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Such a function cannot be continuous, or even locally bounded. Otherwise we have some interval $(a,b)$ with $a>1$ on which $v(x)>B$, so $$\int_0^\infty x^ne^{v(x)}>\int_a^b e^B>\frac{1}{n!}$$ for sufficiently large $n$.
This is wrong. According to the original post, a space is relatively extremely disconnected if there is any base $B$ such that disjoint elements of $B$ have disjoint closure. We can modify your base such that this is the case, say by replacing $[0,1/2)$ with $[0,1/3)$ and $(1/2,1]$ with $(2/3,1]$.
@student If it splits, then we have $(X^*)^*\cong X\times \mathrm{coker} \phi$. If we have $(X^*)^*\cong X\times E$ then the canonical injection must take $X$ to $X\times \{1\}$ and so the sequence splits.
Yes, I think basically any function which has sparse, sharp peaks which come close (in the sense of $1/n^{1+\epsilon}$) to their index will provide a counterexample.
I could see this depending heavily on the diameter of the rope, because changing this would change the points of contact and so change the pressure on each point. It could also be that two (topologically) identical knots with identical rope have different points of contact resulting in very different strengths.
Thank you very much, this is quite helpful. I see now my problem was that I was not properly keeping track of the orientations of the copies of $P$ in my unfolding.
Furthermore, in general we have that the genus of a $k$-gon with angles $\pi m_j/n_j$ is $1+\frac{N}{2}(k-2-\sum\frac{1}{n_j})$ where $N$ is the gcd of $n_1,\ldots,n_k$. Thus the sum of the orders of the zeros of the quadratic differential is $2N(k-2-\sum\frac{1}{n_j})$, which is not in general equal to $2(\sum m_j) - k$. Thus the stratum cannot be determined from this alone.
Yes, I guess I need to be more precise. I reflect the polygon $P$ in the complex plane and then identify $z\in\mathbb C$ with $z+b$ so long as the map $w\mapsto w+b$ carries edges of $P$ to the same edges on a different copy.