bio | website | wwwmath.uni-muenster.de/… |
---|---|---|
location | Münster | |
age | 39 | |
visits | member for | 4 years, 9 months |
seen | Jul 16 at 12:21 | |
stats | profile views | 4,721 |
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.
Jun 28 |
comment |
Automorphism group of a fiber bundle surjects onto diffeomorphism group?
If $P$ is a natural fibre bundle, then - by definition - any diffeomorphism is covered by a bundle automorphism. |
Jun 14 |
awarded | Nice Answer |
Apr 13 |
awarded | Nice Answer |
Feb 17 |
revised |
Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant
added 444 characters in body |
Feb 6 |
awarded | Nice Answer |
Feb 4 |
awarded | Necromancer |
Jan 7 |
answered | The periodic values in Bott periodicity |
Jan 7 |
answered | How to show the square root function of a positive semidefinite matrix is differentiable? |
Jan 5 |
revised |
Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant
added 3 characters in body |
Dec 13 |
answered | Equation for non-invertible elements in Clifford algebras |
Dec 2 |
comment |
Local index formula for >ungraded< elliptic operators
Typically, you would replace the operator $P$ by $P \oplus P^{\ast}: E \oplus F \to E \oplus F$. This is graded, and the index (in the graded sense) is the usual index. An ungraded operator has an index in $K^1$, but only if it is self-adjoint. |
Nov 26 |
comment |
Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
I think I agree. What I did not realize was that $F$ is obtained from $(x/2)/\sinh(x/2)$ by a simple scaling in the argument. In that case, the components of the multiplicative seqeunce of $F$ are obtained from the components of the A-hat-genus by multiplication with constants. |
Nov 24 |
awarded | Favorite Question |
Nov 20 |
awarded | Enlightened |
Nov 20 |
awarded | Nice Answer |
Nov 20 |
comment |
Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
yes, the formula has a typo in it. |
Nov 19 |
comment |
Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
@Juan: I could have guessed. I came across the paper, but did not understand what is going on. Can you tell me what the index theoretic significance of the Mayer class is? |
Nov 19 |
answered | Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$? |
Nov 19 |
comment |
Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
And where does the Mayer class come from? I am just curious. |
Nov 19 |
comment |
Is true that $\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$?
What is the Mayer class? I have never heard about it. |