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In the paper Hardness of embedding simplicial complexes in $\mathbb{R}^{d}$ the abstract states that a finite simplicial complex of dimension $k$ embeds in $\mathbb{R}^{2k}$ while on page $856$ he says that it does not embed in $\mathbb{R}^{2k}$ but rather in $\mathbb{R}^{2k+1}$.

Which one is correct?

My understanding from the definition of a simplex $$\Delta^k = \left\{(t_0,\cdots,t_k)\in\mathbb{R}^{k+1}\mid\Sigma_{i = 0}^{k}{t_i} = 1 \mbox{ and } t_i \ge 0 \mbox{ for all } i\right\}$$

is that a finite simplicial complex of dimension $k$ embeds in $\mathbb{R}^{k+1}$, am I right?

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    You are talking about a simplex, not a simplicial complex. The $2k+1$ result is a general result of Menger. The $2k$ embedding problem is an algorithmic problem mentioned in the abstract.2012-08-29
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    if we can embed the simplex of higher dimension $k$ in $\mathbb R^{k+1}$ then we can embed all the simplicial complex in $\mathbb R^{k+1}$ since $\mathbb R^{m}$ embeds in $\mathbb R^{k+1}$ if $m\leq k+1$.2012-08-29
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    Triangulate a moebius strip. Each triangle is a 2-simplex, but the moebius strip is not embedded in $R^2$.2012-08-29
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    Simplex and simplicial complex are different objects. You can use four 2-simplex to obtain a tetrahedron, as a simplicial complex, which embeds in $\mathbb{R}^3$.2012-08-29
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    Both examples embed perfectly in k+1, k=2. According to the (abstract of the) paper you need at least a 3-dimensional complex, to find something that doesn't embed into $\mathbb{R}^{k+1}$.2012-08-29
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    @Sigur : every simplex is a simplicial complex and every simplicial complex is a finite collection of simplices glued together in a "nice" way!!2012-08-29
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    @palio, yes, I agree. So they are different objects. If they would be the same, then a simplicial complex would be a simplex. As you said, this is not the case.2012-08-29
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    @Sigur : so you are saying that after the gluing process we may lose the embeddebiliy in $\mathbb R^{k+1}$?2012-08-29

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