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 Yearling
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Jun
25
comment There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$
@MhenniBenghorbal Cheers, let me have a closer look.
Jun
19
comment Group action on a manifold with finitely many orbits
Presumably you can sharpen the statement by assuming that $G$ acts faithfully, in which case you get equality for the dimensions of $X$ and $G$.
Jun
17
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.
Jun
17
comment Polytopes characterization in $\mathbb R^n$
$P$ looks more like a hyperplane.
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
15
comment Non-orientable 3-manifold has infinite fundamental group
@dfeuer Good question. I guess I can assume that top homology vanishes.
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.
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
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).
Apr
19
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.
Apr
19
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.
Apr
18
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 $\operatorname{Aut}(\mathfrak{g\oplus g})$?
Apr
18
comment Automorphism group of Lie algebra $\mathfrak{g\oplus g}$
Yes, I mean $\operatorname{Aut}(\mathfrak{g\oplus g)/\operatorname{Aut}(\Delta g})$.
Apr
18
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?
Apr
18
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?
Apr
17
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".
Apr
17
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 $G$.
Apr
11
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.