Wilson loops in SYM $N=4$ do not parametrize an orientable space

Abstract

In this paper we explore the geometric space parametrized by (tree level) Wilson loops in SYM N = 4. We show that, this space can be seen as a vector bundle over a totally non-negative subspace of the Grassmannian, W k,cn . Furthermore, we explicitly show that this bundle is non-orientable in the majority of the cases, and conjecture that it is non-orientable in the remaining situation. Using the combinatorics of the Deodhar decomposition of the Grassmannian, we identify subspaces Σ(W ) ⊂ W k,n for which the restricted bundle lies outside the positive Grassmannian. Finally, while probing the combinatorics of the Deodhar decomposition, we give a diagrammatic algorithm for reading equations determining each Deodhar component as a semialgebraic set.

The twistor Wilson loop and the amplituhedron

“…Note added: The paper [31] by Susama Agarwala and Cameron Marcott dealing with the same problem as this paper was posted on the same day.…”

Combinatorics of the geometry of Wilson loop diagrams I: equivalence classes via matroids and polytopes

“…Example 2.3. Let V = [8] and P = {(1, 4), (2,4), (5,8)}. Then W = (P, [8]) is the Wilson loop diagram…”

“…As noted above, Agarwala and Marin-Amat uncovered a connection between the Feynman integrals developed in this literature and positroids, defined as matroids that are realized as an element of the positive Grassmannian G R,≥0 (k, n) [6]. In separate papers with Fryer and with Marcott, this work was extended to study these Feynman integrals in terms of positroid cells and the positive Grassmannians [3,5]. Others have tried different approaches to define a geometric space associated to these integrals in a manner similar to the Amplituhedron [14,16].…”

“…More generally, it has been shown that the spaces defined by the Amplituhedron and by Wilson loop diagrams are not the same [14]. In fact, it is in some ways more natural to consider Deodhar cells of G(k, n + 1) for Wilson loop diagrams; in this case, one gets that the manifold represented by the diagrams is not orientable [5].…”

“…Combining this with equation (5) gives We first show that none of the vertices defining g can lie in the cyclic interval [x i , x j ]. Indeed, if g is a triangle then we immediately see that its third vertex cannot lie in the interval [x i , x j ], as we could obtain a configuration as described in the statement by replacing d with one of the other edges of g.…”