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Homotopy theory is an important sub-field of algebraic topology. It is mainly concerned with the properties and structures of spaces which are invariant under homotopy. Chief among these are the homotopy groups of spaces, specifically those of spheres. Homotopy theory includes a broad set of ideas and techniques, such as cohomology theories, spectra and stable homotopy theory, model categories, spectral sequences, and classifying spaces.
4
votes
1
answer
153
views
Homotopy equivalence of maps with compact support and maps which vanish at infinity
Let $X$ be a nice space, maybe a manifold, and let $Y$ be a based space.
What sort of conditions must we impose on $Y$ (and $X$ if need be) to get a homotopy equivalence
$$
\mathcal C_c(X,Y) \simeq …
5
votes
2
answers
582
views
Group of units of a ring spectrum vs of its connective cover
Let $R$ be a commutative ring spectrum (interpret this as you will; as an $E_\infty$-ring or as a commutative $S$-algebra etc.) and $\operatorname{GL}_1(R)$ as usual denote its space of units. If $\t …
12
votes
1
answer
477
views
$QS^0$ isn't a product of Eilenberg-Mac Lane
I am reading Lewis' paper "Is there a convenient category of spectra?". To prove the main result on the non-existence of such a nice category, he shows that otherwise the unit component of $QS^0= \var …
8
votes
2
answers
289
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{R …
10
votes
Accepted
Grading ring spectra over the sphere spectrum
One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads!
1. $S$-grading on …
6
votes
Accepted
Is there a "spectral exterior algebra" construction in higher algebra?
Interesting question! I can't give a real answer, but here are some idle musings:
Note that one way to encode exterior powers is as $\Lambda^i_R(E) = \Sigma^{-i}(\mathrm{Sym}^i_R(\Sigma(E)))$ (since p …
4
votes
0
answers
152
views
Preorientation of additive formal group
In "A Survey of Elliptic Cohomology", Section 3.2, Lurie asserts that the preorientations of the additive formal group $\widehat{\mathbf G}_a$ over $\mathbf Z$ are classified by the $\mathbb E_\infty$ …
3
votes
(Pre)orientation vs. formal completion
Okay, I think I may have figured out how B) $\Rightarrow$ A). Because nobody else has given an answer yet, I'm spelling it out (I hope that's not considered bad form). And if what I wrote doesn't make …
2
votes
1
answer
166
views
Are morphisms of parametrized spectra themselves parametrized morphisms of spectra?
Let $X$ be a fixed parametrizing space. Let $E$ and $E'$ be two spectra and let $E_X$ and $E'_X$ be their trivial parametrized versions. Intuitively I imagine that the morphisms of parametrized spectr …
3
votes
1
answer
235
views
Morphisms of parametrized ring spectra
This is a follow-up to this question, in which Denis Nardin nicely explained that
$$
\operatorname{Map}_{\operatorname{Fun}(X, \operatorname{Sp})}(E_X, E'_X)
\simeq
\operatorname{Map}(X, \operatorname …
5
votes
1
answer
444
views
Stable homotopy groups of $QX$
If $X$ is a space, we can form $QX=\varinjlim \Omega^n\Sigma^nX$ which is an infinite loop space with homotopy groups $\pi_i(QX)=\pi^{s}_i(X)$ the stable homotopy groups of $X.$ But these are the unst …
18
votes
1
answer
915
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) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ …
11
votes
1
answer
347
views
(Pre)orientation vs. formal completion
Let $\mathbb G$ be an abelian vatiety over an $\mathbb E_\infty$-ring $A$. That is to say, it consists of an abelian group object in the $\infty$-category of relative schemes $\mathbb G\to \operatorna …
5
votes
Interpolating between the flat and smooth affine lines in spectral algebraic geometry
I believe there are no exterior algebras in sight. To see this, let us think through the $1$-categorical case carefully. We have a commutative ring $R$ and the ordinary category of $R$-modules $\mathr …