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 Jun25 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? Jun25 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. Jun25 asked There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$ Jun19 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 $G$. Jun18 asked Weight spaces of Verma modules 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. Jun17 awarded Yearling Jun16 accepted Non-orientable 3-manifold has infinite fundamental group Jun16 accepted Why are the integers appearing in lens spaces coprime? 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? Jun16 asked Why are the integers appearing in lens spaces coprime? Jun15 comment Non-orientable 3-manifold has infinite fundamental group @dfeuer Good question. I guess I can assume that top homology vanishes. Jun15 asked Non-orientable 3-manifold has infinite fundamental group Jun9 accepted Help with inequality for real numbers: $||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$ Jun6 asked Help with inequality for real numbers: $||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$ May21 accepted Space modelled on ring 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.