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Consider a triangulation of some bounded region of $R^3$ with a (finite) set of tetrahedra (like in Regge calculus). It can be thought of as a simplicial 3-complex with specified lengths of edges. The other way around - a 3-complex with specified lengths of edges can sometimes be isometrically embedded in $R^3$. Now let's modify some of the lengths in an arbitrary way. Chances are - this triangulation can no longer be embedded in $R^3$. My question is: what is the minimal $n$ such, that our triangulation can always be isometrically embedded in $R^n$? My hypothesis is 6.

It is of course possible to modify the lengths in such a way that there is no embedding in any dimension - this happens if some triangle inequality is broken. I'm only interested in modifications small enough that it doesn't happen. Also: I'm only interested in the generic, non-degenerate case.

If there is a proof that $6V-E\geq 0$ (or $nV-E\geq 0$)* for such triangulations, where $V$ is the number of vertices and $E$ is the number of edges, I would be satisfied with that. It would mean that there is more degrees of freedom than equations, when one tries to embed it in $R^6$ (or $R^n$).

*in reality the equations have $O(n)$ symmetry, so there should be some constant on the r.h.s. other than $0$, but I want to know if such inequality makes sense at all for now

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    $\begingroup$ "It is of course possible to modify the lengths in such a way that there is no embedding in any dimension - this happens if some triangle inequality is broken": this can also happen without breaking any triangle inequality (think of a bipiramid of which you increase the length of the long diagonal). $\endgroup$
    – André Henriques
    Commented Jun 22, 2011 at 0:02
  • $\begingroup$ Not entirely sure what you mean. Please explain the example in more detail. Note that by "triangle inequality" I mean any inequality on the lengths of edges that one gets by combining elementary $d(x,y)\leq d(x,z)+d(y,z)$ inequalities. Anyway, what is important is that for every legal, non-degenerate triangulation (in $R^3$) there is a neighbourhood that consists only of legal, non-degenerate triangulations (that probably don't fit in $R^3$). I want to know, how many-dimensional such a triangulation generically is. $\endgroup$
    – nadbor
    Commented Jun 22, 2011 at 9:38
  • $\begingroup$ @unknown(google) I believe André Henriques may be referring to the "tetrahedral inequality" one gets from requiring that the volume of any tetrahedron is positive - this is independent from triangle inequalities. There's a discussion here math.niu.edu/~rusin/known-math/98/tetrahedral_ineq $\endgroup$
    – j.c.
    Commented Jun 22, 2011 at 15:21
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    $\begingroup$ @jc There's an analogous inequality one dim higher. Take the complete graph on the 5 vertices $(2,0,0)$, $(-1,\sqrt 3,0)$, $(-1,-\sqrt{3},0)$, $(0,0,1)$, $(0,0,-1)$. Fix all other edge lengths, and expand $(0,0,1)-(0,0,-1)$ a bit: the result does not embed in ℝ^n. What does this have to do with triangluations of domains in ℝ^3? You can triangulate an $\epsilon$-neighborhood of that complete graph by tetrahedra s.t. all the edges are either very small, or very close to an edge of the complete graph. Now, expand the edges close to $(0,0,1)-(0,0,-1)$: the result does not embed in ℝ^n. $\endgroup$
    – André Henriques
    Commented Jun 22, 2011 at 18:52
  • $\begingroup$ @unknown: "I'm only interested in modifications small enough that it doesn't happen": as you can see from my example, there are ways of deforming so that, regardless of how small the deformation is, it won't embed any more. $\endgroup$
    – André Henriques
    Commented Jun 22, 2011 at 18:55

1 Answer 1

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The ratio $E/V$ has no upper bound. Moreover, for every $n\ge 4$ there exists a triangulation of a tetrahedron in $\mathbb R^3$ with $V=n$ and $E=n(n-1)/2$ (so that the edges form a complete graph).

To construct such a triangulation, consider a curve $\gamma:\mathbb R\to\mathbb R^4$ defined by $$ \gamma(t) = (t,t^2,t^3,t^4) . $$ Pick $n$ distinct points $p_1,\dots,p_n$ on this curve and let $P$ be their convex hull. It is a 4-dimensional convex polytope. It is easy to see (and well-known) that every segment $[p_ip_j]$ is an edge of this polytope. It follows that all faces of $P$ are simplices. So the boundary of $P$ is a triangulated 3-sphere with $n$ vertices and $n(n-1)/2$ edges. Now a standard construction (the stereographic projection from a point lying outside $P$ near one of its faces) turns the face complex of $P$ into a triangulation of one of its faces.

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