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Feb
27
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Feb
27
accepted The semidihedral group of order 16 and ko
Feb
24
comment The semidihedral group of order 16 and ko
Interesting! Is there an analogous family for odd primes?
Feb
24
awarded  Nice Question
Feb
23
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Feb
23
asked The semidihedral group of order 16 and ko
Jan
9
awarded  Enlightened
Jan
9
awarded  Nice Answer
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awarded  Nice Answer
Mar
29
answered Examples for non-naturality of universal coefficients theorem
Mar
29
comment Examples for non-naturality of universal coefficients theorem
Sorry; that should be $X = M \vee \Sigma M$
Mar
29
comment Examples for non-naturality of universal coefficients theorem
@Mariano, thanks for pointing out the subtlety I missed. I think you can produce a counterexample stably by looking at $X = M \wedge \Sigma M$ where $M$ is the mod $2$ Moore spectrum. Let $f:X \to X$ be the map which on $\Sigma M$ is the inclusion of the $\Sigma M$ summand and on $M$ is the sum of the inclusion of the $M$ summand and the essential composition $g:M\to S^1 \to \Sigma M$. The map $f$ is the identity on $H\mathbb{Z}$, but since $g$ is nonzero on $H\mathbb{Z}/2$, the map $f$ is not the identity on $H\mathbb{Z}/2$.
Mar
29
comment Examples for non-naturality of universal coefficients theorem
BTW, the simplest example of the failure of the UCT to split naturally that I know of is the map from a mod $n$ Moore space to a sphere that collapses the bottom cell.
Mar
29
comment Examples for non-naturality of universal coefficients theorem
Doesn't the UCT imply that no such counterexample exists? The splitting isn't natural, but the short exact sequence is. More simply, the multiplication by $2$ map on $\mathbb{Z}$ induces a LES in the homology of $X$ and if $f$ is a homology iso, the 5 lemma shows it is an iso with mod $2$ coefficients. Or am I missing something?
Mar
9
comment Reference request for manifold learning
Have you looked at the surveys by Carlsson or Harer and Edelsbrunner? There are a lot of resources at comptop.stanford.edu
Feb
23
comment What is the intuition behind the Freudenthal suspension theorem
Did you have a specific proof in mind? Also, I believe there is a typo in your statement (consider $n = 1$, $k = 1$). A more general statement is that if $X$ is $n-1$ connected and $n \ge 2$, then the suspension map $\pi_k(X) \to \pi_{k+1}(\Sigma X)$ is an iso if $k\le 2n-2$ and an epimorphism if $k\le 2n-1$; in the case of spheres, $\pi_{n+k}(S^k) \to \pi_{n+k+1}(S^{k+1})$ is an iso if $n+2 \le k$.
Feb
23
comment Do homotopy groups “always” commute with filtered colimits?
@Harry, combinatorial usually means "locally presentable" and "cofibrantly generated"; did I not make that clear in my comment? Regarding model structures on symmetric spectra, you might add Schwede's excellent "Untitled book project ..." (found on his website) to a list of references. The abundance of model structures on symmetric spectra is a good thing.
Feb
23
comment Do homotopy groups “always” commute with filtered colimits?
@Harry, the underlying category of symmetric spectra is not precisely "simplicial symmetric sequences" (rather it is modules over the sphere spectrum symmetric sequence). However, the category of symmetric spectra is locally presentable and the model structures constructed in, e.g. HSS, MMSS, Shipley's "A Convenient ..." paper are all cofibrantly generated, hence combinatorial. Although the combinatorial condition is the "right" condition for this sort of thing, for spectra, you don't need all this machinery. You might just want section 2 of Hovey-Palmieri-Strickland.