most recent 30 from http://mathoverflow.net 2013-05-22T15:30:13Z http://mathoverflow.net/feeds/user/1477 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4069/bivectors-in-3-and-4-dimensions/4116#4116 Pedro 2009-11-04T16:52:55Z 2009-11-04T16:52:55Z

<p>Actually I found the explanation of Baez more attractive than the one of Barrett, the big question is which constraints the 10 bivectors have to obey in order to define uniquely the 4-simplex: the answer of Baez is clear:</p> <blockquote> <p>What constraints do these 10 bivectors satisfy? They can't be arbitrary! First, for any four triangles that are all the faces of the same tetrahedron, the corresponding bivectors must sum to zero. Second, every bivector must be "simple" - it must be the wedge product of two vectors. Third, whenever two triangles are the faces of the same tetrahedron, the sum of the corresponding bivectors must be simple. </p> </blockquote> <p>The second just says that each bivector defines a surface. The third is a consequence of [b' is the common edge] (b' ∧ b) + (b' ∧ b") = b' ∧ (b + b") [simple] For the first I'm still looking...</p> <p>but there is something titillating me: if I have two triangles that lives in the same hypersurface (3D) and that do not share a common edge, the sum of their two bivectors will be simple (because it is a bivector which lives in 3D) so the third condition of Baez is not bijective!!</p>

http://mathoverflow.net/questions/4069/bivectors-in-3-and-4-dimensions/4111#4111 Pedro 2009-11-04T16:16:13Z 2009-11-04T16:16:13Z

<p>Ok so condition (2) simply says that each simple bivector (by its simplicity) will define two vectors thus will define a 2D surface, thus we have constructed the 10 surfaces of the future 4-simplex by this requirement.</p>