Marek
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 Dec16 awarded Caucus Nov26 awarded Popular Question Nov7 awarded Yearling Oct28 awarded Notable Question Oct1 awarded Popular Question Jul2 awarded Curious Jun6 comment Why does differentiating a polynomial reduce its degree by $1$? This is exactly Mitchell's answer above. Why post the same thing again? Jun3 comment Decomposition of cohomology group on $S^{n}$ Can you at least please fix the typos? The question is almost unreadable as it stands. Jun2 revised Poincare dual of unit circle Added another explanation Jun2 revised Poincare dual of unit circle Added definition of Poincare dual Jun2 comment Poincare dual of unit circle @PeterM: I see, I didn't get that part. I think you misunderstood the definition of Poincare dual then, let me edit it into my answer. Jun2 answered Poincare dual of unit circle Feb6 comment What are necessary and sufficient conditions for the product of spheres to be paralellizable? But otherwise this is a very nice and probably classical question that hopefully some local expert will answer soon. To add my gut feeling -- you can't expect parallelizability in general, tangent bundles of spheres are quite complicated and there's no a priori reason for their direct sums to be trivial besides the trivial reason of adding the normal bundle. Feb6 comment What are necessary and sufficient conditions for the product of spheres to be paralellizable? I don't think clutching functions are helpful here, at least when used naively. That is because the contractible neighborhoods for $S^i \times S^k$ will be products of hemispheres, so there's four of them in total with mutual intersections being homotopic to either $S^{i-1}$, $S^{k-1}$ or $S^{i-1} \times S^{k-1}$. Correspondingly, there will be multiple clutching functions depending on which overlap we're talking about -- the one you mention is for passing from southern to northern hemispheres in both spheres simultaneously. Jan16 comment existence of double covering $GL(n, {\bf C})$ has the homotopy type of $U(n)$ which has the same fundamental group as $U(1)$ (this isomorphism is induced by $\det: U(n) \to U(1)$ and backwards by the inclusion $U(1) \to U(n)$ as scalars). Jan15 comment Dolbeault cohomology and analytic regularity Indeed, nobody can stop you from doing that. But in that case you should change the question's title and body since it has nothing to do with complexes in general and Dolbeault cohomology in particular. Jan15 comment Dolbeault cohomology and analytic regularity I don't really understand the question, even with the edit. When you have a complex $M^{\bullet}$, all the spaces are fixed ($C^1$, say) beforehand. You can't just decide that some of them will be $C^2$ when it suits you because you need to quotient. That doesn't make any sense. Jan14 revised How do I maximize $|t-e^z|$, for $z\in D$, the unit disk? Replaced the picture Jan14 answered How do I maximize $|t-e^z|$, for $z\in D$, the unit disk? Jan14 comment How do I maximize $|t-e^z|$, for $z\in D$, the unit disk? There are obviously other solutions even for $y_0 = 0$ because also $\phi = \pi$ works and moreover normal to the curve at every point intersects real axis somewhere. Of course, one needs to check which of the these three candidate points is the true maximizer of the distance.