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Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $\mathbb{R}^d$ for some $d$. Is it possible to find a simplicial complex inside $\mathbb{R}^d$ that triangulates $X$?

As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn

123...
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    What do you mean by "simplicial complex inside $\mathbb{R}^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here. – freakish Jul 25 '18 at 09:21
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    @freakish OP has a topological embedding and wants a piecewise linear embedding. –  Jul 25 '18 at 10:35
  • @MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something? – freakish Jul 25 '18 at 10:48
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    @freakish This is a space that does not come equipped with a piecewise linear embedding into $\Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $\Bbb R^{2n+1}$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $\Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.) –  Jul 25 '18 at 11:00
  • Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $\Bbb R^k$. But other than this correction I think everything else in my comment is correct. –  Jul 25 '18 at 11:02

2 Answers2

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As it was mentioned in the question any planar graph has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $d\geq 2$ and every $k$, $d+1\leq k\leq 2d$, there exist a finite $d$-dimensional simplicial complexes $K$ that can be embedded in $\mathbb{R}^k$ but not linearly. For more information, please see

"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."

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I think what you are looking for is the Simplicial Approximation Theorem:

If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial complex, then any map $f :K\to L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$. You can find the proof in Hatcher Algebraic Topology book.

The proof makes it clear that the simplicial approximation $g$ can be chosen not just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.

Elad
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  • Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4. – 123... Jul 29 '18 at 12:10
  • Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results – Elad Jul 29 '18 at 19:28