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As opposed to tracking any individual point, what I'm concerned with is a manifold that occupies the same set of points as its image under the smooth map.

Suppose $\Phi(x,y)=(x+1,y)$. Consider an arbitrary horizontal line $y=a$.

While no individual points end up where they were before, it's nonetheless true that the horizontal line simply shifts to the right. Thus, the new horizontal line entirely coincides with the old horizontal line.

Consider the map $\Phi(x,y)=(y^3,x^3)$. Under this transformation, most individual points will not end up where they were before. However, the line $y=x$ will rearrange itself once again into the same $y=x$ line. The same is true for the line $y=-x$. The degenerate hyperbola $x^2=y^2$ could be said to be the most general set of points that will rearrange themselves into the same exact curve as before.

This is what I mean by a manifold (or, in these two examples, a curve) that occupies the same set of points as its image, or - in other words - entirely coincides - or overlaps - with its image under $\Phi$.

It seems analogous to the eigenvectors of a linear transformation.

Simon M
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  • Careful: $x^2=y^2$ does not define a manifold. – Randall Jul 14 '23 at 23:50
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    To avoid potential for $\Phi$ to squish all of space down to a point, it may be good to at least suppose that the rank of $D\Phi$ is positive (full rank everywhere may be a good place to start). – Alex Ortiz Jul 14 '23 at 23:55
  • @AlexOrtiz under that circumstance, can I just say that the zero-dimensional point will coincide with its image – Simon M Jul 15 '23 at 00:06
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    I believe your question can be rephrased as follows: given a smooth map $\Phi : M \to M$, does there always exist a proper submanifold $S$ such that $\Phi(S) = S$? – Michael Albanese Jul 15 '23 at 12:30
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    One possibility is to try $S = \cup_{i=1}^{\infty} \Phi^i(T)$ for some $T$. Does this work for you ? Here $\Phi^i$ is $i$ times composition of $\Phi$. You have to make sure $S \neq M$. – Balaji sb Jul 16 '23 at 06:10
  • @Balajisb I think that is a good place to start. I have no idea how to proceed with a proof, though – Simon M Jul 17 '23 at 05:02

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No, this is not allways the case.

Consider for example an irrational rotation $\Phi$ on the unit circle $\mathbb S^1$. For all $s\in\mathbb S^1$

$$ \bigcup_{n\in\mathbb N}\Phi^n(\{s\}) $$ is dense in $\mathbb S^1$. This implies that each non-empty submanifold $S\subset \mathbb S^1$ with $\Phi(S)=S$ is dense in $\mathbb S^1$ and hence contains a $1$-dimensional submanifold $S_\varepsilon\subset S$. But then $S=\mathbb S^1$ since $$ \bigcup_{n\in\mathbb N}\Phi^n(S_\varepsilon)=\mathbb S^1 $$

Claire
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