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So, per the answers in this post, it appears that if a self-diffeo $h$ has $h = \Phi_1$ for $\Phi_t$ the flow of some differential equation $\xi$ on $M$, then $h$ must be isotopic to the identity and "infinitely rootable ('divisible'?) to the identity", that is, there must be a sequence of self-diffeos $(g_b)$ with each $g_b$ isotopic to the identity, $g_b^b = h$, and $\lim\limits_{b \to \infty} g_b = \text{id}_M$ $\left(\text{together with some kind of coherency condition, such as }(g_b)^a = \left(g_{\frac{b}{gcd(a,b)}}\right)^{\frac{a}{gcd(a,b)}}?\right)$.

  1. Are these conditions sufficient? That is, given a self-diffeo $h$ that is isotopic to the identity and "infinitely rootable to the identity", is $h = \Phi_1$ for $\Phi_t$ the flow of some differential equation $\xi$ on $M$?

[Notes: a) The paper, among other sources, shows that the flow uniquely determines the differential equation, $\displaystyle \xi(p) = \left.\frac{\partial}{\partial t}\Phi_t(p)\right\vert^{t=0}$, and, of course, it is well-known that the differential equation uniquely determines the flow: to the extent that $(g_b)$ is unique, $\Phi_t$ and $\xi$ should be unique; otherwise, one should get some sort of inverse limit set of differential equations, all leading to the same $h$ - the various flows should all agree for integral values of t but would likely disagree between these integral values of t. b) One should only need to determine $(g_b)$ on some cofinal subset of the naturals that leads to a dense subset of the rationals, e.g., $b = 2^c$, leading to the dyadic rationals. c) This isn't intended to be a research question; I'm assuming the answer is known.]

Also, for three other questions,

  1. What would be an example of a self-diffeo $h$ of a (closed?) connected Riemannian manifold $M$ that is isotopic to the identity with a unique square root, that is, with exactly one self-diffeo $g$ with $g^2 = h$?

  2. What would be an example of a self-diffeo $h$ of a (closed?) connected Riemannian manifold $M$ that is isotopic to the identity with two or more different square roots, that is, with two (or more) self-diffeos $g_{1,2} \ne g_{2,2}$ but $g_{1,2}^2 = h = g_{2,2}^2$?

  3. What would be an example where $h$ has multiple square roots $g_{i_1,2}$ and each $g_{i_1,2}$ has multiple square roots, $g_{i_1,i_2,4}$ ($g_{i_1,i_2,4}^2 = g_{i_1,2}$ and $i_j$ is an index to the possible roots), and so forth, so one gets some sort of bifurcation in the sequences and gets some sort of "interesting" (?) inverse limit for the set of differential equations?

1 Answers1

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  1. With the coherency condition, one can determine some sort of "rational powers" of $h$, $\Psi_{\pm\frac{a}{b}}(p) = h^{\pm\frac{a}{b}}(p) = (g_b)^{\pm a}(p)$, and then "extend these by continuity" to be a flow $\Psi_t$.

  2. (I think this is a solution to 2), but I don't yet have a proof of uniqueness.) With $M = \mathbb{R}$, if $h(p) = p+1$, there is a unique sequence of self-diffeos $\displaystyle g_b(p) = p + \frac{1}{b}$ all isotopic to the identity, all with $g_b^b = h$, converging to the identity, and satisfying the coherency condition. Hence, we have a case where we have a unique differential equation $\xi$ with $\Phi_1 = h$.

  3. (a) With $M = \mathbb{R}^2$ (or $S^1$), if $h$ is rotation by $\pi$, then $g_{1,2}$ is rotation about the origin by $\displaystyle \frac{\pi}{2}$ while $g_{2,2}$ is rotation about the origin by $\displaystyle -\frac{\pi}{2}$. Continuing in this manner, we see we obtain $g_{1,4}$ is rotation by $\displaystyle \frac{\pi}{4}$ whereas $g_{2,4}$ is rotation by $\displaystyle -\frac{\pi}{4}$, and so we have two distinct sequences of $2^{c \text{ th}}$ roots of $h$ with each element in each sequence isotopic to the identity, where each sequence converges to the identity, and where the coherency condition holds. Hence, we have a case where we have at least two different differential equations $\xi_1 \ne \xi_2$ with $\Phi_{1,t=1} = h = \Phi_{2,t=1}$.

3)(b) (From Jason DeVito) With $S^3$, thinking of $S^3$ as a Lie group, the antipodal map (left multiplication by −1, $L_{-1}$) has uncountably many square roots: left multiplication by any purely imaginary unit quaternion. Per this post, every imaginary quaternion has exactly two quaternion square roots, $u_3$ and $-u_3$ with $(\pm u_3)^2 = q$. Only one of the $\pm u_3$'s at level $3$ will have a angle smaller that $q$ with 1, the other one will be $-u_3$ and will have a smaller angle than $q$ with -1. This pattern continues with $u_{c-1}$ has exactly two quaternion square roots, $u_c$ and $-u_c$ with $(\pm u_c)^2 = u_{c-1}$. Only one of the $\pm u_c$'s at level $c$ will have a angle smaller that $u_{c-1}$ with 1, the other one will be $-u_c$ and will have a smaller angle than $-u_{c-1}$ with -1. If $g_{2^c} = L_{u_c}$, then $(g_{2^c})$ is a sequence of $2^{c \text{ th}}$ roots of $h$ defined on a cofinal subset of the naturals with each $g_{2^c}$ isotopic to the identity and satisfying the coherence condition. Hence, we have a case where we have uncountably many different differential equations $\xi_q$ with $\Phi_{q,t=1} = h$.

  1. I don't have an example for 4) yet.
  • For 2) Uniqueness is not clear to me, is the argument complicated? For 3b), you want left multiplication by $-1$, not $-i$. – Jason DeVito - on hiatus Sep 25 '20 at 13:38
  • a) I'll try to come up with a proof of uniqueness for 2). I was hoping for such comments before I posted this, this is all still really up in the air for me

    b) Done; terribly sorry about the error.

    – Jeffrey Rolland Sep 25 '20 at 13:51
  • (Not sure if I should go forward with this half-baked idea) (Regarding the uniqueness of 2)) If $g^2(p)=h(p)$, $r_1=g′(0)$, and $r_2=g′[g(0)]$, then $h′(0)=1$, so $r_2r_1=1$, so if $g$ has constant, positive derivative, $\displaystyle g(p)=p+\frac{1}{2}$. I think this finishes the proof if $g′$ is constant and positive. $\displaystyle g(p)=−p±\frac{1}{2}$ don't work, so I don't think $g'$ constant and negative will work. I don't know nearly enough about $\text{Diff}_0(\mathbb{R})$ to say what happens in $g′$ is not constant right now. – Jeffrey Rolland Sep 26 '20 at 02:40
  • (Another half-baked idea in the comments) So, for 3)(a) and (b), it appears in all the cases, there is a self-diffeo of $M$ that throws either $\xi_1$ onto $\xi_2$ or $\xi_{q_1}$ onto $\xi_{q_2}$, that is, the differential equations are topologically conjugate; maybe this is generally the case: maybe, if we have multiple $\xi$'s leading to the same $h = \Phi_{t=1}$, the differential equations are always topologically conjugate? – Jeffrey Rolland Sep 28 '20 at 20:18
  • (Answer to second half-baked idea) Per the answer https://math.stackexchange.com/questions/3867495/square-roots-of-diffeomorhpisms-of-manifold-conjugate/3867659#3867659 by @QiaochuYuan , none of the self-diffeos in Part 4) above are topologically conjugate, as can be shown by considering the eigenvalues of the differentials at the fixed point the origin, so the differential equations should not be topologically conjugate. – Jeffrey Rolland Oct 16 '20 at 09:19