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 Jun17 comment Polytopes characterization in $\mathbb R^n$ Then you have a 0-dimensional hyperplane, i.e. a point, which you can interpret as a 0-dimensioal polytope if you like. Your formula for $P$ only gives you polytopes when $n=1$, though. For $n>1$, the formula defines a hyperplane, which won't be a polytope. Jun17 comment Polytopes characterization in $\mathbb R^n$ $P$ looks more like a hyperplane. Jun16 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? Jun16 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? Jun15 comment Non-orientable 3-manifold has infinite fundamental group @dfeuer Good question. I guess I can assume that top homology vanishes. May21 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"... May11 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. Apr28 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. Apr28 comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ I did the calculation. As you said it was very boring—it turns out to be a 10-dimensional subgroup of $GL(6,\mathbb R)$ times $\mathbb Z_2$ (semi-direct). Apr19 comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ "Just compute it" isn't what I hoped for when I asked the question, but maybe the only sensible answer. Thank you for the lead, I'm giving it a try. Apr19 comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ @MarianoSuárez-Alvarez I gave up trying to write it as a quotient and edited the question. Apr18 comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ True. This seems to be less trivial than I imagined. Is $\operatorname{Aut}(\mathfrak{g\oplus g})\cap\operatorname{Aut}(\Delta\mathfrak g)$ normal in $\operatorname{Aut}(\mathfrak{g\oplus g})$? Apr18 comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ Yes, I mean $\operatorname{Aut}(\mathfrak{g\oplus g)/\operatorname{Aut}(\Delta g})$. Apr18 comment Choosing central generator for nilpotent group generated by 3 elements I found a comment on this question on MO, which claims that one generator can be chosen to be central. However, I do not understand the reasoning by which the commenter deduced this fact. Could there be a simple proof for the case when the number of generators is 3? Apr18 comment Choosing central generator for nilpotent group generated by 3 elements Thanks. I guess that means that even in fairly simple cases (nilpotence class 2, 3 generators, torsion-free), these conditions imply very little about the group. I also guess it's a measure of my mathematical ignorance that makes me state overly strong conjectures. Do you have any idea what happens if I add the condition that the centre be generated by one element, i.e. infinite cyclic? Apr17 comment Choosing central generator for nilpotent group generated by 3 elements Yes, really really sorry I didn't remember to put it into the question. By 2-step I mean that $[a,b]$ is in the centre of $G$, i.e. trivial group "in two steps". Apr17 comment Choosing central generator for nilpotent group generated by 3 elements Thanks, that's a neat counter-example. I would like to add the condition "torsion-free" to $G$. Apr11 comment How to show that $\mathrm{Sym}_{n\times n}(\Bbb{R})$ and $\mathrm{Skew}_{n\times n}(\Bbb{R})$ are subspaces of $\mathrm{M}_{n\times n}(\Bbb{R})$ Sorry I hadn't finished reading your proof, which seems complete to me. Apr11 comment When are homology groups free abelian groups? Look up homology of real projective space. Aug20 comment Determine topological properties of a space of matrices Connectedness, homotopy groups, homology groups, cohomology groups. These would all follow from general theory, if one could write down explicitly, what this space is, as for the case when one just imposes the second condition.