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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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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^ …
Oscar Randal-Williams's user avatar
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, …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar