3D Venn diagrams - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:24:28Z http://mathoverflow.net/feeds/question/76564 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76564/3d-venn-diagrams 3D Venn diagrams asri 2011-09-27T21:44:00Z 2012-07-03T00:54:16Z <p>Are there higher-dimensional versions of the concept of rotationally symmetric Venn diagrams, with closed curves replaced by closed surfaces or higher manifolds ? </p> http://mathoverflow.net/questions/76564/3d-venn-diagrams/76567#76567 Answer by psd for 3D Venn diagrams psd 2011-09-27T21:53:19Z 2011-09-27T21:53:19Z <p>"four intersecting spheres form the highest order Venn diagram that is completely symmetric and can be visually represented"</p> <p><a href="http://en.wikipedia.org/wiki/Venn_diagram#Extensions_to_higher_numbers_of_sets" rel="nofollow">http://en.wikipedia.org/wiki/Venn_diagram#Extensions_to_higher_numbers_of_sets</a></p> http://mathoverflow.net/questions/76564/3d-venn-diagrams/76574#76574 Answer by Scott Carter for 3D Venn diagrams Scott Carter 2011-09-27T22:51:22Z 2011-09-27T22:51:22Z <p>The n-simplex as a subset of ${\mathbb R}^{n+2}$ can serve as a stand-in for a Venn diagram. Or if you like you can fatten the vertices until you have over-lapping $n$-balls. </p> <p>More precisely, consider the convex hull of ${e_1, e_2, \ldots , e_{n+1}}$. This is the set $\{ \vec{x} \in R^{n+2}: \sum_i x_i = 1, \ \ \&amp; \ \ 0\le x_i \}$ where $e_i$ represents the $i$th standard unit vector $[0,\ldots, 0,1,0,\ldots, 0]$ that has a $1$ in the $i$th position. Each vertex represents one of your sets. Each edge represents the intersection between two sets. Each triangle represents the intersection among $3$ sets, and each $k$-simplex (which is determined by a subset of size $k+1$ chosen from ${1,2, \ldots , n+2}$ represents the intersection among $(k+1)$ sets. </p> <p>To imagine a model use a protractor and draw the complete graph on $(k+2)$-vertices that are represented by the roots of unity $e^{2\pi i j/(k+2)}$ (The $i$ here is different than the index above).</p>