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seen Oct 12 at 19:14

The Portuguese Language proposal on Area 51 could use your support. Anyone interested in the Portuguese language can follow the proposal here.

Also, the Latin Language proposal is in definition phase. Anyone interested in the Latin language can follow the proposal here.


Jun
17
comment Polytopes characterization in $\mathbb R^n$
$P$ looks more like a hyperplane.
Jun
17
awarded  Yearling
Jun
16
accepted Non-orientable 3-manifold has infinite fundamental group
Jun
16
accepted Why are the integers appearing in lens spaces coprime?
Jun
16
comment Why are the integers appearing in lens spaces coprime?
Thank you very much, I seem to have missed the notation $L(p,q)=L(p;1,q)$. Is a free action really necessary for getting covering map, though? I thought that a covering is a purely topological phenomenon, not depending on the manifold structure. Even in the non-free case I expect the alg. top. considerations still to reduce to those of $S^3$ (and $p$, of course). Am I expecting too much?
Jun
16
comment Why are the integers appearing in lens spaces coprime?
@QiaochuYuan In that case the quotient is a manifold if and only if $p,q$ are coprime. I guess that doesn't matter for homology and homotopy, so for all pure topoloical properties, the space as given in the question is a lens space, then, correct?
Jun
16
asked Why are the integers appearing in lens spaces coprime?
Jun
15
comment Non-orientable 3-manifold has infinite fundamental group
@dfeuer Good question. I guess I can assume that top homology vanishes.
Jun
15
asked Non-orientable 3-manifold has infinite fundamental group
Jun
9
accepted Help with inequality for real numbers: $||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$
Jun
6
asked Help with inequality for real numbers: $||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$
May
21
accepted Space modelled on ring
May
21
comment Radical of $\mathfrak{gl}_n$
Thank you for your answer. This is exactly the reasoning I had in mind. My professor, however, argued that I should not assume the fact that $\mathfrak{sl}_n$ is (semi)simple. In particular, completing the proof in my question would give a proof that $\mathfrak{sl}_n$ is semisimple. The argument you outline does it "the wrong way around"...
May
11
comment Space modelled on ring
@ZhenLin Well, you tell me. In my limited understanding, I think of schemes as functors that give something space-like for each argument, like $GL_n$ spitting out $GL_n(\mathbb C)$ or $GL_n(\mathbb R)$, when feeding it $\mathbb C$ or $\mathbb R$, respectively. In my example, so far I only know how to think about one ring: the ring of (global) functions.
May
11
asked Space modelled on ring
May
6
awarded  Caucus
Apr
28
asked Radical of $\mathfrak{gl}_n$
Apr
28
comment Find $\dim R(T)$ and $\dim N(T)$ from the matrix of a linear map $T$
Look up Gaussian elimination, e.g. on Wikipedia or YouTube.
Apr
28
answered Find $\dim R(T)$ and $\dim N(T)$ from the matrix of a linear map $T$
Apr
28
accepted Automorphism group of Lie algebra $\mathfrak{g\oplus g}$