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Motivational example

Consider a polyhedral graph $G$. A realization of $G$ is given by a convex polyhedron which is - essentially - characterized by the angles between the edges emanating from each vertex. There are $\sum_v d(v) = 2E$ such angles, so the degree of freedom $\mathsf{d}$ of a polyhedron is $\mathsf{d}\leq 2E$ (with $E$ the number of edges). But the angles cannot be choosen freely: for every (flat) face of the polyhedron the interior angle sum must obey a (strict) equality. Thus, $\mathsf{d}\leq 2E - F$ (with $F$ the number of faces). Because the polyhedron is supposed to be convex there is another restriction: for every vertex $v$ of the polyhedron the sum of angles between the edges emanating from $v$ has to be less than $2\pi$. Thus, we have $V$ additional restricting inequalities (with $V$ the number of vertices). But because inequalities are somehow "weaker" than equalities we cannot simply deduce that

$$\mathsf{d}\leq 2E - F - V = E - \chi = E - 2$$

(with $\chi$ the Euler characteristic of the graph/polyhedron). But by which amount $\mathsf{d}$ might be greater than $E-2$?


The original question is (more abstractly) concerned with the question how inequalities enter into the calculation of a "degree of freedom":

How can an (eventually non-integer) degree of freedom be defined when inequalities are involved?

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I'm not sure what you mean by "eventually non-integer degree of freedom". The realization space of a combinatorial type is a very nice semialgebraic set with a well defined notion of dimension. To measure the complexity of such realization spaces one may start looking at their topology. For 3 dimensions the space is contractible, though this phenomenon fails maximally in higher dimensions (cf. Mnev universality). – Gjergji Zaimi Jan 23 '13 at 3:25
Where can I learn more about "realization spaces of combinatorial types"? – Hans Stricker Jan 23 '13 at 7:15

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