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math/0006054
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First note that if MATH is torsion-free then MATH is also torsion-free. Hence its support consists is either empty or consists of the whole relative jacobian surface MATH. Moreover, if MATH is not MATH-MATH then the support of MATH is contained in the support of MATH since MATH for all MATH. However, MATH cannot be supported on the whole surface since it does not contain the whole fibre MATH where MATH is semistable. Thus MATH and MATH is MATH-MATH. Now MATH is the trivial line bundle supported on the zero section of MATH and MATH is a flat line bundle supported on (a divisor in) MATH. Hence, since MATH, we conclude that: MATH as required.
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math/0006054
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Suppose that MATH is an isolated point. It cannot belong to a fibre MATH such that MATH is semistable, since the restriction to every nearby fibre is also semistable and the restriction of MATH to the fibres varies holomorphically. So MATH is unstable. But MATH contains all such fibres, since MATH for all MATH. To obtain the second statement, we must show that MATH has no proper subsheaves supported on points. For a contradiction, suppose it does, and let MATH be a subsheaf of MATH with REF-dimensional support. Then MATH is a torsion subsheaf of MATH, thus contradicting the hypothesis that MATH is torsion-free.
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math/0006054
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Suppose that the support of MATH decomposes as MATH, where MATH is the sum of fibres. When this happens, we have a subsheaf MATH of MATH supported on MATH with degree REF. Assume that MATH is the maximal such effective subdivisor of MATH. Let MATH. Then MATH is MATH-MATH and MATH is MATH-MATH. The resulting sequence after transforming with MATH is then MATH where MATH, for some zero dimensional subscheme MATH.
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math/0006054
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Suppose that MATH is a destabilizing sequence for MATH. We can assume MATH is semistable and MATH has pure dimension REF. Now MATH, for some MATH. Then MATH for all MATH, since MATH. Thus MATH is MATH-MATH by REF . Arguing as in REF , we have MATH hence MATH. So MATH will destabilize MATH unless MATH. For this to be the case we must have MATH. Then MATH is supported on a sum of fibres. But since MATH must be MATH-MATH we see that it has non-negative degree on these fibres and so MATH, thus contradicting the assumption on MATH.
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math/0006054
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The first statement follows from REF , which requires us to show that MATH for all MATH. However, MATH and any map MATH would contradict the stability of MATH. Now suppose that MATH is not MATH-stable and let MATH be the destabilizing subsheaf, so that MATH . Moreover, we may assume that MATH is MATH-stable, thus MATH is MATH-MATH. We may also assume that the quotient MATH is MATH-stable. Then both MATH and MATH are MATH-MATH. Since their transforms must have zero rank, MATH and MATH both have zero fibre degree and so MATH and this contradicts the stability of MATH. The fact that MATH is locally-free follows from the last part of REF .
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math/0006054
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We must first show that MATH is a well defined map of varieties. To see this we use the argument given by NAME and NAME in CITE. The key idea is to observe that MATH coincides (locally in the étale or complex topology) with the projection map from the universal sheaf corresponding to the relative NAME scheme of degree MATH line bundles on families of genus MATH curves. The universal sheaf only exists locally in these topologies but this is enough to show that MATH is holomorphic. Given a spectral curve MATH, observe that its structure sheaf MATH is MATH-MATH since in the short exact sequence MATH, MATH is MATH-MATH and MATH is MATH-MATH. We aim to construct a MATH degree line bundle MATH over the curve MATH, which is stable as a torsion sheaf on MATH. That is, we need to choose MATH points on MATH and guarantee that the restriction of MATH to any proper component MATH of MATH satisfies deg-MATH. We do this by choosing a MATH points on each component MATH. This can always be done because if MATH then MATH intersects MATH at least twice since MATH and MATH are both at least REF and so MATH. Without loss of generality, we assume that the set we have just chosen MATH consists of distinct points away from the singularities of MATH. Therefore, MATH . We pick a class in MATH which is non-zero on each factor. This defines a torsion sheaf MATH on MATH given by this extension class. Then MATH is stable and locally free on its support by the choice of MATH. By REF , MATH is MATH-MATH and MATH is MATH-stable vector bundle such that MATH.
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math/0006054
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By continuity it suffices to prove this when MATH are defined over a point MATH given by an extension: MATH where MATH consists of discrete points and MATH is smooth. Deformations of MATH arising from the fibre of MATH are determined by deformations of MATH along MATH. Then MATH is generated by a non-zero vector in MATH and the fibre MATH over MATH has tangent space given by MATH where MATH generates the tangent space to MATH at MATH. But MATH in MATH for each MATH and so MATH as required.
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math/0006057
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We prove the identities containing the target counital map, the proofs of their source counterparts are similar. Using REF we compute MATH where MATH stands for the second copy of the unit, proving REF . For REF we have MATH . To prove REF we observe that MATH on the other hand, applying MATH to both sides of MATH, we get MATH. REF is immediate in view of the identity MATH. Finally, we show REF : MATH where the antipode REF is used.
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math/0006057
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MATH and MATH are coideals by REF , they commute by REF . We have MATH, conversely MATH, therefore MATH. To see that it is an algebra we note that MATH and for all MATH compute, using REF : MATH . The statements about MATH are proven similarly.
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math/0006057
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Let MATH be the convolution of MATH. Then MATH, and MATH. If MATH is another antipode of MATH then MATH . To check that MATH is an algebra anti-homomorphism, we compute MATH for all MATH, where we used REF and easy identities MATH and MATH. Dualizing the above arguments we show that MATH is also a coalgebra anti-homomorphism : MATH . The proof of the bijectivity of MATH can be found in (CITE).
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math/0006057
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Using results of REF we compute MATH for all MATH. The second identity is proven similarly. Clearly, MATH maps MATH to MATH and vice versa. Since MATH is bijective, and MATH by REF , therefore MATH and MATH are anti-isomorphisms.
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math/0006057
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It is clear that MATH is a homomorphism. If we write MATH with MATH and MATH linearly independent, then MATH. By REF , MATH, that is, MATH is surjective. Since MATH then MATH, therefore, MATH so MATH is bijective. The proof for source subalgebras is similar.
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math/0006057
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For all MATH we compute, using REF : MATH applying MATH to this identity we get MATH. Clearly, MATH, whence MATH is a separability element. The second statement follows similarly.
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math/0006057
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The proof is a straightforward application of REF , and is left as an exercise for the reader.
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math/0006057
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REF implies that the MATH-Hopf module structure on MATH is well defined, and it is easy to check that MATH is a well defined homomorphism of MATH-Hopf modules. We will show that MATH is an inverse of MATH. First, we observe that MATH for all MATH, since MATH so MATH maps to MATH. Next, we check that it is both module and comodule map : MATH . Finally, we verify that MATH and MATH : MATH which completes the proof.
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math/0006057
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Take MATH in REF and use REF .
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math/0006057
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CASE: Suppose that MATH is semisimple, then since MATH is a left ideal in MATH we have MATH for some idempotent MATH. Therefore, MATH and MATH is a left integral by REF . It is normalized since MATH. REF : if MATH is normalized then MATH is a separability element of MATH by REF . REF : this is a standard result CITE.
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math/0006057
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In the notation of REF the element MATH is a normalized integral in MATH.
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math/0006057
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A straightforward (but rather lengthly) computation shows that MATH is a well defined homomorphism, compare (CITE). Let MATH be a basis of MATH and MATH be the dual basis of MATH, that is, such that MATH for all MATH, then the element MATH does not depend on the choice of MATH and the following map MATH is the inverse of MATH.
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math/0006057
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REF for MATH shows that MATH is a projective generating MATH-module such that MATH. Therefore, MATH and MATH are NAME equivalent. Since MATH is semisimple (as a separable algebra), MATH is semisimple.
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math/0006057
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Define a left unit homomorphism MATH by MATH . It is an invertible MATH-linear map with the inverse MATH. Moreover, the collection MATH gives a natural equivalence between the functor MATH and the identity functor. Similarly, the right unit homomorphism MATH defined by MATH has the inverse MATH and satisfies the necessary properties. Finally, one can check the triangle axiom, that is, that MATH for all objects MATH of MATH and MATH (CITE).
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math/0006057
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We know already that MATH is monoidal. One can check (CITE) that MATH and MATH are MATH-linear. To show that they satisfy the identities MATH take MATH. Using the isomorphisms MATH and MATH identifying MATH, MATH and MATH, for all MATH and MATH we have: MATH which completes the proof.
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math/0006057
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Note that MATH is well-defined, since MATH. To prove the MATH-linearity of MATH we observe that MATH . The inverse of MATH is given by MATH therefore MATH is is an isomorphism. Finally, we check the braiding identities. Let MATH, then MATH . Similarly, we have MATH. The third equality of this computation shows that the relations of REF are equivalent to the braiding identities.
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math/0006057
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It follows from the first two relations of REF , that MATH .
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math/0006057
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The proof is essentially the same as (CITEEF).
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math/0006057
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Note that MATH for all MATH. Hence, we have MATH for all MATH. Therefore, MATH . The remaining part of the proof follows the lines of (CITEEF). The results for MATH can be obtained by applying the results for MATH to the quasitriangular quantum groupoid MATH.
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math/0006057
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Take MATH. Then we have : MATH therefore MATH, as required.
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math/0006057
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The proof is a straightforward verification that all the structure maps are well-defined and satisfy the axioms of a quantum groupoid, which is carried out in full detail in (CITE).
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math/0006057
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The identities MATH and MATH can be written as (identifying MATH with MATH and MATH with MATH) : MATH . The above equalities can be verified, for example, by evaluating both sides against an element MATH in the third factor (respectively, against MATH in the first factor), see (CITEEF). It is also straightforward to check that MATH is an intertwiner between MATH and MATH and check that MATH and MATH, see (CITE).
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math/0006057
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One can use the explicit form of the MATH-matrix REF of MATH and the description of the dual MATH to check that in this case the map MATH from REF is surjective.
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math/0006057
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Since MATH is an invertible central element of MATH, the homomorphism MATH is a MATH-linear isomorphism. The twist identity follows from the properties of MATH : MATH for all MATH. Clearly, the identity MATH is equivalent to the twist property. It remains to prove that MATH for all MATH, that is, that MATH where MATH is the canonical element in MATH. Evaluating the first factors of the above equality on an arbitrary MATH, we get the equivalent condition : MATH which reduces to MATH. The latter easily follows from the centrality of MATH and properties of MATH.
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math/0006057
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Follows from REF , and REF.
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math/0006057
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Any endomorphism of an irreducible module is the multiplication by a scalar, therefore, we must have MATH. Applying the counit to both sides and using that MATH, we get the result.
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math/0006057
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Here we only need to prove the invertibility of the matrix formed by MATH where MATH are as above, MATH is a basis in the space MATH of characters of MATH (we used above REF for the quantum trace). It was shown in (CITE, REF) that the map MATH is a linear isomorphism between MATH and the center MATH (here MATH). So, there exists an invertible matrix MATH representing the map MATH in the bases MATH of MATH and MATH of MATH, that is, such that MATH. Then MATH . Therefore, MATH, where MATH. REF below, shows that the existence of a normalized two-sided integral in MATH is equivalent to MATH being semisimple and possessing an invertible element MATH such that MATH for all MATH and MATH for all irreducible character MATH of MATH. Then MATH is an invertible central element of MATH and MATH for all MATH. By REF MATH is invertible central, therefore MATH. Hence, MATH for all MATH and MATH is invertible.
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math/0006057
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The proof is a straightforward verification, see CITE for details.
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math/0006057
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The existence and uniqueness of the solution of REF and the fact that MATH satisfies REF are established by induction on the degree of the first component of MATH, see (CITE).
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math/0006057
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CASE: If MATH, then by MATH one has MATH. Therefore, MATH is a dual pair of non-degenerate left integrals. The property MATH is equivalent to that the quasi-basis of MATH satisfies MATH from where MATH and MATH. Furthermore, MATH. Thus, MATH is a NAME integral. The inverse statement is obvious. CASE: The "only if" part follows from the proof of REF . If MATH is non-degenerate and MATH is its dual left integral, then, as above, MATH; so MATH is a two-sided integral and since MATH, it is normalized.
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math/0006057
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The assumption on MATH is used only to ensure that MATH, once knowing that it is semisimple, so that there is a MATH-basis of matrix units MATH in MATH. MATH : Recall that any trace MATH on MATH is completely determined by its trace vector MATH. Now let MATH be the trace with trace vector MATH. Then MATH is non-degenerate and has a quasi-basis MATH . Notice that MATH. It is straightforward to verify that MATH coincides with MATH, so MATH is non-degenerate and therefore its dual left integral MATH has MATH by MATH. Thus, MATH is a NAME integral. MATH : If MATH is a NAME integral, then MATH is semisimple by REF . Let MATH be a non-degenerate trace on MATH, then there exists a unique MATH such that MATH. One can verify that MATH is a two-sided non-degenerate integral in MATH and that MATH (see CITE, I. REF). To prove that MATH is inner, it is enough to construct a non-degenerate functional MATH on MATH such that MATH for all MATH. But the proof of REF shows that this is the case for the dual left integral to MATH.
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math/0006057
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Clearly, MATH and MATH verify all the conditions of REF , from where the existence of NAME integral follows. Since MATH is non-degenerate, the scalar product MATH is also non-degenerate. By the equality MATH positivity of MATH follows from REF .
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math/0006057
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The result follows from the fact that MATH is a MATH-quantum groupoid, which is easy to verify using the explicit formulas from REF .
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math/0006057
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Since MATH also implements MATH REF , MATH is central, therefore MATH is also central. Clearly, MATH must commute with MATH. The same Proposition gives MATH, which allows us to compute MATH . REF and the trace property imply that MATH . Since MATH and MATH is central, the above relation means that MATH and, therefore, MATH for any irreducible representation MATH, which shows that that MATH (compare CITE).
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math/0006057
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The proof follows from REF .
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math/0006057
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If MATH is such that MATH , then, using CITE, one has MATH therefore, using the properties of MATH and NAME projections, we get MATH so MATH. Similarly for MATH.
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math/0006057
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One can check (CITE) that MATH, hence, since MATH and MATH commutes with MATH, we have MATH which is REF . REF dual to it is obtained similarly, considering the comultiplication in MATH. As a consequence of the ``symmetric square" relations MATH one obtains (CITE) the identity MATH from where it follows that MATH . We use this identity to prove that MATH is a homomorphism : MATH for all MATH, whence MATH. Next, MATH therefore MATH (similarly, MATH). Using REF we have MATH . The antipode MATH is anti-multiplicative and anti-comultiplicative, therefore REF follows from the other axioms. Clearly, MATH and MATH. MATH is biconnected, since the inclusion MATH is connected (CITEEF) and MATH. Finally, from the properties of MATH and MATH we have MATH.
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math/0006057
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The above map defines a left MATH-module structure on MATH, since MATH and MATH . Next, using REF we get MATH . By CITE and properties of MATH we also get MATH . Finally, MATH and MATH iff MATH.
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math/0006057
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If MATH is such that MATH for all MATH, then MATH. Taking MATH, we get MATH which means that MATH. Thus, MATH. Conversely, if MATH, then MATH commutes with MATH and MATH therefore MATH.
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math/0006057
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By definition of the action MATH we have : MATH for all MATH, so MATH is a well defined linear map from MATH to MATH. It is surjective since an orthonormal basis of MATH over MATH is also a basis of MATH over MATH (CITEEF). Finally, one can check that MATH is a NAME algebra isomorphism (CITE).
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math/0006057
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Note that MATH for all MATH. By CITE, the tower of basic construction for MATH is MATH therefore, the depth of this inclusion is equal to MATH, where MATH is the smallest positive integer such that MATH.
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math/0006057
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First, let us check that MATH and MATH are indeed maps between the specified lattices. It follows from the definition of the crossed product that MATH is a NAME subalgebra of MATH, therefore MATH is a NAME subalgebra of MATH, so MATH is a map to MATH. To show that MATH maps to MATH, let us show that the annihilator MATH is a left ideal in MATH. For all MATH, and MATH we have MATH and it remains to show that MATH. By (CITEEF), the square MATH is commuting, so MATH. Since MATH, then MATH, that is, MATH is a left ideal and MATH is a left coideal MATH-subalgebra. Clearly, MATH and MATH preserve MATH and MATH, moreover MATH and MATH, therefore they are morphisms of lattices. To see that they are inverses for each other, we first observe that the condition MATH is equivalent to MATH, and the latter follows from applying the conditional expectation MATH to MATH. The condition MATH translates into MATH. If MATH, then MATH that is, MATH commutes with MATH. Conversely, if MATH, then MATH for some MATH, therefore MATH.
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math/0006057
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If MATH is an object of MATH, then every simple subcomodule of MATH is MATH-dimensional. Let MATH REF be one of these comodules, then all other simple subcomodules of MATH are of the form MATH, where MATH, and MATH . Vice versa, MATH with natural MATH bimodule and MATH-comodule structures is a simple object of MATH.
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math/0006057
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First, we observe that MATH for any MATH. Indeed, since both categories are semisimple, it is enough to check that they have the same set of simple objects. All objects of MATH are also objects of MATH. Conversely, since irreducible MATH subbimodules of MATH are contained in the decomposition of MATH for all MATH, we see that objects of MATH belong to MATH. Hence, by REF , it suffices to consider the problem in the case when MATH has depth MATH (MATH), that is, to prove that MATH is equivalent to MATH. REF gives a functor from MATH to MATH. To prove that this functor is an equivalence, let us check that it yields a bijection between classes of simple objects of these categories. Observe that MATH itself is an object of MATH via MATH and MATH. Since the inclusion MATH has depth MATH, the simple objects of MATH are precisely irreducible subbimodules of MATH. We have MATH, where MATH is a family of mutually orthogonal minimal projections in MATH so every bimodule MATH is irreducible. On the other hand, MATH is cosemisimple, hence MATH, where each MATH is an irreducible subcomodule. Note that MATH and every MATH is a simple submodule of MATH (MATH simple subcomodule of MATH). Thus, there is a bijection between the sets of simple objects of MATH and MATH, so the categories are equivalent.
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math/0006057
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First, let us show that there is a bijective correspondence between right relative MATH . NAME modules and MATH-modules. Indeed, every right MATH . NAME module MATH carries a right action of MATH. If we define a right action of MATH by MATH then we have MATH for all MATH and MATH which shows that MATH and MATH act on MATH exactly in the same way, therefore MATH is a right MATH-module. Conversely, given an action of MATH on MATH, we automatically have a MATH-comodule structure such that MATH which shows that MATH, that is, that MATH is a right relative MATH-module. Thus, we see that the principal graph is given by the connected component the NAME diagram of the inclusion MATH containing the trivial representation of MATH. Since MATH is the basic construction for the inclusion MATH, therefore the NAME diagrams of the above two inclusions are the same.
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math/0006057
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In this case MATH and inclusion MATH is connected, so that MATH (note that MATH is biconnected).
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math/0006063
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We assume that MATH is defined on a coordinate chart MATH, and take MATH. Since MATH, the last condition of the definition of a formal integral takes the form MATH . Equating to zero the coefficient at MATH, of the l.h.s. of REF we get MATH, which can be rewritten as a recurrent equation MATH . Since MATH is a nondegenerate critical point of MATH, the functions MATH generate the ideal of functions vanishing at MATH. Taking into account that MATH for MATH we see from REF that MATH is determined uniquely. Thus the proof proceeds by induction.
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math/0006063
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The condition that MATH is a nondegenerate critical point of the function MATH directly follows from the fact that MATH is a potential of the non-degenerate MATH -form MATH. REF of formal integral are trivially satisfied. It remains to check REF . Replace the pair MATH by the equivalent pair MATH. Put MATH. For MATH REF takes the form MATH . We shall check it by showing that REF MATH; CASE: MATH; CASE: MATH. First, MATH, which proves REF . The function MATH is almost antiholomorphic at the point MATH. Thus, the full jet of the function MATH at the point MATH is equal to zero, which proves REF . The function MATH is almost holomorphic at the point MATH. For a holomorphic function MATH we have MATH. Since MATH and MATH considered modulo MATH depend on the jets of finite order of the functions MATH and MATH at the point MATH taken modulo MATH for sufficiently big MATH, we can approximate MATH by a formal holomorphic function MATH making sure that the jets of sufficiently high order of MATH and MATH at the point MATH coincide modulo MATH. Then MATH. Since MATH is arbitrary, MATH identically. The functions MATH and MATH have identical holomorphic parts of jets at the point MATH, that is, all the holomorphic partial derivatives (of any order) of these functions at the point MATH coincide. Since a left star-multiplication operator of deformation quantization with separation of variables differentiates its argument only in holomorphic directions, we get that MATH. This proves REF . The check for MATH is similar, which completes the proof of the theorem.
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math/0006063
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REF of formal integral are satisfied. It remains to check REF . Let MATH be a vector field on MATH. Denote by MATH the corresponding NAME derivative. We have MATH. Applying MATH we obtain that MATH, which concludes the proof.
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math/0006063
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Using integral representations REF one gets for MATH . Changing variables MATH in REF gives MATH . In order to apply the method of stationary phase to the integral in REF the following preparations should be made. Using REF express the phase function of the integral in REF as follows: MATH . In order to find the critical points of the phase MATH (with respect to the variables MATH; the variables MATH are parameters) consider first the equation MATH . It follows from MATH that MATH. Since MATH for MATH one has MATH for MATH whence it follows that REF holds only if MATH and thus MATH has critical points only if MATH. Since MATH one gets that MATH and MATH. As in the proof of REF from CITE, one shows that for each MATH the only critical point of the phase function MATH is MATH. It does not depend on MATH and, moreover, is nondegenerate. One has MATH since MATH. Finally, a simple calculation shows that the germs of the functions MATH and MATH at the point MATH are equal modulo the ideal generated by MATH and MATH. Applying now the method of stationary phase to the integral in REF one obtains the expansion REF satisfying REF . It follows from REF that MATH and MATH. It remains to show that all MATH are almost analytic along the diagonal of MATH. One has MATH . The function MATH is holomorphic on MATH. Let MATH and MATH be arbitrary holomorphic and antiholomorphic vector fields on MATH, respectively. Then MATH and MATH (the subscripts MATH show in which variable the vector field acts). Thus MATH . Analogously, MATH . Let MATH be a product of MATH derivations on MATH. Then, using integral representation REF , expand MATH to the asymptotic series MATH for some MATH and MATH, and with the norm estimate of the partial sums in the right-hand side term in REF analogous to REF . Since MATH one gets that all MATH. From this fact one can prove by induction over MATH that MATH vanishes to infinite order at the diagonal of MATH. Similarly, MATH vanishes to infinite order at the diagonal. Thus MATH is almost analytic along the diagonal.
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math/0006063
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It was already shown that the phase function MATH of integral REF satisfies the conditions required in the method of stationary phase. Thus REF can be applied to REF . We get that MATH, where MATH is a formal integral at the point MATH associated to the pair MATH and MATH is a nonvanishing formal function on MATH. It follows from REF that MATH, where MATH. The pair MATH is thus equivalent to the pair MATH. Applying REF we get that MATH . It remains to show that MATH is actually a formal constant. Let MATH be an arbitrary point of MATH. Choose a function MATH such that MATH in a neighborhood MATH of MATH. Let MATH be a vector field on MATH. Then, using REF , we obtain MATH . On the one hand, taking into account REF we get for MATH that MATH . The last equality in REF follows from REF . On the other hand, for MATH we have from REF that MATH, from whence MATH. Thus we get from REF that MATH on MATH for an arbitrary vector field MATH, from which the Proposition follows.
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math/0006065
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Clearly, we may assume that a group into which we embedd the amalgam MATH is generated by MATH and MATH. For MATH odd this means the group itself is of exponent MATH, and for MATH even of exponent MATH.
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math/0006065
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Since MATH, MATH, and MATH are of finite exponent, they are the direct product of their MATH-parts, that is, MATH, MATH, and MATH. Clearly, an embedding of MATH provides embeddings for MATH for each MATH; conversely, if we have embeddings into MATH for each MATH, then MATH will give an embedding for MATH.
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math/0006065
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Let MATH be the order of MATH (MATH if MATH is not torsion); then consider the group MATH. Denote the generator of the cyclic groups by MATH, and mod out by the subgroup generated by MATH. Since the MATH are central, the subgroup is normal and this works out.
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math/0006065
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Let MATH be the order of MATH, with MATH if MATH is not torsion. Consider the group MATH and denote the generators of the cyclic groups by MATH and MATH. Then mod out by the subgroup generated by MATH. Again, since the MATH are central, this subgroup is normal. If MATH is of odd exponent MATH, then MATH is generated by elements of exponent MATH, and so it is again of exponent MATH, hence so is MATH. If MATH is of exponent MATH, then MATH is generated by elements of exponent MATH and so MATH is of exponent at most MATH. If, furthermore, the MATH are of exponent MATH, then MATH is generated by a group of exponent MATH and elements of exponent MATH, so MATH is of exponent MATH, and then so is MATH.
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math/0006065
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Since MATH and MATH are commutators in MATH, they are central. Therefore: MATH . Therefore, MATH, as claimed.
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math/0006065
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Suppose that MATH satisfies the conditions, and let MATH be any overgroup of MATH with MATH. Let MATH and MATH be such that MATH for some MATH, MATH. Let MATH, MATH. We want to verify that MATH. Note that since MATH and MATH are MATH-th powers modulo the commutator in MATH, then there is an extension of MATH in MATH with MATH-th roots for both MATH and MATH; therefore REF cannot hold for MATH, MATH, and MATH. Likewise, REF does not hold: if MATH, then MATH, since MATH is of exponent MATH; and if MATH, then MATH. We also have that MATH, MATH, and MATH cannot satisfy REF . For suppose that MATH and for some MATH, we have the congruences MATH with MATH. Then by REF , we have MATH. However, for all such MATH, MATH is a multiple of MATH, hence MATH in MATH. Since we fail REF - REF must be satisfied, and then by REF it follows that MATH . So if the conditions hold, then MATH is absolutely closed, as claimed. Necessity is more complicated. We assume that MATH, MATH, MATH, and MATH do not satisfy REF - REF . We will proceed as follows: first we construct an extension where MATH and MATH have MATH-th roots MATH and MATH, respectively, and where MATH. Then, if necessary, we make MATH trivial. Then we verify that MATH and MATH are central in this extension, so we can adjoin central MATH-th roots MATH and MATH, respectively, to each. Next, we make MATH and MATH into commutators. Then MATH and MATH are of exponent MATH, and their MATH-th powers are, modulo commutators, MATH and MATH. Finally, we select a suitable subgroup which is generated by elements of exponent MATH. If we ensure that throughout the process MATH, we will have an example where MATH is not equal to its dominion. We now proceed with this program: CASE: To facilitate notation, we write MATH and MATH rather than MATH and MATH (respectively) in this step. Let MATH, where we denote the generators of the two copies of MATH by MATH and MATH, and MATH is the normal subgroup generated by MATH, MATH. We need to verify that MATH and that for every MATH we have MATH. The argument is the same as that found in the proof of REF, so we only sketch it here: A general element of MATH can be written as MATH where MATH, MATH is a positive integer, MATH, and MATH. By rewriting the element and expanding brackets, we get that the element equals: MATH where MATH. Say that this is equal to an element of MATH of the form MATH for some integer MATH and some MATH. Writing this as MATH with MATH, MATH, MATH, we obtain equations: MATH where MATH, and MATH. This means that MATH . For MATH, we have MATH, so the failure of REF shows that MATH. By setting MATH, we get that MATH, so we are in the situation of REF ; the fact that REF fails in MATH means that no such solution exists, so no element of the form MATH lies in MATH. Therefore, in MATH we have MATH lies in the dominion of MATH but not in MATH. We change the names of MATH and MATH to MATH and MATH respectively, and we again write MATH and MATH instead of MATH and MATH, respectively, to reduce the number of indices. So we have that MATH is an extension of MATH, generated by MATH, MATH and MATH; that MATH, MATH, and that MATH. We also note that MATH, with MATH as defined in REF . Indeed, MATH, and MATH . However, MATH since MATH fails REF . And also MATH and similarly with MATH. So MATH. We also claim that if MATH and MATH with MATH, then MATH. Indeed, if we write MATH we note that MATH, and so by construction of MATH we must have MATH, and MATH integers with MATH with MATH. However, the failure of REF means that in this case, MATH, as claimed. CASE: Let MATH where MATH is the normal subgroup generated by MATH. We claim that MATH, and that MATH for each MATH. Indeed, if MATH, then MATH is actually trivial already, since MATH; if MATH, then we know that MATH for MATH again by the observations above; thus MATH; and we can verify by inspection that MATH for each MATH, since MATH is central. So MATH is still in the dominion of MATH but not in MATH. CASE: We note that MATH and MATH are central in MATH. That they are central in MATH follows from the fact that MATH and MATH are MATH-th powers in MATH modulo MATH, and MATH is of exponent MATH. Thus, if MATH, then MATH and analogous with MATH. Since MATH is generated by MATH, MATH, and MATH, it suffices to show that MATH and MATH commute with both MATH and MATH. That MATH commutes with MATH is obvious. To verify that it commutes with MATH we note that: MATH since we moded out by MATH. An analogous calculation holds for MATH. CASE: We adjoin central MATH-th roots to both MATH and MATH. We use REF and adjoin roots, which we call MATH and MATH, respectively. Call the resulting group MATH. In MATH we still have MATH in the dominion of MATH but not in MATH, as can be easily verified. CASE: We make MATH and MATH into commutators. We can use REF , since MATH and MATH are central. Denote the resulting group by MATH, and write MATH, MATH. Moreover, since MATH, and likewise MATH, we can choose the MATH to be of exponent MATH each. Again, we still have MATH in the dominion of MATH but not in MATH. Note, moreover, that MATH and MATH are still central in MATH. CASE: Let MATH be the subgroup of MATH generated by MATH, MATH, MATH, MATH, MATH, MATH, and MATH. Note that MATH and MATH are of exponent MATH. Indeed, since MATH and MATH are central, we have: MATH . Since each MATH and each element of MATH is also of exponent MATH, MATH lies in MATH; moreover, it is an overgroup of MATH, and MATH . So MATH lies in MATH. Since MATH and MATH are central in MATH, MATH, which we know does not lie in MATH; therefore, MATH cannot be absolutely closed in MATH. This establishes necessity, and proves the theorem.
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math/0006065
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CASE: As per the remark above, if MATH is cyclic or trivial, then MATH is absolutely closed in MATH, and hence also in MATH. If MATH is not cyclic, then let MATH and MATH be elements of MATH which project to distinct cyclic summands of MATH. Then note that a product MATH is a MATH-th power in MATH if and only if MATH and MATH. Suppose MATH is absolutely closed. Since MATH is abelian, MATH, MATH, MATH, and MATH fail REF . Hence, it must satisfy REF , so there exist integers MATH, MATH, and MATH, and elements MATH such that MATH . But that means that MATH and MATH which is clearly impossible. So MATH is not absolutely closed. That this implies that MATH is cyclic follows because MATH is a sum of cyclic groups, and has the same number of summands as MATH, but since MATH is of exponent MATH the former group is equal to MATH itself. CASE: Again, if MATH is cyclic or trivial, then MATH is absolutely closed in MATH and hence also in MATH. If not, let MATH and MATH be elements of MATH which project to distinct cyclic generators of MATH (which is an abelian group of exponent MATH). Again, MATH, MATH, MATH and MATH do not satisfy REF - REF , and the same argument as above (with congruences rather than equalities) yields a contradiction to the assumption that it satisfies REF .
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math/0006065
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Note that if MATH and MATH lie in MATH, then so do MATH and MATH, so both commutator brackets lie in the dominion of MATH. Expanding the bracket bilinearly, we have MATH . Since MATH, and MATH, the last three terms on the right hand side lie in MATH, so the left hand side lies in MATH if and only if MATH lies in MATH, as claimed.
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math/0006065
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If MATH, the result is easy: the group described is a special amalgamation base of MATH (since everything there is a special amalgamation base), but not a special amalgamation base in MATH or MATH for MATH, as it is abelian but not cyclic. So we may assume that MATH. Consider the group MATH . Then MATH is isomorphic to the subgroup generated by MATH and MATH; however, MATH, so MATH cannot be absolutely closed in MATH. We only need to show now that MATH is absolutely closed in MATH. Let MATH be a MATH-overgroup of MATH, and suppose that MATH for some MATH, MATH. We may assume that: MATH . We want to show that MATH lies in MATH. We write MATH and MATH. Since MATH, we must have that MATH is central in MATH, and likewise with MATH. By calculating the commutators with MATH and MATH, we get that MATH divides MATH, MATH, MATH, and MATH. Therefore MATH divides MATH, MATH, MATH, and MATH. So we rewrite the elements as MATH, MATH. Since we can perturb MATH and MATH with elements of MATH by the Perturbation Argument, we can perturb MATH and MATH by MATH-th powers of MATH and MATH, so we may assume that MATH. If MATH, then we have that MATH since MATH is of exponent MATH. Substituting the values of MATH and MATH, we have MATH . Therefore, MATH. If we consider the vectors MATH and MATH as being vectors over MATH, this means that the two vectors are proportional. If one of them is the zero vector, then either MATH or MATH are the identity element, and there is nothing to do, for example : MATH . So by perturbing MATH and MATH by elements of MATH we may assume that MATH for some MATH. But that means that MATH actually lies in the dominion of the subgroup generated by MATH, which is cyclic and hence absolutely closed; thus, MATH, as desired. (In essence, since MATH is a power of MATH, we can always find exponents to satisfy REF ; see REF). If, on the other hand, MATH, then by choosing MATH, MATH, MATH, and MATH, we have: MATH so by REF , MATH. However, MATH so we have: MATH so again we have that MATH, and we proceed as we did in the case when MATH. Therefore, MATH is absolutely closed in MATH, as claimed.
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math/0006065
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We prove the MATH case first. One implication is trivial. So suppose that MATH is not absolutely closed, and let MATH be any nil-REF overgroup of MATH. Let MATH. We want to show that MATH is also not absolutely closed. If MATH, then there is nothing to prove. So we may assume that MATH is properly contained in MATH. By the description of dominions from REF , there exist elements MATH, and MATH such that MATH and MATH lie in MATH, and MATH does not lie in MATH (it lies in MATH, however). We write MATH. Note that the description of dominions also implies that MATH. Choose MATH such that MATH, and write MATH, MATH. We claim that MATH, MATH, and MATH do not satisfy either REF or REF from REF with respect to MATH. This will show that MATH is not absolutely closed, as these conditions describe the nil-REF absolutely closed groups. REF is the easiest to handle: REF is equivalent to the statement that no nil-REF extension of MATH contains MATH-th roots for both MATH and MATH. However, MATH and MATH are both MATH-th powers modulo the commutator in MATH, so there is a nil-REF extension of MATH with MATH-th roots for both MATH and MATH. In particular, there is such an extension of MATH as well. So MATH, MATH, and MATH do not satisfy REF is somewhat more difficult, and we proceed by contradiction. Suppose that REF is satisfied in MATH. Then there exist integers MATH, MATH, and MATH, and elements MATH such that MATH . By REF , we can write MATH as MATH for some MATH, MATH, MATH, with MATH. In particular, MATH. By REF , the congruences above imply that MATH. However, since MATH, this means that MATH. But the right hand side clearly lies in MATH, and this contradicts the choice of MATH and MATH. So MATH, MATH, MATH, and MATH cannot satisfy REF either. Therefore MATH is not absolutely closed. We conclude that MATH has no absolute closure, which proves the MATH case. For the MATH case, we proceed exactly as above, replacing MATH with MATH, and restricting MATH to the variety in question. The argument above shows that MATH, MATH, MATH and MATH do not satisfy REF nor REF . If MATH, then MATH, which contradicts the fact that MATH is of exponent MATH, since MATH. Thus, they cannot satisfy REF . For REF , if MATH and there is a solution to the congruences with MATH, then by REF we have that MATH. But MATH, which again contradicts the fact that MATH is of exponent MATH. Therefore, MATH is also not absolutely closed in this case, and this proves the theorem in full.
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math/0006065
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Clearly MATH and MATH. We show that MATH. Suppose that MATH does not satisfy REF . If MATH, then we take an element MATH. We let MATH be a MATH-overgroup in which MATH is a commutator, using REF , and we let MATH be a MATH-overgroup in which MATH is not central, say MATH. Then the amalgam MATH cannot be embedded in a MATH-group, since MATH would have to be central by virtue of being a commutator in MATH, but would have to be noncentral since it is not central in MATH. So MATH cannot satisfy REF . Suppose instead that MATH, and that MATH is such that MATH, but there is some MATH-extension with a MATH-th root for MATH. Note that since there is some extension with a MATH-th root for MATH, MATH must commute with every element of exponent MATH-th in MATH. Since MATH is of exponent MATH, this means that MATH must be central in MATH, hence lies in the commutator of MATH. Let MATH be an overgroup in which MATH does not commute with an element of exponent MATH, for example, consider MATH. Since MATH, MATH does not commute with the generator of the cyclic group. For MATH we want a group of exponent MATH where MATH has a MATH-th root modulo MATH, since then we will have a problem: in any group into which we embedd the amalgam MATH, MATH would be a MATH-th power modulo the commutator, so it would have to commute with everything of exponent MATH, yet it does not do so in MATH. To construct MATH we again have to be a bit careful. First consider a nil-REF group in which MATH has a MATH-th root, and call such a root MATH. We may assume that the group is generated by MATH and MATH. Since MATH lies in MATH, it is central, and we can adjoin a central MATH-th root MATH to that element. Since MATH is of exponent MATH, MATH is of exponent MATH; using REF , we can make MATH into a commutator, MATH, with MATH of exponent MATH. Now we consider the group generated by MATH, MATH, MATH, and MATH. Then MATH . Therefore, this group is of exponent MATH, since it is generated by elements of exponent MATH; moreover, MATH, so MATH is indeed a MATH-th power modulo the commutator subgroup. Make this group MATH, and we are done. Thus, if MATH does not satisfy REF , it cannot be a weak amalgamation base. Now suppose that MATH satisfies REF . We use REF to show that MATH is a strong amalgamation base in MATH. Let MATH and MATH be overgroups of MATH. Then MATH, hence lies in the center of MATH, and likewise for MATH. Suppose that MATH, MATH, MATH, MATH are such that MATH and MATH for some prime power MATH. If MATH, then since MATH is MATH-divisible, we can write MATH and MATH. Then: MATH and likewise with MATH. So we may assume that MATH. If MATH, then MATH and MATH must be central in MATH, and therefore they lie in MATH. So MATH. We may therefore assume that MATH with MATH. Clearly there are overgroups of MATH in which MATH has a MATH-th root, so we must have MATH. Let MATH such that MATH. Likewise, there is a MATH such that MATH. Then MATH and MATH . Thus the amalgam MATH is strongly embeddable into a MATH-group. This proves that MATH is a strong amalgamation base for MATH, and proves the theorem.
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math/0006065
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Note that for any MATH, MATH, as every MATH-th power is central and of exponent MATH, and every commutator is central and of exponent MATH. Clearly MATH. Suppose MATH fails to satisfy MATH. If MATH, let MATH be central of exponent MATH, but not in MATH. Since MATH is of exponent MATH and central, we can extend MATH to a group where MATH is a commutator; moreover, we can make MATH with MATH of exponent MATH, so the resulting group is still of exponent MATH. Let that group be MATH. Then let MATH; MATH is also of exponent MATH, and MATH is not central in MATH. Thus MATH cannot be weakly embedded. If MATH, let MATH be such that MATH, but such that MATH, MATH is central in MATH, and there is some extension of MATH where MATH has a MATH-th root. Let MATH be obtained from MATH by first adjoining a MATH-th root MATH to MATH; then adjoining central MATH-th root MATH to MATH. Note that MATH is central of exponent MATH, since MATH, so we can make MATH into a commutator MATH, with MATH of exponent MATH. Then the group generated by MATH, MATH, MATH and MATH is of exponent MATH (since MATH is also of exponent MATH), and here MATH. Let MATH; then, in MATH, MATH does not commute with an element of exponent MATH, which gives that MATH is not embeddable in MATH. This proves that MATH. To prove that MATH, let MATH be an amalgam of MATH groups, and assume MATH satisfies MATH. We verify NAME 's conditions for strong embeddability. The elements of MATH are central, and of exponent MATH, so they lie in MATH. However, MATH is central in MATH, so MATH. A symmetric argument yields MATH. Now suppose that MATH, MATH, MATH, MATH and MATH is a prime power such that MATH. Clearly we may assume that MATH is a power of REF, and moreover that MATH with MATH (MATH-th powers are central in MATH and of exponent at most REF, so they lie in MATH, and so are central in each MATH). Let MATH, MATH. It is easy to verify that MATH and that MATH and MATH are central in MATH, so we must have MATH. Now we can proceed as before to get that MATH, and so the amalgam is strongly embeddable in MATH.
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math/0006067
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To prove the theorem, we follow CITE in starting with a single peg, which we denote MATH and playing the game in reverse. The first `unhop' produces MATH and the next MATH . (As it turns out, MATH is the only configuration that cannot be reduced to a single peg without using a hole outside the initial set of pegs. Therefore, for all larger configurations we can ignore the MATH's on each end.) We take the second of these as our example. It has two ends, MATH and MATH. The latter can propagate itself indefinitely by unhopping to the right, MATH . When the former unhops, two things happen; it becomes an end of the form MATH and it leaves behind a space of two adjacent holes, MATH . Furthermore, this is the only way to create a MATH. We can move the MATH to the right by unhopping pegs into it, MATH . However, since this leaves a solid block of MATH's to its left, we cannot move the MATH back to the left. Any attempt to do so reduces it to a single hole, MATH . Here we are using the fact that if a peg has another peg to its left, it can never unhop to its left. We prove this by induction: assume it is true for pairs of pegs farther left in the configuration. Since adding a peg never helps another peg unhop, we can assume that the two pegs have nothing but holes to their left. NAME the leftmost peg then produces MATH, and the original (rightmost) peg is still blocked, this time by a peg which itself cannot move for the same reason. In fact, there can never be more than one MATH, and there is no need to create one more than once, since after creating the first one the only way to create another end of the form MATH or MATH is to move the MATH all the way through to the other side MATH and another MATH created on the right end now might as well be the same one. We can summarize, and say that any configuration with three or more pegs that can be reduced to a single peg can be obtained in reverse from a single peg by going through the following stages, or their mirror reflection: CASE: We start with MATH. By unhopping the rightmost peg, we obtain MATH. If we like, we then CASE: NAME the leftmost peg one or more times, creating a pair of holes and obtaining MATH. We can then CASE: Move the MATH to the right (say), obtaining MATH. We can stop here, or CASE: Move the MATH all the way to the right, obtaining MATH, or CASE: Fill the pair by unhopping from the left, obtaining MATH. REF simply states that the set of configurations is the union of all of these plus MATH, MATH, and MATH, with as many additional holes on either side as we like. Then MATH is regular since it can be described by a regular expression CITE, that is, a finite expression using the operators MATH and MATH.
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math/0006067
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Our proof of REF is constructive in that it tells us how to unhop from a single peg to any feasible configuration. We simply reverse this series of moves to play the game.
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math/0006067
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Suppose we are given a string MATH where each MATH. Let MATH be a nondeterministic finite automaton (without MATH-transitions) for MATH, where MATH is the set of states in MATH, MATH is the start state, and MATH is the set of accepting states. We then construct a directed acyclic graph MATH as follows: Let the vertices of MATH consist of all pairs MATH where MATH and MATH. Draw an arc from MATH to MATH in MATH whenever MATH makes a transition from state MATH to state MATH on symbol MATH. Also, draw an arc from MATH to MATH for any MATH and any MATH. Since MATH, MATH. Then any path from MATH to MATH in MATH consists of MATH arcs of the form MATH to MATH, together with some number MATH of arcs of the form MATH to MATH. Breaking the path into subpaths by removing all but the last arc of this second type corresponds to partitioning the input string into substrings of the form MATH, so the length of the shortest path from MATH to MATH in MATH is MATH, where MATH is the minimum number of pegs to which the initial configuration can be reduced. Since MATH is a directed acyclic graph, we can find shortest paths from MATH by scanning the vertices MATH in order by MATH, resolving ties among vertices with equal MATH by scanning vertices MATH (with MATH) earlier than vertex MATH. When we scan a vertex, we compute its distance to MATH as one plus the minimum distance of any predecessor of the vertex. If the vertex is MATH itself, the distance is zero, and all other vertices MATH have no predecessors and infinite distance. Thus we can find the optimal strategy for the initial peg solitaire configuration by forming MATH, computing its shortest path, using the location of the edges from MATH to MATH to partition the configuration into one-peg subconfigurations, and applying REF to each subconfiguration. Since MATH, this algorithm runs in linear time.
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math/0006074
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Though MATH-modules MATH fail to be MATH-modules CITE, one can use the fact that the sheaves MATH are projections MATH of sheaves of MATH-modules. Let MATH be a locally finite open covering of MATH and MATH the associated partition of unity. For any open subset MATH and any section MATH of the sheaf MATH over MATH, let us put MATH. Then, MATH provide a family of endomorphisms of the sheaf MATH, required for MATH to be fine. NAME MATH of MATH also yield the MATH-module endomorphisms MATH of the sheaves MATH. They possess the properties required for MATH to be a fine sheaf. Indeed, for each MATH, there is a closed set MATH such that MATH is zero outside this set, while the sum MATH is the identity morphism.
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math/0006074
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A fiber bundle MATH is a strong deformation retract of MATH. Then, the first isomorphism in REF follows from the NAME - NAME theorem (REF ; REF ), while the second one is a consequence of the well-known NAME theorem.
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math/0006074
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The isomorphism REF also follows from the facts that MATH is a strong deformation retract of MATH and that MATH is the pull-back onto MATH of the sheaf MATH on MATH REF .
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math/0006074
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The proof is obvious. The complex REF is exact due to the NAME lemma, and is a resolution of the constant sheaf MATH on MATH since MATH are sheaves of MATH-modules. Then, REF complete the proof.
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math/0006074
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The exact sequence REF is a resolution of the pull-back sheaf MATH on MATH. Then, by virtue of REF , we have a cohomology isomorphism MATH . REF completes the proof.
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math/0006074
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Due to the relation MATH the horizontal projection MATH provides a homomorphism of the NAME complex REF to the complex MATH . Accordingly, there is a homomorphism MATH of cohomology groups of these complexes. REF show that, for MATH, the homomorphism REF is an isomorphism (see the relation REF below for the case MATH). It follows that a horizontal form MATH is MATH-closed (respectively, MATH-exact) if and only if MATH where MATH is a closed (respectively, exact) form. The decomposition REF complete the proof.
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math/0006074
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Being nilpotent, the vertical differential MATH defines a homomorphism of the complex REF to the complex MATH and, accordingly, a homomorphism of cohomology groups MATH of these complexes. Since MATH, the result follows.
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math/0006074
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Let the common symbol MATH stand for the coboundary operators MATH and MATH of the variational bicomplex. Bearing in mind the decompositions REF , it suffices to show that, if an element MATH is MATH-exact with respect to the algebra MATH (that is, MATH, MATH), then it is MATH-exact in the algebra MATH (that is, MATH, MATH). REF states that, if MATH is a contractible fiber bundle and a MATH-exact form MATH on MATH is of finite jet order MATH (that is, MATH), there exists an exterior form MATH on MATH such that MATH. Moreover, a glance at the homotopy operators for MATH, MATH and MATH CITE shows that the jet order MATH of MATH is bounded for all exterior forms MATH of fixed jet order. Let us call this fact the finite exactness of the operator MATH. Given an arbitrary fiber bundle MATH, the finite exactness takes place on MATH over any open subset MATH of MATH which is homeomorphic to a convex open subset of MATH. Now, we show the following. CASE: Suppose that the finite exactness of the operator MATH takes place on MATH over open subsets MATH, MATH of MATH and their non-empty overlap MATH. Then, it is also true on MATH. CASE: Given a family MATH of disjoint open subsets of MATH, let us suppose that the finite exactness takes place on MATH over every subset MATH from this family. Then, it is true on MATH over the union MATH of these subsets. If REF hold, the finite exactness of MATH on MATH takes place since one can construct the corresponding covering of the manifold MATH REF . Proof of REF . Let MATH be a MATH-exact form on MATH. By assumption, it can be brought into the form MATH on MATH and MATH on MATH, where MATH and MATH are exterior forms of finite jet order. Due to the decompositions REF , one can choose the forms MATH, MATH such that MATH on MATH and MATH on MATH are MATH-exact forms. Let us consider the difference MATH on MATH. It is a MATH-exact form of finite jet order which, by assumption, can be written as MATH where an exterior form MATH is also of finite jet order. REF below shows that MATH where MATH and MATH are exterior forms of finite jet order on MATH and MATH, respectively. Then, putting MATH we have the form MATH equal to MATH on MATH and MATH on MATH, respectively. Since the difference MATH on MATH vanishes, we obtain MATH on MATH where MATH is of finite jet order. Proof of REF . Let MATH be a MATH-exact form on MATH. The finite exactness on MATH holds since MATH on every MATH and, as was mentioned above, the jet order MATH is bounded on the set of exterior forms MATH of fixed jet order MATH.
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math/0006074
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By taking a smooth partition of unity on MATH subordinate to the cover MATH and passing to the function with support in MATH, one gets a smooth real function MATH on MATH which is REF on a neighborhood of MATH and REF on a neighborhood of MATH in MATH. Let MATH be the pull-back of MATH onto MATH. The exterior form MATH is zero on a neighborhood of MATH and, therefore, can be extended by REF to MATH. Let us denote it MATH. Accordingly, the exterior form MATH has an extension MATH by REF to MATH. Then, MATH is a desired decomposition because MATH and MATH are of finite jet order which does not exceed that of MATH.
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math/0006078
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Since MATH is a separability element of MATH, MATH commutes with MATH. The assertion about MATH follows similarly.
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math/0006078
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Define a MATH-linear homomorphism MATH by MATH . This map is MATH-linear, since MATH for all MATH. The inverse map MATH is given by MATH . The collection MATH gives a natural equivalence between the functor MATH and the identity functor. Indeed, for any MATH-linear homomorphism MATH we have : MATH . Similarly, the MATH-linear homomorphism MATH defined by MATH has the inverse MATH and satisfies the necessary properties. Finally, we can check the triangle axiom MATH for all objects MATH of MATH. For MATH we have MATH therefore, MATH.
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math/0006078
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We know already that MATH is monoidal, it remains to prove that MATH and MATH are MATH-linear and satisfy the identities MATH . Take MATH. Using the axioms of a quantum groupoid, we have MATH therefore, MATH is MATH-linear. To check the MATH-linearity of MATH we have to show that MATH, that is ,that MATH . Since both sides of the above equality are elements of MATH, evaluating the second factor on MATH, we get the equivalent condition MATH which is easy to check. Thus, MATH is MATH-linear. Using the isomorphisms MATH and MATH identifying MATH, MATH, and MATH, for all MATH and MATH we have: MATH which completes the proof.
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math/0006078
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Note that MATH is well-defined, since MATH. To prove the MATH-linearity of MATH we observe that MATH . The inverse of MATH is given by MATH where MATH. Therefore, MATH is an isomorphism. Finally, one can verify that the braiding identities MATH are equivalent to the relations of REF , exactly in the same way as in the case of NAME algebras (see, for instance, (CITE, XI, REFEF)).
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math/0006078
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Since we have MATH, the first line is a consequence of the relation MATH, the second line follows from MATH and MATH. The last two identities are proven similarly.
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math/0006078
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It follows from the first two relations of REF , that MATH .
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math/0006078
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First, using the same argument as in CITE, XI, REFEF, we can show that MATH. Next, using REF , we obtain MATH . Let MATH denote multiplication MATH in MATH. Set MATH and MATH. It follows from the above relations that MATH . On the other hand, REF implies that MATH . Therefore, MATH and MATH.
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math/0006078
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Note that MATH for all MATH, by REF . Hence, we have MATH for all MATH. Therefore, using the axioms of a quantum groupoid, we get MATH . The remaining part of the proof follows the lines of (CITEEF). The results for MATH can be obtained by applying the results for MATH to the quasitriangular quantum groupoid MATH.
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math/0006078
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See (CITEEF).
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math/0006078
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Take MATH. Then we have MATH . Therefore MATH, as required.
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math/0006078
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From the observation that MATH we have that the restriction of MATH to MATH is a linear map onto MATH. On the other hand, REF allows to identify MATH with the dual vector space to MATH, from where MATH and the result follows.
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math/0006078
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Associativity of multiplication in MATH and hence in MATH can be verified exactly as in (CITEEF). Let us check that MATH is an ideal. We have : MATH where MATH and we used the identity MATH. Similarly, one checks that MATH therefore for all MATH we have MATH, so MATH is an ideal. We also compute MATH and similarly MATH, so that MATH is a unit. Now let us verify that the structure maps MATH and MATH are well-defined on MATH. We have, using properties of a quantum groupoid and its counital subalgebras: MATH for all MATH. Next, we need to check the axioms of a quantum groupoid. NAME and multiplicativity of MATH are established as in (CITEEF), since the computations given there do not use the unitality of multiplication and comultiplication. For the counit property, we have: MATH where we used the amalgamation property MATH following from REF . Now we verify the remaining axioms of a quantum groupoid. For all MATH and MATH we compute MATH which is REF . For REF we have: MATH where we used the axioms of a quantum groupoid and the definition of MATH. In order to check REF , let us compute the target counital map MATH. We have MATH . Similarly one computes the source counital map : MATH . Using these formulas we have : MATH and MATH . In the above computations we used repeatedly the amalgamation relations in MATH: MATH that follow from REF , the axioms of a quantum groupoid and properties of the counital maps. Finally, we prove the relation which is equivalent to MATH being both algebra and coalgebra anti-homomorphism: MATH . Note that MATH and MATH.
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math/0006078
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The identities MATH and MATH can be written as (identifying MATH with MATH and MATH with MATH): MATH . The above equalities can be verified by evaluating both sides on an element MATH in the third factor (respectively, on MATH in the second factor), see (CITEEF). To show that MATH is an intertwiner between MATH and MATH, we compute MATH where we used MATH for all MATH. Finally, let us check that the element MATH satisfies MATH and MATH. The first property is equivalent to MATH which can be regarded as an equality in MATH : MATH . Evaluating both sides on arbitrary MATH in the second factor, we get MATH where MATH is the source counital map in MATH. The second property is similar.
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math/0006078
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First, observe that for any pair of dual bases MATH and MATH as above and all MATH and MATH the element MATH belongs to MATH. Indeed, MATH for all MATH and, likewise, MATH . Next, we compute MATH where we used the identities MATH that follow from REF and from REF of a quantum groupoid. Therefore, MATH . Thus, we conclude from REF that the map MATH is surjective, that is, MATH is factorizable.
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math/0006078
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Since MATH is an invertible central element of MATH, the homomorphism MATH is a MATH-linear isomorphism. The twist identity MATH follows from the properties of MATH: MATH for all MATH. Clearly, the identity MATH is equivalent to the twist property. It remains to prove that MATH for all MATH, that is, that MATH where MATH is the canonical element in MATH. Evaluating the first factors of the above equality on an arbitrary MATH, we get the equivalent condition : MATH which reduces to MATH. The latter easily follows from the centrality of MATH and properties of MATH.
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math/0006078
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Follows from REF , and REF.
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math/0006078
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Since the trace of an endomorphism MATH, in terms of the canonical element MATH, is MATH, the definition of MATH gives: MATH where we used REF defining MATH and MATH.
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math/0006078
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An endomorphism of an irreducible module is multiplication by a scalar, therefore, we must have MATH. Applying the counit to both sides and using that MATH, we get the result.
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