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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
8
votes
Accepted
Is equivariant oriented cobordism finite?
No. For a $G$-manifold $M$, taking the signature of the fixed points $M^G$ defines a homomorphism $\phi : \Omega_n^G \to \mathbb{Z}$, as if $W : M_0 \leadsto M_1$ is a cobordism then so is $W^G : M_0^ …
5
votes
Almost free actions on simply-connected spaces
EDIT: This is wrong, as Jens explains below.
It is enough to show that the isotropy groups of the 0-cells are trivial: higher dimensional cells must have smaller isotropy.
We may suppose, by restr …
11
votes
Accepted
What is the homotopy type of the poset of nontrivial decompositions of $\mathbf{R}^n$?
Let me write $V$ for a finite-dimensional vector space over some field (the field will not play a role), and $\mathsf{P}(V)$ for the poset described in the question, which I consider as a category. Le …
7
votes
Accepted
The Image of the Mod 2 Homology of BSp in the Homology of BSO
We have $H^*(BO, \mathbb{F}_2) = \mathbb{F}_2[w_1, w_2, \ldots]$, where the $w_i$ are the Stiefel--Whitney classes. If $f: \mathbb{RP}^\infty \to BO$ classifies the reduced universal real line bundle, …
9
votes
Accepted
Calculate the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and $H^d[U(1)\rtimes Z_2...
The group $U(1) \rtimes \mathbb{Z}/2$ you describe is nothing but the group $O(2)$ (as $U(1) = SO(2)$).
As such I think one can see the spectral sequence for the extension does collapse, and one obta …
3
votes
On the homological dimension of a Borel construction
I think $f: M /\!\!/ G \to B\Gamma$ cannot be nontrivial on $\mathbb{Q}$-(co)homology in degrees beyond the dimension of $M$, because I think one can find a factorisation
$$f_* : H_*(M /\!\!/ G ; \mat …