bio  website  math.northwestern.edu/… 

location  Seattle, WA  
age  22  
visits  member for  4 years, 11 months 
seen  45 mins ago  
stats  profile views  4,706 
Graduate student at Northwestern.
Feel free to email me at
dwilson [at] u. northwestern. edu
:)
13h

comment 
Convergence of a sum with the ranks of homotopy groups
Finite generation is a condition not a consequence of your other assumption... Edit: Ok now it is. 
2d

comment 
Which automorphisms on $H_{1}(M^{3})$ are induced by homotopy equivalences?
I agree it should be simpler because of Mostow, but rephrasing it this way didn't seem to help me though I didn't think about it too long. I imagine that if you run the general machine above one will find that you can rephrase the obstruction in elementary terms. 
2d

comment 
Which automorphisms on $H_{1}(M^{3})$ are induced by homotopy equivalences?
(cont'd) so presumably someone who knows more about these manifolds could compute them. That gives you something up to pcompletion, so you'll have to do something rationally as well and then glue. See, e.g., GoerssJardine VIII.4 for this obstruction theory spelled out, and also the work of Lannes etc. 
2d

comment 
Which automorphisms on $H_{1}(M^{3})$ are induced by homotopy equivalences?
I don't know anything about hyperbolic 3manifolds, but there is a general obstruction theory for lifting maps in (co)homology to actual homotopy classes of maps. The first obstruction is that the map on cohomology be a map of modules over the Steenrod algebra, but I don't think that's much of a condition in your case since 3 is a small number. After that there is a sequence of obstructions living in algebraically defined AndreQuillen style groups in the category of unstable modules over the Steenrod algebra. That sounds very complicated and it is, but in this case there's not much to compute 
May 3 
comment 
What if homotopy were expanded to allow any connected space instead of [0,1]?
I think I'm misunderstanding something. If we're working in a world where connected=pathconnected then this notion is the same as homotopy: If two maps are Chomotopic, then pick any path between the points $z_0$ and $z_1$. Precomposing with this path gives an ordinary homotopy. So is the thrust of this question just what happens if we allow weirder connected spaces? 
Mar 15 
comment 
Brandt's definition of groupoids (1926)
is one of your "groupoids" supposed to say "categories"? 
Mar 10 
comment 
How to prove that any perfect complex on an affine scheme is strictly perfect?
(and the reference for my entire comment, almost verbatim, is Thomason's classification of triangulated subcategories paper) 
Mar 10 
comment 
How to prove that any perfect complex on an affine scheme is strictly perfect?
Here's something that is true: On any scheme with an ample family of line bundles (e.g. quasiprojective, or affine, or separated regular Noetherian) every perfect complex is quasiisomorphic to a strictly perfect complex. I'm not sure if that's what you meant by "is" in your claim, but the reference is SGA6 II.2.2.8 or ThomasonTrobaugh 2.3.1. 
Mar 7 
comment 
Lemma 2.1.1.4 in Lurie's HTT
For the first question: I believe the exact thing I need is the aforementioned corollary or 2.1.2.10, I can try to rewrite this more clearly later. 
Mar 7 
comment 
Lemma 2.1.1.4 in Lurie's HTT
For your second question: Applying the last argument for $Y = K$ and $Y = X_s$ shows that in both cases the homotopy class of a lift is determined by $f$, i.e. there is only one homotopy class of lift for $K \times \Delta^1 \rightarrow S$ so it must be the one obtained by composiiton. 
Mar 7 
comment 
Lemma 2.1.1.4 in Lurie's HTT
(Alternatively I can quote Corollary 2.1.2.9 of HTT if you don't like that argument at the end.) 
Mar 7 
answered  Lemma 2.1.1.4 in Lurie's HTT 
Feb 10 
comment 
Anything between vector bundles and sphere bundles?
I thing taking connective covers moves the wrong way. For example, we have a map $BO \rightarrow BhAut(S)$. What you're trying to do (stably) is factor this map through other things. A connective cover, say $SO$, gives a map $BSO \rightarrow BO$ which is not what you want. 
Feb 9 
awarded  Popular Question 
Feb 8 
awarded  Nice Question 
Feb 8 
accepted  Mazur secret Bourbaki report “Analyse padique” 
Feb 8 
comment 
Mazur secret Bourbaki report “Analyse padique”
This is awesome! Thanks for your heroic dedication to the spread of knowledge! Hope the physicists didn't burn down your office though... 
Feb 8 
awarded  Inquisitive 
Feb 7 
comment 
Mazur secret Bourbaki report “Analyse padique”
Apparently it's been too long since I've asked a question and I've forgotten where to find the "Community Wiki" option... Help? 
Feb 7 
asked  Mazur secret Bourbaki report “Analyse padique” 