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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 …
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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 …
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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 …
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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_\ …
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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 …
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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 …
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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 …
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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 …
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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{ …
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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 …
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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 $G = O,SO,U$) …
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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} …
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24 votes
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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 …
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