THE GENERIC EXISTENCE OF A CORE FOR q-RULES

Donald G. Saari

Northwestern University

Abstract. A q-rule is where a winning coalition has q or more of the n voters. An important issue is to understand when, generically, core points exist; that is, to determine when the core exists in other than highly contrived settings. As known, the answer depends upon the dimension of issue space. McKelvey and Schofield found bounds on these dimensions, but Banks found a subtle, critical error in their proofs. The sharp dimensional values along with results about the structure of the core are derived. These values can be identified with the number of issues needed to lure previously supporting voters into a new coalition.

A “q-rule” is where a winning coalition has at least q of the n voters. For instance, the commonly used majority rule is the q = [\frac{n}{2}] + 1 rule where [x] denotes the “greatest integer function.” A standard assumption, asserting that if C is a winning coalition then those voters not in C cannot form another winning coalition, restricts the q values to range between the majority and unanimity (q = n) rules.

When applying rules to economics and spatial voting, we need to understand when they are stable. To illustrate with an historical example, several times when the selection of a pope for the Catholic Church required only a simple majority of the eligible Cardinals (q = [\frac{n}{2}]+1), the precarious instability of the system was manifested by the church erupting into dissension and conflict complete with a pope and anti-pope vying for power. Stability became the issue, so in 1179 the Third Lateran Council changed the selection procedure to the q = [\frac{2n}{3}] + 1 rule that remains in use (Saari, [S1, p 15-16]).

The core is an accepted way to examine “stability.” Recall, x is a core point with a q-rule if, for all other proposals y, it is impossible to find q voters who prefer y to x; the core is the set of all core points. Using the pope selection example, the flavor of this definition is captured by R. Kieckhefer’s (a noted expert on the history of religion) explanation for the two-thirds procedure ([S1, p. 16]). As he argues, for a candidate (y) to replace a pope (x), the pope would have to bungle affairs sufficiently badly on enough issues to alienate at least half of his original supporters. Eventually this happened.

When new concerns emerge, we might expect, as the church example suggests, that the stability of a q-rule might diminish or even vanish. Each issue defines a direction in “issue space,” so the number of new topics is measured by the increase in the dimension, k, of issue space. If these concerns are sufficiently attractive to enough agents, stability may be jeopardized. As developed here, this is the general situation.

Key words and phrases. Core, spatial voting, q-rules.

This research was partially supported by NSF IRI 9103180. My thanks to J. Banks for informing me about this problem, to Banks and R. McKelvey for reading my several working notes preparatory for this paper, and to S. Brams and D. Richards for several suggestions.

To see what can happen, let x_{j} be the ideal point for the jth agent with utility function u_{j}(y) = −∥y − x_{j}∥. In words, with Euclidean distance preferences, the closer a point is to her ideal point, the more she likes it. For n odd, q = [\frac{n}{2}] + 1, and k = 1, the median voter’s ideal point, x_{med}, is the only core point. The standard proof to establish that x_{med} is a core point starts by noting that when the ideal points are placed on a line, [\frac{n}{2}] of these points are on either side of the median voter’s bliss point x_{med} (the global maximum for her utility function). Any other alternative, z, is on one side of x_{med}. Thus, the median voter and all voters on the other side of x_{med} prefer x_{med} to z. This description proving that x_{med} is a core point is robust; the assertion and argument hold even if the voters’ ideal points are altered.

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Even for k ≥ 2, where n is odd and q = [\frac{n}{2}] + 1, core points that are bliss points exist. A trivial example is unanimity where all ideal points have the same position. This situation is highly unlikely, so, to justify the core, robust examples are needed. One approach is to use the Plott [P] construction of pairing voters’ ideal points. Start by placing the first agent’s ideal point, x_{1}, somewhere in R^{k} and then pass [\frac{n}{2}] lines through x_{1}. Each line is divided by x_{1} into two sides; place a voter’s ideal point on each segment. (This is depicted in Fig. 1 for k = 2.) As any line through x_{1} separates the ideal points so that no more than [\frac{n}{2}] of them are on either side, the proof that x_{1} is in the core becomes a minor modification of the “median voter proof.” Moreover, notice that if x is a core point for a q-rule, it is a core point for a q_{1}-rule where q_{1} > q. (If q voters cannot be found to vote against x, then it is impossible to find even more that are willing to do so.) Consequently, Plott’s construction establishes the existence of a core point for all q-rules for odd values of n and k ≤ \frac{n−1}{2}.

While this pairing of ideal points makes x_{1} a core point for the majority rule, the example is not robust. To see this, in Fig. 1 I slightly varied voter five’s preferences as denoted by the dagger on the dashed line. With this new configuration, y is preferred by a majority (voters two through five) to x_{1}. To prove this assertion, construct the convex hull defined by the ideal points of voters two through five, and draw a line perpendicular to an edge of the hull. It follows from elementary trigonometry that when this coalition compares two points on this line, each voter prefers the point closer to the edge. As y and x_{1} are on such a line, the conclusion follows.

Actually, the core is empty for the perturbed alignment. To prove this, draw the convex hulls for all four-voter coalitions and observe that they have no point in common. So, for any proposal x, there is a y preferred by a decisive coalition. Consequently, while Plott’s construction establishes the existence of a core point (for all q-rules for odd values of n and k ≤ \frac{n−1}{2}), it is not indicative of what can happen because this conclusion may not survive even the slightest change in preferences. As this fickle behavior is caused by allowing larger k values, the issue is to find the dimensions of issue space which do, or do not, permit existence assertions to be robust. In mathematical terms, for a given n and q-rule, we seek those k values whereby, generically, the core is empty. Similarly, we seek those k values whereby core points exist for an open set of preferences (i.e., it exists when preferences are slightly perturbed in any manner.) This problem is resolved here for all n, q rules.

The problem of determining the k values where the core is generically empty has received considerable research attention. Using a singularity theory argument (that I found to be insightful with its careful embedding of various classes of singularities within other classes), Schofield [Sc] and McKelvey and Schofield [MS1] published assertions stating that the core is generically empty for certain k values. While the embedding approach simplifies their analysis, the issue space dimension they compute is not for the core, but rather for a different object. Although this means that their conclusion may describe only a subset of the relevant k values, these results were correctly and widely greeted as a major advance.

Recently, Banks [B] found an error in these papers which, unfortunately, invalidated the conclusions. He also showed that the correct estimates may differ from what was previously believed. (Banks also used the embedding approach.) Thus, the core problem was reopened; again, the issue was to determine the appropriate k values. I close this aspect of the problem by finding the correct, sharp values. As shown below, it is interesting how there is a natural interpretation for these sharp k values that is compatible with Kieckhefer’s comments about the pope problem.

My improvements are possible because instead of the embedding argument I characterize the geometry of “core-singularities” (Prop. 1) so that singularity theory can be applied to this geometry. But, while my approach differs significantly from that of [Sc], [MS1], and [B], [Sc] and [MS1] were the first to use singularity theory to analyze the core. (Earlier, however, Smale [Sm], Saari and Simon [SS], and others used the same approach to analyze related issues such as Pareto points.)

As for creating examples, there probably are several papers in addition to [P] and McKelvey and Schofield [MS2]. However, I am unaware of any paper proposing general assertions about robust existence. Therefore, the assertions and the approach developed in Sect. 5, appear to be the first general results which are known to be best possible.

2. Dimensions for the existence of the core

Assume that each agent has C^{∞} smooth, strictly convex preferences. Namely, for each x ∈ R^{k}, the set

(2.1)  M(x) = {y ∈ R^{k} | u(y) ≥ u(x)}

is strictly convex. In the following theorem, “generically” means a residual set (a set that can be expressed as a countable intersection of open-dense sets) of utility functions where the space of these functions has the Whitney C^{∞} topology (see Golubitsky and Guillemin [GG] and Saari and Simon [SS]). (The [SS] arguments improve the conclusions in the sense that when issues are restricted to a compact subset of R^{k}, “generic” now can mean “opendense.”) For a first reading, when “generic” refers to nonexistence, interpret it as meaning “everything except improbable, carefully concocted examples which are not indicative of what can happen because the conclusion can fail with even an arbitrarily slight change in the preferences.” When “generic” describes existence, it means that examples exist where the conclusion holds even after an example is slightly modified; they hold for a “C^{2} open set” in the Whitney topology. (The utility functions can be slightly perturbed along with its first and second derivatives and the conclusions remain.) Also, while the “smoothness” conditions can be relaxed, rather than describing the attendant technicalities I prefer to concentrate on the structure of the core.

To simplify notation, call a core point that also is a bliss point for some agent a “blisscore point.” Other core points are called “nonbliss-core points.” While my results emphasize strictly convex preferences, it is easy to extend them to all smooth preferences. An interpretation and description follows the statement of the theorem.

Theorem 1. a. Generically, for a q-rule where \frac{n}{2} < q < n, bliss-core points exist if and only if

(2.2)  k ≤ 2q − n.

b. For any k and n, there exists a q-rule where core points exist generically. Indeed, if n = q (that is, for the unanimity rule), then, for any dimension k ≥ 1, there exist open sets of preferences with bliss and nonbliss core points.

c. Let the “excess size of issue space dimension” be β = k − [2q − n]. Generically, there exists nonbliss core points for a q-rule if and only if q satisfies

(2.3)  \frac{1}{2β + 4} + n (1 − \frac{1}{2β + 4}) ≤ q.

In terms of the dimension of policy space, generically, a given q-rule with n voters has a nonbliss-core point if and only if β ≤ \frac{4q−3n−1}{2(n−q)}; that is, iff

(2.4)  k ≤ 2q − n + \frac{4q − 3n − 1}{2(n − q)}.

d. Consider the ratio α = \frac{q}{n} where a candidate must receive at least α of the vote to be selected. For a given α rule, \frac{1}{2} < α ≤ 1, and a dimension for issue space, k, there exists a positive integer n_{0} so that for all n ≥ n_{0}, generically, the core is nonempty.

To see what this theorem means, start with the simple majority rule q = [\frac{n}{2}] + 1 where n is even. As this q = \frac{n}{2} + 1 value, when expressed as 2q − n = 2, is Eq. 2.2, it means that robust, simple majority examples with a bliss point exist up to two-dimensional issue spaces. For nonbliss core points, the value of the right-hand side of Eq. 2.4 is 2 − \frac{n−3}{n−1} < 2, so, for even values of n, nonbliss simple majority core points exist (generically) only for a one-dimensional issue space.

Compare these outcomes with the restricted behavior when n has an odd value. Here the q = \frac{n+1}{2} value, expressed as 2q − n = 1, again is Eq. 2.2, so with an odd number of voters robust bliss-core points exist only for a single-dimensional issue space. For nonbliss core points, the dimensional bound is given by Eq. 2.4, or by 1 − \frac{n−1}{n−1} = 0. Therefore, generically, simple majority nonbliss core points fail to exist for odd values of n.

To better understand this behavior, notice from Eqs. 2.2, 2.4 that 2q − n is a stability measure; the larger the value, the more we can expect from core points. (This is further developed in Thm. 2.) This measure has its minimum value when n is odd and q = [\frac{n}{2}] + 1, suggesting, and it is true, that with odd numbers of voters, simple majority core points have a precarious existence.

It seems reasonable to expect someone who voted for an accepted proposal will defect only when presented an offer that cannot be refused. If each defecting voter needs a particular issue, and if the voters have a degree of independence of beliefs, the bound on the dimension of issue space allowing stability should roughly agree with the number of voters that need to defect to change the outcome. To relate this intuition to the 2q − n measure, observe that if k is the maximal admissible dimension of issue space, Eq. 2.2 becomes

(2.5)  q − \frac{n}{2} = \frac{k}{2}

which indicates that the maximum dimension of issue space ensuring a robust core is determined by the difference between q and the simple majority rule.

Applying this argument to the pope-selection problem where q ≈ \frac{2n}{3}, we find that stability can be expected for \frac{k}{2} = \frac{2n}{3} − \frac{n}{2} or k ≤ \frac{n}{3}. In words, without enough issues – one issue per defecting Cardinal – to alienate a third of the voting Cardinals, the original choice could remain stable. Instability occurs with enough issues (k > \frac{n}{3}) to permit half of the pope’s original supporters to defect. Observe how this mathematical argument parallels Kieckhefer’s insightful explanation reported in the introductory paragraphs.

Extending this argument to α rules (as defined in the theorem), we find that the issue space dimension always roughly agrees with the number of voters that need to be alienated, or allured, away from x to support another proposal. To see why, observe for any α rule that stability occurs should k ≤ (2α−1)n. Compare this dimensional limitation with the number of voters, x, a losing coalition of (1 − α)n voters must lure away from a winning coalition of αn voters to form a new winning coalition; i.e., (1 − α)n + x = αn, or x = (2α − 1)n. As these numbers agree, the assertion follows. To illustrate with α = \frac{3}{4}, the number of defecting voters must constitute half of all voters. Again, examples supporting the spirit of this “issue per defecting voter” come from news accounts of the “new issues” (sometimes called “bribes”) that occasionally are introduced to entice Congressmen to abandon the status quo (x) by voting for proposal y.

The \frac{3}{4} rule arises in a different context. As computed below, if a q-rule requires less than a \frac{3}{4} vote, the nonbliss core points generically exist only up to dimension k ≤ 2q − n − 1. Thus, the “excess dimension” assertion is operative only for q-rules starting with the \frac{3}{4} rule. The reason I defined “excess dimension of issue space,” β, in terms of 2q − n is to emphasize (see the proof in Sect. 6) that when β > 0, the gradients of most preferences are restricted to lower dimensional subspaces of issue space. For nonbliss points, this occurs for β ≥ 0.

These comments about β have important consequences for Euclidean preferences. To explain, once a voter’s ideal point x_{j} is specified, we know the gradient and all higher derivative terms of these preferences when evaluated at x. As this forces Euclidean preferences to belong to the excluded set of “non-generic behavior,” Theorem 1 is not applicable. On the other hand, Euclidean preferences are so widely used that it is worth developing a parallel theory for them. To do so, recognize that because the “ideal points” are the only relevant parameters, “robustness” must mean that the positions of these points are not highly restricted. Thus, “generic” for Euclidean preferences refers to conclusions which hold for an open set of locations for each voter’s ideal point. To illustrate, the median voter conclusion for k = 1 is robust; the example of Fig. 1 is not. In particular, the assertion of the last paragraph (that when β ≥ 0, most gradient vectors must be in a lower dimensional subspace) forces most ideal points to be in a lower dimensional setting; this violates robustness.

Corollary 1. For Euclidean preferences, bliss-core points exist generically iff k ≤ 2q − n. Nonbliss core points exist generically iff k ≤ 2q − n − 1.

Additional consequences of Theorem 1 follow.

Examples: a. Part b follows immediately from c. To see why, we only need to determine whether q = n (which is an admissible choice) satisfies Eq. 2.3. This only involves checking whether the equivalent inequality

\frac{1}{2β + 4} ≤ \frac{n}{2β + 4}

is satisfied; it is trivially true for n ≥ 1.

b. By converting the n(1 − \frac{1}{2β+4}) term from Eq. 2.3 to the mixed number a + \frac{x}{2β+4}, it follows that q = a + 1. Therefore, dropping the equality and the first term on the left side of Eq. 2.3, we obtain the equivalent

(2.6)  \frac{2β + 3}{2β + 4} < \frac{q}{n}.

To illustrate Eq. 2.6, all q-rules allowing issue space to exceed the 2q − n dimension by β = 100, are those satisfying

q > n\frac{203}{204}.

This suggests using n values that are integer multiples of 204, say, n = 4(204) = 816. The minimal q value is 1 + 4(203) = 813, so the 813-rule generically admits core points where k ≤ 2q − n + β = 1626 − 816 + 100 = 910.

c. If q is the smallest value solving these inequalities for n = γ(2β +4), then q = n−γ +1, so the restriction on the dimension of issue space is

(2.7)  k ≤ n + (β − 2γ + 2) = (β + 1)(2γ + 1) + 1.

This inequality indicates the k growth rates relative to the q-rule.

d. With Eq. 2.6, we discover the β-bifurcation roles played by different rules. For instance, with β = 0, we have from Eq. 2.6 that the \frac{3}{4} rule is the bifurcation point. (More precisely, “one more than a three-fourths vote”). With β = 1, the bifurcation occurs at the \frac{5}{6} rule. Continuing, we see that the bifurcations arise at the (one more than) \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, …, \frac{odd integer}{1+odd integer}, … of the votes. From this, it follows that for any β, there are supporting n and q values where q < n. But once n and q are specified, Eq. 2.4 bounds k. To illustrate with q = 30 and n = 35, it follows from Eq. 2.2 that bliss core points exist, generically, up to dimension

k ≤ 60 − 35 = 25

and nonbliss core points exist, generically, up to dimension

k ≤ (60 − 35) + \frac{120 − 106}{10} = 25 + \frac{14}{10},

or k = 26. This example where the nonbliss points exist longer than the bliss points suggests finding when (4q − 3n − 1)/2(n − q) is negative (to force the bliss point to persist longer than the nonbliss core points). Again, we recover the earlier assertion that this is when the decision rule is bounded above by the \frac{3}{4} rule.

A related issue is to determine whether, generically, the nonbliss points could vanish several dimensions earlier than the bliss points. If this were possible, it would require q and n values where

\frac{4q − 3n − 1}{2(n − q)} < −1,

or, by solving, where 2q < n + 1. As such a phenomenon requires using less than a majority vote, it is not relevant.

e. To prove part d of the theorem, notice that Eq. 2.4 can be represented as

k ≤ n(2α − 1) + \frac{4α − 3 − \frac{1}{n}}{2(1 − α)} < n(2α − 1) + \frac{2α}{1 − α}, α = \frac{q}{n}.

The right-hand side is a linear equation in n, so the conclusion follows. This assertion means that even α-rules close to the majority rule (i.e., α ≈ \frac{1}{2}) can be supported in, say, a 100 dimensional issue space with enough voters. To illustrate with α = 0.5001, a core point can be supported in a hundred dimensional issue space with around a half million voters. So, for a city about the size of Minneapolis, as long as the number of issues doesn’t exceed a hundred, stability could exist. □

3. Comments on the structure of the core

Theorem 1 tells us when the core exits, but it does not address the structure of this set. For instance, can the core be the union of disjoint sets? (It can when convexity is dropped, but not with strict convexity.) Of the many important issues, I choose to describe some which provide added support for the theme that the stability of a core improves with larger q values. These results describe how the core changes when the values of q, n and k vary.

For intuition about what to expect, start with a R^{1} example of eleven voters, where the jth voter’s ideal point is located at the integer j, j = 1, …, 11. For the simple majority rule (q = 6), x_{6} = 6 is the only core point. However, with q ≥ 7, the core is the interval [11 − q, q] where the endpoints are bliss core points. This suggests that if q_{1} > q, then the q_{1} core probably strictly contains the q core. In general, this is true.

Another stability measure is the topological dimension of the core. For instance, in the eleven voter example with q = 8, the fact the core includes interval (3, 8) introduces a strong sense of stability – near any core point is another one. Similarly, in the seven voter example of Fig. 2, the core is the shaded region, so, near any core point in the interior is another core point. Thus, a larger dimensional core suggests added resilience. To expand on this notion, notice that if a core has an interior with x as one of these points, then x remains a core point even after the preferences are slightly varied. Generically, when the core does not have an interior (such as for bliss-core points), by slightly varying the preferences, the core points might change but the structure remains as asserted. This underscores the importance of Theorem 2b, c.

In Fig. 2, the core is any point in the closure of the shaded region. (There are no blisscore points.) Observe that a boundary line connects the bliss points of two agents, and a vertex is where two bliss-connecting lines intersect, so they represent specialized core points where the preferences must line up appropriately. Singularity theory ensures that this picture characterizes the general situation. Namely, singularities form a “stratified structure” where a restricted setting is in the closure of the previous setting. Thus, while the boundary of a core need not have straight lines, generically it consists of core points of restricted types where preferences line up as indicated.

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Implicit in these comments is that when the core exists for dimensions k − 1 and k, the core for dimension k should be a subset of the larger core allowed with the smaller dimension k − 1. To show this is true, for a k dimensional issue space with a k_{1}-dimension subspace, k > k_{1} > 0, let P_{k,k_{1}} be the natural projection mapping. (For instance, if k = 3 and k_{1} = 2 represents the x-y plane, then P_{3,2}((x, y, z)) = (x, y) where the z component is dropped.) The k_{1}-dimensional preferences are assumed to be the k-dimensional preferences restricted to the lower dimensional plane. If this lower dimensional plane is obtained by dropping k − k_{1} issues, then it is a coordinate plane of the k-dimensional space. Otherwise, the plane represents where certain issues are combined into a single issue. (With minor technical changes, this holds for a k_{1} dimensional smooth manifold.) The following tells us that a core point persists when issues are restricted.

Theorem 2. a. Suppose n and k are such that non-bliss core points exist generically for a q-rule; q < n. The core for the q + 1 rule always contains the q-rule core. It is generically unlikely that the two cores are the same.

b. Generically, bliss-core points are isolated points. Generically for those q, n, and k values that satisfy Eq. 2.3 for β = −1 , the set of nonbliss core points has a nonempty interior.

c. Suppose β ≥ 0 is needed to satisfy Eq. 2.3 for specified k, n, q values. Generically, the core is a union of submanifolds with dimension k − (β + 1)(2q − n + 1).

d. If x is a core point for a q-rule in a k dimensional space and if there is a k_{1} dimensional plane passing through x, then P_{k,k_{1}}(x) is a core point for the k_{1} dimensional issue space.

We now encounter an interesting conflict. This discussion demonstrates that larger q values provide a wider selection of core points with added stability. The message seems to be that larger q-values are better. So, why don’t we always use unanimity as the deciding rule? The reason, of course, is obvious; very little can be accomplished beyond the status quo with an unanimous rule because it is so difficult to design an alternative that makes everyone happy.

This instinct agrees with the technical statements. With unanimity and Euclidean preferences, the core is the convex hull defined by the voter’s ideal points. (To see this, let y be outside of this hull and x the nearest point on this hull. As the line y − x is orthogonal to this edge, it follows from trigonometry that all voters prefer x to y. With minor modifications, the same argument holds when x is a vertex. But, if y is in the hull, then, as any other x is farther from some voter’s ideal point than y, that voter will veto the move to x.) Because the unanimity core usually strictly contains the q-rule core (Thm. 2a), it is easier for the status quo to be in the unanimity core than in the core of any other q-rule. Once the status quo is in a core, change is impossible. So, rather than being a desirable feature, maybe we should avoid procedures allowing a large core with their danger of fixing the outcome at the status quo.

Restating this concern, why care about the core? Well, if the core is empty, then for any x_{1} there is a x_{2} that is supported by a winning coalition over x_{1}. Similarly, there is a x_{3} supported by a decisive coalition over x_{2}. The argument continues — forever. Without core stability, we suffer the “chaos” behavior of spatial voting carefully described by Kramer [K], McKelvey [M] and others. (Richards [R] provides a new explanation.) In fact, McKelvey demonstrates sincere voting situations starting with any initial proposal, after several other proposals passes through another specified proposal, and then, after more offered choices, returns to the original one. In other words, without the stability offered by the core, an outcome cannot be trusted as truly reflecting the views of the voters.

What we need is a compromise between enjoying the stability of the core while preventing gridlock which arises whenever the core contains the status quo. With q-rules, this requires choosing a q value allowing a robust core that is not “too large.” If we know the number of issues typically involved in a decision process, this k value determines the minimal q-rule that permits core stability but, maybe, without gridlock. Furthermore, the containment assertions of Thm. 2 suggest that the core for such a q-rule may be a refinement that more accurately captures the voters’ interest. On the other hand, perhaps the true problem is caused by restricting attention to binary comparisons and q-rules. Elsewhere this question is explored in more detail.

4. A characterization of core points

To motivate the basic technical tool of this paper, start with Euclidean preferences. As shown in the Sect. 3, the core for the unanimity rule is the convex hull defined by the voters’ ideal points. A similar argument shows that with a q-rule and a specified coalition of q-voters, “their core” consists of the convex hull of their q ideal points. So, the core for a q-rule is the intersection of the convex hulls defined by all possible q-voter coalitions. If this intersection is empty, the core is empty.

This geometry explains why the core vanishes with larger k values. For instance, with n = 7 and q = 4, if k = 1, all points must lie on a straight line, so the convex hulls must have a non-empty intersection permitting the core to exist. However, even for k = 2, we cannot expect three or more ideal points to be on any line. Consequently, the extra freedom provided to position ideal points makes it difficult for all convex hulls to have a common intersection point. One remedy is to require the hulls to have more vertices; this is equivalent to increasing the q value.

This intuition extends to describe what happens with strictly convex smooth preferences. (“Smoothness” ensures that the curved utility surfaces can be approximated by the planes defined by the derivative conditions; strict convexity requires all preferred points to be on one side of this plane.) Thus, basic to our arguments is Prop. 1 which characterizes core points for q-rules in terms the derivative properties the utility functions must have in order for this intersection to be nonempty. In this description, Co_{x}({v_{j}}_{j∈C}) denotes the convex hull of the (vertices of the) vectors \{x + v_{j}\}_{j∈C} and \overline{Co}_{x}(\{v_{j}\}_{j∈C}) is this convex hull minus the vertices. For this proposition, “smooth” can be relaxed to C^{2} smoothness.

Proposition 1. Assume the voters have smooth, strictly convex preferences. A necessary and sufficient condition for x to be a core point is if for any set of q agents, C, either

(4.1)  x ∈ \overline{Co}_{x}(\{∇u_{j}(x)\}_{j∈C})

or x is both a vertex of Co_{x}({v_{j}}_{j∈C}) and a bliss point for some voter in C.

This assertion makes sense; if q or more voters prefer alternatives that are in the same general direction from x, then they can block the selection of x. The directional derivative is determined by a scalar product, \frac{∂u(x)}{∂v} = (∇u(x), v), so the sense of “the same general direction” is captured by passing a plane through x where, for all agents in this coalition, the ∇u_{j}(x) vectors are strictly on the same side of the plane. Indeed, choosing v as the normal for this plane where v is on the same side as the gradients, it follows from the positive value of (∇u_{j}(x), v) that all utilities are improved by moving in this direction. Conditions that ensure x is a core point are those that prevent this scenario from occurring. This is the content of the proposition; the proof is a formal expression of this intuition. Incidentally, Prop. 1 is a geometric extension of the argument of Sect. 1 used for the median voter and Plott’s constructions.

The proof uses an important relationship between local and global comparisons of alternatives. Call x an “infinitesimal core point” if it is a core point when the admissible choices are restricted to a sufficiently small open neighborhood of x. While a core point always is an infinitesimal core point, it is easy to construct examples where an infinitesimal core point is not a core point. (The idea is similar to constructing examples where a local maximum is not a global maximum; just use utility functions where the level sets have many “wiggles.”) On the other hand, just as appropriate convexity assumptions force a local maximum of a function to be a global maximum, the following lemma asserts that our strict convexity assumption on utility functions forces infinitesimal core points to be core points.

Lemma 1. If all agents have smooth, strictly convex preferences, then an infinitesimal core point is a core point.

Proof of Lemma 1. Assume that x is an infinitesimal core point, but not a core point because a decisive coalition C prefers y. This means that y ∈ ∩_{i∈C}M_{i}(x). But, by being the intersection of convex sets, ∩_{i∈C}M_{i}(x) is convex. Consequently, any point y_{t} = tx+(1−t)y on the straight line connecting y and x is in this set. Thus, for any t sufficiently close to unity (so y_{t} is sufficiently close to x), this same coalition would prefer y_{t} to x. This means that x is not an infinitesimal core point. The contradiction completes the proof. □

Proof of the proposition. To prove that the stated condition is necessary, it suffices to show that if Eq. 4.1 does not hold, then x is not an infinitesimal core point. If Eq. 4.1 fails to hold, there exists a coalition of q voters, C, where x is not in its convex hull. This geometry permits the construction of plane passing though x with the convex hull strictly on one side. If v is the normal vector for this plane that points toward the side with the convex hull, then \frac{∂u_{j}(x)}{∂v} is positive for all voters in the decisive coalition C. Consequently, x cannot be a core point.

If x is a vertex of the hull for some q-voter coalition C, then, by construction, ∇u_{j}(x) = 0 for some voter in this coalition. Since x is a vertex, a plane can be passed through x which does not meet any other point of the convex hull. Now, if x is not the bliss point for agent j, then, by the assumption of strict convexity, M_{j}(x) contains an open set. (This open condition is not necessary; all we need is that M_{j}(x) contains a point other than x.) By a rotation of the preferences of this agent, if necessary, the set M_{j}(x) can be made to intersect the convex hull. (That is, let Ω be a rotation matrix. Construct a new set of preferences that are defined by u_{j}^{∗}(x) = u_{j}(Ω(x)). According to the chain rule, ∇u_{j}^{∗}(x) = 0 so the conditions of the proposition continue to apply.) This means there are alternatives that are preferable to x for all members of the decisive coalition C. Consequently, if x is not a bliss point, it is possible to find preferences satisfying the conditions of the proposition where x is not a core point. This completes the proof of necessity.

The proof of sufficiency involves three cases; the first one has x as an interior point of Co_{x}({∇u_{j}(x)}_{j∈C}) for all possible decisive coalitions, and the other two are the two possible ways x can be a boundary point of this hull for some decisive coalition. For the interior point situation, any plane passing though x must have vertices of Co_{x}({∇u_{j}(x)}_{j∈C}) strictly on each side of the plane – these vertices are defined by gradient vectors of the agents from this coalition. Now, if x is not a core point, there exists an alternative y that is preferred by this coalition to x. Let v = y − x be a normal vector for a plane passing though x. For those gradient vectors on the side of the plane opposite of v, the directional derivative \frac{∂u}{∂v} = (∇u, v) is negative. As these agents have a lower utility for any such change, they will not vote for such a move. Thus, x is an infinitesimal core point.

The analysis is essentially the same when x is a boundary point. If x is a vertex of Co_{x}({∇u_{j}(x)}_{j∈C}) for a q-voter coalition C, then, by assumption in the Proposition, x must be a “bliss point” for some agent in this coalition. Clearly, whenever x is a bliss point for some voter, this voter does not wish to change. Consequently, this coalition cannot change the outcome.

For the remaining case, if x is on the boundary of \overline{Co}_{x}(\{∇u_{j}(x)\}_{j∈C}) but not a vertex (but, it could be a bliss point), then x could belong to several boundary components. (For instance, if x is on the intersection of two boundary faces, it is on both faces and on the defined edge.) Choose the lowest dimensional boundary component. Because x is not a vertex, it must be in the interior of the convex hull defined by the gradient vectors in this component. The gradient vectors (and agents) on this component are the ones we analyze.

Note that a slight modification of the above argument (where x is an interior point) proves that a change v^{∗} in this boundary component, or a change with a nonzero component in this component, is unacceptable to some voter. This is because, as x is in the interior of the hull defined by vectors on the boundary component, for some voter with a gradient in this component, the directional derivative \frac{∂u_{k}}{∂v^{∗}} is negative. Consequently, the only potential admissible changes for this coalition must be orthogonal to the boundary component, i.e., the changes must be in a direction v that is orthogonal to all of the gradient vectors on this component. Now, by strict convexity, for any of these voters with gradient in the plane, the only point of the set M_{k}(x) that is in the tangent plane passing though x with normal ∇u_{k}(x) is x. In other words, after point x, the level set u_{k}(x) is strictly on the ∇u_{k}(x) side of the tangent plane. So, for each voter with a gradient vector in the boundary plane, any change in the v direction constitutes a change of lower utility. This means that x is an infinitesimal core point. □

Proposition 1, which is critical for what follows, should be viewed as providing an higher order extension of an application of Smale’s [Sm] characterization of Pareto points to core points. The connection with Pareto points is that a core point is a Pareto point for every decisive coalition. The higher order derivative conditions are snuck in with the strict convexity arguments. As it will become clear, these strict convexity properties are critical for our analysis (when β ≥ 0) because it involves analyzing what happens when x is on the boundary of this hull. (Indeed, this insight is what permits me to avoid the embedding argument of Banks, McKelvey, and Schofield.) For this to occur, we need strict convexity. It is important to stress that this assumption does not restrict our conclusions because level sets with flat spots belong to the nongeneric setting.

Proof of Thm. 2d. The projection of a convex set to a lower dimension subspace is a convex set. As the projection of ∇u_{j}(x) is the gradient of u_{j} restricted to the lower dimensional submanifold, the proof follows directly from Prop. 1. □

5. Constructing examples

In Sect. 6, which could be read first, it is shown that you cannot do better than the bounds specified in Thm. 1. I also outline why when singularity theory asserts an event exists “generically,” either it exists with open sets of preferences, or it never occurs. To avoid the danger of carefully describing the empty set, in this section examples are constructed to verify that the “open” conditions of Sect. 6 are not empty. These conditions can be identified with solving a system of linear equations (geometrically identified with the “dividing planes” introduced below), but lower dimensional constructive examples need to be created.

In creating examples, we need not establish robustness; singularity theory does that. To explain with the analogy of an open set U about a point p, if p is a boundary of another open set V, then U ∩ V is nonempty. Our preference “examples” play the role of p while the open set U is ensured by singularity theory. Thus certain perturbations of a boundary example create a robust situation. To demonstrate this important concept, my examples start with “boundary situations” where certain slight changes in preferences can cause the core to vanish. Other choices of perturbations, the ones I use, generate robust settings. For technical reasons (discussed in Sect. 6), the examples are restricted to bliss-core point examples where k ≤ 2q − n and nonbliss core points where k ≤ 2q − n − 1.

With the exception of a bliss-core point, Prop. 1 emphasizes directions, rather than lengths, of the vectors {∇u_{j}(x)}_{j∈C}. Thus, to construct examples it suffices to find appropriate gradient directions at x that satisfy Prop. 1. Once directions are given, preferences are easy to construct. For instance, as a direction defines a ray from x, on each ray place an agent’s ideal point to define her Euclidean preferences. For examples without Euclidean preferences, use Taylor series. To illustrate with x = (1, 2) and the abbreviated expansion

u(x_{1}, x_{2}) = u(1, 2) + ∇u(1, 2) · (x_{1} − 1, x_{2} − 2)+

\frac{1}{2}D^{2}u_{(1,2)}((x_{1} − 1, x_{2} − 2), (x_{1} − 1, x_{2} − 2))

construct examples by choosing ∇u(1, 2) and D_{(1,2)}^{2}u consistent with the strict convexity assumption. For instance, for the direction \frac{1}{5}(−4, 3), we could choose ∇u = (−4, 3) and D^{2}u as the diagonal matrix with entries −1, −1 to define the utility function u(x_{1}, x_{2}) = −4(x_{1} −1)+3(x_{2} −2)−(x_{1} −1)^{2} −(x_{2} −2)^{2}. A similar construction holds for all dimensions.

Thus, once gradient directions are found, examples easily follow. As “directions” are identified with unit vectors, treat them as points {x_{j}} on a unit sphere, S^{k−1}, with center x. (So, x_{j} defines the “vector” or “direction” x_{j} − x.) For bliss-core point settings, the bliss point is positioned at x.

Gradient directions. Coalitions that prefer other alternatives (in k-dimensional issue space) to x can be found by passing a (k − 1)-dimensional “dividing plane” through x. If q points are on a side of this plane, this decisive coalition denies core status for x. Thus, the goal is to analyze all dividing planes passing through x by counting the points on each side.

[1] If no {x_{j}} points are on the (k − 1)-dimensional dividing plane passing through x, then, according to Prop. 1, there are no more than q − 1 points on a side.

Because an arbitrarily small change in the orientation of a plane defined by k − 1 linearly independent vectors can force all k − 1 points onto a specified side of its new position, [1] restricts the number of points on a side of its original orientation.

[2] A dividing plane with k − 1 linearly independent vectors can have no more than

(5.1)  s^{∗} = (q − 1) − (k − 1) = q − k

points on either side. Conditions [1] and [2] are used repeatedly to construct examples and prove the theorems. Indeed, because the number of points on either side can change only when a dividing plane rotates through a point, it follows from Prop. 1 that a sufficient condition for x to be a core point in a k-dimensional space is if it satisfies [1] for one dividing plane without gradient vectors and [2] for all (k − 1)-dimensional dividing planes where a dividing plane has, other than a bliss point, k − 1 of these vectors.

Because x is a core point for a q_{1}-rule if it is a core point for a q-rule, q_{1} > q, it suffices to create examples supporting the minimal q value. From Eq. 2.5, this requires assigning points so that half of the points not on a dividing plane are on each side of the plane. (Such a construction yields the bounds of the theorem.) A natural approach is to symmetrically position the n points on S^{k−1}. It is trivial to symmetrically locate n points on a circle (place them 2π/n radians apart), but it is difficult to do this for higher dimensional spheres. (For instance, even on S^{2}, what is the symmetric choice for n = 5?) To sidestep this problem, which plagues other mathematical concerns, a method is devised to capture the geometry needed for the theorems. Specific cases need to be considered before the general argument can be described.

One-dimensional issue space. I use k = 1 to introduce how [1] and [2] help create examples with a specified x as a core point. Start by positioning m points according to the “alternating rule” where the first point is placed to the right of x, the second to the left, the third to the right, etc. until all points are positioned and no two points occupy the same position. For even integer m, there are \frac{m}{2} points on each side of x. According to [1], this choice supports q ≥ [\frac{m}{2}] + 1 rules. For k = 1, a “dividing plane” is the point x, so [2] is not applicable. Choosing n = m as the number of voters, the example supports x as a nonbliss core point for all q ≥ [\frac{n}{2}] + 1 rules with an even number of voters. (The example is robust as the points can be slightly altered without changing the argument; e.g., the nonbliss core points is the interval defined by the closest point on either side of x.) If n = m + 1, so x is a bliss-core point for an odd number of voters where one voter’s gradient is zero, this construction supports all q ≥ [\frac{n−1}{2}] + 1 = [\frac{n}{2}] + 1 rules. (Because n is an odd integer, we have [\frac{n}{2}] = [\frac{n−1}{2}].)

Let ⌈x⌉ be the “round-up” function with value x iff x is an integer, and value [x] + 1 otherwise. For odd integer m, at most ⌈\frac{m}{2}⌉ points are on a side of x, so this construction supports all q ≥ ⌈\frac{m}{2}⌉ + 1 = [\frac{m}{2}] + 2 rules. If n = m, so x is a nonbliss core point, this example supports all q ≥ [\frac{n}{2}] + 2 rules. When n − 1 = m, so x is to be a bliss core point for an even number of voters, the example supports all q ≥ [\frac{n−1}{2}] + 2 = [\frac{n}{2}] + 1 rules. Observe that when n is even, we have [\frac{n−1}{2}] = [\frac{n}{2}] − 1. Actual preferences are found by using one of the methods described earlier. Because these examples evenly split the points not on a dividing plane, they support all q, n rules allowed by Theorem 1 for k = 1.

Two dimensions. The k = 1 “alternate side” construction is used for all k ≥ 2. This involves perturbing “degenerate” preferences where Plott’s construction is used when possible. The following geometry simplifies the perturbation argument.

For k = 2, we need to place m points on S^{1}, the circle with center x. Whatever the choices, some line through x misses all of them. Let this line be the x_{2}-axis and its orthogonal complement be the x_{1}-axis. (See Fig. 3.) Identify each point on the circle with a point on either the “left line” (the line x_{1} = −1) or the “right line” (x_{1} = 1). Namely, draw a ray from x through the point on the circle (e.g., the dot on the circle in Fig. 3) and determine where and on which line this ray intersects (e.g., the point “1” on the left line of Fig. 3). As a point on either line uniquely identifies a circle point, we can consider only points on the lines.

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• • • •  ⋆ ⋆ x_{1} x_{2} x 1 2  1^{∗} 2^{∗}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}• • • • ⋆ ⋆ ⋆x_{2}^{∗} x_{1}^{∗} x_{1} x_{2} x_{3} Figure 3. The two-dimensional case.

The “dual” for a left line point is where the line through this point and x intersects the right line. Denote each dual point by star. (See Fig. 3.) Because a dual point uniquely determines a regular point on the left side, a configuration can be described by points and dual points placed on the right line. Observe that a point and its dual lie on opposite sides of a dividing plane (e.g., in Fig. 3, 2 and 2^{∗} are on opposite sides of the 1–1^{∗} line) and that if a point is on a dividing plane, then so is its dual.

The points for a two-dimensional example must span a two-dimensional space; this requires m ≥ 3. Arbitrarily place ⌈\frac{m}{2}⌉ of them, \{x_{j}\}_{j=1}^{⌈m/2⌉}, as regular points (dots) on the right line. Plott’s configuration corresponds to where each dual point, denoted by a star, is placed on top of a regular point. Denote the dual point on top of x_{j} by x_{j}^{∗} where the regular point defined by x_{j}^{∗} is x_{⌈m2⌉+j} on the left line. If m is even, each regular point is covered; if m is odd, the uncovered regular point is the top one. For even values of m, a dividing line satisfying [2] is defined by one of these points. Because dual points are on regular points, the same number of starred and regular points are on each side of this dividing line. Thus, after translating dual to regular points, the same number of points are on either side.

To perturb Plott’s configuration (to generate a robust example), translate each x_{j}^{∗} slightly upwards from x_{j}. (See the figure on the right in Fig. 3.) For odd values of m, each bounded interval (defined by regular points) has a starred point. When m is even, each bounded interval has a starred point and one starred point is in the top unbounded interval.

Start the analysis with m odd and the x_{1} = 0 dividing plane. One side has ⌈\frac{m}{2}⌉ points so [1] imposes the restriction q − 1 ≥ ⌈\frac{m}{2}⌉, or q ≥ [\frac{m}{2}] + 2 rules. If a dividing plane passes through the top dot on the right line (see Fig. 3), then, by construction, precisely [\frac{m}{2}] dots and [\frac{m}{2}] stars are below this line. Consequently, there are precisely [\frac{m}{2}] right points below and [\frac{m}{2}] left points above the dividing plane. According to [2] and Eq. 5.1 where k = 2 and s^{∗} = [\frac{m}{2}], so far this example supports all q ≥ [\frac{m}{2}] + 2 rules.

Rotate the plane about x downwards. When the plane hits the next point, the dual point removed from the bottom side (it is on the new dividing plane) is balanced by the new regular point above the plane (the top point that defined the previous orientation of the dividing plane). Thus, s^{∗} keeps the same value. With the alternating assignment, each time the plane passes through a point, the point is removed from one side while a point of the opposite type (defining the previous orientation) now is on the upper side. After converting dual points to regular points, [\frac{m}{2}] points are above and below this dividing plane. Thus this construction provides examples for k = 2 core points for any q ≥ [\frac{m}{2}] + 2 when m is odd. If n = m (so n is odd and x is a nonbliss core point), this construction supports all q ≥ [\frac{n}{2}] + 2 rules. For n = m + 1 is even and x is a bliss core point, this construction supports all q ≥ [\frac{m}{2}] + 2 = [\frac{n−1}{2}] + 2 = ([\frac{n}{2}] − 1) + 2 = [\frac{n}{2}] + 1 rules.

For even values of m, this argument shows that when the dividing plane passes through a point, there are ⌈\frac{m−1}{2}⌉ = [\frac{m}{2}] points on one side and [\frac{m}{2}] − 1 points on the other. According to Eq. 5.1 and [2], the associated bounds are q ≥ s^{∗} + k = [\frac{m}{2}] + 2. In terms of n, when n is even and the core is not a bliss point (so n = m), this construction supports all q ≥ [\frac{n}{2}] + 2 rules. When n is odd and the core is a bliss point (so n = m + 1) then, because [\frac{m}{2}] = [\frac{n−1}{2}] = [\frac{n}{2}], this construction supports all q ≥ [\frac{m}{2}] + 2 = [\frac{n}{2}] + 2 rules.

In this construction, which holds for all k, dual points are rotated off of Plott’s diagram in the same direction. (This is not the only possibility.) Similarly, for all k the four cases are when m is odd or even and x is or is not a bliss point (i.e., when n is even or odd). A factor always changing the outcome is that [\frac{n−1}{2}] = [\frac{n}{2}]−1 when n is even; this computation occurs with an even number of voters and a bliss-core point. A simple argument (starting with Eq. 2.5) shows this perturbation construction provides k = 2 supporting examples for all of the n, q rules admitted by Thm. 1.

Three-dimensions. Let x be at the origin of an axis system. Instead of placing points on the sphere S^{2}, immediately place them on left and right planes defined, respectively, by x_{1} = −1, x_{1} = 1. Restrict the construction by using the unit circle, S^{1}, in the right plane with center (1, 0, 0).

A true k = 3 example requires the convex hull to have a positive three-dimensional volume; i.e., m ≥ 4. As indicated in Fig. 4, place ⌈\frac{m}{2}⌉ regular points in an arbitrary fashion on S^{1}. For boundary (Plott) preferences, place x_{j}^{∗} on x_{j} where x_{j}^{∗} is the dual of x_{⌈ \frac{m}{2} ⌉+j} from the left plane. If m is even, all regular points are covered; if m is odd, x_{⌈ \frac{m}{2} ⌉} is left uncovered. Using the x_{1} axis as an axis of rotation (this axis is orthogonal to the right plane and it passes through the center of S^{1}), rotate all starred points a slight amount in a fixed direction. As the convex hull of the regular points has same number of edges as vertices, the shift defines an “alternating rule” procedure where each dual point is in an arc of the circle defined by an edge. If m is even, there are as many arcs as dual points, so each arc has a dual point. If m is odd (as in Fig. 4), one arc has no dual point. Observe that even after slightly perturbing this new configuration in any direction (even moving points off the circle), this construction holds where dual and regular points alternate as determined by the connecting cords. Also notice that after converting to regular points, m = 4 corresponds to a tetrahedron.

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Figure 4. Points and dual points.

As [1] is satisfied for x_{1} = 0, to analyze this configuration for even values of m ≥ 4, consider a dividing plane defined by two of the positioned points, p_{1}, p_{2}. Choose p_{1} as an axis of rotation for a dividing line. On either side adjacent to p_{1} are two points denoted by n_{1}, n_{2}. By the alternating rule, both points have the type opposite that of p_{1}, say they are dual points. Either n_{1} or n_{2} along with p_{1} defines a face of the convex hull, so all other points, \frac{m}{2} − 1 of each kind, are on one side of this plane. After converting dual points to regular points, this places \frac{m}{2} − 1 points on each side.

Let the “inner” or “lower” side of the n_{1}, p_{1} plane be the side with the convex hull. Rotate the dividing plane (about p_{1}) off of the chosen n_{j} inwards to the next point of the hull; by the alternating rule, it is a regular point. This new orientation removes a regular point from the “inner” side (it now is on the dividing plane), but adds a dual point to the “outer” side to keep the same number of regular points on each side of the new dividing plane. Because points of different type alternate on the circle, this argument continues. Consequently, with the final orientation of the dividing plane at p_{2}, after dual points are converted back to regular points, there are \frac{m}{2} − 1 points on each side.

According to Eq. 5.1, this construction supports q ≥ ([\frac{m}{2}] − 1) + 3 = \frac{m}{2} + 2. Therefore, if n = m (so x is not a bliss point) it supports all q ≥ [\frac{n}{2}]+2 rules in a three-dimensional issue space. If n = m + 1, so n is odd and x is a bliss point, because [\frac{n}{2}] = [\frac{n−1}{2}] this construction supports all q ≥ [\frac{n}{2}] + 2 rules. Again, this configuration supports all q and n values which, according to Thm. 1, admit a three-dimensional issue space with β < 0.

The only change for odd values of m ≥ 5 is to consider the effect of the sector without a dual point. By choosing the dividing plane to pass through these two regular points, all [\frac{m}{2}] dual points and the remaining ⌈\frac{m}{2}⌉ − 2 = [\frac{m}{2}] − 1 of the regular points are on one side. Consequently, one side of this dividing plane has [\frac{m}{2}] points (and the other side has [\frac{m}{2}]−1). Using the alternating point construction, it follows that this is true for any dividing plane defined by two points. Therefore, from [2] we have that q ≥ [\frac{m}{2}] + 3. So, if m = n (that is, when x is not a bliss point and n is odd), this example supports all q ≥ [\frac{n}{2}]+3 rules. Where x is to be a bliss-core point, n is even, and m = n−1, the relationship [\frac{m}{2}] = [\frac{n−1}{2}] = [\frac{n}{2}]−1 shows that this construction supports all q ≥ [\frac{n}{2}] + 2 rules. Again, it is easy to show that these examples support all rules described in Thm. 1 with β < 0..

General case. For k ≥ 4, the larger dimension of the unit sphere in the right plane, S^{k−2}, and added orientations make it more difficult to describe a configuration but easier to show that appropriate perturbations of x_{j}^{∗} off of x_{j} are possible. This is because the argument now is identified with the algebraic comparison of the number of equations and the number of unknowns. I use a m = 2k construction to indicate the source of the equations and one of many possible solutions.

Place k regular points near the vertices of an equilateral k-gon in the right plane; e.g., for k = 4, the points are near the vertices of an equilateral tetrahedron. The regular points are fixed, so the variables are z_{j} = x_{j}^{∗} − x_{j}, j = 1, …, k with initial values z_{j} = 0. Notice that the regular points are nearly on the orbit of the permutation τ = (1, 2, …, k) for an appropriate initial point x_{1} ∈ S^{k−2}. (Hence τ^{j}(x_{1}) = x_{j+1} where x_{1} is chosen so that the angle between x_{1}, τ^{j}(x_{1}) has the correct value; e.g., for k = 3, select x_{1} so its angle with τ(x_{1}) is 120^{o}.) The placement of dual points is similar to how the τ orbit defines the Condorcet triplet from social choice ([Chap. 3.1, S1]).

With k − 1 ≥ 3, the vector cross product is replaced with the wedge product of (k − 1) vectors; e.g., see [W]. Let y_{j} = x_{j} − x, y_{j+ \frac{m}{2}} = x − x_{j}^{∗}, j = 1, …, \frac{m}{2}, (observe the difference in orientation), and let n be the wedge product of k − 1 of these vectors. If n does not involve x_{j}, x_{j}^{∗} pairs, n is a non-zero k − 1 form indicating a normal direction to the dividing plane spanned by these vectors. A x_{j}, x_{j}^{∗} pair used to define n, however, adds a degree of degeneracy in n’s functional form when z_{j} = 0. This degeneracy represents the linear dependency of these vectors, so the space spanned by the vectors drops a dimension.

For those y_{j} not used in n, the sign of the wedge product n∧y_{j} determines the orientation of y_{j} relative to n. The orientation difference in defining y_{j} for a dual point requires sign of n ∧ y_{j} to differ from n ∧ y_{j+ \frac{m}{2}} even for z_{j} = 0. (The orientation is that of x_{j+ \frac{m}{2}} .) This choice is equivalent to assigning −1 to dual points and 1 to regular points. As there are as many dual as regular points, the sum of these signs over all m points is zero. To satisfy [1] and [2], the sums of signs from the wedge product computation for points on each side of a dividing plane can differ by no more than unity. The choices of zero or unity depends on the parity of the k and m values.

For a configuration where n is non-degenerate and all z_{j} = 0, the sum of signs on each size of the dividing plane is zero. (One side has no points and the other has a dual point on its mate.) We need to perturb z_{j} so that all n choices are non-zero and the counting condition is satisfied. In moving dual points, interpret “above” – for a plane containing a face of the hull defined by the regular points – as designating the side of the plane without the hull. To illustrate with an equilateral triangle, the “bottom” side of a line through two vertices is the side with the triangle. Thus, “above” and “below” are relative to the choice of points.

As the available degrees of freedom to move x_{j}^{∗} correspond to the dimension of the tangent space T_{x_{j}}S^{k−2}, perturbing all z_{j} terms defines a vector in ∏_{j=1}^{k} T_{x_{j}}S^{k−2} which has dimension k(k − 2). (We could use T_{x_{j}}R^{k−1} to allow changes inside of the S^{k−2}. While such changes are admissible, add flexibility to the argument (corresponding to more variables), and are expected after determining the basic location of the dual points, we don’t need the extra dimensions.) For n defined by regular points, move the dual points so that [\frac{k−1}{2}] of them are over each face of the k-gon. Over such a plane, the sum of signs is −[\frac{k−1}{2}] (corresponding to the [\frac{k−1}{2}] dual points) and below it the sum is −(k − [\frac{k−1}{2}]) + 1 = −(1+⌈\frac{k−1}{2}⌉)+1 = −⌈\frac{k−1}{2}⌉ corresponding to the k−[\frac{k−1}{2}] dual points and one regular point. So, when k is even, both sums agree; when k is odd, they differ by unity. Because each plane has k − 1 dual points, and because this is an open condition for each dual point, and because each dual point can be over j planes, 1 ≤ j ≤ k − 2, this construction is admissible if k − 2 ≥ [\frac{k−1}{2}], or if k ≥ 3.

One solution is to move x_{1}^{∗} in direction v_{1} (the first order representation for z_{1}) so that x_{1}^{∗} is off the plane defined by n and over [\frac{k−1}{2}] of the planes but below all others. As τ corresponds to a linear mapping, τ^{j}(v_{1}) defines a tangent vector for T_{x_{j+1}^{∗}}S^{k−2}, j = 1, …, k. By symmetry, τ^{j}(v_{1}) moves x_{j+1}^{∗} over and under the specified number of dividing planes defined by regular points. (I show below that the requisite number of dual points are above each plane.) Conversely, as this construction is equivalent to moving the regular points off of the dual points (i.e., move x_{j} off of x_{j}^{∗} in direction −v_{j}), the same condition holds for dividing planes defined solely by dual points.

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To illustrate with k = 4 (Fig. 5), the regular point configuration is, essentially, an equilateral tetrahedron. Dual point x_{1}^{∗} could be moved over any one of three faces x_{1}, x_{2}, x_{3}, or x_{1}, x_{3}, x_{4}, or x_{1}, x_{3}, x_{4}. Choosing the first one places x_{1}^{∗} below the other two. With this choice, τ moves x_{2}^{∗} over x_{2}, x_{3}, x_{4}, point x_{3}^{∗} over x_{3}, x_{4}, x_{4}, and x_{4}^{∗} over face x_{4}, x_{1}, x_{2}.

For computations, instead of computing the orbit {τ^{j}}, treat τ = (1, 2, …, k) as having k equivalent forms where the jth form starts with j and retains the cyclic ordering; e.g., the second form is (2, 3, …, k, 1). Move x_{1}^{∗} over [\frac{k−1}{2}] faces in the following manner. Let the first two τ terms, 1 and 2, identify the x_{j} subscripts for a common edge of several faces. The remaining subscripts defining the first face are the first k − 3 terms of the permutation σ_{1} = (3, 4, …, k) obtained from τ by dropping the first two τ entries. To find the rest of the faces, use the same edge and let the remaining subscripts for the faces come from [\frac{k−1}{2}] of the equivalent forms of σ_{1} starting with σ_{1}, then (4, 5, …, k, 1), (for the second face), …. This places x_{1}^{∗} over the specified faces and below all others.

To move x_{j}^{∗}, use the jth form of τ = (j, j + 1, …, 1, …, j − 1). The first two terms, j, j + 1, define the common edge for the faces; the remaining subscripts are the first k − 3 entries of the first [\frac{k−1}{2}] forms of σ_{j} = (j + 2, j + 3, …, j − 1), (j + 3, …, j − 1, j + 2), . . .. Clearly, this permutation argument represents the v_{j} choices selected earlier and all faces of the k-gon are represented. Also, this procedure recovers the k = 4 construction. To place regular points over the faces of the k-gon defined by dual points, use the inverse permutation τ^{−1} = (k, k − 1, …, 2, 1).

While a group theory exercise can be used to verify that [\frac{k−1}{2}] dual points are above each face, the following geometric argument emphasizes the symmetry and the requirement that each x_{j}^{∗} is above [\frac{k−1}{2}] faces and below all others. (Note for k = 4 (Fig. 5) that only [\frac{4−1}{2}] = 1 dual point is above each plane.) From symmetry, this condition only needs to be checked for one face, say, x_{1}, x_{2}, …, x_{k−1}. Observe that x_{j}^{∗} is above this face only if one of the [\frac{k−1}{2}] admissible forms of σ_{j} has the index k at the end. This is true iff k is either the last or one of the first [\frac{k−1}{2}] − 1 entries of σ_{j}. (For instance, k is the last entry for σ_{1}, so x_{1}^{∗} is above this plane.) The conclusion follows from a simple computation.

To illustrate with k = 6 by positioning x_{2}^{∗} with the σ_{2} forms (4, 5, 6, 1), (5, 6, 1, 4), (6, 1, 4, 5), and (1, 4, 5, 6), observe that x_{2}^{∗} is moved slightly off of x_{2} onto S^{4} so it is above the x_{2}, x_{3}, x_{4}, x_{5}, x_{6} and x_{2}, x_{3}, x_{5}, x_{6}, x_{1} planes, but below all others. To illustrate which [\frac{6−1}{2}] = 2 dual points are above which planes, observe that only x_{1}^{∗} and x_{4}^{∗} are above the x_{1}, x_{2}, …, x_{5} plane. Point x_{2}^{∗} is not above this plane because 6 is neither the first nor last entry of σ_{2} = (4, 5, 6, 1), and x_{5}^{∗} is excluded because 6 is not in σ_{5} = (1, 2, 3, 4) (x_{6} is on the common edge for the faces). Similarly, only x_{3}^{∗}, x_{6}^{∗} are above the x_{3}, x_{4}, x_{5}, x_{6}, x_{1} plane.

Now consider a plane which, before the perturbation, coincides with a face of regular points, and after the perturbation it is defined by ⌈\frac{k−1}{2}⌉ regular points and the [\frac{k−1}{2}] dual points lifted above this face. By simple geometry using the facts that the dual points associated with the remaining regular points are below this plane but near the regular point and by symmetry (given by v_{j} = τ^{j−1}(v_{1})), this dividing plane has no points on one side of it. As this plane contains a face of the hull defined by the regular and perturbed dual points, there are k − ⌈\frac{k−1}{2}⌉ regular and k − [\frac{k−1}{2}] dual points on the other side. This difference between the sums of signs is either zero or −1 depending on whether k is odd or even. Also, any dividing plane obtained by going from the face of regular points to the face of the hull of dual and regular points involves an alternating argument where each regular point is replaced with its dual mate. This adds +1 to the side with the new regular point, but removes a −1 from the side losing a dual point so the difference between sums of signs remains the same.

So far, only one condition is imposed on v_{1}; other conditions (i.e., other equations) result from analyzing dividing planes with dual points below a particular face. Start with an “edge” or axis of rotation consisting of k − 2 regular points. Adjoining one the remaining two regular points to this edge defines a face of the k-gon. There are two kinds of edges; a nonsingular edge is where the total number of dual points above both faces is 2[\frac{k−1}{2}] while a “singular edge” has a smaller number. It is not difficult to show that all edges are nonsingular for k ≤ 5 and that singular edges always occur for k ≥ 6. (Singular edges are created when the same dual points are above both planes sharing that axis.)

A nonsingular edge, has k−2[\frac{k−1}{2}] dual points between (i.e., below) both faces. This value is 1 or 2 depending, respectively, on whether k − 1 is even or odd. The former case, where the sum of signs on each size of a dividing plane agree, has only one dual point between faces. Thus the alternating structure and the “equality of sums” property hold. For the latter setting, where the sums of signs differ by unity, notice that the side of a dividing plane with the larger value of the two sums is “over” a face. Therefore, when the dividing plane rotates, at some position the side with the larger sum changes. This is the role played by the two dual points between the two faces; it preserves the summation property. The only problem (i.e., an extra equation to be satisfied) is to avoid the degeneracy where the plane passing through the axis of rotation and one dual point meets the second dual point. (In the left diagram in Fig. 6, for instance, there are x_{2}^{∗}, x_{3}^{∗} choices where the plane passing through the x_{1}, x_{2} axis and x_{2}^{∗} contains x_{3}^{∗}. According to the assignment process, this degeneracy arises only with an axis containing an edge assigned to a dual point.) The equation (i.e., let n be the wedge product of the k−2 points on the axis and a dual point x_{i} −x+τ^{i−1}(v_{1}) and then set n ∧ (x_{j} − boldx + τ^{j−1}(v_{1})) equal to zero) represents a zero determinant expressed in terms of the k − 1 variables (coordinates) of z_{1}. Thus this lower dimensional degeneracy is avoided by almost all v_{1} values. As a specific choice, move x_{1}^{∗} closer to the x_{1}, x_{2} edge than any other edge. (For instance, this requires x_{2}^{∗} to be over the x_{2}, x_{3} edge so that the line through x_{2}, x_{2}^{∗} is closer to x_{3} than x_{3}^{∗}.) By symmetry arguments, the rotating plane passes through each dual point separately and preserves the sum condition. This situation is displayed in Fig. 6 where the rotation is about the x_{1}, x_{2} and about the x_{2}, x_{4} axis.

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Viewing k = 4 from edges.

To illustrate by rotating a plane about the x_{2}, x_{4} edge (the bottom vertex of the shaded figure on the right in Fig. 6), the succession of points is x_{2}^{∗}, x_{3}, x_{3}^{∗}, x_{1}^{∗}, x_{1}, x_{4}^{∗}. (As the x_{2}, x_{4} edge is not assigned to a dual point, the permutation argument does not allow a degeneracy.) The necessary transition between which side has the larger sum occurs when the dividing plane passes between x_{3}^{∗} and x_{1}^{∗}. The figure on the left shows the transition about an axis corresponding to an edge in the assignment of faces for dual points. (This is where a degeneracy condition occurs because the “next” τ dual point is near the “next” τ edge.) The rotating dividing plane passes through x_{4}^{∗}, x_{4}, x_{3}^{∗}, x_{2}^{∗}, x_{3}, x_{1}^{∗}.

The restrictions (i.e., new equations) imposed by a singular edge occur because with only 2[\frac{k−1}{2}] − α points above the two faces, there are k − 2[\frac{k−1}{2}] + β dual points between them, α, β > 0. But the extra dual points (β > 0) between the faces jeopardize the alternating count procedure. To rectify this problem, we first show that β = α. This equality follows because to have α dual points above both faces, they are in a sector such as the one indicated in Fig. 7a. (The lower vertex of the shaded region is the singular edge. This construction requires k − 2 ≥ 2 because lower dimensions would require a point in this sector to be outside S^{k−2}. This edge is the hypersurface connecting the indicated S^{k−2} points, and the sector is the region between the planes and the sphere – or the surface of the sphere.) The wedges on the left and right sides have [\frac{k−1}{2}] − α dual points. There are the k − [\frac{k−1}{2}] dual points below the face represented by the downward slanting line. Adding the points in the two sectors on this side of the line, we obtain \{k − 2[\frac{k−1}{2}] + β\} + \{[\frac{k−1}{2}] − α\}. The α = β conclusion follows by comparing values.

As there are k −2[\frac{k−1}{2}] more points (either one or two) between the faces than over both, it is clear how to preserve the alternating rule (i.e., solve the added equations); position the α dual points so that when the dividing plane rotates, it alternates passing through a point between the planes and one of the α points above both planes. Because dual points from the two different regions are moved onto different sides of the rotating dividing plane, the alternating structure is preserved. A dimensional (each x_{j}^{∗} is over at most [\frac{k−1}{2}] faces, so these one-dimensional restrictions occur at most [\frac{k−1}{2}] − 1 times) and symmetry argument (each τ^{j}(v_{1}) vector emphasizes different components of v_{1}) shows that such a construction is possible. (Specific choices determine an ordering the faces that x_{j}^{∗} is over in terms of the distance of this point to each plane; this positions the dual point in a sector.) When k − 1 is odd, there are two more dual points between the faces which, as indicated in Fig. 6 for k = 4, are needed to change the side with the larger sum.

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^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} ^{.} •  • • • • • x_{1}, x_{2}, x_{4}, x_{5} x_{2}^{∗} x_{5}^{∗} x_{4}^{∗} x_{1}^{∗} x_{3}^{∗}  x_{6}^{∗} x_{3}  x_{6} Figure 7. A k = 6 singular edge.

To illustrate with the figure to the right in Fig. 7, the plane with indicated axis passes through the dual points in the order x_{6}, x_{5}^{∗}, x_{4}^{∗}, x_{6}^{∗}, x_{3}^{∗}, x_{1}^{∗}, x_{2}^{∗}, x_{3}. (The x_{5}^{∗} point comes before x_{6}^{∗} because the line between x_{5}, x_{5}^{∗} is closer to x_{6} than x_{6}^{∗}. This construction, then, imposes restrictions on other dimensions – distance from planes – between x_{4}^{∗} and x_{5}^{∗}.) Because dual points are chosen from opposite sides of the axis of rotation, this preserves the alternating structure needed for the construction. This argument concerns only those dual points above both faces, there are more variables than conditions to satisfy.

This argument easily is modified to handle an axis of rotation with dual points. If the axis has x_{j}^{∗}, start with x_{j} and obtain the alternating rule established above. By construction, before moving the dividing plane from x_{j} to x_{j}^{∗}, the difference between sums of signs satisfies the desired property. One side of this plane has x_{j}^{∗}; it adds −1 to the summation. By moving the plane off of x_{j} and toward x_{j}^{∗}, +1 is added to side that now has x_{j}. When the plane reaches x_{j}^{∗}, the −1 value of −1 is removed. Thus, the sum on both sides increase by +1, so the difference in sums remains the same. Observe that this argument only works for [\frac{k−1}{2}] dual points. However, using the symmetry between dual and regular points, the same argument applies should we start with an axis consisting of all dual points, rather than regular points. Thus, for any number of dual and regular points on the dividing plane, the conclusion about the difference between the sums of signs remains.

A n degeneracy created by a x_{j}, x_{j}^{∗} pair already is partially resolved because they occur for various orientations of the rotating dividing plane. (For instance, in Fig. 6, one dividing plane has the x_{2}, x_{2}^{∗} pair.) To resolve the remaining situations, use the fact that x_{j}^{∗} is near x_{j} and the line defined by these two points passes closer to x_{j+1} than x_{j+1}^{∗} but over the x_{j}, x_{j+1}, x_{j+2}, … plane. The first assumption means that only a small perturbation is made in changing from the regular to the partner dual point. The second, which provides one dimension for the orientation of a dividing plane, allows x_{j}, x_{j}^{∗} lines to be replaced, initially, by x_{j}, x_{j+1} directions and then perturbed. The only issue is the side containing x_{j+1}^{∗}, and this is decided by the initial assumption. This completes the construction.

From this construction, s^{∗} = ⌈\frac{k+1}{2}⌉, so this configuration supports all q ≥ k + ⌈\frac{k+1}{2}⌉ = ⌈\frac{3k+1}{2}⌉ rules for n = 2k where x is a nonbliss core point and n = 2k + 1 where x is a bliss core point. To illustrate with k = 4, these are all q ≥ 7 rules for n = 8 and x a nonbliss core point, and n = 9 and x a bliss core point. For k = 50, this supports all q ≥ 71 rules for n = 100 and x a nonbliss core point.

The same construction and reasoning holds for all remaining cases, so I emphasize only differences. The first step, moving a specified number of dual points above each plane, is an open condition. The purpose of the remaining equations is to avoid degeneracies; they are related to singular edges. As no more than [\frac{k−1}{2}] − 1 equations are assigned to each dual point, and each equation involves one of the k −1 available coordinates, standard dimension arguments (of the n-equations, m unknowns, m ≥ n type) about algebraic degeneracy conditions ensure the conclusion. In fact, because not all of the available variables are used, k ≥ 4 admit alternative constructions with the desired behavior. (Indeed, this argument essentially uses only k − 2 of the available k(k − 2) dimensions. More important for the alternating behavior are the regions created by the planes and by placing points closer to one plane than another as used for singular edges.) Other settings use some of these extra sectors defined by the various planes.

To start with m = 2k − 1, remove x_{k}^{∗} which is over the [\frac{k−1}{2}] faces with x_{j} subscripts (k, 1, 2, 3, …, k − 2), (k, 1, 3, 4, …, k − 1), …, (k, 1, [\frac{k−1}{2}] + 1, …, [\frac{k−1}{2}] − 1). As we need [\frac{k−1}{2}] dual points over each face, compensate for the loss of x_{k}^{∗} by moving x_{1}^{∗} so it also is over the first x_{k}^{∗} face, x_{3}^{∗} so also it is over second x_{k}^{∗} face, …, and x_{[ \frac{k−1}{2} ]+1}^{∗} so it also is over the last x_{k}^{∗} face. Standard linear algebra arguments ensure that this construction is possible. The analysis differs from m = 2k only in that singular edges start with k = 4 (because x_{1}^{∗} is over two faces sharing the x_{1}, x_{2} edge). This singular edge construction just involves choosing positions (how close to what plane) for dual points that are above both faces so they alternate with the dual points between the surfaces. (This singular edge manifests the loss of the z_{k} variables.)

The analysis remains essentially the same; e.g., a plane passing through all regular points has [\frac{k−1}{2}] dual points above, and (k − 1) − [\frac{k−1}{2}] dual points and a regular point below it. The same dimensional arguments for the singular edges apply to preserve the alternating sign convention. By comparing sums we have, again, that the difference between sums is zero or unity, depending, respectively, on whether k − 1 is odd or even. As this construction defines s^{∗} = [\frac{k−1}{2}] + 1 = ⌈\frac{k}{2} ⌉, it supports all q ≥ k + ⌈\frac{k}{2} ⌉ = ⌈\frac{3k}{2} ⌉ rules for n = 2k − 1 and x a nonbliss core point, or n = 2k and x a bliss core point.

To illustrate with k = 4 and m = 7, to replace x_{4}^{∗} move x_{1}^{∗} above the x_{1}, x_{2}, x_{4} plane so that the x_{1}^{∗}, x_{1}, x_{2} dividing plane separates x_{2}^{∗} and x_{3}^{∗}. (This singular edge construction places x_{1}^{∗} close to the x_{1}, x_{2}, x_{3} plane. In Fig. 5, x_{1}^{∗} is in the wedge defined by extending the bottom and left side planes with x_{1}, x_{2} as a singular edge. In left figure of Fig. 6, x_{1}^{∗} is in the small wedge with dashed lines.) The same “alternating side” argument, or sum of signs, shows that (after dual points are converted) s^{∗} = 2, so this configuration supports all q ≥ ⌈\frac{m−3}{2}⌉ + 4 = 6 rules when n = m = 7 and x a nonbliss core point, and n = m + 1 = 8 where x is a bliss core point.

The argument for m > 2k an even integer mimics the k-gon construction. In the twodimensional subspace of the right plane defined by y_{1} = 1, y_{2}, y_{3}, place c = \frac{m}{2} − (k − 3) points (roughly equally spaced) on the unit circle with center (1, 0, 0); this two-dimensional configuration, C(3), has c vertices and c faces. To define a three-dimensional configuration C(4) by adding the dimension y_{4} (so the configuration is in the three-dimensional subspace y_{1} = 1 of a four-dimensional space), in the unit sphere in this subspace, place a (rescaled) copy of C(3) on the intersection of the y_{4} = t < 0 plane with the sphere, and place an added point near where the positive y_{4} axis passes through the sphere. This figure has c + 1 vertices and faces. The number of vertices is obvious; c faces come from connecting each lower dimensional C(3) face with the new vertex; the last face is the hull C(3) which becomes a face in the higher dimensional setting. Notice that c = 3 and an appropriate t choice returns the equilateral tetrahedron used with k = 4.

To extend C(α − 1) to C(α), increase the subspace dimension by including coordinate y_{α}. Place a rescaled copy of C(α − 1) on the sphere defined by the intersection of the y_{α} = t < 0 plane with the unit sphere. Add a point near where the positive y_{α} axis passes through the sphere. Again, the number of faces and vertices of C(α) is c + α − 3 because the new vertex and each lower dimensional C(α − 1) face defines a C(α) face, and the hull C(α − 1) becomes a C(α) face. Thus C(k) has c + (k − 3) = \frac{m}{2} faces and vertices. When c = 3, the t values at each stage of the construction can be selected to obtain the equilateral k-gon.

A degeneracy, created by placing c > 3 points on a two-dimensional subspace, needs to be broken to ensure robust examples and to introduce additional regions (generated by the intersections of the various planes) for dual points. If t_{j} is the y_{k} coordinate of x_{j} in C(k), we have t_{j} = t for j < \frac{m}{2}. For a small ϵ > 0 value, change the t_{j} values so that t = t_{1} < t_{2} < ⋯ < t_{\frac{m}{2} −1} < t + ϵ < 0, and denote this configuration by D(k). Denote by D(3) the hull defined by the perturbation of the first c points. To show that the dual points can be perturbed off of the vertices of D(k) in the desired manner, dimensional arguments are used (but illustrated with examples). While C(k) has \frac{m}{2} faces and vertices, D(k) has \frac{m}{2} vertices and c + (c − 2)(k − 3) faces. (At each stage in the argument computing the number of faces for C(k), C(α − 1) becomes a face for C(α). The stated number arises by computing how this face divides after the perturbation.) More important than the number is that with a small ϵ value D(k) is essentially the same as C(k).

An alignment with [\frac{k−1}{2}] dual points above each D(k) face has a −[\frac{k−1}{2}] sum for these points. Below the face are the (k − 1) − [\frac{k−1}{2}] = ⌈\frac{k−1}{2}⌉ dual points that are moved off of a vertex of the face, and \frac{m}{2} − (k − 1) pairs of dual and regular points. The sum of signs from each pair is zero, so the total sum of signs is −⌈\frac{k−1}{2}⌉. Thus, the difference between sums is zero when k − 1 is even, and unity when k − 1 is odd.

The argument about moving a dividing plane by replacing regular points with dual points above this face is as given earlier. Thus we only need consider dual points that are below faces. So, consider an axis defined by k − 1 regular points. Adding another regular point defines a plane; for two choices of an added regular point the plane contains a D(k) face. Between these extreme faces are \frac{m}{2} − k regular points and their dual mates. As these points cancel in the sum of signs, the major concern involves dual points with a regular point mate on one of the extreme planes. The earlier singular edge argument shows that with less than 2[\frac{k−1}{2}] dual points over both faces, α = β. Consequently, ignoring pairs of regular and dual points, the difference between the number of dual points between and above both planes is either one or two. Therefore, the earlier alternating structure argument applies. Thus, the construction is possible if [\frac{k−1}{2}] points can be placed over each face.

Notice that each face is defined by k − 1 points and each x_{j}, j ≠ 1, is a vertex for at least k − 1 faces. Point x_{1} is a vertex for k − 1 C(k) faces and all the new D(k) faces obtained by breaking the degeneracy. By choosing a small ϵ value, the dual points above both D(3) and D(k) can be above several faces (recall, for a sufficiently small ϵ value, D(3) is essentially flat) while remaining near the regular point mate. The fact that it is possible to move the dual points so that over each D(k) face there are precisely [\frac{k−1}{2}] dual points is an immediate consequence of the observation that if x_{j} is a vertex for b faces, then x_{j}^{∗} can be moved over any number of faces between one and b − 1. It cannot be over all b faces as this would place the dual point outside of S^{k−2}. This is reflected by the earlier construction where (k −1)−1 faces were identified by considering only those faces passing through the x_{j}, x_{j+1} edge.

With this construction, s^{∗} = ⌈\frac{m−(k−1)}{2}⌉, so it supports all

(5.2)  q ≥ k + ⌈\frac{m − k − 1}{2}⌉

rules when n = m and x is a nonbliss core point, or n = m + 1 and x is a bliss core point.

To illustrate with k = 4 and m ≥ 10, place x_{1}^{∗} above x_{1}, x_{2}, x_{3} face, x_{j}^{∗} above x_{2}, x_{3}, x\frac{m}{2} face, x_{j}^{∗} above both x_{j}, x_{j+1}, x\frac{m}{2} and x_{1}, x_{j}, x_{j+1} faces, j = 3, …, c − 1, x_{c}^{∗} over x_{c}, x_{1}, x\frac{m}{2} , and x∗_{\frac{m}{2}} over the x\frac{m}{2} , x_{1}, x_{2} face. Each face has [\frac{4−1}{2}] = 1 dual point over it. The only singular edge constructions involve the x_{j}, x_{j+1} edges for j = 3, … c − 1. It is easy to see that this condition is satisfied should x_{j}^{∗} also be “above” relative to D(3) the x_{c}, x_{j}, x_{j+1} plane. that this can be done is obvious from the geometry. This is illustrated in Fig. 8 with c = 5 where the shaded region is D(3) from the perspective of the side facing x\frac{m}{2} . The dagger, representing x_{3}^{∗}, is placed in the narrow (because ϵ has a small value) sector between the extensions of the two planes so it is above both of them. (As the differences between D(k) and C(k) arise in D(3), the singular edge arguments involve D(3).)

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•  •  • •  •  • Figure 8. Position of x_{3}^{∗} in D(4) for m = 12 x_{1}  x_{2} x_{5}  x_{3} x_{4}  †  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The case where m > 2k is odd is done by dropping x∗_{\frac{m}{2}} and using the same counting argument. Again, Eq. 5.2 holds, but with the smaller m value.

Finally, we need to consider the special cases where m satisfies k + 1 ≤ m < 2k. These special constructions are in R^{k} rather than in the right plane, and only regular points are used. For m = k + 1, place the points near the vertices of an equilateral (k + 1)-gon on S^{k−1} where x is at the center of the sphere. A dividing plane consists of k − 1 of these points, so s^{∗} = 1. Thus, this construction supports all q ≥ k + 1 rules for n = m (i.e., only the unanimity rule) and x a non-bliss core point, or n = m + 1 = k + 2 (so q ≥ n − 1) and x a bliss core point.

For m = (k + 1) + j where 1 ≤ j ≤ k − 2. On j of the faces of the (k + 1)-gon, attach another (k+1)-gon. (So, for k = 3 and j = 1, this becomes two tetrahedrons glued along one of the equilateral triangle faces.) Thus, j pairs of vertices are opposite each other. Adjust the length of the distances so all points are on S^{k−1} and the distances between opposing vertices is symmetric. Let x be near the center. (So, for the k = 3, j = 1 illustration, x is near the center of the equilateral triangle used to glue the two tetrahedrons together.) If x is not on the plane defined by k points, then it follows that at most ⌈\frac{m−(k−1)}{2}⌉ points are on a side of a dividing plane. Thus, this configurations supports all q ≥ k + ⌈\frac{m−k+1}{2}⌉ rules for n = m and x a nonbliss core point, or n = m + 1 and x a bliss core point.

6. Upper bounds

The examples of Sect. 5, examples support the assertions of Thm. 1 for β = −1. It remains to show that these examples are robust, to construct β ≥ 0 examples, to show that the bounds of Thm. 1 are tight (i.e., that one cannot do better), and to prove Thm. 2a. All are done with singularity theory.

The reader unfamiliar with this important tool can view singularity theory (e.g., [GG] or [SS]) as a sophisticated implicit function theorem. Let f : R^{n} → R^{m}, be a smooth mapping and Σ a smooth b dimensional manifold in R^{m} (so, the codimension of Σ is m − b). According to the implicit function theorem, if f satisfies appropriate conditions, then, at least locally, f^{−1}(Σ) is a codimension m − b (or dimension n − (m − b)) submanifold of R^{n}. The needed transversality condition for x where f(x) ∈ Σ is that the span of the tangent spaces D_{x}f(R^{n}) and T_{f(x)}Σ is R^{m}. (The usual inverse function theorem where Σ is a point only requires the D_{x}f(R^{n}) to span R^{m} – to have rank m – as the tangent space of a point is the zero vector.)

To use this tool to analyze first and second derivative conditions imposed upon a function f, use the “jet” map. This is the mapping j^{2}f(x) = (x, f(x), Df_{x}, D_{x}^{2}f). The domain of this mapping is R^{n}, and the range is J^{2} = R^{n} ×R^{m} ×L(R^{n}, R^{m})×B(R^{n}, R^{m}) where L(R^{n}, R^{m}) are the linear maps from R^{n} to R^{m} and B(R^{n}, R^{m}) ≈ L(R^{n}, L(R^{n}, R^{m})) are the bilinear symmetric maps. If the derivative conditions define a manifold Σ in J^{2} for f = (u_{1}, …, u_{n}) where j^{2}f meets Σ transversely, then the implicit function theorem ensures that the set of points satisfying these conditions is (locally) smooth submanifold with codimension equal to the codimension of Σ. The difficulty is to verify the transversality condition.

This problem is resolved by extending the notion of a transverse intersection to allow j^{2}f to miss Σ. Then the Thom Transversality Theorem ensures for a generic set of functions, the intersection is transverse. (When issue space is restricted to a compact subset, this assertion holds for an open-dense set of functions [SS].) In words, the general situation is that the preferences either fail to satisfy, or do satisfy the core conditions robustly. Thanks to this important assertion, the proof of the theorem reduces to (1) representing the core derivative conditions of Prop. 1 in terms of a manifold Σ in J^{2}, (2) finding the codimension of Σ, and (3) showing that we are not discussing the empty set.

Task (3) requires showing that there exists a f where j^{2}f ∈ Σ. This is important because there are many examples where j^{2}f never meets Σ; they correspond to settings where the alternatives force the voters’ preferences to cluster in ways that do not satisfy Prop. 1. In other words, rather than asserting that the core always exists, Theorem 1 only ensures that there exist robust examples in the indicated issue spaces. It is important to note that we do not need to verify that j^{2}f meets Σ transversely. The assertion that there exist functions that do so, hence robust examples exist (not necessarily this particular f) is a gift from Thom. The idea is that if j^{2}f ∈ Σ is a boundary point, then, as Thom’s result has the conclusion holding for an open set about f, near-by functions satisfy the transversality condition. To illustrate, the examples of Sect. 5 satisfy the conditions, so from singularity theory, we know there is an open set of preferences with this property for the specified dimensions of issue space.

Proof of Thm. 1. For specified n and q, to construct a manifold Σ in L(R^{k}, R^{n}) (the space of linear maps from R^{k} to R^{n}), use the identification

L(R^{k}, R^{n}) ≈ {A = (A_{1}, A_{2}, …, A_{n}) | A_{j} ∈ R^{k}}.

Namely, think of A_{j} as a dummy variable where its range of R^{k} represents all possible choices for ∇u_{k}(x). An advantage of using this identification, rather than discussing properties of matrices, is that it is a simple parametric representation of a manifold in L(R^{k}, R^{n}) in terms of the A_{j} vectors. Since A_{j} is meant to capture conditions on ∇u_{j}(x), these Σ manifold definition follows directly from the core conditions on the gradients. (This representation is another way my method differs significantly from the approach used in [B, Sc, MS1].)

Bliss-core points. To start with bliss-core points, assume that the core point is the bliss point for the first agent. This means we must examine all points x where ∇u_{1}(x) = 0. All such situations are captured by the manifold Σ_{1} = {A | A_{1} = 0}. Specify the core conditions of Prop. 1 as CC_{m} = {A| there are m nonzero vectors A_{j} and the plane in R^{K} defined by any choice of k − 1 of them has no more than ⌈\frac{m−k+1}{2}⌉ vectors on either side \}. This is an open condition, and the examples of Sect. 5 show it is a nonempty condition. Because there are no restrictions on any other J^{2} coordinates, if there is any x ∈ R^{k} where ∇u_{1}(x) = 0, then j^{2}(u_{1}, …, u_{n}) ∈ Σ_{1}. The codimension of Σ_{1} ∩CC_{n−1} is k (reflecting that all k components of A_{1} are completely specified), so, generically, the set of x’s satisfying this condition is k − k, or zero-dimensional; it is a set of isolated points.

Imposing any other condition on Σ_{1} which increases the codimension, so it is larger than k, corresponds to where the set of points satisfying the conditions is generically empty. For instance, if the core point is the bliss point for two or more agents where one is the jth, j ≠ 1, then the manifold capturing this situation is Σ_{2} = {A | A_{1} = A_{j} = 0} with codimension 2k. Thus, the set of points satisfying this condition generically form a union of k − 2k dimensional manifolds. As this dimension is negative, this behavior is generically impossible. Generically, a point is a bliss point for at most one agent.

A condition motivated by Plott’s construction is to suppose that at the bliss-core point the gradients of two other agents lie along the same line. If these agents are, say, 2 and 3, then the manifold in L(R^{n}, R^{m}) capturing these conditions is Σ_{3} = {A | A_{1} = 0, A_{2} = λA_{3}, λ ∈ R} with codimension 2k − 1. (k codimensions come by specifying A_{1} = 0 and k − 1 from the fact that the direction of A_{2} is specified.) As the codimension of Σ_{3} is larger than the dimension of issue space when k ≥ 2, such behavior is generically impossible outside of a one-dimensional issue space.

To generalize Σ_{2}, Σ_{3}, consider where ∇u_{1}(x) = 0 and there are 1 ≤ b ≤ k agents, with index set D, so that {∇u_{j}(x)}_{j∈D} is in a b − 1 dimensional subspace. To see that this is generically impossible, consider the manifold Σ_{1,D} = {A | A_{1} = 0, there exists scalars λ_{j}, not all zero, so that ∑_{j∈D} λ_{j}A_{j} = 0}. As the sum ensures that one A_{j} is determined by the others, another value is added to the codimension; as Σ_{1,D} has codimension k + 1, this defines a generically empty situation.

To use Σ_{1,D} ∩ CC_{n−1} to prove Theorem 1, recall from [2] and Prop. 1 that a plane defined by k−1 linearly independent vectors can have no more than (q−1)−(k−1) = q−k gradient vectors on either side. If no other gradient vectors are on this plane, then by counting the maximum number of vectors that can be on both sides, the number on the dividing plane and the bliss point, we have 2(q − k) + k − 1 + 1 ≥ n or k ≤ 2q − n. That is, if k > 2q − n, there is another gradient vector on this plane. Such a situation requires these core points to be in Σ_{1,D} for some D with k indices.

All possible situations are given by the \binom{n−1}{k} distinct sets of k indices from \{2, 3, …, n\}. If D denotes such a set, j^{2}f must be in some Σ_{1,D}. Thus, the set of points satisfying such situations is generically empty. This condition includes the bliss-core points for k > 2q − n. As all bliss core points are obtained by changing the choice of D and the identity of the voter whose bliss point is the core point, this is a finite condition. Namely, it corresponds to a finite union of submanifolds, all with the same codimension. Each submanifold is obtained from the first by a permutation of the indices. Thus, the assertion follows. □

Nonbliss core points. The argument showing that Eqs. 2.3, 2.4 determine upper bounds for the generic existence of nonbliss core points resembles the bliss-core point setting but the counting argument is more difficult. To help the reader, the ideas are introduced with a special case.

The q = 4, n = 5 rule. The first choice allowing β ≥ 0 is the q = 4, n = 5 rule; Theorem 1 asserts that issue space can have dimension up to k = 3. To show this is an upper bound, let x be a nonbliss ideal point and consider the plane defined by two gradient vectors ∇u_{j}(x). Using the arguments leading to [2], no more than one gradient vector can be on either side of this dividing plane. Counting one for each side, and the two that define the plane, we have accounted for four of the five gradient vectors. Thus the dividing plane also must contain the last gradient vector. Moreover, x is in the convex hull defined by these three gradient vectors. If not, then there is a line passing through x where all three gradient vectors are on the same side. If the dividing plane is rotated about this line, all three gradient vectors will end up on the same side as one of the remaining gradient vectors; this violates the assumption that x is a core point.

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This condition, where a plane defined by two gradient vectors has a third and their convex hull contains x, imposes strict conditions on all five points! To see why, suppose {x_{j}}_{j=1}^{3} are on the plane, x is in the interior of this hull, and x_{5} is off of the plane. (See Fig. 9.) The pairs of points {x_{1}, x_{5}}, {x_{2}, x_{5}}, and {x_{3}, x_{5}} (along with x) define three more planes which intersect along the line connecting x and x_{5}. With this condition, another point must be on each plane. This extra point, x_{4}, cannot be x as x is not a bliss point. If x_{4} is not on the x–x_{5} line, then x_{4} is on one of the three planes, so we need two more points. With n = 5, this is impossible. Consequently, x_{4} is on the x – x_{5} line. Whatever the orientation of the x–x_{5} line, there is a plane though x with two of {x_{j}}_{j=1}^{3} and x_{5} on the same side. If x_{4} were between x_{5} and x, four points would be on the same side of a dividing plane; this violates [1]. Thus, x_{4} is on the other side of the plane.

In J^{2}, these conditions are captured by Σ = {A | {A_{j}}_{j=1}^{3} are linearly dependent; there exists λ < 0 so that A_{4} = λA_{5}}∩CC_{5}. To find the codimension, notice that k−2 dimensions are imposed by the dependency condition on the first three vectors, and k−1 dimensions are obtained from the second. Thus, as the codimension is 2k −3, such examples are generically possible only when 2k − 3 ≤ k or k ≤ 3. If x is on the boundary of the hull, same argument holds except that an extra codimension is added to reflect that two vectors are along the same line. This situation, then, holds generically only for k ≤ 2. As the construction requires k ≥ 3, it is generically empty.

The final possibility is if all five vectors are on the plane. Here, Σ = {A|A_{j}, j = 3, 4, 5, is a linear combination of A_{1} and A_{2}}. There are no restrictions on the choice of A_{1}, A_{2}, but a codimension k − 2 imposed on the choice of all remaining vectors. Thus, the total codimension is 3k − 6. Such behavior is generically possible only if 3k − 6 ≤ k, or if k ≤ 3.

To obtain all possible cases, use all possible permutations of the indices. As this number is finite, the conclusion holds. To create examples, use the gradient directions defined by the construction. Notice, for Euclidean preferences, both cases restrict the positioning of ideal points so, here, the conclusion is not generic.

General case. The proof for the general case extends the above construction. Let x be a nonbliss core point, and choose any k − 1 linearly independent gradient vectors. If no other gradient vectors are on this plane, then, according to Prop. 1 and [2], no more than q − k gradient vectors can be on either side of this plane. Using this maximum as the number on each side and counting all gradient vectors on the plane, we have that 2(q −k)+(k −1) ≥ n, or 2q − n − 1 ≥ k. Thus, if k > 2q − n − 1, another gradient vector must be on this dividing plane. To underscore this assertion, denote the dimension of issue space as k = β + (2q − n) where the admissible values for the “excess dimension” are β = −1, 0, . . .. First we consider what happens should all points be on this hyperplane, then we analyze what happens when some points are off of it.

If all gradients are in the same 2q − n − 1 dimensional space, then this space is defined by a basis; assume it is given by the first 2q − n − 1 vectors. As all other vectors are in this linear space, the J^{2} representation is Σ = {A | A_{i} ∈ Span({A_{j}}_{j=1}^{2q−n−1}) ∀i ≥ 2q − n, the plane defined by any k − 1 vectors from ({A_{j}}_{j=1}^{2q−n−1}) has no more than q − k vectors on either side}. The dependency condition imposes a β + 1 dimensional constraint on the choice of A_{j}, j ≥ 2q − n − 1. (The dividing plane condition is an open one that, from Sect. 5 can occur, so it does not contribute to the codimension.) Thus, the codimension of Σ is (1 + β)(2n − 2q + 1). Consequently, such a condition exists generically only if the inequality

(β + 1)(2n − 2q + 1) ≤ k = 2q − n + β

is satisfied. Collecting terms and solving for q we obtain Eq. 2.3; solving for k leads to Eq. 2.4. Considering all possible permutations of indices completes the proof.

(This construction does not explicitly consider where no set of 2q − n − 1 vectors are linearly independent. But, this condition increases the codimension for each A_{j} so the total codimension of the new version of Σ is larger. Therefore, such situations become generically unlikely with smaller k values.)

Vectors out of the plane. It remains to consider where not all vectors are in the same 2q − n − 1 dimensional subspace. This analysis uses the next statement.

Lemma 2. Suppose x is a nonbliss core point and a set of 2q − n − 1 linearly independent gradient vectors define a plane L passing through x. The convex hull of the gradient vectors on L contains x but not as a vertex. If x is a boundary point of this convex hull, and if there are 2q − n − 1 + e vectors in L, then at least e + 1 of the vectors lie on the same boundary surface of this hull and the hull they define has x as an interior point.

Proof. Replace gradients with points x_{j} where directions are x_{j} − x. As n is finite, we can extend L to a codimension one plane L_{1} which contains no point off of L. If x is not in the convex hull, then, because it is separated from this hull, a hyperplane H passes through x where all points from L are strictly on one side. With 2q − n − 1 + e points on L, and, hence, on L_{1}, e ≥ 1. A slight change in the orientation of L_{1}, using H ∩ L_{1} as as an axis of rotation, can force all points on L_{1} to one side of the new dividing plane. As x is a core point, no more than (q − 1) − (2q − n − 1 + e) = n − q − e points are on either side of L_{1}, so, at most, 2(n − q − e) points can be off of L_{1}. The number of points not on L is n − (2q − n − 1 + e) = 2n − 2q + 1 − e, so 2(n − q − e) ≥ 2n − 2q − e + 1 or −1 ≥ e. This contradiction proves the assertion.

If x were a vertex, it would be a bliss point which violates the assumption. If x is a boundary point of the hull, it may be on one or several boundary components of the hull; choose the one, B, with minimal dimension γ. Because p is not a vertex, the number of points, r, on B satisfies r ≥ γ + 1 where (because B has minimal dimension) x is in the interior of the hull defined by these points. To find a bound on r in terms of e, notice that (n is finite) there is a plane P passing through x and intersecting B ⊂ L_{1} so that at least \frac{r}{2} of the B points are on the same side as the 2q − n − 1 + e − r remaining points of L_{1}. By using the dividing plane argument, where the plane comes from rotating L_{1} about the P ∩ L_{1} axis, and the fact that one side of L_{1} has at least half, n − q − \frac{e−1}{2}, of the remaining points, it must be that (n − q − \frac{e−1}{2}) + (2q − n − 1 + e − r) + (\frac{r}{2}) ≤ q − 1 or e + 1 ≤ r. □

While there exist configurations allowing core points where all points are not on the 2q − n − 1 dimensional space, we show how the rapid growth of the binomial coefficient forces the codimension to grow so fast that this is generically possible only in very special cases. Assume that x_{n} (representing ∇u_{n}(x)) is not in L and that x is in the interior of the hull of the 2q − n points {x_{j}}_{j=1}^{2q−n} on L. Because {x_{j}}_{j=1}^{2q−n} defines a convex hull with interior, any of x_{j} is removed, the remaining directions (i.e., x_{i}−x) are linearly independent. Moreover, because x_{n} ∉ L, when x_{n} replaces any two points from {x_{j}}_{j=1}^{2q−n}, the new set defines a 2q − n − 1 dimensional plane. As there are \binom{2q−n}{2} ways to choose pairs from \{x_{j}\}_{j=1}^{2q−n}, there are \binom{2q−n}{2} planes. From the linear independence statement, each plane does not include the pair dropped to define the basis and these planes meet only along the x–x_{n} line.

This dimension requires each plane to have one more point. One possibility is to place a point, x_{n−1}, on the x-x_{n} line to simultaneously satisfy this condition for all planes. (Again, x must be between x_{n−1} and x_{n}.) The J^{2} manifold containing this condition is Σ = {A | vectors A_{j}, j = 2q − n, …, n − 2 are in the space spanned by {A_{j}}_{j=1}^{2q−n−1}; there exists a scalar λ < 0 so that λA_{n} = A_{n−1}}. Each A_{j}, j = 2q − n, …, n − 2 vector adds (1 + β) to the codimension and only the length of A_{n−1} can be chosen, so the codimension of Σ is (1 + β)(2n − 2q − 1) + (k − 1). To be generically possible, this value must be bounded above by k which means that (1 + β)(2(n − q) − 1) ≤ 1. In turn, this inequality is satisfied only for n = q + 1, β = 0, or q = n. Notice, a larger β value is obtained when all points remain on the plane.

When the extra point for each plane is not chosen to satisfy all planes simultaneously, the codimension increases more rapidly. To minimize the number of extra points is to choose points that satisfy as many planes as possible. As γ = 2q − n ≥ 3, this occurs by choosing a point that is in the intersection of γ − 2 planes; in L, this intersection is the x_{j} – x line for some j. As there are \binom{γ}{2} planes, we need at least δ = [\binom{γ}{2}/(γ − 2)] ≥ 3 points chosen in this way. First assume that all these points are in L. The J^{2} manifold including this situation is Σ = {A | any γ − 1 vectors from Γ = {A_{j}}_{j=1}^{γ} are linearly independent, each vectors {A_{γ+j}}_{j=1}^{δ} is a scalar multiple of a vector from Γ}. The codimension of Σ is bounded above by (1 + β) + δ(k − 1). As δ ≥ 3, k ≥ γ ≥ 3, this value always is larger than k, so this setting never is generically possible.

More generally, if points are chosen to be on γ−s planes, where k = γ+β ≥ γ ≥ s+1, then, as there are \binom{γ}{s+1} ways to choose these planes, there must be at least δ_{s} = [\binom{γ}{s+1}/(γ−(s+1))] points. Thus, the J^{2} set containing this condition is Σ_{s} = {A | any γ − 1 vectors from Γ_{s} = {A_{j}}_{j=1}^{γ} are linearly independent, each vectors {A_{γ+j}}_{j=1}^{δ_{2}} can be expressed as a linear combination of s vectors from Γ_{s}}. The codimension of Σ_{s} is bounded above by (1 + β) + δ_{s}(k − s). Therefore, to be generically possible, this number must be bounded above by k, which, by using k = γ + β, requires satisfying 1 + (δ_{s} − 1)γ + δ_{s}β ≤ δ_{s}s. As δ_{s} > 2, γ > s, this never is satisfied. (I leave it as a simple exercise to show that if points are chosen to be on different number of planes, then the same conclusion holds. This is because the more planes a point is on, the higher its codimension. The other extreme, of choosing more points with lower codimension requires so many more points that the codimension grows faster.)

The remaining possibility is to choose extra points off of L. This modifies Σ by allowing the A_{j} points to include A_{n} its representation. Thus the new codimension is (1+β)+δ_{s}(k− s − 1). To be generically possible, this number must be less than k, or 1 + (δ_{s} − 1)γ + δ_{s}β ≤ δ_{s}(s + 1). Again, because δ_{s} ≥ 3 and γ ≥ s + 1, this never can be satisfied.

Finally, if x is on the boundary of the convex hull of the points on L, then the same argument applies, but the codimension escalates more rapidly because, as Lemma 2 asserts, most of the points are on a lower dimensional subset of L and all added points also are on this set. Thus, the same computations show that this setting is not generically possible.

Constructing examples. The construction of examples is simple. Choose any point x. Using the construction of Sect. 5, find gradient directions for points where these gradient directions define a 2q − n − 1 dimensional space and where they satisfy the alternating rules for the specified n, q values. Now, use these directions to define gradients. The only difference in choosing second derivative terms is that they must include all variables.

To illustrate with n = 4, q = 5 and k = 3, use Fig. 2. Translate and rotate the figure in R^{3} so that a specified x from the shaded region is at the desired location of a nonbliss core point, and the orientation of the plane is consistent with the desired plane of gradient vectors. Define the gradient directions by the directions from x to the vertices of the star. This construction may suggest that a two-dimensional core results; this would be in conflict with Theorem 2c which asserts it should be zero-dimensional – isolated points. The explanation is that in R^{2}, the chosen gradient directions have only one degree of freedom when the base point is varied. Consequently, the same general star figure is defined by the gradients at neighboring points. With the added degrees of freedom from R^{3}, when the base point is varied, the gradient need not lie in the same plane. Once this happens, [1] and [2] are violated, so a core does not exist. Therefore, this construction does, in fact, define the core to be a collection of isolated, zero-dimensional points.

Dimensions. The dimensional statements follow directly from the codimension statements earlier in this section. This is because, generically, the sets are the union of isolated smooth submanifolds with the indicated codimension of appropriate Σ. Another approach is to note that the construction developed in Sect. 5 allows for freedom in the choice of the gradient directions (an open set about each direction), so the construction is robust. The same comment applies to when the construction of Sect. 5 applies to the β ≥ 0 settings.

I stated that, generally, the core has a stratified structure. As this statement is a direct consequences of singularity theory, it is not formally asserted nor proved. Yet, related ideas are in the proof of Thm. 2a. First, however, I need that the core is closed. This is a consequence of continuity of the gradients and the fact that Prop. 1 defines a closed condition. The only difference in the J^{2} representation is that some of the A_{j} vectors are on the same plane. This adds to the codimension which, in turn, reflects that the boundary is a lower dimensional object.

Theorem 2a. Only the proof of Theorem 2a remains. To show that if x is a q core point then it is a q + 1 core point, notice that the convex hull defined by q + 1 gradient vectors includes the convex hull defined by a subset of them. The conclusion follows from Prop. 1.

Next, consider n, k, q values where, generically, the core for the q rule must be in a lower dimensional submanifold. If the generic situation for the q + 1 rule allows the core to have a nonempty interior (e.g., q = n − 1 or where q + 1 admits β = −1), then, generically, the two cores cannot agree as they have different dimensions.

Suppose the core for the q + 1 rule is, generically, in a lower dimensional set. The dimension for the q rule is d_{q} = k − (β_{1} + 1)(2q − n + 1) where β_{1} = k − (2q − n − 1. The excess dimension for the q + 1 rule is β = k − (2(q + 1) − n − 1) = β_{1} − 2. The above argument handles β_{1} < 2, so let β_{1} ≥ 2. The core dimension for the q + 1 rule is d_{q+1} = k−(β +1)(2q−n+3) = k−(β_{1}−1)((2q−n+1)+2). To see that d_{q+1} > d_{q}, so the conclusion will follow because the dimensions disagree, compute d_{q+1} − d_{q} = 2((2q − n + 1) − β_{1}) + 2. As q ≤ n − 2, the value of β_{1} is bounded above by q − \frac{3}{4}n, (see the expression for β found between Eqs. 2.3, 2.4) so d_{q+1} − d_{q} > 0.

The remaining case is where the core for the q rule has a nonempty interior. Assume that the core for the q and q + 1 rules are the same. Generically, the core is a closed set where if x is a boundary point, then x satisfies Eq. 4.1 by being on the boundary of the convex hull for some coalition. Suppose not; suppose x is in the interior of each Co_{x}({∇u_{j}(x)}_{j∈C} for each decisive coalition C and that v is such that x + tv is not a core point for any t > 0. By continuity and the fact that x is an interior point, there is an open neighborhood of x so that any point in this neighborhood also is in the convex hull of the gradients defined at that point. Any such point is a core point. This contradiction proves the assertion.

So, assume both cores agree and they have a nonempty interior. We consider points on the k − 1-dimensional boundary. This corresponds to a q-rule core point x and a coalition C, say C = {1, 2, …, q}, where x is on the boundary of Co_{x}({∇u_{j}(x)}_{j∈C}. This boundary is k − 1 dimensional and it contains k of the gradient vectors. (If it contained more, then this would add to the codimension violating the fact that x is on a k − 1 dimensional component.) To represent this as a J^{2} condition, let Σ_{q} = {A|k vectors from {A_{j}}_{j=1}^{q} are linearly dependent }. As this has codimension one, the set of points satisfying this condition is, generically, a collection of k − 1 dimensional manifolds. Indeed, it includes the boundary of the core. (To make it explicitly the boundary of the core, add the convexity conditions of Prop. 1.)

Because the cores agree, x is on the k − 1 dimensional boundary of the q + 1 rule core. In particular, this means that there is some agent i so that when C is augmented to C_{i} by adding agent i, i > q, then, x is a boundary point of Co_{x}({∇u_{j}(x)}_{j∈C_{i}}. From C_{i}, q + 1 different coalitions of q voters can be constructed; q of them are formed by replacing an agent from C with i. When ∇u_{i}(x) replaces one of the gradient vectors from C, then either there are k − 1 gradient vectors on this plane, or ∇u_{i}(x) is on this plane. The first case means that a plane can be passed through x using this new q voter coalition where x is not in its convex hull. As this means x is not a core point, it must be that ∇u_{i}(x) is in this plane. In J^{2}, this is captured by the manifold Σ_{q,i} = {A ∈ Σ_{q}|A_{i} is in the span of the k linearly dependent vectors }. This adds another codimension, so, generically, x belongs to a k − 2 dimensional manifold. The dimension contradiction proves the theorem. Indeed, with slight extra care, we have that, in general, the q-core is in the interior of the q + 1 core.

Removing strict convexity. If the convexity assumption is removed, the portions of Prop. 1 where x is on the boundary of the convex hull need not hold. These conditions are needed primarily for the β ≥ 0 analysis. Second, an infinitesimal core point need not be a core point. So, in the more general setting, all infinitesimal core points that are not core points must be removed. The main change in the conclusion is that the core may be the union of several submanifolds of the indicated dimensions.

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