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## Registered User

 Name Member for 3 years Seen Jun 11 at 21:43 Website Location Age
 Apr24 revised A_infinity structure on cohomology and the weight filtrationadded more details Apr24 comment A_infinity structure on cohomology and the weight filtrationDan -- you are right: multiplication is a map of MHS's of degree 0! Apr24 answered A_infinity structure on cohomology and the weight filtration Mar16 accepted Euler characteristics and characteristic classes for real manifolds? Feb22 comment Algorithmically unsolvable problems in topologyThanks, Marek, this looks interesting. Feb22 revised Constructible sheaves and dg-modulestag change Feb22 comment Constructible sheaves and dg-modulesquid -- I agree that dg-algebras would be a more appropriate tag. Jan6 comment How to show a certain determinant is non-zero.. but you could try math.stackexchange.com Jan5 comment continuous R^2xR^2xR^2/E^+(2) -> R^3 injection?Leon -- I see, so you don't allow reflections. In this case the space, call it $X$, is an open cone over the set of all triangles with the sum of the sides equal 1. The latter space, call it $Y$, is the union of two triangles with sides of one glued to the sides of the other in a bijective way, i.e., $Y=S^2$, which makes $X$ homeomorphic to $\mathbb{R}^3$. Jan5 comment continuous R^2xR^2xR^2/E^+(2) -> R^3 injection?Todd -- so you did but I didn't see it when I started writing mine. Jan5 comment continuous R^2xR^2xR^2/E^+(2) -> R^3 injection?unknown google -- if I understand you correctly, your quotient space is juet the space of all triangles in $\mathbb{R}^2$ with ordered vertices; it includes degenerate triangles, in which one of the vertices lies in the interior of the segment that joins the other two. This space is indeed a subspace of $\mathbb{R}^3$: each triangle is determined, up to a composition of rotations, reflections and translations, by the lengths of the edges (these are ordered, as the vertices are). If you do not allow reflections, then a triangle is determined by the lenghts of the sides plus orientation. Dec30 comment Morphisms of Spectral Sequences and alternating productsHiro -- I'm afraid I still can't follow your notation. Do you mean that $E_n$ is the $n$-th term of the spectral sequence, i.e., $E_n=H_*(E_{n-1},d_{n-1})$? In that case one would normally expect to see $E^n$, since you write $E^1_{a,b}$, with 1 on top. Also, the term a spectral sequence converges to is denoted $E^\infty$. Dec29 comment Morphisms of Spectral Sequences and alternating productsHiro -- what exactly are $E_n, f_n$ and $F_n$? If $E_n=\bigoplus_{a+b=n}E^1_{a,b}$ and similarly for $F_n$ then, under your conditions, where can infinitely many terms appear from?