Olivier Bégassat
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 Apr 27 comment Properties of the Majorization Order on $\mathbb{Z}^n$ Can you describe the order please? Apr 26 comment If p is a prime number of the form $4n+3$, show that we cannot solve $x^2\equiv -1\mod p$ You probably mean $x^2+1=0$. Apr 24 comment Why the square of ideal in Lie algebra is also ideal? $I^2$ is the subspace generated by all brackets $[i,i']$ with $i,i'\in I$. By definition, it satisfies the subspace axioms. To show that it is an ideal, you need to prove that $[i'',x]\in I^2$ for all $i''\in I^2$ and $x\in L$. By bilinearity of the bracket, you only nedd to prove $[i'',x]\in I^2$ for $i''$ of the form $[i,i']$ with $i,i'\in I$. The result is a direct consequence of the Jacobi identity, and the fact that $I$ is an ideal. This actually shows that $[I,J]$, the subspace generated by all the $[i,j]$ with $i\in I$ and $j\in J$, for $I,J$ ideals, is an ideal. Apr 22 answered I need to show that this sequence is increasing and I'm almost there but I need help on last step. Apr 22 comment The automorphism group of the real line with standard topology @IttayWeiss I was just throwing stuff out there, I'm not convinced this actually works in a compatible way with the group structure. Apr 22 comment The automorphism group of the real line with standard topology @IttayWeiss I'm not talking about deformation retractions that respect the group structure, only about deformation retractions of the underlying topological space, I don't think the dynamic properties of the automorphisms plays any role in that. As for something that is compatible with the group structure, how about scaling in the domain? What about the maps $H:A\times[0,1]\to A, (f,t)\mapsto \lbrace x\mapsto\frac{f(tx)-f(0)}{t}+f(0)\rbrace$? At least when one considers the subgroup of $C^1$ diffeomorphisms, this should give a strong deformation retraction onto the affine functions. Apr 22 comment The automorphism group of the real line with standard topology @IttayWeiss I was only talking about the topology of this space, not the group structure. I remember proving that it is a topological group with the compact open topology years ago. don't you think that the space should strongly deformation retract onto the subset of all affine homeomorphisms with slope $\pm1$? From there it will deformation retract onto $\lbrace \pm1\rbrace$ I think... Apr 22 comment The automorphism group of the real line with standard topology It seems to me, at first sight, reasonable, that there should be two path components, and a strong deformation retraction onto $\lbrace 1,-1\rbrace$, where $1$ is shorthand for the identity of $\Bbb R$. After all, the homeomorphisms of $\Bbb R$ are precisely the surjective stricly monotone functions, and there are obvious "convex homotopies". Apr 21 answered the mapping class group of the disk is trivial proof Apr 21 revised Good topologies on $\mathcal{P}(X)$ added 765 characters in body Apr 21 asked Good topologies on $\mathcal{P}(X)$ Apr 20 comment Bounded linear functionals on $L^\infty$. Are you sure $C^0[0,1]$ isn't the space of continuous functions on the unit interval? The notation is standard, and they are all measurable and bounded. Apr 20 comment Can a cube always be fitted into the projection of a cube? Are you allowing for all projections, or only for the orthogonal ones (i.e. those projections with orthogonal image and kernel)? Apr 20 comment Homology Whitehead theorem for non simply connected spaces Ok, if you say so ^^ I awarded you the bounty, I guess I need to read the articles more in depth to believe their claims and my own explanation ^^ Apr 20 revised Homology Whitehead theorem for non simply connected spaces added 25 characters in body Apr 20 comment Homology Whitehead theorem for non simply connected spaces Also, are the spaces $X$ and $Z$ the actual spaces he uses to find homology equivalent (in $\pi_1$ and $H_*$), non homotopy equivalent spaces? Apr 20 comment Homology Whitehead theorem for non simply connected spaces @studiosus But as I understand it, it is only required that the Euler characteristics be equal and $m=1$, and it seems to me he is saying that $m=1$ whenever $\chi>\chi_{\min}(G,2)$. Since we know by example (the complex associated to the presentation $\langle a,b\mid a^3=b^2\rangle$) that there is a $[G,2]$-complex with $\chi=0$ we have $0\geq\chi_{\min}(G,2)$, and so, for any $[G,2]$-complex with $\chi\geq 1$, $m=1$. My point is that it apparantly doesn't matter that $\chi_{\min}(G,2)=0$ or not, it only matters that it be less than our $\chi$. Am I correctly interpreting the article? Apr 20 revised Homology Whitehead theorem for non simply connected spaces deleted 3 characters in body Apr 20 revised Homology Whitehead theorem for non simply connected spaces added 281 characters in body Apr 20 revised Homology Whitehead theorem for non simply connected spaces edited body