1,033 reputation
511
bio website nullplug.org
location Ratisbon, Germany
age 33
visits member for 4 years
seen Aug 22 at 14:04

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 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
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 non-trivial 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