Algebraic Properties of Positional Voting Systems

A Senior Comprehensive Paper

presented to the Faculty of Carleton College Department of Mathematics and Statistics Peri Shereen, Advisor

by Jordan Cahn John Eckert Margalo Mullaney Dana Neidinger

5/23/2016

Dedication

Thanks to Peri Shereen for advising us, and to Zajj Daugherty, Alexander K. Eustis, Gregory Minton and Michael E. Orrison, and Donald Saari for their excellent sources.

iii

Abstract

In this paper we explore the positional voting system through an algebraic lens, modeling voting procedures as linear transformations. We present results from Orrison et al. in the ﬁrst chapter to set up the framework we use to discuss positional voting. In the second chapter, we present our results on preserving reversal - if everyone in the electorate reverses their votes, will the order of the results be reversed as well? The main theorem in this paper classiﬁes reversal-preserving and reversal-ignoring positional voting procedures.

Contents

Dedication iii

Abstract v

Contents vii

1 Positional Voting 1 1.1 Profiles ...... 1 1.1.1 The Voting Profile ...... 1 1.1.2 Group Action of Sn on Profiles ...... 3 1.2 Weighting Vectors and Voting Procedures ...... 3 1.2.1 The Positional Voting Procedure ...... 3 1.2.2 Partial Ranking Procedures to Full Ranking Procedures ...... 5 1.2.3 QSn Modules and Neutrality ...... 7 1.3 The Importance of the Weighting Vector ...... 8 1.3.1 Decomposition of Weighting Vectors and Equivalent Weighting Vectors ...... 8 1.3.2 Relationships Between Weighting Vectors and Results ...... 9

2 Reversal 13 2.1 Deﬁning Reversal ...... 13 2.2 Preserving Reversal ...... 14

3 Conclusion 19

Bibliography 21

vii

1. Positional Voting

This paper will explore positional voting systems through algebraic means. Saari has shown that many types of voting systems can be modeled geometrically; we aim to study voting systems algebraically. We studied an algebraic framework of positional voting systems explored previously by Michael Orrison et al. In this chapter, we will explore many results from their paper in order to set up our own theorem in the next chapter. A positional voting system has three main deﬁning components:

• Proﬁles: representations of the preferences of the voters in the electorate. We will view proﬁles as being in the vector space over combinatorial tabloids, which represent the voters’ rankings.

• Weighting vectors: the vector that determines the voting procedure on a proﬁle. We will view weighting vectors as w ∈ Qm where m is the number of rows in a given tabloid. Thus, the kth entry of the vector assigns a certain number of points to the kth ranked candidate(s).

• Results: the outcome of an election when the weighting vector is applied to a proﬁle.

1.1 Proﬁles

In an election, the ﬁrst step is to mathematically represent the votes of the electorate. Through the use of combinatorial tabloids, we will illuminate the system we used to represent our voter base.

1.1.1 The Voting Proﬁle

Let λ = (λ1, . . . , λm) be a sequence of positive integers such that the sum λ1 + ··· + λm = n. This λ determines the shape of a Ferrers diagram, which is a left-justified array of dots with λi dots in the ith row. A Young tableau of shape λ is such a shape where every dot is replaced with a box containing a single number from 1 to n with no repetition. Two tableau are said to be row-equivalent if they differ only by a permutation of the entries within the rows of each tableau. An equivalence class of tableaux under this relation is called a tabloid of shape λ. Tabloids are represented as diagrams with m rows, where each row i is an unordered set of λi numbers. From a voting-theoretic standpoint we can view tabloids as rankings, with the candidates listed in the top row being in first place and so on, with the candidates listed in the last row getting last place. Two candidates that are both placed in a voter’s kth ranking position will be treated with the same weight by positional voting, hence the reason we use row equivalences.

1 2 2 1 tableaux: 3 6= 3 4 4

1 2 2 1 tabloids: 3 = 3 4 4

When λ = (1,..., 1) we call this a full ranking of the candidates. Otherwise we say this is a partial ranking. In our results we will be dealing primarily with the full ranking situation, because as we will see in section 1.3.2, all partially ranked positional systems can be modeled using fully ranked cases. Let Xλ denote the set of tabloids of shape λ. When λ = (1,..., 1), we can see that |X(1,...,1)| = n!, as there is a natural bijection between full rankings of n candidates and permutations of n. The bijection

1 2 CHAPTER 1. POSITIONAL VOTING

involves assigning a number to each candidate. If candidate i is in place k we say that in the corresponding permutation k “goes to” i. In general, |Xλ| = n! , as we have all of the possible permutations of n λ1!...λm! candidates but divided by the row-equivalencies because every ordering of a row overcounts. In an election we will consider a ﬁxed λ, and voters will choose from one of the rankings in Xλ based on which candidates they prefer. Thus, to determine the makeup of the entire electorate, we need to know how many people voted for each ranking x ∈ Xλ.

Deﬁnition 1.1. A voting proﬁle p is a function p : Xλ → Q such that p(x) is the number of voters that voted for the ranking x. Note that we do not require that p(x) be a positive integer; it may take any value in Q.

We can consider the set M λ = {p : Xλ → Q}, the set of all proﬁles of shape λ. We see that M λ forms a vector space over the rationals: (p + q)(x) = p(x) + q(x)

(αp)(x) = αp(x)

We can easily ﬁnd a basis of M λ, namely the indicator functions of M λ, where the indicator function of x ∈ Xλ is given by the Kronecker delta: ( 1 x = y δxy = 0 x 6= y

Essentially, the indicator function of x is the function that gives one vote to a ranking x ∈ Xλ. Therefore, since the indicator functions correspond to the elements of Xλ, we can consider this basis as simply the elements of Xλ. We can view a proﬁle as a formal linear combination of the tabloids in Xλ, where the coeﬃcient of each tabloid x is the number of people that voted for the particular ranking, or p(x). For example, in the fully ranked 3-candidate case, we have

1 1 2 2 3 3 p = a 2 + b 3 + c 1 + d 3 + e 1 + f 2 3 2 3 1 2 1

1 1 for a, b, c, d, e, f ∈ Q, where a = p 2 , b = p 3 and so forth. Given a standard ordering, we can then 3 2 λ consider each proﬁle p as a vector in Q|X |. For instance, in the previous example we would have

 a  b c p =  d  e f where |Xλ| = 3! = 6, so p ∈ Q6. For our standard ordering we have chosen to use lexicographic ordering, as shown above, where when we order the rows of the tabloids in order of numerical ascendance. The tabloids fall in “dictionary order.” For instance, 1 2 1 3 2 3 3 4 3 > ··· > 2 > ··· > 4 > ··· > 2 4 4 1 1

We have already seen how there is a natural correspondence between full rankings in X(1,...,1)and elements (1,...,1) of the symmetric group Sn, so we can consider a bijection between fully ranked profiles in M and elements of the group ring S , where the group ring is simply the set of sums P a σ for coefficients a Q n σ∈Sn σ σ in the field Q. 1.2. WEIGHTING VECTORS AND VOTING PROCEDURES 3

1.1.2 Group Action of Sn on Profiles Instead of viewing tabloids as permutations, we can consider a different relation between permutations and tabloids (and therefore profiles), namely the action of a permutation on a profile. λ Let λ = (λ1, . . . , λm). We define the group action of Sn on X as a permutation of the entries of the tabloids. In other words, consider candidate k in tabloid x. In σx the position where k was will now have element σ(k). Example 1.2. Suppose we have n = 8, λ = (3, 2, 3), σ = (1 3 5)(4 2)(6 7), then

 8 7 4  σ(8) σ(7) σ(4) 8 6 2   σ  5 6  = σ(5) σ(6) = 1 7 2 1 3 σ(2) σ(1) σ(3) 4 3 5

λ λ −1 This group action of Sn on X extends to an action of Sn on the profile space M by (σp)(x) = p(σ x) [1]. This seems strange, but we can in fact see that the inverse of σ is required in the definition of this action in order for it be a group action. Specifically it is needed to fill the associative requirement:

(στ)p(x) = p((τ −1σ−1)x) = p(τ −1(σ−1x)) = (τp)(σ−1x) = σ(τp)(x).

This action is key to viewing proﬁles as algebraic objects.

1.2 Weighting Vectors and Voting Procedures

In positional voting, the voting procedure is determined by a vector of weights corresponding to the m rows of the tabloid of shape λ.

1.2.1 The Positional Voting Procedure

λ Given λ = (λ1, . . . , λm) and p ∈ M we would like to assign a score to each candidate in a standard way. A weighting vector w ∈ Qm is the vector that assigns points to each candidate based on what position in a ranking the candidate appears. This weight is multiplied by the number of votes that put the candidate in the corresponding rank. For instance, let   w1  .  w =  .  wm λ For each tabloid x ∈ X , if candidate c is in row j of the tabloid x, then candidate c is given wjp(x) points. We then sum over all tabloids x ∈ Xλ to determine the total number of points assigned to candidate c. Essentially, we give wj points to candidate c for every voter that placed candidate c in jth place in their ranking. In the case of positional voting, since this vector determines the results of any proﬁle, this vector completely determines the voting system, or the positional voting procedure of type λ.

Example 1.3. Consider a 3-candidate full ranking election (λ = (1, 1, 1)) with candidates c1, c2, and c3, weighting vector w = [x, y, z]T and proﬁle

c1 c1 c2 c2 c3 c3

p = 7 c2 + 0 c3 + 4 c1 + 2 c3 + 6 c1 + 1 c2 .

c3 c2 c3 c1 c2 c1

We can display the results in a results vector r listing how many points each candidate received from the positional voting procedure:     rc1 7x + (4 + 6)y + (2 + 1)z

r = rc2  = (4 + 2)x + (7 + 1)y + 6z (1.1)

rc3 (6 + 1)x + 2y + (7 + 4)z 4 CHAPTER 1. POSITIONAL VOTING

For instance, if we chose w = [1, s, 0]T , then we would have results     rc1 7 + 10s

r = rc2  =  6 + 8s 

rc3 7 + 2s

We can see that our choice of weighting vector matters a great deal, and we could get very diﬀerently ranked results if we used diﬀerent values for s. This will be discussed more in section 1.3.2. Note that we can view our results as living in M (1,n−1) as follows:

c1 c2 c3 r = (7 + 10s) + (6 + 8s) + (7 + 2s) c2 c3 c1 c3 c2 c3

We also notice, as in equation 1.1, that from a positional voting procedure we get full results that seem like the results of multiplying our proﬁle vector by a matrix. Thus, let us deﬁne such a linear transformation that gives us the desired positional voting outcome:

λ (1,n−1) Definition 1.4. Let Tw : M → M be the linear transformation from the profile space to the results space such that Tw(p) gives us the result of the positional voting procedure of type λ according to w on profile p. λ The matrix of Tw will be an n × |X | matrix, where each column j corresponds to a permutation that corresponds to the jth lexicographically ranked tabloid: that is, the entry i, j is the weight candidate i receives in basis tabloid j based on its position in the tabloid.

Example 1.5. In the three-candidate case, we have the following 3 × 6 matrix:

x x x y z y z w = y 7−→ Tw = y z x x z y z z y z y x x

In general, for the full ranking case this transformation can be written as an n × n! matrix which can be T created recursively: with a weighting vector w = [w1, w2, . . . , wn] , weight w1 will appear in each row (n − 1)! times, on the diagonal from the upper left to lower right. In each block of n × (n − 1)! with the weights w1 taken out, we will have a copy of the (n − 1) × (n − 1)! transformation matrix with weights w2, . . . , wn. This deﬁnition simpliﬁes notation, for instead of “the positional voting procedure of type λ according to weighting vector w,” we can simply use Tw to represent this concept. Not only is Tw a linear transformation, but the map taking w to the matrix of Tw is also linear. This means that we could say Tw is “linear in its subscript”:

m Proposition 1.6. Let w1 and w2 ∈ Q and α ∈ Q. Then Tw1+w2 = Tw1 + Tw2 and Tαw = αTw.

Proof. First let     u1 v1  .   .  w1 =  .  , w2 =  .  um vm

Now consider Tw1+w2 . Column j of the matrix of Tw1+w2 will be the weighting vector w1 + w2 under the permutation corresponding to the jth tabloid in the lexicographical ordering, let us call this σj. Let us say σj sends entry k of a vector to entry i. Then the ith entry of column j will be σj(uk + vk) = ui + vi.

Entry i of column j of Tw1 + Tw2 will be σj(uk) + σj(vk) = ui + vi. Next consider Tαw. Column j of this matrix will be (σj(αw)) = α(σj(w)). By the rules of scalar multiplication of matrices the matrix of αTw would be the matrix of Tw with every entry multiplied by α. So the jth column of αTw would also be α(σj(w)). 1.2. WEIGHTING VECTORS AND VOTING PROCEDURES 5

1.2.2 Partial Ranking Procedures to Full Ranking Procedures We will see that any transformation on partial rankings can be extended to a transformation on full rankings. This is why the bulk of our research focuses on the fully ranked case: any partially ranked case can be analyzed using fully ranked cases.

First, we see that we can view proﬁles p ∈ M λ as proﬁles p ∈ M (1,...,1). This is because for any partial ranking there is a set of full rankings that may be underlying the partial ranking, as shown below. The idea is that a voter likely has a full-ranking in mind and their partial ranking has an equal chance of representing any of the possible corresponding full-rankings. Within a row i of a shape λ, there are λi candidates. If we consider all possible permutations of these candidates (turning the tabloid back into corresponding tableaux), we can then turn the tableaux into full rankings by reading top to bottom and left to right. For example:   2 2 5 5       2 5 2 5 2 5 5 2 5 2  5 5 2 2      1 3 → 1 3 , 3 1 , 1 3 , 3 1 → 1 , 3 , 1 , 3     4 4 4 4 4  3 1 3 1     4 4 4 4 

We notice because there are λi! ways to permute the elements in one row, for any shape λ there will be Qm i=1 λi! = f corresponding full rankings. The number of total corresponding full rankings is equal to the size of the row-stabilizer.

λ Deﬁnition 1.7. Let the row-stabilizer of x ∈ X , denoted Rx, be the set of permutations that send x to itself: Rx = {π ∈ Sn | πx = x} (For more on the properties of row stabilizers, see [3].)

Deﬁnition 1.8. Let ι : M λ → M (1,...,1) be the inclusion mapping that maps each tabloid x ∈ Xλ (indicator function) to the sum of its corresponding full rankings times 1/f where f = |Rx|.   2 2 5 5     2 5  5 5 2 2  1       ι  1 3  =  1 + 3 + 1 + 3  4   4  3 1 3 1    4 4 4 4

This deﬁnition can be extended linearly to an function on proﬁles, let ι(p) = p.

So we have seen that we can view each profile p ∈ M λ as a profile p ∈ M (1,...,1) that is constant on the equivalence classes that form the tabloids of shape λ. This is because every equivalence class represents a p(x) tabloid, and for a single tabloid x, each member of the equivalence class will have coefficient f . Similarly, we can extend weighting vectors in Qm to weighting vectors in Qn, or weighting vectors acting on shapes λ = (λ1, . . . , λm) to full rankings with λ = (1,..., 1).

w1 . n Deﬁnition 1.9. Let λ = (λ1, . . . , λm), w = . . Let w ∈ Q be the weighting vector whose ﬁrst λ1 wm entries are w1, and in general wi will appear λi times in succession.

In the following proposition, we will see that when we apply w ∈ Qm to a proﬁle p ∈ M λ, we get the same result as applying w ∈ Qn to p ∈ M (1,...,1). Thus, we can represent every election with a partial ranking as an election using full rankings.

Proposition 1.10. Tw(p) = Tw(p) 6 CHAPTER 1. POSITIONAL VOTING

Proof. Let λ = (λ1, . . . , λm). Let

w1    w  w1 1  | |  .  .   .  w = w =  w1  Tw = w ··· wσ ···  .    . | | wm . wm

Tw has a column vector corresponding to each σ ∈ Sn, because there is a natural bijection between full rankings and permutations in Sn. Note that these corresponding vectors are themselves permutations of w, denote the vector corresponding to σ ∈ Sn as wσ. Instead of showing the desired equality holds for any proﬁle p, we will show it holds for an arbitrary basis λ tabloid {tj} ∈ X and extend by linearity to an entire proﬁle. We can view this basis tabloid {tj} as the

proﬁle vector p{tj } with all 0’s and 1 in the jth entry.  0  .  0  .   .  1/f  . . p{t } =  1  −→ p{t } =  .  j  .  j  .  .  1/f  0  .  . 0 Qm Thus p{tj } will have f entries each with value 1/f, where f = i=1 λi! = Rtj . These entries appear in the positions of the lexicographic ordering that correspond to the full rankings that we get from tabloid {tj}. Let {t1} be the tabloid of shape λ ordered ﬁrst under our lexicographic ordering. This can be thought of as the identity tabloid. Note that there exists some σj ∈ Sn such that σj{t1} = {tj}, where the permutation applied to a tabloid permutes the numbers in the tabloid. Also, note that this σj is not unique. This is

because any σj such that σj{t1} = {tj} multiplied on the left by some permutation π ∈ R{t1}, will have the same property. If we have two permutations σ1 and σ2 such that σ1{t1} = σ2{t1} = {tj}, then we will

have wσ1 = wσ2 = wσj , because applying a permutation in the row-stabilizer will simply permute the same weights in w. These wσj ’s correspond to full rankings of {tj}, and we will have |R{tj }| = f permutations corresponding to a tabloid {tj}. So

 0  .    .  | | |  1/f  T (p ) = w ··· w ··· w ···  .  w {tj }  σj σj   .  | | |  1/f   .  . 0

Since the coordinates in the proﬁle vector are ordered lexicographically, and the columns in Tw correspond

to full rankings ordered lexicographically, each 1/f in an entry i will correspond to a wσj in column i. So 1 1 T (p ) = w + ··· w = w w {tj } f σj f σj σj

Now let us show that Tw(p{tj }) = wσj . In the columns of Tw, we also have permutations of the weighting vector, not one for each σ ∈ Sn, but one for each tabloid {tj} – so one for each σ in the “equivalence class” determined by the row-stabilizer. In other words, for each tabloid {tj} of shape λ there exists one column for a single wσ such that σ{tj} = {tj}. These too are ordered lexicographically, but as partial rankings of shape λ rather than full rankings. So

 0   | |  . .  1  Tw(p{tj }) = Tw = w ··· wσj ··· = 1 · wσj = wσj  .  | | . 0 Example 1.11 (Plurality Voting). We shall illustrate the above proposition by considering the case of plurality voting, which is a common voting procedure used in which each voter chooses only their favorite candidate. 1.2. WEIGHTING VECTORS AND VOTING PROCEDURES 7

Essentially, the votes are in X(1,n−1), because each voter ranks a single candidate ﬁrst, and does not rank the rest of the candidates. We then use the 2-dimensional weighting vector weighting ﬁrst place 1 and second place 0.

1 n 1 p = a1 + ··· + an w = 2 ... n 1 ... n − 1 0

However, as per our last proposition, we can consider each vote as a sum of full rankings, and use the generalized weighting vector w to get the same results.     1 1 n n 1     0 a1  2 n  an  1 n-1   p =  . + ··· + .  + ··· +  . + ··· + .  w = . (n − 1)!  . .  (n − 1)!  . .  .     . n 2 n-1 1 0 Equivalently, since the weighting vector w has n − 1 trailing zeroes, we can simply consider each vote as a single full ranking, where the order of the last n − 1 candidates on the ballot does not have an eﬀect on the outcome of the election. Because we can view any partial ranking case in terms of a fully ranked case, in our research we chose to focus on fully ranked elections.

1.2.3 QSn Modules and Neutrality Now let us further consider the algebraic structure of our proﬁle and results spaces in order to illustrate another result about positional voting systems.

λ We can fairly easily extend the group action of Sn on M from section 1.1.2 to an action of the group ring λ QSn on M . If X a = aσσ ∈ QSn, σ∈Sn then (ap)(x) = P a p(σ−1x). By deﬁning a ring action on M λ, we have turned it into a module – σ∈Sn σ speciﬁcally, a QSn-module. (1,n−1) Furthermore, we can similarly consider the results space M to be a QSn-module. It turns out that λ (1,n−1) the voting procedure Tw : M → M is a QSn-module homomorphism.

Proposition 1.12. Tw is a QSn-module homomorphism. That is, for all a ∈ QSn, Tw(ap) = aTw(p). λ (1,n−1) The proof of this fact involves fairly technical manipulations of the action of Sn on M and M . We provide the general idea below.

λ Proof idea. First, we show that Tw preserves the action of Sn – that is, Tw(σp) = σTw(p). Let x ∈ X be −1 the ith tabloid (ordered lexocographically), and ﬁx σ ∈ Sn. Say that σ x occurs jth lexocographically. Then p(x) and σp(x) are the ith and jth basis elements of M λ, respectively.

Tw(σp) = rj σTw(p) = σri

where ri and rj are the ith and jth columns of the matrix representation of Tw, respectively, considered as an element of M (1,n−1). T Let w = [w1, w2, . . . , wn] . Then ri = wπ(1){t1} + ··· + wπ(n){tn} for some permutation π, where

k {tk} = . 1 2 ··· k − 1 k + 1 ··· n 8 CHAPTER 1. POSITIONAL VOTING

Then n X σri = wπ(k)σ{tk}. k=1 But n n X X rj = wσ−1π(k){tk} = wπ(k)σ{tk} = rj k=1 k=1 Now that we have that Tw preserves permutations, it is straightforward to show that it is a QSn-module P homomorphism. If a = aσσ then σ∈Sn ! X −1 X −1 X −1 Tw(ap(x)) = Tw aσp(σ x) = aσTw(p(σ x)) = aσσ Tw(p(x)) = aTw(p(x)).

σ∈Sn σ∈Sn σ∈Sn

Due to the action of Sn on both the proﬁle space and results space, we get the desirable property of neutrality, which in some ways can be considered a property of “fairness.” Deﬁnition 1.13. A voting system is said to be neutral if permuting the labels on the candidates and permuting the results in the same way does not change the outcome of an election.

Because all Tw are QSn-module homomorphisms, all positional voting procedures are neutral:

Tw(σp) = σTw(p) To model a non-neutral voting system algebraically would be challenging, as we would have to intentionally create some transformation that was not a homomorphism. Such a transformation would likely be diﬃcult to study.

1.3 The Importance of the Weighting Vector

Now that we have explored the positional voting system, what can our choice of weighting vector tell us about the results of an election?

1.3.1 Decomposition of Weighting Vectors and Equivalent Weighting Vectors Certainly, the weighting vector is crucial in determining the outcome of an election. However, it is not always the case that two diﬀerent weighting vectors produce diﬀerent outcomes. For instance, simply scaling the weighting vector by a positive number may change the number of points received by each candidate, but it cannot change the order of the candidates. Consider the situation in example 1.3. If we had instead chosen the weighting vector 2w = [2, 2s, 0]T , we’d have found the result 14 + 20s 2r = 12 + 16s 14 + 4s Since every entry in the results vector is doubled, the order of the candidates does not change. This “order of the candidates” is called the ordinal ranking:

n Deﬁnition 1.14. Two results r, s ∈ M (1 ) are said to have the same ordinal ranking if

ri ≥ rj ⇐⇒ si ≥ sj with

ri = rj ⇐⇒ si = sj

ord ord ord for all 1 ≤ i, j ≤ n. In this case, we write r ≈ s. (If this is not the case, we write r 6≈ s.) Note that ≈ is an equivalence relation. 1.3. THE IMPORTANCE OF THE WEIGHTING VECTOR 9

Put more simply, two results vectors have the same ordinal ranking whenever they rank the candidates in the same order.

As we have seen, multiplying a weighting vector by a positive scalar does not change the ordinal ranking of the results. Similarly, adding the same constant to each entry of the weighting vector will not change the ordinal ranking of the results. In order to put this more formally, we must examine how the structure of Qm aﬀects weighting vectors. We start by considering the “all-ones” vector 1 ∈ Qm, the m-dimensional vector with a 1 in every coordinate. Let V = span{1}. By deﬁnition, V ⊕ V ⊥ = Qm, so any weighting vector w ∈ Qm can be written ⊥ uniquely as w = wb + 1w, where wb ∈ V and 1w ∈ V . Let   u1  .  ⊥ wb =  .  ∈ V . um

m P Then, by deﬁnition, 0 = 1 · wb = ui. So the entries in wb always sum to zero. We will call such vectors i=1 sum-zero weighting vectors, and refer to wb and 1w as the sum-zero and all-ones parts of w, respectively. Since V is one-dimensional, we will also write 1w = α1 (for some α ∈ Q) when convenient. Since using a multiple of the all-ones vector as a weighting vector will always result in a tie, in many ways only the sum-zero part of w is of voting-theoretic interest.

Motivated by the discussion at the beginning of this section, we consider when two weighting vectors give the same ordinal results for all proﬁles.

Definition 1.15. We say that two weighting vectors w, x ∈ Qm are equivalent whenever there exist scalars α, β ∈ Q with α > 0 such that x = αw + β1. In this case, we write w ∼ x. It is straightforward to check that ∼ is an equivalence relation. This definition matches our intuition that scaling a weighting vector by a positive constant and adding the same value to every coordinate should not affect ordinal outcomes of an election. We will show this formally in the next section. There is an equivalent definition of equivalent weighting vectors that emphasizes the fact that election outcomes depend only on the sum-zero portion of the weighting vector.

Deﬁnition 1.16. Given two weighting vectors w, x ∈ Qm, w ∼ x if and only if there exists a positive γ ∈ Q such that wb = γxb. Proof of deﬁnition equivalence. (1.15 ⇒ 1.16) If w = αx + β1 (α > 0) then, using the decomposition x = xb + δ1, we see that w = αxb + (β + αδ)1 and, by the uniqueness of the decomposition of w into sum-zero and all-ones parts, wb = αxb. (1.16 ⇐ 1.15) Similarly, if wb = γxb (γ > 0), then (using the same decomposition of x)

w = wb + α1 = γxb + α1 = γx + (α − γδ)1. Therefore, to determine whether two weighting vectors are equivalent, we need only consider their sum-zero parts.

1.3.2 Relationships Between Weighting Vectors and Results The following theorem answers some intuitive questions we have about the importance of the weighting vector, speciﬁcally the relationship between weighting vectors and results:

• Given a certain set of weighting vectors, applied to the same proﬁle, what kind of results can you get?

• If you want speciﬁc results, how restricted are the weighting vectors you can choose? 10 CHAPTER 1. POSITIONAL VOTING

Theorem 1.17. Let n ≥ 2, and let λ = (λ1, . . . , λm) be a partition of n. Suppose that w1,..., wk form a m linearly independent set of weighting vectors in Q such that w1,..., wk are sum-zero vectors. If r1,..., rk n λ are any sum-zero results vectors in Q , then there exist inﬁnitely many proﬁles p ∈ M such that Twi (p) = ri for all 1 ≤ i ≤ k.

We will give a proof for the fully ranked case (so wi = wi). For a full proof, see [1]. (1,n−1) ∼ n Proof outline. Let U denote the set of sum-zero vectors in M = Q . U is a simple QSn-module (this (n−1,1) submodule is often denoted S ; see [3]). Define T : U → U such that T (wi) = ri. Since both the weighting vectors and results vectors are linearly independent, we may extend T to a linear transformation. Now, since T is a linear transformation from a simple module to itself, by a theorem of Burnside there exists a ∈ QSn so that T (u) = au for all u ∈ U. So there exists a such that T (wi) = awi = ri. We can find infinitely many b ∈ QSn such that bu = 0 for all u ∈ U. First let us show that there exists an element b1 of Sn such that b1u = 0 for all u ∈ U. Let X b1 = σ.

σ∈Sn   u1  .  Now consider b1u, u ∈ U. Let u =  .  and Sn = {σ1, . . . , σn!}. Then the ith entry in b1u will be un

σ1(ui) + ··· + σn!(ui).

There are (n − 1)! permutations in Sn that map any ui to a ﬁxed uj. Therefore the ith entry of b1u is

(n − 1)!u1 + ··· + (n − 1)!un = (n − 1)!(u1 + ··· + un) = (n − 1)!(0) = 0 because all u ∈ U are sum-zero. Since any arbitrary entry of b1u is 0, we know b1u = 0 . Note that for any q ∈ Q (qb1)u = q(b1u) = 0. Therefore there are inﬁnitely many such b = qb1 ∈ QSn

such that bu = 0 for all u ∈ U. Set p = a + b. Then Twi (p) = pwi = awi + bwi = ri + 0. Since there are inﬁnitely many suitable b, there are inﬁnitely many p as well.

This surprising result shows us how important the weighting vector is in determining the results of an election. If you have a set of linearly independent weighting vectors then if they act on a fixed profile their results may have no relationship to each other. In other words, people can vote the same way under different positional voting systems and the systems can yield wildly different results. Also interestingly, for any weighting vector and any result, there will be infinite ways that people could vote to achieve the result. This theorem will be used in the proof of our next theorem on weighting vectors, which pertains to the effect of equivalent weighting vectors acting on profiles. As promised, we will now prove that equivalent weighting vectors do indeed give the same ordinal results when applied to any profile. In fact, the only way in which two weighting vectors give the same ordinal result for all profiles is if they are equivalent.

n Theorem 1.18. Let n ≥ 2, and let w and x be weighting vectors in Q . The ordinal rankings of Tw(p) and (1,...,1) Tx(p) will be the same for all p ∈ M if and only if w ∼ x.

Proof. (⇒) First assume w ∼ x. Then by definition w = αx + β1 for some α, β ∈ Qn with α > 0. This (1,...,1) implies that Tw(p) = Tαx+β1(p) for all profiles p ∈ M . Because the transformation T is linear in its subscript, Tw(p) = Tαx+β1(p) = αTx(p) + βT1(p). T1 is a transformation that gives the same result in every row regardless of the profile it is applied to, so ord adding any scalar multiple of it does not change the ordinal ranking of a result. Thus, Tw(p) ≈ αTx(p). ord It is also clear that αTx(p) ≈ Tx(p), since the scalar multiple α > 0 does not change the ordinal ranking: for any rational numbers m and n in the results vector Tx(p), if m > n, then αm > αn. Thus, by the ord (1,...,1) transitivity of ≈ , it must be true that Tw(p) has the same ordinal ranking as Tx(p) for all profiles p ∈ M . 1.3. THE IMPORTANCE OF THE WEIGHTING VECTOR 11

(⇐) Contrapositive. Suppose that w x. Then by deﬁnition 1.16, wb 6= γxb. So, in particular, wb and xb are linearly independent sum-zero weighting vectors. Let us choose r1 and r2 to be two sum-zero results vectors ord such that r1 6≈ r2. By the previous theorem, there exists a proﬁle p such that Tw(p) = r1 and Tx(p) = r2. Thus, b b Tw(p) = T (p) + T1 (p) = r1 + α1 wb w Tx(p) = T (p) + T1 (p) = r2 + β1 bx x ord ord ord It is clear that r1 ≈ r1 + α1, and similarly r2 ≈ r2 + β1. So since r1 6≈ r2,

ord Tw(p) = r1 + α1 6≈ r2 + β1 = Tx(p)

(1,...,1) So the ordinal rankings of Tw(p) and Tx(p) will not be the same for all p ∈ M .

2. Reversal

In our research, we tackled the idea of reversal: in an election where everyone reverses their votes, is the ordinal ranking of the results reversed as well? This idea of reversal preservation is a property of some positional voting systems on full rankings. In this chapter of our paper, we deﬁne what we mean by reversal, as well as present an original theorem classifying all reversal-preserving and reversal-ignoring weighting vectors.

2.1 Deﬁning Reversal

(1,...,1) Let us define a new group action of Sn on X , the set of all tabloids of shape λ = (1,..., 1) (this will be (1,...,1) different from the action defined in section 1.1.2). For π ∈ Sn and t ∈ X ,

π · t = tπ : Rowi 7−→ Rowπ(i) where Rowi denotes the ith row of a tabloid t. That is, π maps t to the tabloid tπ whose π(i)th row is the ith row of t. Essentially, we deﬁne applying a permutation to a tabloid as permuting the rows of the tabloid. It is relatively trivial to prove that π · t is group action:

(1,...,1) (1,...,1) (1,...,1) 1. For all π ∈ Sn and t ∈ X , π ·t will be in X because X is closed under the permutation of the rows, as each row is the same length, 1.

2. For all t ∈ X(1,...,1) e · t = t.

(1,...,1) 3. For all t ∈ X and π, σ ∈ Sn

π · (σ · t) = π · (tσ) = tπσ Rowi 7→ Rowπ(σ(i))

and (πσ) · t = tπσ

Now let us consider permuting a profile. This will be useful when we consider finding the reversal of a profile. Profiles are linear combinations of tabloids, so we will define a permutation acting on a profile as a permutation acting on its component tabloids. If

p = c1t1 + c2t2 + ··· + cn!tn!

then π · p = c1(t1)π + c2(t2)π + ··· + cn!(tn!)π We will use this action of a permutation on a proﬁle mainly to discuss the action of reversal.

Definition 2.1. We define the reversal of a profile as the action of the permutation τ on the profile, where

( n−1 n+3 (1 n)(2 n − 1) ··· 2 2 when n is odd τ = n n+2 (1 n)(2 n − 1) ··· 2 2 when n is even

In other words, for a full-ranked tabloid τ swaps the first and last rows, the second and second-to-last rows, and so on. This results in the coefficients in the profile being swapped accordingly. Note that though all of the component tabloids of a profile are reversed, the profile vector itself will not appear “flipped” because of the nature of lexicographic ordering.

13 14 CHAPTER 2. REVERSAL

Weighting vectors, however, are “ﬂipped” by τ. We deﬁne τw as the weighting vector w as shown below with the subscripts of its entries permuted by τ. So     w1 wn  .   .  w =  .  −→ τw =  .  wn w1

Intuitively one can postulate that reversing a proﬁle accomplishes the same goal as keeping the proﬁle the same and reversing the weighting vector. When you reverse a ranking, the candidate that was getting the top weight, w1, will now get the bottom weight, wn by construction of Tw. Reversing the weighting vector instead would similarly make the candidate that was previously getting the top weight now get the bottom weight. We will prove that this intuition is indeed fact.

Lemma 2.2. Tw(τ · p) = Tτw(p)

Proof. It will be suﬃcient to prove this for an arbitrary basis vector of the proﬁle space, ei. Note that ei represents one ballot, which is a vote for the ith tabloid lexicographically. We will call the tabloid to which ei corresponds ti. Let us say that t1 is the identity tabloid and σi ∈ Sn such that ti = σit1. We have reduced our problem to the proof of

Tw(τ · ei) = Tτw(p)

First let us consider Tw(τ ·ei). Since ei is a profile representing ti = σit1, τ ·ei is representing τ ·ti = τ ·(σit1). Therefore in the profile τ · ei there will be a one in the coordinate that corresponds to the basis tabloid τ · (σit1) and a zero in every other coordinate. The result of Tw(τ · ei) will then be equal to one of the columns of the Tw matrix. By the definition of the function Tw, the column that corresponds with τ · (σit1) will be τ(σiw). Therefore Tw(τ · ei) = τ(σiw)

Next consider Tτw(ei). We know that the ith column of Tw is σiw, and each column in Tτw is the same column from Tw permuted by τ. So the ith column of Tτw is τ(σiw). We know that when ei is multiplied by the Tτw matrix our result will be the ith column of Tτw so,

Tτw(ei) = τ(σiw)

. Therefore

Tτw(ei) = Tw(τ · ei) = τ(σiw) This extends linearly to show that

Tτw(p) = Tw(τ · p)

Note that this applies to other permutations besides τ, but we will be focusing on τ in this paper.

2.2 Preserving Reversal

Instinctively, we would like a voting system to have the property that, when every voter reverses her preferences, the results are similarly reversed. Thus, we would like to examine when a given weighting vector “preserves reversal.” When is it the case that, for any proﬁle p, Tw(τ · p) has the opposite ordinal ranking as Tw(p)? We proceed to deﬁne this more formally.

Deﬁnition 2.3. A weighting vector w ∈ Qn is said to preserve reversal if, for all fully-ranked proﬁles (1n) ord p ∈ M , Tw(τ · p) ≈ −Tw(p). (We then say w is reversal-preserving.) ord Additionally, w is said to ignore reversal if, under the same conditions, Tw(τ · p) ≈ Tw(p). (We then say w is reversal-ignoring.) 2.2. PRESERVING REVERSAL 15

Intuitively, for any results vector, we can think of r and −r as having “opposite ordinal rankings,” since ord ri > rj ⇐⇒ −ri < −rj. Thus, when we say that Tw(τ · p) ≈ −Tw(p), we are really saying that Tw(τ · p) and Tw(p) have opposite ordinal rankings, which is the situation we wish to explore.

To better understand reversal, it is helpful to work with the permutation matrix Aτ corresponding to τ. To this end, we deﬁne the following n × n matrix: 0 1   ..  Aτ =  .  1 0

n It is straightforward to check that τw = Aτ w for all w ∈ Q . By Lemma 2.2, Tw(τ · p) = Tτw(p) = TAτ w(p). Since we wish to compare this to Tw(p) and −Tw(p) = T−w(p), we need to examine the eigensystem of Aτ . In the case that n is even,

−λ 1 0 ......

−λ 1 n |A − λI| = 0 0 = (−λ) − 1 1 −λ ......

1 0 −λ and in the case that n is odd,

−λ 0 1

......

n−1 n−1 |A − λI| = 0 1 − λ 0 = (−λ) (1 − λ) + (1 − λ) = (1 − λ)((−λ) + 1)

......

1 0 −λ

n In both cases, Aτ has two distinct eigenvalues: λ = −1, with multiplicity b 2 c, and λ = 1, with multiplicity of n n n d 2 e. Let V−1 = {w ∈ Q | Aτ w = −w} and V1 = {w ∈ Q | Aτ w = w} be the corresponding eigenspaces. Since Aτ is a real-valued symmetric matrix, it is diagonalizable over R (by the Real Spectral Theorem). n And, since it has only rational eigenvalues, it must be diagonalizable over Q. Thus, V−1 ⊕ V1 = Q . n Furthermore, there are convenient bases for both these spaces: {ei − en−i+1 | 1 ≤ i ≤ b 2 c} is a basis for n n V−1 and {ei + en−i+1 | 1 ≤ i ≤ d 2 e} is a basis for V1 (where ei are the standard basis vectors of Q ). As a consequence of this, all elements of V−1 must be sum-zero and

dn/2e 1 X 1 = e + e 2 i n−i+1 i=1

is an element of V1.

Lemma 2.4. If w ∈ V1 then w ignores reversal. If wb ∈ V−1 then w preserves reversal. ord Proof. Assume that w ∈ V1. Then Tw(τ · p) = TAτ w(p) = Tw(p). So Tw(τ · p) ≈ Tw(p). Assume that wb ∈ V−1. Since w ∼ wb , by Theorem 1.18

ord ord Tw(τ · p) ≈ T (τ · p) = T (p) = T (p) = −T (p) ≈ −Tw(p). wb Aτ wb −wb wb ord So Tw(τ · p) ≈ −Tw(p). 16 CHAPTER 2. REVERSAL

Thus V−1 and V1 contain exclusively weighting vectors that preserve and ignore reversal, respectively. We will show that, in fact, all weighting vectors that ignore reversal are elements of V1, and all weighting vectors that preserve reversal are equivalent to a vector in V−1.

n Theorem 2.5. Let w ∈ Q = V−1 ⊕ V1 be written uniquely as w = w-1 + w1, where w-1 ∈ V−1 and w1 ∈ V1. Then

i) w preserves reversal if and only if w1 = γ1 for some γ ∈ Q and

ii) w ignores reversal if and only if w-1 = 0. Proof. The “if” directions for both statements follow directly from Lemma 2.4. To prove the “only if” direction:

i) If w = w-1 + w1 preserves reversal then, by deﬁnition,

ord Tw(p) ≈ −Tw(τ · p).

Furthermore,

−Tw(τ · p) = −Tw-1+w1 (τ · p) = −TAτ w-1 (p) − TAτ w1 (p) = Tw-1 (p) − Tw1 (p) = Tw-1−w1 (p).

ord Thus, Tw(p) ≈ Tw-1−w1 (p). So, by Theorem 1.18, w ∼ w-1 − w1. By deﬁnition, this means that w = w-1 + w1 = α(w-1 − w1) + β1, where α, β ∈ Q and α > 0. Rearranging yields

(1 − α)w-1 = −(α + 1)w1 + β1.

Notice that (1 − α)w-1 ∈ V−1 and −(α + 1)w1 + β1 ∈ V1. But V−1 ∩ V1 = {0}, so both of these vectors are zero. In particular, (α + 1)w1 = β1. β Since α + 1 > 0, set γ = α+1 . Thus w1 = γ1.

ii) If w = w-1 + w1 ignores reversal then, by deﬁnition,

ord Tw(p) ≈ Tw(τ · p).

ord Similarly to in i), Tw(τ · p) = Tw1−w-1 (p). So Tw(p) ≈ Tw1−w-1 (p). Again applying Theorem 1.18, w ∼ w1 − w-1, so w-1 + w1 = α(w1 − w-1) + β1 with α, β ∈ Q and α > 0. Rearranging yields

(α + 1)w-1 = (1 − α)w1 + β1.

As in i), both sides of this equality must be 0. So (α + 1)w-1 = 0. Since α + 1 > 0, w-1 = 0.

A direct consequence of this theorem is that w ignores reversal if and only if w ∈ V1. Furthermore, since both w = w-1 + w1 and w = wb + α1 are unique decompositions of w (with α1 ∈ V1 and both w-1 and wb sum-zero), w preserves reversal if and only if wb ∈ V−1. A weighting vector of particular interest is the Borda count:

n − 1 n − 2    .  b =  .  .    1  0

The Borda count has many desirable voting-theoretic properties, one of which is that it always preserves reversal; this is shown in the n = 3 case in [2]. For general n, the result follows directly from Theorem 2.5. 2.2. PRESERVING REVERSAL 17

Corollary 2.6. The Borda count weighting vector b preserves reversal.

Proof. It is suﬃcient to show that b ∈ V−1.

n − 1 (n − 1)/2 1 n − 2 (n − 3)/2 1     n − 1    .   .  . b =  .  =  .  + . ,     2    1  (3 − n)/2 1 0 (1 − n)/2 1 so (n − 1)/2 (n − 3)/2    .  b =  .  ,   (3 − n)/2 (1 − n)/2 which is indeed an element of V−1. More generally, Theorem 2.5 is useful in several ways. As exempliﬁed by the Borda count, it allow us to easily determine when a weighting vector preserves reversal. In a similar vein, it makes it straightforward to deﬁne a space of all reversal-preserving weighting vectors: V−1 ⊕ span{1}. Finally, it gives us another way to think about the structure of weighting vectors – in addition to the decomposition into sum-zero and all-ones parts, for any weighting vector there is a unique decomposition into reversal-preserving and reversal-ignoring parts.

3. Conclusion

In our research we have focused on reversal, and when it is preserved by positional voting systems. This leaves us with the further question of when other permutations besides the reversal permutation, τ, are preserved or ignored. The method that we used to prove Theorem 2.5 starts to break down for different permutations. One reason is that in this proof we we use the handy fact that −Tw has the opposite ordinal ranking as Tw. If we are interested in different permutations of profiles besides reversal then we may need another clever way to talk about how a permutation affects an ordinal ranking. Additionally, our methods break down when considering permutations that do not consist of disjoint transpositions because the permutation matrices of such permutations have complex eigenvalues. This leads us to postulate that perhaps only permutations that consist of disjoint transpositions can be preserved. Further extensions of this paper include the question of how replicate Donald Saari’s work using algebraic methods (see [2]), and whether algebra can, or should, be used to model non-neutral systems.

Bibliography

[1] Z. Daugherty, A. K. Eustis, G. Minton, and M. Orrison. “Voting, the symmetric group, and representation theory.” In: The American Mathematical Monthly 116.8 (2009), pp. 667–687. [2] D. Saari. Basic geometry of voting. Berlin, Germany: Springer, 1995. [3] B. E. Sagan. The symmetric group: Representations, combinatorial algorithms, and symmetric functions. New York, NY: Springer, 2001.