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Alex Becker's user avatar
Alex Becker's user avatar
Alex Becker
  • Member for 13 years, 9 months
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Integral transform and $\frac{1}{n!}$.
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 $n$.
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Relative extremely disconnected space
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 $(2/3,1]$.
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Double duals characteristic
Oops, I meant $X\times \{0\}$.
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Double duals characteristic
@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.
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Question about exponential-type limit
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.
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Knot security (When to trust your life with a knot)
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.
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Who named it the Snake Lemma?
Very nice, Andrew!
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2D problems which are easier to solve in 3D
The dynamics of a billiard in certain subsets of the plane are easier to analyze by looking at 2D surfaces which embed in 3D space, if that counts.
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Double duals characteristic
@Yemon I now see why I should refresh my window more often...
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Double duals characteristic
What category does $X$ live in? Is it a Banach space, a topological vector space, or any vector space?
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Strata of quadratic differentials from rational billiards
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.
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When tensor reflects exact sequences?
If a module $N$ both preserves and reflects exact sequences, it is called faithfully flat.
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Strata of quadratic differentials from rational billiards
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.
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Strata of quadratic differentials from rational billiards
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.