Skip to main content

All Questions

Filter by
Sorted by
Tagged with
41 votes
1 answer
10k views

Why not a Roadmap for Homotopy Theory and Spectra?

MO has seen plenty of roadmap questions but oddly enough I haven't seen one for homotopy theory. As an algebraic geometer who's fond of derived categories I would like some guidance on how to build up ...
John Salvatierrez's user avatar
26 votes
1 answer
1k views

From the perspective of bordism categories, where does the ring structure on Thom spectra come from?

To fix ideas, let's consider the Thom spectrum of framed bordism $M$, the spectrum whose homotopy groups are the framed bordism groups. $M$ has a ring spectrum structure inducing the product of ...
Qiaochu Yuan's user avatar
21 votes
1 answer
3k views

Motivation and potential applications of spectral algebraic geometry

Nowadays there is a lot of talk about derived algebraic geometry, but not so much about the related subject of spectral algebraic geometry. Now I'm curious what future is there for spectral algebraic ...
JDou9's user avatar
  • 433
18 votes
1 answer
930 views

When do the polynomial algebra and free algebra coincide in brave new algebra?

Given an $\mathbb E_\infty$-ring (highly structured commutative ring spectrum if you want) $R$, we have the free $R$-algebra (on one generation) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ ...
A Rock and a Hard Place's user avatar
18 votes
1 answer
2k views

Is the $\infty$-category of spectra “convenient”?

A 1991 paper of Lewis, titled “Is there a convenient category of spectra?” proves that there is no category $\mathrm{Sp}$ satisfying the following desiderata$^1$: There is a symmetric monoidal smash ...
Emily's user avatar
  • 11.8k
16 votes
1 answer
608 views

Multiplicative Brown representability?

The Brown representability theorem can be convenient way to construct a spectrum. But to get a ring spectrum of even a very unstructured form seems to be harder. There's even currently a statement on ...
Tim Campion's user avatar
12 votes
2 answers
2k views

Connective spectra and infinite loop spaces

It seems to be standard that connective spectra are "the same" as infinite loop space. However, I do not understand the reason why the associated spectrum is connective. For me, an infinite loop ...
Matthias Ludewig's user avatar
8 votes
2 answers
294 views

Morphisms of $\mathbb E_l$-rings between $\mathbb E_k$-rings for $l<k$

Given two commutative rings $A$ and $B$, any map of rings $A\to B$ will automatically preserve the commutative structure. This is to say, the forgetful functor $\operatorname{CRing}\to \operatorname{...
A Rock and a Hard Place's user avatar
5 votes
1 answer
322 views

Is there a model structure for S-modules such that cofibrant operad-algebras forget to cofibrant S-modules?

In 1997, Elmendorf, Kriz, Mandell, and May wrote a book Rings, Modules, and Algebras in Stable Homotopy Theory in which they introduced the category of $S$-modules as a model for the stable homotopy ...
David White's user avatar
  • 30.3k
5 votes
1 answer
444 views

endomorphisms of modules over symmetric ring spectra

I have a probably very basic question about modules over symmetric ring spectra: Let $R$ be a commutative symmetric ring spectrum and let $M$ and $N$ be module spectra over $R$. Moreover, let $\...
Ulrich Pennig's user avatar