3,811 reputation
935
bio website math.berkeley.edu/~aaron
location Berkeley, CA
age 27
visits member for 3 years, 6 months
seen Apr 9 at 6:23

Apr
2
awarded  Popular Question
Mar
23
comment Refining homotopy commutative maps of spectra to maps of E_{\infty}-ring spectra
The map $MU \to BP$ is a complex orientation, but it's known that it can't be $E$-almost anything. (It's a big open problem whether $BP$ admits an $E_\infty$-structure at all, but even if this exists it won't be compatible with its complex orientation.)
Mar
11
awarded  Popular Question
Feb
27
comment Kan fibrations and surjectivity
Perhaps, if you embed ssets into augmented ssets. Anyways, no matter how you slice it I feel like this is mostly a matter of convention -- although I do like your observation in the comments on the question to which you linked.
Feb
26
comment Kan fibrations and surjectivity
Well, I'd argue that it's certainly defined, exactly as how all the other horns are -- it's just that it's empty!
Dec
30
comment What's the point of spectra?
Thanks Cary! If you work with infinite CW complexes, though, I think this becomes alright again. A phantom map is a nontrivial map which can't be seen by any finite CW complexes.
Dec
18
awarded  Popular Question
Nov
25
awarded  Nice Question
Oct
27
awarded  Popular Question
Sep
27
awarded  Yearling
Jun
8
awarded  Popular Question
May
20
awarded  Enlightened
May
20
awarded  Nice Answer
May
7
awarded  Caucus
May
4
comment What's the point of spectra?
In #2, cofibrations and fibrations aren't quite "the same thing" -- this depends on taking a model for spectra (and in any model these are going to be inequivalent). Nor are pushouts and pullbacks the same thing. The important point here is that a pair of composable arrows is a fiber sequence iff it's a cofiber sequence.
May
2
comment Basic question about definition of Chern classes
Perhaps most generally, for any topological group $G$ and any cohomology theory $E$, you have "characteristic classes" given by elements of $E^*(BG)$.
Apr
30
comment When does a cohomology theory have a ring structure?
Singular cohomology with ring coefficients admits a ring structure, and all the other theories that tend to agree with this derive their ring structure from it (although Cech cohomology for a ring-valued sheaf also has a nice internal definition). As for extraordinary cohomology theories, this isn't really an answer but a restatement: any extraordinary cohomology theory is represented by a spectrum, and to have a product structure on the cohomology theory is precisely to have not just a spectrum but a ring spectrum. (This is analogous to having not just an abelian group but a ring.)
Apr
29
answered What's the point of spectra?
Apr
29
comment Visualize Fourth Homotopy Group of $S^2$
Jason's suggestion does indeed work -- $\eta^2$ is a generator of $\pi_4(S^2)$. More generally, there's a framework due to Sinha and Walter that describes elements of homotopy in terms of linking numbers. I've never tried to understand it, but the relevant papers are here: arxiv.org/pdf/math/0610437.pdf, arxiv.org/pdf/0809.5084.pdf
Apr
18
comment Problems that differential geometry solves
Probably this is just my own ignorance, but I thought the theory of elliptic curves in the smooth setting is rather trivial?