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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?