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Does anyone know how to prove that any $n$-dimensional PL-manifold has an imbedding in euclidean space of dimension $2n$ ? Is this done via dimension theory ? Are there any references ?

Thanks a lot for your help !!!


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I mean not any PL manifold, but any compact PL manifold – onebengaltiger Mar 22 '12 at 6:59

The sketch of the idea is here:

it's called the Strong Whitney Embedding Theorem.

Getting the embedding into $\mathbb R^{2n+1}$ is just general position. Similarly, you can construct an immersion in $\mathbb R^{2n}$ by general position.

At that point, you perturb the immersion so that all double points are regular, meaning intersecting like $\mathbb R^n \times \{0\}^n$ and $\{0\}^n \times \mathbb R^n$ in $\mathbb R^{2n}$.

Then you try to apply the "Whitney Trick". This only works if the manifolds dimension $n$ is at least $3$. When $n=1,2$ you have to resort to other methods. $n=1$ is simple, since the circle is the only compact connected boundaryless 1-manifold. $n=2$ could use the classification of surfaces.

The basic idea of the Whitney trick is explained on the webpage. You try to `cancel' opposite double-points by finding a local model where you can construct the cancelling motion (regular homotopy it tends to be called in that literature -- or a 1-parameter family of immersions).

Sometimes you can't, and for that you modify the immersion by adding a local double point. That's what the function $\alpha_m$ describes on the Wikipedia page, the local model for the double-point introduction.

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But this results concerts only smooth manifolds doesnt it ? My question is about PL manifolds. It is known that every DIFF manifold has a PL structure, but there are PL manifolds which cannot be smoothed, so that we cannot use differentiability on these manifolds in order to imbed them in euclidean space of dimension $2n$ – onebengaltiger Mar 22 '12 at 6:58
All the same arguments work just fine in the PL world. I'm not using smooth manifold theory at all, it's just the PL world and smooth manifold theory have essentially the same constructions available, at least for the purposes of this theorem. – Ryan Budney Mar 22 '12 at 8:02
Ah ! this is good news for my work, thanks a lot for your help !!! – onebengaltiger Mar 22 '12 at 23:12

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