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11h
comment I have no experience with math research
Consider the polynomial $f(x) = 1 - (x/2)$. You should consider asking such questions at math.stackexchange.com .
20h
awarded  Necromancer
Apr
30
comment Looking for “large knot” examples
Dear Ryan - Perhaps you could accept one of the answers? This question pops up as unanswered on my screen. :)
Apr
26
comment Non-Cayley expander graphs
See, for example, pages 5 and 6 of Lubotzky's book "Discrete groups, expanding graphs, and invariant measures".
Apr
24
answered Non-Cayley expander graphs
Apr
20
awarded  Tag Editor
Apr
20
revised configuration-spaces wiki description
Changed "vector space" to something sensible.
Apr
19
suggested approved edit on configuration-spaces tag wiki
Apr
19
reviewed Approve configuration-spaces tag wiki
Apr
19
reviewed Reject traces tag wiki excerpt
Apr
18
comment Are square tiled surfaces dense in the moduli space of translation surfaces?
No. It works because scaling is not relevant. You can't "really" tell the difference between $\omega$ and $r\omega$. In a similar fashion - the union of the lines in $\mathbb{R}^2$ (through the origin and of rational slope) are dense.
Apr
16
answered How to get a polygon from a translation surface $(X,\omega)$
Apr
16
comment Are square tiled surfaces dense in the moduli space of translation surfaces?
there is a forgetful map from $Q(S)$ to Teichmuller space $T(S)$ obtained by forgetting $\omega$. Set $Q(X)$ to be the fiber of this map over the point $[X] \in T(S)$. I claim that $Q(X)$ is naturally homeomorphic to the vector space of one-forms on $X$. Square-tiled surfaces (where I allow scaling by a real number) are dense in that vector space, and thus in $Q(X)$, and thus in $Q(S)$.
Apr
16
comment Are square tiled surfaces dense in the moduli space of translation surfaces?
Ok - I am not using language the way you want me to - for that I apologize. So I will back up a bit. Suppose that $S$ is a topological surface (closed, connected, oriented). Let $\hat{Q}(S)$ be the space of all pairs $(X, \omega)$ where (i) $X$ is a Riemann surface marked by $S$ and (ii) $\omega$ is a one-form. We form a quotient $Q(S)$ by taking $(X, \omega)$ to be equivalent to $(X', \omega')$ if there is a biholomorphic map $h {:}\, X \to X'$ that (a) commutes (up to isotopy) with the markings and (b) pulls $\omega'$ back to $\omega$. Now...
Apr
15
answered Are square tiled surfaces dense in the moduli space of translation surfaces?
Apr
12
revised Why is it true that if two 4-manifolds are homeomorphic then their squares are diffeomorphic?
Edited to fix typos. Grammar.
Apr
12
answered Torsion elements in the mapping class group
Apr
12
answered Symmetry conjecture for minimal dilatation pseudo Anosov mapping classes
Apr
11
comment Classification of elements in mapping class groups
@Yu-ChanChang - If you like this answer, you should accept it (by clicking on the "check" mark.)
Apr
11
comment Elements of infinite order in the topological mapping class group
If $M$ has boundary, then a homeomorphism will induce a mapping class on the boundary, and that can lead to interesting obstructions. (Consider a three-dimensional handlebody and a Dehn twist about a disk). If $M$ is non-compact, then a homeomorphism will induce a permutation of the ends of $M$. (Consider a thickened tree.) So - do you have side conditions on $M$ to eliminate these kinds of "dimension-reducing" techniques?