bio | website | wwwmath.uni-muenster.de/… |
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location | Münster | |
age | 38 | |
visits | member for | 3 years, 10 months |
seen | Jul 31 at 8:47 | |
stats | profile views | 4,266 |
I am a mathematician based in Münster, Germany. A fellow MO user once described me as a ''very homotopy-theoretic geometric topologist'' and I largely approve this description.
Jul 16 |
awarded | Enlightened |
Jul 16 |
awarded | Nice Answer |
Jul 16 |
revised |
Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant
added 7 characters in body |
Jul 16 |
answered | Norms on Clifford algebra (C^* norm) |
Jul 11 |
answered | Euler Class of a vector field |
Jul 11 |
comment |
Euler Class of a vector field
@Henry T. Horton: you can only take the orthogonal complement if the vector field does not have a zero. |
Jul 2 |
awarded | Curious |
Jun 25 |
comment |
Question about the fundamental group of rational homology 3-spheres
If I am not mistaken (!), this shows, together with the poincare conjecture, that $M$ is a connected sum of two lens spaces. So unless $M$ is a sum of lens spaces, it has $\pi_2 (M) =0$. |
Jun 25 |
comment |
Question about the fundamental group of rational homology 3-spheres
I do not know if that helps, but here is a remark. Assume that $M$ is an oriented rational homology sphere such that $\pi_1 (M)$ requires 2 generators. If $\pi_2 (M) \neq 0$, then $M$ contains an embedded $S^2$ which is nontrivial in homotopy (by the sphere theorem). This sphere must be two-sided, and so by cutting $M$ along the sphere, you write $M$ in a nontrivial way as the connected sum of two manifolds; and both of them are rational homology spheres, and none is simply connected (by the Poincare conjecture). By Grushko's theorem, the fundamental groups of both pieces are cyclic. |
Jun 17 |
revised |
Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant
added 156 characters in body |
Jun 16 |
answered | Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant |
May 28 |
comment |
Construction of a Bott manifold
The spin bordism group of 8-dimensional manifolds is $Z \oplus Z$, detected by the $\hat{A}$-genus and the signature. A basis is given by a Bott manifold $B$ and $HP^2$. The K3-surface K has $\hat{A} (K^2) =4$ and $\sign (K^2)=16^2$. Therefore $[K^2] = 16^2 [HP^2]+ 4 [B]$, and this yields Laures formula (note that $[K^2]$ is divisible by 4). |
May 28 |
answered | Construction of a Bott manifold |
May 13 |
awarded | Nice Answer |
May 1 |
revised |
Signature of compact oriented 4-manifold
edited body |
Apr 25 |
awarded | Nice Answer |
Apr 25 |
revised |
When does a cohomology class induce an isomorphism between homotopy groups?
added 1093 characters in body |
Apr 24 |
answered | When does a cohomology class induce an isomorphism between homotopy groups? |
Apr 9 |
awarded | Enlightened |
Apr 9 |
awarded | Nice Answer |