Abstract

Let be a topological space. The semigroup of all the étale mappings of (the local homeomorphisms ) is denoted by . If , then the -right (left) composition operator on is defined by   , . When are the composition operators injective? The Problem originated in a new approach to study étale polynomial mappings and in particular the two-dimensional Jacobian conjecture. This approach constructs a fractal structure on the semigroup of the (normalized) Keller mappings and outlines a new method of a possible attack on this open problem (in preparation). The construction uses the left composition operator and the injectivity problem is essential. In this paper we will completely solve the injectivity problems of the two composition operators for (normalized) Keller mappings. We will also solve the much easier surjectivity problem of these composition operators.

1. Introduction

Let be a topological space. A mapping is called a local homeomorphism of or an étale mapping of if for any point there exists a neighborhood of such that the restriction of to , denoted by , is an homeomorphism. The set of all the étale mappings of , denoted by , is a semigroup with a unit with the composition of mappings taken to be the binary operation. If , then the -right composition operator on is defined by The -left composition operator on is defined by We were interested in the injectivity of these two composition operators in two particular cases. The first is the case of entire functions that are étale (and normalized). The second case is that of the polynomial mappings with determinant of their Jacobian matrix equal (identically) to and whose -degrees equal their total degrees. For the first case we use the following.

Definition 1. Consider the following Thus we use in this case the symbol instead of . Then we have the following.

Proposition 2 (see [1]). Consider the following , is injective.

Theorem 3 (see [1]). Let . Then is not injective if and only if This settled the first case. It should be noted (see [1]) that the proof for the left composition operator is much more involved than the proof for the right composition operator (which follows directly from the Picard little theorem). It is in fact the second case that initiated our interest in the injectivity of the composition operators. It results from a new approach to study étale polynomial mappings and in particular the two-dimensional Jacobian conjecture [2–4]. This approach constructs a fractal structure on the semigroup of the (normalized) Keller mappings and outlines a new method of a possible attack on this open problem (in preparation). The construction uses the left composition operator and the injectivity problem is essential. In this paper we will completely solve the injectivity problems of the two composition operators for (normalized) Keller mappings. We will also solve the much easier surjectivity problem of these composition operators.

2. The Semigroup of Normalized Keller Mappings and a Few Facts on Their Asymptotic Variety

Let be an étale mapping that satisfies the two normalizations(1);(2) and , where .The set of all such mappings will be denoted by . This semigroup (with respect to composition of mappings) is the parallel of the semigroup for entire functions. The -dimensional Jacobian conjecture can be rephrased in each of the following forms: (a);(b) is a group.For the next survey of results we refer to the following paper: [5]. We denote by the asymptotic variety of , that is, the curve of all the asymptotic values of the mapping . The canonical geometric basis of will be denoted by . This basis consists of finitely many rational mappings of the following form: , where , , , and . Also the effective powers in have a gcd which equals . The cardinality of the geometric basis, , equals the number of components of the affine algebraic curve . we have the double asymptotic identity , where the polynomial mapping is called the -dual of . Each generates exactly one component of . This component is normally parametrized by . We will denote by the implicit representation of this component in terms of the irreducible polynomial . There exists a natural number and a polynomial . The affine curve is called the -phantom curve of . The -component of , , is a polynomial curve, which is not isomorphic to , and hence in particular must be a singular irreducible curve. We have the relation . The exponent satisfies the double inequality . In our case of the canonical rational mappings , we have . The following is true: Thus the -preimage of the -component of (which is the -image of ) is the union of two curves: the first is and the second is the so-called -phantom curve of . Even if for a single the -phantom curve is empty, then JC(2) follows. Also if , then is a surjective mapping.

Proposition 4. If , then , .

Proof. . Next we have such that such that .

The proposition tells us that compositions of étale mappings do not decrease the geometric basis of the right factor and consequently do not decrease the left image of its asymptotic variety. We naturally ask under what conditions the geometric basis of is actually larger than that of ? In other words, we would like to know when is it true that ? This happens exactly when . This means that , . Let , . Then We clearly have and so . By we have . This is not necessarily a member of the geometric basis of . The canonical geometric basis of , contains finitely many rational mappings of the form . Since , it follows that (a similar phenomenon as the Picard little theorem). If is an asymptotic tract of , then cannot be a bounded subset of . The reason is that if is compact, then is compact and since this would imply the contradiction that is bounded (and hence cannot be an asymptotic tract). Hence has at least one component, say , that goes to infinity. This is because the number of components of is finite and is not bounded. So has a limit along which equals the above asymptotic value of . This proves the following generalization of the second part of Proposition 4, namely, the following proposition.

Proposition 5. If , then .

This proposition implies that if , then necessarily because, as shown in the proof of Proposition 4  , .

Proposition 6. Let . If such that , then ; that is, is a surjective mapping.

Proof. Since we have because in this case the only points in the complement of the image of are the finitely many Picard exceptional values of which are asymptotic values of . If, as the assumption says, , then by Proposition 5 we must have and so there are no Picard exceptional values of the mapping .

If is not a surjective mapping, then the last proposition implies that ; we must have . In particular . This is the choice . If we choose , we get . Now it is clear that by induction we get the infinite chain of strict inequalities: where times fold composition. Since the cardinality of the geometric basis is the number of components of the asymptotic variety , it follows in this case that the asymptotic varieties of iterates of are of increasing complexity in the sense that the number of components of the curve is strictly larger than the number of components of the curve . By Proposition 5 we always have . Hence = . By induction we get in general:

3. The Composition Operators on Are Not Surjective, but the Right Composition Operator Is Injective

Proposition 7. The mappings , are not surjective if and only if . In fact in this case we have , .

Proof. Consider the following:

Proposition 8. is injective.

Proof. . Since we have and by the assumption . Hence .

We naturally inquire if also is injective. So let us assume that . Then . If we denote , then and also . If , then and so . If, on the other hand, , then there are points for which . By our assumption, , so is not in and it identifies different images of and of the same . We ask the following question: suppose that , , is there a point for which ?

Based on our experience with entire functions we tend to prove that the answer to the question is negative. Indeed this is the case and the proof is almost identical to the entire case; see [1]. Namely, if the answer is affirmative, then we have two types of points in : for which and the complimentary set, where both sets are nonempty. Let us denote by the first subset of ; that is, The subset of is open in the strong topology because , are étale mappings and if , then an open neighborhood of in the strong topology such that . So the complimentary subset of is a closed nonempty subset of . Let be a boundary point of . Let satisfy . Then , , , and . This implies that in any strong neighborhood of there are different points, say for large enough, so that . Hence is not injective in any strong neighborhood of . Thus . This contradiction proves the following.

Proposition 9. Suppose that is not injective for some . Then , such that and we have .

Remark 10. Proposition 7 asserts that such that we have and .

4. The Size, , of the Generic Fiber of a Keller Mapping

We will need the generic size of a fiber of a mapping . If we denote and , then the fiber over is . It is well known that this set is a finite subset of and, by the Bezout theorem, we have . Moreover, there is a number that we will denote by such that generically in we have . This means that is a closed and proper Zariski subset of . In fact . Thus we have .

Definition 11. Let . We will denote . We will call the geometrical degree of the étale mapping .

Proposition 12. Consider the following .

This is a well-known result. We include one of its proofs for convenience.

Proof. . But generically in ,   and generically in , .

Definition 13. An étale mapping is composite if such that . An étale mapping is prime if it is not composite. This is equivalent to for some . The subset of of all the prime mappings will be denoted by . Thus the set of all the composite étale mappings is .

Proposition 14. , is not a prime number. Equivalently, , is a prime number .

Proof. such that (by the definition) (by Proposition 12 and the fact that is a composite integer.

Theorem 15. The following hold true. (1)If  , then .(2) such that .

Proof. If , then are all composite étale mappings. Let , then such that . So such that . Hence . Continuing this we get for any such that and by Proposition 12  . But , and so , , a contradiction to . Thus .
Now part 2 is standard, for if we take and . If we take , , and . If , then for some . So by Proposition 12   and since it follows that and we conclude the proof of part 2 using induction on the geometrical degree. Namely, , for , , and some primes .

5. The Metric Spaces

We will need a special kind of four (real) dimensional subsets of . These will serve us to construct suitable metric structures on .

Definition 16. Let be an open subset of with respect to the strong topology that satisfies the following conditions. (1) ( has no “slits”).(2) is a compact subset of (in the strong topology).(3), .We define the following real valued function: Here we use the standard set-theoretic notation of the symmetric difference between two sets and ; that is, .

Remark 17. It is not clear how to construct an open subset of that will satisfy the three properties that are required in Definition 16. We will postpone for a while the demonstration that such open sets exist.

Proposition 18. is a metric on .

Proof. (1)  the volume of (where the last equivalence follows by the fact that and are local homeomorphisms in the strong topology and because of condition (1) in Definition 16) (by condition (3) in Definition 16).
(2) By , it follows that .
(3) Here we use a little technical set-theoretic containment. Namely, for any three sets , , and we have This implies that from which it follows that Hence the triangle inequality holds.

So far we thought of the volume of as the volume of the open set which is the symmetric difference between the image and the image of the open set . However, the mappings and are étale and in particular need not be injective. We will take into the volume computation the multiplicities of and of . By Theorem 3 on page 39 of  [6] we have the following: given that we define . Then . Thus the Jacobian condition implies that . So the real mapping preserves the usual volume form. In order to take into account the multiplicities of the étale mappings and when computing the volume of the symmetric difference we had to do the following. For any instead of computing, we compute For every we denote by the subset of such that for each point of there are exactly points of that are mapped by to the same image of that point. In other words, . We assume that is large enough so that we have . For our étale mappings it is well known that if , then , so the volume of these ’s contribution equals 0. The dimension claim follows by the well-known fact that the size of a generic fiber equals and that is also the maximal size of any of the fibers of . However, for the sake of treating more general families of mappings, we denote by the volume of the set . Then has a partition into exactly subsets of equal volume. The volume of each such a set is and each such a set has exactly one of the points in for each . We note that by the Jacobian Condition. Thus the volume with the multiplicity of taken into account is given by We note that is a partition, so . Hence we can express the desired volume by We note that this equals and since is a partition we have . As expected, the volume computation that takes into account the multiplicity of is in general larger than the geometric volume . The access can be expressed in several forms: Coming back to the computation of the metric distance the volume of we compute the volume of   with the multiplicity of   while the volume of   is computed with the multiplicity of .

6. Characteristic Sets of Families of Holomorphic Local Homeomorphisms

In this section we prove the existence of sets that satisfy the three properties that are required in Definition 16. The third property will turn out to be the tricky one.

Definition 19. Let be a family of holomorphic local homeomorphisms . A subset is called a characteristic set of if it satisfies the following condition: ,  .

We start by recalling the well-known rigidity property of holomorphic functions in one complex variable. Also known as the permanence principle, or the identity theorem. The identity theorem for analytic functions of one complex variable says that if is a domain (an open and a connected set) and if   is a subset of that has a nonisolated point and if is an analytic function defined on and vanishing on , then for all .

There is an identity theorem for analytic functions in several complex variables, but for more than one variable the above statement is false. One correct statement is as follows: “Let and be holomorphic functions in a domain in . If for all in a non-empty set in , then in . Hence, analytic continuation of holomorphic functions in several complex variables can be performed as in the case of one complex variable. Contrary to the case of one complex variable, the zero set of a holomorphic function in a domain , , contains no isolated points. Thus even if in a set with accumulation points in , it does not necessarily follow that in . For example, in with variables and we can take and ” [7, Chapter 1, page 16].

In spite of the above standard identity theorem for complex variables that requires a thick set (i.e., an open set) on which one can do much better. Let us start with the following. Let be an entire function of two complex variables and . Let us define a subset of as follows. We take a convergent sequence of different numbers. Thus and . For each , let be a convergent sequence of different numbers, such that their limit is . We define . Now we have the following.

Proposition 20. If vanishes on , that is, for , then is the zero function.

Remark 21. We note that is a thin set, in fact a countable set. Even the closure is thin.

Proof of Proposition 20. Since is an entire function, it can be represented as a convergent power series centered at with an infinite radius of convergence. We can sum the terms in the order we please. Let us write as a power series in with coefficients that are entire functions in . Thus we have , where for each , is an entire function in the variable . For a fixed we have by our assumptions the following: for . But so that is an entire function of the single variable , which vanishes on a convergent sequence . By the identity theorem of one complex variable we deduce that , the zero function. Since it follows that the Maclaurin coefficients , vanish. Now, this is valid for each , and converges. Since each is an entire function which vanishes on a convergent sequence it follows, once again, by the identity theorem in one complex variable, that is, , the zero function . Hence we conclude that .

This type of elementary arguments that was used to construct a thin set for identity purpose is not new. For example, see the following.

Theorem 22 (see [8]). Let be a domain, and let be a subset of that has a nonisolated point. Let be a function defined for such that is analytic in for each fixed and analytic in for each fixed . If whenever and both belong to , then for all .

Advancing along the lines of the construction of the thin set in Proposition 20 we note that if is a sequence of different numbers that converges to , and if for each there is a straight line segment of ’s such that two entire functions and agree on the union (a countable union) of the segments , that is, , , then , .

We now will construct characteristic sets of families of holomorphic local homeomorphisms .

Definition 23. Let be a natural number and . An -star at is the union of line segments, so that any pair intersects in .

Definition 24. Let be a line segment and let be a countable dense subset of  . Let be a sequence of different natural numbers and , let be an -star at such that one of the star’s segments lies on , and such that , . Here we denoted . Moreover, we group the stars in bundles of, say 5, thus getting the sequence of star bundles: and for each bundle of five we take the maximal length of its rays to be at most the length of the maximal length of the previous bundle. We define Let be a sequence of different complex numbers that converges to . Let be a partition of the natural numbers, . In fact all we need is the disjointness, that is, . Let us consider the starred segments and define the following countable union of starred segments in : where we assume that the lengths of the star rays were chosen to satisfy disjointness in ; namely, We let or if we need a closed (compact) set, the closure of this union.

Proposition 25. Let be any family of entire holomorphic local homeomorphisms . Then is a characteristic set of  .

Proof. Let satisfy . Then each starred line segment must be mapped onto a curve and each -star on , is mapped onto a holomorphic -star This is because the valence sequences of the stars are pairwise disjoint natural numbers, and , are local homeomorphisms and hence preserve the star valencies . The centers of the holomorphic stars form a countable and a dense subset of the curves . By continuity this implies that the restrictions coincide. Since and are holomorphic, this implies by Proposition 20 (which is a variant of the identity theorem for entire functions ) that .

Remark 26. Proposition 25 holds true for any rigid family of local homeomorphisms. Rigidity here means that So the proposition holds true for holomorphic mappings, for harmonic mappings, and in particular for .

We recall that Definition 16 required also two additional topological properties; namely, the open set should satisfy ; is compact (all in the strong topology). These automatically exclude the set that was constructed in Definition 24. However, we can modify this construction to get at least an open set.

Proposition 27. Let be any family of holomorphic local homeomorphisms . Let be any open subset of 2 with a smooth boundary that contains the compact . Then the open set is a characteristic set of  .

Proof. Since can not be mapped in the smooth by an holomorphic local homeomorphism, we have for any for which that also . Now the result follows by Proposition 25.

Remark 28. We note that if   is a compact then satisfies, at least, the requirement that is compact. However, the “no-slit” condition fails.

Now that we gained some experience with the topological construction of we are going to make one more step and fix its shortcomings that were mentioned above. We need to construct a domain of which has the following three properties.(1) relative to the complex topology.(2) is a compact subset of relative to the strong topology.(3), .The complex topology and the strong topology are the same. Our construction will be a modification of the construction of the domain that was constructed in Proposition 27. We start by modifying the notion of an -star that was introduced in Definition 23.

Definition 29. Let be a natural number and . A thick -star at is a union of triangles, so that any pair intersects exactly at one vertex, and this vertex (i.e., common to all the triangles) is .

Definition 30. Let be the construction of Definition 24 that uses thick -stars.

Proposition 31. Let be any family of holomorphic local homeomorphisms . Then is a characteristic set of  .

Proof. The proof is the same word by word as that of Proposition 25 where we replace -star by thick -star .

We finally obtain our construction.

Proposition 32. Let be any family of holomorphic local homeomorphisms . Let be an open ball centered at with a radius large enough so that (where is the set in Proposition 31). Then the domain is a characteristic set of .

Proof. The proof is the same as that of Proposition 27 where we replace -star by thick -star .

7. Injectivity of the Left Composition Operator

We would like our natural mappings, the right mapping and the left mapping , to be say bi-Lipschitz with respect to the metric (that reflects the fact that our mappings satisfy the Jacobian condition). Considering first the right mapping , it would mean that given three étale mappings and a characteristic set of we need to compare the volume of (multiplicities of and of are taken into account) with the volume of the deformed set . A short reflection shows that the two volumes are not comparable (in the sense of bi-Lipschitz). The situation is completely different when we replace the right mapping, by the left mapping, . For example, we have the following.

Proposition 33. the mapping is an isometry of the metric space .

Proof. For any two mappings and in we need to compare with . We have (using our assumption on ) Since is also (globally) volume preserving we have This proves that .

We now drop the restrictive assumption that . Thus we merely have and we still want to compare with , for any pair . We only know that is a local diffeomorphism of and (by the Jacobian condition) that it preserves (locally) the volume. In this case the geometrical degree of , can be larger than . We have the identity which holds generically (in the Zariski sense) in . Hence the (complex) dimension of the set is at most . The Jacobian condition implies (as we noticed before) that preserves volume taking into account the multiplicity. The multiplicity is a result of the possibility that is not injective and hence the deformation of the characteristic set by convolves (i.e., might overlap at certain locations). However, this overlapping is bounded above by . So if is a measurable subset of and we compare the volume of with the volume of its image , then This can be rewritten as follows: This is the place to emphasize also the following conclusion (that follows by the generic identity ; namely, provided that the set tends to cover the whole of the complex space in an appropriate manner. To better understand why the quotient tends to the lower limit rather than to any number in the interval (if at all) we recall that our mapping belongs to and so is a polynomial étale mapping. So any point for which is an asymptotic value of   and hence belongs to the curve which is the asymptotic variety of  . In other words the identity is satisfied exactly on the semialgebraic set which is the complement of an algebraic curve. We now state and prove the main result of this paper.

Theorem 34. Let . Then one has the following. (i).(ii)Suppose that is a family of characteristic sets of such that , then one has  for being large enough.(iii)Under the assumptions in (ii) one has In particular, the left mapping , , is a bi-Lipschitz self-mapping of the metric space with the constants .

Proof. (i) , such that and . By it follows that and so and . Hence , so , and finally .
(ii) and (iii). Here the proof is not just set theoretic. We will elaborate more in the remark that follows this proof. We recall that . This implies that we have , the extension degree of ; see [3]. This is the so-called Fiber theorem for étale mappings. Moreover the image is cofinite; [3]. Also has a finite set of exactly maximal domains . This means that is injective on each maximal domain , and , and and the boundaries are piecewise smooth (even piecewise analytic). For the theory of maximal domains of entire functions in one complex variable see [9], and for that theory for meromorphic functions in one complex variable see [10, 11]. Here we use only basic facts of the theory which are valid also for more than complex variable. If is a family of characteristic sets of such that , then by the above , where is a finite set, and if , then we have the identity Recalling that we write the last identity as follows: Taking any two points and (as in the defining equation of the set on the right-hand side in the last identity), we note that there are , such that (for !). For a large enough characteristic set of , we will have and and so (since ). Hence will not include the point . We conclude that if and are -equivalent (), then will not belong to for large enough . We obtain the following crude estimate: One can think of as a large open ball centered at the origin of , and with the radius and look at the images of the two polynomial étale mappings and and compare the volume of which is of the order of magnitude , where depends on the algebraic degrees of and , with the volume of the set in the left-hand side of the last equation. Similar estimates are used in the theory of covering surfaces by Ahlfors, see [12, chapter 5]. We conclude that Hence

Remark 35. The facts we used in proving (ii) and (iii) for étale mappings are in fact true in any dimension , that is, in . In dimension it turns out that the codimension of the image of the mapping is and in fact the coimage is a finite set. Also the fibers are finite and have a uniform bound on their cardinality (one can get a less tight uniform bound by the Bezout theorem). Here are few well-known facts (which one can find in Hartshorne’s book on Algebraic Geometry, [13]). (1)The following two conditions are equivalent.(a)The Jacobian condition: the determinant is a nonzero constant.(b)The map is étale (in standard sense of algebraic geometry). In particular it is flat.Let be étale. Let . (2)For every prime ideal (), with residue field the ring is finite over ().(3) is a quasifinite mapping.(4)The set is open in .(5)For every point the fiber is a finite subset of .(6)The ring homomorphism is injective, and the induced field extension is finite.(7)There is a nonempty open subset such that on letting , the map of schemes is finite. For any point we have the equality the geometrical degree of .(8)The dimension of the set is at most .(9)If is affine, then and .Let be the topological space which is the set given the classical topology. Similarly for , the map of schemes induces a map of topological spaces . (10)The map is a local homeomorphism.An immediate conclusion from Theorem 34 is the following,

Corollary 36. the left mapping , is an injective mapping.

8. Extending the Notion of Geometrical Degree

In this section we will outline the fact that some of the notions and results that are related to geometrical degree of an étale mapping originate, in fact, in the more basic topological spaces (no algebraic or holomorphic structure is needed). We will skip most of the proofs (that are elementary).

Definition 37. Let be a topological space. The semigroup of all the continuous mappings, , will be denoted by . Here, as usual, the binary operation is composition of mappings.

Proposition 38. Let be a topological space, , and . Then is injective.
Let be a topological space, , and is injective. Then for any , the property implies that ; that is, any is determined by its restriction .
Let be a topological space that has the following property: for any closed and any point there exist two continuous mappings such that , but .
Let be such that is injective. Then .

Remark 39. Proposition 8 follows.

We are ready to discuss the notion of the geometrical degree, , of appropriate mappings in .

Lemma 40. Let be a topological space, and . If is open, then the set is closed.

Corollary 41. Let be a topological space, , is open, and . Then one has the following. (1), the union of a closed set and an open set.(2), the intersection of a closed set and an open set.(3)If exists, then is a closed set.

Remark 42. .

Definition 43. Let be a topological space, , is open, and the maximum exists. Then we call the geometrical degree of .

Example 44. If with the complex topology and , then we know that exists. We also know that the set is a plane algebraic curve (possibly empty). Thus it is closed in . Moreover it is also small because .

We need one more property to hold for our mappings, namely, that the fiber size will generically be , that is, that the set of all for which will be a large set measured in the topology of  . This leads us to the following.

Definition 45. Let be a topological space. We will denote by the set of all the mappings that have the following properties. (1).(2) is open.(3)The maximum exists.(4).

Proposition 46. Let be an Hausdorff space. Then one has the following. (1) is a semigroup with an identity (where the binary operation is composition of mappings). In fact .(2), .

Proof. Checking that is closed for composition: . Also open is open. If , then and ; we have . Since , it follows that exists and, in fact, that . This gives the first three properties in the last definition. We need to check that . For that matter it will be convenient to denote . Let . This set is open and dense in . By our definitions , a finite set of exactly points. Since is Hausdorff we can find open neighborhoods of , respectively which are pairwise disjoint. The images are open (since is open) neighborhoods of the point . Let us take the intersection Then is an open neighborhood of and consists of neighborhoods of . We define , . Then since is closed and is open and dense in , the ’s are open and dense subsets of the ’s. Since is continuous and , it follows that are open and dense in . We note that each point is such that because is disjoint of . Thus the point has a neighborhood and open and dense subsets of such that each point (this set is still open and dense in ) is such that and each is such that . Hence . This proves both that and that .

Corollary 47. Let be Hausdorff and a semigroup of mappings which are continuous and open and suppose that there is an absolute constant such that , . Then the sets , cannot be dense in , unless , .

Corollary 48. Let be Hausdorff and a family of mappings which are continuous and open and satisfy , (in particular , ). Let be the semigroup generated by (composition of mappings is the binary operation). Then , .

Definition 49. Let be an Hausdorff space. A mapping is called a composite mapping if such that . A mapping is a prime mapping if it is not composite. This is equivalent to the following: if for some , then . The subset of of all the prime mappings will be denoted by . Thus the set of all the composite mappings is .

Proposition 50. , is not a prime integer. Equivalently, , is a prime integer .

Theorem 51. If , then .
  ∀F∈E(X), , , such that .

Maybe few elementary examples are in place.

Example 52. In Definition 45 we take with the complex topology and (with an abuse of notation) take for . Then Proposition 46(2), , is the elementary fact from algebra that we have .

A second example is given in Section 4 of this paper.

Example 53. In Definition 45 we take with the complex topology, and . Then the theory that was outlined in Propositions 12 and 14 and Theorem 15 is a special case of the above more general topological theory.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.