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Stable homotopy theory is that part of homotopy theory (and thus algebraic topology) concerned with all structure and phenomena that remain after sufficiently many applications of the suspension functor.
2
votes
Crafting Suspension Spectra
For large $i_0,\cdots,i_n$ we can realize $BP_\ast/(v_0^{i_0},\cdots,v_n^{i_n})$ as the $BP$-homology of a finite spectrum $S/(v_0^{i_0},\cdots,v_n^{i_n})$. This follows from Devinatz-Hopkins-Smith. S …
6
votes
Accepted
Universal property of $\mathbb S[z]$ and $E_\infty$-ring maps
$\newcommand{\E}{\mathbf{E}}$Dylan answered question 3 (and hence question 1) in the comments, but here's another equivalent way to see it: $\E_\infty$-maps $S^0[z]\to R$ with $R$ a discrete ring (i.e …
3
votes
Calculate homotopy groups of $\mathbb{Z}_2$-equivariant loop spaces of "complex" topological...
$\newcommand{\Z}{\mathbf{Z}}\newcommand{\Map}{\mathrm{Map}}$Let $\sigma$ denote the sign representation of $\Z/2$, and let $S^{d\sigma}$ denote the one-point compactification of $\sigma^{\oplus d}$. L …
17
votes
Accepted
Homology of spectra vs homology of infinite loop spaces
$\newcommand{\H}{\mathrm{H}} \newcommand{\Z}{\mathbf{Z}}$Let $X$ be a space. Then the $E$-(co)homology of $X$ is the same as the $E$-(co)homology of its suspension spectrum, i.e., $E_\ast(X) \cong E_\ …
8
votes
2
answers
596
views
Realizing $\mathcal{A}(2)//\mathcal{A}(1)$ by a finite spectrum
Let $\cal A$ denote the mod 2 Steenrod algebra. Can the $\mathcal{A}(2)$-module structure on $\mathcal{A}(2)//\mathcal{A}(1)$ be enriched to an $\cal A$-module structure? If so, is there a finite spec …
13
votes
0
answers
279
views
A geometric interpretation of the odd-primary Kervaire elements
Let $\Omega^\mathrm{fr}_\ast \cong \pi_\ast S$ denote the graded ring of cobordism classes of framed manifolds, which, by the Pontryagin-Thom construction, is isomorphic (as a graded ring) to the stab …
16
votes
Spectra with "finite" homology and homotopy
Here are two ways of thinking about it. The first comes from the way one proves the final statement you cited: if $X$ has finitely many nonzero homotopy groups which are all finitely generated, then i …
2
votes
Thom spectra, tmf, and Weierstrass curve Hopf Algebroid
$\newcommand{\MU}{\mathrm{MU}} \newcommand{\SU}{\mathrm{SU}} \newcommand{\tmf}{\mathrm{tmf}} \newcommand{\ko}{\mathrm{ko}} \newcommand{\BGL}{\mathrm{BGL}} \newcommand{\ku}{\mathrm{ku}} \newcommand{\GL …
6
votes
Accepted
Spectral and derived deformations of schemes
In general, these are incredibly hard questions. It seems to me that one natural question to ask (if you are interested in $\pi_0$ of ring spectra) would be about understanding even periodic $\mathbf{ …
16
votes
Accepted
Sphere spectrum, Character dual and Anderson dual
The Anderson dualizing spectrum $I_\mathbf{Z}$ can be defined as follows. Consider the functor $X\mapsto \mathrm{Hom}(\pi_{-\ast} X,\mathbf{Q/Z})$ from the homotopy category of spectra to graded abeli …
4
votes
Sphere spectrum, Thom spectrum, and Madsen-Tillmann bordism spectrum
I'm a little confused by your question. You seem to be implying that the Madsen-Tillmann spectra are not Thom spectra, but this is not true: the definition of the spectrum $MTG(n)$ (for $G = O,SO,U$) …
1
vote
Abelian versions of straightening and unstraightening functors
Let $\mathcal{C}$ be a presentable $\infty$-category, and let $X$ be a Kan complex. Let $\mathcal{C}_{/X} = \mathrm{Fun}(X, \mathcal{C})$. Then $\mathrm{Sp}(\mathcal{C}_{/X}) = \mathrm{Sp}(\mathcal{C} …
24
votes
Accepted
Has anyone seen a nice map of multiplicative cohomology theories?
I'm not sure I understand what "the" map is here, but I'll attempt to answer
the questions that were asked in the body of the question. Sorry if I'm just
saying things that you already know.
$\newcomm …