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math/0005265
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CASE: Using REF we get that MATH . So we get for every MATH that MATH . It follows that MATH. Moreover, we have for all MATH that MATH implying that MATH and hence MATH . CASE: This follows immediately from REF. CASE: Using the estimate MATH, this is an easy consequence of REF.
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math/0005265
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One implication is trivial. We will prove the other one. Therefore suppose that there exists a non-zero element MATH such that MATH. Let MATH denote the MATH-weak closure of MATH in MATH. Then MATH is a MATH-weakly closed left ideal in MATH so there exists a projection MATH in MATH such that MATH. Choose MATH. By the previous lemma, we know that MATH for every MATH. Therefore the normality of MATH implies that MATH for all MATH. In particular, we find that MATH which implies that MATH. From this all we conclude that MATH, thus MATH. Therefore REF implies that MATH or MATH. But the assumption at the beginning of this proof tells us that MATH, hence MATH and MATH.
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math/0005265
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Let MATH. Then MATH. Thus MATH and MATH. By REF , we get that MATH and MATH . By the left invariance of MATH, we know that MATH, thus MATH, so MATH. So we have proven that MATH is a two-sided ideal in MATH. Using the techniques of the proof of REF, we arrive at the conclusion that MATH is MATH-weakly dense in MATH.
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math/0005265
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Because MATH belongs to MATH, we have that MATH belongs to MATH and MATH . On the other hand, since MATH, the left invariance of MATH implies that the element MATH belongs to MATH and MATH . Therefore MATH belongs to MATH and MATH . Since MATH is a unitary corepresentation, we get that MATH . Combining this with REF , we see that MATH belongs to MATH.
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math/0005265
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By REF, we get for every MATH that MATH . REF tells us that MATH belongs to MATH and MATH . Combining these two facts, we see that MATH belongs to MATH and MATH .
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math/0005265
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Take an orthonormal basis MATH for MATH. Choose MATH. Then the MATH-weak lower semi-continuity of MATH implies that MATH so the previous lemma implies that MATH . By the lower semi-continuity of MATH, this implies that MATH for all MATH from which we conclude that MATH belongs to MATH. By result REF, We have moreover for all MATH that MATH .
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math/0005265
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We know that there exists a bounded net MATH in MATH such that MATH converges strongly-MATH to MATH and MATH is a bounded net that converges strongly-MATH to MATH. Then MATH is surely a bounded net that converges strongly-MATH to MATH. It follows easily from the previous lemma that MATH belongs to MATH and MATH . Therefore the net MATH is bounded and converges strongly-MATH to MATH . Using the MATH-strong-MATH-strong closedness of MATH, the lemma follows.
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math/0005265
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The element MATH belongs to MATH. Hence the previous lemma implies that MATH belongs to MATH, implying that MATH belongs to MATH.
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math/0005265
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By REF we get that MATH belongs to MATH implying that the element MATH belongs to MATH. Because MATH, the result follows.
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math/0005265
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Choose MATH and MATH. Notice that MATH. Using REF , we see that MATH . Let us now get hold of some interesting elements: CASE: Using REF , we get the existence of a net MATH in MATH such that CASE: MATH for all MATH, CASE: The net MATH converges strongly to REF. CASE: REF guarantees that the MATH-algebra MATH is a non-degenerate sub-MATH-algebra of MATH (MATH is defined in REF). Hence NAME 's density theorem implies the existence of a net MATH in MATH such that CASE: MATH for all MATH, CASE: MATH converges strongly-MATH to REF. Since MATH is dense in MATH, this implies easily the existence of a net MATH in MATH such that CASE: MATH for all MATH, CASE: The net MATH converges strongly-MATH to REF. Thus we get that CASE: MATH for all MATH, CASE: The net MATH converges strongly to REF. We will use these nets to construct the nets whose existence is claimed in the statement of the proposition. So let MATH be the directed set of all finite subsets of MATH. Set MATH and put the product ordering on this set. Choose MATH. We get first of all the existence of an element MATH such that MATH for all MATH. By the chain of equalities in REF , We have for MATH that MATH . This implies that CASE: We have for MATH that MATH . CASE: The net MATH converges strongly to MATH which is equal to MATH. From this we infer the existence of an element MATH such that MATH for all MATH. Now set MATH and MATH. Then CASE: MATH CASE: For all MATH, MATH . It is also clear that the last inequality implies that the net MATH converges strongly to REF.
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math/0005265
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Choose MATH and MATH. The previous lemma guarantees the exstence of a directed set MATH, a net MATH and a net MATH in MATH such that CASE: MATH for all MATH, CASE: The net MATH converges strongly to REF. By result REF, we have for every MATH that MATH belongs to MATH and MATH . From this it follows that MATH is a bounded net that converges strongly to MATH. Referring to result REF, we conclude from this that the net MATH converges to MATH.
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math/0005265
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Let us first prove the first statement. Choose MATH and MATH. Then we have for all MATH that MATH which belongs to MATH. This implies that MATH belongs to MATH. Let MATH. Using the inequality MATH, the above considerations imply that MATH belongs to MATH. Now we turn to the second statement. Therefore choose MATH, MATH and MATH an orthonormal basis for MATH. We have that MATH . Since MATH is an increasing net that converges strongly to MATH, the above equality and the MATH-weak lower semi-continuity of MATH implies that MATH . Take an orthonormal basis MATH of MATH. Then result REF implies that MATH . On the other hand, referring to result REF once again, we get that MATH . Comparing both expressions, we conclude that MATH .
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math/0005265
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Choose MATH, MATH, MATH such that MATH, MATH and MATH. REF tells us that MATH is the closed linear span of elements of the form MATH . Take a basis MATH of MATH. There exists MATH such that MATH for all MATH. Since MATH is in standard form, we can find an element MATH such that MATH for all MATH. From the previous lemma, we know that MATH belongs to MATH for all MATH and that the net MATH converges to MATH. CASE: Fix MATH for the moment. Then MATH . Therefore the convergence in REF implies that the net MATH converges to MATH. Therefore the proposition follows from the considerations in the beginning of the proof.
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math/0005265
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Choose MATH, MATH, MATH and MATH. Then MATH . Choose an orthonormal basis MATH for MATH. Choose MATH and MATH. By REF , we know that the net MATH converges to MATH . In the next part we will show that each of the sums in this net belongs to MATH. Therefore fix MATH. Because MATH belongs to MATH and MATH belongs to MATH, we get that MATH belongs to MATH and that MATH belongs to MATH and MATH . This implies that the net MATH is a bounded net that converges to MATH . Therefore result REF guarantees that the net MATH converges to MATH . But, using REF , we have for all MATH that MATH which clearly belongs to MATH. Therefore the convergence in REF implies that MATH belongs to MATH. Consequently, by referring to the convergence in REF , we see that the element MATH . The lemma follows now from REF .
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math/0005265
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Define the unitary element MATH such that MATH for all MATH. Then the following holds CASE: MATH for all MATH, CASE: MATH , CASE: MATH for all MATH. Call MATH the norm closure of MATH, then MATH is a MATH-weakly dense sub-MATH -algebra of MATH such that MATH belongs to the multiplier algebra MATH (for once, the tensor product is here the minimal MATH -algebraic tensor product). These last properties follows from the fact that MATH is a manageable multiplicative unitary in the sense of CITE. First we prove some small technical results CASE: Consider a NAME space MATH, MATH, MATH and MATH. Then the element MATH belongs to MATH and MATH . If MATH, we denote by MATH the element in MATH defined by MATH for all MATH. We have that MATH . Therefore MATH . Since MATH, this implies immediately that the element MATH belongs to MATH and MATH implying that the element MATH belongs to MATH and MATH . CASE: Consider MATH and MATH. Then MATH belongs to MATH and MATH . This is just a special case of the result proven in the previous part. CASE: Let MATH. Then MATH and MATH . We have that MATH. Because MATH belongs to MATH, this implies that MATH belongs to MATH and MATH . Having dealt with these elementary technical issues, we can start with the essential part of the proof. Let MATH denote the NAME of MATH. By REF above, we get that MATH . Therefore, MATH . Referring to REF above, we infer from this that MATH . Hence, invoking REF above, we conclude that MATH so that REF us lets conclude that MATH . Now, MATH . For MATH, we define the element MATH as MATH . By REF , we have for MATH, MATH and MATH that MATH . Similarly, using REF , we have for MATH and MATH that MATH . Therefore inclusion REF implies that MATH . Take a bounded net MATH in MATH such that MATH converges strongly-MATH to REF. Let MATH and MATH. Then MATH so that the normality of MATH implies that MATH is a bounded net that converges strongly-MATH to MATH . Using this fact and REF , we get for all MATH and MATH that the net MATH is bounded and converges strongly-MATH to MATH . Therefore result REF and inclusion REF above imply that MATH . Because MATH is supposed to be in standard form, this becomes MATH .
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math/0005265
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Define MATH to be the projection on MATH. Also define MATH, then MATH implying that MATH is a partial isometry in MATH. Define MATH to be the final projection of MATH. Thus, since MATH, we get that MATH . Because MATH, we have that MATH, implying that MATH. Hence MATH . Arguing as in the proof of REF, we conclude from this that MATH. So we get for every MATH that MATH and thus MATH by result REF This implies that MATH. Therefore there exists a closed subspace MATH of MATH such that MATH is the orthogonal projection on MATH, thus MATH. Hence the previous proposition implies that MATH thus MATH and the lemma follows.
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math/0005265
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By REF and the previous lemma, we know that MATH . Since MATH is an isometry, this implies that MATH and hence MATH. Using the previous lemma once more, we get that MATH so that MATH is unitary. Because MATH, the proposition follows.
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math/0005265
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We may assume that MATH and MATH. We have for MATH that MATH . Therefore the right invariance of MATH implies that MATH. Using REF, we conclude from this that MATH belongs to MATH. By right invariance of MATH we have moreover for all MATH that MATH implying that MATH.
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math/0005265
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Let MATH denote the densely defined closed linear map from within MATH into MATH such that MATH is a core for MATH and MATH for all MATH. Recall that the pair MATH,MATH is by definition the polar decomposition of MATH, that is, MATH. Choose MATH and MATH. Let MATH. Then MATH belongs to MATH and MATH. Moreover, MATH which implies that MATH belongs to MATH. It follows that MATH belongs to MATH and MATH . Since such elements MATH form a core for MATH, we conclude that MATH . Taking the adjoint of this inclusion we find that MATH . Since MATH, we get for every MATH, that the element MATH belongs to MATH and MATH. Since MATH, this implies easily that MATH for all MATH and MATH. So we get for MATH and MATH. MATH or in other words, MATH . Since MATH is a core for MATH and MATH is a core for MATH, it follows easily that MATH for all MATH and MATH. By REF, we conclude that MATH . Taking the adjoint of this equation of this equation and multipliying it with MATH from the left and the right, we arrive at the other inclusion.
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math/0005265
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Apply the previous proposition with MATH and some NAME MATH for MATH. In this case , MATH and MATH is the modular operator for MATH in this NAME. Then we get for MATH and MATH that MATH and the corollary follows.
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math/0005265
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Call MATH the modular operator of the pair MATH, MATH. By REF we know that MATH for all MATH. Applying the previous proposition to the pair MATH, MATH (in which case MATH and MATH), we get that MATH. If we apply the previous proposition to the pair MATH, MATH in which case MATH, we get that MATH. So we get for all MATH that MATH .
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math/0005265
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By the previous proposition, we have for MATH that MATH from which we conclude that MATH belongs to MATH. By REF we also get that MATH . Therefore REF implies the existence of a n.f.s. weight MATH on MATH such that MATH for all MATH. REF tells us that MATH for all MATH. Hence, MATH.
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math/0005267
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Partially ordering the up-sets and (separately) the down-sets in MATH by set inclusion, let MATH be a minimal up-set in MATH containing the vertices of some path from MATH to MATH. (By assumption, MATH is an up-set containing such a path, so MATH exists.) Let MATH be a minimal down-set in MATH such that MATH contains the vertices of some path from MATH to MATH. (Again, MATH is a down-set satisfying this condition, so MATH exists.) Let MATH be a path from MATH to MATH. We can label minimal and maximal elements of MATH and count the segments of MATH alternating upward and downward as follows: As the path MATH is traversed from MATH to MATH, the path traces out either an upward or a downward path from MATH to MATH, either a downward or an upward path from MATH to MATH, etc., alternatingly, as illustrated in MATH . Then we can find a path MATH in MATH, with ground set MATH, from MATH to MATH so that MATH is as small as possible, that is, a path MATH from MATH to MATH satisfying MATH such that the number MATH of segments alternating upward and downward is smallest among such paths. We first claim that any two minimal elements in MATH are incomparable in MATH and that any two maximal elements in MATH are incomparable in MATH. To see this, suppose that MATH for two minimal elements MATH and MATH in MATH. Then there is some downward path MATH in MATH from MATH to MATH. Let MATH be the first element among the MATH's with MATH which belongs to MATH. So MATH. Since MATH, MATH are in MATH, all the vertices of the downward path MATH are also. Replacing the part of MATH between MATH and MATH by the path MATH (possibly in reverse order), we get a path from MATH to MATH with a lower value of MATH, which is impossible. Thus, any two minimal elements in MATH are incomparable in MATH. The same holds for any two maximal elements in MATH. To finish the proof, we claim that the path MATH is the induced subposet of MATH on MATH, that is, that no pairs of elements of MATH are comparable in MATH, beyond those specified by the poset MATH. To see this, suppose that there are elements MATH and MATH of MATH which are comparable in MATH but not in MATH. By a replacement scheme similar to that employed in the preceding paragraph, we can get a path MATH (with ground set MATH) from MATH to MATH in MATH which bypasses some MATH. If we define MATH to be the down-set in MATH generated by MATH and (dually) MATH to be the up-set MATH then MATH and MATH, and MATH. If MATH is minimal in MATH, then (by the preceding paragraph) MATH is incomparable in MATH with any minimal element of MATH, and therefore MATH. Similarly, if MATH is maximal in MATH, then MATH. Thus, we have contradicted the minimality of MATH or MATH according as MATH is minimal or maximal in MATH, which completes the proof.
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math/0005267
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Let MATH be the poset on ground set MATH obtained by deleting the edge joining MATH and MATH from the NAME diagram of MATH. Since MATH is a path from MATH to MATH in MATH, by REF there is a path MATH from MATH to MATH which is an induced subposet of MATH. Then, the cycle MATH is as desired.
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math/0005267
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Let MATH and MATH be two unequal upward paths from MATH to MATH in MATH. Without loss of generality, the two paths differ in their first step, that is, MATH. Clearly MATH and MATH are incomparable. Let MATH. Then MATH is nonempty because MATH. Let MATH be a minimal (in MATH) element of MATH. Let MATH and MATH be upward paths from MATH to MATH and from MATH to MATH, respectively. Then, for any MATH and MATH, MATH and MATH are incomparable; otherwise, the minimality of MATH is contradicted. Thus, the ground set MATH gives the desired induced subposet.
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math/0005267
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If MATH has a diamond as an induced subposet, then by REF we can find a subdivided diamond MATH which is an induced cyclic subposet of MATH. Clearly MATH has height at least MATH. Suppose now that MATH has no diamond as an induced subposet. Let MATH be a cycle with height at least MATH such that MATH. Let MATH and let MATH with ground set MATH be the subposet of MATH via induced cover graph. So MATH is a path in MATH. By REF , there is a path MATH in MATH from MATH to MATH which is an induced subposet of MATH. If MATH is an upward path from MATH to MATH, then MATH has two distinct upward paths from MATH to MATH which, by REF , contradicts our assumption. Thus, MATH is not an upward path and in particular MATH and MATH are incomparable in MATH but comparable in MATH. This implies that MATH is not an induced subposet of MATH. Let MATH be a minimal segment of the path MATH which is not an induced subposet of MATH, that is, a segment MATH such that REF MATH is not an induced subposet of MATH, and REF any proper segment of MATH is an induced subposet of MATH. Then MATH and MATH must be incomparable in MATH but comparable in MATH. Without loss of generality, we may assume that MATH in MATH. Then there is a downward path MATH in MATH from MATH to MATH. Let MATH be a cycle in MATH. Since MATH is an induced subposet of MATH, we must have MATH and in particular MATH has height at least MATH. We claim that MATH is an induced subposet of MATH; establishing the claim will complete the proof of the lemma. To prove the claim, suppose that MATH and MATH are comparable in MATH but incomparable in MATH for some MATH and MATH. Then MATH is comparable with MATH or MATH in MATH according as MATH or MATH. If the pair (either MATH or MATH) are comparable in MATH, then there are two unequal upward paths (the one through MATH and the other consisting of a segment of MATH) with common ends in MATH. By REF , this contradicts our diamond-free assumption. If the pair is incomparable in MATH, then some proper segment of MATH is not an induced subposet of MATH, and this contradicts the minimality of MATH.
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math/0005267
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CASE: If MATH is the diamond as in REF , then there are at least two unequal upward paths from MATH to MATH. By REF , MATH has a cycle (namely, a subdivided diamond). CASE: Suppose that MATH is a subdivision of the MATH-crown displayed and labeled in REF . Since by REF has height at least MATH, without loss of generality we may assume that there exists MATH such that MATH. Then we can find an upward path MATH in MATH from MATH to MATH with height at least MATH. Since MATH is incomparable in MATH with each of MATH and each of MATH, no upward path in MATH from any MATH to any MATH contains the directed edge MATH unless MATH. Let MATH be the cover graph of MATH. Then MATH and MATH are connected in the graph MATH, which implies that MATH is non-acyclic. CASE: Suppose that MATH is the MATH-crown for some MATH, as displayed and labeled in REF . Since the set MATH is nonempty, we can find a maximal element MATH in the set MATH. Then MATH is incomparable with MATH and therefore the subposet of MATH induced by the ground set MATH is again a MATH-crown. Thus, we may without loss of generality assume that the MATH-crown has no element MATH satisfying MATH and MATH. Let MATH (with MATH possible) be an upward path from MATH to MATH. By our assumption, an upward path from MATH to MATH does not contain the directed edge MATH. Furthermore, no upward path from any MATH to any MATH [except when MATH] contains the directed edge MATH either; otherwise, MATH is not an induced subposet of MATH. Thus we see that MATH and MATH are connected in the graph MATH, which implies that MATH is non-acyclic. CASE: Suppose that MATH is the poset as in REF . Since the set MATH is nonempty, we can find a maximal element MATH in the set MATH. If MATH and MATH are comparable, then we have MATH. Noticing that MATH is incomparable with MATH and MATH, the cycle MATH is an induced subposet of MATH which is a subdivided MATH-crown with height MATH. Then REF implies that MATH is non-acyclic. So we may assume that MATH and MATH are incomparable. But then an upward path MATH in MATH from MATH to MATH (with MATH possible) does not share the directed edge MATH with any upward path from MATH to either MATH or MATH. Moreover, an upward path from MATH to MATH does not contain the directed edge MATH; otherwise, maximality of MATH is contradicted. Thus, we see that MATH and MATH are connected in the graph MATH, which implies that MATH is non-acyclic.
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math/0005267
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We prove the claim of REF by induction over the cardinality of MATH. The claim is vacuous when MATH. We now suppose that the claim is true for an acyclic poset MATH when MATH for fixed MATH, and consider an acyclic poset MATH with cardinality MATH. Let MATH be a leaf in MATH. Without loss of generality, we assume that MATH is maximal in MATH; thus, there is a unique element MATH which is covered by MATH. We consider the subposet MATH of MATH induced by the ground set MATH. Let MATH be a stochastically monotone system of probability measures on MATH. Then the subsystem MATH is stochastically monotone. Since MATH is an acyclic poset and MATH, by the induction hypothesis there exists a probability measure MATH on MATH which realizes the monotonicity. Since MATH, by REF there exists an upward kernel MATH satisfying REF for the pair MATH of probability measures. We can define a probability measure MATH on MATH by MATH where MATH denotes the projection from MATH to MATH and MATH denotes the MATH-projection from MATH to MATH for each MATH. In words, this says simply that we couple together the probability measures MATH, MATH using MATH and then extend the multivariate coupling to MATH using the upward kernel MATH from MATH to MATH. Observe that MATH couples the probability measures MATH: MATH correctly, and that MATH in MATH implies MATH in MATH. So coupling MATH to MATH correctly automatically couples MATH to each MATH REF correctly. Thus, MATH realizes the monotonicity of MATH, and therefore the claim holds for MATH.
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math/0005267
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Notice that for any MATH, MATH . Since MATH, we deduce MATH. Thus we obtain MATH which completes the proof.
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math/0005267
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Since MATH, we see that the system MATH is stochastically monotone. To see that it is not realizably monotone, suppose that there exists a system MATH of MATH-valued random variables which realizes the monotonicity of MATH. By applying REF repeatedly, we (almost surely) have MATH. But then (after perhaps taking care of null sets) MATH realizes the monotonicity of MATH, which is a contradiction.
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math/0005267
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We have already seen that a poset MATH of Class B has either the diamond or a crown as an induced subposet. If MATH has an induced MATH-crown, then MATH is in Class B by definition. If MATH has an induced subposet which is either the diamond or a MATH-crown for some MATH, then, by REF , MATH is non-acyclic and therefore MATH is in Class B, again by definition.
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math/0005267
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Let MATH be a non-acyclic poset and MATH be either the diamond or a MATH-crown for some MATH. We will construct a stochastically monotone system MATH of probability measures on MATH which is not realizably monotone, by dividing the construction into two cases. CASE: Suppose that MATH has a diamond MATH as an induced subposet. Let MATH be labeled as in REF . By REF , there exists a stochastically monotone system MATH of probability measures on MATH which is not realizably monotone. It then suffices by REF to show that the system can be enlarged to a stochastically monotone system MATH of probability measures on MATH. For this, define a partition MATH, and MATH of MATH by MATH . Then we can extend MATH to MATH by putting MATH for each MATH. It is routine to check that this extension maintains stochastic monotonicity, that is, that if MATH, then MATH. This is true if MATH for some MATH and also if MATH or MATH. If MATH and MATH, then MATH. If MATH and MATH, then MATH. So stochastic monotonicity is clear. CASE: Suppose that MATH has no diamond as an induced subposet. By REF , MATH has an induced cyclic subposet MATH. In the same way as what we did in REF , we can label the cycle MATH as illustrated in MATH . By our REF assumption, MATH must be a subdivided MATH-crown for some MATH. Let MATH be the MATH-crown MATH. By REF (and then further using REF , if necessary), there exists a stochastically monotone system MATH of probability measures on MATH which is not realizably monotone. Let MATH and MATH be defined as in REF so that MATH for all MATH. Consider the partition MATH: MATH of MATH, where MATH and MATH for MATH. By letting MATH and MATH, we can define a system MATH of probability measures on MATH by MATH this system extends MATH. We claim that MATH is stochastically monotone. Let MATH. If MATH or MATH, then MATH. This is also trivial if MATH. If MATH, then MATH, so MATH. If MATH, then MATH. We need only show that it is impossible to have both MATH and MATH. Indeed, then MATH, but for some MATH we have MATH. But then there are two distinct upward paths from MATH to MATH in MATH, namely the one using edges in the cover graph MATH and one containing MATH. This violates REF , since we are assuming that MATH has no induced diamond. Thus, we have established the claim and, by REF , MATH cannot be realizably monotone.
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math/0005267
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If MATH, then MATH is both a down-set and an up-set in MATH. If MATH, then MATH and there is a unique predecessor of MATH; otherwise, the uniqueness of the path is contradicted. Let MATH be the predecessor of MATH. Clearly, MATH belongs to the cover graph of MATH. Suppose that MATH covers MATH in MATH. We claim that MATH is a down-set in MATH, that is, that MATH whenever MATH in MATH for some MATH. [Since we have the same rooted tree MATH for the dual MATH, in proving the claim we will also settle the case that MATH covers MATH in MATH.] To see this, look at the paths from the root MATH to MATH and MATH, say, MATH from MATH to MATH and MATH from MATH to MATH. For some MATH, the two paths descend the same vertices until the MATH-th vertex, then split at the MATH-st vertex. The path from MATH to MATH is then MATH, which is downward in MATH by assumption. That MATH implies that MATH for some MATH. Furthermore, we have MATH; otherwise, the downward path from MATH to MATH contains the directed edge MATH, which is impossible. Thus, the path from MATH to MATH contains the vertex MATH, which implies that MATH.
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math/0005267
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By REF and its trivial consequence REF , MATH clearly implies REF - REF . We proceed to the converse. Since any up-set MATH in MATH is the disjoint union of the components MATH of the subgraph of MATH induced by MATH and MATH are all up-sets in MATH, to prove MATH it suffices to show REF for every up-set MATH which induces a connected subgraph of MATH. If a set MATH induces a connected subgraph of MATH, then we can find MATH and incomparable elements MATH of MATH in MATH so that MATH where (as usual) the union is empty if MATH. Furthermore, suppose that MATH is an up-set in MATH. If MATH, then MATH is trivially an up-set in MATH; otherwise, MATH covers its predecessor MATH in MATH and, by REF , MATH is an up-set in MATH. Similarly, MATH is a down-set in MATH for each MATH. Therefore, we have MATH which establishes the sufficiency of REF - REF .
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math/0005267
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Suppose that the hypotheses in REF hold. If MATH, then MATH and the inequality clearly holds. Otherwise, MATH for every MATH. Since the path MATH is upward and MATH covers MATH in MATH, by REF MATH is a down-set in MATH. Therefore we have MATH for all MATH, as desired. REF is reduced to REF by considering the dual MATH.
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math/0005267
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We first claim that for every MATH, there are incomparable elements MATH of MATH in MATH which satisfy both the hypotheses of one of REF and also MATH . We will show this by induction over the cardinality of MATH. If MATH, then MATH is a minimal element in MATH and indeed MATH. Suppose that the claim holds for any MATH such that MATH. Let MATH satisfy MATH. If MATH, then recall that MATH is a leaf in MATH so that MATH is a singleton, say MATH. By the induction hypothesis, we can find incomparable elements MATH of MATH in MATH which satisfy both the hypotheses of one of REF for MATH. If REF obtains, then MATH A similar derivation concludes that MATH when REF obtains. Therefore, we have MATH, which proves REF . Furthermore, the claim holds for MATH. If MATH, then MATH has a predecessor MATH and successors MATH for some MATH (recalling our assumption MATH). Without loss of generality, we may assume that MATH covers MATH in MATH. (We can treat the case that MATH covers MATH in MATH in exactly the same way by considering the dual MATH.) Then, by REF , MATH is a down-set in MATH. Since MATH cannot have a bowtie as an induced subposet, only the following three cases can occur: CASE: MATH covers MATH in MATH, or REF MATH cover MATH in MATH, or, with MATH, REF MATH cover MATH, and MATH covers MATH, in MATH. The induced subposet of MATH on MATH for each of these three cases is illustrated in the following figure. CASE: If MATH, the claim is obvious. Otherwise, we have MATH. For each MATH, the section MATH is a down-set in MATH by REF , and therefore by the induction hypothesis we have incomparable elements MATH of MATH in MATH satisfying REF for MATH. Thus we have MATH and MATH for all MATH. Since the MATH's are incomparable in MATH, the claim holds for MATH. CASE: For each MATH, MATH is an up-set in MATH by REF , and therefore by the induction hypothesis we have incomparable elements MATH of MATH in MATH satisfying REF for MATH. By applying REF to REF , we have MATH which implies that MATH. Thus the claim holds for MATH. CASE: By REF , MATH is a down-set in MATH and therefore by the induction hypothesis we can find incomparable elements MATH of MATH in MATH satisfying REF for MATH. Since MATH has no bowtie as an induced subposet and MATH in MATH, we have MATH. Thus, we can find some MATH so that MATH . For each MATH, the section MATH is an up-set in MATH by REF , and therefore by the induction hypothesis we have incomparable elements MATH of MATH in MATH satisfying REF for MATH. By applying REF to REF , we obtain MATH which implies that MATH. Thus, we have established the claim. In order to show REF , it suffices to show that if MATH is an up-set in MATH then we have MATH . (Again, we can verify the case that MATH is a down-set in MATH by considering the dual MATH.) Suppose that MATH is an up-set in MATH. Then we can find incomparable elements MATH of MATH in MATH satisfying REF . By applying REF to REF , we have MATH . This completes the proof.
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math/0005267
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Suppose that there exists an induced subposet MATH of MATH which is poset-isomorphic to one of the posets REF - REF . If there is an acyclic poset MATH which has MATH as an induced subposet, then MATH is also an induced subposet of MATH; by REF , this is impossible. We have thus shown that REF . To prove REF , observe that a cycle is simply a subdivision of either the diamond or a crown. So if MATH has an induced cyclic subposet MATH which is not a MATH-crown, then we can find an induced subposet of MATH (automatically, of course, an induced subposet of MATH) that is one of the posets REF - REF . Thus, the failure to satisfy REF implies the existence of an induced subposet which is one of the posets REF - REF .
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math/0005267
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Let MATH be the collection of all subsets MATH of MATH such that the subposet via induced cover subgraph of MATH on the ground set MATH is a poset of the form REF for some MATH. Note that if a subposet MATH via induced cover subgraph is of the form REF then MATH is an induced subposet. By REF , MATH has an induced cyclic subposet which is necessarily a MATH-crown; thus, MATH is nonempty. Choose a maximal subset MATH from MATH. Then let MATH be the subposet of MATH induced by MATH and label it as in REF . Consider the NAME diagram of MATH as represented in the plane. First remove the arcs from each of MATH to each of MATH. Since the elements MATH are all drawn above the elements MATH, we can insert a new vertex MATH above the elements MATH but below the elements MATH, and then add new arcs from MATH to MATH and from MATH to MATH. This creates a new poset MATH with ground set MATH, as illustrated in MATH . The subposet of MATH induced by MATH introduces the arc from each of MATH to each of MATH, thus restoring the NAME diagram of MATH. We claim that MATH does not have any cycle which contains an upward path MATH with MATH. Granting the claim for the remainder of this paragraph, we define MATH . By convention, we include MATH both in MATH and in MATH. Let MATH and MATH be the subposets of MATH induced by MATH and MATH, respectively. Clearly, MATH and MATH are both connected posets. By observing that MATH is connected, we find MATH. The claim implies that MATH and that there are no edges between MATH and MATH in the cover graph of MATH. Therefore, MATH, which implies REF . Since MATH, MATH, and MATH, we have MATH, as desired in REF . To see REF for MATH (the same argument works for MATH), suppose that MATH has an induced subposet MATH which violates REF . If MATH, then let MATH; otherwise, let MATH. Then the subposet MATH of MATH induced by MATH is poset-isomorphic to MATH. Furthermore, if MATH is a cycle in MATH, then MATH is so in MATH. But the existence of such an induced subposet of MATH contradicts the assumed REF . So REF holds, and REF is established modulo a proof of the claim. Now we show the claim. To prove this by contradiction, we further enlarge the poset MATH as follows. We first remove the arcs from MATH to each of MATH from the NAME diagram of MATH. Then a new element MATH is drawn above MATH but below each of MATH, and the arc from MATH to MATH and the arcs from MATH to each of MATH are introduced in the diagram. This creates a new poset MATH, as illustrated in MATH . Clearly, MATH is an induced subposet of MATH. If we can show that MATH has no cycle which contains the edge MATH, then MATH has no cycle which contains an upward path MATH for some MATH, establishing the claim and REF . Thus, to obtain a contradiction, suppose that MATH has a cycle which contains the edge MATH. By REF , we may assume that such a cycle, say MATH, is an induced cyclic subposet of MATH. Then the cycle MATH in MATH is an induced subposet of MATH, and therefore by REF the induced cyclic subposet MATH of MATH must be a MATH-crown, and therefore MATH. Note that MATH for MATH; otherwise, the cycle MATH cannot be an induced subposet of MATH. Write MATH and MATH. Now consider the comparability in MATH between MATH and MATH. If MATH is comparable with some MATH of MATH, then either MATH or MATH, contradicting our knowledge that MATH and MATH are edges in the cover graph of MATH. Thus MATH is incomparable with each of MATH. Similarly we can see that MATH is incomparable with each of MATH. If MATH is comparable with some MATH of MATH with MATH, then MATH; otherwise, MATH, contradicting the assumption that MATH and MATH are incomparable. Suppose that there is an upward path MATH in MATH from MATH to MATH with MATH. Then it is not hard to see that the cycle MATH in MATH is an induced subposet of MATH with height MATH, contradicting REF . Therefore, MATH must cover MATH in MATH. Note that MATH cannot be comparable with all of MATH, since MATH. Similarly, MATH cannot be comparable with all of MATH (but MATH may cover some of them). Thus, we can find some elements MATH, MATH of MATH so that MATH is incomparable with MATH and MATH is incomparable with MATH. We have now found the subposet of MATH induced by MATH to be poset-isomorphic to the poset REF . This contradicts REF .
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math/0005267
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Suppose that a poset MATH satisfies REF . We will make an inductive argument over the cardinality of MATH. But if MATH is acyclic, then the argument is vacuous. In particular, MATH with cardinality at most MATH is acyclic. Now let MATH be a connected non-acyclic poset with cardinality MATH. By REF , there exists a pair MATH and MATH of connected posets satisfying REF - REF . Then, by the induction hypothesis and REF - REF , MATH and MATH can be enlarged to acyclic posets MATH and MATH, respectively. Since MATH, the ground sets MATH and MATH can be given so that MATH. Let MATH. Then MATH is acyclic. Furthermore, MATH is an induced subposet of MATH. By REF , MATH is an induced subposet of MATH, as desired.
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math/0005267
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Suppose first that a non-acyclic poset MATH is not enlargeable to an acyclic poset. Then, by REF , MATH has an induced subposet MATH which is one of the posets REF - REF . If MATH is the diamond, then by REF MATH has a cycle with height at least MATH. Thus REF implies that monotonicity equivalence fails for MATH. If MATH is either the MATH-crown for some MATH or the double-bowtie poset REF , then by REF - REF monotonicity equivalence fails for MATH. Suppose now that MATH is a subdivision of the MATH-crown as displayed and labeled in REF and has height at least MATH. Then we may assume that there exists MATH such that MATH in MATH. So we find an upward path MATH in MATH from MATH to MATH with height at least MATH. Since MATH is incomparable in MATH with each of MATH and each of MATH, no upward path in MATH from any MATH to any MATH contains either MATH or MATH as an edge unless MATH. Let MATH be the cover graph of MATH. Then there exists a path MATH from MATH to MATH in the graph MATH. Thus, MATH has a cycle MATH with height at least MATH. Therefore, by REF monotonicity equivalence fails for MATH. Suppose now that a poset MATH is enlargeable to an acyclic poset. We will prove that monotonicity equivalence holds for MATH by induction over the cardinality of MATH. If MATH is acyclic, then, by REF , MATH is a poset of monotonicity equivalence. Thus, if MATH, then MATH is acyclic and therefore a poset of monotonicity equivalence. Now let MATH be a non-acyclic poset with cardinality MATH and let MATH be a stochastically monotone system of probability measures on MATH. By REF , there exists a pair MATH and MATH of posets satisfying REF - REF . Let MATH be all the elements covered by MATH, and let MATH be all the elements covering MATH in MATH. Since MATH is an induced subposet of MATH, we have MATH for all MATH and all MATH. By REF , we can find a probability measure MATH on MATH such that MATH for all MATH and all MATH. Let MATH. Then we can enlarge the system MATH to a system MATH, maintaining stochastic monotonicity. Note that the subsystems MATH and MATH are also stochastically monotone. Since [by REF ] MATH [respectively, MATH] is enlargeable to an acyclic poset, by the induction hypothesis there is a system MATH [respectively, MATH] of MATH-valued random variables which realizes the monotonicity of MATH [respectively, of MATH]. Let MATH and MATH. We can define a probability measure MATH on MATH by MATH where MATH . Then MATH realizes the monotonicity of MATH. By REF , MATH is realizably monotone; thus, monotonicity equivalence holds for MATH.
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math/0005267
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Suppose first that MATH. Let MATH and MATH denote the distribution functions of MATH and MATH, respectively. Let MATH be fixed, MATH, and MATH. If MATH, then REF obviously holds. If MATH, then we have MATH for all MATH such that MATH. By REF , the section MATH is a down-set for every MATH, which implies, by REF , that MATH in MATH. If MATH, then we have MATH for all MATH such that MATH. Again by applying REF , we obtain MATH in MATH. In any case, REF holds. Now suppose that REF holds. Then we can construct a pair MATH of MATH-valued random variables satisfying REF - REF via MATH with a single random variable MATH uniformly distributed on MATH. By REF , we have MATH. This completes the proof.
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math/0005267
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Let MATH be a stochastically monotone system of probability measures on MATH. Let MATH be a random variable uniformly distributed on MATH. Then we can construct a system MATH of MATH-valued random variables satisfying REF via MATH . By REF , the system MATH satisfies REF ; thus, MATH is realizably monotone.
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math/0005268
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The dimension statement is straightforward. To prove the relation, we pull MATH apart along MATH, and study the corresponding moduli spaces (see REF for more discussion on such matters). Let MATH denote the complement MATH, given a cylindrical-end metric modeled on the product metric MATH, where MATH is given its standard, round metric. Note that this metric can be extended over both MATH and MATH to give metrics with non-negative scalar curvature. Consequently, the moduli spaces of solutions over MATH, MATH, and MATH consist entirely of smooth reducibles (that is, the moduli spaces are identified with MATH, MATH, and a point respectively). Let MATH denote the moduli space of finite energy solutions to the NAME equations over MATH in the MATH structure MATH. Thus, we can think of the boundary map as a map MATH . Gluing theory gives a diffeomorphism for all sufficiently large MATH: MATH where MATH denotes the metric on MATH with neck-length MATH and MATH corresponds to the unique reducible on MATH which extends to MATH. Consequently, MATH since MATH is represented by the holonomy class around MATH (see REF ). Similarly, gluing gives a diffeomorphism of MATH and consequently MATH . Together, REF prove the proposition.
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math/0005268
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This is a combination of REF when MATH, and REF in the remaining case.
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math/0005268
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This follows from REF (see also the proof of REF in the perturbed case).
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math/0005268
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According to REF , the space MATH is identified with the space of degree MATH line bundles over MATH with non-trivial MATH. The forgetful map MATH which takes a degree MATH divisor, thought of as a complex line bundle with section, to the underlying complex line bundle gives the surjection to this locus.
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math/0005268
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When MATH, this is proved in REF, where it appears as REF . The remaining case is covered by REF .
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math/0005268
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This is a dimension-counting argument. Suppose we have a sequence MATH, for some increasing, unbounded sequence MATH of real numbers. By local compactness, there is a subsequence which converges in MATH to a pair of configurations MATH and MATH over MATH and MATH respectively. By the usual compactness arguments (see CITE), the total variation of the NAME functional of MATH over the cylinder MATH remains globally bounded (independent of MATH), so MATH and MATH both have finite energy. First, we prove that either Hypothesis (H REF) or (H REF) is satisfied. There are a priori four cases, according to which critical manifolds MATH and MATH lie in. CASE: The case where MATH while MATH is excluded because MATH. CASE: The case where MATH while MATH is excluded by a dimension count, as follows. In this case, we see that MATH. But MATH so MATH . It follows from REF that MATH is smooth of the expected dimension, so from the usual transversality results, the above intersection is generically empty. CASE: Suppose that MATH and MATH. Then we see that MATH but, according to REF MATH which is generically empty. Thus, it follows that MATH must be reducible. Moreover, according to REF , MATH which is also generically empty. Hence, MATH and MATH satisfy Hypotheses (H REF). CASE: If MATH and MATH both lie in MATH, then the Hypotheses (H REF) are satisfied. The assertion at the beginning of the proposition follows easily.
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math/0005268
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The compactness of MATH and MATH follows from the usual compactness arguments, together with the facts that the NAME functional is real-valued, MATH, and there are no other critical manifolds. Compactness of MATH follows from this, together with a straightforward dimension count (see the discussion above in the proof of REF , part (P REF)).
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math/0005268
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Recall that MATH consists entirely of reducibles all of which are smooth, according to REF ; thus, gluing theory identifies the moduli spaces MATH for large MATH with MATH. (See also CITE, where this result appears as REF .)
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math/0005268
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The moduli space MATH comes equipped with an obstruction bundle MATH, defined by MATH. (whose MATH-theory class canonically extends over all of MATH). The dimension count in REF guarantees that each solution in MATH extends uniquely to a smooth reducible over MATH. Thus, gluing theory gives that MATH where MATH denotes the NAME class of a bundle (or MATH-theory element). The NAME formula says MATH. Using the index theorem for families, together with the holomorphic interpretation of the obstruction bundle MATH given in REF , it is a straightforward computation that the total NAME class of MATH is MATH (see also REF ); thus, MATH . Putting all this together, we have that MATH where MATH. Since MATH and MATH, the proposition then follows from REF .
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math/0005268
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Gluing shows that MATH . According to REF , MATH either has degree MATH, or MATH is empty. The latter case would force MATH (for the given genus and self-intersection number). To rule out this latter case, we need only look at an example where the irreducible term is non-zero. Let MATH be a ruled surface MATH over MATH associated to the line bundle with NAME number MATH. Let MATH denote the section with self-intersection number MATH, and fix any MATH. Let MATH denote the MATH structure over MATH given by MATH, where MATH is the canonical MATH structure on MATH associated to the NAME structure, and MATH denotes a fiber in the ruling. It is easy to see that MATH, as the corresponding space of divisors is empty (see REF ). Moreover, we know that MATH (compare REF ). Thus, in light of REF , we have examples where MATH, forcing the degree to be non-zero.
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math/0005268
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This a standard argument from NAME theory. A general discussion of compactness results for the anti-self-duality equation can be found in CITE (see especially REF); so we sketch the argument here only briefly. A sequence MATH converges in MATH, after passing to a subsequence, to some solution MATH to the NAME equations on MATH. Since each of the MATH have finite energy, so does MATH; thus, it has a boundary value. If MATH, then the length-energy estimates of NAME CITE can be used to show that the convergeance is MATH as in CITE. If, on the other hand, MATH, it must be the case that MATH. Now, let MATH be the number so that MATH . Clearly, MATH. After passing to a subsequence, we can find a configuration MATH so that the sequence MATH, viewed as a sequence of configurations over MATH, converges in MATH to MATH. In fact, MATH must solve the NAME equations and it must have finite energy, so MATH. The usual length-energy estimates then guarantee that the boundary values match up.
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math/0005268
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If a sequence MATH converges to an ideal point MATH then there is a divergent sequence MATH of real numbers so that MATH where MATH is the map induced by translation by MATH on the first coordinate. Since each path has finite energy, continuity of the restriction maps (see CITE) guarantees that MATH .
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math/0005268
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Gluing describes the end of MATH as a fibered product MATH where the superscript denotes based versions of the moduli spaces. This gives the space of ideal solutions a disk-bundle neighborhood in MATH.
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math/0005268
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We must show that MATH is homologous to MATH. It suffices to verify this over the subset MATH, since the complement has codimension two, and the classes in question are one-dimensional. Over the subset, now, the claim is easy to verify. On MATH, MATH is represented by MATH, the holonomy around a representative of MATH ``at infinity" (see REF ); while MATH is represented by MATH, where MATH. Now, the cylinder MATH provides a homotopy between MATH and MATH.
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math/0005268
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Clearly, the difference MATH is the first NAME class of the circle bundle MATH. Here, MATH denotes the moduli space based at MATH; see REF. To prove the proposition, we must verify that this bundle admits a section MATH in the complement of MATH (that is, over MATH) and that, with respect to a trivialization of the circle bundle over a disk transverse to the submanifold, the restriction of the section to the boundary induces a map from the circle to the circle which has degree one. The section MATH is induced by parallel transport, as follows. Let MATH be a half-infinite arc formed by joining MATH to any arc which connects MATH to MATH. Over the point MATH, parallel transport via MATH along MATH induces a homomorphism in MATH. We now verify that the trivialization induces a degree one map around circles transverse to the submanifold. For any point in the submanifold MATH fix fibers MATH . These choices induce a trivialization of MATH over a disk in MATH transverse to MATH (obtained by varying the gluing and translation parameters). Calculating the desired degree amounts to seeing how the holonomy along MATH varies as the gluing parameter is rotated. But holonomy along any path which crosses the gluing region once varies as a degree one function of the gluing parameter.
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math/0005268
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We can reduce to a corresponding statement for configurations over MATH, as follows. Let MATH be a line bundle over MATH, so that MATH. Then, pull-back induces a map MATH to the configurations where the fiber-wise holonomy of the connection is constant, and the section is covariantly constant around each fiber. The identification between the critical manifolds and the symmetric powers MATH described in CITE is obtained by proving that MATH lies in the image of this pull-back map, and indeed that it lies in the pull-back of the vortex moduli space, which, according to CITE (see also CITE), is in turn identified with the space of divisors, by looking at the zero-set of the section. The key points we need presently are that MATH lies in MATH, and that configurations are the pull-backs of configurations MATH, where MATH is MATH-holomorphic section. Over MATH, there is a universal line bundle MATH, defined in the usual manner. Note that MATH so MATH for MATH agrees with MATH, where the former MATH-map is induced from MATH, and the latter from MATH. We have thus reduced the proof of the proposition to a statement purely over MATH; so for the duration of the proof, MATH will refer to MATH, MATH will refer to MATH, and all MATH-maps will be calculated over MATH. To facilitate the proof over MATH, we pause for a discussion about the canonical section MATH of the universal line bundle MATH, which takes the configuration MATH to the based configuration MATH. This section has the property that, under the canonical identification of MATH (where MATH is the bundle over MATH with NAME number MATH), the restriction MATH is identified with the section MATH of MATH. In particular, if MATH is a holomorphic section, then MATH has at most MATH zeros; moreover, if it has MATH zeros, then each is transverse. Now, if MATH is a point (that is, a generator of MATH), then by definition, MATH is the element of MATH whose pairing against any homology class MATH is given by MATH where, as usual, MATH denotes the restriction of MATH to MATH. Choose points MATH which are distinct from MATH. Recall that MATH is generated by the surface MATH (where MATH are points on MATH), and the tori of the form MATH, where MATH are closed curves in MATH, which we can choose to miss MATH. The canonical section MATH restricted to a torus of the form MATH clearly vanishes nowhere (as all MATH of the zeros have been constrained to lie in the set MATH which does not include the point MATH); thus, MATH . Over MATH the canonical section vanishes at the single point MATH. We verify transversality of this zero, as follows. View MATH as a section over MATH; we know that MATH, and that MATH induces an isomorphism from MATH to MATH (that is, that the zero of MATH at MATH is transverse). Differentiating the equation that MATH, we see that MATH . Thus, the section MATH of MATH has a single, transverse zero, which shows that MATH . Moreover, the sign is positive since the section is holomorphic. Hence, we have proved the result when MATH. Proving the result for classes coming from MATH amounts to proving that, if MATH and MATH closed curves in MATH which meet transversally, then MATH . Note first that the zeros of the canonical section MATH, restricted to MATH are the points MATH (a zero of MATH corresponds to a point where the section MATH vanishes at some point of MATH, but the zeros of MATH lie in MATH, and MATH is empty). We must now consider the local contribution of each zero (and check transversality). Consider the map MATH defined by MATH. Suppose for notational simplicity that MATH. We can view MATH as a section of MATH pulled back to this torus. Now, evaluated on a typical tangent vector to the torus MATH, the derivative of MATH at the intersection point is given by MATH . (We have used the chain rule and the derivative of the relation that MATH.) Transversality of the intersection of MATH and MATH at MATH ensures that the image of this differential is MATH so transversality of the section corresponding to MATH at its zero MATH ensures that the image of the differential surjective onto the fiber of MATH over MATH; that is, the canonical section is transverse. The sign is correct, as one can see by inspecting REF .
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math/0005268
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The vector space MATH is generated by homogeneous elements of the form MATH where MATH is a subset of MATH, and MATH, MATH, MATH, MATH are integers with MATH. Clearly, it suffices to prove the proposition for homogeneous generators of degree MATH. Modulo MATH, such an element is equivalent to the element MATH . Indeed, in light of the fact that MATH, we can arrange (after possibly simultaneously permuting the indices of the MATH and MATH) that for each MATH, MATH. Moreover, the original homogeneous element would automatically lie in MATH unless we had that MATH for all MATH. Put together, must consider elements of the above form which satisfy the constraint that MATH for all MATH. If the degree of such an element is MATH, it must vanish in MATH. This vanishing can be seen geometrically: MATH is NAME dual to the subset (identified with MATH) of MATH where one point is constrained to lie in a specified point on MATH: MATH (respectively, MATH) is NAME dual to the cycle where one point is constrained to lie on MATH (respectively, MATH). Thus, (if one chooses the point representing MATH to be MATH), then MATH is NAME dual to the locus where two distinct points are constrained; one is to lie on MATH, the other on MATH. Similarly, the manifold NAME dual to MATH gives a constraint on two distinct points in the symmetric power. Finally, the remaining MATH and MATH give additional, disjoint constraints (these are disjoint, if one chooses that representing curves to be disjoint from the MATH and MATH for MATH, which can be arranged since MATH for all MATH). Thus, since the total degree of the expression considered is MATH, we have put constraints on MATH distinct points, forcing the intersection to be empty.
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math/0005268
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By REF , MATH lies in the ideal generated by MATH. Now the proposition follows from REF .
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math/0005268
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This follows immediately from the fact that MATH.
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math/0005268
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We prove that both moduli spaces MATH and MATH are empty. Suppose there were some finite energy solution to the NAME equations in a MATH structure with MATH. We know that the spinor lies entirely in one of the two summands in the splitting of the spinor bundle MATH (that is, it is a MATH- or a MATH-spinor, in the notation of REF). By conjugating if necessary (which switches the two summands and sends the MATH structure MATH to another one MATH with MATH), we can assume without loss of generality that the solution is a MATH-solution. According to REF (and REF , when the boundary value is reducible), we can extend the data MATH over the associated ruled surface MATH, obtained by attaching the curve MATH at infinity. The fact that MATH is an extension of MATH says that MATH where MATH is the curve in MATH with self-intersection number MATH (which is identified with MATH). By our hypothesis, then, MATH . On the other hand, REF says that MATH converges to the pullback of a form over MATH whose integral is MATH, so REF guarantees that the NAME number of restriction to the other section of the ruling satisfies the bound MATH . Now, since the NAME dual of a fiber is MATH, we see that the evaluation of MATH on a fiber is given by MATH . According to REF , it follows that MATH (and hence also MATH) must vanish identically, contradicting the irreducibility hypothesis on MATH. The fact that the reducibles are smoothly cut out in this range follows in an analogous manner, using REF .
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math/0005268
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This follows exactly as in REF (for irreducible boundary values) and REF (for reducible boundary values) of CITE. The key observation at this point is to note that MATH is an isomorphism, which allows one to ``unroll" parts of the NAME deformation complex to identify it with the deformation theory of divisors in MATH. As in CITE (see also CITE), we can identify MATH with the operator over the cylindrical end with MATH where MATH. According to the theory of CITE, the operator MATH is NAME for all weights MATH. In particular, it has the same index for all small MATH as it has on the weight MATH, where it can be connected via NAME operators to the manifestly self-adjoint operator MATH where MATH denotes the formal MATH-weighted adjoint of MATH. It follows from the homotopy invariance of the index that MATH has index zero on MATH. From the maximum principle, it has no kernel, so it induces an isomorphism as claimed, identifying the deformation theory of the NAME equations with the deformation theory of divisors in MATH. Passing to the ruled surface then follows from REF.
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math/0005268
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The deformation theory around a solution MATH is identified the deformation theory around a corresponding divisor in the line bundle MATH with MATH; that is, with divisors in a line bundle which (topologically) pulls back from MATH. According to REF , all such divisors actually pull back from the base MATH; and indeed, the deformation theory corresponds to deformation theory of degree MATH divisors in the base MATH, which is unobstructed. Thus, MATH is a manifold of real dimension MATH, transversally cut out by the NAME equations. The above transversality applies to MATH as well, except that the expected dimension is greater by one, as we saw in REF . This identification of deformation theories of MATH proves that MATH is an orientation-preserving local diffeomorphism onto its image in MATH. In fact, it is injective, as follows. As we saw, any two solutions with the same boundary values actually vanish over the same disks (with the same multiplicities). By the usual analysis of the vortex equations, any two such solutions must differ by a complex gauge transformation; that is, a function MATH which satisfies MATH where MATH is a function which decays on the cylinder. By the maximum principle, such a function must vanish identically.
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math/0005268
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By a standard excision argument, we have MATH which calculates MATH, given REF . Similarly, we have MATH which gives us MATH. Smoothness of MATH and MATH follows from adapting methods of CITE.
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math/0005268
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Most of this is a straightforward adaptation of CITE. We begin with the identification of the moduli spaces over MATH. As in CITE, the equations over MATH reduce to vortex equations over MATH. More specifically, the components of the moduli spaces MATH correspond to line bundles MATH over MATH with the property that MATH the spinor bundle of MATH (here MATH denotes the canonical line bundle over MATH). The vortex equations are are equations for MATH, MATH, which, in the case at hand, take the form MATH . Thus, one of MATH or MATH must vanish. In fact, in our case, MATH . In fact, if MATH then the solution space to these equations REF is empty. More specifically, letting MATH, we see that when MATH, then by integrating REF over MATH against MATH, we get MATH which forces MATH (since MATH). Since in this case MATH, MATH, so MATH must vanish. If, on the other other hand, it is MATH, then we obtain in the same manner that MATH which forces MATH (since MATH). Since MATH represents a class in MATH, it follows that MATH. So, all irreducibles correspond to MATH-vortices in the line bundle MATH with MATH. The identification of this space of vortices with the symmetric product follows from CITE (see also CITE). NAME of the irreducible manifold MATH follows exactly as in CITE. To see non-degeneracy of MATH, we appeal to results of REF. Consider the NAME operator on the MATH structure with spinors MATH with connection induced from a connection MATH whose curvature pulls up from MATH. It is shown in REF that this NAME operator admits no harmonic spinors unless the holonomy around a fiber circle in MATH is trivial. In fact this holonomy is trivial when the following integral is congruent to MATH modulo MATH: MATH (we have used here REF ). Since MATH, this holonomy is non-trivial, so the reducibles admit no harmonic spinors, that is, MATH is smoothly cut out by the equations. We now perform the NAME calculations (see the proof of REF). Suppose MATH, and MATH. Then, we have MATH where by MATH, we mean the integral MATH, which when MATH is induced from a line bundle over MATH, agrees with the degree of that line bundle. So, MATH which is negative; while MATH. The smoothness of the space of flows, and its identification with the symmetric product, follows exactly as in the unperturbed case (see REF).
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math/0005268
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We begin by proving MATH is empty. Note that MATH consists entirely of MATH-solutions, hence so must any section in MATH. Thus, a solution MATH induces a non-zero element in MATH with MATH . But MATH, according to REF . The same argument, now appealing to REF , shows that MATH is empty, and that MATH is smooth.
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math/0005268
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The proofs of REF apply directly in this perturbed context.
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math/0005268
|
We begin by proving the second claim. Note that MATH comes with a tautological connection along the MATH factor, with the property that for any path MATH and connection MATH, MATH . MATH . Now, fix a path in MATH, MATH . We need to show that for all paths in the configuration space MATH we have that MATH . This follows from the fact that for a line bundle MATH over the the torus MATH, the first NAME number is the degree of the map from MATH defined by MATH (a map which makes sense only after one puts a connection on MATH, but the degree is independent of this connection, so we left it out of the notation), together with the universal property of REF . Thus, we have identified MATH on any one-dimensional homology class. The rest of the proposition is established, once we see that for a point MATH, MATH generates MATH of the configuration space. But this follows easily from the fact that MATH acts freely on the space of irreducible configurations.
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math/0005268
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Fix a compactly-supported cut-off function MATH in MATH, and consider the section MATH of MATH, thought of as a line bundle over MATH, giving MATH the MATH topology. This transversality follows from the fact that, for any MATH, as we vary MATH, the integral MATH can take on any complex value. This, in turn, follows from the unique continuation theorem for elliptic differential operators, which guarantees that the section MATH cannot vanish identically over MATH.
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math/0005271
|
The set of MATH-fixed points in MATH constitutes a circle, denoted by MATH, containing MATH and MATH. Then MATH divides the sphere MATH into two hemispheres, say MATH and MATH. Let MATH be a complex MATH-vector bundle over MATH. Note that the restriction of MATH to the subspace MATH decomposes into the NAME sum of MATH-invariant sub-line bundles. Choose a MATH-invariant sub-line bundle, say MATH, over MATH. Then it is always possible to extend MATH to a non-equivariant sub-line bundle over MATH, since the restriction of MATH to MATH is non-equivariantly trivial and the set of one-dimensional subspaces of the fiber is homeomorphic to the NAME manifold MATH, the fundamental group of which is trivial. We now extend it over the other hemisphere MATH using the MATH-action on MATH to get a resulting MATH-invariant sub-line bundle of MATH.
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math/0005271
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It suffices to show that every complex MATH-line bundle MATH over MATH is trivial by REF . Choose two MATH-invariant hemispheres, denoted by MATH and MATH of MATH containing MATH and MATH, respectively, such that MATH and MATH is a MATH-invariant circle. Since MATH (respectively, MATH) is equivariantly contractible to MATH (respectively, MATH), MATH restricted to MATH (respectively, MATH) is equivariantly isomorphic to the product bundle MATH (respectively, MATH) where MATH (respectively, MATH) denotes the fiber of MATH at MATH (respectively, MATH). Then MATH induces a MATH-equivariant clutching map of MATH into the set MATH of linear maps between the two fibers MATH and MATH. Note that MATH and MATH are isomorphic as complex representations of MATH since MATH and MATH are connected by MATH-fixed points, and thus the induced MATH-action on MATH is trivial. Since MATH acts on the circle MATH as a reflection, the clutching map is equivariantly null-homotopic, that is, MATH is equivariantly trivial.
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math/0005271
|
Given a MATH-vector bundle isomorphism MATH, the map MATH sending MATH to MATH gives a MATH-vector bundle isomorphism.
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math/0005271
|
The construction of induced bundles is the same as that of induced representations.
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math/0005271
|
For MATH and MATH, the map MATH sending MATH to MATH gives a (non-equivariant) vector bundle map. Given MATH we have MATH for some representative MATH and MATH. Then the equalities MATH imply that MATH is MATH-equivariant. On the other hand, the inverse bundle map is given by the map sending MATH to MATH for MATH and MATH.
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math/0005271
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The necessity is obvious since a MATH-vector bundle isomorphism MATH restricts to a MATH-vector bundle isomorphism MATH, and the sufficiency follows from the fact that MATH is functorial.
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math/0005271
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It is easy to check that the map MATH sending MATH to MATH gives the inverse of the map in the lemma.
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math/0005271
|
We now consider the map MATH sending each generator MATH of MATH to MATH. Since both MATH and MATH are free abelian groups, MATH is a well-defined group homomorphism. Moreover it is injective, since the set of the elements MATH for all MATH such that MATH can be extended to an additive basis of MATH. For each MATH, either MATH or MATH. In case that MATH, we have MATH. Thus the image of MATH is generated by the elements MATH for all MATH, that is, the MATH-submodule of MATH consisting of the elements MATH for all characters MATH of MATH. Given a character MATH of MATH, as in REF , we use the same notation MATH for the product MATH-vector bundle over MATH with MATH as the character of the fiber representation. Then REF implies that MATH . Since MATH, we have the equalities MATH showing that MATH is a MATH-module homomorphism. It remains to show the ring structure on MATH. It suffices to show that the tensor product of any two generators in MATH is zero. Note that, given an induced bundle MATH, the image of MATH mapped by the MATH-action, that is the MATH-isotypical component of MATH, is isomorphic to MATH where MATH denotes the dual bundle of MATH. Indeed, the pull-back bundle MATH is isomorphic to MATH, since the action MATH is a reflection so that it induces the multiplication by MATH on the second cohomology level. It follows that MATH since MATH in MATH. Therefore, by REF , we have the equalities MATH since MATH in MATH (see CITE).
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math/0005271
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By assumption there exists an element MATH such that MATH for all MATH. It follows that MATH for any character MATH of MATH, which completes the proof.
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math/0005273
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REF are clear. We only check REF . For REF , let MATH . By REF above, MATH, so we check MATH. Let MATH. If MATH, then let MATH, otherwise we must have MATH, so we can write MATH as a disjoint union MATH with MATH, MATH. In either case we have MATH, MATH, MATH. So there is a function MATH with MATH, so MATH while MATH. Hence MATH. We now turn to the proof of REF . Call a function MATH ``conservative" if it satisfies MATH for all MATH. Clearly all conservative functions are in MATH. Let MATH, MATH. So there is some set MATH with MATH. Let MATH, MATH. So MATH, MATH. Now let MATH be arbitrary. We have to show that MATH is in the clone generated by MATH and MATH. Pick two distinct elements MATH in MATH. The function MATH is conservative, hence in MATH. Let MATH, MATH, and define two ``approximations" MATH, MATH to MATH as follows: MATH . Let MATH if MATH, MATH if MATH. By definition of MATH, MATH, so all we have to show is that MATH are all in the clone generated by MATH and MATH. We already know that CASE: MATH (because MATH is conservative), CASE: MATH (because the range of MATH is in MATH), CASE: MATH (because MATH takes only REF values) It remains to show MATH. Let MATH be an ``inverse" of MATH, that is, MATH (MATH can be arbitrary function with range MATH.) Define MATH by MATH. Note that the range of MATH is MATH, MATH, so MATH. Now we have, for all MATH, MATH, so MATH.
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math/0005273
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The upper bound follows from MATH. For the lower bound: it is known that there are MATH many maximal ideals, and we have just shown that the function MATH maps them injectively to precomplete clones.
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math/0005273
|
Let MATH, MATH: MATH be two REF functions such that the ranges of MATH, MATH, MATH are disjoint. Since MATH contains a function which is not almost unary, there is some MATH with MATH for all MATH in the range of MATH. Then the function MATH is a pairing function.
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math/0005273
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Choose MATH and MATH in MATH with MATH. Using unary functions, we will construct a binary function in MATH which is not in MATH. Let MATH, so MATH. So there is some MATH and a sequence MATH of elements of MATH such that all values MATH are different. Similarly, there is some MATH and a sequence MATH of elements of MATH such that all values MATH are different. Now for MATH define functions MATH as follows: Fix some element MATH. MATH . Formally, the functions MATH are in MATH, but each of them is essentially unary: MATH depends only on MATH for MATH, and only on MATH for MATH. This implies that MATH. Now the function MATH, that is, MATH, will be in MATH but not in MATH, since the values MATH are all different, as are the values MATH.
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math/0005273
|
Since each MATH, we can find a decomposition MATH, MATH, and a function MATH mapping each MATH to a small subset MATH such that: CASE: For all MATH, all MATH: MATH. CASE: For all MATH, all MATH: MATH. By the previous lemma, we cannot have both MATH and MATH in MATH, so without loss of generality assume MATH. Now fix any element MATH. We will show that the set MATH is small. Consider MATH. Identifying MATH with MATH, we can write the tuple MATH as MATH, MATH, MATH. Now note that for MATH we have MATH, so MATH, which is a small set. Hence MATH . For each fixed MATH the set MATH is small (since MATH), so, since MATH is small, also MATH is small.
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math/0005273
|
We will show how to construct a pairing function from MATH and MATH. Define MATH . We claim that for all distinct MATH: MATH. Indeed, if MATH, then MATH if MATH, then MATH and if MATH, then MATH . Hence MATH is a pairing function.
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math/0005273
|
See CITE or CITE.
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math/0005273
|
Let MATH be an independent family of subsets of MATH. Write MATH for MATH. For each MATH we let MATH the clone generated by MATH. We will now show that CASE: MATH, for all MATH CASE: Whenever MATH, then MATH already generates MATH. This will conclude the proof, because REF together with REF implies that each MATH can be extended to a precomplete clone, and REF implies that no single precomplete clone can contain MATH for distinct MATH, MATH. Proof of REF : NAME there is some MATH. By independence, MATH cannot be covered by any finite union from MATH. So by the lemma, MATH is not in the clone MATH. Proof of REF : If MATH, then there is without loss of generality some MATH. Now MATH, MATH, and by REF , MATH generates MATH.
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math/0005273
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Proceed by induction on the complexity of the terms. We have to thin out the set MATH finitely many times in order to make finitely many functions REF or constant.
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math/0005273
|
Let MATH be the set of subterms of MATH (including MATH itself). Collect all REF functions appearing in MATH for MATH, that is : MATH . The set MATH is finite, the identity function is in MATH, and all functions in MATH are REF. We may thin out the set MATH so that the family MATH is pairwise disjoint. So since MATH witnesses MATH, we can find MATH such that For all MATH: MATH (and MATH). This implies MATH. Now we can prove by induction on the complexity of the subterms MATH of MATH that MATH.
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math/0005273
|
REF are easy. For REF, assume that MATH, with MATH, MATH. We have to distinguish several cases: CASE: MATH, MATH. Since MATH, and MATH is canonical, we have MATH for all MATH, so MATH is symmetrical. CASE: MATH. So MATH. Pick any MATH with MATH. Then MATH, so MATH implies MATH, this means MATH. Similarly we find MATH. So MATH is neither REF on MATH nor REF on MATH. CASE: Similar to REF .
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math/0005273
|
By REF, it is enough to find a function MATH which is REF on MATH. If MATH is symmetrical and REF on MATH (and also REF on MATH, of course), then we may assume (replacing MATH by MATH for some appropriate MATH, if necessary), that MATH for all MATH. We claim that the function MATH is REF on MATH. Indeed, if MATH, then we have: either MATH, MATH, or MATH, MATH. In the first case we get either MATH directly, or MATH, MATH, so again MATH. The second case leads to a contradiction: MATH, MATH. So we assume now that MATH is canonical but not symmetrical. By REF , we know that MATH. Replacing MATH by MATH for an appropriate MATH, we may assume that CASE: MATH is constantly MATH. CASE: MATH takes only even values MATH, and is REF. Since MATH contains a heavily binary function, MATH contains some function MATH with MATH for all MATH. Now check that the map MATH is a pairing function.
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math/0005274
|
We let MATH and put MATH so that we have a filtration of subspaces MATH with MATH, for all MATH by REF . Let MATH and let MATH be the minimal integer such that MATH. Since MATH implies that MATH, this setting puts us in the situation of REF, from which we conclude that MATH is a finite-dimensional vector space over MATH. Of course MATH and hence it follows that MATH is finite-dimensional as well.
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math/0005274
|
We will continue to use the notation defined earlier. First we show that MATH. Suppose that MATH. It is easy to see that MATH is invariant under MATH. Now there exits a basis MATH of MATH together with non-zero complex number MATH such that MATH, where MATH is the element of REF . Since MATH is a finite-dimensional vector space it follows in particular that MATH acts nilpotently on MATH for all MATH. But MATH and so the action of the MATH's on MATH commutes. Therefore there exits a non-zero MATH such that MATH. But in this case MATH, which contradicts the minimality of MATH. Thus MATH. In the case when MATH, there exists an epimorphism of MATH-modules MATH, with MATH finite-dimensional due to REF. By irreducibility of MATH it follows that MATH is an irreducible MATH-module. Now if MATH, then there exists a non-zero vector MATH invariant under the action of MATH, for MATH. Again we have an epimorphism of MATH-modules MATH.
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math/0005274
|
By REF every finite irreducible MATH-module is a homomorphic image of MATH. Now the usual argument for highest weight representations implies that given a finite-dimensional irreducible MATH-module MATH the MATH-module MATH contains a unique maximal submodule, from which the bijection then follows.
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math/0005274
|
Note that MATH is singular if and only if MATH. We compute MATH . But then MATH, since otherwise REF would imply that MATH. However, MATH together with REF implies that MATH . Now REF gives MATH . Now if MATH, then REF gives MATH and MATH. But then MATH by REF . Hence MATH so that by REF we have MATH. Now if MATH, we have from REF MATH and hence MATH. The first equation of REF then implies that MATH. On the other hand if MATH, the first equation of REF gives MATH.
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math/0005274
|
The lemma follows immediately from the following two equations: MATH .
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math/0005274
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If MATH and MATH, then by REF MATH contains no proper singular vector and hence is irreducible. Suppose that MATH and MATH. In this case consider the submodule of MATH generated by the singular vector MATH. This module is precisely MATH above and hence MATH is freely generated over MATH by MATH and MATH. We claim that MATH is irreducible. The even part of MATH is isomorphic to the semi-direct sum of MATH (generated by MATH) and MATH (generated by MATH), where MATH is the one-dimensional NAME algebra. We first consider MATH as a module over the MATH. The vectors MATH and MATH have MATH-weights MATH and MATH, respectively, and furthermore are both annihilated by MATH and MATH, for MATH. Now since MATH and MATH, we have MATH and MATH. From this it follows that MATH as a module over MATH is a direct sum of two non-isomorphic irreducible modules, namely MATH and MATH (see REF for notation). But we have MATH which implies that as a MATH-module MATH is irreducible. The case when MATH and MATH is completely analogous and we leave it to the reader. Finally in the case when MATH, both MATH and MATH are proper singular vectors. Now the submodule in MATH generated by these two vectors contains MATH, and hence has codimension MATH over MATH. So the resulting quotient is the trivial module.
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math/0005274
|
By REF MATH is a singular vector in MATH of MATH-weight MATH. Consider MATH, the MATH-submodule generated by MATH. Then we have MATH, where MATH is the irreducible MATH-submodule generated by MATH. Note that the map MATH extends uniquely to an epimorphism of MATH-modules from MATH to MATH. In particular it is a MATH-module epimorphism. Now both modules are completely reducible MATH-modules and hence this map sends MATH-invariants onto MATH-invariants. Since MATH is generated over MATH by MATH, it follows that MATH is generated over MATH by MATH. Now MATH is a MATH-submodule of MATH, since MATH. Thus it is a free MATH-submodule generated by MATH. We compute MATH . By inspection it is clear that the following is a set of MATH-generators for MATH. MATH . First consider the case when MATH. It follows from the description of MATH above that MATH is a MATH-basis for the MATH-invariants of the quotient MATH. Since MATH and MATH both have MATH-weight MATH, they generate two copies of the irreducible MATH-module of dimension MATH. On the other hand MATH and MATH both have weight MATH, and so they generate two copies of the irreducible MATH-module of dimension MATH. Thus MATH is a free MATH-module of rank MATH. So in order to complete the proof it remains to show that MATH is irreducible. Note that MATH, MATH, together with MATH generate a copy of MATH and so we may consider MATH as a module over MATH. By parity consideration MATH is a direct sum of two MATH-modules, namely MATH and MATH, where MATH is the irreducible MATH-module generated by MATH. It is subject to a direct verification that MATH, for MATH, annihilates the vectors MATH (in fact one only needs to check that MATH, others being trivial) and hence MATH as a MATH-module is a direct sum the following four non-isomorphic irreducible modules: MATH, MATH, MATH and MATH, where we denote by MATH the irreducible MATH-module of highest weight MATH. Now we compute MATH from which it follows that we may go from each irreducible MATH-component of MATH to the irreducible component containing the highest weight vectors, and hence the module MATH is irreducible. Now if MATH the vectors MATH and MATH are both zero. Therefore the quotient MATH. But then the first identity in REF shows that MATH is irreducible. The rank of MATH is then MATH in the case MATH. Finally when MATH, the vectors MATH, so that MATH reduces to MATH. Therefore MATH is the trivial module and so has rank MATH.
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math/0005274
|
By REF MATH is a singular vector of MATH of MATH-weight MATH. Let MATH denote the MATH-submodule generated by MATH. Consider first the case MATH. As in the proof of REF MATH is a free MATH-module generated over MATH by MATH. We compute MATH . This implies that the set MATH generates MATH over MATH and so MATH is a MATH-basis for MATH in the case when MATH. Next consider the case MATH. In this case, letting MATH be as before, MATH is generated over MATH by MATH. Hence it follows from the above formulas that again MATH is a MATH-basis for MATH. In the case when MATH we let MATH denote the module generated by MATH. It follows that the vectors MATH generate MATH over MATH so that MATH generate MATH. Hence MATH contains in addition a one-dimensional (over MATH) subspace spanned by MATH. However, MATH in MATH and hence it is a MATH-invariant by REF. In this case we set MATH so that the quotient module MATH is again generated over MATH by MATH. Now MATH and MATH have MATH-weight MATH, while MATH and MATH have MATH-weight MATH. Thus MATH has rank MATH over MATH. So it remains to show that MATH is irreducible. Again we consider MATH as a module over MATH. It is easy to check that MATH, MATH, annihilates MATH. (Again one really only needs to check that MATH.) Thus it follows that in the case of MATH that MATH is a direct sum of the following four non-isomorphic irreducible MATH-modules: MATH, MATH, MATH and MATH, where as before MATH stands for the irreducible MATH-module of highest weight MATH and MATH is the MATH-submodule generated by the vector MATH. Now we compute MATH from which again it follows that we may go from any irreducible MATH-component of MATH to the component containing the highest weight vectors, and hence MATH is irreducible. As for the case MATH we have MATH as a MATH-module is also a direct sum of the MATH. The first three modules, as in the case of MATH are irreducible. However, MATH contains a unique irreducible submodule generated by the vector MATH, which is isomorphic to MATH. But then REF together with the fact that MATH shows that MATH is irreducible in this case as well. In the case when MATH we have by REF that MATH is the unique (up to a scalar) singular vector inside MATH. Let MATH denote the MATH-submodule generated by MATH. Since MATH has MATH-weight MATH, MATH is the free MATH-module generated by MATH. We have MATH from which it follows that MATH is generated over MATH by MATH. Since MATH in this situation, we see that MATH is generated over MATH by the vectors MATH, just as in the case MATH. Now the exact same argument as in the MATH case shows that MATH is irreducible and has rank MATH over MATH.
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math/0005274
|
By REF MATH and MATH are singular vectors in MATH. Consider MATH and MATH, the MATH-submodules generated by MATH and MATH, respectively, and let MATH. Then we have MATH and MATH, where MATH and MATH are the irreducible MATH-submodules generated by MATH and MATH, respectively. Let's first compute MATH. Since the MATH-weight of MATH is MATH, we know that MATH is a free MATH-module generated over MATH by MATH. We have MATH . Next we find MATH-generators of MATH. Similarly MATH generates MATH over MATH: MATH . It follows that MATH is generated over MATH by the following set MATH . Suppose that MATH. From the description of MATH we see that MATH is a MATH-basis for the MATH-invariants of the quotient MATH. Since MATH has MATH-weight MATH, it generates a copy of the irreducible MATH-module of dimension MATH. Now MATH and MATH both have weight MATH, and so they generate two copies of the irreducible MATH-module of dimension MATH. Finally MATH has weight MATH, and so it generates a copy of the irreducible MATH-module of dimension MATH. Thus MATH is a free MATH-module of rank MATH. So we need to show that MATH is irreducible. As in REF MATH, MATH, together with MATH generate a copy of MATH, which thus allow us to study the MATH-module structure of MATH. By parity consideration MATH is a direct sum of two modules, namely MATH and MATH, where MATH is the irreducible MATH-module generated by MATH. We can easily check that MATH, for MATH, annihilates the vectors MATH. (Again the only non-trivial part is to check that MATH.) Thus MATH as a MATH-module is a direct sum of the following four irreducible modules: MATH, MATH, MATH and MATH, where as usual MATH is the irreducible MATH-module of highest weight MATH. Note that, contrary to the MATH case, the odd part here is a sum of two isomorphic modules. To conclude that MATH is irreducible, we show again that one may go from each irreducible MATH-component to the irreducible component containing the MATH-highest weight vectors. But this follows from the following computation. MATH . Now if MATH the vector MATH is zero. Therefore the quotient MATH. But then REF and the first identity in REF show that MATH is irreducible. The rank of MATH is then MATH, which equals to MATH, in the case MATH. Finally when MATH, the vectors MATH so that MATH reduces to MATH. Hence MATH is the trivial module and so has rank MATH.
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