**4**

votes

**1**answer

300 views

### Values of the multiplicative group over a ring spectrum

In his notes on elliptic cohomology, Lurie defines the multiplicative group $\mathbb{G}_m$ over a ring spectrum $A$ as $\operatorname{Spec} A[\mathbb{Z}]$. What is the value $\mathbb{G}_m(B)$ of the ...

**4**

votes

**1**answer

537 views

### A Model Structure on Symmetric Monoidal Categories

The recent article found here revisits Thomason's proof that symmetric monoidal categories model all connective spectra, but stops short of showing that there is a full closed model structure on this ...

**5**

votes

**1**answer

699 views

### Serre spectral sequence with spectra

A friend recently asked me if i had heard anything about a stable Serre Spectral Sequence or one constructed with spectra, has any one else ever heard of this? is there any reason other than ...

**13**

votes

**1**answer

820 views

### Complex orientations on homotopy

I am wondering if there is a more "geometric" formulation of complex orientations for cohomology theories than just a computation of $E^*\mathbb{C}$P$^{\infty}$ or a statement about Thom classes. It ...

**6**

votes

**1**answer

642 views

### Spectra and localizations of the category of topological spaces

Can we construct the category of spectra (or maybe just its homotopy category) from the category of pointed topological spaces
using some kind of localization combined with other categorical ...

**3**

votes

**3**answers

356 views

### Are there universe-indexed spectra over simplicial sets?

In "Rings, Modules, and Algebras in Stable Homotopy Theory" Elmendorf, Kriz, Mandell and May introduce a notation of spectra indexed over an universe $\mathcal{U}$ as a collection of pointed ...

**13**

votes

**1**answer

696 views

### Stable ∞-categories as spectral categories

Let C be a stable ∞-category in the sense of Lurie's DAG I. (In particular I do not assume that C has all colimits.) Then C does have all finite colimits, the suspension functor on C is an ...

**13**

votes

**3**answers

1k views

### complex cobordism from formal group laws?

Reading Ravenel's "green book", I wonder about his question on p.15 "that the spectrum MU may be constructed somehow using formal group law theory without using complex manifolds or vector bundles. ...

**6**

votes

**2**answers

309 views

### Reference for iterated homotopy fixed points?

What are (good) references for results about iterated homotopy fixed points? That is, suppose G is a topological group acting on a space (or spectrum) X, and H is a normal subgroup of G. Then one ...

**11**

votes

**2**answers

666 views

### squares in stable homotopy

I noticed that the generator of the second stable stem b is the square of the generator of the first stable stem a, in the sense that if take two copies of a and smash product them together you get b ...

**8**

votes

**1**answer

351 views

### Stable presentable categories as module categories

There is a theorem of Schwede and Shipley which classifies categories of modules over an A∞ ring spectrum as those stable presentable (∞,1)-categories with a compact generator. Suppose I ...

**12**

votes

**3**answers

908 views

### Classifying triangulated structures on a graded category

I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology). However, it's never been clear to me ...

**14**

votes

**2**answers

906 views

### Exotic spheres and stable homotopy in all large dimensions?

Given that the kervaire invariant one problem has been solved in (almost) all dimensions....my question is whether there exists an exotic sphere in all sufficently lagre dimensions? Given the Kervaire-...

**11**

votes

**2**answers

726 views

### Are generalized cohomology theories a homotopy category of some category of invariants?

I was taught to think of generalized cohomology theories as the homotopy category of (symmetric) spectra. But is there also a category of 'invariants', that is, some category of contravariant functors ...

**43**

votes

**9**answers

5k views

### understanding Steenrod squares

There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called Steenrod squaring: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from ...