paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0101007 | This will be proved by the computation in the proof of REF . |
math/0101007 | REF is straightforward by observing that MATH descends to MATH and so defines an isometric isomorphism MATH by REF , applied to MATH instead of MATH (thus MATH, and MATH). Since MATH is irreducible, it is enough to compare the inner products on MATH in order to prove REF . Apply REF . |
math/0101007 | Before we embark on this computation we establish some useful relations. First recall that REF MATH. Also recall REF . We see that MATH where MATH denotes the longest element of set of minimal coset representatives MATH. Next we observe that for any MATH, MATH where MATH. This formula follows in a straightforward way from REF . If MATH is a simple reflection and MATH an equivalence class, we check that (recall that MATH is MATH-generic) MATH . Since we assume that MATH is MATH-generic, we have MATH for all MATH and all MATH such that MATH. From this, REF and induction to the length of MATH we see that MATH for all MATH. Observe that we also have MATH . We note that MATH. Since MATH and MATH are involutions, this implies that MATH, where MATH (also a standard parabolic subsystem). Let MATH, with MATH and MATH. Keeping in mind the above preliminary remarks, and the fact that MATH is central, we now compute (with MATH and MATH): MATH . The result is independent of MATH, implying the first equality of the theorem. The second equality follows because MATH is central. Indeed, this implies that we have MATH where the second MATH of course refers to the MATH-structure on MATH. In REF we have used the evaluation MATH . This is allowed because the element MATH is MATH-invariant, and thus belongs to the center of MATH. At several places in REF we have freely used formulae of CITE (see REF ) for the structure of MATH. For example, MATH when MATH. By this we easily see that for all MATH, MATH and hence we may conclude by REF that MATH . The proof is finished. |
math/0101007 | We have, with MATH, MATH, where MATH is defined by MATH . Both these respect the pairing MATH, by REF . |
math/0101007 | CASE: The representation MATH is cyclic and generated by MATH, with MATH an arbitrary vector in MATH. By the intertwining property it is therefore enough to show that MATH is meromorphic in MATH, and regular outside the indicated set. Using REF , we have MATH . Since MATH is a regular function on MATH for all MATH, this is a rational expression. The generalized MATH-weights in MATH are of the form MATH. So the inverse of MATH can have poles only at the indicated set. CASE: In order to see the unitarity, we first note that by REF , the statement is equivalent to the unitarity with respect to the inner products on MATH and MATH defined by the embedding of these spaces in MATH as in REF . Choose an embedding MATH as in REF . By REF , the map MATH is an isometry. By REF we see that this isometry satisfies, for MATH, MATH. Hence if we identify MATH with MATH, we can define MATH. Then the map MATH defines a unitary map satisfying REF . Now it is clear, in the notation of REF , that we can identify the space MATH with MATH, and the map MATH is then identified with the right multiplication with MATH, thus with MATH. This is unitary on MATH, by REF . By the irreducibility of MATH and MATH this concludes the proof of REF . CASE: This last assertion of the Theorem is now obvious, since these maps are clearly inverse to each other. |
math/0101007 | By the unitarity on MATH, the meromorphic matrix entries of MATH are uniformly bounded for MATH in the open set of MATH where MATH is well defined and regular. This is the complement of the collection of real codimension MATH cosets in MATH as described in REF . This implies that the singularities of the matrix entries which meet MATH are actually removable. |
math/0101007 | Let MATH and let MATH be such that MATH. Then MATH, MATH fixes MATH, and MATH. We have MATH, thus MATH for some MATH. Hence MATH. Since MATH is MATH-generic this implies that MATH. Hence MATH, and thus also MATH. |
math/0101007 | By making a suitable choice of MATH in the orbit MATH we may assume that MATH for some MATH. We write MATH instead of MATH and MATH instead of MATH. Thus it is assumed that MATH is MATH-generic. Define MATH. Clearly, MATH by the MATH-invariance of MATH. Conversely, suppose that MATH. Hence there exists a MATH such that MATH. This is an element of the tempered residual subspace MATH, so that MATH. Since MATH is MATH generic, we have MATH by REF . Because MATH, we obtain MATH. This implies that MATH. By REF we see that MATH. Hence we have MATH, MATH and MATH. We conclude that MATH. There exists a unique MATH such that MATH. One easily checks that MATH. Note that it follows that the intersection MATH is contained in MATH (for any choice of MATH and MATH). |
math/0101007 | By REF , MATH is well defined on MATH, and thus MATH. By the previous lemma, the set MATH is MATH-saturated. Since MATH is closed, this implies that MATH is open (and obviously dense) in MATH. Finally, again by the previous lemma, MATH is injective on MATH. Thus, being a closed map, MATH is homeomorphic onto its image. |
math/0101007 | The topology on MATH is second countable since MATH is separable. Thus, in order to check the continuity of MATH, it suffices to check that MATH maps a converging sequence MATH to a converging sequence in MATH. We check this using the NAME description of the topology of MATH (see CITE). By REF , restriction to the dense subalgebra MATH is a homeomorphism with respect to the Fell topologies. Let MATH be the representation space of MATH (with MATH). We equip MATH with the inner product MATH of REF (which is independent of MATH), and we choose an orthonormal basis MATH of MATH. In order to check that MATH in the Fell topology with respect to MATH, we need to check that for all MATH, MATH for all matrix coefficients. This is clear since the matrix coefficients are regular functions of the induction parameter. To see that the map MATH is closed, assume that we have a sequence MATH converging to MATH. Since MATH is a homeomorphism and MATH is a finite quotient of MATH, we may assume that MATH converges, to MATH say, by possibly replacing the sequence by a subsequence. Since MATH is finite for each MATH and MATH, we may assume that MATH, again by taking a subsequence. Then MATH is independent of MATH, and lower semi-continuity of MATH on MATH implies that MATH. Choose an orthonormal basis MATH for MATH. Convergence in the NAME means that there exists for all MATH an orthonormal subset MATH of size MATH in the representation space MATH of MATH, such that the matrix coefficients of MATH with respect to MATH converge to the matrix coefficients of MATH with respect to MATH. By the independence of the inner product MATH of the induction parameter REF we may assume, by further restricting to a subsequence, that the sets MATH converge in MATH to an orthonormal set MATH. It follows that the matrix of MATH with respect to MATH equals a principal block of the matrix of MATH with respect to a suitable orthonormal basis MATH of MATH for MATH. Since MATH is irreducible it is easy to see that this is impossible unless MATH. The map MATH is injective by REF . Finally, by REF we see that the complement of MATH has measure MATH in MATH with respect to the NAME measure of the representation MATH of MATH. The support of the NAME measure is equal to MATH, since MATH is a faithful representation of MATH (by definition of MATH). Thus the closure of MATH is MATH. But MATH is closed as we have seen above, implying that MATH is surjective. |
math/0101007 | This is a simple extension of REF , with a similar proof. |
math/0101007 | CASE: See REF . The complement of MATH has measure zero by the argument in the last part of the proof of that theorem. The density follows since MATH is the support of the NAME measure (compare REF ). CASE: By formula of REF we have MATH . We decompose MATH according to REF to obtain MATH . By REF we have MATH. Thus (by REF ) we can rewrite the integral as integral over MATH. When we use parameters and notations as explained in REF , and we express MATH according to REF , we obtain MATH . According to our definition of MATH this is equal to MATH . This is a decomposition of MATH in characters of inequivalent irreducible representations of MATH (see REF ). Hence this uniquely determines the NAME measure (by CITE, NAME REFEF) on MATH. We conclude that MATH is equal to the NAME measure of MATH. CASE: The equivalence of REF is well known, see the proof of REF . It is allowed to use the formulation with MATH-equivariant sections because of the unitarity and the regularity of intertwining operators REF and by REF . |
math/0101007 | By the isometry property of MATH, MATH for all MATH. Whence the first assertion. It is clear that MATH. If MATH then MATH for some MATH. Thus MATH for MATH-equivariant MATH. Hence MATH. |
math/0101007 | On the connected component MATH the map MATH is equal to MATH where MATH is the multiplication in MATH by MATH, and MATH is the multiplication in MATH by MATH. These are all analytic diffeomorphisms, because of REF . The MATH equivariance follows from the fact that MATH is well defined (and thus equivariant, since MATH is equivariant) from MATH to MATH, and that MATH is MATH-equivariant. This implies that for MATH, MATH. It follows that MATH . |
math/0101007 | For MATH in the connected component MATH of MATH we write MATH with MATH. We have MATH . We remind the reader of the convention REF ; in particular, the expression MATH occurs only if MATH, in which case this expression stands for MATH. If MATH, we should reduce REF to MATH . By REF it is clear that poles and zeroes of these functions will only meet MATH if MATH when MATH or MATH if MATH. In these cases the statement we want to prove reduces to the statement that the function MATH is holomorphic and invertible on the domain MATH, where MATH is a real number and MATH. By REF both the denominator and the numerator of MATH have a zero in this domain only at MATH (if this belong to the domain), and this zero is of order MATH both for the numerator and the denominator. The desired result follows. |
math/0101007 | The map MATH as defined above is clearly a MATH-linear isomorphism by REF . It is an algebra homomorphism because we have MATH . What remains is the proof that MATH. Notice that MATH is the subalgebra generated by MATH and the elements MATH where MATH with MATH. The MATH-image of MATH equals MATH since MATH is an analytic diffeomorphism. To determine the image of MATH we use formula REF, applied to the situation MATH. This tells us that MATH . Hence we see that MATH . By REF it is clear that this is indeed in MATH, and that these elements together with MATH generate MATH. |
math/0101007 | For all MATH we have MATH showing that the left hand side has the correct eigenvalue for multiplication of MATH by MATH on the left. For the multiplication of MATH by MATH on the right a similar computation holds. This shows, in view of REF that, for regular MATH and outside the union of all residual cosets, the left and the right hand side are equal up to normalization. But both the left and the right hand side are equal to MATH if MATH. Hence generically in MATH, we have the desired equality. Since both expressions are holomorphic in MATH, the result extends to all MATH. |
math/0101007 | Take a neighborhood MATH with MATH satisfying REF relative to MATH. Let MATH denote the MATH-cycle defined by MATH. In view of REF we see that, for all MATH, MATH . Let MATH. The scaling operation sends the root labels MATH to MATH, and follows the corresponding path of MATH of the residual point MATH. Obviously the position of MATH (in REF ) relative to MATH is independent of MATH. And also, the position of MATH relative to the facets of the dual configuration MATH is independent of MATH, since the effect of the scaling operation on MATH simply amounts to the application of the map MATH. In view of REF , we can take the cycle MATH in order to define the state MATH of MATH. In other words, we have, for MATH, MATH where MATH . By REF , the function MATH extends, for all MATH, to a regular holomorphic function on MATH. Clearly, MATH is MATH-invariant. In other words, MATH is an element of MATH. Its value in MATH can be computed easily, if we keep in mind that the index MATH (by REF applied to the residual coset MATH). We obtain, by a straightforward computation: MATH . We now continue the computation which we began in REF , using the fact that MATH and the fact that MATH extends uniquely to MATH in such a way that for all MATH and MATH, MATH. We get MATH . This gives the desired result. |
math/0101007 | This is an immediate consequence of the previous lemma. |
math/0101007 | CASE: As was explained in the proof of REF , it is sufficient to prove this statement for MATH. Recall that in this case MATH. We have MATH where MATH denotes the spectral radius of MATH. According to REF , the spectrum MATH equals the union of the compact sets MATH. This set is by definition the closure in MATH of the set MATH. It is well known that for any MATH, the map MATH is lower semi-continuous as a function of MATH (compare CITE, VII, REF ). Since there are only finitely many triples MATH, it is sufficient to show that there exist constants MATH, MATH such that the spectral radius of MATH is bounded by MATH, uniformly in MATH. The roots of a monic polynomial are bounded by the sum of the absolute values of the coefficients of the equation (including the top coefficient MATH). Hence the spectral radius of a MATH matrix MATH is bounded by a polynomial of degree MATH in MATH. Since MATH it is sufficient to show that there exists a suitable basis for the parameter family MATH (with MATH) of representations, in which the matrix coefficients of the MATH (with MATH) are uniformly bounded by MATH for suitable constants MATH and MATH. As in the proof of REF , there exists a basis MATH of MATH, the representation space of MATH, such that the MATH REF simultaneously act by means of upper triangular matrices in this basis. Moreover, by REF it is clear that the diagonal elements are bounded in norm by MATH when MATH. By the compactness of MATH we conclude that there exists, for each MATH, an unipotent upper triangular matrix MATH with positive coefficients such that every matrix coefficient of MATH in the above basis is uniformly (in MATH) bounded by the corresponding matrix coefficient of MATH. Let us denote by MATH the set of MATH matrices with non-negative entries, and introduce the notation MATH for complex matrices MATH. Introduce a partial ordering in MATH by defining MATH if and only if MATH. Since MATH is a semigroup for matrix multiplication, it is clear that if MATH and MATH are in MATH and MATH, then MATH. In addition we have the rule MATH for arbitrary complex matrices MATH and MATH. Let MATH denote a set of MATH-generators for the cone MATH, and let moreover MATH be a basis of MATH. Put MATH for MATH, MATH with MATH if MATH. We can thus find an upper triangular unipotent matrix MATH such that for all MATH, MATH, MATH: MATH. If we write MATH with MATH if MATH, then MATH with MATH if MATH. From this we see that there exists a constant MATH independent of MATH such that MATH. Thus, with MATH the nilpotent logarithm of MATH, and MATH: MATH where MATH is a constant depending on MATH only, and MATH is the degree of the polynomial function MATH. This finishes the proof. CASE: As in the proof of REF , but we restrict ourselves to the (finitely many) discrete series representations. This implies that we can find MATH unipotent and MATH such that for all MATH, MATH. Inserting this in REF we find that the matrix entries of MATH are bounded by MATH, with MATH independent of MATH. Hence the spectral radius of MATH is uniformly bounded for MATH, proving the desired estimate. |
math/0101007 | The character MATH of MATH is a positive trace, and thus satisfies the inequality MATH by REF . By NAME 's criterion REF this implies that MATH is tempered. |
math/0101007 | This is well known, and follows easily by the remark that the restriction to MATH of the trivial representation is equal the square root of the NAME modulus MATH (see for example, REF ): MATH. Now apply the NAME criteria REF . Similar remarks apply to the case of the NAME representation. |
math/0101007 | Use REF . |
math/0101007 | The continuity of MATH and MATH is immediate from the definitions. So let us look at the multiplication. Let us write MATH . It is easy to see that MATH and that MATH if MATH. Therefore MATH and by REF , there exist constants MATH such that for all MATH and MATH: MATH . We put MATH. It is easy to see that there exists a MATH such that MATH converges. By REF we have that MATH for all MATH. Given MATH, let MATH. Using these remarks we see that for all MATH and MATH in MATH the following holds: MATH . This finishes the proof. |
math/0101007 | CASE: The dimension of MATH equals MATH on an open dense subset, and the measure MATH is absolutely continuous with respect to the NAME measure on MATH. Hence, using the fact that MATH is an isometry, we find MATH . CASE: As in REF we get MATH . |
math/0101007 | The expansion follows as in the above Theorem. It is convergent in MATH because of REF . |
math/0101007 | This follows from MATH. Note that MATH is indeed constant (that is, independent of MATH) by REF . Also note Conjecture REF. |
math/0101007 | The main step is to show that NAME 's bound of REF becomes uniform in MATH under the assumption. Let MATH and choose MATH such that MATH. Let us first fix MATH. Consider the isomorphism of localized NAME algebras MATH where MATH, with MATH a suitably small ball around the origin in MATH. We have, by the assumption that MATH: MATH . On the other hand, for all MATH, the eigenvalues of the self adjoint operator MATH are of the form MATH. Hence the operator norms of the operators MATH are bounded by MATH, for a suitable constant MATH. Given MATH we can write MATH with MATH, MATH and MATH, where MATH denotes the set of shortest length representatives of the left cosets of the stabilizer MATH of MATH in MATH. If we write MATH and MATH, we can thus choose signs MATH and MATH such that MATH . Let us simply denote this decomposition by MATH. Now by REF , and the remark that MATH intertwines the action of MATH on MATH with that of MATH on MATH, we have (MATH is a constant, not necessarily the same as above): MATH where MATH is independent of MATH and MATH. In particular, this implies that the highest power of MATH in MATH tends to MATH with MATH (with MATH). Hence the left hand side of the equality REF is a NAME series in MATH with coefficients in MATH. On the other hand, according to REF , MATH can be expanded as a NAME series in MATH with coefficients in MATH. The desired result follows. |
math/0101007 | By induction on the codimension. In a residual coset MATH of codimension MATH we find only finitely many cosets MATH of codimension MATH in MATH with MATH. The invariance is obvious from the invariance of the index function MATH. |
math/0101007 | By induction on MATH we may assume that the assertions of REF hold true for MATH in REF . From the definition we see that MATH is not the empty set. An element MATH of MATH can not be constant on MATH, and hence MATH. Thus MATH . Since also MATH equality has to hold. Hence MATH and MATH spans MATH. Since MATH is parabolic, we conclude that MATH. This proves REF . The subgroup MATH is isomorphic to MATH, which is a torus because MATH is free. By REF then, its dimension equals MATH. It contains MATH, hence is equal to MATH. It follows that MATH is the connected component of the group of fixed points for MATH, proving REF . REF are trivial. |
math/0101007 | It is clear from the definitions that MATH contains MATH and MATH, and hence has maximal rank. Given a full flag of MATH-residual subspaces MATH, satisfying REF at each level, we see that the sets MATH, MATH are contained in MATH. It follows by reverse induction on MATH (starting with MATH) that each element of the flag is MATH-residual. |
math/0101007 | To see this we may assume that MATH is of type MATH. Then MATH is of type MATH, MATH, and MATH, being of maximal rank in MATH, is of type MATH for some MATH. In particular, MATH for all MATH. Moreover the index of MATH in MATH is at most MATH. Thus MATH takes values in MATH on MATH, and is trivial on elements of the form MATH. It follows that MATH is of order at most MATH on MATH. |
math/0101007 | Straightforward from the definitions. |
math/0101007 | In the compact torus MATH, the MATH-orbits of such points correspond to the vertices of the fundamental alcove for the action of the affine NAME group MATH on MATH. Now we have to restrict to MATH. |
math/0101007 | Unfortunately, I have no classification free proof of this fact. With the classification of residual subspaces at hand it can be checked on a case-by-case basis. By the previous subsection REF the verification reduces to the case of residual points for graded affine NAME algebras. In CITE (compare REF ) this matter was verified. |
math/0101007 | We construct the sequence inductively, starting with MATH. Suppose we already constructed the flag up to MATH, with MATH. Let MATH denote the set of elements of MATH of dimension MATH contained in MATH, and let MATH denote the cardinality of MATH. By assumption, MATH. Since every MATH either contains MATH or intersects MATH in an element of MATH, we have MATH . Assume that MATH. Then, because MATH and MATH, MATH contradicting the assumption MATH. Hence there exists a MATH with MATH, which we can define to be MATH. |
math/0101007 | Define MATH to be the (multi-)set of codimension MATH cosets of MATH arising as connected components of the following codimension MATH sets: MATH . Here MATH, and MATH when MATH. We give the components of MATH, MATH the index MATH, and we give the components of MATH, MATH index MATH. Suppose that MATH is any coset of a subtorus MATH in MATH. Then MATH is equal to the sum of the indices the elements of MATH containing MATH. Assume that MATH. By REF there exists a sequence MATH of components of intersections of elements of MATH such that MATH and MATH (we did not assume that MATH is a component of an intersection of elements in the multiset MATH, hence MATH may occur). If MATH is the smallest index such that MATH, then MATH is by definition residual, and thus violates REF . Hence such MATH does not exist and we conclude that MATH for all MATH. This proves that MATH and that MATH is residual. |
math/0101007 | CASE: If MATH allows a nontrivial diagram automorphism then MATH is simply laced. So we are in the situation of the NAME classification of distinguished weighted NAME diagrams. A glance at the tables of REF shows that this fact holds true. CASE: This is a consequence of REF , since MATH acts on MATH by means of an automorphism (see also REF ) which acts trivially on the set of indecomposable summands of MATH. CASE: For this REF have also no other proof to offer than a case-by-case checking, using the results of this section and the list of real residual points from CITE. The amount of work reduces a lot by the remark that it is well known in the simply laced cases (see REF of the appendix REF). In the classical cases other than MATH, it follows from a well known theorem of CITE that the index of MATH is at most MATH. Hence the desired result follows if we verify that for real residual points of the classical root systems, the values MATH are in the subgroup of MATH generated by the root labels, which is direct from the classification lists in CITE. For the real points of MATH it is also immediate from the above and REF . For nonreal points MATH we look at (the proof of) REF . If MATH has order MATH, then MATH is of type MATH. We need to check the values of the roots MATH and MATH on MATH in this case. But the roots MATH are in MATH, and take rational values in the labels MATH REF on MATH. The real residual points of MATH are all rational in the root labels (see CITE). Again using REF and NAME, we need to check in addition the nonreal residual points MATH with MATH (generating a lattice of index MATH in MATH) and MATH (index MATH). These cases can be checked without difficulty. In the case of MATH, there are generically MATH real residual points, two of which have rational coordinates and one has rational coordinates only in the square roots of the labels. In addition there are two nonreal residual points MATH for MATH, which are easily checked. (We need to check only the case with MATH (index MATH in MATH)). |
math/0101007 | According to REF , MATH. Hence the proof reduces to REF, or can be proved directly in our setup in the same way, compare REF . |
math/0101007 | CASE: We already saw in REF that the rank of MATH is indeed maximal. So we are reduced to the case MATH. Let MATH denote the group of integer powers of MATH, and denote by MATH the root subsystem of roots MATH such that MATH. Now MATH is a root subsystem of rank equal to MATH, with the property that MATH such that MATH we have MATH. Of course, MATH is a residual point of MATH. By an elementary result of NAME and NAME there exists a finite subgroup MATH such that MATH is semisimple with root system MATH. We claim that for every simple root MATH of MATH we have MATH or MATH. To see this, observe that all the roots MATH with MATH are in the parabolic system obtained from MATH by omitting the simple roots MATH such that MATH with MATH. If this would be a proper parabolic subsystem, MATH would violate REF in this parabolic. This proves the claim. Define the element MATH. Note that MATH belongs to MATH by the previous remarks. Consider the grading of MATH given by this element, and define a standard parabolic subalgebra MATH of MATH by MATH . Its nilpotent radical MATH is MATH and by the definition of residual points we see that MATH is a distinguished parabolic subalgebra (see REF .). According to (REF .) we can choose MATH in the NAME class associated with MATH, and MATH, such that MATH form a MATH-triple in MATH. By MATH representation theory it is now clear that MATH. Consider the grading of MATH and MATH induced by MATH. By definition of MATH we see that MATH and MATH. Hence MATH is distinguished in MATH by REF , proving the desired result. Note also that, by REF , in fact MATH, and hence that MATH. CASE: Is immediate from the defining property MATH of the grading with respect to the NAME diagram of a distinguished class. CASE: Is clear by the well known MATH correspondence between distinguished classes and their NAME diagrams. CASE: The result follows from the well known MATH correspondence between unipotent classes and nilpotent classes for connected semisimple groups over MATH. |
math/0101007 | The maps between these two sets as defined in REF are clearly inverse to each other. |
math/0101017 | First let us note that for this pair of equations to be independent equations on the first derivatives of the two functions MATH of the two variables MATH, we need to ask that, if we write these two equations as MATH with fictitious variables MATH replacing MATH, then MATH . The characteristic variety is then defined by the equations MATH on some new variables MATH (see CITE). We are asking that these equations have no real solutions except MATH . Write MATH and write MATH a row vector. If we have such a solution MATH then we obtain a linear relation between the row vectors MATH and MATH which we can suppose, switching subscripts if needed, is MATH . But then MATH has rank one. Conversely, if a linear combination of MATH and MATH has rank one, then we can pick such a MATH and the characteristic variety is not empty. Thinking now of our differential equations as defining a submanifold of the NAME bundle, we think of REF-planes MATH as our family of REF, with the equations MATH determining REF constraints on those REF-planes, that is, as the equations of our subbundle. Therefore the tangent space to this family of REF is given by the equations MATH . Taking the NAME of REF at a fixed point MATH, the tangent plane to our family of REF is precisely cut out by the equations MATH with the added conditions that MATH forcing us to stay at a particular point. But then MATH so that the equations on the tangent space are precisely MATH . The identification of tangent vectors MATH with linear maps is carried out as follows: a curve in MATH is a one parameter family MATH of REF: MATH . Taking any linear map MATH with kernel MATH, assuming that MATH depends smoothly on MATH, and letting MATH be the induced linear map on MATH we can identify the vector MATH with the linear map MATH . We then have the conformal quadratic form defined by taking determinant, which depends on a choice of volume element in MATH and in MATH, so that without choosing a volume element, the quadratic form is only defined up to conformal transformations. Returning to our partial differential equation, in our local coordinate formulation, we can take MATH to have coordinates MATH and let MATH and take MATH any tangent vector to our fiber, that is, so that MATH and find that using MATH the matrix representing the vector MATH is precisely MATH. Note that we have assumed that the tangent vector is tangent to a point MATH with MATH the requirement that our point have MATH can be arranged by a change of coordinates, or we can use MATH . It is clear that a REF-plane is positive definite precisely when its dual space is positive definite. Note that the dual space of MATH is MATH and is equipped with the same quadratic form. We have seen that the absence of real points in the characteristic variety is precisely the definiteness of the tangent planes to the fibers of our subbundle. We can assume that they are positive definite by choice of orientation of MATH. |
math/0101017 | Indeed we will see that the MATH conformal structure on MATH is just MATH. The result is surprising since the notion of ellipticity of a line congruence is invariant under arbitrary linear transformations of MATH, while the splitting of MATH into self-dual and anti-self-dual REF-forms is not. To see the result, note that a REF-plane MATH can be represented by a REF-vector MATH with MATH a basis for MATH, and this REF-vector is unique up to scaling. With orientation taken into account, it is unique up to positive rescaling. Conversely, every REF-vector MATH with MATH is of this form for a unique REF-plane MATH. Taking a metric on MATH we identify REF-vectors and REF-forms, and split our REF-vector into self-dual and anti-self-dual pieces: if MATH are orthogonal coordinates on MATH, and MATH we can express a REF-form uniquely as MATH so that MATH . The space of REF satisfying MATH is just the space of MATH with MATH. To modulo out by rescaling, we can always take MATH . Then it is clear that MATH where MATH provides coordinates on MATH and MATH on MATH. We will now see that in these coordinates the conformal quadratic form is MATH. Take the family of REF MATH given by MATH and consider REF-vector MATH which represents REF-plane MATH . Calculate that MATH . |
math/0101017 | The set of REF containing the MATH axis is given in our coordinates by MATH since these REF-planes must be represented as REF-vectors by MATH. But this is the graph of an isometry MATH and therefore cannot have any tangent vectors in common with a strictly contracting map. |
math/0101017 | The intersections are all negative, but the intersection number is MATH . |
math/0101017 | Suppose that there are two antipodal points MATH that get mapped to the same point MATH. Then any point in the image must be the image of a point MATH lying in a hemisphere about either MATH or MATH. So any point in the image lies in the interior of a hemisphere about MATH. Therefore we need only show that some pair of antipodal points MATH get mapped to the same point MATH. The image cannot contain two antipodal points, because their preimages could be no further apart than antipodal points, and the map is strictly contracting. Therefore, the image misses some point of MATH and the result is clear from the following lemma. |
math/0101017 | Suppose otherwise. Let MATH . This is a continuous map MATH so that MATH . Because MATH is simply connected, there is some continuous map MATH so that MATH . But then MATH for some integer MATH. However, plugging in MATH: MATH which is a contradiction. |
math/0101017 | Rotation by MATH and dilation into a map to a point takes us into the NAME - NAME equations. |
math/0101017 | Smoothly pick a positive definite inner product on each fiber of MATH . The ``center of mass" of the image of each strictly contracting map MATH provides a target to deform to. |
math/0101017 | Suppose that the line congruence is written as the graph of a map MATH . Suppose that the image of MATH is contained in the hemisphere about MATH. Suppose also that we have selected on MATH not just a conformal class of positive definite quadratic form, but an actual quadratic form. Then our MATH can be thought of as a linear function on MATH which is positive on the image of MATH. At the same time, MATH is a REF-form. Take any MATH, that is, MATH . We have MATH so that MATH because MATH is positive on the image of MATH. Therefore MATH is a taming symplectic structure. |
math/0101017 | If MATH on every REF-plane in MATH a compact elliptic line congruence, then pick MATH with MATH . Then MATH so that MATH and MATH on both MATH and MATH so MATH on MATH. If two REF-forms MATH both tame a compact elliptic line congruence MATH then clearly MATH does as well for any MATH . |
math/0101017 | We can arrange, by linear transformation, that MATH is an element of MATH. Then the taming condition on an elliptic line congruence is that, thinking of it as a map MATH its image is contained in the hemisphere of MATH containing MATH. But then we can ``shrink" MATH for example using the Riemannian geometry exponential map about MATH in MATH, which contracts the image and preserves the ellipticity (the strict contractivity) of MATH. |
math/0101017 | We can pick these almost symplectic structures locally, and glue them together using convex positive combinations. |
math/0101017 | As is well known, and proven in CITE, a manifold admits an almost symplectic structure precisely when it admits an almost complex structure, and that almost complex structure can be chosen to be tamed by the almost symplectic structure. Then the NAME - NAME equations of pseudoholomorphic curves for the almost complex structure provide a tamed proper pseudocomplex structure. |
math/0101017 | As for REF . |
math/0101017 | The positivity of intersection holds for NAME spheres, while transversality holds for all compact elliptic line congruences, and we can deform any compact elliptic line congruence into a NAME sphere. |
math/0101017 | Take such a basis MATH . We may rescale it to ensure that MATH . Think of each MATH as a MATH matrix. Using the action MATH we can first arrange that MATH. Now we have to work with the subgroup of MATH fixing MATH which is the group of transformations MATH . Calculate that MATH so that MATH are now traceless. But then MATH must be traceless, have determinant REF, and so has minimal polynomial MATH, that is, is a complex structure. We can therefore arrange by MATH that MATH where MATH . Let MATH be complex conjugation, that is, MATH . Check that MATH and MATH are perpendicular to MATH and MATH, and to one another. So MATH and MATH must be obtained from MATH and MATH by an orthogonal transformation of the plane they span. But we still have the freedom to employ the subgroup of MATH fixing MATH and MATH, that is, the transformations MATH where MATH is complex linear. This enables us to rotate MATH into MATH and then get MATH . But now the orientation of the basis forces the two MATH signs to be equal. |
math/0101017 | Given any positive definite REF-plane MATH we can take a basis MATH for MATH which is orthonormal, up to a positive scaling factor, with MATH spanning MATH and we can use the group MATH to get it to the form MATH . |
math/0101017 | If REF-planes MATH and MATH contain a common nonzero vector, we can take MATH and MATH to be that vector. Conversely, if MATH then by NAME 's lemma MATH for some vectors MATH and MATH . Squaring both sides MATH . Without loss of generality, we take MATH in a basis MATH . We find then that the matrix MATH has vanishing determinant, so has kernel. Then take MATH and MATH so we can pick MATH to make this vanish. |
math/0101017 | Suppose that MATH has a singular point. Pick a complex structure on MATH tangent to MATH at that singular point. Suppose that MATH is represented in the splitting MATH by MATH while the line congruence MATH is represented, at least locally, by a strictly contracting map MATH . Then a point MATH on the graph of MATH represents a REF-plane intersecting REF-plane MATH in at least a line precisely when MATH . We can arrange that the NAME sphere tangent to MATH at this point be MATH . Taking the gradient with respect to MATH, we find that singular points will be precisely those where MATH where MATH indicates transpose. But by the tangency condition, MATH. So therefore MATH . But then MATH so that MATH. This shows that MATH is MATH up to reorientation, so that intersection occurs on more than a line. Hence MATH is not totally real. |
math/0101017 | We can deform MATH to any of its osculating complex structures while keeping MATH totally real. We may have to move MATH while we move MATH but it is easy to arrange that MATH with either orientation, never belongs to MATH during the deformation of MATH since MATH is just a point in the NAME, while MATH is a surface. In the process, we never generate a singular point on MATH. The result is now a calculation for the standard NAME sphere on MATH. |
math/0101017 | The characteristic variety of the tableaux MATH for MATH is the same for MATH as for MATH because MATH. But by definition, MATH . Calculating in complex coordinates, we find that the characteristic variety is the union of the two eigenspaces of MATH, with eigenvalues MATH. This splitting of the characteristic variety (as a real algebraic variety) determines MATH up to sign on MATH, and so MATH on MATH. If we had a minus sign here, that would give MATH the opposite orientation. Similarly, by taking transposes, we find MATH on MATH. |
math/0101017 | Recall that we defined MATH to be the real linear isomorphisms MATH matching orientation. There is a map MATH defined by inclusion. The NAME - NAME REF-form is just MATH. Take MATH a NAME sphere tangent to MATH at MATH, and MATH the associated complex structure. Let us assume that MATH, with complex linear coordinates MATH for simplicity of notation, so that MATH is the standard complex structure, and consider the quantity MATH so that MATH . We calculate that MATH and that again at MATH so that indeed MATH precisely along the directions of MATH, where MATH is the NAME sphere of our complex structure. Consequently, every NAME sphere MATH tangent to MATH at a point MATH must satisfy MATH . This shows us that the relations among MATH near the locus MATH cannot be MATH since these relations would also have to hold on MATH . Indeed, on MATH we still have MATH, since MATH there, but MATH is a submersion, and MATH is semibasic for it, providing a basis for the semibasic forms (check this in complex coordinates). Therefore, near MATH we must find that the relations among MATH can be written MATH . We can follow how these relations behave as we travel up the fibers of MATH. We find that the coefficients MATH can be made to vanish, on a principal left MATH bundle. Moreover, it is easy to see that MATH is precisely this bundle, following the general formalism presented in CITE. |
math/0101017 | Let MATH be a compact homogeneous elliptic line congruence. Then by homogeneity, REF-forms MATH and MATH (which one can calculate are invariantly defined on MATH itself) must be either everywhere vanishing or everywhere not vanishing. Moreover, if one vanishes everywhere, then by REF they both do. But if neither vanishes anywhere, then there is a distinguished subbundle of MATH on which MATH. Using the NAME - NAME equations, we see that this renders MATH well defined on MATH, and thereby forces MATH to have a globally defined nowhere vanishing REF-form, in fact a parallelism, so that MATH cannot be topologically a sphere. |
math/0101017 | We only have to prove that starting from some such MATH we can define MATH as above, and it turns out to be an elliptic line congruence. It is clear that MATH is compact, since it is a subset of a NAME (which is compact) and defined by a closed condition. Define a vector field MATH on the sphere MATH by MATH . This is a nowhere vanishing vector field defining a circle action on the sphere. Moreover, on REF-plane MATH containing a vector MATH and the vector MATH this vector field is just MATH, where MATH is the osculating complex structure. So the flow of MATH on MATH is MATH and in particular the flow curves on the sphere MATH are great circles. The quotient space is clearly MATH. By properness and freedom of the action, MATH is a smooth compact surface. We have to show that MATH is elliptic. This follows from the result of CITE that great circle fibrations of REF-sphere correspond precisely to elliptic line congruences. We won't need this result, so the reader will forgive the author for not providing a complete proof of the results of NAME and NAME. |
math/0101017 | First, take MATH a MATH curve. The prolongation MATH is the set of tangent planes of MATH. We find that MATH restricts to MATH. So MATH and thus MATH . Therefore if MATH and MATH with MATH, then MATH. We have MATH so MATH . We have to show that MATH is MATH-holomorphic. Choose (locally) a coframing MATH from our MATH structure. Since MATH is MATH, we have MATH on MATH, and so MATH. But from the structure equations, this gives MATH . Therefore the MATH forms MATH restrict to MATH to satisfy some complex linear relations, and thus MATH is MATH-pseudoholomorphic. Now take MATH any MATH tangent surface transverse to the fibers of MATH. Since our desired result is local, we can suppose that MATH is a diffeomorphism. Let MATH. By definition, MATH so that MATH and since MATH is a diffeomorphism: MATH . Therefore, MATH, and so MATH is the prolongation of MATH. |
math/0101017 | REF-manifold is constructed by taking the foliation MATH which is locally a fiber bundle, and producing a base space for that fiber bundle. The rest is elementary, following the general pattern of the equivalence method. |
math/0101017 | We have seen in REF that the vanishing of the microlocal invariant MATH forces the vanishing of MATH, and vice-versa, and that vanishing of both forces an elliptic line congruence to be a NAME sphere (in other words, flat). Applying our identification of the elliptic line congruence invariants on a pseudocomplex structure, the result is immediate. |
math/0101017 | The diffeomorphism MATH is constructed first by taking any complex valued coordinates MATH on MATH, for which MATH is a complex line, and then using elementary coordinate changes and changes of MATH adapted coframing to arrange the required equations. To arrange the remaining equations, which are only required to hold at MATH, we write out an exterior differential system for the first equations (which we have already solved), prolong it, check that it is torsion free, and see thereby that there is no obstruction to solving the prolongations of all orders, at least at any required order (but not solving it as a PDE system - we are only solving at, say, second order). The system will look like the above equations together with equations like MATH and so on (quite complicated), describing relations between the various REF-forms. These coefficients MATH can be chosen at will, and of course we choose them to vanish. Then we take a solution at whichever order we like, arranging it to solve all the required equations at MATH, arranging all of these arbitrary coefficients of the prolongations to vanish at MATH, and then repeat the manipulations we used previously to obtain new equations which do not disturb the equations at MATH, as is easy to see. |
math/0101017 | This is identical to the proof in CITE. |
math/0101017 | As in the theory of almost complex manifolds, we have holomorphic polynomials as lowest order terms, in suitable complex coordinates (see CITE). Moreover, the prolongations force the order of these polynomials down at each step, eventually reaching order one. |
math/0101017 | A simple application of NAME 's lemma. |
math/0101017 | See CITE. The basic idea is as follows. First write out the condition on MATH being MATH-holomorphic in a local coordinate MATH on MATH and adapted coordinates MATH on MATH in the form MATH where MATH is a complex MATH matrix, vanishing at MATH. This is possible because the adapted complex coordinates impose a complex structure which agrees at MATH with the almost complex structure MATH on MATH. We apply NAME 's theorem to see that in any disk about MATH in MATH: MATH a triple of holomorphic functions. These intuitively represent the holomorphic part of MATH in these coordinates. One needs to show that the map MATH given in this way is a smooth map of appropriate NAME spaces if the disk MATH is made small enough since MATH is quite small in appropriate norms. |
math/0101017 | Choose any adapted coframing MATH. Now pull them back with MATH. By definition, you get MATH and MATH and MATH are MATH forms, say MATH where MATH is a local holomorphic coordinate on MATH. Then taking differential, we find MATH where MATH . Therefore MATH satisfies a linear first order determined elliptic equation with smooth coefficients, and we can apply NAME 's lemma to show that if MATH vanishes to infinite order at a point, then it vanishes everywhere. But points where MATH are precisely points where MATH is tangent to the fibers of MATH since MATH span the semibasic REF-forms for this map and MATH. Thus MATH can only have a discrete set of zeros, and the set of critical points is discrete. |
math/0101017 | The equation for the lift of the projection MATH is easy to prove, and holds except at critical points. Therefore, matching of MATH and MATH to all orders at MATH implies matching of MATH and MATH to all orders at MATH, and therefore we can apply uniqueness of jets of pseudoholomorphic curves in almost complex manifolds. |
math/0101017 | The projections must be asymptotic to all orders near the critical point MATH. This implies that the prolongations of any order of the projections must be asymptotic to all orders as we approach the critical point. As above MATH at noncritical points. This implies, because the critical points are discrete, that the MATH must be asymptotic at MATH to all orders. Therefore, they must agree by NAME 's lemma. |
math/0101017 | The relevant equivalence problem concerns the pullback of MATH to the lift of MATH, MATH. We can further adapted the coframes MATH of this pullback by asking that MATH vanish on tangent spaces of MATH. This forces structure equations MATH which gives us as structure equations on this bundle MATH which are the structure equations of a complex structure on a surface, and so local equivalence with the flat example follows by the NAME - NAME theorem. Then MATH can be arbitrarily extended off of MATH so that MATH are functions on MATH. |
math/0101017 | If we were to try to prove this result by imitating the almost complex case, we would simply look in coordinates of the type guaranteed by the previous lemma, so that MATH is cut out by MATH. We can use MATH as a local holomorphic coordinate on MATH, and arrange that MATH is the point MATH. Take MATH a local holomorphic coordinate on MATH, so that MATH is the point MATH. Now we have MATH described by functions MATH with MATH having infinitely many zeros near MATH. The lowest order terms of MATH at MATH must be holomorphic polynomials, unless they vanish to all orders. In the almost complex case, this actually determines that MATH itself has holomorphic lowest order term. But then, MATH cannot vanish at infinitely many points approaching MATH, unless this lowest order term vanishes, because it will dominate: MATH would give MATH near MATH. Therefore MATH vanishes to infinite order at MATH. Because of the equation MATH we have MATH and MATH vanishing to all orders in MATH. Now this gives us MATH at MATH, up to infinite order, and since MATH is a MATH form, this tells us that MATH is holomorphic to all orders. We can MATH to show that the formal holomorphic series given by the NAME expansion of MATH must converge (because the terms are controlled, by differentiating this equation, in terms of values of MATH). The result then follows by uniqueness of continuation, using NAME 's lemma. However, in the pseudocomplex case, it is unclear that MATH must have holomorphic lowest order term, simply because MATH does; it might look like MATH with MATH, and then we would have lowest order terms (constant terms) of MATH holomorphic. The problem is essentially to show that MATH, so that in our coordinates the two curves strike in MATH. But this is not too hard: each real line at a point of MATH is contained in a unique MATH plane in MATH. Now take each pair of points MATH and draw a line between them in some local coordinates. Then the MATH planes containing these lines must converge to MATH, the MATH plane tangent to MATH at MATH. But then, we could have used points close to the MATH instead of the actual MATH, or points close to the MATH as well. We can pick points of MATH which are not critical for MATH, and carry out the same construction. This shows that the values of MATH must tend to MATH as MATH. Therefore MATH and MATH have the same osculating complex geometry at MATH, so their lowest order terms are holomorphic there. The story is now the same as in the almost complex case. |
math/0101017 | The proof is essentially as in CITE. |
math/0101017 | See CITE. |
math/0101017 | See CITE. |
math/0101017 | As in CITE. |
math/0101017 | This follows from any of the myriad proofs of NAME compactness, by applying the canonical almost complex structure on the MATH. |
math/0101017 | Use prolongation to all orders. |
math/0101017 | Apply the symmetry to various MATH curves. |
math/0101017 | The proof is due to CITE. The idea is to carry out the above construction of MATH for the initial curve, which is governed by the function MATH, say. Then we invert the construction for the functions MATH corresponding to each MATH. This can only be done locally. |
math/0101017 | By results of NAME, there is only one symplectic structure on MATH up to rescaling, so that we can suppose that our symplectic structure is the usual one on MATH, and from here the rest of the arguments are the same as in NAME 's CITE. |
math/0101017 | This follows immediately from the identification of the equations on microlocal invariants of almost complex structures: we would need MATH which (when we unwind these equations and differentiate the structure equations) forces all of our invariants to vanish; hence local equivalence with the flat case. |
math/0101017 | Suppose that MATH is an embedding. Then we can see from the structure equations that the normal bundle to MATH in MATH, call it MATH is the pullback by MATH of the vector bundle associated to the principal bundle MATH and the representation MATH while the canonical bundle of MATH with respect to MATH is associated to the representation MATH and the canonical bundle of MATH is associated to MATH . Therefore MATH which gives the result immediately: MATH . The problem in the nonembedded case comes from MATH, the self intersection number, not being identical to MATH. We might imitate the complex proof and carry out blowup, but the blowup is not a pseudocomplex manifold. To deal with this, we can follow NAME 's arguments, making a deformation of pseudocomplex structure, say as MATH, near each singular value of MATH, and deforming MATH as we do, say as MATH, so that MATH and MATH is complex near each singular point of MATH. Then we can employ blowup at each singular point. |
math/0101017 | Suppose that our pseudocomplex manifold fails to be NAME hyperbolic, so that we have a basic parameterized MATH curve MATH . Take a sequence of concentric disks in MATH of arbitrarily large radius. Impose the NAME metric on each, conformally matching the induced metric. Then any pair of points of MATH in the image of MATH are joined by NAME chains of arbitrarily small length. |
math/0101017 | The argument is identical to NAME 's in CITE. We sketch it. Suppose that MATH is our proper pseudocomplex structure on a compact manifold MATH. Take any Riemannian metric on MATH. Given a sequence of parameterized MATH disks, MATH with arbitrarily large velocity vectors, CITE presents an argument that shows that we can replace these MATH with new parameterized MATH disks MATH so that MATH is largest at the origin of the disk. and so that the norm of the velocity of MATH becomes arbitrarily large. We can then reparameterize to produce, just as NAME does, a sequence of parameterized MATH disks MATH, using disks MATH in the complex plane of arbitrarily large radii, so that the velocities of these MATH disks are bounded. Since their first derivatives are largest at the origin, this ensures that their first derivatives are bounded, so that some subsequence converges uniformly. The uniformity ensures by REF that they converge uniformly with all derivatives to a MATH line MATH . |
math/0101017 | Take an infinitesimal symmetry MATH . The flow of MATH is complete, by compactness, for MATH any infinitesimal symmetry. Not every flow curve of MATH is actually a parameterized MATH curve for MATH. However, if we pick a flow curve MATH of MATH, passing through a point where MATH, it is easy to see that it will be a MATH curve, and that its lift MATH will be a flow curve of MATH. The completeness ensures that its domain will be the entire complex line. Therefore MATH must vanish everywhere. By ellipticity of the MATH structure on MATH, the group of symmetries of a proper pseudocomplex structure on a compact manifold is a finite-dimensional NAME group (see CITE). But without any nonzero infinitesimal symmetries, it must be zero dimensional, so a discrete group. By the uniform convergence theorem for symmetries REF , we known that a uniform limit of symmetries converges with all derivatives. Suppose that we have a sequence of symmetries. We can bound their derivatives by taking a MATH disk, and looking at how its velocity gets stretched under a symmetry. The stretching factor must be bounded, or we will be able to recursively apply the symmetry to generate MATH disks of arbitrarily large velocity. This bounds the derivative of the symmetry, since there is a MATH disk with any velocity close enough to MATH, by continuity. Therefore any sequence of symmetries have a convergent subsequence, so the symmetry group is compact. Because it is also discrete, it must be finite. |
math/0101017 | The proof consists in an elementary limit argument. See CITE. |
math/0101017 | The proof is the same as in CITE, using the quasiminimality of MATH curves, the pseudoconvexity of small balls (see REF), and NAME compactness. |
math/0101017 | Nearly tangent means that the tangent plane of the MATH curve should be close to MATH, where MATH is the foliation. Without loss of generality, suppose that the MATH curve actually passes through a point of MATH. It is easy to calculate the Laplacian of MATH in terms of the subbundle structure equations: MATH so that if MATH is a holomorphic coordinate on a MATH curve, and MATH, then MATH on the MATH curve. |
math/0101017 | As in the complex case. |
math/0101017 | Take a function MATH, and pull it back to MATH. Now differentiate it: MATH and then take its second derivative: MATH . If MATH then there is a REF-plane in MATH on which MATH and on which MATH and another on which MATH . By local solvability of elliptic partial differential equations, we can then find MATH curves on which the Laplacian takes either sign. Therefore we will need MATH at each point to have plurisubharmonicity of MATH. But it is easy to calculate that MATH forces MATH to be constant. Therefore we need MATH. By REF (assuming MATH proper), this occurs precisely for MATH almost complex. Conversely, if MATH is almost complex, then in adapted coordinates the function MATH is plurisubharmonic near the origin. |
math/0101017 | Take adapted complex coordinates MATH as in CITE, so that the totally real submanifold becomes Lagrangian (indeed, we can ask that it become the set of real points of MATH). Then integrate the NAME form MATH which must be positive on a ``small" pseudoholomorphic curve, because the curve is nearly holomorphic. But the boundary sits in a Lagrangian manifold, which (if we use a small enough neighborhood) contradicts the quasi-minimality of the pseudoholomorphic curve. |
math/0101017 | Looking upstairs in MATH, a MATH curve with boundary in a totally real surface MATH becomes a pseudoholomorphic curve in MATH with boundary in MATH. |
math/0101017 | This requires only calculating examples of surfaces with arbitrary REF-jet, since it depends only on REF-jet of any surface. For more details, see CITE. |
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