# 3D Venn diagrams

Are there higher-dimensional versions of the concept of rotationally symmetric Venn diagrams, with closed curves replaced by closed surfaces or higher manifolds ?

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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.

More precisely, consider the convex hull of $\{e_1, e_2, \ldots , e_{n+1}\}$. This is the set $\{ \vec{x} \in R^{n+2}: \sum_i x_i = 1, \ \ \& \ \ 0\le x_i \}$ 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.

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).

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Specifically, you can choose a ball of radius $\sqrt(2)$ around each point in the hyperplane that goes through $e_1,e_2,...,e_{n+1}$ – Will Sawin Sep 27 '11 at 23:50
The trick is to do this with $2^{n+1}$ distinct regions (counting the exterior region) in a dimension (potentially much) lower than $n+2$. At least that would be the analogue of the well-known 2-dimensional problem that has been solved in the case of $n$ prime by Griggs-Killian-Savage. – Patricia Hersh Jul 3 '12 at 1:25

"four intersecting spheres form the highest order Venn diagram that is completely symmetric and can be visually represented"

http://en.wikipedia.org/wiki/Venn_diagram#Extensions_to_higher_numbers_of_sets

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The same page shows a two-dimensional Venn diagram of five ellipses that is rotationally symmetric, so I think that sentence is referring to a stronger condition. – Jonah Ostroff Sep 27 '11 at 22:22
The two-dimensional diagram is only symmetric up to a reflection, no? – Will Sawin Sep 27 '11 at 23:48