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Aug
27 |
awarded | Yearling |
Feb
23 |
awarded | Nice Question |
Sep
3 |
awarded | Enlightened |
Sep
3 |
awarded | Nice Answer |
Sep
2 |
answered | What is the geometric fixed points of an (equivariant) Eilenberg Maclane Spectrum? |
Aug
27 |
awarded | Yearling |
Aug
20 |
comment |
How does the kernel of the map $\Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)$ relate to equivariant cohomology?
Alternatively one can take the simplicial $G$-space $G^{\cdot+1}$ all of whose face maps are projection maps and whose degeneracies are diagonal maps. You can then take the product of this simplicial $G$-space with the constant simplicial $G$-space $X$. To obtain a simplicial $G$-space which levelwise looks the same as the one in the previous comment, but with different structure maps and different $G$-actions (this one is diagonal and realizes the Borel construction). When $X$ is $G$-free the augmentation admits an extra degeneracy and the the associated cochain complex is exact. |
Aug
20 |
comment |
How does the kernel of the map $\Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)$ relate to equivariant cohomology?
I don't think the projection maps will define a simplicial diagram. For example, if your diagram was simplicial, there should be an equality between the two composite face maps $(g_1,g_2,x)\mapsto g_1 g_2 x$ and $(g_1,g_2,x)\mapsto g_1 x.$ The standard construction uses the group multiplication on $G$ not the projection. Using this construction one obtains an augmented simplicial $G$ space $G^{\cdot+1}\times X\rightarrow X$. This is the bar construction on $X$ and the geometric realization is $G$-homeomorphic to the Borel construction. |
Aug
19 |
revised |
RO(G) grading of Mackey functors
added the keyword rational |
Aug
19 |
revised |
RO(G) grading of Mackey functors
added 41 characters in body |
Aug
19 |
answered | RO(G) grading of Mackey functors |
Jun
9 |
revised |
The homotopy of universal Thom spectrum
There was a mistake: Neil Strickland's observation about the contractibility of holds for odd primes |
Jun
8 |
awarded | Enlightened |
Jun
8 |
awarded | Nice Answer |
Jun
8 |
comment |
The homotopy of universal Thom spectrum
@Prasit: I did not fully understand your comment, but I will reiterate how my answer relates to your question. The case in your question is $MSL_1(S_p)$ whose $\pi_0$ is torsion-free. Since I suspected you meant to ask about $MGL_1(S_p)$ I included that case. That spectrum is universal in the sense that it is contractible and hence terminal. I am claiming that the homotopy groups of $MG$ are $\bZ/p$-modules and calculating them is equivalent to calculating the homology as as a comodule over the dual Steenrod algebra. I did not claim that the latter problem was easy; it is just algebraic. |
Jun
8 |
comment |
The homotopy of universal Thom spectrum
@NeilStrickland: This is an unpublished result of Hopkins and Mahowald. It appears as Thm 4.12 here: nullplug.org/publications/p-torsion.pdf . |
Jun
8 |
comment |
The homotopy of universal Thom spectrum
@Prasit: $R$ is a wedge of suspensions $H\mathbb{Z}/p$'s so you get one $\mathbb{Z}/p$ in the homotopy groups for each summand and for each such summand you get a copy of the dual Steenrod algebra in the homology. |
Jun
7 |
revised |
The homotopy of universal Thom spectrum
Further fill out the answer following helpful comments from Neil Strickland. |
Jun
7 |
comment |
The homotopy of universal Thom spectrum
Thanks Neil! I was in a bit of a rush. I will complete my response and add your comments. |
Jun
7 |
answered | The homotopy of universal Thom spectrum |