Polynomials constant on a hyperplane and CR maps of hyperquadrics
We prove a sharp degree bound for polynomials constant on a hyperplane with a fixed number of distinct monomials for dimensions 2 and 3. We study the connection with monomial CR maps of hyperquadrics and prove similar bounds in this setup with emphasis on the case of spheres. The results support generalizing a conjecture on the degree bounds to the more general case of hyperquadrics.
Key words and phrases:Polynomials constant on a hyperplane, CR mappings of spheres and hyperquadrics, monomial mappings, degree estimates, Newton diagram
2000 Mathematics Subject Classification:14P99, 05A20, 32H35, 11C08
We are interested in the relationship between the complexity of real algebraic expressions on the one hand, and geometric properties on the other. Our main goal is to generalize a result of D’Angelo, Kos and Riehl. In [DKR] they considered a class of real polynomials in two variables, and gave a precise quantitative description of the relationship between the degree of the polynomials, and the number of their nonzero coefficients. We are interested in generalizing their results to larger classes of polynomials, and in particular to higher dimensions. The classes of polynomials we will consider arise naturally in CR geometry, and our results give new degree bounds for holomorphic mappings between balls. We will discuss the relationship to CR geometry in more detail later.
The number of non-zero coefficients of real polynomials is known to bound the geometry in different ways as well. For example, Descartes’ rule of signs implies that the number of real roots of a real polynomial is bounded by a function depending only on the number of its nonzero coefficients. For a discussion of these types of results, we refer the reader to the book Fewnomials [fewnomials] by Khovanskiĭ. Although the results in this article are not directly related those discussed in [fewnomials], important techniques that are used in this article can be found in the book as well.
Let us describe our results more precisely. Let be real polynomial of degree in the variables such that when . Let be the number of distinct monomials in . In general there exists no function such that the inequality
holds for all polynomials. For example, the polynomial is of degree for arbitrary, but and on .
A condition that arises naturally in CR geometry, where this problem originated, is to require all coefficients of to be positive. In this case the problem has been completely solved when : for each the best possible was given by D’Angelo, Kos and Riehl [DKR]. We note that when then whenever for arbitrary . Thus when , no function such that (1) holds can exist even when the coefficients are required to be positive. D’Angelo has conjectured the best possible bound for . In this paper we prove his conjecture for , and we give further reasons to suggest that the conjecture also holds for .
We will give a condition of indecomposability, which is weaker than positivity of the coefficients, and show that for this weaker condition implies the same estimate. In future work we hope to show that indecomposability gives the same estimates as positivity for all . To define indecomposability we will work in a projective setting. This setting, which we will discuss in the next section, is more pleasant to work in because of added symmetry, and we will prove all our results in the projective setting.
When we have not yet been able to prove that the bound achieved for polynomials with positive coefficients is also the best bound for polynomials that satisfy indecomposability. Under the additional mild assumption of no overhang, defined in § 2, we are able to achieve the same bound for . We note that positivity of the coefficients implies the no-overhang condition.
Let be a polynomial of degree with positive coefficients such that whenever . Let .
(D’Angelo-Kos-Riehl [DKR]) If we have the sharp estimate
(D’Angelo-Lebl-Peters [DLP]) If then
The second estimate is not sharp. Furthermore, in [DLP] we also showed that for sufficiently larger than (or alternatively when is sufficiently small) we have the sharp bound
In fact, we have classified all that give equality in (4) for and as above.
We extend the results above in two directions. First, we prove similar estimates for polynomials under the weaker assumption of indecomposability. Second, we prove a sharp estimate for polynomials with positive coefficients when .
Let be a polynomial of degree such that whenever . Let .
If and is indecomposable then we have the sharp estimate
If and all coefficients of are positive then we have the sharp estimate
If , is indecomposable, and satisfies the no-overhang condition then we have the sharp estimate
If and is indecomposable then
With the results of this paper we conjecture that (4) holds not only under the positivity condition, but also under the weaker condition of indecomposability for . It is not hard to construct examples that achieve equality in (4) and hence the bound is sharp if true.
We will study the connection with CR geometry in § 8. Our main results can be stated as follows. Let be the hyperquadric with positive and negative eigenvalues as a subset of complex projective space. A monomial map is a map whose components are single monomials (in homogeneous coordinates). We say a CR map is monomial-indecomposable if it is not of the form , in homogeneous coordinates, for two monomial CR maps and . Note that is the sphere. We say that a map has linearly independent components if the components of the map in homogeneous coordinates are linearly independent.
Let be a monomial CR map with linearly independent components that is monomial-indecomposable. Then we have the sharp estimate
Let be a monomial CR map. Then we have the sharp estimate
Let , , be a monomial CR map with linearly independent components that is monomial-indecomposable. There exists a number depending only on and such that
Note that both conditions of monomial-indecomposable and linear independence of the components are needed. The value we obtain for in item (iii) of Theorem 1.3 is far from optimal. See § 7 for the precise bound we proved.
We conjecture that the sharp bounds for and hold not only for monomial, but also for all rational CR maps of hyperquadrics that are indecomposable (into rational CR maps) for all . For we conjecture the sharp bound (4) holds for the same class of functions.
Acknowledgement: The authors would like to thank John D’Angelo for introducing this problem to us in 2005, and for his help and guidance since then. Part of the progress was booked during workshops at MSRI and AIM, the authors would like to thank both institutes.
2. Projectivized problem
We will first formulate a projectivized version of the problem. Take a polynomial such that on . Write for homogeneous polynomials of degree . Now homogenize and to obtain that
We will generally replace with . Then, changing notation, we look at homogeneous polynomials , , such that
This substitution makes the problem more symmetric and easier to handle. On the other hand, in this formulation the condition that has had only positive coefficients is harder to see. We therefore need a substitute.
Let denote the set of homogeneous polynomials such that on .
We say has p-degree if is the smallest integer such that there exists a monomial (where is a multi-index) and a polynomial of degree such that . Note that if the monomials of have no common divisor then we can take and the p-degree of is equal to the degree of . Let denote the polynomials in of p-degree .
We will say that is indecomposable if we cannot write , with , such that and are two nontrivial polynomials with no monomials in common.
We define be the number of distinct monomials in .
Let denote the set of polynomials of degree that have only non-negative coefficients, and such that on .
Just as in the previous definition, we define be the number of distinct monomials in .
Suppose we start with . We homogenize with and then let . The induced polynomial is indecomposable. Were it decomposable, the monomial could belong only to one of the summands. The other summand would then correspond (after changing back to ) to a polynomial with nonnegative coefficients that is zero on . It is not hard to see that such a polynomial must be identically zero.
Let us state some obvious results.
The only indecomposable are polynomials for , in particular and .
There exist indecomposable with and arbitrarily large p-degree.
To construct examples for the second item, note that is indecomposable, , and p-degree is . Matters become more complicated for . We will prove the following theorem.
If is indecomposable and of p-degree , then
Instead of directly proving this result, we will prove a stronger lemma, which we will also need for , and which will imply the result. To state this lemma, we need to first give the discrete geometric setup.
We define the homogeneous polynomial by:
We will generally work with rather than . In fact, we will even forget the sizes of the coefficients of and deal only with the signs in a so-called Newton diagram.
The condition of indecomposability will turn out to be a connectedness condition on the Newton diagram, when the problem is translated into a problem in discrete geometry below. The p-degree will then be the diameter of the set of nonzero signs in the Newton diagram.
For , we will write .
Let as above, where is of degree . Let be the map
We define the function as follows. Let be the coefficient of in when we treat as a multi-index. Then
We call the Newton diagram of , and we will say that is the Newton diagram corresponding to .
We will call the points of , and we will call a -point if , a -point if and an -point if . We say that the monomial is the monomial associated to and vice-versa. We will often identify points of with the associated monomials.
Therefore, the Newton diagram of is an array of s, s, and s, one for each coefficient of . For example when , the monomial corresponds to the point with coordinates . In the Newton diagram we let if , we let if , and we let if . Note that we include negative powers in the Newton diagram, even though any time is not in the positive quadrant.
We will generally ignore points where . We can give graphical representation of by drawing a lattice, and then drawing the values of in the lattice. For convenience, when drawing the lattice, we put at the origin, and then let be directed at angle and at angle . Similarly we depict the diagram for . In the figures we will use light color circles and spheres for for positive coefficients and dark color circles and spheres for negative coefficients. We will not draw the circles and spheres corresponding to the zero coefficients. See Figure 1 for sample diagrams.
We say two points in the Newton diagram are adjacent if the corresponding monomials and are such that there exist and such that . Between any two points (or any two monomials) we can define the distance as 1 if the corresponding monomials are adjacent, and if they are not adjacent as the length of the shortest path along adjacent monomials.
The support of a Newton diagram is the set of nonzero points. That is, . As cannot contain any negative coordinates, we will generally consider a subset of (where ).
We say that the support of a Newton diagram is connected if any two points of are joined by a path of adjacent points of .
For a set , we let be the smallest simplex that contains . More precisely, is the smallest set such that and for some and some we have
When then the diameter of in the distance defined above is easily seen to be . We define the size of as the diameter of plus one.
The adjacency between points in the figures is marked by drawing a line connecting the two circles or spheres. If , then we will show that p-degree of is equal to the size of where is the support of the Newton diagram corresponding to .
Let be a Newton diagram. Let be a monomial of degree . Let be the set of points in defined by
We call a node if the direct image , , , or . Do note that we allow negative indices in . For a diagram let
See Figure 2 for examples of diagrams with nodes marked. In dimension 2, the nodes are marked with a triangle where each vertex of the triangle touches a point of the node. In dimension 3 we mark the nodes similarly with a simplex.
The number of nodes functions as a lower bound for the number of non-zero coefficients for a polynomial in , or rather, as a generalization thereof. Let and let be the induced homogeneous polynomial with corresponding Newton diagram . If the diagram has a node at the points of the form , with , then it is easy to see that for the polynomial , the coefficient of the monomial cannot be zero. The non-negativity of the coefficients of puts a restriction on the kind of nodes that can occur in each mutually connected collection of nonzero points, but as we are also interested in allowing coefficients of the polynomials to be negative we will ignore this restriction as much as possible.
In two dimensions it will be sufficient to obtain only the very mild restrictions on the support that are given by the next two lemmas.
Let . Write . Let be the Newton diagram of , with support .
If is indecomposable, then is connected.
The size of is equal to the p-degree of .
Let us prove the first item. Suppose that is disconnected. Then take be the connected components. Define the by taking those monomials of that are in . As if , then no points in is adjacent to any point of . Therefore has no monomials in common with . Thus define . We see that and is decomposable.
Now let us prove the second item. By taking in the definition of and the size of , it is not hard to see that the size of is at least the p-degree of . Suppose that does not contain any points with for some . Then is divisible by . The result follows by iterating through all the variables. ∎
Let , let be the induced homogeneous polynomial and the Newton diagram of , with support . Then is connected, contains and is of size .
First we note that is indecomposable. Suppose that were decomposable. That would in fact imply that there would exist two nontrivial polynomials and with positive coefficients such that , and are constant on the hyperplane , and have no monomials in common. In fact, the constant term in can only be in one of or and hence one of them is zero on the hyperplane. So suppose that on . As all coefficients of are positive, then they are all zero. Hence . Therefore is indecomposable.
Thus by Lemma 2.8, is connected. If we let . If , then we obtain a contradiction as above, would have all positive coefficients and be zero on . Therefore . Then it is not hard to see that must contain a nonzero constant. Therefore so does and hence .
Once , we see that the size of cannot be any smaller than . Since for all points we have , then is of size . ∎
Our proof in three dimensions requires stronger restrictions on . We make the following definition.
A set has left overhang if there exists a point , such that and for all .
A point is right overhang if it is a left overhang after swapping variables. We say simply that is an overhang if it is a left or a right overhang.
Let . Define , , by , , and . Then we say that is an overhang if is an overhang of for some .
When no overhang exists in we say that satisfies the no-overhang condition. We will sometimes say that a Newton diagram satisfies the no-overhang condition, which means that the support of has no overhang. Similarly, when we say that satisfies the no-overhang condition we mean that the support of the corresponding Newton diagram satisfies the no-overhang condition.
The projection is equivalent to setting the 2nd and 3rd variable equal to each other in the underlying polynomial. Similarly for and . The Figure 3 illustrates a diagram with an overhang. The overhang is marked with a circle in the figure. The second view is along one axis signifying the projection of the support by a for some . The overhang in 2 dimensions is then easy to see.
Let , , and be as in Lemma 2.9. Then has no overhang.
We start with . Let . Suppose that be an overhang. Without loss of generality assume is a left overhang. Since , then we have a node and contains the monomial . As , the coefficient of has to be positive. Let , the coefficient of must be negative. Because we are multiplying by , we note that the coefficient of in must be negative as well, otherwise we would get a negative coefficient of in . But that means that the coefficient of in for all must be negative. But is a polynomial and hence for some , the coefficient is zero. Then we get a negative term in which is a contradiction. Therefore the lemma holds for .
Now let . Let . Suppose for contradiction that is an overhang. Without loss of generality suppose that is an overhang of . For some , let . Define . We can pick such that the coefficient of in is nonzero, which we can certainly do. Let if we homogenize to create , find and the Newton diagram , then if is the support of , then it is easy to see that . Furthermore, . If was an overhang of , then it is an overhang of . The fact that leads to a contradiction. ∎
In the next two sections we will often refer to directions of a support , imagining the point at the bottom and the points with at the top. Note that points at the top correspond to monomials of degree in nonhomogeneous coordinates obtained by setting . Similarly by up (resp. down) we will mean the direction of increasing (resp. decreasing) (that is, degree in nonhomogeneous coordinates). By a side of we mean a collection of points for which a fixed coordinate is equal to zero. In the two dimensional case we will use left and right without worrying about which is which. By the th level we will mean points where .
3. Two-dimensional results
Let us start by recalling some ideas of the proof of the -dimensional result by D’Angelo, Kos, and Riehl. We start with a polynomial with non-negative coefficients that satisfies when . We assume that is of degree and we wish to estimate from below the number of non-negative coefficients of .
D’Angelo, Kos and Riehl then write for some polynomial , but instead we will use the induced homogeneous polynomial and the notation as was introduced in the previous section. We obtain the Newton diagram of and as we discussed in the previous section, the number of nodes of is a lower estimate for the number of positive coefficients of . The positivity of puts strong restrictions on the diagram , but for now we only require that the support of is connected. We will at first require that but we will remove this restriction later.
By manipulating the Newton diagram without increasing the number of nodes, D’Angelo, Kos and Riehl were able to show that the number of nodes of must be at least . Since the node involving the points , , and does not correspond to a positive coefficient of but all other nodes do, we see that must have at least positive coefficients. Considering that generally many of the coefficients of the polynomial are ignored when counting nodes, it is truly remarkable that this argument gives the sharp bound for .
Their method is the following: they make small local changes to such that after every step the number of nodes has not increased and the lowest and left most node has moved to the right or up. After a finite number of steps all nodes must be at the top level and it is then easy to give a lower estimate on the number of nodes.
Instead of following their argument we give a slightly different proof. The reason is that the above argument cannot possibly work in dimensions and higher. It is not hard to see that if all nodes are at the top level then the estimate on the number of nodes would be at least quadratic in . But since it is easy to give examples where the number of coefficients of (and thus also the number of nodes of ) is at most linear in , it cannot always be possible to push the nodes to the top level without increasing their number. Fortunately the argument that we will give here is also a bit simpler than the original proof.
Instead of manipulating to remove the lowest nodes, we will repeatedly eliminate the lowest zero, and end up with a Newton diagram that consists entirely of s and s.
Let be a Newton diagram with support in two variables. Suppose that is of size , is connected, and contains . Then
The proof of Proposition 3.1 will consist of two short lemmas. We will say that a diagram is an extension of if the support of contains the support of and such that .
Let and be as in Proposition 3.1. Let and suppose that contains all points with . Then there exists an extension of that has equal or fewer nodes, and whose support contains all points with .
We change the -points at level into alternating - and -points, one connected group of -points of level at a time. Since there must be at least one - or -point at the -level, there are two cases that we should consider: one is that the group has - or -points on both sides, and the other is that the group extends to one of the boundaries. In the former case no new nodes can be created by changing the points and we are done. In the latter case it is possible to create a node at the boundary, but by choosing the boundary point correctly one boundary node is eliminated as well and we are done. See Figure 4. The circles with the dark borders are the nonzero points that were added in a single step. ∎
Let be a Newton diagram in two variables with support of size and suppose that is maximal: it contains all with . Then has at least nodes.
Let us consider a horizontal row at level . Suppose that, starting with and ending with , the sign changes times from to or vice-versa. Also suppose that there are sign changes on the next higher row . Then there are at least nodes involving points only on these two rows.
Now suppose that there are sign changes on the highest row . Then there are exactly nodes on this highest row. Since there are zero sign changes on the lowest row , it follows from the observation in the previous paragraph that there must be at least nodes involving some point with , as well as the one node involving the points and . Adding the three numbers gives a total of at least nodes. The minimum number of nodes can therefore only be achieved when , the maximum number of sign changes on the row . Inserting shows that there are at least nodes. ∎
Let be a Newton diagram with connected support of size in two variables. Then
Without loss of generality we may assume that the support contains points on both sides, that is point and a point . By connectedness must contain a connected path from the left side to the right. By the proof of Lemma 3.2 we can change all the -points of above this connected path into - and -points without increasing the number of nodes. By switching the roles of and we then obtain a new Newton diagram with the same number of nodes whose support is connected, of size and contains , which by Proposition 3.1 completes the proof. ∎
4. Preparing for three dimensions
The main idea that we use for estimating the number of nodes of a -dimensional Newton diagram is to ignore the nodes that might occur in the interior of the diagram and to only count the number of nodes that occur on each of the faces of the support . For each of these faces we can then use two dimensional estimates. Of course, by counting the number of nodes on each face separately and then adding the estimates we are in danger of counting some nodes more than once, namely nodes involving points that lie on more than one face. We therefore adjust the node count.
Let be a -dimensional Newton diagram. A node that involves no zeroes is called an interior node, a node with exactly one zero is called an edge node, and a node with two zeroes is called a vertex node. When both zeroes of a vertex node lie at the bottom of the node, the node is called a bottom node.
The weighted surface count of is the number of interior nodes, plus half the number of edge and vertex nodes, but ignoring the number of corner nodes:
Essentially the same proof as that of Lemma 3.3 gives us then the following, perhaps somewhat surprising result. While edge nodes have weight only , we still get the same estimate.
Let be a Newton diagram in two variables with support of size and suppose that its support is maximal, that is, contains all with . Then
As we said, the proof is similar to proof of Lemma 3.3. Again, we consider two adjacent levels in the diagram. If the higher level has sign changes and the lower level has sign changes then there must be at least nodes involving points only on these two levels. This weighted count now includes half nodes at the edges.
What is left is to count the edge nodes that only involve the top level. There are always 2 vertex nodes, plus if there sign changes at the top level, then there are edge nodes. An induction argument completes the proof. ∎
The result is surprising because even though the nodes on the edges and at the vertices have weight only half, we still obtain the same estimate as in Lemma 3.3, except for the obvious two times that we lose for the two vertex nodes on the top level and the that we lose for the bottom node. Thus Lemma 3.3 follows immediately from Lemma 4.2. Do note that we now allow more half nodes at the edge. Therefore for a diagram with edge nodes and 3 vertex nodes we again obtain .
The result in Lemma 4.2 is necessarily also sharp but we even have this stronger sharpness result that will be useful later:
Let be a maximal Newton diagram with support of size . Then we can find a Newton diagram with support of size that is maximal, agrees with for all with and such that is exactly equal to .
Construct the example level by level, starting with the top level, which is already determined by . Each next lower level is obtained by removing the leftmost entry of the previous level, and then changing all the signs (so becomes and vice verse). Using this construction there cannot be any interior nodes, and the number of edge nodes on the sides is equal to the number of sign changes on the top level. The number of edge nodes on the top level is equal to minus the number of sign changes. Adding these two numbers plus the two vertex nodes on the top level gives nodes that each have weight . ∎
In the three dimensional proof we will perform several different operations on -dimensional Newton diagrams that change the shape of one or more faces. We end the section with three -dimensional lemmas that will be needed to estimate how the number of nodes on the faces change with these operations.
Let be a two-dimensional Newton diagram, and suppose that its support contains all points with and at least one point with . Then there exist a Newton diagram that is an extension of (i.e. the - and -nodes remain unchanged) and the support of contains all points with and satisfies
Finally the set only contains points with .
The proof is similar to that of Lemma 3.2 except that we need to be slightly more careful because different kinds of nodes receive different weights. This time we change only one -point at the time, starting with a -point that lies to the left or to the right of a - or -point. By assigning to this point a or an , there are a number of nodes of that have weight but that by the change either receive weight or are no longer a node, depending on the sign we assign to the -point. The situation for these nodes is completely symmetric: if assigning a adds to the weighted surface count then assigning an subtracts to the weighted surface count, where is a multiple of . There is also at most one bottom node of that has no weight, but that can turn into an edge node of , adding a node to the weighted surface count depending on the choice of sign. Now it is easy to see what can be done:
If then we can choose the sign such that no bottom node is changed into an edge node. If then we choose the sign such that the weighted surface count is decreased by at least one half from the old -nodes, and since the gain from the potential bottom node that changes into an edge node is at most , the weighted surface count cannot increase. ∎
Let be a Newton diagram with support . Suppose that for some , whenever and when . Then define by
Then there exists a Newton diagram with support such that
We can define naturally by assigning . The estimate follows immediately. ∎
Let be a Newton diagram with support such that for some , whenever . Suppose that contains at least one point with . Define
Then there is a Newton diagram with support such that .
By Lemma 4.3 we know that given any choices for the level , there exist a Newton diagram of degree with the chosen top level and maximal such that the . We can choose the top level of such that by merging with no new interior nodes are created, and such that at least one top level edge node of is canceled by a point of at level . If is the Newton diagram obtained by merging with then it follows that
5. Three-dimensional results
Before we start proving our main result, where we use two-dimensional estimates to do a weighted count of the number of nodes that must occur on each of the faces of a Newton diagram, we should first carefully define what we mean by a face.
A face point of a Newton diagram with support is a point of that is adjacent to a -point. A vertical face of is a maximal connected family of face points that lie in a plane . A horizontal face is a maximal connected family of face points that lie in a plane .
We can think of a simplex centered at each point of . Then the union of the simplices form a polyhedron. Then faces of the polyhedron correspond to the faces of as defined above. See Figure 5 where a vertical face is shaded.
For each face of we define as the minimal number of nodes that must occur for any possible configuration , but counting only nodes that involve only points of and points that do not lie in . In other words, if it is possible for a node in to be cancelled by - or -points outside the face , we assume that they are cancelled.
As before, we write for the number of nodes of (as above) but weighting edge nodes as , weighting vertex nodes in vertical faces as except for bottom nodes, and not counting bottom nodes in all faces and vertex nodes in horizontal faces. We get the following inequality:
Where ranges over all the faces of . Since bottom nodes are not counted for we obtain that
We prove the following result:
Let be a 3-dimensional Newton diagram with support of size , and assume that has no overhang. Then has at least nodes.
We will actually prove a stronger statement. Instead of estimating the number of nodes of , we estimate for each face of the number of nodes that must at least occur on that face of , and add those numbers. Of course by doing so a node that occurs on more than one face will be counted more than once.
A node that occurs at an edge (which means two of the involved points must be ) will be counted twice, and a node that occurs at a vertex (which means three of the involved points are ) is counted three times. In order to get the right count we will give weight to the edge nodes for each face. Vertex nodes are only counted (with weight ) on vertical faces, and only when they are not bottom nodes (otherwise they have weight zero).
Let be the support of a dimensional Newton diagram , and let be the two dimensional faces of . For a vertical face we define
where the minimum is taken over all -dimensional Newton diagrams with support . For a horizontal face we use the same definition except that none of the vertex nodes are counted at all (all have weight zero), so then
Finally we define
Of course the number we obtain is a lower estimate and does not have to be equal to the actual number of nodes of for several reasons. First, interior nodes are not accounted for. Second, the -dimensional Newton diagrams do not have to correspond to the diagrams on the faces of . In fact, we don’t require that the Newton diagrams “match” at all, so there may not be any Newton diagram that has optimal as faces. Finally, bottom nodes all have weight 0, so for each bottom node we count one node too few. Nevertheless we will be able show that the sum is at least , and since there is at least one bottom node (the node involving the point ), this will conclude the proof of the theorem.
Before giving the specifics, we note that we can assume that the p-degree of a polynomial corresponding to is . That is, we could translate to assume that for each , there is a point such that , but still assume that . Now by requiring that has no overhang, it is not hard to see that must contain the point .
For the purpose of contradiction we assume that there exists a with . Let the be the smallest size for which such exists. Further assume that for this the number of points of , is maximal, i.e. the number of s and s of are maximal.
Now look at the largest so that whenever . Either or .
If then we can estimate the number of surface nodes of by applying the estimates obtained in § 3 to each of the sides of the tetrahedron . For each of the three vertical faces we obtain the estimate by Lemma 4.2 and for the horizontal face only since the three vertex nodes have weight zero. In total we obtain that , which contradicts our assumption.
Therefore we may assume that . Now we look at the plane . Then there are two possibilities. Either contains at least one element on each of the edges , and , or does not contain an element on one of these edges. Let us deal with the latter case first.
One of the edges is empty: Slice off a face.
In this case we may without loss of generality assume that does not contain elements of the form . Then it follows from the no overhang condition that does not contain any elements of the form with . Therefore if we remove from all elements of the form and we subtract from the first coordinate of all remaining elements of , we obtain a new subset that satisfies the hypotheses of the theorem, except that it now has size . See Figure 6. By the assumption on the minimality of we have that the number of surface nodes of must be at least . The contradiction therefore follows if we show that is at least smaller than .
Fortunately almost all the faces of are equal to those of . There are two side faces that lose one line, namely the faces in the - and -planes. Both of these faces lose a surface node by Lemma 4.5. The horizontal face in the -plane also loses one line but it may or may not lose half a node. If the intersection is empty then it is clear that the horizontal face does lose half a node. To see this, we refer back to the proof of Lemma 4.2. If the intersection is empty, we have removed a full line of points from the face, thus removing at least half a node. However, if this intersection is non-empty then the horizontal face may not lose surface nodes (keep in mind that vertex nodes have weight zero for the horizontal faces).
One direction remains to be considered. If is empty then the face is replaced by the face , which is just a triangle with side length one smaller. Therefore the weighted surface count for this smaller triangle is one-half smaller and the estimate is complete. When is not empty the situation is slightly more complicated. Again we lose the surface nodes on the face , but we gain nodes when the faces of in the plane are glued to the triangle . In this case it follows from Lemma 4.6 that we lose at least node which completes the estimate.
None of the edges is empty: Fill .
The only case that is left to deal with is when contains at least one element on each of the edges , and . The idea now is to take enlarge by defining . See Figure 7. By our assumption on we have that is strictly larger than , and satisfies the no-overhang condition as well. If we can show that the number of surface nodes of is not larger than that of , then we get a contradiction with our assumption that is maximal.
We apply Lemma 4.4 to see that the number of surface nodes on each of the side faces , and does not increase when we extend to . All the other vertical faces can only become smaller when we replace by , so the number of surface nodes on those faces only become smaller.
The only faces that are left are the horizontal faces on the levels and . Let us say that we start with some optimal -dimensional Newton diagrams with supports corresponding to the horizontal faces at levels and for set . We create a configuration for the horizontal face of at level as follows.
We keep the same Newton diagrams for the faces at level that we already had. We replace every at level of and coordinate , where , by the sign of the point of the Newton diagram containing that point, although it is possible that we need to reverse all the signs for a face , as we shall see below. After we have done this procedure, we will see that we can assign signs to the remaining -points of the form with in a way that does not increase the number of nodes.
By sliding the faces at level upwards in this way there can only be fewer internal nodes, as the faces can become smaller but not larger. There are vertex (or edge) nodes that remain vertex (or edge) nodes so we do not need to worry about those. However, when ‘gluing’ these configurations to the configurations of the old faces at level it is also possible for edge and vertex nodes to become internal nodes, which have weight one. This is possible in two different ways.
The first way is for an vertex node of an old -level face to become an internal node. Since the horizontal vertex had weight zero, this gives us one extra node, but as we remarked earlier, we lose twice half a node on vertical faces when such a vertex becomes internal, so we do not need to worry about this case.
The other way is for an edge node of an old -level face to become an internal node. We can deal with this case by flipping the signs of old level configurations. By doing this we get that at most half of the edge nodes of old configurations that actually become internal nodes, and the other edge nodes disappear, which solves this case as well.
That leaves us with the remaining -points on the edge . As we only have to worry about the horizontal faces (as the weighted count for the vertical faces is already taken care of) and vertex-nodes in horizontal faces have weight zero, it is easy to assign or to each point one by one, starting with points adjacent to non- points in the edge (which we assumed exist), in such a way that no interior nodes are created and the number of edge nodes remains the same. This completes the proof. ∎
6. Sharp polynomials in three dimensions
In the previous section we have obtained some information of what a Newton diagram with minimal number of nodes looks like. While it is not hard to construct a polynomial corresponding to a given Newton diagram, in general we will get many more nonzero terms in the corresponding than there are nodes in the diagram.
We will say an indecomposable polynomial is sharp if is minimal among all indecomposable polynomials . We make a similar definition for ; is sharp if is minimal in . We will concentrate on in this section since we have proved sharp bounds only for and not for . Also using the nonhomogeneous notation will make the computations and the exposition simpler. Of course, if it turns out that indecomposable polynomials in posses the same degree bounds as , then the sharp polynomials we describe in this section will also be sharp in .
It is an easier problem to list all possible Newton diagrams with the sharp number of nodes, but it is harder to decide which diagrams correspond to actual sharp polynomials in or . For the classification of sharp polynomials is not trivial even in , see [DL:complex, LL]. We will see that classification in dimension 2 is related to classifying the sharp polynomials in dimension 3.
It is not hard to construct certain sharp polynomials . That is, achieves equality in the bound
It is not hard to see that and that , and therefore . Thus, are sharp. These polynomials are known (see for example [DLP]) as generalized Whitney polynomials. At each step, instead of we can also pick an arbitrary monomial of of degree . We then obtain all the other generalized Whitney polynomials.
The Newton diagram for the Whitney polynomials is easy to see. In the nonhomogeneous setup, where we just take the Newton diagram of , the diagram is simply a connected path of s going from to where . In the homogeneous setup, the signs are not all , but instead alternate between and .
It is natural to ask what all the sharp polynomials look like. In particular, one might wonder if there are there any examples that are not generalized Whitney polynomials. The following construction was given in [DLP]. Start with
which is sharp and is in . Now replace with to obtain