the respective MATH norms of MATH on the corresponding constant MATH slice of MATH. Then REF yields the differential inequalities MATH . Here, MATH is the distance between MATH and the spectrum of MATH. With REF understood, a simple comparison argument establishes the following: When MATH, then the inequalities in REF require both MATH to decay exponentially from the ends of MATH. This exponential decay for the MATH norms of MATH on the constant MATH slices of MATH can then be bootstrapped to give REF using standard elliptic regularity techniques. |
math/0105209 | The argument here for the structure of MATH is completely analogous to that derived in CITE for the MATH self-dual moduli spaces on manifolds with cylindrical ends. In this regard, observe that the lower two components of the image of MATH in REF are nothing more than the linearization of the equations in the first two points of REF. Meanwhile, the vanishing of the first component of the image of MATH in REF only asserts that the given section of MATH is MATH - orthogonal to the space of tangents to the orbit of the gauge group. The fact that MATH is NAME with the MATH domain and MATH range follows from REF by standard arguments; for example, by invoking REF to control the behavior of MATH on MATH, the fact that MATH is NAME follows almost directly from results in CITE. The formula in REF for the index of MATH can be derived with the help of the excision properties of the index from the following input: First, the formula in REF holds when MATH is compact, see for example, CITE. Second, take MATH and MATH to be a constant, non-zero self-dual MATH - form. Then, take the solution MATH to be the pull-back via the projection to MATH of a solution in REF 's space MATH. Here, the form MATH in REF is the pull-back to MATH of MATH. In this case, the kernel and the cokernel of MATH are both trivial. (The operator has constant coefficients, so is straightforward to analyze.) |
math/0105209 | The argument here is, modulo some notational changes, almost identical to that which proves the analogous assertion in the case where MATH is compact; see, for example the books CITE or CITE. The largest modifications to the compact case argument are needed to address the orientation assertion, and in this regard, the reader can refer to the proof of REF. |
math/0105209 | As NAME pointed out CITE, the key to compactness theorems for the NAME - NAME moduli spaces is the NAME - NAME formula in REF. Of course, if MATH, then the left hand side of REF is zero; thus contracting both sides with MATH using the hermitian metric on MATH yields a differential inequality for the function MATH. Moreover, when the first point in REF also holds, then, as noted in CITE, the maximum principle applies to this differential inequality and provides a uniform upper bound for MATH in terms of MATH, MATH and an asymptotic bound for MATH on the ends of MATH. REF provides such an asymptotic bound for MATH, so MATH on the whole of MATH. Here, MATH depends only on the Riemannian metric. Note, by the way, that this constant MATH is independent of both the MATH structure MATH and MATH. This bound on MATH together with REF provide a uniform upper bound on the MATH norm of MATH. This upper bound is also independent of both MATH and MATH. This last bound on MATH provides a MATH bound on the anti-self dual part, MATH, of MATH's curvature MATH - form via the string of identities MATH . In particular, notice that the resulting upper bound for the MATH norm of MATH is independent of the class MATH. Standard elliptic regularity techniques can be invoked to bootstrap these upper bounds on MATH and MATH into uniform and MATH independent upper bounds for all MATH norms for a suitable point on the gauge orbit of MATH. (See CITE or CITE to see how this is done.) In particular the latter imply that any sequence of gauge orbits in MATH is defined by a corresponding sequence MATH of solutions to REF which converges in the MATH topology on compact subsets of MATH. Meanwhile, the uniform bounds that are provided by REF imply that a sequence of gauge orbits in MATH is actually defined by a sequence MATH of solutions to REF which converges in the strong MATH topology on the whole of MATH. This last fact implies convergence in each MATH and it implies that there can be only finitely many MATH for which the corresponding MATH is not empty. |
math/0105209 | The invariance assertions in the theorem are a standard consequence of the third point in REF . Indeed, the intervals in MATH pair each point in MATH either with another point in this space, but one with the opposite sign for the MATH count, or else with a point in MATH which has the same sign for the MATH count. Meanwhile, each point in the latter space which is not paired by a component of MATH to one in MATH is paired by a component of MATH with another such point, but one with the opposite MATH count sign. |
math/0105209 | Except for two assertions, the theorem follows directly from REF via arguments which are standard fair in gauge theories. (Arguments of this sort were first given by NAME in the context of MATH gauge theories, see for example, CITE.) The assertions which do not follow immediately from REF concern the vanishing of all but finitely many of the numbers MATH and MATH or MATH as the case may be. However, these last assertions follow from REF . |
math/0105209 | To obtain the proposition's first assertion, start with the NAME - NAME formula in REF, take the inner product of both sides of the latter with MATH, and then integrate the result over MATH. Integration by parts (which REF justifies) and input from the NAME - NAME equations can then be used to derive the equality MATH . Next, fix a closed MATH - form MATH which represents the class MATH and equals MATH on MATH. Because both MATH and MATH are supported on MATH, this last equation with REF implies the inequality MATH . Here, MATH depends only on the metric, MATH and MATH, while MATH is a positive, universal constant. With regard to REF, note that when MATH is self-dual, the integrand of the last term in REF can be replaced by MATH. The first integral on the right side of REF is equal to MATH. Meanwhile, REF relates the integral in REF of MATH to that of MATH. With these relations understood, the inequality in the first point of the proposition follows directly from REF. The second point's inequality follows from the first and REF. Meanwhile, the proposition's third point follows directly from REF given that the integrand in the last term on the right is replaced by MATH. |
math/0105209 | The first bound on MATH follows from the MATH bounds in REF using standard elliptic regularity techniques. To obtain the bound in REF, let MATH denote the distinct integer points where the bound in REF is violated for the cylinder MATH. REF guarantees that there are no more than MATH such points where MATH is given in REF . The estimate in REF then follows from REF . |
math/0105209 | First, note that MATH is bounded by some number MATH on MATH which depends only on the metric and MATH. To find the bound MATH, first use the NAME - NAME equations to rewrite MATH in REF in terms of MATH and so derive the following differential inequality for MATH: MATH . Here, MATH is a universal constant. The maximum principle applies to REF and bounds MATH in terms of MATH and its size near the ends of MATH. Meanwhile, REF bounds the size of MATH near the ends of MATH in terms of MATH. For the next step in the proof, take MATH, set MATH and then multiply both sides of REF by a standard, smooth function on MATH which is MATH on MATH and MATH on MATH and on MATH. Integrate the result over MATH, and then integrate by parts to remove the operator MATH from MATH. With the bound on MATH by MATH, a simple manipulation gives the inequality MATH here, MATH is another constant which depends only on MATH. The next task is to control the MATH norm of MATH. Here, the vanishing of MATH is employed to conclude that MATH. Now, note that by differentiating REF, the form MATH can be expressed in terms of MATH and MATH and as a result, its norm is bounded by a uniform multiple of MATH, where MATH is a universal constant. Thus, REF provides a MATH bound on MATH. With this bound in hand, it now proves useful to rewrite the equation MATH by separating out MATH - derivatives from derivatives along the tori MATH. There result two equations, MATH . Here, MATH is the exterior derivative along the torus, and MATH denotes the NAME star along the torus. Also, MATH is the MATH - dependent MATH - form on MATH which is obtained from MATH by contracting with the unit vector in the MATH - direction. To analyze the preceding equations, write MATH, where MATH and where MATH is a time dependent function on MATH obeying MATH. Letting MATH denote the MATH norm over MATH, it follows by standard arguments that there is a solution MATH which obeys MATH for all MATH. Next, consider the projection of the top line in REF onto the kernel of MATH. The result is an equation for MATH which reads MATH where MATH is the MATH - orthogonal projection of MATH onto the kernel of MATH. (Thus, MATH.) To analyze REF, consider the projections MATH, MATH and MATH onto the respective positive, negative, and zero eigenspaces of MATH acting on the kernel of MATH. Likewise, introduce the analogous projections of MATH, namely MATH. Then REF implies MATH . Here, MATH is the smallest non-zero absolute value of an eigenvalue of MATH on MATH. The first and second lines in REF can be integrated to yield MATH . The final line can be integrated to find that MATH . The explicit formula for MATH must be used to derive REF. For this purpose, introduce the MATH - dependent MATH - form MATH on MATH by writing MATH. Then, up to a sign, MATH is the time derivative of the projection of MATH onto the space of harmonic MATH - forms on MATH. These last two equations with REF, the equation in the first point of REF, the bound MATH and REF have the following consequence: The MATH norm of MATH is uniformly bounded on each MATH slice of the cylinder MATH in terms of MATH and the assumed MATH bound of MATH on MATH and on MATH. With the latter understood, standard elliptic regularity techniques find a uniform pointwise bound for MATH at points where MATH. Of course, the bound on the MATH norm on MATH slices provides one on all length MATH cylinders in MATH, and with this understood, the bound in REF follows from REF . |
math/0105209 | Use MATH to decompose the bundle MATH on MATH as MATH, where MATH is a complex line bundle. Note that MATH must be topologically trivial since the restriction of its first NAME class to MATH must vanish. Now consider MATH. Write MATH to correspond with the splitting of MATH. Then, NAME 's arguments from CITE for compact NAME manifolds (using REF to justify integration by parts) can be employed here to prove that MATH and that MATH is holomorphic with respect to that complex structure on MATH that is defined by the flat metric and the self-dual form MATH. In addition, MATH. Note that the maximum principle insures that MATH everywhere unless MATH. In this case, the solution is gauge equivalent to the constant solution on MATH. With this understood, assume below that MATH. To proceed, note that MATH so MATH. Differentiate this last identity to find an equation of the form MATH, where MATH are universal constants. A similar equation holds for MATH and comparing these two equations with the help of the maximum principle gives the pointwise bound MATH. This last inequality, with the condition MATH, implies that MATH which is equivalent to MATH. In fact, it follows now that MATH, where MATH is a constant, anti-self dual MATH - form with norm equal to that of MATH. Furthermore, MATH is proportional to MATH everywhere. (See REF.) It also follows that MATH is holomorphic with respect to the complex structure on MATH that is defined by the given flat metric and the self-dual form MATH. Let MATH. This form has square zero, and its kernel defines a MATH - dimensional distribution on MATH on which MATH and MATH both vanish. This distribution is also invariant under the complex structure on MATH because MATH has zero projection onto the corresponding MATH. Furthermore, since MATH is constant, the resulting foliation is the image in MATH of a linear foliation of the universal covering space MATH by complex lines. Moreover, as MATH has finite integral over MATH, the time coordinate (the MATH factor in MATH) is constant on each leaf of the foliation. Furthermore, as MATH is holomorphic, its zero set, a union of leaves of the foliation, is a smooth, compact, codimension MATH, complex submanifold of MATH. This implies that each leaf of the foliation is a closed, linear torus in MATH. The proposition follows directly from the preceding remarks. |
math/0105209 | If the assertion were false, then REF would find a solution on MATH with MATH identically. The latter is outlawed by REF . |
math/0105209 | If this assertion were false, REF would find a non-trivial, MATH - invariant solution on MATH which is not obtained from a vortex solution via a map which factors through the projection to MATH. This is impossible, for if MATH has a MATH - invariant orbit under MATH, then MATH must be a union of MATH orbits in MATH. |
math/0105209 | REF describes each MATH on the ends of MATH. In particular, for each component MATH and each such MATH, there is the corresponding set MATH that appears in REF. If there is no convergent subsequence, then REF requires at least one end MATH for which the sequence of sets MATH is not uniformly bounded. NAME so, a subsequence of MATH can be found for which the sets MATH for each fixed end MATH all have the same number of elements as MATH ranges over the subsequence. (Henceforth, all subsequences will be implicitly relabled by consecutive integers starting from MATH.) By passing again to a subsequence, these sets can be assumed to converge on compact domains in each end MATH. With this last point understood, then a limit, MATH, of a subsequence of MATH is obtained using relatively standard compactness arguments. The sequence MATH for a component MATH is obtained by translating along MATH to follow elements in MATH which do not stay bounded as MATH ranges through the sequence MATH. In this regard, REF together with the translation invariance of the equations in REF on MATH play the key role. Indeed, the construction of MATH begins by obtaining a finite set of centered, limit vortices for each end by translating each MATH on the end and then, after passing to a subsequence, one follows each `clump' of energy. Here, the centers of these `clumps' on the end MATH for the element MATH are, by definition, obtained by first partitioning the set MATH which appears in REF into subsets whose elements are much closer to each other than to the other subsets of the partition. Then, the center of the clump subset is declared to be the average of the MATH coordinates of the elements in the subset. After passing again to a subsequence, these `clump' partitions of the sets MATH can be assumed to produce the same number of subsets as MATH ranges through MATH and to be labeled for each such MATH so that the following is true: First, the distance between subsets with different labels diverges as the index MATH on MATH tends to infinity. Second, the subsets with the same label have a fixed number of elements as MATH ranges through MATH, and these elements can be themselves labeled so that the resulting ordered sets converge as the index MATH tends to infinity. One then translates the restriction of each MATH on an end MATH so that the average MATH - coordinate of a given, labeled clump subset of MATH is MATH. After passing to a subsequence, the resulting translated sequence of NAME - NAME solutions will then converge strongly in the MATH topology on compact domains in MATH to a solution to REF on MATH. This limit is equivalent to a vortex solution as described by REF . No generality is lost by translating this vortex in MATH so it is centered. By the way, the bound provided in REF on the size of MATH explains the bound in REF on the number of vortex solutions that arise. In any event, a component, MATH provides a well defined, finite set of centered vortex solutions, each corresponding to one of the labels of the clump partition just described. This labeled set of limit vortex solutions can be characterized by the corresponding data MATH, with each MATH for MATH. This characterization of the vortices is then used to construct, for each MATH, a vortex solution on MATH which is obtained by gluing with the map MATH the translated (via the MATH factor in MATH) versions of the vortices from the limit set. Here, the particular translation for a vortex depends on the particular MATH and the particular clump label. To be precise, the translation is chosen so that the center in MATH of the translated vortex agrees with the average MATH - coordinate of the corresponding clump subset in MATH. After the application of the gluing map MATH, the result is a vortex which gives a NAME - NAME solution on MATH that is close to MATH, where the latter differs substantially from the limit MATH and which is close to the trivial solution elsewhere. With the preceding understood, the convergence assertion in REF follows with standard elliptic regularity arguments for compact domains together with REF to predict the form of each MATH on those parts of MATH where the connection component of MATH has small curvature. The proof of REF ends with an explanation for the fact that the solution MATH sits in a moduli space defined by the original MATH structure MATH but with a class MATH with MATH. The explanation starts with the observation that the convergence behavior described by REF insures that the original MATH is also the MATH structure for MATH. Keep in mind here that each MATH gives solutions to REF on MATH whose MATH structure is sent to zero by the map MATH in REF. The explanation ends with the observation that the difference between the cup products of MATH and MATH with MATH is a consequence of the final line in REF when the convergence behavior in REF is noted. |
math/0105209 | These assertions are all derived using the NAME formula MATH or the corresponding formula for MATH. The latter switches MATH with MATH everywhere and adds an extra term which is proportional to MATH. As this term is small at large distances from MATH (see REF), the formula for MATH implies that MATH is NAME on the MATH completion of its domain as a map into the MATH completion of its range. With this understood, REF implies that the cokernel of MATH (which is the kernel of MATH) is empty. Furthermore, REF plus the last point in REF implies the estimate in the third point of the lemma. The latter estimate plus the NAME alternative implies the estimate in the fourth point. In this regard, remark that when MATH is MATH - orthogonal to the kernel of MATH, then MATH for some element MATH. Thus, MATH . The fact that the kernel dimension is MATH can be proved as follows: Differentiate the map which associates to MATH a vortex MATH. (See REF.) Each such directional derivative gives an independent element in MATH. Conversely, every element in MATH can be integrated to give an element in the tangent space to MATH. This follows from the vanishing of the kernel of MATH. For the exponential decay estimate, consider first the MATH case. If MATH, then REF and the NAME formula for MATH imply the following: Where the distance from MATH is greater than MATH, the norm of MATH obeys MATH where MATH is a function on MATH which obeys MATH with MATH a universal constant. The control of MATH where the distance to MATH is less than MATH comes via standard elliptic regularity which finds a constant MATH such that MATH at all points. Here, MATH depends only on MATH. With the preceding understood, let MATH denote the time coordinates of the MATH points in MATH. An application of the comparison principle to REF now yields the following: Fix MATH and there exists MATH which is independent of MATH and such that MATH . To obtain the analog of REF for the case where MATH, introduce the NAME 's function MATH for the operator MATH with a pole at MATH. An application of the comparison principle to the operator MATH finds a constant MATH which makes the following assertion true: If the MATH coordinate of MATH is MATH, that of MATH is MATH, and also MATH, then MATH . Furthermore, the absolute values of the derivatives of MATH at such MATH enjoy similar upper bounds. With the NAME 's function in hand, multiply both sides of REF by MATH and integrate both sides of the result over the region, MATH, where the distance to the set MATH is at least one. This operation produces an integral inequality which can be further manipulated to yield the following bound: MATH . Here, MATH is a universal constant, while MATH depends only on MATH. To explain, the first term in REF is due to the integration by parts boundary term that arises when the operator MATH in REF is moved from MATH to MATH. Meanwhile, the second term in REF comes from the integral of MATH. In this regard, REF is used to bound MATH, REF with MATH very close to MATH is used to bound MATH, and REF is used to bound MATH. The MATH version of REF follows directly from REF since the second term in this equation is not greater than MATH, with MATH depending only on the parameter MATH. The proof for the asserted bounds on the higher derivatives of MATH is obtained by first differentiating the equation MATH say, MATH times, and then writing the latter as an equation of the form MATH. The preceding argument for the MATH bound can be copied to obtain the desired estimates. |
math/0105209 | Let MATH. Then, by virtue of the vortex equations in REF, this function obeys MATH thus MATH everywhere. The first consequence of REF comes via the maximum principle, this being the previously mentioned fact that MATH as long as MATH. This last point can be parlayed to give a universal lower bound for MATH at points at fixed distance from MATH. In particular, the following is true: MATH . Accept REF for the moment to see its application first. In this regard, REF is applied here to supply a uniform, positive lower bound, MATH, for MATH on the constant MATH circles where MATH for at least one MATH. With such a bound in place, reapply the maximum principle to REF but use MATH and restrict to points in MATH where MATH for all MATH. The resulting conclusion (that MATH) and REF together give REF. To justify REF, consider the ramifications were the claim false. In particular, there would exist a sequence MATH, and a corresponding sequence of pairs of points MATH, where MATH, MATH and MATH. After translating each MATH appropriately, all of the points MATH can be taken to be a fixed point MATH. Meanwhile, the sequence MATH, has a convergent subsequence with limit MATH whose distance is MATH or less from MATH. Also, a subsequence MATH converges strongly in the MATH topology on compact domains in MATH to a vortex solution, MATH, although the latter may lie in some MATH for MATH. Indeed, the MATH convergence of MATH and MATH convergence of MATH follows directly from REF by appeal to the NAME - NAME theorem; convergence in MATH can then be deduced by differentiating the vortex equations. In particular, the MATH convergence here implies that MATH and MATH. However, these two conclusions are not compatible; as argued previously, MATH everywhere if MATH is less than MATH at any point. |
math/0105209 | To consider the assertion about the kernel of MATH, write MATH in MATH - component form with respect to the splitting in REF as MATH, where each of MATH and MATH also have two components. Then, MATH . Integration by parts along the fibers of MATH equates MATH, and MATH. Now, write MATH, where MATH is constant along each fiber of MATH, and where MATH is MATH - orthogonal to the constants along each fiber of MATH. It then follows that MATH . The assertions that concern the kernel of MATH follow from REF using REF . An analogous argument proves the assertions in the lemma that concern the kernel of MATH. |
math/0105209 | Let MATH denote the MATH - orthogonal projection (on MATH) onto the span of the eigenvectors of MATH with eigenvalue MATH. Meanwhile, let MATH denote the corresponding projection onto those eigenvectors with eigenvalue MATH. Let MATH denote the MATH norm of the time MATH version of MATH, and likewise define MATH. Then MATH obey a pair of coupled differential inequalities of the form MATH . Here, MATH is the second smallest positive eigenvalue of MATH. This last equation can be integrated (after some algebraic manipulations) to obtain the bound MATH where MATH and MATH depends only on MATH. Meanwhile, write MATH and consider the equations for MATH and for the corresponding MATH. In particular, these equations have the form MATH where MATH and MATH with MATH. Integrating these last two equations gives MATH where MATH with MATH and MATH depending only on MATH. |
math/0105209 | The assertions of REF about MATH follow directly from REF . Meanwhile, the assertions about MATH follow from REF after changing MATH to MATH in REF. Then, given these assertions, the bounds in REF follow from REF . |
math/0105209 | According to REF , there exists a pair MATH on the gauge orbit of MATH which obeys MATH at all points MATH where MATH. Thus, on some smaller length cylinder, this pair differs very little from MATH and so can be analyzed by treating REF as a perturbation of a linear equation. Moreover, as explained below, there is a fiducial choice of such a gauge equivalent pair MATH, where MATH obeys REF and also MATH on a subcylinder of the form MATH, where MATH depends only on the metric and MATH. With the preceding understood, view MATH as a section of REF. By virtue of REF, this section obeys an equation of the form in REF, where MATH is a linear function of the components of MATH. In particular, REF is applicable here with MATH replaced by MATH because the condition in REF insures that MATH obeys the requisite bounds on MATH. Thus, REF can be seen to follow from REF. The refined derivative bounds for MATH in REF are obtained from the MATH bounds via standard elliptic techniques via REF. In particular, to obtain the MATH bounds, simply differentiate REF and, remembering that MATH is a linear functional of the components of MATH, observe that the result has the same schematic form as REF. Thus, a second appeal to REF provides the MATH bounds on MATH. The derivation of the MATH bound is only slightly more complicated. It remains now to justify the asserted existence of the point MATH on the gauge orbit of MATH for which both REF hold. For this purpose, use REF to conclude that MATH is gauge equivalent to MATH, where MATH obeys MATH at all points MATH where MATH. Here, MATH is a constant which is independent of MATH. The pair MATH is guaranteed to come from the same gauge orbit as MATH if MATH and MATH are related via the identity MATH where MATH is a smooth function on MATH. Thus, the goal is to find MATH in REF so that REF holds on an appropriate subcylinder. In this regard, REF should be viewed as an equation for the function MATH. In particular, if MATH has a suitably small MATH norm, then this equation has the schematic form MATH where MATH with MATH a constant which is independent of MATH and MATH. The analysis of REF is facilitated by the following observation: Because of REF, the pair MATH is uniformly small on uniformly large subcylinders of MATH. That is, given MATH, there exists MATH which is independent of MATH and MATH such that at all points MATH, all derivatives from orders MATH through MATH of MATH are bounded in size by MATH. Meanwhile, the NAME 's function, MATH, for MATH defines a bounded operator from MATH to MATH and satisfies the pointwise bound in REF. The preceding observations suggest a contraction mapping construction of a solution MATH to REF on a uniformly large subcylinder of MATH. For this purpose, fix MATH and introduce a smooth, non-negative function MATH on MATH which equals MATH on MATH, vanishes where MATH and has first and second derivatives bounded by MATH. Then, consider the map from MATH to itself which sends MATH to MATH . Here, the fact that MATH defines a self map on MATH is insured by the right hand inequality in REF. Moreover, REF imply the following: There exist constants MATH and MATH which are independent of MATH and MATH and such that MATH is a contraction mapping on the radius MATH ball in MATH. For such MATH, the map MATH has a unique fixed point, MATH, in this ball. Of course, REF insures that this MATH decays to zero exponentially fast as MATH on MATH. Moreover, REF in conjunction with REF can be used to prove that MATH and its derivatives obey the required norm bounds throughout MATH. These last derivations are straightforward and omitted. |
math/0105209 | Use REF to find MATH and a gauge for MATH on the cylinder MATH of the form MATH, where MATH come from REF and obey REF, while MATH is the pull-back from MATH of a pair that defines MATH. Now, choose a smooth, non-negative function MATH on MATH which equals MATH where MATH and MATH where MATH. Use MATH and the function MATH to define a configuration MATH on MATH as follows: MATH . Note that MATH vanishes except when the coordinate MATH. Moreover, when MATH, then REF guarantees that MATH where MATH is independent of MATH; it depends only on the constants MATH and MATH. Thus, for large MATH, the pair MATH is very close to solving REF on MATH. The following lemma makes this notion precise: Under the assumptions of REF , there exists MATH, MATH and, given MATH, there exists MATH and these have the following properties: First, an upper bound for MATH, the numbers MATH, and a lower bound for MATH depend only on MATH and MATH. Second, when MATH, then the pair MATH has MATH distance less than MATH from gauge orbits of solutions to REF on MATH that come from vortex solutions with vortex number MATH. Third, such an orbit contains a unique pair MATH for which MATH obeys MATH . Here, MATH depends only on the vortex number MATH; in particular, it is independent of MATH, MATH, and the original pair MATH. The proof of REF is given below. REF enters the proof of REF in the following way: Fix some positive MATH and then MATH as in REF . An upper bound for MATH is derived in the subsequent arguments. In any event, suppose that MATH. Let MATH denote the open set of elements which provide gauge orbits of solutions to REF on MATH that have MATH distance less than MATH from the pair MATH. Given a vortex solution from MATH, let MATH denote the corresponding solution to REF that is provided by REF . Introduce the resulting MATH and observe that MATH obeys an equation of the form MATH . Here, MATH denotes the operator MATH from REF as defined using MATH, MATH is a universal, quadratic, fiber preserving map from REF to REF and MATH is interpreted as a section of REF. With MATH and MATH understood, consider: Under the assumptions of REF , there exists MATH which depends only on MATH and MATH and which has the following significance: If MATH, then there exists MATH as described in the preceding paragraph for which the corresponding MATH is MATH - orthogonal to the kernel of the operator MATH. The proof of this lemma is also given below. Given the statement of REF , the proof of REF continues with the following observation: There exists MATH and MATH which depend only on MATH and MATH and are such that when MATH, then the MATH norm of REF 's section MATH satisfies MATH . Indeed, the existence of such a pair MATH is guaranteed by REF and the third point in REF. As is explained below, this upper bound on the MATH norm of MATH implies the pointwise bounds: MATH . Here, MATH depends only on MATH and MATH; in particular, it is independent of MATH. Indeed, the first line in REF follows from REF using standard elliptic estimates, while the second follows from the first after an appeal to REF . Notice that the first point of REF implies the first point in REF. The second point in REF is obtained as follows: Start with the second line of REF and so conclude that MATH . Here, MATH is independent of MATH and is determined solely by MATH and MATH. This last bound is used to derive an upper bound for the MATH norm of the section MATH of REF that is defined by the rule CASE: MATH where MATH, CASE: MATH where MATH. Note that a suitable upper bound for the MATH norm of MATH will yield, with REF, the second point in REF. To obtain such a bound, use REF to conclude that MATH obeys an equation of the form MATH where MATH except when MATH. Moreover, where MATH is not zero, it obeys the bound MATH by virtue of REF; here MATH is independent of MATH as it is determined solely by MATH and MATH. With REF understood, write MATH, where MATH is the MATH - orthogonal projection of MATH onto the kernel of MATH. In this regard, MATH enjoys the following upper bound: MATH where MATH is independent of MATH, being determined solely by MATH and MATH. Indeed, remember that MATH is orthogonal to the kernel of MATH and so the projection of MATH onto this kernel is the same as that of MATH. In particular, the size of the latter projection obeys REF for the following reasons: First, MATH enjoys the bound in REF. Second, REF guarantees that any MATH is bounded by MATH at the points where MATH. Finally, any MATH obeys MATH at all points. Here, again, MATH is independent of MATH; it depends only on the vortex number MATH and hence only on MATH and MATH. Given REF, the previously mentioned bound on MATH, and the fact that REF can be written as MATH, another appeal to REF finds MATH and MATH which are independent of MATH, depend only on MATH and MATH and are such that when MATH, then MATH. This last bound, REF directly yield the final point in REF. |
math/0105209 | All of the arguments for REF closely follow arguments previously given, and so, except for the outline that follows, the details are left to the reader. The outline starts with the observation that a slightly modified version of the argument for REF establishes the existence of an upper bound for MATH and MATH such that when MATH, then MATH has MATH distance MATH or less from a solution to REF on MATH. The bound on the vortex number comes from the MATH constraint in the statement of REF . Given that MATH is close to a solution to REF, then the points in REF are proved by arguments which are essentially the same as those used above to prove REF . Note that these may require an increase of the lower bound for MATH. The arguments for the lemma's uniqueness assertion are basically those used to prove the slice theorem for the action of the gauge group on the space of solutions to REF. |
math/0105209 | As explained previously, REF provides the nonempty, open set MATH of elements that determine solutions to REF on MATH whose MATH distance is less than MATH from MATH. The assignment to a point in MATH of the corresponding MATH defines a map from MATH into the space of MATH sections of REF. This map is smooth; the proof is straightforward so its details are left to the reader. The assignment to a point in MATH of the square of the MATH norm of the corresponding MATH then defines a smooth function, MATH, on MATH. As is explained momentarily, the differential of MATH vanishes at precisely the points where MATH is MATH - orthogonal to the kernel of MATH. Indeed, let MATH denote a tangent vector to MATH at the point that corresponds to MATH. Then, the differential of MATH in the direction defined by MATH has the form MATH, where MATH and MATH is tangent to the orbit through MATH of the gauge group MATH. This is a consequence of the fact that MATH is fixed and only MATH is moved by MATH. Meanwhile, MATH is MATH - orthogonal to MATH by virtue of the first point in REF and so the differential of MATH vanishes in the direction of MATH if and only if MATH is MATH - orthogonal to MATH. As REF guarantees that the kernel of MATH is the span of the MATH, so MATH is orthogonal to MATH if and only if MATH comes from a critical point of MATH. With the preceding understood, it remains only to demonstrate that MATH has critical points. For this purpose, a return to REF is in order. In particular, with REF and in conjunction with standard elliptic regularity arguments, REF leads to the following conclusion: There exist constants MATH and MATH which are independent of MATH and MATH and such that if MATH, then the MATH norm of MATH is bounded by MATH. Since the MATH norm of MATH is bounded by its MATH norm, this last point and the second point in REF guarantee that MATH is proper when MATH is small and MATH is large. NAME precisely, MATH and MATH must exist such that when MATH and MATH, then the map MATH is a proper map on the set MATH. As a proper function has at least one critical point so there exists at least one MATH where the corresponding MATH and MATH are orthogonal. |
math/0105209 | The proof that MATH is smooth is straightforward and will be omitted. The subsequent discussion addresses the question of whether MATH has critical points. For this purpose, it proves convenient to identify the tangent space to a given stratum MATH of MATH at a point of interest, MATH, with a NAME space of smooth, square integrable pairs MATH that obey the equation MATH. Here, MATH is an imaginary valued section of MATH, MATH is a section of MATH, and MATH is an imaginary valued section of MATH with compact support on MATH. It is a consequence of REF that the restriction of MATH to each component of MATH has the form MATH where MATH are complex numbers and MATH with MATH a universal constant which is greater than MATH. With this last equation understood, then the statement of REF follows with the verification that any value for the complex number MATH can be obtained by a suitable choice of pairs MATH which obey MATH. This verification is the next order of business. REF constant MATH in REF is given by MATH . With REF understood, let MATH be a non-zero complex number and let MATH denote the section of MATH which is zero except where MATH on the given component of MATH in which case MATH . Thus, MATH where MATH denotes the MATH inner product over MATH and MATH denotes the real part. This section MATH is introduced for reasons which should be clear momentarily. Now, as MATH, the section MATH can also be written as MATH where MATH holds and MATH maps the MATH - orthogonal complement in MATH of MATH to the MATH - orthogonal complement of MATH in MATH. With the help of this decomposition, REF becomes MATH . To proceed, introduce MATH to denote the MATH - orthogonal projection of MATH onto the kernel of MATH and then introduce MATH. REF can be rewritten with the help of MATH as MATH . In the meantime, it follows from REF (after changing MATH to MATH) that there exists MATH which is independent of MATH and such that MATH on MATH. Moreover, there exists MATH which is independent of MATH and such that MATH where MATH on the given component of MATH. In this regard, note that REF insures that MATH enjoys a MATH and MATH independent bound on MATH when MATH. Now, there are two possibilities to consider. The first is that there exists a complex number MATH such that MATH . The second possibility is that there is an unbounded subsequence of values for MATH such that the corresponding sequence MATH has a non-zero limit for each unit length MATH. Now, in this second case, it follows from REF that all complex numbers can be realized by MATH in REF by tangent vectors at MATH of the form MATH with MATH. With the preceding understood, suppose that REF holds for some unit length complex number MATH. Then, as MATH except where MATH on the given component of MATH, the sequence MATH converges as MATH to a non-zero section, MATH, of the bundle MATH which obeys MATH . (As MATH has compact support, the integral in REF is well defined.) REF to the preceding discussion, if MATH is a critical point of MATH, then there exists such unit length MATH such that MATH for all sections MATH of MATH which have compact support on MATH. To make use of REF, it is also important to realize that MATH is also MATH - orthogonal to all compactly supported sections of the MATH summand in MATH. To see that such is the case, let MATH denote a compactly supported section of the summand in question. Given MATH, there exists a unique, MATH function MATH on MATH which obeys MATH. With MATH in hand, then MATH where MATH. Thus, MATH . To see that the left hand side of REF vanishes in the limit as MATH, note that MATH lies in the last two summands of REF. Meanwhile, REF implies that the projection of MATH into these summands is MATH where MATH is independent of MATH. Given this last bound, the vanishing as MATH tends to infinity of the right hand side of REF follows directly from REF. To summarize: If MATH is a critical point of the map MATH, then there exists a non-zero section, MATH, over MATH of MATH with the following properties: MATH . Now, the first point here implies that MATH restricts to MATH as a section of MATH only. With this understood, then the projection of the equation in the second point of REF onto the MATH summand of MATH asserts that MATH for all sections MATH of MATH with compact support on MATH. This last condition can hold only if MATH is identically zero on MATH. Having established that MATH vanishes identically on MATH, it then follows that MATH on the whole of MATH since there is a version of NAME 's unique continuation principle CITE which holds for elements in the kernel of MATH. The preceding conclusion establishes that MATH has no critical points as claimed. |
math/0105223 | Let MATH be a form in MATH. Here MATH stands for NAME forms on MATH. Notice that MATH. Hence MATH . The first term in the right-hand side belongs to MATH and the second term in the right-hand side belongs to MATH. It follows that the induced map MATH is given by the formula MATH . |
math/0105223 | Let a coboundary MATH be a covariant MATH-Lagrangian. Then it must have weight MATH, because it is linear in variables MATH. Let us show that MATH is of first order. Consider the identity REF for the case when index MATH is equal to MATH. The variational derivative MATH does not depend on variables MATH if at least one of indices MATH is equal to MATH and MATH. Hence it follows from REF that MATH does not depend on variables MATH if MATH, thus, the Lagrangian MATH is of first order. |
math/0105224 | As the second equation clearly implies the first one, we only have to prove the second statement. First let us suppose that the braid MATH is positive, that is, it can be written as a product MATH of generators MATH for certain indices MATH. Then MATH. Let MATH denote the closure of MATH. Since a positive braid is in particular strongly quasipositive, the slice NAME characteristic MATH of MATH equals MATH by CITE. The link MATH can be obtained as the transverse intersection of a complex plane curve MATH and REF - sphere MATH CITE. As a slight perturbation of the coefficients of MATH does not change the link type of MATH, we can assume that the projective closure MATH of MATH is a smooth algebraic curve. Note that MATH by CITE. Let MATH. Observe that MATH is still connected. Now suppose that MATH is a proper immersion of a surface having no closed component with boundary MATH which has MATH positive double points. Blowing up the negative double points of MATH and replacing the positive double points by handles, we obtain a nullhomologous embedded surface MATH, where MATH is the number of negative double points of the immersion, and MATH, see for instance CITE for a detailed description of the construction. Now consider the union MATH. The NAME characteristic of MATH is given by MATH . Now MATH is a connected smoothly embedded surface representing the same homology class as the algebraic curve MATH (with respect to a suitable structure as a rational surface on MATH), hence MATH by CITE or CITE. Therefore MATH and the claim follows. As to the general case, we can write MATH as a product of generators and their inverse elements, that is, MATH with indices MATH and signs MATH. Let MATH denote the number of positive exponents and let MATH denote the number of negative exponents. Then MATH. Consider the positive braid MATH. Clearly MATH. The braid MATH can be obtained from MATH by MATH positive crossing changes, in particular MATH and MATH have the same number of components. As observed in CITE, there exists a union of immersed annuli in MATH connecting MATH and MATH which has exactly MATH self - intersection points, all of them being positive. Now suppose that we are given an immersion MATH as above. Gluing this immersed surface with the annuli connecting MATH and MATH yields an immersed surface MATH with boundary MATH which has MATH positive double points, note that of course MATH. By the first part of the proof we can conclude that MATH . As MATH, this proves our claim. |
math/0105224 | If all the crossings of the diagram are negative, inspection of the NAME surface obtained by NAME 's algorithm from MATH shows that the right hand side of REF is a non - positive number, so we can assume that there is at least one positive crossing. The proof is a slight modification of an argument used in CITE to prove a similar inequality for the slice genus. Let MATH denote the number of positive respectively negative crossings of MATH. Consider the surface MATH which is obtained by gluing disks corresponding to those NAME circles which are not strongly negative with bands corresponding to all the positive crossings of MATH. Then MATH is a quasipositive surface CITE, and its boundary MATH is a strongly quasipositive link, having a positive diagram MATH with MATH . NAME circles and MATH crossings. By CITE, this implies that MATH . Now suppose that MATH is a proper immersion of a connected surface of genus zero with boundary MATH and MATH positive double points. Then MATH. Let MATH denote the NAME surface for MATH constructed by applying NAME 's algorithm to MATH. Then MATH is obtained from MATH by adding additional disks and bands, in particular MATH. By gluing MATH and MATH along MATH and removing MATH, we therefore obtain a proper immersion MATH with boundary MATH which has MATH positive double points. Moreover MATH. Now it is known CITE that one can deform MATH into a diagram which is the closure of a braid MATH while preserving the writhe and the number of NAME circles, that is, MATH and MATH has MATH strings. As MATH we obtain from REF that MATH which implies the desired inequality, note that MATH and MATH. |
math/0105224 | As the kinkiness and the slice genus are concordance invariants, we can asssume that MATH itself is strongly quasipositive. By definition, a strongly quasipositive knot MATH can be obtained as the closure MATH of some strongly quasipositive braid MATH (here we use the notation from CITE). Then MATH and as shown in CITE, the slice genus MATH is given by MATH . Moreover the quasipositive braided NAME surface MATH for MATH has NAME characteristic MATH, so MATH. Now the assertion follows from REF . |
math/0105224 | Suppose MATH is the closure of the positive braid MATH. Let MATH denote the number of letters in MATH. It is well known that MATH can be unknotted using MATH crossing changes, see for instance CITE. However, since MATH is positive, MATH, and therefore we can conclude, using once more the results from CITE, that MATH . Since it is always true that MATH, these two numbers must coincide. Now, by REF , we also have MATH. Hence MATH and we obtain that also MATH. Finally the inequality MATH implies that MATH, and as MATH we have MATH, as claimed. |
math/0105224 | The first part of the statement is an immediate consequence of the results in CITE. As shown there, the mirror image of such a pretzel knot is strongly quasipositive (again note that in CITE, a slightly different definition of pretzel knots is used, in fact the knot which is there called the pretzel knot of type MATH is the mirror image of the knot which we denote by MATH). The sum of two strongly quasipositive knots is again strongly quasipositive, and hence every non - trivial strongly quasipositive knot has infinite order in the smooth concordance group by CITE. Using REF , we can also conclude that the negative kinkiness MATH is not zero and that MATH cannot be unknotted by using only positive crossing changes. |
math/0105224 | Consider the subgroup MATH of the smooth knot concordance group generated by all twist knots MATH, MATH. Recall (see REF ) that MATH. This is clearly also true for every knot which can be written as a connected sum of twist knots and mirror images thereof. As the kinkiness is a concordance invariant, this implies that for every knot MATH whose concordance class is in MATH, MATH. If such a knot is strongly quasipositive, then REF shows that MATH, in other words MATH is slice and the concordance class represented by it is the trivial one. In CITE, NAME modified the arguments of CITE to show that MATH has infinite rank, this is even true for the subgroup of MATH generated by all twist knots MATH where MATH is a square. Hence MATH has all the required properties and the proof is complete. |
math/0105225 | Given a subspace MATH, we will write MATH for the annihilator of MATH in MATH (as opposed to MATH, the annihilator in MATH) with respect to the Killing form on MATH. The statement of the Lemma is equivalent to saying that the adjoint action map MATH is an isomorphism. First, note that MATH representation theory implies: MATH and that the map MATH is a bijection. It follows easily that MATH. Thus, there is a direct sum decomposition MATH . Next, define a MATH-action on MATH by MATH . Note that for any MATH, we have MATH. Also, the action of MATH on MATH satisfies MATH for any MATH. The action map MATH is MATH-equivariant. Moreover, by REF , MATH induces an isomorphism between the tangent spaces of the MATH-fixed points MATH and MATH. Thus, REF follows from the following general result: An equivariant morphism MATH of smooth affine MATH-varieties with contracting MATH-actions which induces an isomorphism between the tangent spaces of the MATH-fixed points must be an isomorphism. To prove this, let MATH be the MATH-fixed point of MATH, let MATH be the tangent space of MATH at MATH, and write MATH for the formal character of the coordinate ring of a MATH-variety MATH. Since MATH induces an isomorphism MATH, the pullback on coordinate rings MATH is injective. The surjectivity of MATH follows from the equation: MATH, see REF . |
math/0105229 | We have MATH . Also, calculate MATH which is the required formula. |
math/0105229 | Calculate, by using the defining relations of a pre-operad: MATH which is the required formula. |
math/0105229 | Use REF . Note that MATH and calculate, MATH which is the required formula. |
math/0105229 | Two proofs can be found in CITE. |
math/0105229 | See REF. |
math/0105229 | See REF. |
math/0105229 | See REF. |
math/0105229 | Combine the Main Theorem with REF . |
math/0105229 | Use the ground identities from the previous subsection REF. |
math/0105229 | We calculate these vertex values in a standard way, by using REF . First calculate MATH. Use REF and MATH from the Boundary Lemma to note that MATH . We must compare it term by term with MATH . Now, recall the sign MATH and use composition relations to note that MATH which lead one to the required formula for MATH. Next calculate MATH. Use REF and MATH from the Boundary Lemma to note that MATH . We must compare it term by term with MATH . Now, recall the sign MATH and use composition relations to note that MATH which lead one to the required formula for MATH. At last calculate MATH. Use REF and MATH from the Boundary Lemma to note that MATH . We must compare it term by term with MATH . Now, recall the sign MATH and use composition relations to note that MATH which lead one to the required formula for MATH. |
math/0105231 | Use the NAME identity. |
math/0105231 | Combine REF with REF . |
math/0105231 | See REF. |
math/0105231 | See REF. |
math/0105231 | See REF. |
math/0105231 | Use the ground identities from the previous REF. |
math/0105233 | Since MATH, MATH, and MATH are of finite exponent MATH, they are the direct product of their MATH-parts. Thus MATH, MATH and MATH. Clearly, an embedding of MATH into MATH provides embeddings for MATH into MATH for each prime MATH. Conversely, if we have embeddings into MATH for each MATH, then MATH gives an embedding for MATH. |
math/0105233 | It suffices to show that MATH lies in MATH and has the corresponding universal property. That it lies in MATH follows because both MATH and MATH are of exponent MATH, and all commutators are either in MATH, MATH, or the cartesian. The former two are already of exponent MATH, and the latter is of exponent MATH since we have killed the MATH-th powers explicitly. That MATH and MATH are embedded into the quotient follows from the description given above. Given maps MATH and MATH, with MATH, the universal property of MATH gives a unique map to MATH, which will factor through the quotient MATH, since in MATH all MATH-th powers of commutators are central. It is now easy to verify that this map is indeed unique from the quotient, giving that this is in fact isomorphic to the coproduct in MATH. The final statement now follows. |
math/0105233 | By REF , the cartesian of MATH is isomorphic to the cartesian of MATH modulo its MATH-th power. Using NAME 's isomorphism, this gives, in MATH: MATH . |
math/0105233 | MATH . For any MATH-overgroup MATH of MATH, we have MATH . MATH . Note that MATH is a MATH-overgroup of MATH, so MATH is central there, and thus MATH. MATH . Assume that MATH. Since the cartesian of MATH is isomorphic to MATH it follows that MATH is nontrivial in MATH, as desired. |
math/0105233 | Set MATH, MATH, MATH, and apply REF . That they both lie in MATH follows because they are obtained as quotients of products of groups in MATH. |
math/0105233 | Since MATH and MATH lie in MATH, they are central in MATH. So, MATH . Therefore, MATH, as desired. If MATH, then MATH, so MATH. |
math/0105233 | Let MATH be the order of MATH (MATH if MATH and MATH is of infinite order). In particular, MATH, and also MATH. Let MATH be the group: MATH . Note that MATH necessarily, and that MATH. Now let MATH . |
math/0105233 | For necessity, assume that MATH is an overgroup of MATH with MATH, MATH, and MATH. If MATH, then MATH, since MATH, and similarly with MATH. If MATH, then MATH since MATH, and again similarly for MATH, yielding REF . For REF , note that since the congruences hold modulo MATH, they also hold modulo MATH. We have MATH, so by REF MATH . Since MATH, and MATH, the left hand side must be trivial, so MATH, as desired. Moreover, if MATH, we can adjoin central MATH-th roots to MATH and MATH to get MATH-th roots for MATH and MATH. This gives necessity. For sufficiency, we will construct MATH, along the lines of NAME 's proof of REF . Let MATH where MATH and MATH are infinite cyclic groups. Let MATH be the smallest normal subgroup of MATH containing MATH and MATH. We claim that MATH. Indeed, a general element of MATH may be written as MATH with MATH, MATH, MATH, and MATH. Rewriting, MATH where MATH. We do this by first replacing the conjugate MATH by MATH, and then expanding the commutator brackets bilinearly. We write it as MATH, with MATH, MATH, and MATH. Assume that this element lies in MATH, and equals MATH. Then by uniqueness we have MATH, MATH. On the other hand, MATH for some integer MATH, so we must have MATH. Therefore, we have that: MATH . Since MATH, we get MATH . Let MATH, MATH, and let MATH. We rewrite REF , and we get MATH . Since the cartesian equals MATH, this means that MATH so we must have MATH . The congruences also hold modulo MATH, and from REF , we have that MATH so MATH. Thus, MATH, so by REF we must have MATH; but since MATH by REF , this means that MATH. So MATH, as claimed. Let MATH. Note that MATH is a subgroup of MATH, and that, in MATH, MATH, MATH. Note as well that MATH in MATH, and hence also in MATH. If MATH, consider the subgroup of MATH generated by MATH, MATH, and MATH. We already have that MATH. Note also that: MATH by REF ; to show that the subgroup MATH lies in MATH, it will now suffice to show that both MATH and MATH are central. First, note that both MATH and MATH are central in MATH: for given any MATH, MATH so by REF , MATH. A similar calculation holds for MATH. Therefore, MATH centralizes MATH; it trivially centralizes MATH, so we just need to know that MATH centralizes MATH to get that it is central in the subgroup in question. But MATH in MATH; so MATH is central in MATH, and an analogous calculation holds for MATH. Thus, if we let MATH, we have MATH have MATH-th roots in MATH, and MATH. Assume now that MATH. We still have that both MATH and MATH are central in MATH, by the same argument as above; however, the orders of MATH and MATH may be greater than MATH. Let MATH be the result of adjoining to MATH central MATH-th roots MATH and MATH to MATH and MATH, respectively. Note that both MATH and MATH are of exponent MATH: MATH by REF , and analogously with MATH. Let MATH be the result of adjoining to MATH elements MATH, of exponent MATH, with MATH, MATH, as in REF , so that MATH commutes with all elements of MATH, except MATH, etc. Let MATH be the subgroup of MATH generated by MATH, MATH, MATH, and the MATH. We claim MATH. Note that MATH and analogously with MATH. And MATH is central, since MATH is central here, and so is MATH; the same is true of MATH, so MATH. Finally, note that in MATH, we have MATH and similarly MATH; thus MATH, as desired. |
math/0105233 | For the case MATH, we can proceed one prime at a time by working in the MATH-parts of MATH. If we simply use the analogous construction directly when MATH, we will always be in the situation analogous to the case MATH in the previous theorem, so everything works out. Note that REF is trivially true if MATH. |
math/0105233 | Decompose MATH into a sum of cyclic groups, MATH. If MATH, then for each MATH let: MATH . Then let MATH. It is easy to verify MATH satisfies the given condition. If MATH, then simply adjoin a central MATH-th root to the generator of each cyclic subgroup, to obtain an abelian group MATH with MATH. |
math/0105233 | If MATH, let MATH . If MATH, we let MATH be the split metacyclic MATH-group: MATH . It is not hard to verify these definitions will work. |
math/0105233 | Since MATH is of bounded exponent, we can decompose it as a sum of cyclic groups, each of order MATH for some MATH, with MATH. For each direct summand, use REF , and then take the direct sum of the groups so obtained. |
math/0105233 | Decompose MATH into a sum of cyclic groups of prime power order and deal with the MATH-parts separately. The condition that MATH guarantees that we can apply REF in each case, and we are done. |
math/0105233 | Say that MATH, with MATH of order MATH modulo MATH (MATH if MATH and MATH is of infinite order modulo MATH), MATH. Write: MATH . Let MATH, where MATH is a cyclic group of order MATH (infinite cyclic if MATH). Note that MATH. Let MATH be the normal subgroup of MATH generated by the elements MATH, MATH, and MATH for MATH. First we claim that MATH. Note that MATH, MATH are central for each MATH, and that if MATH, MATH is central and MATH. If MATH, then MATH. A general element of MATH can be written as: MATH where MATH, MATH, MATH, MATH, MATH, and each MATH is a finite set of elements of MATH. We can use a single finite set MATH by setting the necessary MATH to zero. We first look at the factor MATH. Ordering the factors lexicographically, we obtain MATH where MATH . Since MATH, we have MATH. So we have that MATH . Expanding the second factor in REF , we have MATH . If we expand the third factor in REF , and writing MATH with MATH, MATH, MATH, we have MATH . Looking back to the second factor in REF , we have MATH . And the third factor in REF yields: MATH by setting MATH and MATH. Making the substitutions into the expression for MATH given in REF , we obtain: MATH . Assume that MATH, and we will prove that MATH. The maps MATH and MATH induce a unique map MATH by the universal property of the coproduct. Since MATH, MATH; on the other hand, all MATH map to MATH, so MATH . Let MATH be the product of MATH copies of cyclic groups of order MATH (infinite cyclic if MATH), and denote the generators of the factors by MATH. Mapping MATH by mapping identically to the first coordinate, and mapping MATH, we get another induced map MATH. On the one hand, MATH maps to itself, but on the other hand we have: MATH . Since we already know that MATH, this yields that the first factor is trivial, so MATH for each MATH; thus, MATH for each MATH. Therefore, we also have that MATH; hence, since MATH, MATH . In addition, since MATH, and MATH, we have that either MATH, or MATH and MATH. If MATH, then MATH is trivial anyway, and if MATH, then MATH. So several factors in the original expression are actually trivial, and we have: MATH . In particular, all the other terms in the original expression for MATH are actually trivial, so MATH . Fix MATH. Pick MATH with a commutator of order MATH (of infinite order if MATH), for example the relatively free group of rank REF generated by MATH and MATH. Map MATH, MATH, and MATH for MATH, and map MATH; this induces a map MATH, which necessarily maps MATH to MATH; it also maps MATH to MATH . Therefore, MATH for each choice of MATH, so each of these commutators in MATH are trivial. Let MATH be a cyclic group of order MATH (infinite cyclic if MATH) and consider the group MATH. Fix MATH, and map MATH to itself via the canonical inclusion, send MATH to MATH, and MATH for MATH. This induces a map MATH. The image of MATH is again trivial, and also equal to MATH . By REF , we must have MATH . The MATH all lie in MATH, as does MATH; so we must have MATH. We know that MATH, and moreover, that MATH, so by REF in the statement of the Proposition, MATH. Therefore, MATH . Looking at MATH, we must have (since the second factor lies in MATH, it makes sense to calculate this in MATH): MATH . Taking the product over all MATH, we have MATH . From REF , we have that MATH so MATH. Using this in the expression above, we have MATH . We also have that: MATH and that MATH . Therefore, MATH . Since we already have that MATH this gives that MATH, as claimed. Therefore, MATH. Let MATH. Note that, since MATH, MATH embeds into MATH. What we want to do now is to embed MATH into MATH, so that MATH in MATH. Since MATH, and MATH is generated by MATH modulo MATH, we may use REF . Map MATH via the inclusion into MATH; define MATH by MATH. Note that since MATH, the MATH are well defined. We need to check the conditions of REF to ensure a map MATH is induced, and that this map is injective. First, we need to verify that MATH. As MATH, and MATH, the image of MATH in MATH is equal to MATH, as desired. Second, we need to check that MATH. But again, MATH lies in MATH, and MATH. Finally, to verify that MATH, which we already noted. Thus, MATH and the MATH induce a morphism MATH. Since MATH, MATH is contained in MATH. Thus, in order to prove that MATH realizes a (weak) embedding of MATH, we only need to show that MATH is injective. Since MATH is the identity, it is injective. The maps MATH and MATH induce a morphism MATH. The kernel contains MATH, and also contains MATH, so it factors through MATH. This gives an induced map MATH by MATH. Note that MATH, so we may look at MATH, which is a subgroup of MATH. So we have MATH . Also, MATH maps to MATH by MATH . Since MATH maps again to MATH, we get an endomorphism MATH which is the identity on each MATH. So the first map, which is the map induced by MATH, must be injective, and together with the injectivity of MATH we conclude from REF that MATH is injective. Thus, MATH realizes a (weak) embedding of the amalgam. Finally, we strengthen this embedding into a strong embedding. Since MATH, we have MATH. Let MATH. We construct embeddings MATH by mapping to MATH via MATH and to MATH via the canonical projection; and MATH by mapping to MATH via the inclusion and to MATH via the zero map. Now assume that MATH, MATH map to the same element of MATH. Then, MATH, so MATH, proving that their intersection in MATH is contained in MATH. Since the embeddings into both coordinates agree on MATH, their intersection contains at least MATH; thus MATH realizes the strong embedding of MATH, giving the result. |
math/0105233 | First we prove necessity. Suppose that MATH contains MATH and MATH, and MATH in MATH. Any element in MATH is central in MATH, so MATH, giving REF by symmetry. For REF , note that both MATH and MATH are central in MATH and commutator brackets act bilinearly, so we have MATH and this equals MATH by symmetry. Since MATH and MATH, their common value lies in MATH, hence in MATH. For sufficiency, note that both conditions are inherited to any finitely generated subamalgam MATH, so we may assume that both MATH and MATH are finitely generated (and finitely generated over MATH) by REF . Let MATH and MATH be central amalgams. By REF , the subgroups MATH of MATH and MATH are isomorphic. Note that if MATH, then MATH is of exponent MATH, so MATH is of exponent MATH, and the same with MATH. By REF (for MATH) or REF (for MATH), there exists MATH such that MATH. Let MATH and MATH. Then MATH since MATH, MATH, and MATH; and analogous for MATH. Let MATH in both MATH and MATH. We want to apply REF to MATH. Considering MATH as a subgroup of either MATH or MATH, by REF we have that: MATH . Again, it is straightforward to verify that MATH. This gives REF . To test REF , let MATH, MATH, MATH, MATH with MATH. We want to show that MATH. By construction of MATH, we may write MATH with MATH, MATH, and MATH. Since all three commute with one another, MATH. Since MATH, this says that MATH. We know MATH, and MATH, MATH, so we may write MATH . Therefore, MATH. Note that MATH, with MATH. Symmetrically, we may write MATH with MATH, MATH, MATH, and MATH with MATH, MATH, and MATH. Since MATH, MATH, and MATH satisfy REF in the statement, we must have that MATH . Now, in MATH we have: MATH . Symmetrically, in MATH we have MATH . Since MATH, we conclude that MATH, as claimed. Thus MATH satisfies the conditions of REF , so there exists a MATH which realizes a strong embedding of MATH. We claim that this embedding, via the inclusions of MATH and MATH into MATH and MATH, respectively, yields a strong embedding of MATH. Since MATH in MATH, we must have that MATH. Since MATH is itself a strong amalgam of MATH and MATH, MATH (since MATH in MATH, and MATH in MATH). Symmetrically, MATH. So MATH in MATH, and therefore, MATH provides a strong embedding for MATH, as desired. |
math/0105233 | Necessity follows along the same lines as the necessity in REF . Simply note that for each MATH, in any MATH-group realizing the amalgam, we have MATH. For sufficiency, we will construct an overgroup MATH of MATH, which is a subgroup of both MATH and MATH, and which satisfies REF . That will yield that MATH can be strongly embedded, and thus that MATH can be weakly embedded using the same group that realizes MATH. Let MATH be the normal closure of MATH in MATH, and analogous for MATH. Let MATH . Trivially, MATH. Order MATH by inclusion; the union of any chain in MATH again lies in MATH, as there will only be finitely many indices involved in checking REF . By NAME 's Lemma, MATH has maximal elements. Let MATH be a maximal element of MATH. We claim that MATH satisfies REF there is the same as REF here, so MATH satisfies it by virtue of lying in MATH. To show that MATH satisfies REF , let MATH with MATH, MATH, MATH, MATH, MATH with MATH. We want to prove that MATH. Let MATH and MATH. Then MATH if and only if MATH for each MATH, and if so MATH. Indeed, this follows from the fact that MATH satisfies REF : set MATH, and all terms equal to MATH or MATH. So we conclude that MATH and MATH with the same MATH and the same MATH. Since MATH, and MATH, we have that MATH . Since MATH, MATH; we denote the common subgroup of MATH and MATH by MATH. Then MATH, MATH, and we claim that in fact MATH. If MATH, then MATH for some MATH, MATH. Since MATH, we also have MATH, and by REF for MATH, MATH; so MATH lies in MATH. Symmetrically for MATH, yielding that MATH satisfies REF . For REF , let MATH, MATH, MATH, MATH, MATH, MATH, MATH with MATH, and let MATH. We want to prove that MATH . Write MATH, for some MATH, MATH. In addition, we may write MATH, so MATH; so we have MATH . Since MATH, we may apply REF to MATH to obtain that MATH which means that MATH . Therefore, MATH satisfies REF . Thus, MATH, MATH, so by maximality of MATH, MATH. Therefore, MATH. Which in turn shows that MATH satisfies REF from REF . Thus MATH is strongly embeddable, and therefore MATH is weakly embeddable. |
math/0105233 | First we prove that all such MATH must lie in the dominion. Indeed, say MATH, and we have morphisms MATH which agree on MATH. Then MATH and by symmetry this equals MATH. So MATH. To prove equality, it suffices to show that MATH is strongly embeddable in MATH. We check REF from REF is trivial, since MATH. For REF , assume that MATH satisfy MATH for some MATH, MATH, MATH. We want to prove that MATH. The equality of the two follows by bilinearity of the commutator bracket, which we can now apply since we are working inside a single group MATH: MATH . To prove that it lies in MATH, note that every element of MATH can be written as an element of MATH times some commutators. Thus, if MATH, then there exist MATH such that MATH. Since MATH, there is some MATH with MATH, and analogously there exists an element MATH with MATH. By construction of MATH, we must have MATH; but this is equal to MATH, giving REF . Thus, MATH is strongly embeddable, so MATH. |
math/0105233 | We may assume that MATH is finitely generated. Suppose first that MATH; by looking at the MATH-parts separately, which we can do since dominions respect finite direct products, we may assume that MATH is a prime power, and MATH; for if MATH, then MATH is abelian, and MATH. Suppose that MATH. Looking at REF , we need only consider MATH and MATH. If MATH, then MATH, so it lies in MATH. If MATH, then, choosing MATH so that MATH, we have MATH so again MATH. Therefore, MATH, and the dominion is trivial. Now suppose that MATH, and MATH. Let MATH. We will prove that MATH. Let MATH be a normal subgroup of MATH, with MATH finite and such that MATH (the existence of such a MATH follows from the fact that in a finitely generated nilpotent group, every subgroup is closed in the profinite topology). Let MATH be such that MATH is of exponent MATH, and MATH. Since MATH is squarefree, the dominion of MATH in MATH in MATH is trivial by the previous case, so MATH so MATH, and since MATH is not in that dominion, we conclude that MATH either. This gives the result. |
math/0105233 | Since all dominions are trivial in MATH, every group is a special amalgamation base. Say MATH gives a weak embedding of MATH in MATH. Then we may strongly embed MATH in MATH. By embedding MATH into the first copy of MATH and MATH into the second copy of MATH, we get a strong embedding for MATH. |
math/0105233 | The inclusion of the left hand side into the right hand side is immediate, since MATH. To prove the reverse inclusion, say that we have MATH, MATH with MATH, and MATH such that MATH. We want to prove that MATH. Note that MATH contains all MATH with MATH and MATH: no restriction on MATH dividing MATH, but also a smaller group MATH rather than MATH. Write MATH with MATH, MATH, and analogously MATH. Since MATH, we may write MATH for some integer MATH. Then MATH . So MATH, where MATH. In a similar manner, MATH, with MATH. Therefore, MATH. But since MATH, MATH-th powers of commutators are trivial. So: MATH . Therefore, MATH, as desired. This proves the equality. |
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