Given a complex Lie algebra $\mathfrak{g}$, a choice of Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ and a dominant integral weight $\lambda$ of $\mathfrak{g}$, there is a natural construction of a complex valued function $$D_\lambda: \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta \to \mathbb{C},$$ where $\mathfrak{t}$ is the real slice of $\mathfrak{h}$ obtained by intersecting the compact real form of $\mathfrak{g}$ with $\mathfrak{h}$ and $$\Delta = \bigcup_{\alpha \in \Phi} \Delta_\alpha$$ with $\Phi$ being the set of roots of $\mathfrak{g}$ and $\Delta_\alpha$ is the kernel of $\alpha \otimes 1: \mathfrak{t} \otimes \mathbb{R}^3 \to \mathbb{R}^3$, with $1$ denoting the identity map on $\mathbb{R}^3$.
Given a root $\alpha \in \Phi$ and a point $\mathbf{x} \in \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$, note that, by the definition of $\Delta$, we have $$ (\alpha \otimes 1)(\mathbf{x}) \in \mathbb{R}^3 \setminus \{ \mathbf{0} \}.$$ Given a nonzero $v \in \mathbb{R}^3$, we denote by $\hat{v} \in S^2$ its normalization with respect to the Euclidean inner product on $\mathbb{R}^3$. We denote the Hopf map by $$ h: S^3 \to S^2,$$ where $S^3 \subset \mathbb{C}^2$ is the sphere in $\mathbb{C}^2$ with respect to the hermitian inner product on $\mathbb{C}^2$. Given $v \in S^2$, we say that $\psi \in S^3$ is a Hopf lift of $v$ if $h(\psi) = v$. Note that a Hopf lift is well defined only up to a phase factor.
We denote by $\psi_\alpha$ a choice of Hopf lift of the normalization of $(\alpha \otimes 1)(\mathbf{x})$. We are omitting $\mathbf{x}$ from the notation to make it easier to read, but of course, $\psi_\alpha$ depends on $\mathbf{x}$ too.
Denote by $\Phi_+$ a choice of positive roots of $\mathfrak{g}$. Note that by definition of $\lambda$ being an integral dominant weight of $\mathfrak{g}$, given any $\alpha \in \Phi_+$, $$ m_\lambda(\alpha) := 2 \frac{B(\alpha, \lambda)}{B(\alpha, \alpha)} $$ is a nonnegative integer, where $B(-, -)$ is an $\operatorname{ad}$-invariant symmetric bilinear form on $\mathfrak{h}$, which we assume fixed (and which induces an $\operatorname{ad}$-invariant symmetric bilinear form on $\mathfrak{h}^*$, also denoted by $B(-,-)$).
Choose a Hopf lift $\psi_\alpha \in S^3$ for any root $\alpha \in \Phi$. Consider the multiset (as in, a set, but allowing for multiplicities): $$ S_\lambda = \{ (\psi_\alpha, m_\lambda(\alpha), \lambda) ; \alpha \in \Phi_+ \}.$$
Given an element $w \in W$, we define $$ w.S_\lambda = \{ (\psi_{w^{-1}.\alpha}, m_{w^{-1}.\lambda}(w^{-1}.\alpha), w^{-1}.\lambda) ; \alpha \in \Phi_+ \}.$$ We define $$ S_{w^{-1}.\lambda} = w.S_\lambda. $$ If $w_1$, $w_2 \in W$, it is a proposition that $w_1.S = w_2.S$ iff $w_1.\lambda = w_2.\lambda$. Let $w_1, \ldots, w_k \in W$ such that $$W/W^\lambda = \{ w_1 W^\lambda, \ldots, w_k W^\lambda \}.$$ The orbit of $S_\lambda$ under the Weyl group $W$ is then $$W.S_\lambda = \{ S_{w_1.\lambda}, \ldots, S_{w_k.\lambda} \}.$$ Note that $$ k = \lvert W/W^\lambda\rvert $$ where $W^\lambda$ is the stabilizer of $\lambda$ in $W$, by the proposition I have mentioned a bit before. Define $$ \mathcal{S} = \bigcup_{i = 1}^k S_{w_i \lambda}. $$
With respect to $\Phi_+$, we can write $$ \mathcal{S} = S_+ \sqcup S_-,$$ as the disjoint union of multisets consisting of positive and negative roots respectively.
We are now ready to define the denominator, say $b_\lambda(\mathbf{x})$, of $D(\mathbf{x})$, by $$ b_\lambda(\mathbf{x}) = \prod_{(\alpha, m) \in \mathcal{S}_+} \omega(\psi_\alpha, \psi_{-\alpha})^m, $$ where $\omega \in {\bigwedge}^2(\mathbb{C}^2)^*$ is the "standard" complex symplectic form on $\mathbb{C}^2$.
Given an element, say $w_iW^\lambda$ of $W/W^\lambda$, we define $$\phi_{w_iW^\lambda} = \bigodot_{(\alpha, m) \in S_{w_i\lambda}} \psi_\alpha^{\odot m}.$$ Note that we have dropped the $\lambda' = w_i.\lambda$ from our notation in order to simplify it.
We then form the tensor product $$ \phi = \bigotimes_{w_iW^\lambda \in W/W^\lambda} \phi_{w_iW^\lambda}. $$ Note that we are assuming that $W/W^\lambda$ has been ordered in some fashion, but in the end, $D_\lambda$ will not depend on this ordering. Given $(\alpha, m, \lambda') \in \mathcal{S}_+$ with $m > 0$, we first note that $(\alpha, m, \lambda') \in S_{\lambda'}$ and $\lambda' = w_i.\lambda$ for some unique $i$, with $1 \leq i \leq k$. We let $s_\alpha \in W$ denote the reflection with respect to the hyperplane orthogonal to $\alpha$. Note that $s_\alpha$ interchanges $\alpha$ with $-\alpha$. We can then write $s_\alpha(\lambda') = w_j.\lambda$, for some unique $j$, with $1 \leq j \leq k$. Note that $j \neq i$, for otherwise $B(\lambda', \pm \alpha)$ are both positive, which is absurd. Given $(\alpha, m, \lambda') \in \mathcal{S}_+$, we contract $\phi$ using $\omega$ $m$ times, by contracting one of the (remaining) indices from $\phi_{w_iW^\lambda}$ with one of the (remaining) indices from $\phi_{w_j W^\lambda}$ using $\omega$.
We successively perform these $\omega$-contractions on $\phi$ for each $(\alpha, m, \lambda') \in \mathcal{S}_+$. At the end, we are left with a scalar, i.e. a complex number, which we denote by $a_\lambda(\mathbf{x})$.
We finally define $$ D_\lambda(\mathbf{x}) = \frac{a_\lambda(\mathbf{x})}{b_\lambda(\mathbf{x})}. $$
It can be shown that $D_\lambda$ is invariant under the Weyl group $W$ and invariant under the action of $\operatorname{SO}(3)$ which acts trivially on $\mathfrak{t}$ and via its natural action on $\mathbb{R}^3$, which thus induces an action on the domain of $D_\lambda$. Also, $D_\lambda(\mathbf{x})$ gets complex conjugated upon applying an improper orthogonal transformation onto $\mathbb{R}^3$ (improper here is equivalent to having determinant equal to $-1$). Moreover, $D_\lambda$ is invariant under scaling of $\mathbf{x} \in \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$.
I wonder if this construction could be related to the geometric Langlands program. I mean, if we think of $\lambda$ as associated to a representation of a Lie algebra and if we think of $D_\lambda$ as a complex-valued function on the space of regular Cartan triples of the "dual" algebra (in the sense of Langlands), then could there be some connection with the geometric Langlands? Note that in the simplest case where $\mathfrak{g} = \mathfrak{sl}(n, \mathbb{C})$ and $\lambda = e_1$, we then get that $D_\lambda$ is a normalized determinant (actually nothing but the Atiyah–Sutcliffe determinant).
Conjecture: given $\mathfrak{g}$, $\lambda$ as above, I conjecture that $$\operatorname{Re}(D_\lambda(\mathbf{x})) > 0$$ for any $\mathbf{x} \in \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$.
In particular, if true, the previous conjecture implies that $D_\lambda$ is nowhere vanishing on $\mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$.
Edit 1: I fulfilled my promise by describing the function $D_\lambda$ above in detail. It came out of my attempt of generalizing the Atiyah problem on configurations to a Lie-theoretic setting. I am trying to find a context for it where it would be natural, so to speak (its natural habitat).
Edit 2: I would like to add some comments in order to answer David Ben-Zvi's questions below. Sir Michael Atiyah had asked for some Lie-theoretic generalization of the Atiyah or Atiyah--Sutcliffe problem on configurations of points. I initially found, without much trouble, a similar construction for the symplectic groups $Sp(m)$ rather than the unitary groups $U(n)$ (the latter corresponding to the original Atiyah problem on configurations of points).
I then tried to generalize these 2 constructions to a Lie-theoretic setting. A variant of the construction above appeared in J. of Exp. Math. One can think of the $\phi_{w_iW^\lambda}$ as homogenenous polynomials in 2 complex variables, or equivalently as non-homogeneous polynomials in 1 complex variable. But to my dismay, the matrix containing the coefficients of these polynomials was not in general square (i.e. the number of such $\phi$s, which is what I denoted by $k$ in the description above, is not always equal to the degree of the polynomials plus 1). So one cannot in general simply take the normalized determinant of the matrix. However, one could take finitely many $\omega$-contractions and obtain a complex number, which we can then normalize. And this function $D_\lambda$ does generalize both the Atiyah-Sutcliffe normalized determinant, and the normalized determinant appearing in my symplectic variant.
I did not write a computer program to evaluate the above and test numerically, but I do suspect that both the real part of $D_\lambda$ and its absolute value are bounded below by a positive bound, which is nothing but the evaluation at a collinear configuration. If true, this would be an interesting and vast generalization of the original problem.
As to whether or not $D_\lambda$ is related to an HK moment map, well, there are some triholomorphic isometric actions on flat quaternionic vector space for which the level set of the moment map at the origin (of the moment space, so to speak) is nothing but the conditions that $\psi_\alpha$ is a Hopf lift of $(\alpha \otimes 1)(\mathbf{x})$. But I am not really sure how nicely $D_\lambda$ fits in this picture. I think that it does descend to the HK quotient, but that it is not holomorphic, if I remember well (I have unsuccessfully tried to apply twistor methods to the Atiyah problem on configurations). I have learned about the links with the moment map from Sir Michael Atiyah himself, by the way. I am sharing the info to answer your question and to hopefully bring the community closer to a solution!
Edit 3: if we let $$ d = \sum_{\alpha \in \Phi_+} m_\lambda(\alpha), $$ then the map $$ \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta \to \prod_{k \, \text{times}} \mathbb{C} P^d$$ which maps $\mathbf{x}$ to $(\phi_{w_iW^\lambda})$ is smooth $SU(2) \times W$-equivariant, where the action of the $SU(2)$ factor on the domain factors via $SO(3)$ (where $SU(2) \to SO(3)$ is the adjoint action) and the action of the $SU(2)$ factor on $\mathbb{C}P^d$ is via a regular homomorphism $SU(2) \to SU(d+1)$. It may be that such a map and its associated function $D_\lambda$ may have an interpretation when for instance $\lambda$ corresponds to a minuscule representation. Just a thought...
Edit 4: numerically, for $\mathfrak{g} = \mathfrak{sl}(4)$ and $\lambda = e_1 + e_2$, it turns out that, while $D_\lambda(\mathbf{x})$ seems to have a positive real part, thus somewhat confirming the conjecture above, it is not in this case minimized at a collinear configuration (i.e. at $4$ collinear points in $\mathbb{R}^3$).
The way we evaluate $D_\lambda(\mathbf{x})$ reminds me of quantum theory, where we add all (complex) probability amplitudes corresponding to different paths, with the probability itself being the absolute value squared of the previous sum (after proper normalization).
Edit 5: I will try to answer @David Ben-Zvi's question again. If $G = U(n)$ (or $SU(n)$ if you prefer) and $\lambda$ corresponds to the standard $n$-dimensional representation of $G$, we then recover the Atiyah problem on configurations of points. More specifically, $k = n$ in this case (i.e. the orbit of $\lambda$ under $W$ has $n$ elements) and $d = n-1$ ($d$ was defined in edit 3). This means we have $k = n$ $\phi_{\lambda'} \in \operatorname{Sym}^{n-1}(\mathbb{C}^2)$, i.e. we have $n$ elements in a complex $n$-dimensional vector space (with each $\phi_{\lambda'}$ defined up to a complex scaling). So the matrix containing the coefficients of the $\phi_{\lambda'}$ is $n$ by $n$, and thus is a square matrix.
In this special case, my $D_\lambda(\mathbf{x})$ is, up to a positive rational factor, nothing but a normalized determinant of that matrix, where the normalization factor is actually nothing but $1 / b_\lambda(\mathbf{x})$.
In general though, $k$ is not always equal to $d + 1$, so that the $k$ by $d + 1$ matrix containing the coefficients of the $\phi_{\lambda'}$ is not in general square, so that a determinant interpretation of $D_\lambda(\mathbf{x})$ is not available to us in general.
That being said, back to the special case $G = U(n)$ with $\lambda$ corresponding to the standard $n$-dimensional representation of $G$, i.e. back to the Atiyah problem on configurations, if $D_\lambda$ is nowhere vanishing on $\mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$, then by the determinant interpretation of $D_\lambda(\mathbf{x})$ in this case, we get that the $\phi_{\lambda'}$ are linearly independent over $\mathbb{C}$. It is then possible to construct a smooth map
$$ \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta \to GL(n, \mathbb{C}) / U(1)^n, $$
which, after an orthogonalization process which preserves Weyl equivariance (essentially coming from the Cartan decomposition of a non-singular matrix as a product of an hermitian positive definite matrix times a unitary matrix), leads to a smooth map
$$ \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta \to U(n) / T^n, $$
which is $SU(2) \times W$ equivariant.
Thus, if the non-vanishing assumption on $D_\lambda$ is true in this case, it would then lead to a positive solution of the Berry-Robbins problem.
Basically, Sir Michael Berry and Jonathan Robbins have an interesting geometric interpretation of the sign factor arising in the spin-statistics theorem in the case of 2 identical quantum particles. Their idea is very closely related to the famous "belt" trick and thus to spinors. But they had some technical issues to extend their construction to $n > 2$ identical particles. Their work thus led to the formulation of the Berry and Robbins problem which, if true, would then lead to a generalization of their construction for any $n$ identical particles.
The Berry and Robbins problem asks if there exists, for any $n \geq 2$ (integer), a continuous map from $$C_n(\mathbb{R}^3) \to U(n) / T^n$$ which is equivariant under the symmetric group $S_n$.
Back to the general case though, I am not sure exactly what the non-vanishing of $D_\lambda$ leads to in the general case, since a determinant interpretation of $D_\lambda(\mathbf{x})$ is not readily available. I will have to think... I hope this (at least partially) answers your question.
