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 Jul 7 comment $\pi_0$ in the long exact sequence of a fibration and quaternionic projective space Oh, you're right. I mixed up Sp(1) with Spin(1). A simple dimension count should have shown this. Sorry for the confusion and thank you for your answer. The long exact sequence then allows me to calculate the homotopy groups (properly this time). Jun 25 comment There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$ @MhenniBenghorbal I understand that $T\colon f\mapsto g$ is a contraction mapping, if it is well-defined, which is only more or less obvious if $g$ is unique in the first place. Jun 25 comment There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$ I'm not even sure how to define the contraction operator without assuming that $g$ is unique, though. Jun 25 comment There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$ @MhenniBenghorbal I have a Cauchy sequence converging to a unique limit, which satisfies the property. I do not know, whether there are other functions which satisfy this property, though. The unique limit doesn't show uniqueness of the function satisfying the property. Jun 25 comment There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$ @MhenniBenghorbal I like your answer here. Successive iteration gives existence (provided one checks that the sequence is Cauchy). Any suggestions for uniqueness? 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 $\operatorname{Aut}(\mathfrak{g\oplus g)/\operatorname{Aut}(\Delta g})$.