bio  website  nullplug.org 

location  Ratisbon, Germany  
age  33  
visits  member for  4 years 
seen  Aug 22 at 14:04  
stats  profile views  501 
I am a postdoc (Akademischer Rat auf Zeit) at the University of Regensburg. I dig all things homotopical and algebraic.
57m

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 torsionfree. 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/ptorsion.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 
Jun 6 
comment 
Group actions in a homotopy category
. whose composite is multiplication by $G$ and which factors through $\mathrm{Ho}(M)^G(X,Y)$. From here one can see that $\mathrm{Ho}(M^G)(X,Y)\cong\mathrm{Ho}(M)^G(X,Y)$ when $G$ acts invertibly. It appears the nontrivial part is seeing that $\mathrm{tr}$ descends to the homotopy category. 
Jun 6 
comment 
Group actions in a homotopy category
As often happens, behind a vanishing spectral sequence argument lies a more elementary argument. My more involved method is probably overkill in this case. Nonetheless, the argument indicates why it is true. For $X,Y\in M^G$ we have a map $$ \mathrm{tr}\colon M(X,Y)\rightarrow M^G(X,Y)$$ given by $f()\mapsto \sum_{g\in G} (g\cdot f)()=\sum_{g\in G} gf(g^{1} )$. I suspect that $\mathrm{tr}$ descends to homotopy categories, perhaps by an argument with right homotopies. If this is the case, then precomposing with the restriction functor gives a self map on mapping sets in $\mathrm{Ho}(M^G)$.. 
Jun 6 
revised 
Group actions in a homotopy category
More precisely and correctly answered the question. 
Jun 6 
revised 
Group actions in a homotopy category
More precisely and correctly answered the question. 
Jun 6 
revised 
Group actions in a homotopy category
deleted 1 character in body 