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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
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 …
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 …
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^ …
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, …
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 …
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 …