All Questions
8 questions
5
votes
1
answer
185
views
Decompose $MT(E(d)\times_{\mathbb Z_2} SU(2))$ as the wedge sum or smash product of spectra
Consider the extension
$$1\to SU(2)\to X\to O\to1,$$
there are 4 possibilities for $X$:
$X=O\times SU(2)$ or $E\times_{\mathbb{Z}_2}SU(2)$ or $Pin^+\times_{\mathbb{Z}_2}SU(2)$ or $Pin^-\times_{\...
- 2,147
5
votes
0
answers
161
views
Twisted spin-bordism invariant and a possible Postnikov square from $d=2$ to $d=5$
This is a follow up more advanced question following Twisted spin bordism invariants in 5 dimensions. We follow the definitions in the earlier post.
I had discussed my computation of
$$
\Omega_5^{...
- 10.5k
7
votes
1
answer
470
views
Twisted spin bordism invariants in 5 dimensions
[Note]: My question will be a bit long. So, first, thank you for your careful reading, generous comments, helps and answers, in advance!
The spin $G$-bordism invariant can be twisted in the way that ...
- 10.5k
4
votes
0
answers
97
views
Decomposition of bordism groups for $BG$ where $G$ is a product of two groups
Let a group $G=G_1 \times G_2$, where
$G_1$ is a discrete group (can be finite or infinite),
$G_2$ be any compact Lie group or finite group.
Question: Is there some simple result that we can ...
- 10.5k
6
votes
0
answers
142
views
Pin cobordism v.s. "KO" theory in low or in any dimensions
Fact: The spin cobordism is equivalent to "KO" theory in low dimension if we only consider the 2-torsion.
This is related to a question and an answer supports the claim.
Here we denote the $p$-...
- 10.5k
6
votes
0
answers
224
views
Arf-Brown-Kervaire invariant and a surjective map $G \to Pin^-$
We know that the Arf-Brown-Kervaire (abk) invariant is a bordism invariant of
$$
\Omega_2^{Pin^-}(pt)=\mathbb{Z}/(8\mathbb{Z}),
$$
where the $\mathbb{Z}/(8\mathbb{Z})$ is generated by a 2-manifold $M^...
- 2,147
5
votes
1
answer
278
views
Manifold generators of O-bordism invariants
If I understand correctly, I can obtain the $O$-cobordism group of
$$
\Omega^{O}_3(BO(3))=(\mathbb{Z}/2\mathbb{Z})^4,
$$
The 3d cobordism invariants have 4 generators of mod 2 classes, are generated ...
- 2,147
9
votes
2
answers
641
views
Künneth formulas/theorem for bordism groups and cobordisms?
We are familiar with Künneth theorem:
The Kunneth formula is given by $R$ as a ring, $M,M'$ as the R-modules, $X,X'$ are some chain complex. The Kunneth formula shows the cohomology of a chain ...
- 10.5k