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The Portuguese Language proposal on Area 51 could use your support. Anyone interested in the Portuguese language can follow the proposal here.


Mar
19
comment Can one exchange fibre and base space in a fibre bundle?
@geodude My question was about "general fibre bundles", I'm not really interested in direct products. Thank you for the link.
Jan
23
comment All almost complex structures on a manifold
@JimBelk $\mathcal O(k)$ is the complex line bundle (real rank 2 bundle) of Chern class $k$ over $\mathbb P^1\cong S^2$ (see the notation on Wikipedia). I loosely expect the space to be finite-dimensional, because the space of complex structures on $\mathcal O(k)$, $k$ neg., is finite-dimensional (this one has to calculate). Even if it isn't finite-dimensional, though, one should be able to construct at least a one-parameter family of almost complex structures for the theorem to be of any practical value.
Jan
22
comment Intuitively what is it if making a modification of a torus?
As far as I can tell, your modification is not an equivalence relation.
Dec
13
comment Derived functors and coboundary operator
@ZhenLin I'm not exactly sure what your point is, but as far as I understand, you can calculate group cohomology either from an injective resolution (which is difficult to find) or from a projective resolution with the Hom functor applied to the whole resolution.
Dec
13
comment Derived functors and coboundary operator
@ZhenLin In principle, yes. I added more details in the question body.
Dec
12
comment Derived functors and coboundary operator
@ZhenLin Edited. Is this more clear?
Sep
7
comment Cohomology of $\mathcal O(k)$
@atricolf The total space of the bundle $\mathcal O(k)$.
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
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.