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Bertram Arnold

Feb 1, 2021 16:02

For p = 2, n = 1, the $E(1)$-local sphere can be identified with the fiber of $\psi_3-1: bspin\to bo$, so there is a long exact sequence which sandwiches the $E(1)$-local homotopy groups between homotopy groups of ko. The image of the classical J-homomorphism $pi_*(O)\to \pi_*(S)$ is detected in the source, and the mu-family is detected in the target.

Bertram Arnold

Feb 1, 2021 15:48

The Adams mu-family generates Z/2-summands in degrees 8k+1 and 8k+2 which are detected in KO_*, so they should lie in filtration 1. They are not in the image of J (excluding the special case 8k+1 = 1).

Bertram Arnold

Aug 27, 2020 12:19

@ArunDebray dimensional reduction in the Stolz-Teichner program is a bit subtle because of the rigid Euclidean super-geometries that the "spacetime" manifolds are equipped with. To reduce from 1d to 0d, one can indeed compactify over a circle bundle since there is a quotient map from the 1d to the 0d super Poincare group (cf arXiv:0711.3862). But from 2d to 1d, such a map just doesn't exist. Dan Berwick-Evans has some ideas about this in arXiv:1311.6836.

Bertram Arnold

Dec 20, 2019 13:04

There defininitely are some restrictions: Segal constructs the Leray spectral sequence when the target (in this case X/G) is paracompact, and the stalks of the derived direct image can be identified with lim H^*(U_{hStab(x)};Q), where the limit is over Stab(x)-invariant neighbourhoods of x, so the vanishing result might use some property of X (local contractibility should be enough).

Bertram Arnold

Dec 19, 2019 14:06

.. @BrunoStonek The canonical map p: X_{hG} = EG x_G X -> X/G gives rise to a Leray spectral sequence (constructed for instance in Segal's
Classifying spaces and spectral sequences, Proposition 5.2), with E_2^{p,q} = H^p(X/G,R^q p_*(Q))$ converging to H^{p+q}(X_{hG},Q). The stalk of R^q p_*(Q) at the orbit [x] is H^q(Stab(x);Q), so the spectral sequence is concentrated in the line q = 0 and the edge homomorphism corresponding to H^*(X/G;Q) -> H^*(X_{hG};Q) is an isomorphism.

Bertram Arnold

Aug 9, 2018 12:12

For the simply-connected case, this follows from the construction in Proposition 4C.1 of Hatcher's book

Bertram Arnold

Jul 4, 2017 13:41

No, for instance the threefold product of the periodic (i.e. non-bounding) spin structure on S^1 should have nonvanishing kernel (just tensor up sections in the kernel of the Dirac operator on S^1). However since the Atiyah orientation MSpin -> KO is an iso in this dimension the Â-genus of every three-dimensional spin manifold vanishes, so the kernel and cokernel have the same dimension (more precisely, a perturbation of the Dirac operator has no kernel)

Bertram Arnold

Mar 2, 2016 17:40

The statement is about a single quaternionic structure: If V is a complex vector space and J an antilinear isomorphism of V satisfying J^2 = -1, there is a real structure (i.e. an antilinear isomorphism which squares to +1) on V+V given by
0 J
-J 0

Bertram Arnold

Jan 30, 2016 17:34

And this retraction is given by also collapsing all $n$-cells except the one you are considering. In your writeup it looks as if you want to collapse only this $n$-cell (which won't work). Look also at the definition of $X^{n-1}_\beta$ after Definition 10.10

Bertram Arnold

Jan 30, 2016 17:12

Essentially because the category of pointed spaces has a zero object

Bertram Arnold

Jan 30, 2016 17:12

But this admits a retraction - it's just the inclusion of a wedge summand

Bertram Arnold

Jan 30, 2016 16:55

These "incidence numbers" should just be the degree of the map constructed by Switzer. The n-cells of X and Y are related by f: X^n/X^{n-1} -> Y^n/Y^{n-1}, and you use the characteristic maps of the n-cells to map from and into these bouquets of spheres to/from S^n

Bertram Arnold

Jan 9, 2016 18:29

The Z/2-cohomology of $S$ is just the $S$-homology of HZ/2, so evaluating both theories on HZ/2 should do the trick

Bertram Arnold

Jan 9, 2016 18:15

@lenticcatachresis In the exercise it says "Let A be a symmetric spectrum of abelian groups", that should imply that the corresponding spectrum splits as a sum of Eilenberg-MacLane spectra

Bertram

Jan 9, 2016 17:49

@lenticcatachresis $H^1(S;Z/2) = 0,H^1(\bigvee_r \Sigma^r H\pi_r^s; Z/2) = Z/2$ coming from $\pi_1 = Z/2$

Bertram

Dec 3, 2015 18:01

sigma1|1=dirac

Bertram Arnold

Homotopy Theory