{ "id": "1301.0973", "version": "v1", "published": "2013-01-06T04:23:48.000Z", "updated": "2013-01-06T04:23:48.000Z", "title": "Which Exterior Powers are Balanced?", "authors": [ "Devlin Mallory", "Abigail Raz", "Christino Tamon", "Thomas Zaslavsky" ], "comment": "14 pages, 2 figures", "journal": "Electronic Journal of Combinatorics, 20 (2013), no. 2, article P43", "categories": [ "math.CO" ], "abstract": "A signed graph is a graph whose edges are given (-1,+1) weights. In such a graph, the sign of a cycle is the product of the signs of its edges. A signed graph is called balanced if its adjacency matrix is similar to the adjacency matrix of an unsigned graph via conjugation by a diagonal (-1,+1) matrix. For a signed graph $\\Sigma$ on n vertices, its exterior k-th power, where k=1,..,n-1, is a graph $\\bigwedge^{k} \\Sigma$ whose adjacency matrix is given by \\[ A({$\\bigwedge^{k} {\\Sigma}$}) = P^{\\dagger} A(\\Sigma^{\\Box k}) P, \\] where P is the projector onto the anti-symmetric subspace of the k-fold tensor product space $(\\mathbb{C}^{n})^{\\otimes k}$ and $\\Sigma^{\\Box k}$ is the k-fold Cartesian product of $\\Sigma$ with itself. The exterior power creates a signed graph from any graph, even unsigned. We prove sufficient and necessary conditions so that $\\bigwedge^{k} \\Sigma$ is balanced. For k=1,..,n-2, the condition is that either $\\Sigma$ is a signed path or $\\Sigma$ is a signed cycle that is balanced for odd k or is unbalanced for even k; for k=n-1, the condition is that each even cycle in $\\Sigma$ is positive and each odd cycle in $\\Sigma$ is negative.", "revisions": [ { "version": "v1", "updated": "2013-01-06T04:23:48.000Z" } ], "analyses": { "subjects": [ "05C50", "05C75" ], "keywords": [ "signed graph", "adjacency matrix", "k-fold tensor product space", "exterior k-th power", "k-fold cartesian product" ], "tags": [ "journal article" ], "note": { "typesetting": "TeX", "pages": 14, "language": "en", "license": "arXiv", "status": "editable" } } }