For some smooth path-connected manifold $M$, is there any diffeomorphism which could not be represented by the translation of the integral curve for some vector field $X$ on $M$?
Also, is there an 'if and only if' condition of the $M$ where $Diff(M)$ is precisely equivalent into the set of translation?
Best Answer
For any manifold $M$ of positive dimension, there are diffeomorphisms which are not given by flows.
If a diffeomorphism is given by a flow, then flowing for less and less time gives a homotopy (in fact, an isotopy) from your diffeomorphism to the identity. Hence, one obstruction is that a diffeomorphism could be homotopically distinct from the identity.
For example , spheres (in any dimension) all have self-diffeomorphisms which are orientation reversing. These maps are not even homotopic to the identity map, and hence, cannot be given as flows of vector fields.
So, a better question to ask is "Are all elements of $Diff^0(M)$, the subgroup of diffeomorphisms isotopic to the identity, given by flows?"
This still has a negative answer, as shown by Gabowski. In fact, Gabowski proves the following:
In the very special case $M = S^1$, it's easy to explicitly write down elements of $Diff^0(S^1)$ which are not given by flows. Here's one. (I am expanding on a special case found in notes of Neeb, though I believe the originally proof is due to Milnor).
Thinking of $S^1 = \mathbb{R}/2\pi\mathbb{Z}$, consider the function $f:S^1\rightarrow S^1$ given by $f(x) = x + \pi + \frac{1}{3} \sin^2(x)$.
Proposition The map $f$ is a diffeomorphism of $S^1$ which is isoptic to the identity map.
Proof: Note first that $f(x + 2\pi) = x + 2\pi + \pi + \frac{1}{3} \sin^2(x + 2\pi) = f(x) + 2\pi$, so this really is a function on $S^1$.
Viewed on $\mathbb{R}$, the function $f(x)$ has $f'(x) = 1 + \frac{1}{3} \sin(x)\cos(x) = 1 + \frac{2}{3}\sin(2x)$, so obviously $\frac{5}{3} \geq f'(x) > 0$. This implies that, on $\mathbb{R}$, $f$ is a diffeomorphism. Thus, $f$, viewed as a map on $S^1$, is a local diffeomorphism. Further, since $S^1$ is compact, $f$ must be a covering map.
Now, since, viewed on $\mathbb{R}$, $|f'| < 2$, it must be a one-sheeted covering map. Thus, it is a diffeomorphism.
Finally, to see that $f$ is isotopic to the identity, first think of shrinking the $\frac{1}{3}$ factor down to $0$. All we needed to show $f$ as a diffeomorphism was that the factor was not too big, so all of these is a diffeomorphism of $S^1$. This provides an isotopy to the function $x \mapsto x + \pi$. But this latter map is just a rotation by $\pi$, so is obviously isotopic to the identity. $\square$
Now, we consider iterates $f^n:= \underbrace{f\circ ... \circ f}_{n\text{ times}}$. Recall that point $x$ is called a period of $f$ of order $n$ if $f^n(x) = x$, but $f^k(x) \neq x$ for any $1 \leq k < n$.
Proposition: The points $0,\pi\in S^1 = \mathbb{R}/2\pi\mathbb{Z}$ are the only periodic points of $f$, and the period of both is $2$.
Proof: First, $f(0) = \pi$ and $f(\pi) = 2\pi = 0\in \mathbb{R}/2\pi\mathbb{Z}$, so these points have period $2$.
So now we show there are no other periodic points. First, if $x_0$ is a periodic point with $\pi < x_0 < 2\pi$, then $f(x_0)$ is also periodic, but $f(\pi) < f(x_0) < f(2\pi)$, i.e., $2\pi < f(x_0) < 3\pi$. Thus, $f(x_0) - 2\pi$ is a periodic point between $0$ and $\pi$. Thus, if there is a periodic point, there is one between $0$ and $\pi$.
So, let $x_0$ be a point with $0 < x_0 < \pi$.
We will prove this below. For now, believing the claim, it follows that $0 < (x_k - x_0) - k\pi < \pi - x_0 < \pi$, so, if $x_k - x_0 = 2s\pi$ for some $s$, then $x_k - x_0 - k\pi = (2s-k)\pi$ so $0 < (2s-k)\pi < \pi$. Dividing through by $\pi$, we get $0 < 2s-k < 1$, so $2s-k$ is an integer between $0$ and $1$, giving a contradiction. Thus, $x_k - x_0$ is not a multiple of $2\pi$, so $x_0$ is not a periodic point. $\square$
Proof of claim: When $k = 1$, the inequality reads $x_0 < x_1 - \pi < \pi$, so let's prove that. Since $x_1 = f(x_0) = x_0 + \pi + \frac{1}{3}\sin^2(x_0)$, and since $\sin^2(x_0) \neq 0$ since $0 < x_0 < \pi$, it follows that $x_1 > x_0 + \pi$, i.e., that $x_0 < x_1 - \pi$. Further, since $f$ is increasing, the max of $f$ on $[0,2\pi]$ is $f(2\pi) = 2\pi$, so $f(x_0) < 2\pi$ as $x_0 < \pi$. Thus, $x_1 < 2\pi$, so $x_1 - \pi < \pi$. This concludes the base case.
Now, assume inductively that $x_0 < x_k - k\pi < \pi$. Note that $f(x_k - k\pi) = f(x_k) - k\pi$. Then applying $f$ to this inequality (and recalling that $f$ is increasing), we get $x_1 < x_{k+1} - k\pi < 2\pi$. So, $x_0 + \pi < x_{k+1} - k\pi < 2\pi$, so $x_0 < x_{k+1} - (k+1)\pi < \pi$, concluding the inductive step for the first inequality. $\square$
Now, we are ready show that $f$ is not the flow of any vector field. The main idea is that if it was a time $t$ flow, then $f$ is a square of the time $t/2$ flow. Thus, the next proposition will show that $f$ is not the flow of any vector field.
Proposition: There is no diffeomorphism $g$ for which $f = g\circ g$.
Proof: Assume for a contradicting that such a $g$ exists. Note that $g^4(0) = f^2(0) = 0$, so $0$ is a periodic point of $g$. Further, the period must be a divisor of $4$, so it is $1,2,4$. We analyze each of these in turn.
If the period divides $2$, then $g^2(0) = 0$. This gives the contradiction $0 = g^2(0) = f(0) = \pi$.
If the period is $4$, then consider $g(0)$. Then $f^2(g(0)) = g^4(g(0)) = g(0)$, so $g(0)$ is a periodic point of $f$. Thus, $g(0) = 0$ or $g(0) = \pi$. If $g(0) = 0$, the $0$ is order $1$ (not $4$), if $g(0)= \pi$, then $g(\pi) = g^2(0) = f(0) = \pi$, so $\pi$ is an order $1$ period $g$, which implies $\pi$ is an order $1$ period of $f$, giving the final contradiction. $\square$