Reputation
785
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
6 18
Impact
~8k people reached

Dec
12
comment Derived functors and coboundary operator
@ZhenLin Edited. Is this more clear?
Dec
12
revised Derived functors and coboundary operator
added 652 characters in body
Dec
12
asked Derived functors and coboundary operator
Nov
15
revised Cohomology of $\mathcal O(k)$
added 197 characters in body
Nov
15
awarded  Custodian
Nov
15
revised Cohomology of $\mathcal O(k)$
added 28 characters in body
Nov
15
reviewed Reviewed Is there a difference between the support and the support space of a d-dimensional complex matrix?
Nov
15
revised Solvable Lie algebra with codimension 1 ideal
added 56 characters in body
Oct
20
answered Orbit and Stabilizer
Sep
7
revised Cohomology of $\mathcal O(k)$
added 36 characters in body
Sep
7
comment Cohomology of $\mathcal O(k)$
@atricolf The total space of the bundle $\mathcal O(k)$.
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.