Johannes Ebert
|
Registered User
|
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.
|
|
22h |
answered | The “right” $C^*$ algebraic proof of Bott Periodicity |
|
1d |
awarded | ● Nice Question |
|
May 22 |
comment |
Karoubi versus Kasparov K-theory Yes, that is a convenient picture, but not the one I wish to consider (I have a concrete application in mind). |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory @Paul: I do not see where you use the information from a Karoubi element here. |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory This cannot possibly be correct, because there is an important piece of structure missing: there is no $Z/2$-grading on your $H$. This is absolutely essential. Here is the reason: if $X$ is compact and if $(E,\phi)$ is a \emph{finite-dimensional} $Cl^{p,q}-C(X)$-bimodule, then $\mathcal{K}(E)=\mathcal{L}(E)$. Thus EACH graded operator $F$ yields a Kasparov module $(E,\phi,F)$, and they all represent the same element in $KK$. If one wants to represent a KK-class by a finite-dimensional Hilbert module, all the information is contained in the grading. -1; sorry about that. |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory Still, it feels too complicated. What I imagined is that to a Karoubi triple $(E, \eta_0,\eta_1)$, one can associate a Kasparov element $(H,F)$, where $H$ is the space of sections in a \emph{finite-dimensional} graded $Cl^{q,p}$-vector bundle and $F$ some operator. Or is this too naive and there exists an obstruction against this? |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory @Paul: ''unbounded multiplier''? We are on a finite-dimensional space, and all $C^{\ast}$-algebras in sight are unital. Moreover, $c^2 =-1$, whence $(1+c^2 )=0$. Also, if it is well-defined at all, the element you define is zero, since you have an invertible operator. |
|
May 21 |
revised |
Karoubi versus Kasparov K-theory added 1051 characters in body |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory Alternatively, I could rephrase my question: How do I make $K^{p,q}(X) \to [X; \mathcal{F}^{p,q}]$ explicit (in this direction; otherwise it is useless in the context where this question arose) |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory Mapping from the space $\mathcal{F}^{p,q}$ to $KK$-theory is much simpler. First of all, it should be the space of $Cl^{p,q+1}$-antilinear Fredholms. When you consider the last basis vector as a grading, you get a graded Hilbert space, with a graded $Cl^{p,q}$-action. The operator $F$ anticommutes with that action, hence it defines a Kasparov module for $KK(Cl^{p,q};R)$. |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory This is not what I would call an explicit map. |
|
May 21 |
comment |
Karoubi versus Kasparov K-theory @Zhaoting: Theorem III.4.22, loc. cit, states that the definition I gave in my question agrees with Def III.4.11. This is not the source of my confusion. |
|
May 20 |
asked | Karoubi versus Kasparov K-theory |
|
Apr 25 |
comment |
Relative index theorem for Clifford linear Dirac operators Ulrich Bunke has written a paper "A K-theoretic relative index theorem...", available on his webpage mathematik.uni-regensburg.de/Bunke. He treats the case of an arbitrary (complex or real) C^*-algebra as coeffient algebra. |
|
Apr 22 |
awarded | ● Popular Question |
|
Apr 21 |
answered | How we do actually compute the topological index in Atiyah-Singer? |
|
Mar 24 |
awarded | ● Notable Question |
|
Mar 24 |
answered | What are your favorite instructional counterexamples? |
|
Mar 24 |
answered | is there any algebraic function that has a specific relation to transcendental one? |
|
Mar 21 |
revised |
Is every ‘'group-completion’' map an acyclic map? added 394 characters in body |
|
Mar 8 |
awarded | ● Necromancer |
|
Mar 6 |
revised |
Nearby homomorphisms from compact Lie groups are conjugate added 271 characters in body |
|
Mar 6 |
answered | Nearby homomorphisms from compact Lie groups are conjugate |
|
Mar 4 |
revised |
Parallelizability of the Milnor’s exotic spheres in dimension 7 added 5 characters in body |
|
Feb 21 |
awarded | ● Necromancer |
|
Feb 21 |
revised |
What is the exterior derivative intuitively? deleted 1 characters in body |
|
Feb 14 |
awarded | ● Popular Question |
|
Feb 5 |
comment |
Nice proofs of the Poincaré–Birkhoff–Witt theorem Lie's Third Theorem is hard, but there is a wonderful differential geometric proof whose Lie algebraic part is really simple. I found it in a paper by Van Est, "Une demonstration de E. Cartan du troisieme theoreme de Lie". Another proof of Lie III without even less Lie algebra theory is in the book by Duistermaat-Kolk. Both proofs use that $H^2(G;\mathbb{R})=0$ for simply-connected Lie group. |
|
Feb 5 |
awarded | ● Necromancer |
|
Jan 31 |
accepted | Equivariant Cohomology for actions with finite stabilizers |
|
Jan 30 |
comment |
Awfully sophisticated proof for simple facts @Ryan: yes, in a sense it is a nice proof. Lefschetz fixed point theorem is a hard result, which depends either on Poincare duality or on simplicial approximation. Most topological proofs I know are considerably more elementary (and use the topology of the complex plane, which is more obviously related to the problem than self-maps of $CP^n$). |
|
Jan 30 |
awarded | ● Civic Duty |
|
Jan 30 |
comment |
Equivariant Cohomology for actions with finite stabilizers @Demin: it does not collapse, unless some condition holds. |
|
Jan 30 |
revised |
Equivariant Cohomology for actions with finite stabilizers added 229 characters in body |
|
Jan 30 |
answered | Equivariant Cohomology for actions with finite stabilizers |
|
Jan 30 |
comment |
Equivariant Cohomology for actions with finite stabilizers You cannot apply Vietoris-Begle here, properness of the map is a crucial assumption. In the present case, the fibres are spaces $BG_x$, and if $1 \neq G_x$, there is no compact model for $BG_x$. |
|
Jan 30 |
awarded | ● Enlightened |
|
Jan 30 |
awarded | ● Nice Answer |
|
Jan 29 |
comment |
How to write down explictly the isomorphism of two finite dimensional representation of compact groups? But the differentials $d$ and $e$ preserve the number $p+q$ (make a picture of the bigraded algebra to see this). |
|
Jan 29 |
answered | How to write down explictly the isomorphism of two finite dimensional representation of compact groups? |
|
Jan 29 |
comment |
How to write down explictly the isomorphism of two finite dimensional representation of compact groups? ""isomorphic as repn of GL(V)" is equivalent to isomorphic as repn of SU(V) (when chosen a metric) since GL(V) and SU(V) generate the same subalgebra in End(V), by density theorem." This is not true: you can twist any rep of $GL(V)$ by a power of the determinant; without changing the restriction of the representation to $SL(V)$. |
|
Jan 28 |
accepted | Relation between groups and classifying spaces |
|
Jan 28 |
revised |
Relation between groups and classifying spaces deleted 16 characters in body |
|
Jan 25 |
awarded | ● Nice Answer |
|
Jan 25 |
answered | Relation between groups and classifying spaces |
|
Jan 25 |
revised |
third stable homotopy group of spheres via geometry? deleted 81 characters in body |
|
Jan 24 |
awarded | ● Necromancer |
|
Jan 21 |
comment |
On the Universality of the Riemann zeta-function Yes, now the question is clear, I was misunderstanding it. |
|
Jan 21 |
comment |
Naturality of the transfer in group cohomology The problem is that the pullback of $EG'\to BG'$ via $Bf$ is only connected if $f$ is surjective. If $f$ is surjective, the naturality holds. |
|
Jan 21 |
answered | On the Universality of the Riemann zeta-function |
