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 Yearling
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Sep
7
asked Cohomology of $\mathcal O(k)$
Jul
7
comment Cell decomposition for connected sum
@DanielRust Thanks. Is the idea to use another Mayer-Vietoris sequence for the "punctured" manifolds, or is there another way to calculate their homology?
Jul
7
asked Cell decomposition for connected sum
Jul
7
accepted $\pi_0$ in the long exact sequence of a fibration and quaternionic projective space
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).
Jul
7
asked $\pi_0$ in the long exact sequence of a fibration and quaternionic projective space
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
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?