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seen Dec 15 at 18:18

Jul
2
awarded  Curious
Dec
4
accepted Mayer-Vietoris for a cover without triple intersections
Oct
26
asked Mayer-Vietoris for a cover without triple intersections
Sep
22
accepted Reflection along subspace
Sep
13
revised Reflection along subspace
added 17 characters in body
Sep
13
asked Reflection along subspace
Aug
1
awarded  Nice Question
Jul
28
asked When is a group ring an integral domain
Jul
25
comment Seifert manifolds and Fuchsian group
Thank you. I somehow still had cocompact in mind. I see that $\pi_1(M)$ is an extension of $\pi_1(surface)$ by $\pi_1(circle)$. If $G$ is the fundamental group of a hyperbolic surface $\Sigma$ I can take the universal covering and obtain an embedding of $\pi_1(\Sigma)$ in $PSL(2,\mathbb{R})$ by the monodromy action. How does this work with orbifolds?
Jul
25
revised Seifert manifolds and Fuchsian group
added 18 characters in body
Jul
24
awarded  Editor
Jul
24
revised Seifert manifolds and Fuchsian group
added 106 characters in body
Jul
24
asked Seifert manifolds and Fuchsian group
Jul
12
accepted Irreducible representation of dimension $5$ of $S_5$
Jul
12
comment Irreducible representation of dimension $5$ of $S_5$
en.wikipedia.org/wiki/… thinks that the full icosahedral symmetry group is not $S_5$.
Jul
12
awarded  Supporter
Jul
12
comment Irreducible representation of dimension $5$ of $S_5$
It seems to me wikipedia suggests that there is a 5-dim irreducible representation of $S_6$ and that the restricted representation which is a representation of $S_5$ is also irreducible.
Jul
12
asked Irreducible representation of dimension $5$ of $S_5$
Jul
15
comment Spinning construction
$\partial (A \times D^{p+1})$ itself is not interesting it is just $S^{n+p}$. What is interesting is the embedding of this in $\partial (B \times D^{p+1}) \cong S^{m+p}$. The construction is used to get higher dimensional knots from lower dimensional.
Jul
15
comment Spinning construction
The boundary is here meant in the sense of manifolds. So $\partial (A \times D^{p+1})$ is $A \times S^p \cup \partial A \times D^{p+1}$ and similar with $\partial (B\times D^{p+1})$. But each part of the first decomposition has an embedding in the corresponding part in the B case as $A \subset B$ and $\partial A \subset \partial B$.