The general position theorem asserts that any $M$ $m$-manifold unknots in $R^n$ provided $n\geq 2m+2$. The general position theorem assumes a smooth setting. Is unknotting still hold in the PL setting? what is the lower bound on n in that case? what is the argument for unknotting a general $m$-manifold (compact, closed, not necessarily connected ) in the PL setting?
1 Answer
Yes, if two PL-embeddings $f,g:M^k\hookrightarrow N^n$ of a compact $PL$ manifold of dimension $k$ are homotopic and $n\ge 2k+2$, then they are PL-isotopic.
This is Corollary 5.9 in
Rourke, C. P.; Sanderson, B. J., Introduction to piecewise-linear topology, Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 69. Berlin-Heidelberg-New York: Springer-Verlag. VIII,123 p. with 58 fig. Cloth DM 42.00; $ 13.40 (1972). ZBL0254.57010
and a strengthened version is Theorem 4.1.1 in
Daverman, Robert J.; Venema, Gerard A., Embeddings in manifolds, Graduate Studies in Mathematics 106. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-3697-2/hbk). xvii, 468 p. (2009). ZBL1209.57002.
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$\begingroup$ Can corollary 5.9 be generalized to: if two PL-embeddings $f,g : M_1^k1 \sqcup M_2^{k2} \longrightarrow N^n$ are homotopic and $n\geq k_1+ k_2+2$ then they are PL-isotopic ? (here $M_i^{k_i}$ is a compact PL $k_i$ manifold ) $\endgroup$– SteveMay 24, 2019 at 14:09
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$\begingroup$ @Steve: I would say no. Note that the dimension of the disjoint union is $\operatorname{max}\{k_1,k_2\}$, and in general $k_1+k_2<2\operatorname{max}\{k_1,k_2\}$. In particular if $M^1=S^1$ and $M_2=\varnothing$ (regarded as a $0$-manifold, say) then with $N=S^3$ you get classical knot theory. $\endgroup$ May 24, 2019 at 14:39
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$\begingroup$ thank you. So the result holds with $n\geq max\{k_1,k_2\}+2$. Just to clarify, by PL-isotopic, do you mean PL-ambient isotopic? or something else ? $\endgroup$– SteveMay 24, 2019 at 18:31
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$\begingroup$ @Steve: I think you need a factor of 2 in front of the max. And I mean whatever is stated in those textbooks, I think it's ambient PL-isotopy. $\endgroup$ May 24, 2019 at 19:12
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$\begingroup$ Sure, dropping the 2 was a typo. thanks again! $\endgroup$– SteveMay 24, 2019 at 19:14