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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 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
25
asked There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$
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
18
asked Weight spaces of Verma modules
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
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.