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Hugo Chapdelaine's user avatar
Hugo Chapdelaine's user avatar
Hugo Chapdelaine's user avatar
Hugo Chapdelaine
  • Member for 14 years
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complex deformations of abelian varieties
Thanks Donu for the quick answer. So could you give me more details on how you get the $g^2$?
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Homotopy groups of a Bouquet of n-spheres
@Ryan, yes I meant that the complement of the unknot in $\mathbf{R}^3$ deformation retracts to $S^2\vee S^1$.
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Homotopy groups of a Bouquet of n-spheres
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Homotopy groups of a Bouquet of n-spheres
You are right completely right Ryan, I forgot the point at $\infty$, it is the unknot complement in $\mathbf{R}^3$ which is homotopy equivalent to a countable wedge of $S^2$. I'll remove my motivation.
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A bounded homogeneous space which fails to be symmetric?
Thanks Prof. Helgason for the short genealogy of the problem. By the way, I discovered a nice paper of Borel: Les fonctions automorphes de plusieurs variables complexes, where on p. 177 he mentions Cartan's result in the case n=3 and points out that the problem was still open (in 1952) for n=4.
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When does grading pass to (co)-homology?
Thanks a lot Ralph for the reference on Cohen Macaulay rings.
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On the determination of a quadratic form from its isotropy group
It is possible though to save my idea about using unicity of $\rho_v^F$ by saying that is is the unique $F$-isometrie such that $s^2=1$, $s(w)=-w$ for all $w\in\mathbf{C}v:=L$ and such for all $w$ which are $F$-perp to $L$ one has $s(w)=w$.
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On the determination of a quadratic form from its isotropy group
Ok I was being stupid, one simply unfolds the equality $\rho_v^F(w_1)\cdot_G\rho_v^F(w_2)=w_1\cdot_G w_2$ in order to get the equality. I finally got it!
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