Configurations of Points and the Symplectic Berry-Robbins Problem

We present a new problem on configurations of points, which is a new version of a similar problem by Atiyah and Sutcliffe, except it is related to the Lie group $\operatorname{Sp}(n)$, instead of the Lie group $\operatorname{U}(n)$. Denote by $\mathfrak{h}$ a Cartan algebra of $\operatorname{Sp}(n)$, and $\Delta$ the union of the zero sets of the roots of $\operatorname{Sp}(n)$ tensored with $\mathbb{R}^3$, each being a map from $\mathfrak{h} \otimes \mathbb{R}^3 \to \mathbb{R}^3$. We wish to construct a map $(\mathfrak{h} \otimes \mathbb{R}^3) \backslash \Delta \to \operatorname{Sp}(n)/T^n$ which is equivariant under the action of the Weyl group $W_n$ of $\operatorname{Sp}(n)$ (the symplectic Berry-Robbins problem). Here, the target space is the flag manifold of $\operatorname{Sp}(n)$, and $T^n$ is the diagonal $n$-torus. The existence of such a map was proved by Atiyah and Bielawski in a more general context. We present an explicit smooth candidate for such an equivariant map, which would be a genuine map provided a certain linear independence conjecture holds. We prove the linear independence conjecture for $n=2$.


Introduction
We introduce the relevant manifolds.
C n = {(x 1 , ..., x n ) ∈ (R 3 \{0}) n ; x i = x j and x i = −x j for all 1 i < j n} We denote by F n the following flag manifold F n = Sp(n)/T where T is the diagonal n-dimensional torus in Sp(n). The following finite group will be needed W n = (Z/(2)) n ⋊ Σ n where Σ n is the permutation group on n elements. We refer to it as the Weyl group, for it is the Weyl group of the Lie group Sp(n). The element (1, . . . , −1, . . . , 1) ∈ (Z/(2)) n , with a −1 in the ith position only, acts on (x 1 , . . . , x n ) by replacing x i with −x i , and leaving all other x j invariant. An element σ ∈ Σ n acts by permuting the n points x j .
On the other hand, the action of W n on F n can be described as follows: an element in (Z/(2)) n having −1 only in the ith position, multiplies the ith column of each point in F n by the quaternionic structure j (leaving the other columns invariant), while a permutation σ simply permutes the columns of each point gT ∈ F n .
Having described the main players in our story, we ask the following question, which was asked by Berry and Robbins for the group U (n) ( [5]), and solved positively by Atiyah in [1], and then later generalized (and solved positively) to an arbitrary compact Lie group G by Atiyah and Bielawski in [3]: Question: Is there for each n 2, a continuous map f n : C n → F n which is intertwining for the corresponding actions of W n ?
Actually, as we wrote earlier, Atiyah and Bielawski have already posed and solved in [3] a more general problem for any compact Lie group G (in our case, G = Sp(n)). But their solution is non-elementary, as it relies on an analysis of the Nahm equations. Here we propose, similar to Atiyah ([1] and [2]), and Atiyah and Sutcliffe ( [4]), a more elementary construction in the same spirit as those papers, but for the case G = Sp(n), instead of G = U (n).

The Main Construction
We first associate to each configuration x ∈ C n , n polynomials p 1 to p n of t ∈ CP 1 of degree less than or equal to 2n − 1, each defined up to a factor only. Namely, we let p i be a polynomial having as roots the stereographic projections of the normalizations of − Similarly, we introduce n other polynomials q 1 to q n , with q i having as roots the antipodals of the roots of p i , namely the stereographic A key observation is that the space of polynomials of degree less than or equal to 2n − 1 has a natural quaternionic structure j, this having to do with 2n − 1 being odd. Writing such a polynomial as Moreover, since the roots of q i are the antipodals of those of p i , it follows that q i is a factor times jp i . If we think of the space of polynomials above with its quaternionic structure as H n , then the p i define n column vectors v i in H n , each defined up to a C * ambiguity.
We observe that x i → −x i has the effect of mapping v i → jv i (i.e. p i → q i ). On the other hand, permuting the x i corresponds to permuting the v i in the same way.
Hence, assuming the v i are always H-linearly independent, for any x ∈ C n , we would have defined a W n -equivariant map from C n into GL(n, H)/(C * ) n . Following this map by an orthogonalization procedure which respects the action of W n , we finally get a smooth map from C n into F n . But this is so provided the following conjecture is true: Conjecture 1: given any x ∈ C n , the 2n polynomials p i and q i (1 i n) are linearly independent over C.
In the following section, we define a natural determinant function, and then, we prove the conjecture above for n = 2.

A Determinant Function
Let t ± i and t ±± ij ∈ CP 1 be the stereographic projections of the normalizations of the following vectors, respectively ±x i (1 i n) and ± x i ± x j (1 i, j n and i = j) We then choose lifts u ± i = (u ± i , v ± i ) and u ±± ij = (u ±± ij , v ±± ij ) ∈ C 2 \{0} of these roots under the Hopf map: h : We then form the polynomials p i having as roots the chosen lifts (u Thus in particular, once the lifts are chosen, the polynomials p i are determined uniquely, in other words, the factor of each p i gets fixed. Similarly, form the polynomials q i having as roots the chosen lifts (u + i , v + i ) and (u ++ ij , v ++ ij ) and (u +− ij , v +− ij ) for all j = i. We now form the complex 2n by 2n matrix M = (p 1 , q 1 , . . . , p n , q n ) having the coefficients of p i and q i as column vectors. One then defines the quantity Then the determinant function D(x 1 , . . . , x n ) = det(M )/P is independent of the choices of lifts, and thus well defined. Similar to the Atiyah-Sutcliffe determinant, D is actually invariant under the action of the Weyl group W n on C n , and is also invariant under scaling, and rotations in R 3 . However, unlike the Atiyah-Sutcliffe determinant, it is always real-valued, because it is the determinant of a 2n by 2n complex matrix, which represents an n by n quaternionic matrix, and thus is always real (indeed, the complex conjugate of such a 2n by 2n complex matrix can be shown to be in the same conjugacy class as the complex matrix itself, so they must have equal determinants).

The case n = 2
We consider here the case n = 2. We have two points x 1 , x 2 ∈ R 3 such that x 1 = x 2 and x 1 = −x 2 . Using a rotation in R 3 , we can assume that they both lie on the xy-plane. We think of the xy-plane as the complex plane. Using a rotation in the xy-plane and scaling, we can further assume that x 1 = 1 and we then let z be the complex number representing x 2 in the xy-plane. Thus z = 1 and z = −1. We let We then have We then multiply the second column by −AB and add it to the first column, and we multiply the second column by −ABg and add it to the third column, and finally subtract the second column from the fourth one, and get, after expanding the determinant along the first row: Taking a 2 out from the first column, and using elementary column operations using the first column in order to make the entries in the (3,2) and (3,3) positions vanish, we get we can then write where ℑ(g) denotes the imaginary part of g. Therefore, using the definitions of A, B and g in terms of z, we get Writing z = re iθ , and after simplification, we get 4D = 2 + (1 + r)(1 − cos(θ)) |z − 1| + (1 + r)(1 + cos(θ)) |z + 1| + 2r(1 − cos(θ))(1 + cos(θ)) |z − 1||z + 1| Using 1 + r |z + 1| and 1 + r |z − 1|, and that 1 + cos(θ) and 1 − cos(θ) are both nonnegative, 4D 4 + 2r(1 − cos(θ))(1 + cos(θ)) |z − 1||z + 1| Thus D 1 + r sin 2 (θ) 2|z − 1||z + 1| This proves the inequality D 1, which in turns implies the linear independence conjecture, for n = 2. Moreover, it is not too difficult to see that equality D = 1 occurs if and only if sin(θ) = 0, or, in other words, if the two points x 1 and x 2 lie on the same line through the origin.

A Conjecture
Similar to conjecture 2 in [4], we make the following conjecture Conjecture 2: for any n > 2 and for any x ∈ C n , we have D(x) 1.
The author wrote a small code in Python in order to test this conjecture numerically. For instance, for n = 4, by generating pseudo-randomly 5 configurations, with points in R 3 inputted row-wise: automatically get that f n is equivariant for the actions of the Weyl group W n of Sp(n), so that f n would then be a solution of the generalized Berry-Robbins problem, in the sense of Atiyah and Bielawski, for the group Sp(n).