Can the full shift be embedded in a flow?

Question

Write $I=[0,1]$, and let $S$ be the shift on $X=\{ (x_n)_{n\in\mathbb Z} : x_n\in I^k \}$. Is there a flow $\phi_t$ on $X$ with $\phi_1=S$? Here I require that $\phi_t$, for fixed $t$, is at least a homeomorphism on $X$, with respect to the product topology.

I'm really mainly interested in $k=2$ (though I'd also be interested in the variant where we replace the cube by a torus of that dimensions, which should make it easier to flow around). For $k=1$, the answer is no, for the trivial reason that $p$-periodic sequences must be invariant under $\phi_t$, so we won't be able to evolve $(\ldots 010101\ldots)$ towards its shift without crossing through a constant sequence, but these are invariant.

For $k\ge 2$, this specific problem disappears, but of course there could be other obstructions, even of this nature, since $\phi_t$ has to preserve any property that can be described dynamically in terms of $S$ (such as being an almost periodic sequence).

This sounds like the kind of question someone must have thought about already, so will perhaps be easy for the experts.

Answer 1

For odd $k$, the shift in $X^k$ has no square root (and hence does not lie in a 1-parameter subgroup). Indeed, the the set of 2-periodic points can be identified with $(I^k)^2$, where the shift acts as $(x,y)\mapsto (y,x)$. It would then preserve the set of fixed points, namely the diagonal $(x,x)$, so any root should preserve the subset of points of period 2, the set $M^k_2$ or pairs of distinct points of $\mathbf{R}^n$. There is an obvious homotopy retract of the latter to the set of pairs $(x,-x)$ where $x$ ranges over the unit sphere $\mathbf{S}^{k-1}$, commuting with the flip (first homotope the center to 0 and then renormalize...). Here the flip can be identified with the antipode map. Since $k$ is odd, the antipode of the $(n-1)$-sphere is orientation-reversing and hence its action of the antipode on the real cohomology $H^{k-1}(\mathbf{S}^{k-1},\mathbf{R})\simeq H^{k-1}(M^k_2,\mathbf{R})\simeq\mathbf{R}$ is given by sign change and hence has no square root. So the flip of $M^k_2$ has no square root for odd $k$.

For even $k$, $I^k$ is a square and hence the shift map has an obvious square root (and we can even find a 1-parameter subgroup interpolating the flip in $M^k_2$), so another argument would be needed.