paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9911172 | First, on the points MATH the relative NAME inequality is MATH. On the other hand, a generalized NAME band algorithm similar to the generalized NAME algorithm for the standard generators described in REF produces a surface MATH with MATH. Hence, we know that the relative NAME inequality is an equality for all points MATH and in addition for MATH. Thus, by REF , the NAME norm is additive on all points MATH with MATH. Since we know by REF that the relative NAME number is also additive on these points, we are done. |
math/9911172 | We will apply the following elementary equations: MATH . If the proposition is proven for the terms on the right hand side, then it is also true for the ones on the left hand side. First, the proposition is true in MATH, that is, for braids MATH or MATH by induction on MATH. For the representative of the unknot MATH it is also true. It is enough to prove the claim, that every homogeneous braid in MATH can be brought with the transformations REF and the braid relations MATH into one in which MATH occurs at most once, modulo words of lesser length. For this let the claim already be proven for MATH. Thus, we can assume that a braid MATH, MATH, modulo terms of lower word length, has in MATH at most one term MATH. If MATH is empty then with REF or REF we can reduce the word length. Otherwise, since MATH commutes with MATH for MATH, we can apply REF or the braid relations to bring MATH into a form in which MATH occurs less and which is still homogeneous. |
math/9911172 | By REF and with the notations there, we know that the minimum degree MATH of the HOMFLY polynomial is at most MATH. Since MATH we have MATH . The last equation follows, since MATH. |
math/9911172 | Let again MATH and MATH. With these band generators, the braid group MATH has a presentation MATH . It is known by work of CITE and of Xu CITE that a NAME surface with maximal NAME characteristic is realized by a shortest word in these three generators. This means, if a closed MATH-braid MATH is represented by a shortest word of length MATH then a minimal spanning surface MATH has MATH. It is the surface that we get by the band NAME algorithm described in REF. Now NAME 's conjecture claims that the NAME number of MATH and MATH of a minimal spanning surface for MATH differ at least by the minimum degree of MATH in MATH. By the computation of MATH it is clear that for a MATH-braid the only possible powers of MATH in the monomials of MATH are MATH, MATH, MATH, and MATH. Furthermore, MATH is divisible by MATH and thus the minimum degree of MATH in MATH can be at most MATH. Now, let MATH be a MATH-braid, which is of minimal length, with exponent sum MATH over all negative exponents and MATH over all positive exponents. Thus, the difference between the NAME number and MATH is MATH. So, we only have to check NAME 's conjecture for braids MATH with MATH. First assume, that MATH is a positive word in the generators. Our claim is that the highest degree in MATH is MATH where MATH is the word length. Moreover, we claim that coefficient of MATH which is a polynomial in MATH has constant term MATH. If the word length is one then, as one might easily check, MATH. If the word length is two, the verification of the claim is an easy case-by-case check as well. Assume that the word length is at least MATH. If a square of a generator occurs, then for any of the generators MATH we have MATH and the claim holds by induction. Since we have the relations REF we can thus assume that any subword of length three is of the form MATH or a cyclic permutation of it. Now MATH . By the induction hypothesis the claim follows. By cyclic permutation the claim also follows for the other two words of length three. Now assume, that MATH. So, the closed braid can be assumed as MATH where MATH is a positive braid, with respect to the band generators, and MATH is one of these generators. Again the braid is assumed to be of minimal length. Let MATH, MATH and MATH be minimal spanning surfaces for the closed braids MATH, MATH and MATH, where MATH is a generator of MATH. Essential to our arguments is REF and CITE. Two of the three values MATH are the same and the third one is not bigger than the other two. We use the theorem in the following way: In each application of the skein relation REF one of the three terms in REF is strictly less than one of the others. Hence, the third one has to be equal to the larger one. Our claim is that the highest degree in MATH is MATH, where MATH is the minimal word length; this equals MATH of a minimal spanning surface for the braid. Its coefficient has a non-trivial term in MATH. So, NAME 's conjecture follows from this. If the braid MATH contains a square in MATH, then we can reduce the braid with the help of REF . Otherwise, we can assume that every positive subword of length three is of the form MATH or a cyclic permutation of it. If the word length is one then MATH. If the word has length two then an easy computation shows that MATH. Now assume, that the word starts with MATH . We have for some positive word MATH . By the result of NAME and NAME the word in the second term in the last equation has to have minimal braid length and thus the claim follows by induction. With the same argument the claim follows if the word starts with MATH or MATH. If the word starts with MATH it has to be of length at least MATH, otherwise it wouldn't be minimal. Since we already assumed that it is square-free, it has to start with MATH, thus MATH . By induction and using REF and NAME, it follows now that the highest term is coming from MATH. If MATH or MATH are the initial subwords then the claim follows in the same way. This completes the proof. |
math/9911172 | For MATH odd let MATH be MATH. Since MATH is homogeneous the closure of MATH has as maximal NAME characteristic over all spanning surfaces MATH. Obviously MATH is equivalent to its mirror image. It follows immediately from the last section that the difference between the maximal NAME number among any representative of MATH and the NAME number of the special MATH is at most MATH. Thus, MATH is less than or equal to MATH. The claim follows. It can actually be easily shown that if the braid has the form MATH with positive MATH and MATH then the maximal possible NAME number among all representatives of the closure of the braid is MATH. For this we show that the polynomial MATH does not vanish for MATH. Thus, with REF we get the desired result. We already know by the proof of REF that the claim is true for the closed MATH-braid MATH. By REF it follows that MATH . Hence, using the defining relations for MATH we get MATH . By induction we can conclude that every MATH is a polynomial MATH . For MATH odd, all coefficients in MATH are negative, and for MATH even, all coefficients are positive. Thus, cancellation cannot occur and MATH is nontrivial for all MATH. |
math/9911174 | REF is obvious. CASE: The only defining relation of MATH that depends on the orientation of MATH is the skein relation. Reversing the orientation of MATH interchanges the parts MATH and MATH of a link which can be compensated by replacing MATH by MATH. This way we obtain REF by using REF . |
math/9911174 | The MATH-module MATH can be described by non-commutative generators MATH REF and the Relations MATH for all MATH. By the Relation MATH we have MATH for all MATH. The Relation MATH then implies that MATH is well-defined. If MATH and there exists MATH with MATH and MATH, then MATH implying MATH. If MATH, then MATH implies MATH. Therefore MATH is surjective. Since we know a presentation of MATH by generators and relations, it is easy to verify that an inverse map of MATH is well-defined. |
math/9911174 | Let MATH be a link and let MATH, MATH be ordered based links obtained from MATH by choosing basepoints MATH and MATH on the components of MATH that are pulled into MATH along paths MATH and MATH respectively. We first assume that all the paths MATH and MATH are disjoint. Then we can pull the points MATH on MATH into MATH along MATH and the points MATH on MATH into MATH along MATH. This shows that we can pass from MATH to MATH by isotopies of ordered based links and by application of the Relations MATH and MATH. The Relations MATH and MATH only influence signs and MATH changes signs simultaneously with MATH. If the paths MATH and MATH are not disjoint, then we choose basepoints MATH of MATH and paths MATH disjoint to MATH and MATH and apply the argument above to the pairs MATH and MATH. This shows that the maps MATH and MATH are well-defined isotopy invariants of links. It follows from the skein relation by the same arguments as for the usual NAME polynomial of links that the maps MATH are NAME invariants of degree MATH (see CITE). |
math/9911174 | Strongly positive amphicheiral knots in MATH are in one-to-one correspondence with knots MATH in MATH with MATH. To see this, recall that the set of fixed points of an orientation-reversing involution MATH of MATH is either MATH or MATH (see CITE). For strongly positive amphicheiral knots MATH it is sufficient to consider involutions MATH with fixed point set MATH and MATH. By REF (see CITE, CITE), we then have MATH. The corollary follows by applying REF to the image of MATH in MATH. |
math/9911174 | Let MATH be a diagram of an ordered based link. For the proof we use the following strategy: we will never increase the number of crossings of MATH, and make computations modulo diagrams with fewer crossings. This allows us to make crossing changes. We will prove the lemma by induction, where the induction base is given by the following arguments for link diagrams without crossings, and the induction step is given by the same arguments for link diagrams with crossings. Case MATH, REF : We apply an isotopy that does not increase the number of double points in MATH such that the projection of MATH to MATH consists of parallel strands connecting points in MATH with points in MATH. Then we pass from MATH to a diagram of a product of knots by crossing changes. Case MATH, REF : Each component of MATH represents a unique word in the generators MATH of MATH. We claim that we can replace each of these components by a diagram of a based knot MATH representing a cyclically reduced word in MATH: for words that are not cyclically reduced we find a segment MATH of MATH corresponding to a cancellation such that MATH connects two points MATH for some MATH and all the parts of MATH that go into MATH between MATH and MATH intersect MATH (see REF ). The part MATH can first be moved to a suitable height by making crossing changes and then be pulled back by applying an isotopy that does not increase the number of crossings as shown in REF . This will remove one cancellation from the word in the generators MATH corresponding to the knot projection. We continue by induction on the number of cancellations to pass to a cyclically reduced word. Case MATH, REF : Recall the definitions of MATH and MATH for MATH REF . We continue with the modification of each based knot MATH in MATH by moving the basepoint of MATH until MATH where MATH. By the Relation MATH this only contributes a factor MATH. We can further achieve that the projection MATH of the basepoint of MATH to MATH lies in MATH. Assume that there are points in MATH to the left of MATH in our pictures of MATH. Let MATH be the left neighbor of MATH in MATH. Notice that there may be points of MATH between MATH and MATH. Let MATH be the element of MATH that is represented by MATH with basepoint MATH and with an orientation such that at MATH the curve MATH enters the band. By the definition of the representative MATH we have MATH. Assume first MATH. We have MATH. Therefore, following the two strands of MATH starting by entering a band at MATH and MATH, we follow paths MATH and MATH respectively that pass through the same bands until we find a crossing MATH between them. Denote that part of MATH by MATH whose projection lies in a small neighborhood of the triangle bounded by MATH, MATH, and MATH with corners MATH, MATH, and MATH. By crossing changes we pass to a suitable height on MATH and then pull back this part by an isotopy as shown in REF until the points MATH and MATH are interchanged. The projection of MATH arrives very close to MATH in MATH such that no crossings with the remaining part of MATH can appear. If MATH, then we push the basepoint along MATH until its projection arrives at MATH. Therefore in any case this modification does not increase the number of crossings of MATH. This implies that by induction we can assume that there are no points in MATH to the left of MATH. Then we pass from MATH to a descending knot by crossing changes. Since this works for all components of MATH we have proven the lemma in the case where MATH. CASE: We choose an open disc MATH. Then MATH is the NAME strip. It is easy to see that every link in MATH can be represented by a link in MATH and that a descending link in MATH is isotopic to a descending link in MATH. Therefore we conclude by using this lemma for MATH. |
math/9911174 | For MATH there is nothing to prove. Assume that MATH. Let MATH be a link in MATH. By the first two steps of the proof of REF we can assume that the connected components of MATH represent cyclically reduced words in MATH. Applying the following REF to all components MATH of MATH will imply the lemma by induction on the number of crossings. REF ("Collect all starting points on the left"): Choose a maximal MATH such that MATH where MATH and MATH. Assume that MATH (otherwise there is nothing to do). Consider the MATH points MATH on MATH such that MATH and MATH with each of these points represents MATH. We want to move all the points MATH to the left side in MATH without increasing the number of crossings of MATH: if this is not already the case, then choose MATH such that the point MATH to the left of MATH in MATH is not one of the points MATH. Let MATH be the element of MATH represented by MATH with respect to MATH and oriented such that it leaves MATH at MATH. Since we use this orientation of MATH the word MATH is obtained from MATH by a cyclic permutation. The minimality of MATH and the maximality of MATH implies MATH. We have MATH. Therefore, we can follow MATH on two strands along MATH REF starting at MATH and MATH until we find a crossing MATH between these two strands. As in REF of the proof of REF we make crossing changes and then pull back this crossing and possibly some other parts of MATH by an isotopy until the points MATH and MATH are interchanged (see REF ). With this modification we do not increase the number of crossings of MATH and the point MATH has moved one step to the left in MATH. We will now verify that no other point MATH has moved to the right during this operation: assume that we reach MATH after starting at MATH and traveling along MATH with MATH. Then for MATH we either have MATH or MATH. We can exclude the second case because this would imply MATH which is impossible. By the minimality of MATH and the maximality of MATH we have MATH and MATH. From MATH we deduce MATH. Then MATH and MATH implies MATH. Therefore, starting at MATH and MATH and traveling along MATH, the first strand reaches MATH to the right of the second one. This means that by pulling back the crossing MATH we move the two points MATH and MATH to the left (see REF for an example). We can continue by induction until all points MATH are on the left side of MATH. REF ("move component into cabling position"): Successively for MATH, we consider all parts MATH of MATH such that the projections MATH connects some interval MATH with MATH inside of MATH. By crossing changes we move all the MATH to a suitable height and then push the part MATH containing all crossings between these strands across the band at MATH. This isotopy can be chosen in the following way such that the total number of crossings of MATH does not increase: after being pushed across the band the part MATH arrives very close to MATH and the strands that now enter into MATH without intersections in their projection to MATH can be moved into a position such that they coincide with original parts of MATH except in neighborhoods of the crossings and except for orientations. After having done the previous modification for MATH we make crossing changes such that MATH, where MATH is a knot in MATH. Applying the arguments in the proof of REF for MATH to MATH we see that MATH is equal to MATH modulo diagrams with fewer crossings. |
math/9911174 | The proof of REF shows that the MATH-module MATH is generated by the links MATH with MATH, MATH and MATH. Consider links MATH of the form MATH where MATH and MATH are ordered based links and MATH with MATH. For MATH we always have MATH. As shown in the proof of REF we may assume that in a diagram of MATH the projections of the components MATH and MATH of MATH represent cyclically reduced words in MATH. Furthermore, we may assume that the projections MATH and MATH of the basepoints of MATH and MATH lie in MATH. When we follow the two strands of MATH starting by entering a band at MATH and MATH, we follow paths MATH and MATH respectively that pass through the same bands until we find a crossing MATH between them. We pull this crossing back as shown in REF . We continue with the modification of MATH as in REF of the proof of REF . We obtain that MATH is equal to MATH modulo diagrams with fewer crossings, where we have MATH and where we can choose the embedding MATH such that MATH is a knot descending with respect to its basepoint (but in general not a descending knot). The link MATH lies in the ideal MATH. We obtain the lemma by induction on the number of crossings. |
math/9911174 | CASE: Since MATH is even and the diagram of MATH is as shown in REF we can pass from MATH to a product MATH with MATH by crossing changes and by Relation MATH. If a crossing of MATH is spliced, then we obtain again a closure of a tangle as in REF . By induction on the number of crossings of MATH we can expand MATH as a linear combination over MATH of monomials MATH with MATH. Under the hypotheses of the lemma all these monomials equal MATH in MATH. CASE: Let MATH be as in REF. When we push the basepoint of MATH along the generator MATH of MATH and MATH is odd, then Relation MATH contributes the sign MATH. If MATH is even, then we will see in the following that we will push the basepoint along MATH an even number of times, because the basepoint is on one of the first MATH strands of MATH. This will not give a sign contribution either. If the basepoint passes through a crossing, then by changing this crossing the knot will become descending with respect to the new basepoint. If the crossing is spliced, then we obtain a product of two knots that are descending with respect to their basepoints. One of these knots is isotopic to MATH or satisfies the conditions from REF. Therefore we do not change MATH if we push the basepoint along MATH and change crossings until the basepoint is close to the left boundary of MATH in our pictures. This implies MATH if MATH is homotopic to MATH (the number MATH satisfies MATH and MATH). There exists a diagram of MATH as a closure of a tangle as in REF on MATH strands having MATH crossings, such that by changing arbitrary crossings and by splicing one arbitrary crossing we obtain a product of two knots (see REF for MATH). Using this we obtain MATH by similar arguments as above. |
math/9911174 | We prove the lemma by induction. We have MATH. Assume that the lemma is true for all MATH with MATH and MATH. We have to prove the lemma for MATH. Since inversion of the orientations of all components of a link induces a MATH-linear involution of MATH, we can assume without loss of generality that MATH and MATH. We will prove the following two equations: MATH REF imply that MATH for some MATH. Therefore MATH. This together with REF implies also MATH. We prove REF for MATH in REF and explain afterwards why this equation holds for arbitrary MATH. The first equality in REF follows in general from REF for MATH and from Relation MATH. The new knot is descending with respect to its basepoint. The second equality follows by applying the skein relation successively MATH times to crossings that appear when the new basepoint is pulled to the left. When all crossings are changed, then we obtain again MATH. When one of the MATH crossings is spliced, then we obtain a product of two descending knots each of which satisfies the conditions of REF for odd MATH. This gives us the MATH product links MATH where MATH runs successively over the numbers MATH. Looking at signs we obtain REF . Proof of REF : The link MATH is the closure of a tangle as shown in REF . In this diagram the crossings between the two components of MATH are numbered from MATH to MATH. The position of the basepoints on MATH is not important. Let MATH REF be the knot obtained by changing the crossings MATH and by splicing the crossing MATH in REF (see REF ). For later use some crossings of the diagram of MATH are marked by the symbol MATH in REF . Since we can permute the two components of MATH by changing all crossings between them we obtain by the skein-relation and by Relation MATH that MATH for some basepoint on MATH and some signs MATH. REF follows from REF , the induction hypothesis and the subsequent lemma. |
math/9911174 | A binary tree with root MATH is given by the following recursive description: CASE: We start at the top left of the diagram of MATH and follow the orientation of the strand until we come back to the point where we started or until we travel along an undercrossing strand of a crossing where we did not travel along the overcrossing strand of that crossing before. REF If we come back to the starting point on MATH in REF , then the tree for MATH consists of a leaf labeled by MATH. REF If we reach a crossing first as an undercrossing in REF , then we apply Relations MATH and MATH and then a skein relation to this crossing. We obtain MATH for some MATH. The tree for MATH consists of a root vertex connected to a tree for MATH and to a tree for MATH. The knot MATH is equal to a weighted sum of the labels MATH of the leaves in this tree with weights MATH, where MATH is equal to the number of crossings spliced on the path in the tree leading from the root to MATH. The leaves are labeled by links MATH that are products MATH, where MATH is a descending knot. We will further examine the paths in the tree leading from the root to a leaf. There is a unique path in the tree where no crossing is spliced and the label of the leaf at the end of this path is MATH. If a crossing of MATH is spliced, then the first one has to be one of the crossings marked by a MATH in REF . Let MATH be the resulting link diagram (see REF ). For MATH (respectively, MATH) let MATH be a point on MATH near the MATH-th endpoint at the bottom (respectively, at the top) of MATH. Traveling along MATH as in REF of the description of the tree for MATH we first pass through a strand of the crossing that was spliced by passing from MATH to MATH. We label a point on this strand by MATH and a point on the other strand of this spliced crossing by MATH (see REF ). When we continue to travel along MATH we will arrive first at the point MATH and then at the point MATH on MATH without the need of a further modification of MATH. Forgetting the component MATH of MATH containing the points MATH, we obtain a knot MATH. The upper and lower boundary points of the diagram of MATH consist of a subset of the corresponding boundary points of MATH in REF , the MATH-th upper and lower boundary point does not belong to MATH because MATH lies on MATH, and the strands of MATH to the right of the point MATH do not intersect. This implies that MATH is the closure of a tangle on MATH strands as in REF . The link MATH is not the product of MATH and MATH, but the point MATH lies between the MATH-th and MATH-st strand of MATH and when we travel along MATH from MATH back to the starting point we cross MATH only in strands to the right of MATH. This implies that no matter how MATH is modified in the passage from MATH to a leaf MATH, the component of MATH containing the points MATH will always be homotopic to MATH for some MATH. We have MATH for MATH. If the label MATH of the leaf of the tree is of the form MATH for some link MATH, then MATH. |
math/9911174 | The cabled descending knots MATH with MATH are indexed by MATH. By definition they are of the form MATH for some MATH with MATH and MATH. By REF we have MATH implying MATH. Also by REF , we have MATH implying MATH for an arbitrary embedding MATH. This shows that the ideal MATH of REF is contained in MATH. It is easy to see that for all MATH-manifolds MATH we have MATH. For non-orientable MATH we have MATH and MATH. The three arguments from above together with REF imply the proposition. |
math/9911174 | Changing the order of two circles or pushing a label around a circle in a standard embedding of labeled circles as in REF gives the same sign contribution to MATH and to MATH. So MATH is well-defined for chord diagrams. We have to verify that MATH respects the defining relations of MATH: The map MATH respects the relation shown in REF , where any order of the shown parts of the diagram is possible, because one can slide one thickened chord as in REF along the other one (see REF). This relation implies the Relation REF . We can easily verify the compatibility of MATH with the Relations (Rep), (Bas), and (Ord). The compatibility of MATH with the Relation (MATH-Nat) follows from REF which shows an equation between elements of MATH defined by immersed labeled circles as in the definition of MATH. The formula in REF holds because when we push the labels MATH to their new positions, then in the projection to the horizontal line they are commuted with the same labels from the remaining part of the diagram and the commutation between the MATH on the right side with the MATH on the left side gives the sign contribution MATH. |
math/9911174 | Let MATH be a skein triple of ordered based links. Let MATH and MATH. Define the composition of REF by placing MATH onto the top of MATH. In the recursive description of MATH (see REF ) a chord MATH between two parallel parts of the oriented circles is replaced by MATH. By the explicit description of MATH and by REF we have for some MATH: MATH . It is obvious by Relations MATH and MATH for ordered based links and chord diagrams that MATH is well-defined on MATH. Since MATH has no non-trivial torsion elements as a MATH-module, MATH descends to MATH. REF follows because for ordered based links MATH whose MATH-th component is homotopic to MATH we have MATH where MATH is a product of MATH oriented circles with a single label MATH and MATH. |
math/9911174 | CASE: The uniqueness of MATH follows because the MATH-module MATH has a trivial torsion submodule and the condition MATH for descending links MATH with MATH prescribes the image of the generators MATH of MATH (see REF ). Let us prove the existence of MATH. REF implies that the set MATH is a basis of the MATH-module MATH. Define a MATH-linear map MATH on basis elements by MATH . The map MATH is surjective. Consider the composition MATH . By REF we have MATH for all ordered based links MATH. This implies that MATH is mapped injectively to the basis MATH of the MATH-vector space MATH. Considering inductively the sets MATH we see that MATH is linearly independent over MATH (where MATH). This implies that MATH is injective. Hence MATH is an isomorphism and the map MATH has the property MATH for all MATH-descending links MATH with MATH. We obtain MATH as the composition of the canonical projection MATH with MATH. |
math/9911174 | CASE: Let MATH be the NAME strip MATH. Let MATH be a MATH-descending link. If MATH, then MATH by REF implies MATH because MATH factors through MATH. It follows easily from REF that MATH for all MATH-descending links MATH. REF implies this formula for all ordered based links MATH. The case MATH of the theorem can be deduced from the case MATH. For MATH and MATH . REF is trivial. CASE: It is enough to show MATH for MATH. Let MATH be a diagram of an ordered based link in MATH. Let MATH be the number of crossings of MATH and let MATH be the minimal number of crossing changes needed to pass from MATH to a diagram of a MATH-descending link. We prove the theorem by induction on the lexicographical order on MATH. Assume that MATH. Then MATH is a product of descending knots. By REF we have MATH if MATH. Assume that MATH. Then REF imply MATH . If MATH, then we also have MATH and we are back in the first case. Now let MATH and MATH. Choose a crossing of MATH such that MATH, where MATH is obtained from MATH by changing this crossing and let MATH be obtained by splicing the crossing. Then for a MATH we have by induction MATH . In this computation we used that for a skein triple MATH of links we have MATH and MATH is also a skein triple. |
math/9911174 | Applying the MATH-relation to the wheel with a dot we obtain the two diagrams MATH and MATH shown in REF . By REF we then have MATH because MATH. |
math/9911174 | The proof is devided into four steps. CASE: Assume that a component of MATH is a wheel with a dot. Then MATH (by REF it is sufficient to verify this for a degree-MATH wheel with a dot. This case follows from a straightforward compuation). CASE: We show now that when the lemma holds for MATH, then it also holds for all diagrams MATH with MATH where MATH is connected and MATH is not a tree. If MATH is a wheel with a dot then we have MATH by REF of this proof and REF . If MATH is not a wheel, then REF follows from REF . If MATH is a wheel of odd degree, then REF is implied by the MATH-relation, the previous case (where MATH was not a wheel), and REF . Finally, if MATH is a wheel of even degree without a dot, then we have MATH by REF which implies MATH . CASE: Now assume that the lemma holds for a diagram MATH and consider MATH such that MATH is a tree. We call the tree MATH a comb if we cannot apply REF to the part MATH of the diagram MATH. By REF we only need to consider combs of degrees MATH, MATH, MATH, and MATH. By the MATH-relation and REF of this proof, we may arrange the univalent vertices of MATH on the skeleton MATH in any order we want to. We then apply Relation REF to reduce the configuration of labels we need to consider. In the following we investigate the remaining possibilities. MATH . By Relation MATH and REF there is nothing to prove in this case (alternatively, there is a direct proof similar to the case MATH). MATH . We only need to consider the case shown in REF , where we apply Relations MATH and MATH to reduce this case to REF of this proof. MATH . All possible configurations of MATH can be reduced the case shown in REF , where we prove that MATH. MATH . By REF we have MATH because MATH is a linear combination of elements MATH such that MATH or MATH has an isolated part of odd degree. MATH . It is sufficient to consider a comb as shown on the left side of REF . MATH . By the MATH-relation and Part MATH of this proof, we can equivalently consider the linear combination of two diagrams shown on the right side of REF . But by Relation MATH this element equals MATH. CASE: Conclusion: For diagrams of degree MATH the lemma is obvious. Using REF of this proof the lemma follows by induction. |
math/9911174 | Define MATH-valued maps MATH, MATH by MATH and MATH. Let MATH be a based knot with MATH and let MATH be MATH without its basepoint. Then we have MATH where the first equality follows from REF , the second equality follows from REF , and the last one follows because MATH is group-like. It remains to verify that REF are compatible with the replacement of parameters in the proof of REF : by REF there exist MATH such that MATH . By REF we have MATH . As a simple consequence of the construction of MATH we have MATH . REF - REF imply MATH what we wanted to prove. |
math/9911175 | We first determine the naive scattering function associated with the pair MATH. Let MATH be the corresponding intertwiner to the outgoing spectral representation. The map MATH given by MATH is a unitary embedding (commuting with multiplication by MATH). It is in fact onto, as by a well-known result the zero set of MATH has NAME measure MATH. It sends MATH to MATH which by REF is also equal to MATH, with MATH the inner part of MATH (an inner function on the circle is a measurable function of modulus MATH which is almost everywhere the non-tangential boundary value of an analytic function in the interior of the unit disc, itself bounded in modulus by MATH). Division by MATH is a unitary, so the operator MATH is given as MATH from MATH to MATH. In the same manner the ``incoming spectral representer" MATH is just multiplication by MATH from MATH to MATH. The naive scattering function is thus MATH. And the looked-for MATH is its inverse. The conclusion follows . |
math/9911175 | First of all, let us recall the expression for the zeta function MATH which exhibits it as a rational function (see CITE) MATH and a straightforward calculation then shows the following identity in MATH: MATH . The spectral measure of MATH is thus (note that MATH) MATH . It can be written as MATH for MATH which belongs to MATH as the only pole of MATH in the open disc is at MATH. We now compute the naive scattering function as MATH where MATH and MATH were used. The functional equation reads MATH which translates into MATH . We obtain, for MATH: MATH . Let us assume that MATH is in fact an inner function. The rational function on the right - hand side can be inner only if it has no poles in the open disc. But this means that MATH has no zeroes (as MATH and MATH are not acceptable candidates). This is just a way of phrasing the NAME Hypothesis. And conversely under the NAME Hypothesis the naive scattering function is MATH which is indeed an inner function . |
math/9911181 | This follows from the fact that if MATH is NAME split and MATH is a line bundle on MATH, then there is an injective map MATH of abelian groups. |
math/9911181 | Assume that MATH and MATH are two different prime ideals in MATH lying over the prime ideal MATH in MATH. Let MATH and MATH denote the corresponding subvarieties of MATH. Then MATH. Choose MATH. Since MATH, MATH. But MATH and MATH both map to the same point in MATH, which is a contradiction. This proves the first part of the statement. Let MATH denote the irreducible subvariety of MATH corresponding to the prime ideal MATH in MATH lying over MATH. Let MATH be an element generating the maximal ideal in the local ring MATH. Choose MATH such that MATH . As the product MATH is MATH-invariant, we can find MATH and a positive integer MATH such that MATH . Hence, we get MATH from which we obtain MATH (observe that MATH and MATH). Assume, if possible, that MATH. Then replacing MATH by MATH and MATH by MATH, we can assume that MATH. Take a nonzero MATH such that MATH (for example, MATH for an element MATH not invariant under MATH). Since MATH is acting trivially on MATH, it acts trivially on MATH and hence MATH belongs to MATH (here we are using the assumption that MATH). Write MATH for MATH. Applying MATH we get, MATH . But, by choice, MATH and hence MATH. In particular, MATH. A contradiction, proving that MATH . |
math/9911181 | That there exists a unique (reduced and irreducible) divisor MATH in MATH mapping onto MATH follows from the corresponding local statement in REF . Let MATH be a section of MATH with scheme theoretic divisor of zeros MATH equal to MATH. We want to show that MATH has divisor of zeros equal to MATH. But this can be checked locally, and the local statement follows from REF . |
math/9911181 | Let MATH denote the MATH-th ( where MATH) exterior power of the differential MATH of MATH, and let MATH denote the corresponding global section of the line bundle MATH. We want to show that the scheme theoretic divisor of zeros MATH of MATH is equal to MATH: Let MATH denote the complement of MATH in MATH. Then MATH is an open subset of MATH on which MATH acts freely. The restriction of the quotient map MATH to MATH is hence étale. In particular, the support of MATH must be contained in MATH. As MATH is irreducible and MATH is effective, there exists a non-negative integer MATH such that MATH. We have to show that MATH: This can be done locally around a point in MATH, so we may assume that MATH and MATH are affine with coordinate rings MATH respectively. By REF , the image MATH (with the reduced closed subscheme structure) is isomorphic to MATH. We may therefore think of MATH as a closed (irreducible) subvariety of both MATH and MATH (of codim. MATH). Let MATH (respectively, MATH) denote the prime ideal of height REF in MATH (respectively, MATH) corresponding to MATH. Choose MATH (respectively, MATH) generating MATH (respectively, MATH) in the local ring MATH (respectively, MATH). By REF , we know that there exist MATH such that MATH . Replacing MATH by MATH and MATH by MATH, we may assume that MATH. Hence MATH. Now choose a point MATH in MATH such that CASE: MATH is smooth at MATH. CASE: MATH. CASE: MATH (respectively, MATH) is generated by MATH (respectively, MATH) in the local ring MATH (respectively, MATH), where MATH (respectively, MATH) is the maximal ideal corresponding to MATH in MATH (respectively, MATH). (Since all these three conditions are separately valid on dense open sets in MATH, such a MATH indeed exists.) As MATH (by the choice of MATH) is smooth at MATH, the local ring MATH is regular. We can therefore choose elements MATH generating the maximal ideal in this local ring. Hence MATH (respectively, MATH) is a generator of MATH (respectively, MATH) at MATH. Let MATH be the element such that MATH . Then MATH . Noticing that MATH is a unit in MATH (by the choice of MATH), it follows that MATH (since MATH is the exponent of MATH on the right side of REF above). |
math/9911181 | This follows easily since MATH is an étale map (in particular, a smooth morphism) and the set theoretic inverse image of MATH under MATH is exactly equal to MATH. |
math/9911181 | As any line bundle on a smooth variety is the quotient of two effective line bundles, we may assume that MATH is effective. Let MATH be a global section of MATH, and let MATH denote the divisor of zeros of MATH. By the Moving Lemma REF, there exist a divisor MATH rationally equivalent to MATH such that MATH meets properly with MATH. In other words, the complement MATH of the support of MATH contains MATH. Since rationally equivalent divisors give rise to isomorphic line bundles (compare REF ), MATH is trivial. This proves the lemma. |
math/9911181 | Let MATH be a point of MATH. By the above Lemma, there exist an open subset MATH of MATH containing MATH and such that the line bundle MATH is trivial. As the fibre over MATH (under the quotient map) is contained in MATH and as the assertion of the lemma is local, we may assume that MATH. In particular, we can assume that MATH is trivial. Let MATH be a generating global section of MATH. Then MATH is a generating global section of MATH. As MATH is an even form, the section MATH is MATH-invariant, and hence also a global generating section of MATH. This proves that MATH is a line bundle. |
math/9911181 | Taking MATH, the existence of line bundle MATH follows from the above lemma. Since the map MATH restricted to MATH is étale, the canonical bundle MATH pulls back to the canonical bundle MATH. Hence, MATH restricts to the canonical bundle on MATH. The uniqueness of MATH follows since the codimension of MATH in the normal variety MATH is MATH. Since MATH acts freely on (smooth) MATH, the quotient MATH is smooth (and hence NAME). Further, MATH and hence it is NAME (since MATH). Now, the assertion that MATH is NAME, follows from REF . |
math/9911181 | As MATH is an isomorphism over MATH and the restriction of MATH to MATH is isomorphic to the canonical bundle, we see that MATH . As MATH is irreducible, this clearly implies the result. |
math/9911181 | Follows from REF . |
math/9911181 | Choose MATH with the property as given in REF . We need to show that MATH: By REF - REF , we know that MATH . We want to compare this with an alternative way of calculating the left side of the equation above. Since MATH, MATH . As MATH is étale over MATH, the canonical bundle on MATH pulls back to the canonical bundle on MATH. In particular, MATH restricts to the canonical bundle on MATH. But the complement of MATH in MATH has codimension MATH, which forces MATH to be the canonical bundle on MATH (as MATH is smooth, in particular, normal). By REF , we therefore get MATH . Combining MATH, we get MATH (since MATH is a nontorsion element of Pic MATH), which forces MATH to be equal to zero as desired. |
math/9911181 | The assertion that MATH is NAME follows by the same argument as for MATH (compare the proof of REF ) . Now, since the codim. of MATH in MATH is at least two, the corollary follows from the above theorem. |
math/9911181 | Let MATH be a NAME splitting of MATH. Then MATH is a NAME splitting of the MATH-fold product of MATH. As MATH is equivariant with respect to the natural actions of the symmetric group MATH, it takes MATH-invariant functions on MATH to MATH-invariant functions on MATH. As MATH is the subsheaf of MATH consisting of MATH-invariant functions, MATH induces a NAME splitting of MATH. |
math/9911181 | By REF , the MATH-th symmetric product MATH is NAME split. In particular, MATH is NAME split. Let MATH be a splitting section of MATH on MATH. Thinking of MATH as a section of MATH over MATH, as MATH is normal and codim. of MATH in MATH is two, we can extend MATH to a global section MATH of MATH over MATH (compare REF ). Consider the section MATH of MATH over MATH (compare REF ), and extend it to a section MATH of MATH over MATH. (This is possible since MATH is smooth, in particular, normal and the codim. of MATH in MATH is at least two.) We claim that MATH is a splitting section of MATH over MATH. To see this, it is enough to prove that the restriction MATH of MATH to MATH is a splitting section over MATH. But MATH is isomorphic to MATH under MATH, and moreover MATH corresponds to MATH under this isomorphism. As MATH, by definition, NAME splits MATH, the result follows. |
math/9911182 | A direct application of REF . |
math/9911186 | MATH . Since the graph norm of MATH with respect to MATH is MATH, the result follows. MATH . Immediate by the closed graph theorem. MATH . Suppose MATH is strongly standard and let MATH be a standard extension. For any MATH, MATH, hence MATH, which implies MATH, that is, MATH, hence MATH. |
math/9911186 | If MATH is strongly standard and MATH, then MATH cannot be standard by REF , MATH. Now let MATH be not strongly standard, MATH and MATH. If MATH then MATH for some MATH, MATH. However MATH and MATH should be zero, otherwise MATH, hence MATH. Finally, if MATH is standard then MATH is not strongly standard hence MATH is unbounded. Since MATH, MATH is unbounded, hence MATH is not strongly standard. |
math/9911186 | If MATH is strongly standard and MATH, then MATH cannot be standard by REF , MATH. Now let MATH be not strongly standard, MATH and MATH. If MATH then MATH for some MATH, MATH, MATH. However MATH should be zero, otherwise MATH, hence MATH. Finally, observe that MATH is properly extended by MATH, hence MATH cannot be bounded, which means that MATH is not strongly standard. |
math/9911186 | MATH . By REF , MATH, REF, MATH, there exists a standard subspace MATH iff MATH, hence MATH, is unbounded. Since MATH, this is is equivalent to MATH, which in turn corresponds to MATH by spectral mapping. MATH . Let MATH be the selfadjoint operator defined by MATH, MATH, and let MATH be the polar decomposition of MATH. By MATH one obtains MATH and MATH, and by REF , one gets MATH. This implies that there exists a subspace MATH such that MATH is irreducible MATH is unbounded, MATH, MATH, MATH. MATH . By the relations among MATH, MATH and MATH one gets that MATH and MATH, therefore if MATH and MATH, then MATH, hence MATH has the required properties by REF . MATH . As shown before, MATH is a factor if and only if MATH. By REF , this implies that MATH is an injective type MATH factor. Since MATH is an irreducible inclusion, the same reasoning applies to MATH. |
math/9911186 | This proof closely follows the proof of REF. The spaces MATH are both reduced by the spectral projections relative to MATH, where MATH is the modular operator relative to MATH, hence it is sufficient to study the problem on the fiber of the representation of MATH as direct integral with respect to MATH. Then, as in REF , it is sufficient to study the case in which MATH is a standard subspace of MATH, hence MATH can be seen as generated by the vectors MATH and MATH is represented by the matrix MATH where MATH is the complex conjugation. Therefore MATH and it is easy to see that MATH which implies MATH namely the graph norm of MATH is equivalent to the NAME norm, and the strongly standard property follows by REF , MATH. |
math/9911186 | By REF , any vector in MATH can be (uniquely) decomposed in sum of a vector in MATH and a vector in MATH. Iterating this argument we get MATH. Now we note that, by the global MATH-invariance of MATH, MATH . In particular, MATH commutes with MATH for each MATH. Therefore, applying the decomposition MATH to a vector in MATH, we get MATH. Let us take MATH. Since MATH is MATH-antiinvariant, MATH which implies MATH. Now observe that MATH . Since the elements in the center of MATH are MATH invariant (see REF , CITE), it turns out that MATH. Then notice that MATH iff MATH and MATH . Since any MATH may be uniquely written as MATH, MATH, and MATH commutes with MATH if MATH, REF for MATH implies MATH. Then REF for MATH implies MATH and, iterating, MATH for any MATH. Analogously we show that MATH vanishes, and therefore MATH for MATH. If MATH is even this shows that MATH, that is, MATH. When MATH is odd, we proved that MATH. Now, taking MATH, MATH, and making use of the MATH-anti-invariance of MATH, we have MATH hence MATH, and this implies MATH. Finally we observe that the inclusion MATH is irreducible iff MATH, and therefore MATH, are trivial. Then MATH follows from MATH and MATH and MATH follows from MATH and MATH. |
math/9911186 | MATH . First we note that, by REF , MATH that is, each closed commutative subspace of MATH considered as an additive group acts on any second quantization algebra via MATH. Then we only have to check that REF Theorem CITE are satisfied. By REF we get MATH that is, MATH is generated by MATH and a unitary representation of MATH which is strongly continuous by the finite-dimensionality of MATH. Now we claim that the action MATH is dual to the action MATH of MATH on MATH. On the one hand MATH where the last equality follows from REF MATH, MATH. On the other hand, by REF , MATH acts as the dual group on MATH iff the pairing MATH is a duality between real NAME spaces. The factoriality of MATH implies that the pairing is non degenerate, the finite-dimensionality of MATH and MATH implies that it is continuous, and the thesis follows. MATH follows immediately by REF . |
math/9911186 | Let MATH be a sequence of irreducible inclusions of standard subspaces such that the codimension of MATH in MATH is one, and let MATH, respectively, MATH the direct sum of the MATH, respectively, MATH in MATH. As it is well known, direct sums of complex orthogonal real subspaces give rise to tensor products at the second quantization level, therefore the decomposition MATH gives rise to an ITPFI decomposition. Since the codimension of MATH in MATH is one, MATH is two-dimensional and the angle of MATH with MATH is a number MATH. Therefore, by CITE and REF , the type of MATH depends only on the sequence MATH. Choosing isomorphic inclusions MATH one gets a constant sequence MATH hence the factor has type MATH, with MATH. |
math/9911189 | Denote the orbits over MATH in MATH and MATH by MATH and MATH. A neighborhood of MATH in MATH and a neighborhood of MATH in MATH are each isomorphic to a neighborhood of MATH in the same local model MATH, by assumption and by the local normal form theorem. Hence a neighborhood MATH of MATH is isomorphic to a neighborhood MATH of MATH. Since the moment maps are proper, if MATH is a small enough neighborhood of MATH, the preimages MATH and MATH are contained in MATH and MATH, and are hence isomorphic. |
math/9911189 | Let MATH be a MATH - diffeomorphism. By REF , there exists a basic one-form MATH on MATH (see REF ) such that MATH. We now apply NAME 's method: Define MATH for all MATH. By REF below, the MATH are nondegenerate. Let MATH be the vector field determined by MATH. The vector field MATH preserves the level sets of MATH, because for every MATH, MATH. Since MATH is proper, the time-dependent vector-field MATH integrates to a flow, MATH. Let MATH. Then MATH . The vector field MATH is invariant because MATH and MATH are invariant. Consequently, MATH is equivariant, and hence is a MATH - diffeomorphism. Finally, MATH . Therefore, MATH, since MATH is the identity. Then, MATH. In other words, MATH is an (equivariant) symplectomorphism. |
math/9911189 | Let MATH. Since MATH and MATH are closed invariant symplectic forms on MATH with the same moment map, MATH for all MATH. Since MATH is also invariant, it is basic. By REF it suffices to show that the restriction of MATH to the fiber MATH is exact for some regular value MATH of MATH. Since this restriction is the pull-back of a differential form MATH on the orbifold MATH, it is enough to show that MATH is exact. Since MATH is two dimensional, it is enough to show that the integral of MATH over it is zero, that is, that the integrals of MATH and MATH are equal. But this follows from the fact that the NAME measures for MATH and MATH are the same, because the density functions for these measures are given by the symplectic volumes of the symplectic quotients; see CITE. |
math/9911189 | First, let a compact abelian NAME group MATH act on MATH as a codimension one subgroup of MATH with isotropy weights MATH. The vector fields for this action are MATH . Let MATH and MATH be invariant symplectic forms on MATH with constant coefficients that have the same moment map, MATH, and that induce the same orientation. We will show that MATH is non-degenerate. Because MATH is real valued, it can be written in the form MATH where MATH are real, MATH and MATH are complex, MATH, and MATH. By the definition of moment map, MATH is the coefficient of MATH in MATH. Hence MATH . Differentiating again, MATH. Because MATH is MATH-invariant, MATH unless MATH. In this case, MATH . Finally, MATH unless MATH; in this case, MATH. Therefore, MATH, MATH, and MATH are determined by MATH. By what we have shown, if MATH, the coefficients MATH, MATH, and MATH are determined by MATH. Thus, if no weight is zero, MATH is independent of MATH, hence it is non-degenerate. Since the action is effective and the dimension of MATH is MATH, the only other possibility is that exactly one of the weights - let's say the first one - is zero, and the others form a basis of MATH. In this case, MATH for all MATH and MATH, and so the top power of MATH is MATH times the standard volume form. Since MATH and MATH induce the same orientation, and since MATH is determined by MATH and hence independent of MATH for MATH, the signs of MATH and MATH are the same. Therefore, MATH is never zero, and MATH is non-degenerate. Now let MATH be any point with stabilizer MATH. By the local normal form theorem, a neighborhood of the orbit MATH in MATH with the symplectic form MATH is equivariantly symplectomorphic to a neighborhood of the orbit MATH in the model MATH. The tangent space at MATH splits as MATH where MATH is the tangent to the orbit. By the definition of the moment map, MATH is given by a block matrix of the form MATH where MATH is the natural pairing between the vector space MATH and its dual, MATH, and where MATH and MATH are linear symplectic forms on MATH with the same moment map and the same orientation. By the above argument, MATH is nondegenerate. Consequently, MATH is nondegenerate. |
math/9911189 | Let MATH be the local model for a non-exceptional orbit in MATH, with MATH. Inside the moment fiber MATH, the set of points with stabilizer MATH is MATH where MATH is the subspace fixed by MATH. By the definition of exceptional orbit, this subspace is not trivial. Therefore, the local model becomes REF, where the group MATH acts trivially on MATH and acts on MATH through an inclusion into MATH. By a dimension count, this inclusion must be an isomorphism. |
math/9911189 | Consider the slice representation at any non-exceptional orbit. By REF above, the stabilizer MATH is connected, Since the action is effective, the isotropy weights MATH generate the weight lattice. The stabilizer and these weights are determined by the image of the moment map for a local model; this image is the NAME cone MATH. Finally, by the stability of the moment map, the image of the moment map is the same for every local model in MATH. |
math/9911189 | Assume MATH; the general case is similar. Since, by REF , MATH, we can identify MATH with MATH. Since MATH preserves the moment map, it necessarily has the form MATH for some MATH. Similarly, its inverse sends MATH to MATH, where MATH. Since both MATH and its inverse are smooth, MATH and MATH must themselves be smooth. We define MATH by MATH. Then MATH is a smooth equivariant lift of MATH, and it has a smooth inverse given by MATH. |
math/9911189 | Let MATH be a non-exceptional orbit in MATH. By REF , any MATH-diffeomorphism sends it to a non-exceptional orbit MATH in MATH. By REF , the local models for MATH and MATH are the same MATH. By REF and the local normal form theorem, the map lifts to a MATH - diffeomorphism from a neighborhood of MATH to a neighborhood of MATH. By REF , it lifts globally. |
math/9911189 | Suppose that the moment map MATH is onto. Then every element of MATH is in the non-negative span of the MATH. In particular, there exist MATH such that MATH, that is, MATH. Let MATH. Conversely, suppose that there exist positive MATH's so that MATH. Let MATH be any element. Because the action is effective, its weights, MATH, span MATH, so there exist MATH, MATH, such that MATH. Because MATH, we also have MATH for any MATH. Because MATH for all MATH, if we take MATH large enough we get that MATH is in the positive span of the MATH. |
math/9911189 | Suppose that MATH for some MATH, not all zero. Since the moment fiber MATH contains the line MATH, MATH, the map is not proper. Conversely, suppose that MATH whenever MATH are not all zero. Then MATH is positive. Since MATH is quadratic, MATH for all MATH, which implies that MATH is proper. |
math/9911189 | Given MATH, let MATH be the slice representation. Consider the local model MATH with moment map MATH where MATH is the moment map for MATH and where MATH. By REF , the moment map MATH is proper if and only if the moment fiber MATH consists of a single orbit, and otherwise MATH contains infinitely many orbits near MATH. The lemma now follows from the local normal form theorem and the connectedness of moment fibers. |
math/9911189 | Let MATH be the corresponding model. By REF , the moment map MATH is proper. By REF , the preimage in MATH and in MATH of a sufficiently small neighborhood MATH of MATH are isomorphic. Thus, we may work purely inside MATH. Choose a neighborhood of MATH of the form MATH, where MATH and MATH are convex, and where we identify MATH. Because MATH is homogeneous, MATH is contractible. |
math/9911189 | Because the quotient MATH is a one dimensional compact connected NAME group, there exists a homomorphism MATH such that the sequence REF is exact. Such a homomorphism must be of the form MATH for some integers MATH. Let MATH denote the weights for the MATH-action on MATH. Differentiating the identity MATH from REF, we get MATH . By REF , because the moment map MATH is not proper, there exist non-negative numbers MATH, not all zero, such that MATH. Since MATH, by a dimension count the vector MATH must be a multiple of the vector MATH. Therefore, after possibly replacing the vector MATH by the vector MATH, all the MATH's are non-negative. By REF and a similar dimension count, the MATH are strictly positive exactly if the moment map is onto. |
math/9911189 | Consider the defining polynomial, MATH . Let MATH be the number of MATH's such that MATH, and let MATH. We can assume that MATH for MATH and MATH for MATH. Then MATH defines a polynomial MATH. Let us identify MATH with its image in MATH. Then MATH . Let MATH and MATH. By REF , the moment map for MATH is onto. |
math/9911189 | The first claim follows immediately from REF , the local normal form theorem, and REF below. The fact that the symplectic quotients which contain more than one orbit are topological surfaces follows immediately from REF , the local normal form theorem, and REF below. These surfaces are closed because the moment map is proper. They are connected by the connectedness of moment fibers. The symplectic structure on the symplectic quotient induces an orientation on the complement of a discrete set of points (namely, the exceptional orbits) and hence on the symplectic quotient itself. |
math/9911189 | To show that MATH is a homeomorphism, it is both necessary and sufficient to prove that the map MATH is onto and proper and that its fibers are exactly the MATH-orbits. (This follows from the formulas for MATH and MATH.) We will begin by assuming that the moment map MATH is onto MATH. By REF , this implies that the MATH's are positive. Consider the commuting diagram MATH where MATH and MATH . Let MATH be the boundary of the positive orthant MATH. Since MATH for all MATH, the map MATH is a homeomorphism which identifies the product MATH with the orthant MATH. The map MATH then becomes a map from MATH to MATH, given by the formula MATH where in the first coordinate we used the equality MATH. The map MATH is one-to-one and onto, because the function MATH is a homeomorphism from MATH onto MATH, and because for each MATH, the function MATH from MATH to MATH is one to one and onto. The function REF approaches infinity uniformly in MATH as MATH. Therefore, MATH is proper. The properness of MATH follows from that of MATH and MATH. Let us now show that MATH is onto MATH. Since MATH is onto, for any MATH there exists MATH such that MATH and MATH. Choose MATH so that MATH. Since the map MATH is onto, there exists MATH such that MATH. Then MATH. Let us now show that the level sets of MATH are the orbits of MATH. Suppose that MATH and MATH for some MATH and MATH in MATH. Since MATH is one to one, there exists MATH such that MATH. We must show that MATH can be chosen to be in MATH. If all the coordinates of MATH are non-zero, MATH implies that MATH, which further implies that MATH, by REF. If one of the coordinates of MATH, say MATH, is zero, then it is enough to show that the MATH-orbit of MATH coincides with the MATH-orbit of MATH. By a dimension count, it is enough to show that the MATH-stabilizer of MATH is not contained in MATH. Because MATH, the MATH-stabilizer of MATH contains the circle MATH. Since MATH, the polynomial MATH is not constant on this circle. By exactness of REF, this circle is not contained in MATH. For the general case, we may let MATH be the splitting into a surjective part and a toric part, as described in REF . Then MATH . The map MATH is proper, its fibers are the MATH orbits, and it is onto MATH, as we have shown above. The map MATH is a moment map for a toric action, so it is proper and its level sets are MATH orbits. Thus, the map MATH is proper, onto MATH, and its level sets are the MATH-orbits. This is precisely what we needed in order to deduce that MATH is a homeomorphism. |
math/9911189 | Since the moment map is onto, by REF the MATH-orbit of MATH is non-exceptional if and only if the stabilizer of MATH in MATH is trivial. We identify MATH with the subgroup of MATH by which it acts. The stabilizer of MATH then consists of those elements MATH such that MATH whenever MATH and such that MATH . |
math/9911189 | Since MATH acts freely on MATH, the quotient MATH is naturally a smooth manifold (with the quotient differential structure). The map REF is smooth, and, by REF , it is a homeomorphism onto its image. Since a smooth homeomorphism between two smooth manifolds of the same dimension is a diffeomorphism exactly at those points where it is a submersion, we need to show that MATH is onto for all non-exceptional MATH. To show that MATH is onto, it is enough to find MATH such that MATH and MATH. To see this note that, since MATH acts freely, MATH is onto MATH. Additionally, since MATH is holomorphic, MATH and MATH form a real basis to MATH. Recall that MATH and MATH. Hence MATH . CASE: In this case, MATH . Let MATH. Then MATH by REF, and MATH whereas MATH CASE: In this case, MATH . Let MATH and MATH for all MATH. Then MATH and MATH whereas MATH . |
math/9911189 | Let MATH and MATH be the splitting into a surjective part and a toric part, as described in REF . With this splitting, the local model is MATH and its quotient is MATH . The union of the non-exceptional orbits in this quotient is MATH where MATH is the union of the free orbits in MATH. Under the identification MATH, the trivializing homeomorphism MATH on REF is MATH where MATH is the defining polynomial. REF implies that the map MATH pulls back the sheaf of smooth functions on MATH onto the sheaf of smooth functions on MATH. Therefore, it is enough to show that the map MATH pulls back the sheaf of smooth functions on MATH onto the sheaf of smooth functions on MATH . By REF, any invariant smooth function can be expressed as a smooth function of real invariant polynomials. Since MATH acts on MATH through an isomorphism with MATH, the ring of MATH-invariant polynomials in MATH is generated by the coordinates of MATH and MATH, the real and imaginary parts of MATH, and MATH. Finally, note that MATH where MATH is the linear isomorphism dual to the map MATH. Hence, every smooth invariant function is the pullback via REF of a smooth function. |
math/9911189 | By REF , for every exceptional orbit MATH over MATH, the corresponding local model has a non-proper moment map, MATH. By the local normal form theorem, we may choose a grommet MATH such that MATH. By REF , the moment map MATH restricts to a homeomorphism of the exceptional sheet MATH with the image of MATH. Hence there exists a neighborhood MATH of MATH such that MATH. For MATH, the intersection MATH is a closed subset of MATH which does not meet the fiber MATH. Since the moment map is proper, there exists a neighborhood MATH of MATH which does not meet the image under the moment map of any of these intersections. If we define MATH of MATH, and replace MATH by MATH and MATH by MATH, the grommets MATH become wide. Also, the exceptional sheets MATH are then closed and disjoint, so we can shrink each MATH to a smaller neighborhood of MATH to obtain wide grommets whose images have pairwise disjoint closures. |
math/9911189 | Let MATH be wide grommets such that MATH are the exceptional orbits in the moment fiber MATH and such that the images MATH have disjoint closures in MATH. These grommets exist by REF . This will not be ruined if we further restrict to a smaller neighborhood of MATH. Recall that the standard flattening of the local model MATH is MATH where MATH. Replace MATH by its intersection with MATH . Then the restriction MATH is well defined. After this, the grommets determine a unique map MATH on the images of MATH in MATH such that the following diagram commutes. MATH . We need to extend MATH to the rest of MATH, perhaps after shrinking the MATH-s to smaller neighborhoods of MATH. Using the stability of the moment map, REF implies that on the complement of the exceptional sheets in the quotient MATH, the map MATH induced by the moment map is a submersion. Namely, for each point MATH in the domain of this map there exists a neighborhood MATH of MATH in MATH such that a neighborhood of MATH is diffeomorphic to the product of a disk with MATH with the map MATH being the projection map. The partial flattening REF determines an NAME connection for this submersion, defined on the open subset MATH: we declare the horizontal tangent vectors to be those whose push-forward by MATH is tangent to the sheets MATH for MATH. We extend this to an NAME connection on the entire complement of the exceptional sheets, MATH, perhaps after shrinking the MATH-s; this is easily done with a partition of unity. Then for a point MATH in MATH, any path MATH in MATH which starts at MATH can be lifted to a horizontal path in MATH. We proceed as in the proof of NAME 's lemma. Let us assume that MATH and that MATH is a ball centered at MATH. We can choose coordinates on MATH such that MATH becomes MATH . (This is possible by REF .) Denote by MATH the standard vector fields on MATH that are parallel to the coordinate axes, let MATH be their horizontal liftings to MATH, and let MATH, for MATH and MATH, be the flows which the MATH generate. For MATH, define MATH where, if MATH are the coordinates of MATH, then MATH is given by MATH. |
math/9911189 | Let MATH and MATH be complexity one spaces with flattenings about the point MATH. Let MATH and MATH be the associated marked surfaces, as in REF . If the spaces have the same isotropy data, there exists a bijection from the marked points in MATH onto the marked points in MATH which respects the isotropy labels. If the spaces have the same genus, this bijection extends to an orientation preserving diffeomorphism from MATH to MATH; this follows from standard differential topology. Moreover, this diffeomorphism can be deformed near the marked points into a rigid diffeomorphism MATH. This type of result is standard in differential topology; see, for example, CITE. The images of the moment maps MATH and MATH are the same; this follows from REF and from the fact that the isotropy data are the same. Let MATH and MATH be maps given in the flattenings. Under the identification of the symplectic quotients MATH and MATH with MATH and MATH, respectively, the rigid diffeomorphism MATH extends to a map MATH . We will show that the map MATH defined by MATH is a MATH-diffeomorphism. The diffeomorphism MATH fixes an identification between exceptional orbits in MATH and MATH with the same isotropy representation. Thus, we can unequivocally denote by MATH the local models for the exceptional orbits over MATH in both MATH and MATH. Let MATH denote the grommets, with MATH and MATH, and let MATH and MATH denote the exceptional orbits in MATH and in MATH. Our first claim is that the restriction MATH is a MATH-diffeomorphism. This is easy: by the definition of flattening, the restrictions MATH and MATH are both diffeomorphisms. Moreover, the map MATH induced by MATH is a diffeomorphism, since the smooth structures on MATH and MATH agree off the exceptional orbits. It remains to show that MATH is a MATH-diffeomorphism in a neighborhood of each exceptional sheet MATH. Let MATH and MATH denote the grommets of the associated surfaces. Since MATH is rigid, there exist MATH such that MATH is given by rotation by MATH on some neighborhood of the origin in MATH . Let MATH be the defining polynomial for the exceptional orbit MATH. Since MATH is surjective, we may choose MATH so that MATH. This defines an equivariant symplectomorphism from the local model MATH to itself as follows: MATH . This map induces a MATH-diffeomorphism on the quotient, MATH. It remains to show only that the MATH and MATH agree in some neighborhood of MATH. Indeed, when we use the trivializing homeomorphism MATH to identify MATH with MATH, the map MATH sends MATH to MATH. |
math/9911189 | Since the case that the moment fiber MATH contains just one orbit is covered by REF , we may assume that the moment fiber contains more than one orbit. By REF , after possibly restricting to the preimage of a smaller neighborhood of MATH, we may assume that MATH and MATH are equipped with flattenings. By assumption, the spaces MATH and MATH have the same genus and isotropy data. Therefore, by REF , there is a MATH-diffeomorphism MATH. Since the spaces have flattenings, REF is satisfied. (See REF .) By assumption, the NAME measures of MATH and MATH are the same. Hence, we can apply REF . The first implies that the map MATH lifts to a MATH - diffeomorphism from MATH to MATH. The second then guarantees that there exists MATH - symplectomorphism from MATH to MATH. |
math/9911189 | CASE: the moment fiber is a single orbit. By REF , there exists a convex sub-neighborhood MATH of MATH and a MATH - diffeomorphism (in fact, symplectomorphism) from MATH to MATH. By REF we can choose MATH so that MATH and MATH satisfy REF . By REF , this implies that there exists a MATH - diffeomorphism from MATH to MATH, and that MATH and MATH themselves also satisfy REF . The NAME measures coincide; hence we may apply REF , which completes the proof. CASE: the moment fiber contains more than one orbit. By REF , there exists a convex sub-neighborhood MATH of MATH so that MATH and MATH are equipped with flattenings. By assumption, the spaces MATH and MATH have the same genus and isotropy data. Therefore, by REF , there is a MATH-diffeomorphism MATH. Since these spaces have flattenings, REF is satisfied. (See REF .) By assumption, the NAME measures of MATH and MATH are the same. Hence, REF implies that the map MATH lifts to a MATH - diffeomorphism from MATH to MATH. REF completes the proof. |
math/9911189 | By the local normal form theorem, it is enough to construct the vector field MATH on the local models. We can then patch together the pieces by an invariant partition of unity. Notice that if a map MATH between vector spaces is homogeneous of degree MATH, then MATH where MATH and MATH are the NAME vector fields on MATH and MATH; this follows from the equality MATH. In particular, the NAME vector field on MATH lifts to a vector field on MATH. Similarly, if MATH, MATH, are homogeneous (possibly of different degrees), and MATH, then the NAME vector field on MATH lifts to a vector field on MATH. Consider a local model in MATH, namely, MATH, with a moment map MATH. The stratum fixed by MATH is MATH. Because the space is centered, we must have that MATH. Without loss of generality we may assume that MATH. To lift the NAME vector field on MATH to a MATH-invariant vector field on MATH, it is enough to lift it to a MATH-invariant vector field on MATH. This is possible by the previous paragraph, because MATH is bihomogeneous. |
math/9911189 | Without loss of generality we may assume that MATH. Choose MATH so that a MATH-ball about MATH is contained in MATH. Let MATH for MATH be an isotopy such that MATH is the identity map, the image of MATH is contained in MATH, MATH for all MATH and all MATH, and MATH for all MATH near zero and all MATH. Take MATH for all MATH and MATH; let MATH. Let MATH be the vector field on MATH which generates this isotopy: MATH. Since MATH vanishes near MATH, we can write MATH, where MATH is a smooth function, and MATH is the NAME vector field on MATH. By REF , there exists a smooth invariant vector field MATH on MATH such that MATH. So MATH is a smooth invariant vector field on MATH which is a lifting of MATH. Because MATH is proper, the vector field MATH generates an isotopy, MATH. Take MATH. |
math/9911189 | First, we show that MATH is centered. The closure MATH of an orbit type stratum in MATH is itself a compact symplectic manifold with the restricted MATH action and moment map. By the convexity theorem, its moment image is the convex hull of the moment images of its fixed points. Either MATH contains MATH, or its moment image would be contained in MATH, and therefore disjoint from MATH. Let MATH act on MATH with weights MATH and with the moment map that sends MATH to MATH. The moment preimage of MATH in MATH is a ball. Hence, this ball is also a centered complexity one space over MATH. Since both spaces are centered about MATH, and the preimages of MATH are both single orbits with the same isotropy data, by REF the spaces are equivariantly symplectomorphic. |
math/9911189 | The NAME algebra of the maximal torus of MATH can be identified with MATH with the standard metric. The roots are MATH, MATH, MATH. The NAME group acts by permuting the coordinates and by flipping their signs. The orbit through the point MATH is naturally identified with the NAME MATH. The NAME group orbit of this point consists of the points MATH, MATH, and MATH. The image of this moment map is a diamond. The isotropy weights at MATH are MATH, MATH, and MATH. By REF , the preimage of the half space MATH is a ball as required. A similar argument shows that the preimage of the opposite half space is again a ball. |
math/9911189 | The NAME algebra of the maximal torus of MATH can be identified with MATH with the standard metric. The NAME group acts by permuting the coordinates and by flipping the signs of two coordinates at a time. The roots are MATH. The orbit through the point MATH is naturally identified with the NAME MATH. The NAME group orbit of this point consists of the points MATH, MATH, and MATH. The image of this moment map is an octahedron. The isotropy weights at MATH are MATH. By REF , the preimage of the half space MATH is a ball as required. A similar argument shows that the preimage of the opposite half space is again a ball. |
math/9911198 | Assume that MATH is nontrivial. Let MATH be the corresponding morphism of sheaves. Consider the subsheaves MATH of MATH and MATH of MATH. Assume MATH. By commutativity of the diagramme MATH we have that MATH, and then by stability MATH which is a contradiction. Then MATH and MATH is an isomorphism. Now let MATH be a point, and let MATH be an eigenvalue of MATH at the fibre over MATH. Then MATH is not surjective at MATH, hence MATH cannot be an isomorphism and then MATH. |
math/9911198 | Let MATH be a point of the curve MATH and MATH. The exact sequence MATH gives that if MATH for all MATH, then MATH is generated by global sections and MATH. Assume that MATH. Then by NAME duality there is a nonzero morphism MATH, where MATH is the canonical divisor. This gives an effective divisor MATH on MATH and an exact sequence MATH . Let MATH. We have MATH. Then MATH . On the other hand, by REF, and combining both inequalities we get MATH . Then if we take MATH, for any MATH and MATH we will have MATH, thus MATH is generated by global sections and MATH. By standard methods using the Quot scheme, this implies that MATH is bounded. |
math/9911198 | By REF we have that for every subsheaf MATH of MATH . Take MATH and apply the proposition. |
math/9911198 | By semistability, the slope of a subsheaf of a sheaf in MATH is bounded. On the other hand there are only a finite number of possibilities for rhe rank and degree of a sheaf in MATH, then we can apply REF . |
math/9911198 | We will apply the NAME criterion: a point MATH is (semi)stable iff for all one-parameter subgroup REF MATH of MATH we have MATH, where MATH is the minimum weight of the action of MATH on MATH. Let MATH. A REF-PS MATH of MATH is equivalent to a basis MATH of MATH and a weight vector MATH with MATH, MATH, and MATH. The set MATH of all weight vectors is a cone in MATH. If a basis of MATH has been chosen, then by a slight abuse of notation we will denote MATH by MATH, where MATH. We will choose a set of one-parameter subgroups, calculate MATH, and then imposing MATH we will obtain necessary conditions for MATH to be (semi)stable. Then we will show that the chosen set of one-parameter subgroups is sufficient, in the sense that if we check that MATH for all one-parameter subgroups in this set, then the same will hold for any arbitrary one-parameter subgroup in MATH. We have MATH, where MATH (respectively, MATH) is the minimum weight of the action of MATH on MATH (respectively, MATH). Fix a basis MATH of MATH. Define MATH, where MATH is the quotient corresponding to the point MATH. We have (see CITE) MATH . On the other hand MATH . Note that MATH is linear on MATH, but MATH is not. GIT (semi)stable implies REF Let MATH be a (semi)stable point in MATH. Let MATH be a basis of MATH. Define MATH . Note that if MATH is ``good", then the map MATH is an isomorphism (in particular injective), and then MATH. Later on we will see that for sufficiently large MATH, a semistable point is ``good", but now we won't assume that MATH is ``good". Define a filtration of MATH . Let MATH be the restriction of MATH to MATH. To calculate MATH we distinguish seven cases. MATH . Note that in all cases, except REF is a linear function of MATH. First we will consider weight vectors of the form MATH and define MATH (it is clear that any subspace of MATH can be written in this form, after choosing an appropriate bases for MATH). We have MATH. To obtain a formula for MATH we have to analyze each of the seven cases. We will only work out the details for REF , the remaining cases being similar to REF . In REF we have MATH. Then, according to the value of MATH we have MATH . In REF we have MATH, hence MATH . Doing the calculation for the seven cases we check that in every case we have MATH . Then MATH gives MATH . If we vary MATH (allowing MATH), the submodules MATH are bounded, so we can take MATH large enough such that MATH. We have MATH, and MATH, and then we obtain REF . To obtain REF , assume that we have subspaces MATH giving a critical filtration. Let MATH and MATH. Take a bases MATH of MATH adapted to this filtration, that is, such that MATH and MATH. Consider the weight vector MATH . An easy computation then shows MATH. On the other hand MATH, and then MATH gives REF imply GIT (semi)stable Now we have to show that the one-parameter subgroups that we have used are sufficient. As we did before, we will fix an arbitrary base MATH, and we consider the seven different cases. In all cases except REF, MATH, and hence MATH, is a linear function of MATH, and then to prove that MATH for all MATH it is enough to check it on the generators MATH defined above REF . In REF we have MATH, hence it is no longer linear on MATH, and it is not enough to check the condition on the generators MATH. But it is a piecewise linear function. The cone MATH of weights is divided in two cones MATH . Observe that MATH is linear on each of these cones. We will use the following lemma. Let MATH be a cone in MATH, let MATH be a set of generators of MATH, that is, MATH. Let MATH be a linear function such that MATH. Let MATH be the subcone MATH. Then the set of vectors MATH generate MATH. MATH . We apply this lemma with MATH (and then with the negative of this, for MATH), and we obtain a set of generators for MATH and MATH. But all these vectors are either of the form MATH with MATH, or of the form MATH with MATH and MATH, and we have already considered them. |
math/9911198 | We rewrite REF using MATH . We obtain MATH . We have MATH and MATH, hence MATH and the result follows. Now we rewrite REF , using MATH, MATH and MATH. MATH and the result follows. |
math/9911198 | We will proof the three items in three REF . (Semi)stable conic bundle MATH GIT (semi)stable MATH . We will use REF . We will start checking REF . Let MATH be the set of vector bundles MATH that are subsheaves of bundles MATH occurring in semistable conic bundles. It satisfies REF with MATH and MATH. Let MATH, MATH large enough, so that REF hold, and let MATH be the subset of MATH consisting of bundles MATH that satisfy REF . Then the set MATH is bounded. Taking MATH large enough we then have MATH for MATH. In other words, MATH . On the other hand, we still have MATH . Let MATH be a subspace of MATH, and MATH the corresponding sheaf. If MATH belongs to MATH, we get that REF implies REF , because MATH and MATH, because MATH. On the other hand, if MATH belongs to MATH, REF implies REF MATH . In both cases, if REF is strict, then REF is also strict. But assume that there is a semistabilizing subsheaf MATH of MATH (that is, giving equality in REF ). By REF , MATH is generated by global sections. Let MATH. Then MATH, and we have MATH and MATH. Now we will check REF . Let MATH be the set of vector bundles of the form MATH such that MATH gives a critical filtration of a (semi)stable conic bundle-MATH. REF are satisfied with MATH and MATH. Let MATH, and MATH large enough. Let MATH be the subset of MATH consisting of vector bundles MATH satisfying REF . Then MATH is bounded, and taking MATH large enough we have MATH for MATH. Hence for MATH, MATH . On the other hand, for MATH we still have MATH . Let MATH be a critical filtration of MATH. If MATH, we get that REF implies REF , because MATH and also MATH. On the other hand, if MATH, REF implies REF MATH . In both cases, if REF is strict, also REF is strict. But assume that we have subsheaves MATH giving a critical filtration of a semistable conic bundle MATH. By REF MATH and MATH are generated by global sections and MATH. Taking MATH and MATH we have MATH and MATH, hence MATH and also MATH REF . MATH GIT (semi)stable MATH MATH and MATH good If MATH, then by NAME duality MATH. Take MATH. The composition MATH gives a linear map MATH . Let MATH be the kernel of MATH. We have MATH. Then by (semi)stability of MATH we have MATH . We have MATH. Then if MATH is large enough the inequality forces MATH. By definition of MATH we have MATH, then MATH, MATH, and then MATH because MATH is torsion free. We conclude that (for MATH large enough) MATH. Then MATH, and to show that MATH is ``good" it is enough to show that the induced linear map MATH is injective. Let MATH be the kernel. Then we have MATH. By semistability we have REF MATH but MATH, and then MATH must be zero. To show that MATH is torsion free, let MATH be the torsion subsheaf. We have MATH, and then MATH is a subspace of MATH. The associated sheaf MATH has rank equal to zero, and arguing as above we get MATH. CASE: GIT (Semi)stable MATH MATH (semi)stable conic bundle By the previous step we know that we can choose MATH large enough so that MATH is ``good". We will check first REF . Let MATH be a subsheaf of MATH. Define MATH. We have MATH, MATH, MATH, and MATH. Then MATH . Note that if REF is strict, then also REF is strict. But assume that there is a subspace MATH that is semistabilizing, that is, both conditions in REF are equalities. Then MATH and we get that REF for MATH also gives equality. Now we are going to check REF . As in REF , consider the set MATH of vector bundles of the form MATH such that MATH gives a critical filtration. We have already proved REF , thus REF are again satisfied. Then, as in REF , we define the subset MATH, and taking MATH large enough we can assume that the vector bundles MATH and MATH are generated by global sections if MATH. Let MATH be a critical filtration of MATH. Let MATH and MATH. If MATH, then MATH and MATH are generated by global sections and then MATH, MATH, and MATH is a critical filtration of MATH and REF holds MATH . On the other hand, if MATH, REF implies REF MATH . Note that if REF is strict then REF is also strict. But assume that there is a semistabilizing critical sequence MATH, that is, a critical sequence giving equality in both conditions of REF . Then MATH and we also get an equality in REF . |
math/9911198 | of REF. Let MATH (respectively, MATH) be the GIT quotient of MATH (respectively, MATH) by MATH. First we construct a universal family on MATH using the universal families of the Quot scheme MATH and on MATH (we think of MATH as the NAME of one dimensional subspaces of MATH, and hence the universal subbundle of subspaces is MATH). Recall that MATH is in MATH. The universal quotient MATH on MATH pulls back to a vector bundle MATH on MATH. On the other hand the universal subbundle on MATH gives a morphism MATH of sheaves over MATH. By the definition of MATH, there is a line bundle MATH on MATH such that this last morphism factors and gives MATH. Note that the line bundle MATH is needed because the factorization on MATH is only up to scalar multiplication. The triple MATH is a universal conic bundle. Given a family MATH of conic bundles parametrized by MATH, and using the universal family on MATH, we obtain a morphism MATH. This is done in the following way: Let MATH be large enough so that REF holds. Given a family MATH of conic bundles parametrized by MATH, consider the locally free sheaf MATH, and note that MATH is a surjection. NAME MATH with open sets MATH such that there are isomorphisms MATH. Then we have quotients MATH and families of subspaces MATH, and these give maps MATH. On the intersections MATH this maps in general will differ by the action of MATH, then they combine to give a well defined morphism MATH. It is straightforward to check the universal property for MATH, and then MATH is a coarse moduli space. Now we will show that the universal family restricted to MATH descends to MATH, making it a fine moduli space. Applying Luna's étale slice theorem CITE, we can find an étale cover MATH of MATH over which there is a universal family MATH. Consider MATH and take an isomorphism MATH with the condition MATH. This isomorphism exists and is unique by REF , and then it satisfies the cocycle condition of descend theory CITE, and hence the family MATH descends to MATH. |
math/9911198 | We start with a general observation about GIT quotients. Let MATH be a projective variety with a linearized action by a group MATH. Two points in the open subset MATH of semistable points are NAME (they are mapped to the same point in the moduli space) if there is a common closed orbit in the closures (in MATH) of their orbits. Let MATH. Let MATH be the unique closed orbit in the closure MATH in MATH of its orbit MATH. Assume that MATH is not in MATH. Then there exists a one-parameter subgroup MATH such that the limit MATH is in MATH. Note that we must have MATH (otherwise MATH would be unstable). Note that MATH, and then MATH. Repeating this process with MATH we then get a sequence of points that eventually stops and gives MATH. Two points MATH and MATH will then be NAME if and only if after applying this procedure to both of them the orbits of MATH and MATH are the same. We will use the notation introduced in subsection REF. We will prove the proposition using the previous observation. The fact that choosing a ``semistabilizing object" of MATH induces a one parameter subgroup with MATH (where MATH is the corresponding point on MATH) follows from REF and the proof of REF . The fact that the limit point MATH corresponds to MATH is an easy calculation (see CITE). The conic bundle MATH is semistable by REF . It is easy to check that MATH corresponds to MATH, and then REF follow from the fact that MATH is in MATH. |
math/9911198 | This proof was given to us by NAME. We can assume, without loss of generality, that the polarization MATH of MATH is very ample, and then MATH embedds in MATH and MATH acts on MATH. A point MATH is (semi)stable iff its image in MATH is (semi)stable, and then we can assume MATH, with MATH acting on MATH. Let MATH be the projection. Let MATH be a semistable point and MATH a REF-PS with MATH. Let MATH be a point in the fibre MATH, and let MATH . This limit exists and it is not the origin because MATH. We have MATH (by continuity of MATH). Assume that the point MATH is unstable. Then the closure of the orbit of MATH contains the origin, but this closure is included in the closure of the orbit of MATH, and this doesn't contain the origin because MATH is semistable. Then MATH is semistable. Furthermore, MATH cannot be stable because MATH, then MATH is strictly semistable. |
math/9911198 | Let MATH and MATH be large enough so that MATH is locally free, MATH is a surjection and REF holds. Note that the universal family that was constructed on MATH in the proof of REF can be extended to the set MATH of ``good" points. Arguing as in the proof of REF, there is a finite open cover MATH of MATH and morphisms MATH. These morphisms depend on the choices made (the choice of local trivializations of MATH), but the MATH orbit of MATH are independent of the choices. In particular, the property of MATH belonging to MATH only depends on MATH. By REF , MATH lies in MATH (respectively, MATH) iff the conic bundle MATH is stable (respectively, semistable). Then MATH and the openness of MATH in MATH proves that MATH is open (the same argument works for MATH). |
math/9911198 | We will construct a flat family of conic bundles parametrized by an irreducible scheme MATH with the property that every semistable conic bundle of type MATH belongs to the family. Then there is a surjective morphism MATH, where MATH is the open subset representing semistable points, and this proves that MATH is irreducible. Repeating this with the open subset MATH corresponding to stable points, we prove that MATH is also irreducible. Let MATH be large enough so that for any semistable conic bundle MATH in MATH, the vector bundle MATH is generated by global sections REF , and such that MATH . Note that MATH only depends on MATH, MATH, MATH and MATH, but not on MATH. If we choose MATH generic sections of MATH, we have an exact sequence MATH where MATH is a line bundle of degree MATH. By standard methods we can construct a universal family MATH of extensions of line bundles of degree MATH by MATH. This will be parametrized by a scheme MATH that has a morphism to MATH, and the fibre over a line bundle MATH is naturally isomorphic to MATH. Note that to construct this family we need that the dimension of this MATH group is constant when we vary MATH, but this is true thanks to REF . Each point MATH corresponds to an extension of the form MATH . It follows from the argument in the previous paragraph that all vector bundles in semistable conic bundles do occur in this family. Note that, if MATH is a conic bundle, MATH can be thought of as an element of MATH. Now choose MATH large enough so that for any line bundle MATH of degree MATH the following holds MATH for any MATH. Then MATH is constant when we vary MATH, and we can construct a (flat) family of conic bundles parametrized by MATH, and every semistable conic bundle of type MATH belongs to this family. |
math/9911198 | CASE: Follows from the exact sequence MATH and the fact that MATH. CASE: Assume that MATH. Then, in a basis adapted to MATH and then MATH, contradicting the fact that the conic bundle is smooth. CASE: If MATH, then MATH and then MATH, again contradicting the fact that the conic bundle is smooth. |
math/9911198 | Let MATH be a smooth MATH-semistable conic bundle. Let MATH be an isotropic vector subbundle. By REF , MATH. We have MATH, MATH (by REF ), and we check that MATH is a critical filtration. Then MATH, but MATH REF , and then MATH, which proves that MATH is semistable as an orthogonal bundle. Furthermore, if MATH is MATH-stable, then MATH, MATH and MATH is stable as an orthogonal bundle. Conversely, let MATH be an orthogonal semistable bundle. Let MATH be any vector subbundle. Following CITE let MATH, and let MATH be the vector subbundle generated by MATH. We have an exact sequence MATH where MATH is the subbundle of MATH generated by MATH. We have MATH. If MATH, then MATH, MATH, and MATH REF . Then MATH . If MATH, then MATH (by REF and the exact sequence REF ), and then MATH (because MATH is orthogonal semistable and MATH is isotropic). If MATH, then MATH (by REF ), and if MATH, then MATH (by REF ). In any case MATH . Now let MATH be a critical filtration. Then MATH is isotropic, MATH, and then MATH because MATH is isotropic and MATH is orthogonal semistable. This finishes the proof that MATH is MATH-semistable. Furthermore, if MATH is orthogonal stable, the last inequality is strict, and we obtain that MATH is MATH-stable. |
math/9911199 | In the definition of stack associated with a scheme we see that the only automorphisms are identities. |
math/9911199 | Follows from NAME lemma . |
math/9911199 | The implications MATH and MATH follow easily from the definitions. MATH . Assume that MATH is representable. We have to show that MATH is representable for any MATH and MATH. Check that the following diagram is Cartesian MATH . Then MATH is representable. MATH . First note that the Cartesian diagram defined by MATH and MATH factors as follows MATH . Both squares are Cartesian and by REF is representable, then MATH is also representable. |
math/9911199 | The morphism MATH is the composition of the natural morphism MATH (sending each category to the set of isomorphism classes of objects) and the morphism MATH given by the fact that the scheme MATH corepresents the functor. The morphism MATH was constructed in REF . The key ingredient needed to define MATH is the fact that the MATH action on the Quot scheme lifts to the universal bundle, that is, the universal bundle on the Quot scheme has a MATH-linearization. Let MATH be an object of MATH. Since MATH is a subscheme of a Quot scheme, and this universal bundle has a MATH-linearization. Let MATH be the vector bundle on MATH defined by the pullback of this universal bundle. Since MATH is MATH-equivariant, MATH is also MATH-linearized. Since MATH is a principal bundle, the vector bundle MATH descends to give a vector bundle MATH on MATH, that is, an object of MATH. Let MATH be a morphism in MATH. Consider the vector bundles MATH and MATH defined as before. Since MATH, we get an isomorphism of MATH with MATH. Furthermore this isomorphism is MATH-equivariant, and then it descends to give an isomorphism of the vector bundles MATH and MATH on MATH, and we get a morphism in MATH. To prove that this gives an equivalence of categories, we construct a functor MATH from MATH to MATH. Given a vector bundle on MATH, let MATH be the MATH-principal bundle associated with the vector bundle MATH on MATH. Let MATH be the pullback of MATH to MATH. It has a canonical MATH-linearization because it is defined as a pullback by a principal MATH-bundle. The vector bundle MATH is canonically isomorphic to the trivial bundle MATH, and this isomorphism is MATH-equivariant, so we get an equivariant morphism MATH, and hence an object of MATH. If we have an isomorphism between two vector bundles MATH and MATH on MATH, it is easy to check that it induces an isomorphism between the associated objects of MATH. It is easy to check that there are natural isomorphisms of functors MATH and MATH, and then MATH is an equivalence of categories. The morphism MATH is defined using the following lemma, with MATH, MATH the subgroup consisting of scalar multiples of the identity, MATH and MATH. |
math/9911199 | Let MATH be an object of MATH. There is a scheme MATH such that MATH factors MATH . To construct MATH, note that there is a local étale cover MATH of MATH and isomorphisms MATH, with transition functions MATH. Since these isomorphisms are MATH-equivariant, they descend to give isomorphisms MATH, and using this transition functions we get MATH. This construction shows that MATH is a principal MATH-bundle. Furthermore, MATH is also a principal MATH-bundle CITE, and in particular it is a categorical quotient. Since MATH is MATH-invariant, there is a morphism MATH, and this gives an object of MATH. If we have a morphism in MATH, given by a morphism MATH of principal MATH-bundles over MATH, it is easy to see that it descends (since MATH is equivariant) to a morphism MATH, giving a morphism in MATH. This morphism is not faithful, since the automorphism MATH given by multiplication on the right by a nontrivial element MATH is sent to the identity automorphism MATH, and then MATH is not injective. |
math/9911206 | If MATH for some projection MATH on the vertex MATH, then MATH so MATH is not fractal. We now suppose MATH is a subdirect embedding and prove by induction that MATH for all MATH. The induction basis, for MATH, is equivalent to the hypothesis. Now by induction MATH is a subdirect embedding, and each factor MATH maps to MATH by MATH. The composition of two subdirect embeddings is again subdirect, so MATH is subdirect. |
math/9911206 | Assume by induction that MATH is fractal and transitive on MATH, the induction starting at MATH. Since MATH is subdirect, MATH is transitive on MATH for all MATH, and since it is transitive on MATH, it is also transitive on MATH. |
math/9911206 | By the congruence property, MATH is a basis of neighbourhoods of the identity in MATH. |
math/9911206 | Assume MATH is a regular branch group over its subgroup MATH. Clearly MATH can be viewed, through MATH, as a subgroup of MATH, and is of finite index in MATH, so MATH is of finite index in MATH. The second implication holds because branch groups are infinite, and `finite index in an infinite group' is stronger than `non-trivial'. |
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