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math/0004047
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By REF we can find a sequence MATH of complete lattices such that: For all MATH, every monotone MATH is represented by a polynomial with coefficients in MATH. Moreover, MATH is an end extension of MATH. Let MATH. It is easy to see that MATH is complete. [If MATH, let MATH. Since the sequence MATH is weakly decreasing, and since MATH is an end extension of MATH, there must be some MATH such that MATH. Now let MATH, then clearly MATH. ] Let MATH. For simplicity assume MATH. For each MATH define MATH by MATH . MATH is total (since MATH is complete), and clearly monotone. Let MATH be a polynomial such that MATH for all MATH. Note: if MATH, then MATH. Now define a MATH-polynomial MATH by MATH or slightly less formally: MATH . It is easy to check that for all MATH the sequence MATH is eventually constant with value MATH, and this implies MATH.
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We only give a sketch of the main ideas of the proof. The details will appear elsewhere. Start with a lattice MATH. Let MATH that is, MATH is the ``horizontal sum" of the disjoint ortholattices MATH and MATH, where we of course identify the two top elements of the two lattices, and also the two bottom elements. The functions MATH, defined by MATH are clearly monotone. Let MATH. This set is an antichain in MATH, so the functions MATH, defined by MATH is trivially a (partial) monotone function from MATH to MATH. Note that MATH so MATH can be written as a composition of monotone functions and the orthocomplement function. By REF , there is a lattice extension MATH of the lattice MATH such that MATH can be represented by lattice polynomials with coefficients in MATH. A bit of work is now needed to find an ortholattice MATH which is both a lattice extension of MATH and an orthoextension of the original ortholattice MATH. Once this ortholattice is found, the function MATH can be represented as an orthopolynomial with coefficients in MATH.
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The implications REF are trivial. To show REF assume that MATH and fix a base MATH of zero neighborhoods. If MATH is nb-bounded then MATH is bounded for some MATH. Note that MATH is another base of zero neighborhoods. For each MATH in MATH we have MATH. But MATH is bounded and so MATH for some positive MATH, that is, MATH is nn-bounded. Finally, if MATH is nn-bounded, then there is a base MATH such that for every zero neighborhood MATH there is a positive scalar MATH such that MATH. Let MATH be an arbitrary zero neighborhood. Then there exists MATH such that MATH, so that MATH for some MATH. Taking MATH we get MATH, hence MATH is continuous.
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The first equality in REF follows immediately from the definition of MATH. We obviously have MATH . In order to prove the opposite inequality, notice that if MATH, then MATH. Thus, it is left to show that MATH implies MATH. Pick any MATH with MATH, then MATH so that MATH for MATH. Further, since MATH converges to MATH we have MATH . Finally, REF follows directly from the definition if MATH. In the case when MATH, again pick any MATH with MATH, then MATH and MATH .
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math/0004049
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Fix a bounded set MATH. Since MATH converges to zero uniformly on bounded sets then for every base zero neighborhood MATH there exists an index MATH such that MATH whenever MATH. This yields MATH since MATH is bounded. Thus, MATH is bounded for every bounded set MATH, so that MATH is bb-bounded.
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math/0004049
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Fix a zero neighborhood MATH, there exist zero neighborhoods MATH and MATH and an index MATH such that MATH and MATH whenever MATH. Fix MATH. The continuity of MATH guarantees that there exists a zero neighborhood MATH such that MATH. Since MATH, we get MATH, which shows that MATH is continuous.
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math/0004049
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Suppose MATH. If MATH, then MATH for some MATH and some subsequence MATH, so that MATH as MATH goes to infinity. It follows that MATH. On the other hand, if MATH is finite and MATH then MATH for some MATH and for all sufficiently large MATH. Then MATH.
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Let MATH be a linear operator on a topological vector space MATH. Since every singleton is bounded then MATH. Next, assume MATH, fix MATH such that MATH, then the sequence MATH converges to zero equicontinuously. Take a bounded set MATH and a zero neighborhood MATH. There exists a zero neighborhood MATH and a positive integer MATH such that MATH whenever MATH. Also, MATH for some MATH, so that MATH for all sufficiently large MATH. It follows that the sequence MATH converges to zero uniformly on MATH and, therefore, MATH. Thus, MATH. To prove the inequality MATH we let MATH. Then for some base MATH of zero neighborhoods and for every MATH and MATH there exists a positive integer MATH such that MATH for every MATH. Given a zero neighborhood MATH, we can find MATH such that MATH. Then MATH for every MATH, so that the sequence MATH converges to zero equicontinuously, and, therefore, MATH. Finally, we must show that MATH. Suppose that MATH, we claim that MATH. Take MATH so that MATH. One can find a zero neighborhood MATH such that for every zero neighborhood MATH there is a positive integer MATH such that MATH for every MATH. Fix a base MATH of zero neighborhoods, and define a new base MATH of zero neighborhoods via MATH. Let MATH and MATH. Then MATH for some positive integer MATH and MATH. Then MATH and for every sufficiently large MATH, so that MATH, for each sufficiently large MATH, which implies MATH.
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To prove REF let MATH . Since every convergent sequence is bounded, we certainly have MATH. Conversely, suppose MATH, and take any positive scalar MATH such that MATH. Then for every MATH the sequence MATH is bounded, and it follows that the sequence MATH converges to zero, so that MATH and, therefore MATH. To prove REF , suppose MATH is bb-bounded, and let MATH . NAME show that MATH. If MATH converges to zero uniformly on every bounded set, then for each bounded set MATH and for each zero neighborhood MATH there exists a positive integer MATH such that MATH whenever MATH. Also, since MATH is bb-bounded, then for every MATH we have MATH for some MATH. Therefore, if MATH, then MATH for every MATH, so that the sequence MATH is uniformly bounded on MATH. Thus MATH, so that MATH. Now suppose MATH. There exists MATH such that MATH. The set MATH is bounded for every bounded set MATH, so that for every zero neighborhood MATH there exists a scalar MATH such that MATH for every MATH. Then MATH for all sufficiently large MATH. This means that the sequence MATH converges to zero uniformly on MATH, and it follows that MATH. Further, if MATH is bb-bounded, then any finite initial segment MATH is always uniformly bounded on every bounded set, so that a tail MATH is uniformly bounded on every bounded set if and only if the whole sequence MATH is uniformly bounded on every bounded set. REF , and REF can be proved in a similar way.
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It follows from the definition of MATH and REF that MATH . Similarly, since the balanced convex hull of a bounded set is bounded, MATH . Let MATH for every MATH. Then, rephrasing the definition of MATH and applying REF , we have MATH . Similarly, MATH . Finally, MATH .
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To prove the first equality it suffices to show that MATH if and only if MATH whenever MATH is an ultimately bounded net. Suppose that MATH, and let MATH be a zero neighborhood. One can find a zero neighborhood MATH such that for every MATH there exists MATH such that MATH for each MATH. Let MATH be an ultimately bounded net. There exists an index MATH and a number MATH such that MATH whenever MATH. Then for MATH one can find MATH such that MATH for each MATH, so that MATH whenever MATH and MATH. This means that MATH. Conversely, suppose that MATH for each ultimately bounded net MATH, and assume that MATH does not converge equicontinuously to zero. Then there exists a zero neighborhood MATH such that for every zero neighborhood MATH one can find MATH such that for every MATH there exists MATH with MATH. Then there exists MATH such that MATH . The collection of all zero neighborhood ordered by inclusion is a directed set, so that MATH is an ultimately bounded net. Indeed, if MATH is a zero neighborhood then MATH for each zero neighborhood MATH and every MATH. But it follows from REF that the net MATH does not converge to zero. To prove the second equality, let MATH . Since every net which converges to zero is necessarily ultimately bounded, it follows that MATH. Now let MATH, and let MATH be an ultimately bounded sequence. Suppose that MATH, then MATH is ultimately bounded, that is, for each zero neighborhood MATH there exists an indices MATH and MATH and a positive MATH such that MATH whenever MATH and MATH. It follows that MATH for MATH and all sufficiently large MATH. This implies that MATH so that MATH.
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Suppose MATH and MATH and let MATH be an ultimately bounded net in MATH. Then the net MATH is ultimately bounded by REF . By applying REF again we conclude that MATH converges to zero. In particular, MATH, and applying REF one more time we get MATH.
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Assume without loss of generality that both MATH and MATH are finite. Suppose that MATH and take MATH and MATH such that MATH. Let MATH be an ultimately bounded net in MATH. By REF it suffices to show that MATH. Notice that the net MATH is ultimately bounded. This implies that the net MATH converges to zero. Fix a seminorm MATH, then there exist indices MATH and MATH such that MATH whenever MATH and MATH. Also, notice that we can split MATH into a product of two terms MATH such that MATH while still MATH. Further, if MATH and MATH then we have MATH . Since MATH and MATH, we have MATH . Notice that MATH and that MATH. Since MATH is continuous, the net MATH is ultimately bounded for every fixed MATH, so that MATH. It follows that for every MATH between MATH and MATH the expression MATH is uniformly bounded for all sufficiently large MATH and MATH. Similarly, for every MATH between MATH and MATH the expression MATH is uniformly bounded for all sufficiently large MATH and MATH. Therefore there exist indices MATH and MATH such that the finite sum MATH is uniformly bounded for all MATH and MATH. It follows that MATH, so that MATH.
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math/0004049
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Suppose MATH is a fast null sequence in a topological vector space and MATH. Let MATH, the sequence MATH converges to zero, hence is ultimately bounded, then by REF we have MATH .
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For any MATH such that MATH one can find MATH such that MATH and MATH. Consider a point MATH and a base zero neighborhood MATH. Since by the definition of MATH the sequence MATH converges to zero, there exist a positive integer MATH, such that MATH whenever MATH. Therefore, MATH because MATH is balanced. Thus, if MATH, then MATH because MATH is convex. Since MATH, we have MATH for sufficiently large MATH and MATH, and so MATH because MATH is balanced. Therefore MATH is a NAME sequence and hence it converges to some MATH because MATH is sequentially complete. Clearly, MATH is a linear operator. Notice that MATH for every MATH. As MATH goes to infinity, the left hand side of this identity converges to MATH, while the right hand side converges to MATH. Thus it follows that MATH. Finally, notice that MATH commutes with MATH for every MATH. Therefore, if MATH is continuous, then MATH for every MATH. This implies that MATH, so that MATH is the (left and right) inverse of MATH. This means that MATH and MATH. Thus, MATH.
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math/0004049
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Suppose that MATH, then the sum MATH of the NAME series exists by REF . As in the proof of REF we denote the partial sums of the NAME series by MATH. Fix MATH such that MATH and MATH, and consider a bounded set MATH and a closed base zero neighborhood MATH. Since MATH converges to zero uniformly on MATH, there exits MATH such that MATH for all MATH. Also, since MATH, we can assume without loss of generality that MATH. Then MATH whenever MATH and MATH. Since MATH is closed, we have MATH for each MATH and MATH, so that MATH whenever MATH. This shows that MATH converges to MATH uniformly on bounded sets. By REF this implies that MATH is bb-bounded. Further, if MATH is continuous, then by REF we have MATH, so that MATH, whence it follows that MATH.
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math/0004049
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Let MATH. It follows from REF that the NAME series converges to MATH. Again, we denote the partial sums of the NAME series by MATH. Let MATH be such that MATH and MATH. For a fixed closed zero neighborhood MATH there exists a zero neighborhood MATH such that MATH for every MATH. Let MATH, then MATH for some MATH. Then MATH whenever MATH and MATH. Since MATH is closed, we have MATH for each MATH and MATH, so that MATH whenever MATH. This shows that MATH converges to MATH equicontinuously, and REF yields that MATH is continuous.
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math/0004049
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Let MATH. By REF the NAME series MATH converges to MATH. Again, we denote the partial sums of the NAME series by MATH. Fix some MATH such that MATH and MATH. There exists a base MATH of closed convex zero neighborhoods such that for every MATH there is a scalar MATH such that MATH for all MATH. Fix MATH, then for each MATH we have MATH for some MATH, so that MATH whenever MATH. It follows that MATH . Then MATH, so that MATH, which implies that MATH is nn-bounded, and, therefore, MATH holds. Fix MATH. Then MATH for some MATH. Then for every MATH we have MATH whenever MATH and MATH. Since MATH is closed, we have MATH for each MATH and MATH, so that MATH whenever MATH. This shows that MATH nn-converges to MATH.
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math/0004049
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Let MATH. By REF the NAME series MATH converges to MATH. Since MATH then MATH is bb-bounded by REF . But then MATH. Notice that MATH is nb-bounded as a product of a bb-bounded and an nb-bounded operators (see REF). Suppose that MATH. Fix MATH such that MATH and MATH, then the sequence MATH converges to zero uniformly on some base zero neighborhood MATH. We will show that the NAME series converges uniformly on MATH. As in the proof of REF , we denote the partial sums of the NAME series by MATH. Fix a closed base zero neighborhood MATH. Since MATH converges to zero uniformly on MATH, there exits MATH such that MATH for all MATH. Also, since MATH, we can assume without loss of generality that MATH. Then MATH whenever MATH and MATH. Since MATH is closed, we have MATH for each MATH and MATH, so that MATH whenever MATH.
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math/0004049
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If MATH is locally bounded then the identity map is an nb-bounded bijection. Suppose that MATH is an nb-bounded bijection on MATH. Then there exists a closed base zero neighborhood MATH in MATH such that MATH is bounded. Let MATH, then MATH is convex, bounded, balanced, and absorbing. It follows that the space MATH is a locally convex and locally bounded, denote it by MATH. Notice also that the topology of MATH is finer than the original topology on MATH because MATH is bounded. In particular, MATH is NAME. We claim that MATH is complete. Indeed, if MATH is a NAME sequence in MATH, then it is also NAME in the original topology of MATH, which is complete, so that MATH converges to some MATH. Fix MATH, then there exists MATH such that MATH whenever MATH. Let MATH, since MATH is closed we have MATH, that is, MATH in MATH. Thus, MATH is complete, hence NAME. Since MATH is bounded, we can find MATH such that MATH. Then MATH, so that MATH is bounded in MATH. Then MATH is also bounded in MATH by the NAME Theorem, so that MATH for some MATH, hence MATH is bounded.
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math/0004049
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If every zero neighborhood of MATH contains a non-trivial linear subspace, then MATH cannot be one-to-one by REF. Suppose now that MATH and MATH are NAME and assume that MATH is a bijection. Let MATH be the linear inverse of MATH. The Open Mapping Theorem implies that MATH is continuous and hence bb-bounded. It follows that the identity operator of MATH is nb-bounded being the composition of the nb-bounded operator MATH and the bb-bounded operator MATH. But the identity operator is nb-bounded if and only if the space is locally bounded, a contradiction.
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math/0004049
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Define a linear map MATH from MATH to MATH via MATH. Then the dimension of the range MATH is at most MATH. Define also a linear map MATH from MATH to MATH via MATH. It can be easily verified that MATH is well-defined. Then the range of MATH coincides with the range MATH, which is of dimension at most MATH.
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math/0004049
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Suppose MATH maps some weak base zero neighborhood MATH (MATH), to a weakly bounded set. Since the weak topology is NAME, it follows from REF that MATH. In particular, MATH whenever MATH for every MATH. Then REF implies that MATH is a finite rank operator.
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math/0004049
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If MATH is locally bounded then the result is trivial by REF. Suppose that MATH is not locally bounded, then, in view of REF, it suffices to show that MATH. Let MATH, then MATH is bb-bounded. If MATH, then it follows from MATH that MATH. Thus, MATH is a sum of an nb-bounded operator and a multiple of the identity operator, which yields MATH. To finish the proof, it suffices to show that MATH necessarily belongs to MATH (and, therefore, to MATH, MATH, and MATH). Indeed, if the resolvent MATH were bb-bounded, then MATH would be nb-bounded, which is impossible in a non-locally bounded space, a contradiction.
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By REF it suffices to show that MATH. Since MATH is nb-bounded, then MATH is a bounded set for some zero neighborhood MATH. Let MATH and fix a zero neighborhood MATH. Then MATH is again a zero neighborhood. In particular, since the sequence MATH converges to zero uniformly on bounded sets, we have MATH for all sufficiently large MATH. Then MATH, so that MATH converges to zero uniformly on MATH. Therefore MATH, so that MATH.
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math/0004049
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Suppose MATH is bounded for some base zero neighborhood MATH. It follows from REF , and REF, and REF that it suffices to show that MATH. We are going to show that MATH induces a continuous operator MATH on some NAME space such that MATH while MATH, and then appeal to the fact that the spectral radius of a continuous operator on a NAME space equals the geometrical radius of the spectrum. Consider MATH as a continuous operator on the locally bounded space MATH. Then MATH is defined by REF and MATH is defined by REF. We claim that MATH. To see this, suppose MATH, then MATH for all sufficiently large MATH. Let MATH be a base zero neighborhood, then MATH for some MATH, so that MATH for sufficiently large MATH. This implies that MATH, and it follows that MATH. On the other hand, we claim that MATH. Suppose MATH, then MATH is nn-bounded with respect to some base MATH of zero neighborhood. We can assume without loss of generality that MATH, so that MATH for some MATH. It follows that MATH. Since MATH is convex, the the space MATH is, in fact, a seminormed space. We can assume without loss of generality that it is a normed space, because otherwise we can consider the quotient space MATH and the quotient operator MATH on this quotient space instead of MATH. Indeed, since MATH by REF, we conclude that the quotient space MATH is NAME. It follows then that MATH is a normed space, and MATH is norm bounded. The spectrum MATH becomes even smaller when we substitute MATH with MATH. Indeed, suppose MATH, then the resolvent MATH exists in MATH and is continuous. If MATH, then MATH, so that MATH leaves MATH invariant, and, therefore, induces a quotient operator MATH on MATH via MATH. Clearly, MATH is continuous: if MATH in MATH then MATH in MATH for some MATH in MATH, so that MATH. On the other hand, MATH, because if MATH then MATH for all sufficiently large MATH, then MATH, so that MATH for some MATH. It follows that MATH and, therefore, MATH. Finally, we consider the completion MATH of MATH, and extend MATH to a continuous linear operator MATH on the completion. The spectrum of MATH is smaller that the spectrum of MATH, because if MATH then the resolvent MATH can be extended by continuity to MATH on MATH, and MATH is a continuous inverse to MATH, so that MATH. On the other hand, MATH because if MATH then MATH for all sufficiently large MATH, which implies MATH since MATH is a restriction of MATH on MATH.
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math/0004049
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Assume that MATH. Without loss of generality (by scaling MATH) we can assume that MATH. Since MATH is compact, there is a closed base zero neighborhood MATH such that MATH is compact. In particular MATH is bounded, so that MATH for some MATH. We can assume without loss of generality that MATH. We define the following subsets of MATH: MATH . Notice, that MATH is compact because MATH is compact and MATH is closed. Also, if MATH is compact, then MATH is compact as the image of a compact set under a continuous operator. Therefore, every MATH for MATH is compact. Using induction, we can show that the sequence MATH is decreasing. Indeed, MATH by definition, MATH, and if MATH, then MATH. It follows also that MATH is compact and contains zero. Notice that MATH maps every balanced set to a balanced set. Since MATH is balanced, MATH is balanced for each MATH. If MATH is a balanced subset of MATH, then obviously MATH, and when we apply the same reasoning to MATH instead of MATH (which is also a balanced subset of MATH), we get MATH. We use this to show by induction that MATH for every MATH. Indeed, for MATH we have MATH. Suppose MATH for some MATH, then MATH which proves the induction step. Next, we claim that there exists an open zero neighborhood MATH and an increasing sequence of positive integers MATH such that MATH is nonempty for every MATH. Assume for the sake of contradiction that for every open zero neighborhood MATH we have MATH for all sufficiently large MATH. Since MATH contains an open zero neighborhood, then there exists a positive integer MATH such that MATH whenever MATH. This implies that MATH for all MATH. Indeed, this holds trivially for MATH. Suppose that MATH for some MATH. Then MATH, and this implies that MATH because MATH. Now take any open zero neighborhood MATH, then MATH is again a zero neighborhood, and by assumption there exists a positive integer MATH such that MATH whenever MATH. Let MATH, then MATH which contradicts the hypothesis MATH. It follows from MATH for every MATH that MATH for all sufficiently large MATH because MATH is a decreasing sequence. Since MATH is a decreasing sequence of nonempty compact sets, then MATH, so that MATH. For every MATH we have MATH, it follows that MATH and, therefore, MATH. Actually, the reverse inclusion also holds. To see this, let MATH. Then MATH, where MATH. Since MATH is compact, the sequence MATH has a cluster point, that is, MATH for some subsequence MATH and some MATH. Since MATH is continuous we have MATH. On the other hand, since every MATH is closed we have MATH, so that MATH. Thus MATH. Next, we claim that MATH. Indeed, MATH . On the other hand, since MATH are decreasing and MATH, we have MATH and MATH, so that MATH, and this implies MATH for every MATH. Thus MATH. Since MATH is compact, hence bounded, then MATH is also bounded. Then there is a positive constant MATH such that MATH. Without loss of generality we can assume MATH. It follows that MATH . We use induction to show that MATH. Indeed, since MATH for any four sets MATH, MATH, MATH, and MATH, then MATH . Finally, MATH. Next, consider the set MATH. This set is closed under multiplication by a scalar, and MATH implies that MATH is a linear subspace of MATH. We consider the locally bounded topological vector space MATH with multiples of MATH as the base of zero neighborhoods. Since MATH is balanced by definition, this topology is linear, and it is NAME because MATH is compact. Also, it is finer than the topology on MATH inherited from MATH because MATH is compact and, therefore, bounded in MATH. We claim that MATH is complete. Indeed, if MATH is a NAME sequence in MATH then there exists MATH such that MATH for each MATH. Since MATH is compact, the sequence MATH has a subsequence which converges to some MATH in the topology of MATH. Moreover, MATH because the sequence MATH is NAME in MATH. Fix MATH, then there exists MATH such that MATH whenever MATH. Let MATH, since MATH is is closed we have MATH, that is, MATH in MATH. Thus, MATH is complete and, therefore, quasi-Banach. It follows from MATH that MATH is invariant under MATH and the restriction MATH is continuous. We claim that MATH. Suppose that MATH and MATH, then MATH is a homeomorphism, so that MATH is a closed zero neighborhood, and MATH for some positive real MATH because MATH is bounded. Further, MATH. Therefore MATH and since MATH is one-to-one we get MATH. Similarly, we obtain MATH for all MATH, and then MATH. This implies that the restriction of MATH to MATH is onto, invertible, and the inverse is continuous. Thus, MATH. In particular this implies that MATH. On the other hand, it follows from MATH that MATH for all MATH, so that MATH does not converge to zero uniformly on MATH, whence MATH. This produces a contradiction because it was proved in CITE that the spectral radius of a continuous operator on a quasi-Banach space equals the radius of the spectrum.
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math/0004049
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It can be easily verified that MATH is a bijection from MATH onto MATH and, therefore, the restriction MATH is a bijection. Notice that since MATH for each MATH and MATH for each MATH, then MATH so that MATH and MATH commute on MATH. It also follows that for each MATH we have MATH . It follows that MATH for each MATH, so that MATH where MATH. Since MATH it follows that MATH for some MATH, so that MATH.
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math/0004049
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Suppose that MATH and consider the resolvent operator MATH on MATH. Then MATH hence MATH is nn-bounded and MATH. Suppose now that MATH is dense in MATH, MATH is the smallest closed extension of MATH, and MATH. Then there exists an nn-bounded operator MATH such that MATH for each MATH. Then there is a constant MATH such that MATH for each MATH. It follows that MATH can be extended to a bounded operator MATH on MATH. Fix MATH and pick MATH in MATH such that MATH. Then MATH and MATH. Since MATH is closed we have MATH. It follows, in particular, that MATH is onto. Since MATH for each MATH, it follows that for every nonzero MATH the pair MATH doesn't belong to the closure of the graph of MATH. But the closure of the graph of MATH is the graph of MATH because MATH is the smallest closed extension of MATH. It follows that MATH is one-to-one, hence MATH.
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math/0004051
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Given a functor MATH, we define MATH and MATH. Since MATH is a left adjoint, it preserves colimits. The structure maps of the colimit are then the composites MATH . Although MATH does not preserve limits, there is still a natural map MATH for any functor MATH. Then the structure maps of the limit are the composites MATH .
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math/0004051
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Given MATH, we define MATH, with structure maps MATH. Given a map MATH, we define MATH by MATH.
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math/0004051
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To prolong MATH, take MATH to be the identity in REF . To prolong its right adjoint MATH, take MATH to be the composite MATH of the counit and unit of the adjunction. It is a somewhat involved exercise in adjoint functors to verify that the resulting prolongations MATH and MATH are still adjoint to each other. Like all the exercises in adjoint functors in this paper, the simplest way to proceed is to write the adjoint MATH of a map MATH as the composition MATH and to use the standard properties of the unit and counit of an adjunction.
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The main point is that MATH commutes with colimits. We leave the remainder of the proof to the reader.
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By adjunction, a map MATH has the left lifting property with respect to MATH if and only if MATH has the left lifting property with respect to MATH. Since a map is a cofibration (respectively, trivial cofibration) in MATH if and only if it has the left lifting property with respect to all trivial fibrations (respectively, fibrations), the lemma follows.
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math/0004051
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Since MATH is a left NAME functor, every map in MATH is a level cofibration. By REF , this means that MATH for all MATH and all trivial fibrations MATH. Since a map in MATH has the left lifting property with respect to every map in MATH, in particular it has the left lifting property with respect to MATH. Another application of REF completes the proof for MATH. The proof for MATH is similar.
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math/0004051
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The retract and two out of three axioms are immediate, as is the lifting axiom for a projective cofibration and a level trivial fibration. By adjointness, a map is a level trivial fibration if and only if it is in MATH. Hence a map is a projective cofibration if and only if it is in MATH. The small object argument CITE applied to MATH then produces a functorial factorization into a projective cofibration followed by a level trivial fibration. NAME implies that a map is a level fibration if and only if it is in MATH. We have already seen in REF that the maps in MATH are level equivalences, and they are projective cofibrations since they have the left lifting property with respect to all level fibrations, and in particular level trivial fibrations. Hence the small object argument applied to MATH produces a functorial factorization into a projective cofibration and level equivalence followed by a level fibration. Conversely, we claim that any projective cofibration and level equivalence MATH is in MATH, and hence has the left lifting property with respect to level fibrations. To see this, write MATH where MATH is in MATH and MATH is in MATH. Then MATH is a level fibration. Since MATH and MATH are both level equivalences, so is MATH. Thus MATH has the left lifting property with respect to MATH, and so MATH is a retract of MATH by the retract argument CITE. In particular MATH. Since colimits and limits in MATH are taken levelwise, and since every projective cofibration is in particular a level cofibration, the statements about properness are immediate.
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math/0004051
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We only prove the cofibration case, leaving the similar trivial cofibration case to the reader. First suppose MATH is a projective cofibration. We have already seen in REF that MATH is a cofibration. Suppose MATH is a trivial fibration in MATH, and suppose we have a commutative diagram MATH . We must construct a lift in this diagram. By adjointness, it suffices to construct a lift in the induced diagram MATH where MATH is the right adjoint of MATH. Using the description of MATH given in REF , one can easily check that the map MATH is a level trivial fibration, so a lift exists. Conversely, suppose that the map MATH satisfies the conditions in the statement of the proposition. Suppose MATH is a level trivial fibration in MATH, and suppose the diagram MATH commutes. We construct a lift MATH, compatible with the structure maps, by induction on MATH. There is no difficulty defining MATH, since MATH has the left lifting property with respect to the trivial fibration MATH. Suppose we have defined MATH for MATH. Then by lifting in the induced diagram MATH we find the required map MATH.
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math/0004051
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The functor MATH obviously takes level fibrations to fibrations and level trivial fibrations to trivial fibrations. Hence MATH is a right NAME functor, and so its left adjoint MATH is a left NAME functor. Similarly, the prolongation of MATH to a functor MATH preserves level fibrations and level trivial fibrations, so its left adjoint MATH is a NAME functor.
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math/0004051
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Suppose first that MATH does induce a NAME equivalence on the localizations, and suppose that MATH is MATH-local. Then MATH is also MATH-local. Let MATH denote a fibrant replacement functor in MATH. Then, because MATH is a NAME equivalence on the localizations, the map MATH is a weak equivalence in MATH (see CITE). But both MATH and MATH are MATH-local, so MATH is a weak equivalence in MATH. Hence MATH is weakly equivalent in MATH to MATH, where MATH is the MATH-local object MATH. Before proving the converse, note that, since MATH is a NAME equivalence before localizing, the map MATH is a weak equivalence for all fibrant MATH. Since the functor MATH does not change upon localization, and every MATH-local object of MATH is in particular fibrant, this condition still holds after localization. Thus MATH is a NAME equivalence after localization if and only if MATH reflects local equivalences between cofibrant objects, by CITE. Suppose MATH is a map between cofibrant objects such that MATH is a MATH-local equivalence. We must show that MATH is a weak equivalence for all MATH-local MATH. NAME implies that MATH is a weak equivalence for all MATH-local MATH, and our condition then guarantees that this is enough to conclude that MATH is a weak equivalence for all MATH-local MATH. We still need to prove the last statement of the proposition. So suppose MATH is MATH-local. Then MATH is also MATH-local, and, in MATH, we have a weak equivalence MATH. Our assumption then guarantees that MATH is MATH-local, and MATH is in indeed weakly equivalent to MATH.
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math/0004051
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The only if half is clear. Since every cofibrant object is a retract of a cell complex (that is, an object MATH such that the map MATH is a transfinite composition of pushouts of maps of MATH), it suffices to show that MATH, or, equivalently, MATH, is a weak equivalence for all cell complexes MATH. Given a cell complex MATH, there is an ordinal MATH and a MATH-sequence MATH with colimit MATH. We will show by transfinite induction on MATH that MATH is a weak equivalence for all MATH. Since MATH, getting started is easy. For the successor ordinal case of the induction, suppose MATH is a weak equivalence. We have a pushout square MATH where MATH is a map of MATH. Factor the composite MATH into a cofibration MATH followed by a trivial fibration MATH. In the terminology of CITE, MATH is a cofibrant approximation to MATH. By CITE, there is a cofibrant approximation MATH to MATH which is a pushout of MATH. This is where we need our model category to be left proper. For any fibrant object MATH, the functor MATH converts colimits to limits and cofibrations to fibrations CITE. Hence we have two pullback squares of fibrant simplicial sets MATH where MATH and MATH, respectively. Here the vertical maps are fibrations. There is a map from the square with MATH to the square with MATH induced by MATH. By hypothesis, this map is a weak equivalence on every corner except possibly the upper left. But then NAME 's cube lemma (see CITE, where the dual of the version we need is proved, or CITE) implies that the map on the upper left corner is also a weak equivalence, and hence that MATH is a weak equivalence. Now suppose MATH is a limit ordinal and MATH is a weak equivalence for all MATH. Then, for MATH or MATH, the simplicial sets MATH define a limit-preserving functor MATH such that each map MATH is a fibration of fibrant simplicial sets. There is a natural transformation from the functor with MATH to the functor with MATH, and by hypothesis this map is a weak equivalence at every stage. It follows that it is a weak equivalence on the inverse limits, as one can see in different ways. The simplest is probably to note that the inverse limit is a right NAME functor CITE. Thus MATH is a weak equivalence, as required. This completes the transfinite induction and the proof.
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math/0004051
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By definition, MATH is MATH-local if and only if MATH is level fibrant and MATH is a weak equivalence for all MATH and all domains and codomains MATH of maps of MATH. By the comments preceding REF , this is equivalent to requiring that MATH be level fibrant and that the map MATH be a weak equivalence for all MATH and all domains and codomains MATH of maps of MATH. By REF , this is equivalent to requiring that MATH be a MATH-spectrum. Now, by definition, MATH is a stable equivalence if and only if MATH is a weak equivalence for all MATH-spectra MATH. But by adjointness, MATH can be identified with MATH. Since MATH is a weak equivalence between fibrant objects, so is MATH.
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math/0004051
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We know already that the cofibrations are the same in the stable model structure on MATH and the stable model structure of CITE. We will show that the weak equivalences are the same. In any model category at all, a map MATH is a weak equivalence if and only if MATH is a weak equivalence of simplicial sets for all fibrant MATH. Construction of MATH requires replacing MATH by a cofibrant approximation MATH and building cosimplicial resolutions of the domain and codomain of MATH. In the case at hand, we can do the cofibrant replacement and build the cosimplicial resolutions in the strict model category of spectra, since the cofibrations do not change under localization. Thus MATH is the same in both the stable model structure on MATH and in the stable model category of NAME and NAME. Since the stably fibrant objects are also the same, the corollary holds.
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math/0004051
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In view of NAME 's localization REF , we must show that MATH is a stable equivalence for all MATH. Since the domains and codomains of the maps of MATH are already cofibrant, it is equivalent to show that MATH is a stable equivalence for all MATH. Since MATH, we have MATH. In view of REF , this map is a weak equivalence whenever MATH, and hence MATH, is cofibrant. Taking MATH, where MATH is a domain or codomain of a map of MATH, completes the proof.
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math/0004051
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There is a a natural map MATH which is a weak equivalence when MATH is a stably fibrant object of MATH. This means that the total right derived functor MATH is naturally isomorphic to the identity functor on MATH (where we use the stable model structure). On the other hand, MATH is naturally isomorphic to MATH and also to MATH, since MATH and MATH commute with each other. Thus the natural isomorphism from the identity to MATH gives rise to an natural isomorphism MATH and a natural isomorphism MATH, displaying MATH and MATH as adjoint equivalences of categories. It follows that MATH and MATH are both NAME equivalences, as required. Since MATH is adjoint to MATH and MATH is adjoint to MATH, we must also have MATH naturally isomorphic to MATH and MATH naturally isomorphic to MATH.
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math/0004051
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The map MATH is the adjoint of the structure map MATH of MATH, where MATH denotes the counit of the adjunction, MATH denotes the unit, and MATH denotes the structure map of MATH. Thus MATH is the composite MATH . Since MATH is the identity, it follows that MATH.
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math/0004051
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The map MATH is the colimit of the vertical maps in the diagram below. MATH . Since the vertical and horizontal maps coincide, the result follows. For the second statement, we note that if MATH is level fibrant, each MATH is level fibrant since MATH is a right NAME functor (with respect to the projective model structure). Since sequential colimits in MATH preserve fibrant objects, MATH is level fibrant, and hence a MATH-spectrum.
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math/0004051
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By assumption, the map MATH is a level equivalence between level fibrant objects. Since MATH is a right NAME functor, MATH is a level equivalence as well. Then the method of CITE completes the proof. Recall that this method is to use factorization to construct a sequence of projective trivial cofibrations MATH with MATH and a level trivial fibration of sequences MATH. Then the map MATH will be a projective trivial cofibration. Since sequential colimits in MATH preserve trivial fibrations, the map MATH will still be a level trivial fibration.
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math/0004051
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By definition, MATH is a stable equivalence if and only if MATH is a weak equivalence for all MATH-spectra MATH. But we have a level equivalence MATH by REF , and so it suffices to know that MATH is a weak equivalence for all MATH-spectra MATH. But, by REF , MATH is an isomorphism.
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math/0004051
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Suppose MATH is a MATH-spectrum such that the map MATH is an isomorphism. We will show that MATH as a retract of MATH; this will obviously complete the proof. We first note that there is a natural map MATH obtained by precomposition with MATH and postcomposition with MATH. Here MATH is the inverse of the map MATH, which is an isomorphism since MATH is so. On the other hand, we claim that there is also a natural map MATH obtained by applying the total right derived functor of MATH. This is not obvious, since we are asserting that the total right derived functor of MATH preserves the enrichment of the projective homotopy category of MATH over the homotopy category of simplicial sets, even though MATH is not a right NAME functor. Nevertheless, if we assume that this natural map exists, it follows easily that the composite MATH is the identity (in the homotopy category of simplicial sets), and therefore that MATH is a retract of MATH. It remains to show that the total right derived functor of MATH preserves the enriched structure. We first point out that MATH preserves level fibrations between level fibrant objects and all level trivial fibrations, because MATH is almost finitely generated. It follows from NAME 's lemma CITE that MATH preserves level equivalences between level fibrant objects. Because sequential colimits in MATH preserve finite products, MATH also preserves finite products. We claim that, for any functor MATH on a model category MATH that preserves fibrations between fibrant objects, weak equivalences between fibrant objects, and finite products, the total right derived functor of MATH preserves the enriched structure over MATH. Indeed, analysis of the definition of this enrichment CITE shows that it suffices to check that such a functor MATH preserves simplicial frames on a fibrant object MATH. A simplicial frame on MATH is a factorization MATH in the diagram category of simplicial objects in MATH, where MATH is the constant simplicial diagram on MATH, the MATH-th space of MATH is the product MATH, MATH is a level equivalence, and MATH is a level fibration. We further require that MATH is an isomorphism in degree MATH. Since MATH preserves products, weak equivalences between fibrant objects, and fibrations between fibrant objects, it follows that MATH is a simplicial frame on MATH.
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math/0004051
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One can easily check that MATH is an isomorphism, using REF .
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math/0004051
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The first statement follows immediately from REF . By the first statement, if MATH is a stable equivalence, so is MATH. Since MATH is a map between MATH-spectra, it is a stable equivalence if and only if it is a level equivalence. The converse follows from REF .
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math/0004051
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We have MATH, by REF , where MATH denotes the left homotopy relation. We can use the cylinder object MATH as the source for our left homotopies. Then adjointness implies that MATH. Since MATH and MATH are finitely presented, we get the required result.
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math/0004051
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We wil actually show that, if MATH is a level fibration and MATH is a stable equivalence, the pullback MATH is a stable equivalence. The first step is to use the right properness of the projective model structure on MATH to reduce to the case where MATH and MATH are level fibrant. Indeed, let MATH, MATH, and MATH. Then factor the composite MATH into a projective trivial cofibration MATH followed by a level fibration MATH. Then we have the commutative diagram below, MATH where the vertical maps are level equivalences. Then REF, which depend on the projective model structure being right proper, imply that the induced map MATH is a level equivalence. Hence MATH is a stable equivalence if and only if MATH is a stable equivalence, and so we can assume MATH and MATH are level fibrant. Now let MATH denote the pullback square below. MATH . Then MATH is a pullback square for all MATH, and there are maps MATH. Since pullbacks commute with sequential colimits, MATH is a pullback square. Furthermore, MATH is a level fibration, since sequential colimits in MATH preserve fibrations between level fibrant objects. Since MATH is a stable equivalence between level fibrant spectra, MATH is a level equivalence by REF . So, since the projective model structure is right proper, the map MATH is a level equivalence, and thus MATH is a stable equivalence. The characterization of stable fibrations then follows from CITE.
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math/0004051
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We first point out that the right adjoint MATH reflects weak equivalences between fibrant objects. Indeed, suppose MATH and MATH are MATH-spectra, and MATH is a map such that MATH is a weak equivalence. Then, because MATH and MATH are MATH-spectra, MATH is a weak equivalence for all MATH. Since MATH is a NAME equivalence, MATH reflects weak equivalences between fibrant objects by CITE. Thus MATH is a weak equivalence for all MATH, and so MATH is a level equivalence and hence a stable equivalence, as required. In view of CITE, to complete the proof it suffices to show that MATH is a weak equivalence for all cofibrant MATH, where MATH denotes a stably fibrant replacement functor in MATH. Let MATH denote a fibrant replacement functor in the strict model structure on MATH. Then MATH is certainly a weak equivalence. We claim that MATH is already a MATH-spectrum. Suppose for the moment that this is true; then by lifting we can construct a stable equivalence MATH, and since MATH is a MATH-spectrum, this map is in fact a level equivalence. Hence the map MATH is a weak equivalence, as required. It remains to prove that MATH is a MATH-spectrum. Since MATH is a NAME equivalence, the map MATH is a weak equivalence. By lifting, we can factor the weak equivalence MATH through the trivial cofibration MATH. This implies that the map MATH is a weak equivalence, and hence that MATH is a MATH-spectrum, as required.
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math/0004051
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Suppose MATH has right adjoint MATH, MATH has right adjoint MATH, and MATH has right adjoint MATH. The natural transformation MATH induces a dual natural transformation MATH. Define MATH by MATH, with structure maps adjoint to the composite MATH where MATH is adjoint to the structure map of MATH. The functor MATH is analogous to restriction in the theory of group representations, and we must now define the analog to induction MATH. To do so, first note that MATH defines natural transformations MATH for all MATH, by iteration. Define MATH to be the coequalizer of the two maps MATH where the top map is induced by MATH and the bottom map is induced by MATH. To define the structure map of MATH, note that the coequalizer diagram for MATH is just the subdiagram of the coequalizer diagram for MATH consisting of all terms with a positive power of MATH. The inclusion of diagrams induces the desired structure map MATH. We leave to the reader the exercise in adjointness required to prove that MATH is left adjoint to MATH. The functor MATH clearly preserves level fibrations and level trivial fibrations, so MATH is a left NAME functor with respect to the projective model structures. Also, since MATH, we have MATH. To show that MATH is a left NAME functor with respect to the stable model structures, we must show that MATH is a stable equivalence for all cofibrant MATH, by REF . Using the fact that MATH and the fact that MATH is a weak equivalence, we reduce to showing that MATH is a stable equivalence in MATH. This follows from REF , so MATH is a left NAME functor with respect to the stable model structures. the map MATH is a stable equivalence. Thus MATH is a left NAME functor with respect to the stable model structures. We define MATH by defining its adjoint MATH. Indeed, MATH is just the prolongation of the adjoint MATH of MATH. Since MATH is a weak equivalence on all cofibrant objects of MATH, MATH is a weak equivalence on all fibrant objects of MATH. To see this, note that MATH induces a natural isomorphism in the homotopy category. NAME implies that MATH also induces a natural isomorphism in the homotopy category, and it follows that MATH is a weak equivalence on all fibrant objects of MATH. Thus MATH will be a level equivalence on all level fibrant objects of MATH, so MATH is a level equivalence on all cofibrant objects of MATH. We leave it to the reader to check compatibility of MATH with compositions and identities.
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math/0004051
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By REF there is a map of pairs MATH induced by MATH. By REF , MATH is a NAME equivalence. Define MATH to be the composite MATH.
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math/0004051
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We will first show that MATH is a NAME equivalence on the projective model structures. Use the same notation as in the proof of REF , so that MATH denotes the right adjoint of MATH. Then, since MATH is a NAME equivalence, MATH reflects weak equivalences between fibrant objects, by CITE. It follows that MATH reflects level equivalences between level fibrant objects. Hence to show that MATH is a NAME equivalence of the projective model structures, it suffices to show that MATH is a level equivalence for all cofibrant MATH, where MATH is a fibrant replacement functor in the projective model structure on MATH. Thus, we need only show that MATH is a weak equivalence for all MATH and all cofibrant MATH, where now MATH is a fibrant replacement functor in MATH. Since MATH is cofibrant and MATH is a NAME equivalence, it suffices to show that MATH is a weak equivalence for all MATH and all cofibrant MATH. In fact, we can assume that MATH is a MATH-cell complex. Write MATH as the colimit of a MATH-sequence MATH where each map MATH is a pushout of a map of MATH. We will prove that, for all MATH, MATH is a weak equivalence for all MATH, by transfinite induction on MATH. Getting started is easy. The limit ordinal part of the induction follows from CITE, since each of the maps MATH and each of the maps MATH is a cofibration of cofibrant objects. For the succesor ordinal part of the induction, suppose MATH is a pushout of the map MATH of MATH. Then we have a pushout diagram MATH and another pushout diagram MATH in MATH. Note that MATH, where we interpret MATH to be the initial object if MATH. Similarly, MATH. Thus the natural transformation MATH induces a map from the first of these pushout squares to the second. If MATH (and hence also MATH) is cofibrant, then this map of pushout squares is a weak equivalence at both the upper left and upper right corners. Or, if MATH is a weak equivalence for all MATH, then again this map is a weak equivalence at both the upper left and upper right corners. It is also a weak equivalence at the lower left corner, by the induction hypothesis. Since the top horizontal map is a cofibration in MATH, NAME 's cube lemma CITE implies that the map is a weak equivalence on the lower right corner. This completes the induction. We have now proved that MATH is a NAME equivalence with respect to the projective model structures. In view of REF , to show that MATH is a NAME equivalence with respect to the stable model structures, we need to show that if MATH is level fibrant in MATH and MATH is a MATH-spectrum, then MATH is a MATH-spectrum. To see this, note that, since MATH is a MATH-spectrum, the natural map MATH is a weak equivalence for all MATH. There is a natural transformation MATH dual to MATH. Furthermore, MATH is a weak equivalence for all fibrant MATH, as we have seen in the proof of REF . Thus, the natural map MATH is a weak equivalence for all MATH. Since MATH reflects weak equivalences between fibrant objects, it follows that MATH is a MATH-spectrum, as required.
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math/0004051
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We define the action of MATH on MATH levelwise. That is, given MATH and MATH, we define MATH. The structure map is given by MATH . One can easily verify that this makes MATH tensored over MATH. Similarly, define MATH, with structure maps MATH adjoint to the composite MATH where MATH is the evaluation map, adjoint to the identity of MATH. This makes MATH cotensored over MATH. Finally, given MATH and MATH in MATH, define MATH to be the equalizer of the two maps MATH where MATH is the product of the maps MATH induced by the adjoint of the structure map of MATH, and MATH is the product of the maps MATH . Here the first map exists since MATH preserves the MATH action, and the second map is induced by the structure map of MATH. This functor makes MATH enriched over MATH. We must now check that these structures are compatible with the model structure. We begin with the projective model structure on MATH. One can easily check that MATH. Thus, if MATH is one of the generating cofibrations of the projective model structure on MATH, and MATH is a cofibration in MATH, then MATH is a cofibration in MATH. It follows that MATH is a cofibration for MATH an arbitrary cofibration of MATH (see CITE and CITE). A similar argument shows that MATH is a projective trivial cofibration in MATH if either MATH is a projective cofibration in MATH or MATH is a trivial cofibration in MATH. Finally, if MATH is a cofibrant approximation to the unit MATH in MATH, and MATH is cofibrant in MATH, then each MATH is cofibrant in MATH, so the map MATH is a level equivalence as required. Thus MATH with its projective model structure is a MATH-model category. To show that MATH with its stable model structure is also a MATH-model category, we need to show that, if MATH is a stable trivial cofibration and MATH is a cofibration in MATH, then MATH is a stable equivalence. It suffices to check this for MATH one of the generating trivial cofibrations of MATH. In this case, by hypothesis, MATH and MATH are cofibrant in MATH. Thus the functor MATH is a NAME functor with respect to the projective model structure on MATH, and similarly for MATH. Furthermore, if MATH is an element of the set MATH, then MATH, since MATH preserves the MATH-action. In view of REF , the map MATH is a stable equivalence. REF then implies that MATH is a NAME functor with respect to the stable model structure on MATH, and similarly for MATH. Thus, if MATH is a stable trivial cofibration, so are MATH and MATH. It follows from the two out of three property that MATH is a stable equivalence, as required.
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math/0004051
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Given a left proper cellular MATH-model category MATH, we have seen in REF that MATH is a MATH-model category. Just as in REF , a MATH-Quillen functor MATH induces a functor MATH, as does its right adjoint MATH. Since MATH is defined levelwise, it preserves the action of MATH. It is easy to check that MATH preserves level fibrations and level trivial fibrations, so that MATH is a MATH-Quillen functor with respect to the projective model structures. Furthermore, we have MATH, so REF imply that MATH is a MATH-Quillen functor with respect to the stable model structures as well.
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math/0004051
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We first show that the pushout product MATH is a (trivial) cofibration when MATH is a cofibration in MATH, and MATH is a cofibration in MATH (and one of them is a level equivalence). As explained in CITE. we may as well assume that MATH and MATH belong to the sets of generating cofibrations or generating trivial cofibrations. In either case, we have MATH and MATH. But then MATH. Since MATH is a NAME functor, the result follows. Now let MATH denote a cofibrant replacement for the unit MATH in MATH. Then MATH is a cofibrant replacement for MATH in MATH. Indeed, MATH is cofibrant, and MATH is just MATH. Since MATH is cofibrant and MATH is a monoidal model category, the map MATH is a level equivalence. Now, if MATH is cofibrant in MATH, then each MATH is cofibrant. Hence the map MATH is a weak equivalence for all MATH, and so the map MATH is a level equivalence, as required.
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math/0004051
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We only prove the cofibration case, as the trivial cofibration case is analogous. If each map MATH is a cofibration, then we can show that MATH is a projective cofibration by showing MATH has the left lifting property with respect to level trivial fibrations. Indeed, we construct a lift by induction, just as in the proof of REF . To prove the converse, it suffices to show that MATH is a MATH-cofibration for MATH, since MATH is itself a left NAME functor. Then we can write MATH, and we find that MATH is an isomorphism when MATH, and is the map MATH when MATH. This is a MATH-cofibration, as required.
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math/0004051
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We prove this theorem in the same way as REF . Since the cofibrations in the stable model structure are the same as the cofibrations in the projective model structure, the only thing to check is that MATH is a stable equivalence when MATH and MATH are cofibrations and one of them is a stable equivalence. We may as well assume that MATH is a generating cofibration in MATH and MATH is a stable trivial cofibration in MATH; the argument for MATH a stable trivial cofibration and MATH a generating cofibration in MATH is the same. Then, by hypothesis, MATH and MATH are cofibrant in MATH. We claim that MATH is a NAME functor MATH to MATH with their stable model structures, and similarly for MATH. Indeed, in view of REF , it suffices to show that MATH is a stable equivalence for all MATH and all domains or codomains MATH of the generating cofibrations of MATH. But one can easily check that MATH. Then REF implies that this map is a stable equivalence, as required. Thus, both functors MATH and MATH are NAME functors in the stable model structures. A two out of three argument, as in REF , then shows that MATH is a stable equivalence, as required.
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math/0004051
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Note that MATH is certainly enriched, tensored, and cotensored over MATH. Now use the equivalence of categories MATH to transport this structure back to MATH.
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math/0004051
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The functor MATH induces a MATH-functor MATH, which takes the symmetric sequence MATH to the symmetric sequence MATH. It follows that MATH induces a MATH-functor MATH, that takes the symmetric spectrum MATH to the symmetric spectrum MATH, with structure maps MATH . Let MATH denote the right adjoint of MATH. Then the right adjoint of MATH is MATH, which takes the symmetric spectrum MATH to the symmetric spectrum MATH, with structure maps adjoint to the composite MATH . Since MATH, it follows that MATH. It is clear that MATH preserves level fibrations and level equivalences, so MATH is a NAME functor with respect to the projective model structure. In view of REF , to see that MATH defines a NAME functor with respect to the stable model structures, it suffices to show that MATH is a stable equivalence for all domains and codomains of MATH of the generating cofibrations of MATH. But one can readily verify that MATH, which is a stable equivalence as required. Thus MATH is a NAME functor with respect to the stable model structures. If MATH is a NAME equivalence, one can easily check that MATH is a NAME equivalence with respect to the projective model structure. To see that it is still a NAME equivalence with respect to the stable model structures, we need only show that MATH reflects stably fibrant objects, in view of REF . But, if MATH is level fibrant and MATH is a MATH-spectrum, this means that the map MATH is a weak equivalence for all MATH. Since MATH reflects weak equivalences between fibrant objects, this means that MATH is a MATH-spectrum, as required.
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math/0004051
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The map MATH induces a map of commutative monoids MATH. This induces the usual induction and restriction adjunction MATH . That is, if MATH is in MATH, then MATH. Restriction obviously preserves level fibrations and level equivalences, so is a NAME functor with respect to the projective model structure. One can easily check that MATH. It follows that MATH is the map MATH in MATH. The weak equivalence MATH induces a level equivalence MATH. Since the map MATH is a stable equivalence, so is the given map. Thus induction is a NAME functor with respect to the stable model structures. We now prove that induction is a NAME equivalence between the projective model structures. The proof of this is similar to the proof of REF . That is, restriction certainly reflects level equivalences between level fibrant objects. It therefore suffices to show that the map MATH is a level equivalence for all cofibrant MATH. The argument of REF will prove this without difficulty. To show that induction is a NAME equivalence between the stable model structures, we need only check that restriction reflects stably fibrant objects. This follows from the fact that the map MATH is a weak equivalence for all fibrant MATH.
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math/0004051
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We take MATH, where the functor MATH used to form MATH is defined by MATH and is part of the MATH-model structure of MATH (see the comment following REF ). This means that the structure map of MATH involves the twist map. By REF , MATH is a MATH-Quillen equivalence. On the other hand, consider MATH, where now we use the action of MATH on MATH that comes from REF . As pointed out in REF , this means that we are using the functor MATH to form MATH, not the functor MATH. By hypothesis, this functor is already a NAME equivalence, so REF implies that MATH is a MATH-Quillen equivalence. It remains to prove that MATH and MATH are isomorphic as model categories. This is mostly a matter of unwinding definitions. An object of MATH is a set MATH of objects of MATH, where MATH. There is an action of MATH on MATH, and there are MATH-equivariant maps MATH and MATH. In addition, the composite MATH is MATH-equivariant, and there is a compatibility between MATH and MATH, expressed in the commutativity of the following diagram. MATH . An object MATH of the category MATH has the same description if we switch MATH and MATH. There is then an isomorphism of categories between MATH and MATH which simply switches MATH and MATH. The model structures are also the same. Indeed, they are both the localization of the evident bigraded projective model structure with respect to the maps MATH and MATH, where MATH is left adjoint to the evaluation functor MATH. The natural isomorphism MATH follows from the fact that the composites MATH and MATH are equal.
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math/0004051
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We let MATH be the mapping cylinder of MATH. That is, we take MATH to be the pushout in the diagram below. MATH . The map MATH is then the composite MATH, and the map MATH is the map that is MATH on MATH and MATH on MATH. It follows that MATH, as required. The map MATH is defined to be the identity on MATH and the composite MATH on MATH. Since MATH is a trivial cofibration (as a pushout of MATH), it follows that MATH is a weak equivalence, and it is clear that MATH. We must now construct the homotopy MATH. First note that MATH is cofibrant, since MATH is so and MATH is a trivial cofibration, and so MATH is a cylinder object for MATH. In fact, MATH is the pushout of MATH and MATH over MATH. Define MATH to be the constant homotopy MATH on MATH and the homotopy MATH on MATH. The fact that MATH guarantees that MATH is well-defined, and the other conditions on MATH guarantee that MATH is a left homotopy from MATH to MATH. We leave the naturality of this construction to the reader.
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math/0004051
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The reader is well-advised to draw a picture in the topological or simplicial case, from which the proof should be clear. We think of MATH as the interval whose left half is MATH and whose right half is MATH. In particular, MATH is a cylinder object for MATH, where MATH and MATH. Then, because the tensor product preserves pushouts, we can think of MATH as a square consisting of a copy of MATH in the lower left, a copy of MATH in the upper left, a copy of MATH is the lower right, and a copy of MATH in the upper right. We define the necessary map MATH by defining MATH on each subsquare. On the lower left, we use the composite MATH, where MATH is the homotopy making MATH into a unit interval. Similarly, on the upper right square, we use the composite MATH. On the upper left square we use the constant homotopy MATH, and on the lower right square we use the constant homotopy MATH. We leave it to the reader to check that this makes MATH into a unit interval.
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math/0004051
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We define MATH, MATH, MATH and a homotopy MATH from MATH to MATH, where MATH is a unit interval, inductively on MATH, using REF . To get started, we take MATH, MATH to be the identity, MATH to be MATH, and MATH to be the constant homotopy (with MATH). For the inductive step, we apply REF to the diagram MATH and the homotopy obtained as follows. We have a homotopy MATH from MATH to MATH. On the other hand, we also have the homotopy MATH from MATH to MATH. We can amalgamate these to get a homotopy MATH from MATH to MATH, and MATH is still a unit interval by REF . Hence REF gives us an object MATH, a map MATH, and a map MATH such that MATH. It also gives us a map MATH such that MATH and a homotopy MATH from MATH to MATH. This completes the induction step and the proof (we leave naturality to the reader).
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math/0004051
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We first reduce to the case where MATH is itself symmetric. So suppose the generating cofibrations of MATH have cofibrant domains, and suppose MATH is symmetric and weakly equivalent to MATH; this means there are weak equivalences MATH, where MATH denotes a fibrant replacement functor. This means the total left derived functors MATH and MATH are naturally isomorphic on the homotopy category of any MATH-model category. In particular, it suffices to show that MATH is a NAME equivalence on MATH. On the other hand, by REF , there are MATH-Quillen equivalences MATH . It therefore suffices to show that MATH is a NAME equivalence of MATH; that is, we can assume that MATH itself is symmetric. Let MATH denote the given homotopy from the cyclic permutation to the identity of MATH. Let MATH be a cofibrant spectrum, let MATH denote the structure map of MATH, and let MATH denote the structure map of MATH. These two structure maps differ by the cyclic permutation, and therefore we are in the situation of REF , with MATH, MATH, MATH equal to the identity map, and MATH. It follows that we get a functor MATH defined on cofibrant objects of MATH and natural level equivalences MATH and MATH, where the latter map is a level equivalence since MATH is homotopic to MATH. Thus the total left derived functors of MATH and MATH are naturally isomorphic. Since we know already that MATH is a NAME equivalence, so is MATH, and hence so is MATH.
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math/0004051
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For MATH, this follows immediately from the definition of MATH, REF , and REF . The proof for MATH is similar.
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math/0004051
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This is immediate, since limits in MATH and MATH are taken levelwise.
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math/0004051
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We will prove the proposition only for MATH, as the MATH case is similar. Throughout this proof we will use REF , which guarantees that subcomplexes in MATH are determined by their cells. Choose an infinite cardinal MATH such that the domains and codomains of MATH are all MATH-compact relative to MATH. When dealing with relative MATH-cell complexes, we can assume that we have a presentation as a transfinite composition of pushouts of maps of MATH, rather than as a transfinite composition of pushouts of coproducts of maps of MATH, using CITE or CITE. A similar comment holds for relative MATH-cell complexes. We will proceed by transfinite induction on MATH, where the induction hypothesis is that for every presented relative MATH-cell complex MATH whose presentation ordinal is MATH, and for every map MATH where MATH is an integer and MATH is a domain or codomain of MATH, MATH factors through a subcomplex with at most MATH-cells. Getting the induction started is easy. For the induction step, suppose the induction hypothesis holds for all ordinals MATH, and suppose we have a presentation MATH of MATH as a transfinite composition of pushouts of maps of MATH. Then the boundary of each MATH-cell of this presentation is represented by a map MATH, for some MATH, some MATH, and some domain MATH of a map of MATH. This map factors through a subcomplex with at most MATH-cells, by induction. It follows that the MATH-cell itself is contained in a subcomplex of at most MATH-cells, since we can just attach the interior of the MATH-cell to the given subcomplex. Now suppose we have an arbitrary map MATH, where MATH is a domain or codomain of a map of MATH. Such a map is determined by a map MATH in MATH. The map MATH is the transfinite composition of the cofibrations MATH. For each MATH, there is a MATH and a map MATH of MATH such that MATH is the pushout of MATH, where we interpret MATH as the identity map if MATH is negative. Apply CITE to write the MATH-sequence MATH as a retract of a MATH-sequence MATH where each map MATH is a relative MATH-cell complex. We denote the retraction by MATH, noting that the restriction of MATH to MATH factors (uniquely) through MATH. We can think of the entire map MATH as a relative MATH-cell complex, each cell MATH of which appears in the relative MATH-cell complex MATH for some unique ordinal MATH, and so has associated to it the MATH-cell MATH of MATH used to form MATH. The composite MATH then factors through a subcomplex MATH with at most MATH-cells. The proof will be completed if we can find a subcomplex MATH of MATH with at most MATH-cells such that the restriction of MATH to MATH factors through MATH. Take MATH to be a subcomplex of MATH containing the MATH-cells MATH as MATH runs through the cells of MATH. Then MATH contains MATH for every cell of MATH, so MATH contains MATH, as required. Furthermore, since each MATH-cell MATH lies in a subcomplex with no more that MATH-cells, and MATH has no more than MATH cells, there is a choice for MATH which has no more than MATH cells. This completes the induction step and the proof.
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math/0004052
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If the action is MATH-filling, let MATH be nonempty open subsets of MATH. There exist elements MATH, with MATH, such that MATH, MATH. By hypothesis there exist MATH such that MATH. Then if MATH there exists MATH such that MATH. Therefore MATH, that is, MATH. Thus MATH. Conversely, suppose the stated assertion holds. Fix MATH, with MATH, and let MATH. For each MATH, the set MATH is a nonempty and open. Choose MATH such that MATH. If MATH, then MATH for some MATH and so MATH. Therefore MATH.
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math/0004052
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(Inspired by CITE.) Denote by MATH the canonical conditional expectation. Fix MATH. In order to prove the result it is enough to show that there exist MATH such that MATH. Put MATH. Let MATH. There exists MATH such that MATH. Write MATH, where MATH and MATH is finite. Note that MATH, and so MATH. It follows that MATH . Choosing MATH so that MATH then replacing MATH by MATH shows that we can assume that MATH. Since MATH is properly outer for each MATH, it follows from CITE that there exists MATH, MATH such that MATH and MATH for all MATH. Using REF below, we see that there exists MATH such that MATH and MATH. Then MATH . Therefore MATH is invertible since MATH and MATH . Setting MATH we have MATH .
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math/0004052
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There are two cases to consider. CASE: Suppose that MATH is not an isolated point of MATH. Then there exist pairwise disjoint nonempty open sets MATH contained in MATH. Let MATH be the MATH-subalgebra of MATH generated by MATH. By functional calculus, there exist MATH, MATH with MATH and MATH. CASE: Suppose that MATH is an isolated point of MATH. Then there exists a nonzero projection MATH such that MATH. By hypothesis the hereditary MATH-subalgebra MATH is infinite dimensional. Therefore every masa of MATH is infinite dimensional CITE. Inside such an infinite dimensional masa of MATH we can find positive elements MATH, MATH with MATH. Then MATH and MATH for MATH.
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math/0004052
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By REF , there exist MATH, with MATH, MATH, MATH for MATH, and MATH. Since the action is MATH-filling, there exist MATH such that MATH. Therefore MATH . Put MATH. Now MATH and so MATH. Finally, we have MATH .
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math/0004052
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Denote by MATH the canonical map from MATH onto MATH. We first show that the action of MATH on MATH is not MATH-filling. Choose a linear subspace MATH of MATH of dimension MATH. Let MATH, which is a nonempty open subset of MATH. If MATH REF then MATH. For the subspace MATH of MATH has dimension at least one, and so contains a nonzero vector MATH. Then MATH. Thus the action-MATH is not MATH-filling. It remains to show that it is MATH-filling. For this we use ideas from CITE. We claim that there exists a basis MATH for MATH, elements MATH, and (compact) sets MATH with MATH, and with the following property: for any open neighbourhood MATH of MATH REF there exists a positive integer MATH such that MATH for all MATH. It follows that the action is MATH-filling. For let MATH be nonempty open subsets of MATH. Since the action of MATH on MATH is minimal, we may assume that MATH (MATH). Let MATH, so that MATH (MATH). Then MATH. It remains to verify our claim. Fix a positive integer MATH and let MATH, MATH. Consider the matrices MATH and MATH in MATH. These matrices have eigenvalues MATH, MATH, which satisfy MATH. The corresponding eigenvectors for MATH are MATH and MATH; for MATH they are MATH and MATH. If MATH let MATH where A occupies the MATH and MATH rows and columns and the nonzero entries of the vectors are in rows MATH and MATH. Also let MATH . Let MATH. For MATH let MATH . Direct computation shows if MATH then MATH, where MATH. Therefore MATH. Let MATH and consider the basic open neighborhood MATH of MATH defined by MATH . Let MATH. Recall that MATH. Therefore MATH. For MATH and MATH, we have MATH . Now MATH, and for MATH, MATH. This means that MATH for all MATH.
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math/0004052
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Choose MATH with MATH and an open neighbourhood MATH of MATH such that MATH. Since the action is minimal, the family MATH forms an open covering of MATH. By compactness, there exists a finite subcovering MATH. Let MATH be nonempty open subsets of MATH. Since the action of MATH on MATH is minimal, we may choose elements MATH such that MATH (MATH). For MATH, choose an integer MATH such that MATH. Then MATH, where MATH. Therefore MATH.
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math/0004052
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This is an easy consequence of the definitions.
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math/0004053
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Apply NAME 's three term recurrence relation CITE with its parameters specialized as follows: MATH . This gives an expansion of the form REF for explicit coefficients MATH and MATH, which at a first glance still depend on MATH. Using the functional equation MATH for theta functions, it is easily seen that the coefficients are independent of MATH and that they coincide with the dual MATH-functions MATH as defined above.
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math/0004053
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Direct verification.
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math/0004053
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We substitute the definition of the dual NAME function transform MATH into the left hand side of the desired identity, and we use that MATH for MATH and MATH, see REF for the second equality. We arrive at MATH . Since MATH, all integrations are over compact subsets, so we may use NAME 's theorem to interchange the order of integration. This yields the desired result.
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math/0004053
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The proof of the desired identity simplifies when we slightly perturb the parameters MATH. The proof for parameters MATH can then be derived from the perturbed case using the fact that the left and right hand side of the identity depend continuously on the parameters MATH. Let us indicate one class of possible perturbations of the parameters which is sufficient for our purposes. Let MATH with MATH. Let MATH be sufficiently small, then we set MATH while we keep MATH undisturbed. Observe that MATH, MATH etc., and that MATH. Note that the weight function MATH (see REF) corresponding to the parameters MATH only has simple poles. We can now define the (in general complex) measure MATH with respect to the parameters MATH by the same REF as before. We also keep the same notations for the other parameter-dependent objects we have encountered (such as for example, MATH, MATH etc.). Then for spectral parameters MATH with MATH REF we can write for MATH, MATH where MATH is a continuous, rectifiable NAME curve in the complex plane, satisfying the following additional conditions: CASE: MATH has a parametrization of the form MATH for MATH with positive radial function MATH (we orientate MATH according to this parametrization); CASE: MATH is invariant under inversion, that is, MATH, where MATH; CASE: The sequences MATH are in the interior of MATH; CASE: The intersection of the MATH-interval MATH with the interior of MATH is given by the sequence MATH. The existence of MATH is easy to prove with our choice of perturbed parameters, and REF is then a direct consequence of NAME 's theorem. Now recall that the NAME function MATH is an eigenfunction of the second-order MATH-difference operator MATH with eigenvalue MATH, see REF. This implies that MATH . In view of the explicit expression for MATH (see REF), we can write MATH with MATH . Furthermore, by the explicit expression for the weight function MATH (see REF) and for the function MATH (see REF), we have the identity MATH . Substitution of REF into the right hand side of REF then shows that MATH . Using the explicit form of the contour MATH, we can shrink MATH back to MATH in the right hand side of REF while picking up residues of the integrand at the simple poles MATH and MATH. By REF and by the identity MATH, we can pull the two residues together, yielding MATH as desired.
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math/0004053
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We first observe that the asymptotic behaviour of the discrete weights MATH REF is given by MATH where MATH is the positive constant MATH . This is a direct consequence of the explicit REF for MATH (MATH). For the moment we assume MATH and MATH in order to be able to apply the MATH-function expansion for MATH (MATH). We fix MATH, so that MATH for MATH sufficiently negative by REF . Observe that the apparent poles of MATH as function of the parameters MATH are removable. Hence we can extend the definition of the MATH-function MATH to parameter values MATH by continuity and we obtain MATH for all MATH if MATH is sufficiently negative. Since MATH and MATH for MATH, it follows from REF that MATH for all MATH. Fix now MATH with MATH. Then MATH by REF, hence by REF and by the previous paragraph, MATH . Here we have used the fact that MATH is real valued on MATH in order to get rid of the complex conjugate in the integrand. It remains to prove that the limit in the right hand side of REF tends to zero. Let MATH and MATH be two functions in the NAME space MATH. By the asymptotics REF, we have MATH as MATH, and similarly for MATH. Furthermore, we have the asymptotics MATH for the coefficient MATH of the NAME second order MATH-difference operator MATH (see REF). Combined with REF, it follows from REF of the Wronskian that MATH . In particular, the limit in the right hand side of REF tends to zero, hence MATH is orthogonal to MATH in MATH.
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math/0004053
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As in the proof of REF , we assume for the moment that MATH and that MATH. In view of REF, these generic conditions can be removed at the end of the proof by applying the dominated convergence theorem. We fix MATH, then by REF , MATH where we used that MATH is real valued on MATH for the first equality. It remains to evaluate the limits of the Wronskian in the last equality of REF. We use the MATH-function expansion for the NAME functions (see REF ) to rewrite the Wronskian as MATH for MATH. Now there exists open neighbourhoods MATH of MATH in the complex plane such that MATH where MATH admits a convergent power-series expansion around MATH with coefficients depending analytically on MATH and with constant coefficient equal to zero. Furthermore, for sufficiently small neighbourhoods MATH, differentiation with respect to MATH may be interchanged with summation in the power series expansion of MATH around MATH when MATH for some MATH independent positive constant MATH. Combined with REF, we see that MATH as MATH, with MATH uniform in MATH. Now we substitute REF, and we use the fact that MATH has a simple zero at MATH and that MATH then using REF and the fact that MATH, we derive that MATH with MATH and with a remainder term MATH satisfying MATH . Using REF and the fact that MATH has a simple zero at MATH, we derive MATH where MATH and MATH are the constants MATH (see REF) and MATH (see REF) respectively with respect to dual parameters. Here we have used REF for the second equality. We can evaluate now the limits in the last equality of REF by substituting REF for the Wronskian and by using REF together with the fact that MATH. This gives MATH . By a direct computation one verifies that MATH, which completes the proof of the lemma.
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math/0004053
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The proof is similar to the proof of CITE and of CITE, where the analogous statement was derived for the big and the little MATH-Jacobi function transforms, respectively. Since some care has to be taken in order to match the constants, we repeat here the proof in some detail. We prove the proposition with respect to dual parameters (compare REF ). It suffices to prove it for functions MATH which are continuous on MATH and supported within MATH. We use REF and the explicit form of the measure MATH (see REF), together with the invariance of the NAME function MATH and the measure MATH under MATH, to write MATH for MATH sufficiently large. It suffices to prove that the right hand side of REF tends to MATH as MATH. In order to compute the limit MATH of the right hand side of REF, we first observe that the factor MATH occuring in the integrand can be rewritten as MATH compare REF. Similarly as in the proof of REF , we derive now from the first equality in REF, the MATH-function expansion (see REF ) and the mean value theorem, that MATH with MATH and with a remainder term MATH satisfying MATH for all MATH. We substitute REF for the Wronskian in the right hand side of REF. In view of the asymptotics REF, it then suffices to calculate the limit MATH of MATH . The asymptotic behaviour REF shows that the limit MATH of the integral REF of the remainder term MATH gives zero by dominated convergence. By the NAME lemma, the integral REF of the two terms from MATH corresponding to MATH in REF tend to zero as MATH. It thus remains to calculate the limit MATH of REF in which the factor MATH of the integrand is replaced by MATH . It follows now by yet another application of the NAME lemma that it remains to calculate the limit MATH of MATH . Now by the well known MATH-properties of the NAME kernel, the limit MATH of REF exists, and it equals MATH . The proposition follows now directly from the fact that MATH (compare with the proof of REF ).
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math/0004053
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We establish the desired identity with respect to dual parameters. Let MATH and MATH. By REF we have MATH and MATH respectively. In particular, we may assume without loss of generality that MATH is continuous on MATH, and supported within MATH. For MATH we now define MATH by MATH. Observe that MATH is continuous on MATH and supported within MATH. By REF we can now write MATH . Now we observe that REF and the asymptotic behaviour MATH imply that MATH as MATH, where the second equality is a consequence of REF. But MATH for MATH, hence we may use NAME 's dominated convergence theorem to interchange limit and integration in the right hand side of REF. It follows that the right hand side of REF tends to zero, which completes the proof of the lemma.
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math/0004054
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We use the explicit expression of MATH to perform an estimate of MATH: MATH . We introduce the notation MATH and the change of variable MATH . The integral MATH is now given by MATH . We use the obvious inequalities MATH to infer that MATH . It is now clear that MATH is bounded independently of MATH, that is, of MATH by a certain number MATH.
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math/0004054
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By an integration by parts, MATH . We estimate MATH: we first observe that MATH . We estimate MATH from below by MATH; we also observe that MATH is bounded, thanks to estimates REF; therefore we have the following estimate, where we have used again REF : MATH . Next step is to calculate MATH: we use REF and we find that MATH where the MATH's are respectively given by MATH . Our aim is now to estimate the expressions MATH . The first expression MATH is rewritten with the help of the change of variable REF and becomes MATH which we estimate as follows: MATH . Since MATH for MATH large enough, we can see that MATH . Therefore MATH . With the change of variable REF, we may write the expression MATH as MATH . We use REF again together with MATH and we find MATH . But for MATH large enough and MATH, MATH indeed, if MATH, MATH, and the inequality is clear; on the other hand, if MATH, for MATH large enough MATH, and the inequality also follows. Therefore, there exists a number MATH such that for all large enough MATH and all MATH in MATH the following inequality holds: MATH . The third expression is handled as follows: MATH . If MATH, we use the inequality MATH and we obtain MATH since MATH for MATH large enough. On the other hand, for MATH and for MATH large enough MATH and therefore, using the inequality MATH, we obtain MATH . Thus, we have shown that MATH . There remains to estimate MATH . We rewrite REF as MATH and we find that thanks to REF MATH . Summarizing REF with REF, we find estimate REF.
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math/0004054
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The easiest part is the estimate on MATH . We can see that the absolute value of this expression is estimated by MATH . We recognize expressions which have already been estimated in REF. Therefore, it is immediate that MATH . Next comes the slightly more complicated expression MATH . For all MATH, MATH is at most equal to MATH; therefore, if MATH, then the absolute value of REF is estimated by MATH . Therefore, we can see that MATH . The last and most complicated term contains the expression MATH which can be rewritten thanks to NAME 's formula with integral remainder as MATH . The absolute value of this expression is estimated by MATH . Therefore, we find that MATH .
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math/0004054
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The ball of radius MATH about MATH will be invariant by the mapping MATH provided that MATH . Choose MATH small enough for the following inequality to hold MATH . Choose then MATH so large that REF holds. Let us prove now that for an adequate choice of MATH and MATH and for all MATH, the mapping MATH is a strict contraction: the easiest part of the estimate pertains to MATH and it is clear from the proof of estimate REF that MATH . The second easiest term involves the difference MATH which we estimate thanks to NAME 's formula: MATH . Therefore, the corresponding term in MATH contributes an estimate MATH . The last and most complicated term involves the expression MATH . We use a NAME expansion twice to rewrite this expression as MATH . Therefore, the corresponding term is estimated by MATH . Thus, the norm of the corresponding contribution in MATH is estimated by MATH . Therefore, if we summarize the estimates REF, we find that on a ball of radius MATH about MATH, the NAME constant of MATH is estimated by MATH . If we choose MATH small enough for MATH to be less than or equal to MATH, it is clear that we can choose MATH large enough for the sum of the remaining terms in REF to be less than or equal to MATH. Together with the conditions found above for the invariance of the ball of radius MATH about MATH, we have shown the first part of REF . We also infer from this proof that MATH . Its last assertion is an immediate consequence of the equivalence of REF with REF, the definition of the norm, and the fact that the initial data coincide.
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math/0004054
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The first statement is an almost immediate consequence of REF: we have MATH so that on MATH and REF follows. In order to compare MATH and MATH, we write the differential equations that they satisfy: MATH . Therefore, if we subtract the second of these equations from the first, we deduce that MATH . The initial data vanish. We integrate because we want to estimate MATH: MATH . The next step is to estimate the integral on the right hand side of REF. We decompose this integral into three terms: MATH . The first two integrals are very easy to estimate: for MATH an integration by parts gives MATH . Thanks to REF, on MATH, MATH. Therefore MATH . For MATH, the situation is even simpler since it can be readily seen that MATH where MATH. There remains to estimate MATH; a straightforward calculation shows that MATH here MATH is estimated at REF, so that MATH . Therefore, MATH where MATH is defined as MATH . But MATH can be more conveniently rewritten as MATH which we integrate by parts. We find that MATH . We can see now that MATH . Therefore MATH . Since MATH is nonnegative, for MATH, MATH can be estimated from below by MATH which is equal to MATH we infer from relations REF that MATH which concludes the proof.
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math/0004054
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The differential equation satisfied by MATH defined by REF is deduced from REF and is given by MATH . Recall that the principal part MATH is defined by REF. Let us estimate MATH: MATH satisfies the differential equation MATH . Therefore, if we let MATH, and if we denote MATH we find that MATH . Hence, for all MATH, MATH . According to REF and the definition of MATH, there exists MATH such that for all MATH: MATH . Let MATH and MATH be defined by the relations MATH . Thanks to REF MATH and MATH are well defined, and are given by MATH . Therefore, as MATH tends to infinity, both MATH and MATH are equivalent to MATH. The function MATH is strictly increasing with respect to time; thanks to REF , there is a unique MATH such that MATH. We know an equivalent of MATH and MATH as MATH tends to infinity: MATH . Together with REF, the above relation implies MATH and from REF that MATH . REF implies MATH and MATH. We change coordinates now, taking the axis MATH along the second side of the convex cone MATH and the axis MATH perpendicular to MATH, and going out of MATH. The new time variable is a translation of the natural time, denoted MATH, and we set its origin at the time when the representative point enters region MATH. We also let MATH. With these conventions, MATH . We use now the equivalents obtained previously: MATH . The second component MATH of MATH satisfies the ordinary differential equation MATH so that MATH . The first component MATH of MATH satisfies the following ordinary differential equation MATH as long as MATH. The explicit solution of REF with initial data REF is given by MATH . Since MATH is non negative, MATH stays non negative for all MATH and we have the following estimate on the first component of MATH: MATH .
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math/0004054
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The initial part of the motion is described thanks to REF. Estimate REF proves that MATH tends to MATH uniformly on compact sets of MATH; relation REF enables us to conclude.
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math/0004054
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We will show in REF that the expression MATH tends to MATH as MATH tends to infinity, uniformly on MATH. If MATH, then MATH and in consequence, MATH . Let us estimate MATH when MATH is at most equal to MATH. We can see that MATH . Therefore, for MATH large enough, MATH is a strict contraction from the ball of radius MATH about MATH to itself.
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math/0004054
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The integral MATH is analogous to the one defined in REF. We define MATH by MATH and we consider three cases: CASE: MATH: in this case MATH cannot be neglected relatively to MATH. CASE: MATH: in this interval, the dominant term in MATH will be MATH and an elementary computation shows that this expression vanishes for MATH. Thus MATH crosses MATH approximately at a time MATH. CASE: MATH: the last leg of the journey, since MATH is dominant and in MATH, the term involving MATH is dominant. Before proving these estimates, we observe that there exist positive numbers MATH and MATH such that MATH . We tackle now the three separate sub-cases in detail. CASE: We remark that MATH . Therefore, we can estimate MATH as follows: MATH . We observe that over MATH, MATH and we use REF. These observations imply the following inequalities: MATH and thanks to REF of MATH and REF , we obtain MATH . CASE: We cut the integral MATH into two pieces: one piece from MATH to MATH on which we work essentially as in the previous sub-case, and a piece from MATH to MATH on which we work differently. More precisely, on MATH, we observe that MATH, and on MATH, MATH. Therefore, MATH . We estimate MATH from below by arguing that MATH increases from MATH to a maximum, and then decreases exponentially fast to MATH. Therefore, for all small enough MATH, there exists MATH tending to infinity such that MATH. Moreover on the interval MATH is greater than or equal to MATH. Thus, for all large enough MATH, MATH on the interval MATH, and therefore MATH . Thus, we obtain thanks to REF MATH . For the other piece, we estimate MATH from below by MATH, and we obtain MATH . We use REF in the integral term of the right hand side of REF , and we infer that MATH . The combination of REF yields MATH . CASE: We cut now MATH into three pieces, relative to the subintervals MATH, MATH and MATH. On the last two pieces, we observe that MATH is far from MATH, and we use the estimate from below MATH . Moreover there exists MATH such that for MATH and MATH large enough MATH . On the first subinterval, that is, MATH, we use REF ; relations REF imply that MATH thanks to REF we can see that MATH . Thanks to REF and the asymptotics REF, we obtain MATH . Since MATH and MATH, we get finally MATH . Relations REF imply that MATH . We observe that MATH and we use estimates REF; therefore MATH . Now, thanks to REF, we obtain MATH . Let us consider the third piece: we use now estimate REF on the denominator of integrand; since MATH, and thanks to REF, we have MATH and we conclude that the following estimate holds: MATH . We have to keep the two terms in the above expression, since we have no way to ascertain the order of the exponents of MATH. When we compare the exponents in REF, we find that the exponent of MATH in REF is the smaller; when we look at the exponents in REF to the exponent in REF we find that MATH is strictly larger than MATH, and this leads to the conclusion REF.
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math/0004054
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REF implies the uniform equivalence REF, and REF is an immediate consequence of REF. Let us prove an estimate of the derivative MATH at MATH: MATH . We observe that MATH . Therefore, MATH . There exists a constant MATH such that for all MATH . We use the method which gave estimate REF: we cut the integration interval into the three subintervals MATH, MATH and MATH, and on each of these subintervals, we estimate MATH from below exactly as in this calculation. Details are left to the reader, and we obtain MATH . The equivalent of MATH is obtained immediately from the explicit REF for MATH and the equivalents REF. Hence we infer that REF holds.
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math/0004054
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Denote MATH . With these notations, REF can be written MATH . We observe that in the domain MATH, REF has exactly one critical point at MATH . This critical point is attractive, as an examination of the linearization of REF around MATH shows. Moreover, there is a NAME functional given by MATH . Therefore, given MATH with MATH, we can see that for all MATH, MATH is at most equal to MATH, and in particular, MATH remains bounded. We see that when MATH tends to infinity, MATH tends to the critical point MATH. The spectrum of MATH is MATH; therefore, the matrix MATH is well defined, symmetric, positive and definite. In particular, if MATH is the smallest eigenvalue of MATH and MATH is the largest eigenvalue of MATH, MATH . If we let MATH, we observe that MATH . As MATH and MATH are arbitrary in the above calculation, we have proved indeed that for all MATH . Since MATH we have the inequality MATH . We seek a number MATH such that if MATH, then MATH . Indeed, in order to satisfy REF, it suffices to have MATH or equivalently, MATH . But MATH and MATH, so that, with the help of REF, it suffices to satisfy MATH that is, MATH . We shall show now that if we choose MATH such that MATH then there exists MATH such that MATH . Indeed, we know from REF that MATH, and that the limit of MATH as MATH tends to infinity is MATH; therefore, MATH must cross MATH. We denote by MATH the smallest time in MATH such that REF holds. On the interval MATH, the differential REF implies MATH whence MATH . But MATH, and we obtain the inequality MATH . In particular, there exists MATH such that MATH . We also need an estimate on MATH. We first show that it is less than or equal to MATH. By REF we know that MATH. Denote by MATH the connected component of MATH whose boundary contains MATH. If MATH, it is clear that MATH. Assume that MATH and that MATH; then MATH and MATH vanishes. We infer from differential REF that MATH but MATH, because MATH; therefore MATH . On the other hand, as MATH is negative on MATH and vanishes at MATH, a straightforward sign argument shows that MATH which contradicts REF. Now, we prove that MATH . This will be a consequence of the following inequality for all MATH and for all large enough MATH: MATH . We observe that MATH . When we integrate this differential relation, we find that MATH . For MATH large enough, the equivalences REF show that MATH is strictly positive, and REF follows immediately. We infer now from REF and the sign condition on MATH that MATH . Since the NAME functional decreases along trajectories of the system, we obtain for all MATH the inequalities MATH . This is the final estimate we needed before the conclusion.
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math/0004054
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We know from REF that MATH is an increasing function of MATH; if there is a time MATH for which MATH, the conclusion is clear. Assume otherwise; then, with the notations of REF, we can see that MATH and the conclusion is also clear.
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