uniqueness of regular/tubular neighborhood with equivariant boundary

Question

Let $N$ and $N'$ be regular neighborhoods of a subpolyhedron $P$ in a closed PL manifold $M$, and suppose that $t$ is a free PL involution on $M$ such that each of $\partial N$, $\partial N'$ is invariant under $t$. Does there exist an equivariant PL isotopy of $M$ taking $N$ onto $N'$?

Edit: Beware that $N$ is not assumed to be $t$-invariant. The case where $N$ is $t$-invariant is of course trivial.

(Example: $M=S^m$ with $t$ the antipodal involution, $P$ is a point, $N$ is a hemisphere and $N'$ something fancier.)

I also do not know the answer to the similar question in the smooth category, where $N$ and $N'$ are tubular neighborhoods of a submanifold.

It does not seem like usual arguments about regular neighborhoods work to prove this, but also I don't immediately see a counterexample.

(As a side remark, I'm ultimately interested in relative regular neighborhoods and a non-free involution whose fixed point set is forced to lie in $\partial N$ by the choice of the relativization.)

Answer 1

I learned from Piotr Akhmetiev that $S^6$ contains two smoothly embedded $5$-spheres invariant under the antipodal involution that are not equivariantly PL isotopic, and the reference is Lopez de Medrano's "Involutions on Manifolds". Of course they are boundaries of regular neighborhoods of a point (by the higher-dimensional Poincare conjecture) and hence also of tubular neighborhoods of a point (since there are no exotic $6$-balls). A more recent source that Akhmetiev mentioned is Yu. Muranov's survey.

Answer 2

Let's consider the smooth case. The basic idea is to solve the problem on the free quotient, and then to lift the solution by covering space theory. For the external reference below, I'm going to assume $N \subset N'$. Let $G = \lbrace 1,t \rbrace$ be the cyclic group of order two; the argument works for any finite group $G$.

Let $\pi: E \to P$ be the abstract $G$-vector bundle in common. Let $f: E \to M$ be a $G$-equivariant embedding whose image is $N$ and which restricts on the 0-section to the inclusion $P \hookrightarrow M$.

Since the action of $G$ on $M$ is free, the quotient $M/G$ is a manifold and $P/G$ is a submanifold.

There exists an isotopy $\bar{F}: E/G \times [0,1] \to M/G$, constant on the 0-section, from $f/G$ to an embedding $g: E/G \to M/G$ whose image is $N'/G$.

The map $\bar{F}$ is the composite of "aligning", see for example [Lang, Fundamentals of Differential Geometry, Proposition IV.6.1] (freely searchable online through Google Books), followed by "fiberwise dilation" where the amount of dilation can vary from fiber to fiber.

By the Homotopy Lifting Property for covering spaces, there is a unique $G$-map $F: E \times [0,1] \to M$ that covers $\bar{F}$ and that restricts to $f$ at time $0$. Observe that $F$ is a $G$-homotopy deforming $N$ onto $N'$.

It remains to check that $F$ is an isotopy, that is, each restriction $F_t: E \to M$ is an embedding. But this follows immediately from a simple diagram chase, using the facts that $\bar{F}_t: E/G \to M/G$ is an embedding and that the action of $G$ on $M$ is free.

In order to obtain the hypothesis $N \subset N'$, since $P/G$ is assumed to be compact, one can uniformly fiberwise shrink $N/G$ into the sub-neighborhood $N/G \cap N'/G$. Without this compactness hyothesis, I believe that one can non-uniformly fiberwise shrink using a partition of unity on $P/G$.