Introduction

Given an n ⇥ n matrix A with entries in a field F, let A S denote the determinant of the submatrix of A indexed by the set S on the rows and columns. If we set A ; = 1, the principal minors of a matrix form a vector of length 2 n . The principal minor map is the map that assigns to each matrix the vector of its principal minors, namely

given by A ! (A S ) S✓ [n] .

One of the motivating goals of this paper is to characterize the image of this map. This problem dates back to the 19th century [39], [40]. In the cases n = 2 and n = 3, this image is characterized by A ; = 1 over C. In the case n = 4, Lin and Sturmfels [35] give an explicit list of 65 polynomials that cut out the image and they conjectured that it is cut out by equations of degree 12 for any n.

The image of the space of real and complex symmetric matrices was studied by Holtz and Sturmfels [27], who show that the image is closed and invariant under an action of the group SL 2 (R) n o S n and conjectured that the vanishing of polynomials in the orbit of the hyperdeterminant under this group cuts out the image of the principal minor map over C. This conjecture was resolved by Oeding [41]. In [3], we build of techniques in [33] to generalize this result to hold over arbitrary unique factorization domain. Here we use similar techniques to characterize the image of Hermitian matrices.

The principal minor map problem appears in many di↵erent fields and applications, including statistical models, machine learning, combinatorics and matrix theory. One fundamental application is the study of determinantal point processes (DPP). These are probabilistic models that arise naturally in the study of random matrix theory [29] and machine learning [13,21]. Symmetric DPPs have attracted a lot of attention as they reflect the repulsive behavior in modeling, see [1,8,18,32,45]. Non-symmetric kernels are also of interest for modeling both repulsive and attractive interactions [4,12,20]. Learning the parameters of such a model from data leads to the computation problem of reconstructing a matrix from the vector of its principal minors. Gri n and Tsatsomeros [22,23] give a numerical algorithm that reconstructs a preimage of a matrix, if it exists, over C. Rising, Kulesza and Taskarc [43] provide an e cient algorithm for reconstruction in the symmetric case.

In this paper, we study the principal minor map via determinantal representations of an associated multivariate polynomial. Explicitly, to each vector a = (a S ) S⇢[n] 2 F 2 n , we assign a multia ne polynomial f a where f a = P S⇢[n] a S

A is symmetric (Hermitian), we call f a symmetric (Hermitian) multia ne determinantal polynomial. In [3] we prove that the class of symmetric determinantal multia ne polynomials is characterized by their Rayleigh di↵erences being squares. The Rayleigh di↵erence of a polynomial f with respect to i, j 2 [n] is defined to be ij (f ) = @f @x i @f @x j f @ 2 f @x i @x j .

(1)

Here we explore the consequences of factorizations of Rayleigh di↵erences for non-symmetric determinantal representations. As described below, the relationship is more subtle than in the symmetric case. While in [3] we were able to work over arbitrary unique factorization domains, here we mainly restrict ourselves to working over fields in order to deal with this increased complexity. While we do recover some of the results from [3], we do not recover them in full generality. In follow up work, the second author also uses factorizations of Rayleigh di↵erences to recover the full fiber of the principal minor map [2].

Using factorizations of Rayleigh di↵erences into Hermitian squares, we characterize Hermitian determinantal multia ne polynomials over any field K with an automorphism of order two. In particular, this gives a characterization of Hermitian determinantal multia ne polynomials over C.

Main Result 1 (Corollary 5.4, Theorem 5.6). A real multia ne polynomial f has a linear Hermitian detertminantal representation if and only if all of its Rayleigh di↵erences ij (f ) factor as Hermitian squares.

One of the themes of this paper is that factorizations of Rayleigh di↵erences can capture subtle behavior of the principal minor map.

Example (Example 4.8). For example the Rayleigh di↵erences of the polynomial f a (x 1 , x 2 , x 3 , x 4 ) = x 1 x 2 x 3 x 4 x 1 x 2 x 1 x 3 x 1 x 4 x 2 x 3 x 2 x 4 x 3 x 4 + 1 factor into Hermitian squares in multiple ways, e.g. 34 (f a ) = (x 1

)(x 1 + )(x 2 )(x 2 + ). These di↵erent choices of factorization capture some non-generic behavior in the fiber of the principal minor map and lead to three inequivalent determinantal representations of f :

;

.

It is natural to ask about the fibers of the principal minor map. In the symmetric case, the fibers were characterized by Engel and Schneider [19]. Given a matrix A, both its transpose and D 1 AD, for any diagonal matrix D, have the same principal minors as A. In 1984, Loewy and Hartfiel [25] and then Loewy [36] gave su cient conditions on a matrix for its fiber under the principal minor map to be a single point, up to diagonal equivalence and taking transposes. As the example above shows, the fiber in general can be larger. Using the techniques of this paper, the second author establishes a converse to the theorem of Loewy in order to classify the fibers of the principal minor map [2].

Here we use the classical theory of determinantal representations to understand the principal minor map, including ideas from Dixon [16] on the construction of symmetric determinantal representations of plane curves. The study of symmetric and Hermitian determinantal representations is also closely related to the theory of hyperbolic and real stable polynomials, which are multivariable generalizations of real-rooted univariate polynomials. Hyperbolic and stable polynomials have found wide-spread applications in combinatorics [24,37], convex analysis [6], operator theory [26,38], probability [7], and theoretical computer science [5,34,44]. The question of which stable polynomials have definite Hermitian determinantal representations has implications in operator theory and the theory of convex optimization. See [46] for more. In general, the existence of definite Hermitian representations does not follow from the existence of general representations over C. In this paper we show that this is not the case for multia ne polynomials.

Main Result 2 (Theorem 6.4). If a multia ne real stable polynomial f has a linear determinantal representation over C, then it has a definite Hermitian determinantal representation.

From the classification above, we characterize the image of Hermitian matrices under the principal minor map by characterizing the set of real multiquadratic polynomials that factor as Hermitian squares. This leads to explicit equations and inequalities defining the image.

Main Result 3 (Corollary 7.14). The image of the set of n ⇥ n Hermitian matrices under the principal minor map is cut out by the orbit under SL 2 (R) n o S n of two inequalities and three degree-12 equations defined by polynomials in Q[a S : S ✓ 4].

An explicit description of the image of general n ⇥ n matrices remains mysterious. Huang and Oeding [28] give description of the image in the special case where all principal minors of same size are equal (the symmetrized principal minor assignment problem) where they use the cycle sums in their approach. They provide a minimal parametrization of the respective varieties in the cases of symmetric, skew symmetric and square complex matrices. Kenyon and Pemantle [30] adjust the principal minor map by adding the almost principal minors to the vector in the image and they showed that the ideal of the variety in this case is generated by translations of a single relation using the rhombus tiling.

Arguably one of the most surprising results in this paper is that, unlike the case of symmetric and Hermitian matrices, one cannot hope to define the image general n ⇥ n of the principal minor map for n by the orbit of some fixed set of equations under the group SL 2 (F) n o S n . Using factorizations of Rayleigh di↵erences, we found a family of examples that shows that such a finite description is impossible.

Main Result 4 (Theorem 8.1). For any field F, there is no finite set of equations whose orbit under SL 2 (F) n o S n cuts out the image of the principal minor map for all n.

Example (Example 8.1). For instance, we show that the coe cient vector of the polynomial

does not belong to the image of the principal minor map but satisfies the orbit of equations vanishing on the principal minors of 4 ⇥ 4 matrices under the group SL 2 (F)

This paper is organized as follows. In Section 2, we introduce terminology and the basic properties of determinantal representations and the action of SL 2 (F) n o S n . In Section 3, we give a characterization of multia ne determinantal polynomials involving the factoring of Rayleigh di↵erences. For Hermitian determinantal representations, this condition simplifies and we give an algorithm for constructing such representations from a factorization, as described in Section 4 and Section 5. In Section 6 we give a characterization of multia ne stable determinantal polynomials and prove Theorem 6.4. In Section 7, we translate these conditions into explicit equations and inequalities whose orbit under SL 2 (R) n o S n cuts of the image of Hermitian matrices under the principal minor map. Finally, in Section 8, we conclude by presenting a family of examples that disproves the existing of such a finite description for the image of general n ⇥ n matrices under the principal minor map.

Background and notation

For a commutative ring R, we use R[x] to denote the polynomial ring R[x 1 , . . . , x n ] and for f 2 R[x], we use deg i (f ) to denote the degree of f in the variable x i . For d = (d 1 , . . . , d n ) with d i 2 Z 0 , let F[x] d denote the set of polynomials with degree at most d i in x i for each i = 1, . . . , n. These form an R-module of rank

Of particular interest are multia ne polynomials, with degree  1 in each variable, and multiquadratic polynomials, with degree  2 in each variable. These are denoted by R

We use Mat n (F) to denote the set of n ⇥ n matrices with entries in F.

2.1

The action of SL 2 (R) n o S n and homogenezations

One way to interpret this action is with the multihomogenezation of f . Let f d hom in R[x 1 , . . . , x n , y 1 , . . . , y n ] d denote the polynomial

The induced action of on f d hom is just a linear change of coordinates:

Restricting to y 1 = . . . y n = 1 gives back • f . We will also use the usual homogenization of a polynomial to some total degree d, using a single homogenizing variable y. That is, for

Suppose that K is a field with an automorphic involution a 7 ! a with fixed field F. This extends to an involution on K[x] by acting on the coe cients. We will say that a polynomial q 2 F[x] is a Hermitian square if q = gg for some g 2 K[x]. To end this section, we remark that for f 2 F[x], the condition that ij (f ) is a Hermitian square is robust to homogenization. Proposition 2.1. Suppose that K is a field with an automorphic involution a 7 ! a with fixed field F.

Proof . If ij (f hom ) is a Hermitian square, then specializing to y = 1 gives a representation of ij (f ) as a Hermitian square. For the converse, let f 2 F[x] with total degree d and suppose that ij f = gg for some

y] is homogeneous of degree 2d 2. Its restriction to y = 1 equals ij f . Therefore ij (f hom ) equals y 2d 2 2m ( ij (f )) hom , which is the Hermitian square hh where h is the homogenezation of g to total degree d 1.

A description of the action of SL 2 (F) n on determinantal polynomials and matrices

One can describe the rational action of SL 2 (F) n on Mat n (F) via its action on multia ne polynomials in

Let A i denote the ith column of A and e i the vector whose ith entry is one and zero otherwise. By using the factor (c i x i + d i ) to scale the ith column, we see that

where C is the matrix with ith column C i = (a i e i + c i A i ) and B is the matrix with ith column

When the matrix C is invertible, this gives

Up to the scalar multiple det(C), the coe cients of • f are the principal minors of the matrix

where the matrices B and C are defined as above. To avoid confusion, we reserve the notation • for the action of SL 2 (F) n on the polynomial ring F[x 1 , . . . , x n ]. We see that this rational action of SL 2 (F) n on F ⇥ Mat n (F) is compatible its action on determinantal polynomials and that it commutes the action of (F ⇤ ) n on Mat n (F) by diagonal conjugation. More precisely:

Moreover, for any invertible diagonal matrix D,

Determinantal representations and the image of the principal minor map 5

Proof . We take , A as above. The first claim follows from the arguments above. More specifically, the calculations above give that

where B and C are defined as above and E denotes the matrix diag(c 1 x 1 + d 1 . . . , c n x n + d n ). Since diagonal matrices commute, we see that

Resultants

For two univariate polynomials a = P d j=0 a j t j with a d 6 = 0 and b = b 1 t + b 0 with b 1 6 = 0, the resultant of a, b with respect to the variable t is defined to be

This polynomial vanishes if and only if the univariate polynomials a and b have a common root, namely t = b 0 /b 1 . See, for example, [15, §3.5] for more on resultants. We will focus on multia ne polynomials and so focus on resultants in degree d = 1. For k = 1, . . . , n, define

In particular, if g and h both have degree one in x k , this agrees with Res x k (g, h). The benefit of this degreedependent definition is that it is invariant under the action of SL 2 (R). If f 2 R[x] has degree  1 in both x i and x j , then

Proof . By assumption we can write f

where acts of on res x k (g, h) as polynomial of multidegree  d + e 2 • 1 k with 1 k is the vector with kth entry is 1 and zero otherwise.

Proof . Write g = g 1 x k + g 0 and h = h 1 x k + h 0 where g 1 , g 0 , h 1 , h 0 are polynomials in the polynomial ring R[x j :

acting on the jth coordinate. If j = k, then

Taking coe cients with respect to {1, x k }, we see that the res

Since acts on R[x] d+e 2•1 k as the identity, this equals • res x k (g, h).

, where acts on g 1 , g 0 and h 1 , h 0 as elements of multidegree d 1 k and e 1 k , respectively. It follows that

From (4), this gives the following:

Determinantal Representations and Rayleigh Di↵erences

Let R be a unique factorization domain and denote by Mat n (R) the set of n ⇥ n matrices with entries in R.

Theorem 3.1. Let f 2 R[x 1 , . . . , x n ] be multia ne in the variables x 1 , . . . , x n with its coe cient of x 1 • • • x n equals one. Then f = det(diag(x 1 , . . . , x n ) + A) for some A 2 Mat n (R) if and only if for every i 6 = j 2 [n], the polynomials ij (f ) factor as the product g ij • g ji where (a)

In this case, we can take g ij to be the (i, j)th entry of (diag(x 1 , . . . , x n ) + A) adj where M adj represents the adjugate matrix of M .

Proof of ()). This follows from a classical equality on the principal minors of an n ⇥ n matrix, used by Dodgson [17] as a method for computing determinants. This is also known as the Desnanot-Jacobi identity or more generally as Sylvester's determinantal identity. For any i 6 = j and

where e i denotes the i coordinate unit vector in R n and M adj denotes the adjugate matrix of M . See also [33,Prop. 4.6]. For M = diag(x 1 , . . . , x n ) + A,

The last equality follows from (5) with k = i and `= j after rearranging terms.

Determinantal representations and the image of the principal minor map 7

For every i, j 2 [n], let g ij denote (M adj ) ij . Then g ij 2 R[x k : k 6 = i, j] is multia ne in x 1 , . . . , x n and ij (f ) = g ij g ji . Under an appropriate choice of indices, (5) gives

Note that g kk = @f @x k is the coe cient of x k in f and q is the coe cient of x k in g ij . Therefore g kk • g ij f • q is the resultant of g ij and f with respect to x k . Example 3.2. For n 5, one cannot remove condition (b) from Theorem 3.1. Consider

One can check that for every i, j 2 [5], ij (f ) factors as the product of two multia ne polynomials in Q[x 1 , . . . , x 5 ]. For example, 12 (f ) = x 3 x 4 x 5 (x 4 x 5 x 3 + x 4 + x 5 ). Since there is an irreducible factor involving all three variables, there is only one possible factorization of 12 (f ) as the product of two multia ne polynomials g 12 • g 21 , up to scalar multiples and switching the factors, namely g 12 = x 3 x 4 x 5 and g 21 = x 4 x 5 x 3 + x 4 + x 5 . Taking the resultant of g 21 and f with respect to x 3 gives

Each of the three quadratic factors are irreducible and so there is no way of writing this resultant as the product of two multia ne polynomials. Therefore there is no choice of polynomials g 23 and g 31 satisfying the conditions in Theorem 3.1.

x n ] be multia ne in the variables x 1 , . . . , x n and its coe cient of

x n ], then g and h are multia ne in disjoint subsets of the variables x 1 , . . . , x n and we can take their leading coe cients in these variables to be one. Moreover, if the polynomials ij (f ) have factorizations satisfying conditions (a) and (b) in Theorem 3.1, then so do ij (g) and ij (h).

Proof . For any i 2 [n], the degree of f in x i must be the sum of the degrees of g and h in x i . Since this sum of nonnegative numbers is one for each i 2 [n], we see that for some subset I ✓ [n], g is multia ne in {x i : i 2 I}, h is multia ne in {x j : j 2 [n]\I}, and deg i (h) = deg j (g) = 0 for any i 2 I and j 6 2 I.

The highest degree term in f with respect to the variables x 1 , . . . , x n , Q n i=1 x i , is the product of the highest degree terms in g and h. Therefore after rescaling, we can assume that both g and h have leading coe cient in these variables equal to 1. For i 2 I and j / 2 I, @(g • h)/@x i = h • @g/@x i and @(g • h)/@x j = g • @h/@x j . From this, one can check that ij (gh) equals h 2 ij (g) for i, j 2 I, g 2 ij (h) for i, j 2 [n]\I and zero otherwise. Suppose that for i, j 2

Since m ij , m ji are multia ne, they both must be divisible by h, leaving mij mji = ij (g), where mij , mji are multia ne in x i for i 2 I. Moreover, for k also in I,

showing that mik mkj = res x k ( mij , g). The desired factorization for ij (h) with i, j 2 [n]\I follows similarly.

Proof of ((). Suppose that f is irreducible and multia ne of degree n. Let G denote the n ⇥ n matrix with (i, j)th entry g ij for i 6 = j and g ii := @f @xi for i = j. We claim that all of the 2 ⇥ 2 minors of G lie in hf i. This is immediate for the symmetric minors, as

@xi@xj . Moreover, since @f @x1 is the coe cient of x 1 in f , the resultant res x1 (g ij , f) has the form @f @x1 g ij qf for some q. This gives g 11 g ij g i1 g 1j = qf . Finally, suppose that i, j, k, `are all distinct. Then

Since f is irreducible and g 11 = @f /@x 1 has smaller degree, g 11 is not a zero-divisor in R[x 1 , . . . , x n ]/hf i.

Therefore the minor g ij g kl g il g kj belongs to hf i.

From this it follows that f k 1 divides the k ⇥ k minors of G for every 2  k  n, see [42,Lemma 4.7]. In particular, f n 2 divides the entries of the adjugate matrix G adj . Let

Also f n 1 divides det(G), and since these both have degree n(n 1), there must be some constant 2 R for which det(G) = • f n 1 . We can see that = 1 by taking top degree terms. Since deg(f i ) = n 1 and for all i 6 = j, deg(g ij )  n 2, the leading degree term of det(G) comes uniquely from the product of the diagonals

On the righthand side, the leading degree term of f n 1 is also (

Note that the entries of M have degree  (n 1) 2 n(n 2) = 1, so we can write M as M 0 + P n i=1 x i M i for some matrices M i 2 R n⇥n . We claim that P n i=1 x i M i = diag(x 1 , . . . , x n ). To see this, first note that a non-principal (n 1) ⇥ (n 1) minor of G involves at most n 2 elements from the diagonal of G and therefore has degree  (n 2)(n 1) + (n 2) = n(n 2), since the o↵-diagonal entries of G have degree  n 2. Therefore the o↵ diagonal entries of M have degree  n(n 2) n(n 2) = 0.

Moreover in the expansion of any principal (n 1) ⇥ (n 1) minor of G, there is a unique term of degree (n 1) 2 , namely the product of the leading terms of the diagonal elements, Q j6 =i LT(g jj ). We can therefore take the leading terms (7) to find that

Finally, for general f , we take a factorization of

By the arguments above, f k has a determinantal representation of the correct form. Taking a block diagonal representation of these representations (and permuting the rows and columns if necessary to reorder x 1 , . . . , x n ) gives a determinantal representation for f . Remark 3.4. In Theorem 3.1 the matrix G = (g ij ) ij and the corresponding determinantal representation diag(x 1 , . . . , x n ) + A of f satisfy

Multia ne algebra for constructing Hermitian factorizations

In this section, we develop an algorithm for constructing factorizations that satisfy the conditions in Theorem 3.1.

To do this, we find it most convenient to work in the following level of generality throughout this section. Let S be a unique factorization domain with an automorphic involution a 7 ! a. We use 0 and 1 to denote the additive and multiplicative identities of S. The map S ! S given by a 7 ! a then must satisfy

for all a, b 2 S. Let R be the subring of elements fixed by this automorphism, that is R = {a 2 S : a = a}.

The example of interest is the ring S = C[x n+1 , . . . , x m ] of polynomials with complex coe cients with the involution given by complex conjugation. In this case the fixed ring is the subring of polynomials whose coe cients are real,

In what follows, we will build up tools to show that under these assumptions Algorithm 1 produces the desired representation of f . We first exploit some properties of multia ne polynomials. For any disjoint subsets

Note that if f is multia ne in x 1 , . . . , x n , then for any 1  i < j  n, we have

From this, one can check that the formula for ij f can be written without x i , x j :

If in addition we assume that f and all its partial derivatives are irreducible, then ij (f ) will have degree exactly 2 in each variable, as the following lemma shows.

Proof . For 1  i < j  n, we write ij (f ) as a quadratic polynomial in the variable x k :

We next use ring maps given by taking resultants with f . For any i = 1, . . . , n, define

as defined in (3). For instance if we restrict to polynomials g = g i x j + g i with degree one in x i , then

More generally, ' i (g) is obtained from g by substituting f i /f i for x i and multiplying the result by f d i where d = deg i (g). First we will start by listing some of the properties of these maps. Lemma 4.3. If f satisfies Assumptions 4.1, then, for all g, h 2 S[x], the maps ' 1 , . . . , ' n satisfy the following:

Proof . We will prove (5), (6), and ( 7) and all the other properties follow similarly by direct computations. To prove property (5), we write g = g ij x i x j + g j i x i + g i j x j + g ij , then

If deg j (g) = 1 and g j / 2 hf j i, then we claims that the coe cient of x 2 j in ' i (g) is nonzero. To see this, suppose to the contrary that coe↵(' i (g),

By assumption, the polynomial

, up to a constant in R. In particular, the greatest common divisor s = gcd(f ij , f i j ) belongs to S. Then 1 s f i j and 1 s f ij are relatively prime and so the equation above gives that they divide g i j and g ij respectively. From this we get that sg j 2 hf j i, contradicting Assumption 4.1.5. Therefore coe↵(' i (g), x 2 j ) 6 = 0. Applying the map ' j to the expression above and simplifying then gives

Similarly, for (6), suppose that deg i (g) = 1 and deg j (g) = 0. The calculation of ' j ' i (g) above holds with g ij = g i j = 0, giving

The second equality comes from the observation that, as in the argument above, the coe cient,

To prove (7), we write g as g = g j x j + g j and we use

showing that ' j (g) ⌘ f j g modulo hf i.

Lemma 4.4. If f satisfies Assumptions 4.1 and ij (f ) = pp for some 1  i < j  n, then for every k 2 [n] with k 6 = i, j, there is a factorization of each ik (f ) and jk (f ) into qq and rr, respectively, such that ' k (p) = qr.

Proof . Since ik (f ) and jk (f ) factor into two conjugates, we can write

where a 1 , . . . , a s , b 1 , . . . , b t are irreducible in S[x 1 , . . . , x m ] and multia ne in x 1 , . . . , x n . Then

After switching a i with a i and b i with b i if necessary, we get

multia ne polynomials such that ik (f ) = qq and jk (f ) = rr as desired.

Lemma 4.5. If f satisfies Assumptions 4.1 and for some distinct

Proof . We will prove the first equality and the second holds similarly. First notice that deg k (p) = 1 and deg j (p) = 0. Also, deg j (r) = deg k (r) = 0. Then using the properties in Lemma 4.3 we get

Since jk (f ) = rr, dividing the above equation by r gives the desired result.

The following algorithm gives the desired factorizations of ij (f ) into g ij g ij that satisfy the hypothesis of Theorem 3.1, which will in turn give the desired Hermitian determinantal representation in Theorem 5.1.

Algorithm 1 Compatible Hermitian factorizations of Rayleigh di↵erences

Proof . (a) This is immediate for k = 2. For 2 < k  n, notice that 1k (f ) has degree two in x 1 , . . . , x n . Let `2 [n]\{1, k} and let p k,j , p k,j be the unique irreducible factors of 1k (f ) with degree one in x `. By construction, g 1k divides Q j , which in turn divides 1k (f )/p k,j . Since this quotient only has degree one in x `, g 1k must have degree  1 in x `.

(b) follows directly from construction.

(c) We proceed by induction on k. It is trivially true for k = 2. For the inductive step, we will prove the claim for 1k (f ) and the other cases follow. By construction, g 1k g 1k divides 1k (f ). To see this, note that for each j = 1, . . . , m k in Algorithm 1, we can take q j = p k,j if p k,j divides g 1k and q j = p k,j otherwise. Then, by construction, g 1k divides q = Q m k j=1 q j and q • q = 1k (f ), showing that g 1k • g 1k divides 1k (f ). Suppose for the sake of contradiction that 1k (f ) 6 = g 1k g 1k . Then there is some irreducible factor p of 1k (f ) such that pp does not divide g 1k g 1k . We claim that for every 1 < i < k, either p or p divides ' k (g 1i ). By induction, for 1

Since p is irreducible and divides 1k (f ), it must divide either ' k (g 1i ) or ' k (g 1i ) = ' k (g 1i ). In the latter case, p divides ' k (g 1i ). Since neither p nor its conjugate divide g 1k , it follows from the construction that neither p nor p divide Q 0 = gcd{ 1k (f ), ' k (g 12 ), . . . , ' k (g 1(k 1) )}. Hence there exists distinct 2  i, j < k such that neither p divide ' k (g 1i ) nor p divide ' k (g 1j ). By switching p and p if needed, we can assume i < j.

By induction (on k), we know that 1i

) is divisible by p while p divides 1k (f ) and this gives the desired contradiction. Therefore 1k (f ) = g 1k g 1k .

For 1 < i < k, we calculate that

x n ] satisfies Assumptions 4.1, then there exists a factorization of ij (f ) into g ij g ji such that g ij 2 S[x 1 , . . . , x n ], g ji = g ij , and ' k (g ij ) = g ik g kj for all distinct 1  i, j, k  n.

Proof . Let {g ij : 1  i < j  n} be the polynomials given by Algorithm 1 and for i < j let g ji = g ij . By Proposition 4.6, ij (f ) = g ij g ij = g ji g ij . Since ij (f ) is quadratic in each variable x 1 , . . . , x n , then g ij is multia ne in x 1 , . . . , x n . We will show that ' k (g ij ) = g ik g kj for all distinct i, j, k. Assuming that i < j < k and using Proposition 4.6 we get res x k (g 1i , f) = ' k (g 1i ) = g 1k g ki = g 1k g ki , and

Multiplying the above two equations and using Properties 4.3 we get

Using Proposition 4.6 again, we know that ' j (g 1i ) = g 1j g ji and Lemma 4.5 implies that ' 1 (g ji ) = g j1 g 1i . Again using Properties 4.3 we find that

g ki g jk = g ik g kj . Using Lemma 4.5, we get that ' j (g ik ) = g ij g jk and ' i (g jk ) = g ji g ik as desired.

Example 4.8.

For any distinct i, j, k, `2 [4], the Raleigh di↵erences of f with respect to x i and x j are

Using Algorithm 1, we can choose g 12 as any multia ne factor of 12 (f ) of degree two. There are four possibilities (x 3 ± )(x 4 ± ) and one can check that every choice works. We choose g 12 = (x 3 + )(x 4 + ) and compute

To choose g 13 we compute gcd( 13 (f ), ' 3 (g 12 )) = (x 2 )(x 4 )(x 4 + ). Thus, up to scalar multiples, we have two choices for g 13 , namely (x 2

)(x 4 + ) or (x 2 )(x 4

). We will choose the first option, giving g 23 = (' 3 (g 12 )/g 13 ) = (x 1 + )(x 4 + ).

To find g 14 , we compute the gcd( 14 (f ), ' 4 (g 12 ), ' 4 (g 13 )) = (x 2 )(x 3 ) and we get g 24 and g 34 similarly. Taking g ji = g ij for j > i and g ii = @f /@x i , the final matrix is G = 0 @

A .

Now we compute the adjugate matrix of G and divide its entries by f 2 we get

where D = diag( 1, 1, 1, 1). Indeed one can check that det(M ) = f . If we instead choose

), which must be g 13 . The other entries are also determined. Running over all choices gives a total of six representations of f , up to diagonal equivalence, namely the following three matrices and their transposes:

; .

The six matrices A obtained by specializing x i = 0 for all i all have principal minors given by the coe cients of f . Since f is symmetric under the action of the symmetric group, S 4 . One could also consider the orbit of these matrices under conjugation by permutation matrices A 7 ! P 1 AP . One can check that, up to diagonal equivalence, these six matrices belong to a single orbit under this action, as predicted by [28,Corollary 3.7] Let K be a field with an automorphism a 7 ! a of order two. Let F be the fixed field of this automorphism. We call a matrix A 2 Mat n (K) Hermitian if A = A T .

Consequences of Algorithm 1

Theorem 5.1. Let f 2 F[x 1 , . . . , x m ] be a polynomial of total degree n  m that is multia ne in x 1 , . . . , x n and coe cient of x 1 • • • x n equal to one. There exist Hermitian matrices A n+1 , . . . , A m , A 0 so that

Lemma 5.2. Let F be an infinite field and f 2 F[x 1 , . . . , x m ] be irreducible, multia ne in the variables x 1 , . . . , x n with n  m and have coe cient of

x n ] up to a constant in R for all j = 1, . . . , n and the coe cient of Q n i=1 x i is nonzero. As in Assumptions 4.1.5, we say that an element

Suppose that the multidegree of f in x 1 , . . . , x m is given by d 2 N m . By assumption, d i = 1 for i = 1, . . . , n. Note that X is contained in the union S e X e where X e = (a, c) 2 F 2 :

and the union is taken over all vectors e 2 N m that are coordinate-wise  d with the property that e i = 1 and e k = 0 for some i, k 2 [n]\{j}. Here F alg denotes the algebraic closure of F. To see this, suppose (a, c) 2 X , meaning af j + cf j = g • h where g, h 2 R[x 1 , . . . , x n ] are not elements of R. In particular, for some i, k 2 [n]\{j}, deg i (g) > 0 and deg k (h) > 0. Since af j + cf j has degree at most one in each of x i and x k , it follows that deg i (g) = deg k (h) = 1 and deg k (g) = deg i (h) = 0. Taking e 2 N m to be the multidegree of g gives g 2 F[x 1 , . . . , x m ] e and h 2 F[x 1 , . . . , x m ] d e . Then (a, c) 2 X e . By the projective elimination theorem, the image F alg [x 1 , . . . , x m ] e • F alg [x 1 , . . . , x m ] d e is Zariski-closed in the vectorspace F alg [x 1 , . . . , x m ] d . Intersecting with the F-subspace spanned by {f j , f j } shows that X e and hence [ e X e is Zariski-closed in (F) 2 . Therefore this union is either all of F 2 or is an algebraic set of codimension 1. Suppose, for the sake of contradiction, that it is all of F 2 . Then there exists some e for which X e = F 2 . By assumption, there are i, k 2 [n] so that e i = 1 and (d e) k = 1. By Proposition 2.3, it follows that for all a, c 2 F, ik (af j + cf j ) is identically zero. For c = 1, this corresponds to the evaluation of ik (f ) at x j = a. It follows that the polynomial ik (f ) is identically zero (see e.g. [3]). Proposition 2.3 then implies that f factors as the product of two elements in R[x 1 , . . . , x n ] neither of which belong to R.

Proof of Theorem 5.1. ()) By Theorem 3.1, for all i 6 = j, ij (f ) factors as g ij g ji where g ij the (i, j)th entry of the adjugate matrix (diag(x 1 , . . . , x n ) + A) adj . Since A is Hermitian,

(() First assume that F is an infinite field and that f is irreducible in F[x 1 , . . . , x m ]. Let R = F[x n+1 , . . . , x m ]. By Lemma 5.2, there exists a generic 2 SL 2 (F) n such that the partial derivatives of • f are all irreducible in R[x 1 , . . . , x n ] and the coe cient of x 1 • • • x n in • f is nonzero. Then by Corollary 4.7, there exists

Acting by 1 on • f and using Proposition 2.4 we get

h kj and thus using Theorem 3.1 we get a determinantal representation of f over K and since h ji = h ij , the matrix will be Hermitian. Now suppose that f is reducible. Let g be an irreducible factor of f with deg i (g) = 1 for i 2 I ✓ [n] and deg j (g) = 0 for j 2 [n]\I where the coe cient of Q i2I x i equals one. By Lemma 3.3, for every i, j 2 I, ij (g) is a Hermitian square. Therefore g has a determinantal representation of the correct form, g = det(diag(x i : i 2

). Taking a block diagonal representation of these representations (and permuting the rows and columns to reorder x 1 , . . . , x n ) gives a determinantal representation for f . Now suppose that F is a finite field. Consider the transcendental extension of F to F(t) and of K to K(t). Then by the arguments above, f = det (diag(x 1 , . . . , x n ) + A(t)) for some Hermitian matrix A(t) 2 Mat n (K(t)). The (i, j)th entry of A(t) can be written as a ij = pij qij where p ij , q ij 2 K[t] are relatively prime and the polynomial q ij is nonzero. Specializing to t = 0 will give a determinantal representation of f over K. To do this, we need to check that q ij (0) is nonzero for all i, j. If a ij = 0, then we can take p ij = 0 and q ij = 1. Suppose that for some i, j 2 [n], p ij is nonzero and q ij (0) = 0. Then t divides q ij and so also divides q ij . Notice that

x k A are both in F and hence p ij p ij = rq ij q ij for some r 2 F ⇤ . We get the desired contradiction by noticing that t 2 divides the right-hand side of the equation, while it does not divide the left-hand side since p ij and q ij are relatively prime. Therefore we can specialize both sides of the equation f = det (diag x 1 , . . . , x n + A(t)) to t = 0, which gives a Hermitian determinantal representation of f .

showing that the coe cient vector of f is not in the image of Mat 3 (F 2 ) under the principal minor map.

Consider the quadratic extension K = F 2 [↵]/h↵ 2 + ↵ + 1i. The map ↵ 7 ! 1 + ↵ extends to an automorphic involution on K that fixes F 2 . Over K, the Rayleigh di↵erences factor into multia ne polynomials, namely ij (f ) = (x k + ↵)(x k + + ↵), for distinct i, j, k. As then guaranteed by Theorem 5.1, f has a Hermitian determinantal representation over K:

Corollary 5.4. Let f 2 R[x 1 , . . . , x m ] be a polynomial of total degree n  m that is multia ne in x 1 , . . . , x n and coe cient of x 1 • • • x n equals to one. There exist Hermitian matrices A n+1 , . . . , A m , A 0 so that

This provides a partial converse to [33,Corollary 4.3], which states that if some power of a polynomial f has a definite determinantal representation, then for all i, j, the Rayleigh di↵erence ij (f ) is a sum of squares. In particular, Hermitian representations of f give real symmetric determinantal representations of f 2 . We might hope for the following.

Conjecture 5.5. If f 2 R[x 1 , . . . , x m ] is multia ne in x 1 , . . . , x n with n  m, and coe cient of x 1 • • • x n is nonzero, then some power of f has a definite real symmetric determinantal representation if and only if for all i, j, ij (f ) is a sum of squares in R[x 1 , . . . , x m ].

Other multia ne determinantal representations

In this section we restrict ourselves to fields and consider the set of multia ne determinantal polynomials of the form

for some 2 F, some matrix V = (v 1 , . . . , v m ) 2 K n⇥m and some n ⇥ n Hermitian matrix W . Note that when we take V to be the n ⇥ n identity matrix and = 1, this is the principal minor polynomial f W . When n < m, the coe cient of x Proof . ()) Without loss of generality, we show that 12 (f ) is a Hermitian square. First suppose v 1 and v 2 are linearly dependent, i.e. let v 1 = ↵v 2 for some ↵ 2 K. Then we get v 1 v ⇤ 1 = ↵↵v 2 v ⇤ 2 and f (x 1 , . . . , x m ) = f (0, ↵↵x 1 + x 2 , x 3 , . . . , x m ). Taking partial derivatives shows that @f @x1 = ↵↵ @f @x2 and that @ 2 f @x1@x2 = 0. Then 12 (f ) = (↵ @f @x2 )(↵ @f @x2 ) is a Hermitian square. If v 1 and v 2 are linearly independent, then there is an invertible matrix U 2 K n⇥n with Uv 1 = e 1 and Uv 2 = e 2 . Then

where e v i = Uv i and f W = UW U ⇤ . These matrices are still Hermitian and so by equation ( 6), 12 (f ) is Hermitian square.

(() Let d = deg(f ). We can assume, without loss of generality, that the coe cient of

Moreover since the set of polynomials of the form ( 8) is invariant under scaling, we can assume that this coe cient equals one. By Theorem 5.1, there are Hermitian matrices A 0 , A d+1 , . . . , A m so that

We take W = A 0 . By Lemma 5.7 below, for every k = n + 1, . . . , m, the rank of A k equals the degree of f in x k , which is one. It remains to show that the matrix A k has the form v k v ⇤ k for some v k 2 K d . By homogenizing and specializing variables to zero, it su ces to consider polynomials of the form f = det (diag(x 1 , . . . , x d ) + x d+1 A) where A 2 K n⇥n is Hermitian and rank-one. Then f can be written as

, where A jj is the jth entry of A. Then for j = 1, . . . , d,

By assumption, j(d+1) (f ) is a Hermitian square, and so we see that A jj = ↵ j ↵ j for some ↵ j 2 K. Since A has rank one, we can write it as uu ⇤ for some 2 F ⇤ and u 2 K n . If u j 6 = 0, then u j u j = ↵ j ↵ j , meaning that = for = ↵ j /u j . It follows that A = vv ⇤ for v = u.

Lemma 5.7. If f = det(diag(x 1 , . . . , x n ) + P m j=n+1 x j A j + A 0 ) where A j 2 K n⇥n are Hermitian. Then the rank of A j equals the degree f in the variable x j .

Proof . The bound deg j (f )  rank(A j ) follows from the Laplace expansion of the determinant. To see equality, it su ces to take j = m = n + 1 and A 0 = 0. Let f A be the polynomial

From this we see that the degree of f in the variable y equals the size of the largest nonzero principal minor of A. By the so-called Principal Minor Theorem [31, Strong PMT 2.9], this coincides with the size of the largest nonzero minor of A, i.e. rank(A). Therefore for a general polynomial f = det(diag(x 1 , . . . , x n ) + P m j=n+1 x j A j + x 0 A 0 ), the restriction to x k = 0 for k 2 {n + 1, . . . , m}\{j} and x 0 = 0 has degree rank(A j ) in x j , showing that deg j (f ) rank(A j ) This immediately gives the invariance of the set of determinantal polynomials.

Determinantal Stable Polynomials

In this section we consider polynomials over R and C and show that any real stable multia ne polynomial with a complex linear determinantal representation has a definite Hermitian determinantal representation (Theorem 6.4). Moreover, if the original polynomial is irreducible, then the matrix is diagonally similar to a Hermitian one (Theorem 6.6).

We build up to the proofs of these statements with a series of useful lemmas.

Lemma 6.1. Let f 2 R[x 1 , . . . , x m ] be multia ne in the variables x 1 , . . . , x n for some n  m with coe cient of

Proof . For each S ⇢ [n], the set of 2 SL 2 (R) n for which @ S ( • f ) is irreducible is Zariski-open. Therefore it su ces to show that this set is nonempty for each S ⇢ [n]. Then the intersection of these nonempty, Zariski-open sets will be nonempty and Zariski open.

We will proceed by induction on |S|. For |S| = 0, this is immediate, so suppose that |S| 1 and let i 2 S. Note that @ S (f ) = @ i @ S\{i} f . By induction, for generic

Therefore, up to a scalar multiple, @ S\{i} ( • f ) satisfies the hypothesis of Lemma 5.2, and hence for generic e 2 SL 2 (R) acting on the ith coordinate,

Here we use that e commutes with the di↵erential operator @ S\{i} , since e acts as the identity in the coordinates indexed by elements of S\{i}. It follows that for a generic element

Since g is globally nonnegative on R m , g(x 1 , p) is nonnegative on R and so its leading coe cient a(p) must be nonnegative. Lemma 6.3. Suppose g, h 2 C[x 1 , . . . , x m ] are multia ne in x 1 , . . . , x n and @ [n] g and @ [n] h are nonzero polynomials in x n+1 , . . . , x m of total degree at most one. If the product g • h has real coe cients and is nonnegative as a function on R m , then h is a positive scalar multiple of g, i.e. h = g for some 2 R >0 .

0 , we conclude > 0, as desired. (n 1) Now suppose n 1 and write g = g n x n + g n and h = h n x n + h n . Since g • h is real and nonnegative, so is its coe cient of x 2 n , g n • h n . In particular, g n , h n satisfy the hypothesis of the theorem and so by induction, h n = g n for some 2 R >0 . Moreover, for every a 2 R m 1 with g n (a) 6 = 0, the roots (in x n ) of the specialization of g • h at x = a come in complex conjugate pairs. It follows that h n /h n = g n /g n as rational functions in C(x k : k 6 = n). Together with h n = g n , this gives that h = g. Moreover, since g • h = • g • g is nonnegative on R n , we see that > 0. Theorem 6.4. Let f 2 R[x 1 , . . . , x m ] be stable and complex determinantal, i.e.

for some n ⇥ n complex matrices A j . Then there exists Hermitian matrices B 0 , B n+1 , . . . , B m for which

If this coe cient is zero, then Lemma 3.3 implies that @ [n]\{i,j} (f ) is reducible. Let i < j 2 [n]. Since f is determinantal, by Theorem 3.1, the polynomial ij (f ) factors as g ij • g ji where g ij , g ji are multia ne in {x k : k 2 [n]\{i, j}} and have total degree  n 1. In particular, the coe cient of Q k2[n]\{i,j} x k in both g ij and g ji has degree  1 in x n+1 , . . . , x m . By the arguments above we can assume this coe cient is nonzero. Since f is real stable, ij (f ) is also globally nonnegative on R n [9]. Therefore by Lemma 6.3, g ji = g ij for some g ij . It follows that ij (f ) factors as a Hermitian square h ij • h ij where h ij = p g ij . Theorem 5.1 then gives the desired Hermitian determinantal representation. Now suppose f is reducible, say f = f 1 • • • f r where each factor f k is irreducible and multia ne in the variables x i for i 2 I k ⇢ [n]. Each factor is stable. Moreover, by Lemma 3.3, ij (f k ) is either zero or factors as a product of two polynomials that are multia ne in {x `: `2 I k } and with total degree  |I k | 1. Since f k is irreducible, the arguments above show that for every i, j 2

`is a Hermitian square. Theorem 5.1 then gives the desired Hermitian determinantal representation. Remark 6.5. Theorem 6.4 cannot hold for arbitrary real stable polynomials. For example, consider f to be the basis generating polynomial of the Vámos matriod, defined in [10]. It was shown by Wagner and Wei [47] that f is stable. By the theory of matrix factorizations, some power f r of f has a complex linear determinantal representation (see [46, §3.3]). This power is necessarily stable, but as shown by Brändén [10], f r does not have a definite Hermitian determinantal representation.

When f is reducible, one can easily construct determinantal representations of f that are not Hermitian by taking block upper triangular representations. For example, x 1 x 2 equals det

. However, when f is irreducible and real stable, we see that all complex linear determinantal representations are Hermitian, up to conjugation by diagonal matrices.

Theorem 6.6. Let f 2 R[x 1 , . . . , x m ] be stable, irreducible, and complex determinantal, i.e.

for some n ⇥ n complex matrices A j . Then there exists a real diagonal matrix D 2 R n⇥n such that D 1 A j D is Hermitian for all j.

Proof . By Lemma 6.1, there exists

. By Proposition 2.2, we can replace f by • f , and thereby assume that all the coe cients of

. By Lemma 6.3, we can conclude that for each 1  i < j  n, there is some ij 2 R >0 such that a ij = ij a ji .

We claim that the scalars ij satisfy ij = ik kj for all 1  i < k < j  n. For simplicity, we show this for i = 1, k = 2, j = 3 and the proof in general is virtually identical. By the arguments above, the starting determinantal representation of f has the form Recall that by (6), the polynomial ij (f ) factors as (M adj ) ij (M adj ) ji where M = diag(x 1 , . . . , x n ) + A(x). These polynomials are a ne in x k for k 2 [n]\{i, j}. In particular, g := @ [n]\{1,2,3} (M adj ) 13 = a 12 a 23 a 13 (x 2 + a 22 ), and h := @ [n]\{1,2,3} (M adj ) 31 = 12 23 a 12 a 23 13 a 13 (x 2 + a 22 ).

These polynomials satisfy the hypotheses of Lemma 6.3, and so there is some µ 2 R >0 for which h = µg. Since a ij is nonzero for all i, j and a 22 is invariant under conjugation, we see that 12 23 = µ = 13 . More generally ij = ik kj for any i < k < j.

Now define D = diag(1, p 12 , . . . , p 1n ). For i < j, 1j = 1i ij we calculate the (i, j)th and (j, i)th entries of D 1 A(x)D as

7 Defining the set of factoring multiquadratic polynomials and the image of the principal minor map

In this section we give a complete characterization of the image of the principal minor map of Hermitian matrices using the characterization of Hermitian multia ne determinantal polynomials from Section 5 and the characterization of multiquadratic polynomials that are Hermitian squares. This set is invariant under the action of SL 2 (R) n o S n and we derive the defining equations and numerical conditions as the orbit of a finite set under the action of this group, where R is a unique factorization domain. In this section, we will restrict to rings and fields of characteristic 6 = 2.

where char(R) 6 = 2. The polynomial g factors into two linear factors in R[x] if and only if its discriminant Discr x (g) is a square in R.

Proof . ()) If g factors, then it has a root in the fraction field of R. By the quadratic formula, this implies that the discriminant is a square in frac(R), and hence in R.

(() Suppose that b 2 4ac = q 2 for some q 2 R. We can rewrite this as (b q)(b + q) = 4ac. Since R is a unique factorization domain, there is some choice of factorization of a = a 1 a 2 and c = c 1 c 2 so that b q = 2a 1 c 1 and b + q = 2a 2 c 2 . If a = 0, then g factors as 1 • (bx + c), so we can assume a 6 = 0. We can then write g as

This lemma does not hold over rings of characteristic two. See [14, Section 2.4, Exercise 6] for further discussion. Note that for g 2 R[x, y] MQ , Discr x (g) is a polynomial of degree 4 in y whose coe cients are quadratic in the coe cients of g.

from which we conclude that (↵x 2 + x + ) 2 = h(x). Similarly, if b 1 is nonzero then so are b 0 and . We can define = b 1 /(2 ) and use 4b 0 b 1 b 2 = b 3 1 + 8b 2 0 b 3 to conclude that (↵x 2 + x + ) 2 = h(x). In either case, evaluating at x = 1 gives that ↵ + + = ± , and = ±

Proof . It su ces to show that we can recover the conditions in Lemma 7.2, which we can do this with three elements of SL 2 ({0, ±1}): the identity,

⌘ , representing the fractional linear transformations x 7 ! 1/x and x 7 ! x + 1, respectively. Note that

we recover that all of these are squares in R. The element 1 induces the transposition b k 7 ! ( 1) k b 4 k for each k. One can quickly check that we recover the two missing cubics from this action of Using the action of S n , we see that every irreducible factor of g must have degree  1 in each variable.

Remark 7.6. For every choice of i 6 = j 2 [n] and 2 I n 3 we obtain three equations by evaluating B xj (Discr xi (g)), C xj (Discr xi (g)) and D xj (Discr xi (g)) at the point , along with additional two polynomials from the two missing analogous polynomials in Lemma 7.2, which can be recovered from the SL 2 -action on x j . This gives a total of 5n(n 1)13 n 3 sextic equations in the coe cients of g. Lemma 7.7. Let S be a unique factorization domain with char(S) 6 = 2 and an automorphic involution a ! a and let R be the fixed ring under this involution. The quadratic polynomial g = ax 2 + bx + c 2 R[x] is a Hermitian square in S[x] if and only if a and c are Hermitian squares in S and the discriminant Discr x (g) = q 2 with q 2 S[x] and q = q.

Proof . ()) If g factors into two conjugates (sx + t)(sx + t), then a = ss and c = tt and Disc x (g) = b 2 4ac = (st + ts) 2 4sstt = (st ts) 2 which satisfies the desired property.

(() Assume that b 2 4ac = q 2 such that q = q. If a = 0, then b = ±q and thus b = b. Since b 2 R, then b = 0 and g = c is a Hermitian square as desired. If a 6 = 0, then (b q)(b + q) = (b q)(b q) = 4ac = 4sstt, where a = ss and c = tt. Thus, after relabeling if needed, we may assume that b q = 2st. Thus, we can write g as

Theorem 7.8. Let S be a unique factorization domain with char(S) 6 = 2 and an automorphic involution a 7 ! a. Let R be the fixed ring of this automorphism with |R| 13. The polynomial To prove that g is a Hermitian square, we will proceed by induction on n. The case n = 2 is trivially satisfied. For the inductive step, write g as g = p 2 x 2 1 + p 1 x 1 + p 0 for some p 2 , p 1 , p 0 2 R = R[x 2 , . . . , x n ]. By induction we see that p 2 and p 0 are both Hermitian squares and as g is a product of multia ne polynomials, then by Lemma 7.1, we see that Disc x1 (g) = p 2 1 4p 2 p 0 = q 2 for some q 2 S[x 2 , . . . , x n ]. Since p 2 1 4p 2 p 0 2 R, then q 2 2 R and so q = q or q = q. In the former case, Lemma 7.7 implies that g is a Hermitian square and we are done. Otherwise we get ( • q)| x=0 = ( • q)| x=0 for all 2 SL 2 (R) n 1 . Notice that by induction on the other hand, ( • g)| x=0 is a Hermitian square and hence

Thus we conclude that (

implies that q ⌘ 0 and thus q = q and we apply Lemma 7.7 again to deduce that g is a Hermitian square. Assumptions 7.9. We take F to be a field of char(F) 6 = 2 with |F| 13 and K to be a degree-two extension field. Let denote the square root of the discriminant of the minimal polynomial of this field extension. Then K = F( ) and the involution ! = extends to an automorphism of K with fixed field F.

Remark 7.10. In the field K, q = q is equivalent to requiring q = r for some r 2 F.

Lemma 7.11. Let F, K satisfy Assumptions 7.9 and let g

Determinantal representations and the image of the principal minor map 21

Proof . Write g as g = p 2 x 2 1 + p 1 x 1 + p 0 where p 2 , p 1 and p 0 are quadratics in F[x 2 ]. Using Lemma 7.7, we see that g is a product of two conjugate factors if and only if p 2 and p 0 are product of two conjugates in K[x 2 ] and Disc x1 g = q 2 where q = q for some q 2 K[x 2 ]. Notice that by Remark 7.10, this condition is equivalent to q = r where r 2 F[x 2 ] and thus requiring that . By Corollary 7.3, this discriminant being a square is equivalent to conditions (ii) and (iii). Now we are ready to give a complete characterization of the image of the principal minor map of Hermitian matrices using the characterization of Hermitian multia ne determinantal polynomials from Section 5 and the characterization of multiquadratic polynomials that are Hermitian squares.

Recall that to each element a = (a S ) S✓[n] in F 2 n we associate the multia ne polynomial

For n = 3, the discriminant of the Rayleigh di↵erence

. This is equivalent to the three hypotheses of Theorem 7.12, which in turn are equivalent to the three hypotheses of the theorem.

Taking K = C with the action complex conjugation then gives the following. Let F be a field and for n 2, consider the multia ne polynomial f 2n+1 2 F[x 1 , . . . , x 2n+1 ] given by

where we take x 2n+2 = x 2 . We show that this polynomial is not determinantal, i.e. its vector of coe cients do not belong to the image of the principal minor map, but is determinantal after specializing any one variable:

Theorem 8.1. There is no finite set of equations whose orbit under SL 2 (F) n o S n set-theoretically cuts out the image of the principal minor map for all n.

Let I n ⇢ F[a S : S ✓ [n]] be the homogeneous ideal of polynomials vanishing on the image of n ⇥ n matrices under the principal minor map in P 2 n 1 (F). There is a natural inclusion of

Theorem 8.2. The coe cient vector of the polynomial f 2n+1 belongs to the variety of polynomials in the orbit (SL 2 (F) 2n+1 o S 2n+1 ) • I 2n but not the variety of I 2n+1 .

The proof of this theorem relies on the fact that the coe cient of any generic specialization of f 2n+1 lies in the image of the principal minor map, up to scaling. One key observation is that the Rayleigh di↵erences of f 2n+1 do not all factor as the product of two multia ne polynomials, but do have such factorizations after specializing any one variable. We show this explicitly by writing down the determinantal representations of these specializations.

Lemma 8.3. The rational function 1 1+x1 f 2n+1 can be written as det(diag(x 2 , . . . , x 2n+1 ) + A) where for 2  i, j  2n + 1,

if i is odd, j is even, and i > j, x 1 /(1 + x 1 ) if i is odd, j is even, and i < j,

if i = 2, j = 2n + 1, and 0 otherwise.

Proof . Let D denote the determinant of the matrix M = det(diag(x 2 , . . . , x 2n+1 ) + A). By definition, D is a polynomial in 1 1+x1 , x 1 , x 2 , . . . , x n . Moreover the entries for which x 1 + 1 appears in the denominator form a square submatrix whose rows correspond to odd indices and whose columns correspond to even ones. It has the form

where J is the all ones matrix and U is an upper triangular matrix with U ij = 1 for i < j and U ij = 0 otherwise. Since J has rank one, the exponent of 1 + x 1 appearing in the denominator of any minor of this matrix is at most one. This also shows that despite the many appearances of x 1 in numerator of this matrix, it does not appear in the numerator of any minor. There is only one other entry in M containing x 1 , and so the determinant D can be written as (x 1 + 1) 1 p 1 + p 2 where p 1 and p 2 are multia ne in x 1 , . . . , x 2n+1 . Moreover, the only term in the Laplace expansion of the determinant of M avoiding this submatrix is the product of the diagonal Q 2n+1 j=2 x j . Therefore we can write D as (x 1 + 1) 1 p where p is multia ne in x 1 , . . . , x 2n+1 . To show that p = f 2n+1 it su ces to show that they have the same specialization at x 1 = 0 and the same coe cient of x 1 .

When we specialize x 1 to zero, M becomes a block upper-triangular matrix with diagonal blocks of the form

◆

. Its determinant agrees with the specialization of Consider the rational function g obtained by inverting x 1 in 1 1+x1 f 2n+1 , which is

!

.

Let M 0 be the matrix obtained from M by replacing x 1 by x 1 1 and then multiplying the column indexed by 2 by x 1 1 and the row indexed by 2 by x 1 . The entries are now rational functions in x 1 with only 1 + x 1 appearing in the denominator. After specializing M 0 to x 1 = 0 and cyclic shifting the rows and columns by one, we find another block upper triangular matrix with diagonal blocks of the form

. Therefore the determinant of M 0 restricted to x 1 = 0 is given by Q n j=1 (x 2j+1 x 2j+2 + 1). By definition, the determinant of M 0 equals D(x 1 1 , x 2 , . . . , x n ) = x1 1+x1 p(x 1 1 , . . . , x n ). Restricting to x 1 = 0 gives the coe cient of x 1 in p, which must be Q n j=1 (x 2j+1 x 2j+2 + 1). Therefore p coincides with f 2n+1 .

Lemma 8.4. For every m = 2, . . . , 2n + 1, the coe cients of 1 xm f 2n+1 are the principal minors of a 2n ⇥ 2n matrix with entries in {0, ±1, x ±1 m }. In particular, the rational function 1 x2n+1 f 2n+1 can be written as det(diag(x 1 , . . . , x 2n ) + B) where for 1  i, j  2n,

1 if j = i + 1 and i > 1 or (i, j) = (1, 1) or (i, j) = (2, 1),

if i is even and j = i 1 or (i, j) = (2n, 1), x 2n+1 if i odd, i 3, and j = 1, 1/x 2n+1 if i 2 {1, 2} and j is even, and 0 otherwise.

Proof . Let M = det(diag(x 1 , . . . , x 2n ) + B) and let D denote its determinant. As in the proof of Lemma 8.3, the entries of M with x 2n+1 appearing in the denominator appear in a submatrix of rank-one. The entries with x 2n+1 appearing in the numerator are contained in the first column. Moreover, in the Laplace expansion of the determinant, the only terms avoiding the submatrix of entries x 1 2n+1 must include the (1, 1) and (2, 2) entries, and so will not involve any entries with x 2n+1 . It follows that D can be written as x 1 2n+1 p where p is multia ne in x 1 , . . . , x 2n+1 . Therefore it su ces to check that f 2n+1 and p have the same restriction to x 1 = 0 and same coe cient of x 1 .

We see that the coe cient of x 1 in D is the determinant of the matrix M after removing the first row and column. This minor is a block matrix with one block of the form (x 2 + 1/x 2n+1 ) and the rest of the form ✓

x 2j+1 1 1

x j+2

◆

. Therefore the coe cient of x 1 in p and f 2n+1 agree.

The specialization of M to x 1 = 0 is a matrix has the form

◆

for j = 1, . . . , n 1 and x 2n + 1/x 2n+1 . This shows that the restriction of p to x 1 = 0 agrees with that of f 2n+1 .

For the corresponding statement with arbitrary m 6 = 1, we use the symmetries of f 2n+1 under the action of a dihedral group of order n with the cyclic action j 7 ! j + 2 (identifying 2n + j = j for j 2) and reflection n + 1 j $ n + 2 j. There is some element of this group that moves m to 2n + 1, and we can take the image of the representation above. To show that f 2n+1 does not belong to I 2n+1 , we will use the following:

Lemma 8.5. The set of polynomials

, where F alg denotes the algebraic closure of F.

Proof . The set of multiquadratic polynomials in F alg [x] MQ that factor as the product of two multia ne polynomials is the image of F alg [x] MA ⇥ F alg [x] MA under (g, h) 7 ! g • h. Since this map is bilinear, it follows from the projective elimination theorem that the set {q 2 F alg [x] MQ : q = g • h for some g, h 2 F alg [x] MA } is Zariski-closed in F alg [x] MQ . Pulling back by the map ij , it follows that for each i, j 2 [n], the set of polynomials f 2 F alg [x] MA for which ij (f ) factors as the product of two multia ne polynomials is Zariski-closed, as is their intersection over all i, j 2 [n]. It follows that its intersection with F[x] MA is Zariski-closed in F[x] MA . Theorem 3.1 implies that the image of F n⇥n under the principal minor map is a subset of the variety F n , although as Example 3.2 shows, this containment can be strict. In order to show that f 2n+1 does not belong to the variety of I 2n+1 , it su ces to show that f 2n+1 does not belong to F 2n+1 .

Recall that for f = P S✓[n] a S x [n]\S , the coe cient vector of f is defined to be coe↵(f ) = (a S ) S✓[n] 2 F 2 [n] .

Proof of Theorem 8.2. For convenience, let f = f 2n+1 . Let P 2 I 2n be a homogenous polynomial vanishing on the image of F 2n⇥2n under the principal minor map. Let Q denote the image of P under inclusion into F[a S : S ✓ [2n + 1]]. Note that Q only involves a S with 2n + 1 6 2 S. Since our indexing of coe cients is inclusion reversing, we see that the evaluation of Q at the coe cient vector of f depends only on coe cients of monomials containing x 2n+1 . In particular, its evaluation at the coe cient vector of f equals the evaluation of P at the coe cient vector of derivative of f with respect to x 2n+1 , i.e.

Q(coe↵(f )) = P (coe↵(@f /@x 2n+1 )). (11) If F is finite, it su ces to replace it with any infinite field extension, such as F(t) or F alg . Let ( , ⇡) 2 SL 2 (F) 2n+1 o S 2n+1 , with generic, where F alg denote the (necessarily infinite) algebraic closure of F. Call this polynomial g. The coe cient of Q 2n i=1 x i in g is a + c for m = 1 and a for m > 1. In either case, we can assume it is nonzero by the genericity of .

By Lemma 8.3 for m = 1 and Lemma 8.4 for m > 1, the polynomial 1 g is determinantal and its coe cient vector belongs to the image of the principal minor map. Since the image of the principal minor map is invariant under the action of (SL 2 (F) 2n o S 2n ), by (11) This shows that the coe cient vector of f belongs to the variety of (SL 2 (F) 2n+1 o S 2n+1 ) • I 2n .

On the other hand, we calculate that The polynomial f 2n+1 shows that the orbit of the ideal I 2n under (SL 2 (F) 2n+1 o S 2n+1 ) is not enough to cut the set of polynomials f 2 F[x 1 , . . . , x 2n+1 ] all of whose Rayleigh di↵erences factor as the product of two multia ne polynomials. As Example 3.2 shows, even this is not enough to cut out the image of the principal minor map. This leaves the question of what conditions cut out the image of the principal minor map for arbitrary n wide open.