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I have an oddly specific reference request.

I keep encountering a class of problems where, loosely speaking, I have a geometric figure in $\mathbb{R}^n$ where each dimension is "supported" by a vertex of the figure. In other words, if I wanted to delete a vertex from the figure, I would simultaneously have to project my figure down a dimension.

All dimensions have a vertex, but not all vertices have a dimension. Thus, some vertices can be deleted "for free," but for every dimension there is some vertex whose presence is necessary and sufficient to keep that dimension in play.

I've encountered these figures in economics research (specifically, game theory) and in some computational linear algebra problems, but I haven't found a formal treatment of them in either.

Does this sound at all familiar to you? Do these figures have a name? Has anyone studied their properties?

Thanks in advance.

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Does “all dimension have a vertex” mean there is a uniquely identified vertex for every dimension? How so? I would expect some interplay: those vertices you delete “for free” should be sufficient for some of the dimensions, and only after you delete them, some others would become neccessary. But perhaps I've read your description incorrectly. – MvG Oct 12 '12 at 14:30
Yes, "all dimensions have a vertex" mean that there is a uniquely identified vertex for each dimension. Here's an example: interpret the rows of the adjacency matrix of a directed, weighted, complete graph as points in $\mathbb{R}^n$. If we're to delete a node from the graph, that will delete one of our points in $\mathbb{R}^n$ (representing the edges starting at our deleted node), but it will also collapse one of the dimensions of the figure (representing the dimensions ending at our deleted node). – GMB Oct 14 '12 at 21:36

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