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Given a (closed?) connected Riemannian manifold $M^n$ and a self-diffeomorphism $h: M \to M$, is it necessarily the case that there is a differential equation/smooth, tangent vector field $\xi$ on $M$ so that the flow $\Phi_t$ of $\xi$ has $h = \Phi_1$? That is, can we always extend a discrete-time dynamical system to a continuous-time one?

Thanks in advance.

EDIT: Here is a follow-up question: Follow-Up to given a self-map $h$ of a (closed?) manifold, is there a vector field $\xi$ with flow $\Phi_t$ such that $h=\Phi_1$?

Jeffrey Rolland
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    I apologize for the self promotion, but there are even diffeos which are isotopic to the identity, but which are not flows. See, for example: https://math.stackexchange.com/questions/3656034/diffeomorphism-which-is-not-a-translation-of-the-integral-curve-for-some-vector/3656896#3656896 – Jason DeVito - on hiatus Sep 21 '20 at 18:46
  • Wow, that's a great answer! Thanks so much – Jeffrey Rolland Sep 21 '20 at 20:21
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    @JasonDeVito: Whether it's self-promotion is unclear, but certainly do not apologize!! +1 – Ted Shifrin Sep 23 '20 at 16:17
  • @Ted: Fair enough. The reason I considered it self promotion is that all I did was fill in lots of details on a proof outline I found somewhere else. I don't feel like I deserve any significant portion of the credit for that reason. So, may "self promotion" is wrong phrase, but I still hesitate to point people to that answer. But, I am truly sorry for offering my errant apology ;-) – Jason DeVito - on hiatus Sep 23 '20 at 18:32
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    @JeffreyRolland: I just wanted to mention that typically, multiple edits are discouraged, especially if they somehow invalidate existing answers. So, since you have a new question, I'd recommend asking a new one. You can link to this one in that question, or you can copy/paste whatever relevant things you did. And I can answer your naïve question: thinking of $S^3$ as a Lie group, the antipodal map (left multiplication by $-i$) has uncountably many square roots: left multiplication by any purely imaginary unit quaternion. – Jason DeVito - on hiatus Sep 24 '20 at 13:53
  • @JasonDeVito (I'll implement your suggestions later.) Awesome! If I may go to your well one more time, do you know of an example where each $g_{q_1,2}$ has multiple square roots, $g_{q_1, q_2, 4}$ ($q_i$ is a unit quaternion or index to the possible roots), and so forth, so one gets some sort of bifurcation in the sequences and some sort of "interesting" (?) inverse limit for the set of differential equations? – Jeffrey Rolland Sep 24 '20 at 14:32
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    @JeffreyRolland: I don't know the answer, but I'm happy to think about it (when I have more time!) – Jason DeVito - on hiatus Sep 24 '20 at 15:14

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If $h = \Phi_1$ for some flow $\Phi$ then clearly $h$ is homotopic to the identity. So, one can obtain many counterexamples by simply constructing self-diffeomorphisms that are not homotopic to the identity, and one can detect this using reasonably simple invariants of algebraic topology.

For one example, any orientation reversing diffeomorphism of a closed, oriented manifold $n$-manifold $M$ is not homotopic to the identity, because it induces the "multiplication by $-1$" map on $H_n(M;\mathbb R) \approx \mathbb R$. (This works as well for connected oriented manifolds, but the obstruction is a bit harder to describe).

For another example, the torus $T^2 = S^1 \times S^1$ has fundamental group isomorphic to $\mathbb Z \times \mathbb Z$. For any matrix $M \in SL(2,\mathbb Z) = \text{Aut}(\mathbb Z \times \mathbb Z)$ there exists a diffeomorphism $\phi_M : T^2 \to T^2$ such that the automorphism of $\pi_1(T^2) \approx \mathbb Z \times \mathbb Z$ induced by $\phi_M$ is given by the matrix $M$. Thus $\phi_M$ is not homotopic to the identity if $M$ is not the identity matrix.

Lee Mosher
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Not every diffeomorphism stems from a flow. See for example Section 2 (General results) of Flows and diffeomorphisms by Jaime Arango and Adriana Gómez.

Consider for example the diffeomorphism $x \mapsto -x$.

See also this question and answer for related details.

Jeffrey Rolland
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user76284
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Per a comment by Jason DeVito, there are even self-diffeos which are isotopic to the identity, but which are not flows; see, for example, diffeomorphism which is not a translation of the integral curve for some vector field.

Also, per a suggested post in the right column while viewing this, we have the also interesting question/answers Does the set of diffeomorphisms which are induced by flows form a group?

Jeffrey Rolland
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