paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0007127 | By symmetry, it suffices to show MATH. We may assume MATH is MATH-separable. By combining REF with REF , we may choose a power MATH of MATH, such that each element of MATH is a sum of MATH-separable elements of MATH of smaller degree. Thus, we may assume MATH is MATH-separable (and our bound MATH may depend on MATH). D... |
math/0007127 | Choose MATH as in REF . Now choose MATH. Because MATH we have MATH so, from the choice of MATH and MATH, we conclude that MATH. |
math/0007127 | Because MATH the left-hand inequality is obvious. Let MATH and let MATH be the codimension of MATH. Then REF implies that there exist nonzero MATH with MATH, such that MATH contains a codimension-MATH subspace of MATH for MATH. Then, letting MATH, we have MATH, and MATH contains a codimension-MATH subspace of the ideal... |
math/0007127 | REF shows that, by replacing MATH with some MATH (using REF ), we may assume MATH, for every MATH. The terms MATH and MATH in REF are significant only when MATH is small. On the other hand, MATH can never be small (and nonzero) if MATH for some large MATH. Thus, by replacing MATH with some MATH (using REF ), we may ass... |
math/0007127 | Choose MATH as in REF . By replacing MATH with MATH and replacing MATH with MATH, we may assume MATH. Then, by composing MATH and MATH with MATH, we may assume MATH and MATH. Thus, MATH . We wish to show that there is some MATH, such that, for every MATH, we have MATH. For each MATH, there is some MATH, such that MATH.... |
math/0007127 | Let MATH, MATH be finite-index subgroups of MATH, such that MATH is an isomorphism. Let MATH, MATH be the image of MATH in MATH under the projection MATH with kernel MATH. By passing to a finite-index subgroup, we can assume that MATH. Since MATH, we can identify MATH with MATH, so MATH induces an isomorphism MATH. We ... |
math/0007127 | From REF , we may assume there exist CASE: a standard automorphism MATH of MATH; and CASE: a homomorphism MATH, such that MATH for all MATH. By REF , there exists a finite-index open subgroup MATH of MATH, containing MATH, such that MATH extends to MATH. Let MATH. Define MATH by MATH, so that MATH is a continuous homom... |
math/0007127 | Let MATH be an isomorphism, where MATH and MATH are arithmetic lattices in MATH. Define MATH and MATH . Then MATH. By passing to a finite-index subgroup we may assume that MATH, where MATH and MATH. Let MATH and MATH. Then, by passing to a finite-index subgroup, we may assume MATH and MATH. Let MATH denote the projecti... |
math/0007128 | Suppose that MATH has a nontrivial stabilizer MATH. Obviously MATH if and only if MATH, so suppose that MATH from now on. Since MATH is closed in MATH, it is compact and has a finite number of components. Suppose first that MATH is not discrete. Then the identity component of MATH must be a closed MATH-dimensional subg... |
math/0007128 | By REF , any cubic that has a symmetry of order MATH has a linear factor. Conversely, suppose that MATH has a linear factor and is nonzero. By applying a MATH symmetry, it can be assumed that MATH divides MATH, implying that MATH has the form MATH which clearly has a symmetry of order MATH that fixes MATH. The quadrati... |
math/0007128 | A cubic MATH is linear in a direction MATH if and only if the direction generated by MATH is a singular point of the projectivized curve MATH in MATH. Thus, the first statement follows, since the set of real cubic curves with a real singular point is a semi-analytic set of codimension MATH. If there are two distinct si... |
math/0007128 | Let MATH satisfy the hypotheses of the theorem. If the fundamental cubic MATH vanishes identically, then MATH is a MATH-plane, so assume that it does not. The locus where MATH vanishes is a proper real-analytic subset of MATH, so its complement MATH is open and dense in MATH. Replace MATH by a component of MATH, so tha... |
math/0007128 | Let MATH be a connected special Lagrangian submanifold with the property that its fundamental cubic MATH has a MATH-symmetry at each point. If MATH vanishes identically, then MATH is an open subset of a special Lagrangian MATH-plane, so assume that it does not. Let MATH be the dense open subset where MATH is nonzero. B... |
math/0007128 | Suppose that MATH satisfies the hypotheses of the theorem. If the fundamental cubic MATH vanishes identically on MATH, then MATH is a MATH-plane and there is nothing to show, so suppose that MATH. Let MATH be the open dense subset where MATH. The hypothesis that MATH has MATH-symmetry at every MATH implies that there i... |
math/0007128 | It suffices to assume that MATH is connected, so do this. If any MATH-leaf MATH is planar, even locally, then this plane must be MATH-isotropic and NAME and NAME 's REF implies that MATH itself must contain an open subset of a special Lagrangian MATH-plane. By real-analyticity, it follows that MATH itself is planar and... |
math/0007128 | By assumption, at a generic point MATH, the MATH-stabilizer subgroup of MATH is isomorphic to MATH. Let MATH be the open, dense subset where this holds. Then by REF , there exist positive functions MATH with MATH and a MATH-subbundle MATH over MATH on which the following identity holds: MATH . (Of course, MATH is a dou... |
math/0007128 | By REF , any such MATH must have a reducible fundamental cubic MATH. Thus, the MATH-stabilizer of MATH at each point contains a MATH and so is either isomorphic to MATH, MATH, or MATH. If this stabilizer is isomorphic to MATH at a generic point, then REF applies, showing that MATH is a NAME example. If this stabilizer ... |
math/0007128 | Let MATH satisfy the hypotheses of the theorem. The locus of points MATH for which the MATH-stabilizer of MATH is larger than MATH is a proper real-analytic subset of MATH, so its complement MATH is open and dense in MATH. Thus, I can, without loss of generality, replace MATH by a component of MATH. In other words, I c... |
math/0007128 | This is a straightforward application of the NAME Theorem CITE so I will only give the barest details. This is a local result, so it suffices to give a local proof. Let MATH be the rank of MATH and let MATH be its codimension. For any point MATH, there is an open MATH-neighborhood MATH on which there exist real-analyti... |
math/0007128 | This is immediate from the formulae for MATH and MATH. |
math/0007128 | I will only sketch the proof, since the details are straightforward. First, the easy direction: If MATH is ruled, then the analytic set MATH must have dimension MATH at least. Conversely, if the dimension of MATH is at least MATH, then it contains an immersed analytic arc MATH, which generates a ruled surface MATH for ... |
math/0007128 | Combine REF . |
math/0007128 | First, I will define the almost NAME on MATH and show that it is NAME. Consider the mapping MATH that sends the coframe MATH to the oriented line spanned by MATH that passes through MATH. Since the structure equations give MATH it follows that the ten MATH-forms that appear on the right-hand side of this equation are M... |
math/0007130 | The strategy of proof is the same as in REF-dimensional case. One starts with an arbitrary sequence of asymptotically holomorphic sections of MATH over MATH, and perturbs it first to obtain the transversality properties. Provided that MATH is large enough, each transversality property can be obtained over a ball by a s... |
math/0007130 | The smoothness and symplecticity properties of the various submanifolds appearing in the statement follow from the observation made by NAME in CITE that the zero sets of approximately holomorphic sections satisfying a uniform transversality property are smooth and approximately MATH-holomorphic, and therefore symplecti... |
math/0007130 | We only give a sketch of the proof of REF . As usual, we need to obtain two types of properties : uniform transversality conditions, which we ensure in the first part of the argument, and compatibility conditions, which are obtained by a subsequent perturbation. As in previous arguments, the various uniform transversal... |
math/0007132 | We use local coordinates. If MATH then MATH and the associated hamiltonian vector field is MATH . This expression shows that MATH projects to MATH, On the other hand, if MATH, one computes: MATH so the result follows. |
math/0007132 | If the rank of MATH at MATH is MATH we are done, so we can assume MATH. If MATH we proceed, by induction, straightening out vector fields of the form MATH. So let MATH and assume we have constructed coordinates MATH valid on a domain MATH, and a basis of sections for MATH over MATH, MATH such that MATH where MATH depen... |
math/0007132 | If MATH and MATH, there exists a piece-wise smooth path made of orbits of vector fields of the form MATH, with MATH a section of MATH. Integrating sections we can map MATH to MATH, so we may assume that these points of intersection are actually the same. Around MATH we choose coordinates MATH and sections MATH as in th... |
math/0007132 | We compute using REF : MATH so we have MATH which shows that REF is satisfied. |
math/0007132 | By a result of CITE, a generalized distribution associated with a vector subspace MATH is integrable iff it is involutive and rank invariant. Taking MATH, so that MATH, REF shows that MATH is involutive iff the curvature REF-section vanishes. Hence, all it remains to show is that if the curvature vanishes and MATH is a... |
math/0007132 | From REF for the curvature and REF of the exterior MATH-derivative, we compute: MATH where the symbol MATH denotes cyclic sum over the subscripts. The first and fourth term vanish because of NAME 's identity, while the two middle terms cancel out. |
math/0007132 | Let MATH be a domain of a chart MATH where there exists a basis of trivializing sections MATH. On MATH, we define a linear MATH-connection by MATH and a MATH-connection on MATH by MATH where MATH and MATH denote, as usual, the structure functions for this choice of coordinates and basis. A straight forward computation ... |
math/0007132 | A straightforward computation. |
math/0007132 | Let MATH be a MATH-path in MATH. We can find a time-dependent section MATH of MATH over MATH such that MATH. Using the notation above, we define a time-dependent section MATH over the tubular neighborhood such that for MATH and MATH . The lifts MATH are the integral curves of the vector field MATH defined by MATH so MA... |
math/0007132 | Recall that any piecewise smooth path MATH can be made into an NAME. By REF , it is enough to show that for every MATH there exists a neighborhood MATH of MATH in MATH such that if MATH is a piecewise smooth loop based at MATH and MATH is a piecewise smooth family with MATH then MATH is a inner automorphism of MATH. We... |
math/0007132 | Assume first that MATH has trivial reduced holonomy and fix a base point MATH. We choose an embedding of MATH in MATH, a complementary subbundle MATH and trivialization so we can define the holonomy map MATH. Also, we choose a Riemannian metric on MATH. By compactness of MATH, there exists a number MATH such that every... |
math/0007132 | To check that REF is independent of the extensions considered, we fix a local basis of sections MATH for MATH in a neighborhood of MATH. If we write MATH for some functions MATH and MATH, we compute MATH . This expression shows that MATH only depends on the value of MATH at MATH and the values of MATH along MATH, that ... |
math/0007132 | In fact, let us see that the compatible connection MATH constructed in the proof of REF is a basic connection. We use the same notation as in that proof, so if MATH is a leaf of MATH and MATH, we write MATH and we have MATH . Therefore, for any REF-form MATH, we get MATH since MATH. It follows that for any REF-form MAT... |
math/0007132 | If MATH is any basic connection and MATH, we have MATH, so REF for the curvature tensor, gives MATH . But the right hand side is zero, because of the NAME identity. Similarly, if MATH is a differential form such that MATH, we have MATH. Hence, using MATH and the well known formula for the NAME derivative of the NAME br... |
math/0007132 | We compute MATH where we have used first the linearity and symmetry of MATH, then the MATH-invariance of MATH, and last the NAME identity. |
math/0007132 | Suppose we have two MATH-connections in MATH with connection REF-sections MATH and MATH, and denote by MATH and MATH the MATH-sections they define through REF . We construct a REF-parameter family of connections with connection REF-section MATH, MATH, and we denote by MATH its curvature REF-section. By the transformati... |
math/0007132 | Choose a MATH-connection in MATH which is induced by some covariant connection. Given MATH, this covariant connection gives a closed MATH-form MATH defined by a formula analogous to REF , and which induces the usual NAME homomorphism MATH. We check easily that MATH so the proposition follows. |
math/0007132 | According to REF we have MATH and we claim that MATH if MATH is odd (these are the vanishing primary classes that we mentioned to above). The proof that MATH is standard: we can choose an orthonormal basis of sections for MATH so that the curvature REF-sections take there values in MATH. But if MATH, we have MATH for a... |
math/0007132 | Let MATH and MATH (respectively, MATH and MATH) be basic connections (respectively, riemannian connections). It follows from REF that MATH . Hence, it is enough to show that the cohomology classes of MATH and MATH are trivial. Consider first the basic connections MATH and MATH. The linear combination MATH is also a bas... |
math/0007132 | Choose a basic connection MATH and a riemannian connection MATH relative to some metric on MATH. We consider the transverse measure MATH to MATH associated with this metric. We claim that MATH so REF follows. Observe that it is enough to show that REF holds on the regular points of MATH, since the set of regular points... |
math/0007134 | The proof of REF is straightforward. Also, the claim that MATH is a metric on MATH follows immediately form REF and the equality MATH . Therefore we include only the proof of that equality: If MATH then MATH . Hence we can assume that MATH . Since MATH where MATH can by transformed into MATH by MATH elementary operatio... |
math/0007134 | CASE: We prove MATH first. If MATH has no connection then MATH and the inequality is obvious. Therefore we may assume that MATH has a connection MATH . Consider two kinds of operations on MATH: CASE: If there is an unconnected letter MATH in MATH then we delete this letter and we obtain a new word MATH with a connectio... |
math/0007136 | The sum MATH is congruent, modulo MATH, to the total number of edges in MATH consisting of a black vertex MATH and white vertex MATH or MATH. Given a loop MATH in MATH, the number of edges in it consisting of a black vertex MATH and white vertex MATH or MATH is equal to the number of times the loop crosses a ray with i... |
math/0007136 | The determinant on the left side of REF can be obtained from the determinant on the right side by elementary column operations. |
math/0007136 | The base case MATH is trivial. The inductive step MATH follows directly from REF and from Lemma A (special case). The inductive step MATH requires some more work. MATH where MATH is the NAME polynomial of degree MATH, which has the property MATH. For a definition of MATH, consult CITE. The only property of MATH we need... |
math/0007136 | The special case followed from the fact that the same sequence of column operations transforms the row MATH into MATH and the row MATH into MATH, for any values MATH,MATH. Therefore the same sequence transforms the row MATH into MATH. |
math/0007136 | MATH implies MATH . If we take the value at MATH, we obtain MATH . Similarly, MATH implies MATH . If we let MATH be the operator taking MATH to MATH, we can write MATH . Clearly the coefficient of MATH in MATH is equal to MATH times the coefficient of MATH in MATH which is equal to MATH times the coefficient of MATH in... |
math/0007139 | Reduction of the generic matrix MATH modulo MATH in REF leads to a generic remainder which depends on the parameters MATH. Moreover, since a NAME basis of MATH is parameter-free, this generic remainder has the property that its specialization to a fixed choice of parameters MATH gives the remainder of MATH modulo MATH.... |
math/0007139 | As we have seen, MATH is isomorphic to the nonempty affine variety MATH defined in the variables MATH. Here, a point of MATH with coordinates MATH corresponds to the isomorphism MATH. Now any isomorphism MATH induces an isomorphism from the variety to itself, sending MATH to MATH where MATH. This action is regular in M... |
math/0007139 | A point MATH corresponds to an isomorphism with inverse MATH if and only if the system MATH has exactly one solution for MATH. This is equivalent to the MATH matrix MATH having rank MATH and the augmented matrix MATH also having rank MATH. The matrix MATH will have rank MATH if and only if any one of its MATH minors is... |
math/0007140 | Let MATH. Because MATH is an isolated singularity the exponents at MATH are all equal to MATH. Equivalently, there exists an orthonormal basis of MATH with respect to which the matrix of MATH is the direct sum of MATH copies of MATH. Thus, if MATH, then MATH is even. Let MATH be a MATH-invariant symmetric polynomial of... |
math/0007140 | From REF it follows that we can write MATH where MATH and recall that MATH is a section of MATH (see CITE). In fact MATH is the nowhere zero function on MATH characterised by MATH then REF becomes MATH where MATH - note that with respect to a suitably chosen adapted orthonormal frame we have MATH. Also note that MATH i... |
math/0007140 | Let MATH be the manifold (endowed with a circle action) obtained by blowing-up the isolated fixed points, that is, the points of MATH. Then, since signatures add when taking connected sums, the signature MATH of MATH is given by MATH . Because the exponents of each isolated fixed point are equal by hypothesis, the indu... |
math/0007140 | Because MATH and MATH are oriented, MATH is orientable. Thus we can choose MATH such that MATH . Clearly, MATH is smooth on MATH. Furthermore, when MATH, since MATH as we approach a critical point, MATH extends to a continuous vector field on MATH whose zero set is MATH and the flow of MATH extends to a continuous flow... |
math/0007140 | If MATH then, by REF , there exists a free MATH action on MATH whose orbits are connected components of the fibres of MATH. Hence, by the NAME theorem the NAME number of MATH is zero. Also, as is well-known (immediate consequence of REF), all the NAME numbers of MATH are zero. Suppose that MATH. If the set of critical ... |
math/0007140 | Choose one of the orientations of MATH and let MATH be the corresponding volume-form with respect to MATH. Let MATH be the NAME number of MATH. By the NAME Theorem we have MATH where MATH is the NAME tensor of MATH and MATH , MATH are its self-dual and anti-self-dual components, respectively (see CITE). By REF , MATH a... |
math/0007140 | Let MATH be a regular point of MATH and MATH. Let MATH be a unit vector. Because MATH is of constant curvature, by CITE, there exists an open neighbourhood MATH of MATH and a submersive harmonic morphism MATH with values in some NAME surface MATH such that its fibre through MATH is tangent to MATH. Then for any other u... |
math/0007140 | If MATH is submersive then the proof follows from CITE. Suppose that MATH has critical points. Then, by CITE, MATH has positive scalar curvature and there exists a smooth Killing vector field MATH tangent to the fibres of MATH whose zero set is equal to the set of critical points of MATH. Suppose that MATH is NAME. Its... |
math/0007143 | Let MATH be the NAME closure of MATH, and note that the NAME form is also invariant under MATH. Replacing MATH by a finite-index subgroup, we may assume MATH is NAME connected. Let MATH be an NAME decomposition of MATH. Assume MATH and MATH. From REF, we see that MATH contains codimension-one subspaces of both MATH and... |
math/0007143 | Let MATH be the NAME closure of MATH, and note that the NAME form is also invariant under MATH. Replacing MATH by a finite-index subgroup, we may assume MATH is NAME connected. Let MATH be an NAME decomposition of MATH. For each real root MATH of MATH (with respect to the NAME subalgebra MATH), let MATH be the correspo... |
math/0007147 | Let MATH be the affine proalgebraic group whose existence is guaranteed by REF . The semisimplicity assumption, and the assumption that MATH has only finite number of irreducible representations imply that MATH is finite. |
math/0007147 | See for example, [ES, REF ]. |
math/0007147 | By REF , MATH as desired. |
math/0007147 | By a fundamental result of NAME and NAME REF , MATH and hence by REF , MATH is central. Now, we have MATH CITE, so MATH. This shows that MATH in every irreducible representation of MATH. But MATH and MATH are central, so they act as scalars in this representation, which proves REF . |
math/0007147 | By REF , MATH hence the result follows from REF . |
math/0007147 | Straightforward. |
math/0007147 | Let MATH be the functor defined by MATH . Let MATH and MATH where MATH is defined by MATH . Then it is straightforward to verify that MATH is a symmetric functor. |
math/0007147 | This is a special case of [DM, REF ]. This theorem states that the category of fiber functors from MATH to MATH an affine proalgebraic group, is equivalent to the category of MATH-torsors over MATH . But, MATH is algebraically closed, hence there exists only a unique MATH-torsor over MATH . |
math/0007147 | Let MATH . We have two fiber functors MATH arising from the forgetful functor; namely, the trivial one and the one defined by MATH respectively. By REF , MATH are isomorphic. Let MATH be an isomorphism. By definition, MATH is a family of MATH-linear isomorphisms. By naturality, the diagram MATH commutes for any two obj... |
math/0007147 | Let MATH be an isomorphism of triangular NAME algebras. Then MATH defines an isomorphism of triangular NAME algebras from MATH to MATH . This implies that the element MATH is a symmetric twist for MATH. Thus, for some invertible MATH one has MATH. Let MATH. It is obvious that MATH is a NAME algebra isomorphism, so it c... |
math/0007147 | Let MATH be the category of finite-dimensional representations over MATH of MATH . This is a semisimple abelian MATH-linear category with finitely many irreducible objects, which has a structure of a symmetric rigid category (see REF). In this case, the categorical dimension MATH of MATH is equal to MATH . Since the NA... |
math/0007147 | Let MATH be the minimal triangular sub NAME algebra of MATH . By REF , there exist a finite group MATH and a minimal twist MATH for MATH such that MATH as triangular NAME algebras. We may as well assume that MATH . Let MATH be the inclusion map. Then MATH is an injective morphism of triangular NAME algebras as well. In... |
math/0007150 | Obviously MATH is a parallel frame for each MATH. So writing MATH one easily computes MATH and MATH. But MATH. |
math/0007150 | We define MATH and MATH. REF ensures that MATH and MATH are smooth at MATH and MATH. Using MATH this in turn implies that MATH has the form MATH for some MATH. Since the zeroes of MATH are fixed we know that MATH and MATH are constant. We write MATH. One gets MATH and MATH . This can be used to show MATH. Again REF ens... |
math/0007150 | Let MATH and MATH be the solutions to REF corresponding to MATH and MATH. One has MATH and MATH with MATH and MATH. The ansatz MATH leads to the compatability condition MATH or MATH which gives: MATH . Thus MATH and MATH are completely determined. To show that they give dressed solutions we note that since MATH the zer... |
math/0007150 | Obviously the above transformation coincides with the dressing described in the last section with MATH in REF . This proves the lemma. |
math/0007150 | One has MATH for fixed time MATH and MATH . |
math/0007150 | One can proof the theorem by direct calculations or using the equivalence of the dIHM model and the dNLSE stated in REF . If the curve MATH evolves by rigid motion its complex curvature may vary by a phase factor only: MATH or MATH. Plugging this in REF gives MATH which is equivalent to REF with MATH. |
math/0007150 | Analogous to the smooth case. |
math/0007150 | Literally the same as for REF . |
math/0007150 | Let us look at an elementary quadrilateral: For notational simplicity let us write MATH and MATH. If we denote the angles MATH and MATH with MATH and MATH we get MATH with MATH and MATH as in REF . MATH is the corresponding angle along the edge MATH. Note that MATH and MATH are coupled by MATH . To get an equation for ... |
math/0007150 | Comparing the orders in MATH on both sides in REF gives two equations MATH . The first holds trivially from construction the second gives MATH . This can be checked by elementary calculations using REF for the real part of MATH. |
math/0007150 | We use the notation from REF . Since MATH and since MATH is completely determined by MATH and MATH we have, that MATH. On the other hand on can determine MATH by MATH and MATH. Since MATH REF says that MATH and MATH must lie in MATH. |
math/0007150 | For the NAME transformation MATH look at the vector field MATH given by MATH. This must have a zero. |
math/0007150 | With notations as in REF we know MATH and MATH giving us MATH which proofs the claim since MATH goes to REF if MATH tends to MATH. |
math/0007150 | Evolving by rigid motion means for the complex curvature of a discrete curve, that it must stay constant up to a possible global phase, i. CASE: MATH. Due to REF the evolution equation for MATH reads MATH . Using MATH gives MATH and finally MATH . So the complex curvature of curves that move by rigid motion solve MATH ... |
math/0007151 | Substituting MATH and MATH into REF one gets REF . A right-handed version of the Proposition also holds. |
math/0007151 | By uniqueness of the decomposition REF one can set MATH in an arbitrary basis MATH of MATH. Properties of the MATH-algebra map MATH as well as MATH are to be verified. |
math/0007151 | MATH is a MATH version of REF . Taking into account REF one calculates MATH . Hence MATH. Applying now MATH to the both sides, gives MATH. |
math/0007151 | It is enough to check REF on basis vectors: MATH due to REF and using the properties of the antipode. |
math/0007151 | Let MATH be any basis in MATH and MATH the dual basis in MATH. With respect to the given basis one can define the generalize derivations MATH as MATH . Thus MATH. Substituting MATH into REF and comparing the coefficients in front of the same basis vectors we conclude MATH which is equivalent to REF . |
math/0007158 | In the following the sums over MATH are taken over all pair partitions MATH of the set MATH and sums over MATH are taken over all subsets of the set MATH. Products over MATH are taken over MATH. From REF follows that for any MATH and indexes MATH we have: MATH and furthermore MATH . We define MATH REF shows that the co... |
math/0007158 | We define MATH. In the following sums over MATH are taken over all pair partitions MATH of the set MATH with additional property that there exist MATH and MATH such that MATH. From REF we have that MATH where in the last inequality we used REF . |
math/0007158 | Our goal is to construct a sequence MATH such that MATH holds. However, MATH . The first summand converges to MATH by REF . From REF we have that MATH . From REF it follows that this expression converges to MATH. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.