Lower bound of embedding dimension for finite CW-Complex of dimension $d$

Question

Consider a finite CW-Complex $C$ of dimension $d$. Let $n$ be the smallest integer such that the complex embeds in $\mathbb{R}^n$. From Whitney it follows that $n \leq 2d$. How would you bound the embedding dimension of $C^k$? Is it possible to get a lower bound given the embedding dimension of $C$. What properties of $C$ could be used to give bounds?

Answer 1

For non-manifolds, it is not true that $n\leq 2d$. For example, a graph is a $1$-dimensional CW complex, and non-planar graphs are those for which the embedding into $\mathbb{R}^2$ fails. For smooth manifolds, the Whitney embedding theorem is that $n\leq 2d$. I hadn't heard of this until Mike Miller suggested looking into it, but since a finite $d$-dimensional CW complex is a compact metric separable space of inductive dimension $d$, the Menger-Nöbeling theorem implies it can be embedded in $\mathbb{R}^{2d+1}$.

There are lower bounds on $n$ one can get from Stiefel-Whitney classes for smooth manifolds. See Milnor-Stasheff, "Characteristic Classes."

Invariance of domain implies that $d\leq n$.

There is an obstruction by van Kampen that may show that $C$ cannot be embedded in $\mathbb{R}^{2d}$. For $d\geq 3$, Wu and Shapiro proved the obstruction is zero if and only if $n\leq 2d$. For $d=1$, Sarkaria (and Tutte) showed the obstruction is zero if and only if the graph is planar (i.e., $n\leq 2$). Freedman-Krushkal-Teichner showed there is a $2$-complex that does not embed in $\mathbb{R}^4$ even though its van Kampen obstruction is zero. https://math.berkeley.edu/~teichner/Papers/vanKampen.pdf