User dylan wilson - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:17:40Z http://mathoverflow.net/feeds/user/6936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/134217/strict-applications-of-deformation-theory-in-which-to-dip-ones-toe Strict applications of deformation theory in which to dip one's toe Dylan Wilson 2013-06-20T04:18:25Z 2013-06-20T07:20:47Z <p>I hesitate to ask a question like this, but I really have tried finding answers to this question on my own and seemed to come up short. I readily admit this is due to my ignorance of algebraic geometry and not knowing where to look... Then I figured, that's what this site is for!</p> <p>Here's the short of it:</p> <blockquote> <p>What are some examples of <em>strict applications</em> of deformation theory? That is, what are examples of problems that can be stated without mentioning deformation theory <strong>or moduli spaces</strong> and one of whose solutions uses deformation theory? Please state the problem precisely in your answer, and provide a reference if at all possible :)</p> </blockquote> <p>Here's the long of it:</p> <p>I really want to swim in the Kool-Aid fountain of deformation theory and taste of its sweet, sweet purple love, but I'm having trouble. When I wanted to learn about <em>K</em>-theory, I learned about it through the solution to the Hopf invariant one problem, the solution to the vector fields on spheres problem, and through the Adams conjecture. When I wanted to learn some equivariant stuff, it was nice to have the solution to the Kervaire invariant one problem as a guiding force. I have trouble learning things in a bubble; I need at least a slight push.</p> <p>Now, I know that deformation theory is useful for building moduli spaces, but the trouble is that, aside from the ones that appear in homotopy theory, I haven't fully submerged in this sea of goodness either. The exception would be any example of a strict application that used deformation theory to construct some moduli space and then used this space to prove some tasty fact.</p> <p>To give you all an idea, here are the only examples I have found (from asking around) that fit my criteria:</p> <ol> <li><em>Shaferavich-Parshin</em>. Let $B$ be a smooth, proper curve over a field and fix an integer $g \ge 2$. Then there are only finitely many non-isotrivial (i.e. general points in base have non-isomorphic fibers) families of curves $X \rightarrow B$ which are smooth and proper and have fibers of genus $g$. </li> <li>Given $g\ge 0$, then every curve of genus $g$ has a non-constant map to $\mathbb{P}^1$ of degree at most $d$ whenever $2d - 2 \ge g$. </li> <li>There are finitely many curves of a given genus over a field.</li> <li>The solution to the Taniyama-Shimura conjecture uses deformations of Galois reps.</li> </ol> <p>1, 2, and 3 are stolen from Osserman's really great note: <a href="https://www.math.ucdavis.edu/~osserman/classes/256A/notes/deform.pdf" rel="nofollow">https://www.math.ucdavis.edu/~osserman/classes/256A/notes/deform.pdf</a> </p> <p>I really like the theme of 'show there are finitely many gadgets by parameterizing these gadgets by a moduli space with some sort of finite type assumption, then showing no point admits nontrivial deformations.' Any examples of this sort would be doubly appreciated. (I guess Kovács and Lieblich have an annals paper where they do something along these lines for the higher-dimensional version of the Shaferavich conjecture, but since they end up counting deformation types of things instead of things, it doesn't quite fit the criteria in my question... but it's still neat!)</p> <p>Galois representations are definitely a huge thing, and I'd be grateful for any application of their deformation theory that's more elementary than, say... the Taniyama-Shimura conjecture.</p> <p>So yeah, that's it. Proselytize, laud, wax poetic- make Pat Benatar proud.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example Spectral sequences: opening the black box slowly with an example Dylan Wilson 2010-11-06T06:33:59Z 2013-06-20T02:48:47Z <p>My friend and I are attempting to learn about spectral sequences at the moment, and we've noticed a common theme in books about spectral sequences: no one seems to like talking about differentials.</p> <p>While there are a few notable examples of this (for example, the transgression), it seems that by and large one is supposed to use the spectral sequence like one uses a long exact sequence of a pair- hope that you don't have to think too much about what that boundary map does.</p> <p>So, after looking at some of the classical applications of the Serre spectral sequence in cohomology, we decided to open up the black box, and work through the construction of the spectral sequence associated to a filtration. And now that we've done that, and seen the definition of the differential given there... we want some examples.</p> <p>To be more specific, we were looking for an example of a filtration of a complex that is both nontrivial (i.e. its spectral sequence doesn't collapse at the $E^2$ page or anything silly like that) but still computable (i.e. we can actually, with enough patience, write down what all the differentials are on all the pages).</p> <p>Notice that this is different than the question here: <a href="http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences" rel="nofollow">http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences</a>, though quite similar. We are looking for things that don't collapse, but specifically for the purpose of explicit computation (none of the answers there admit explicit computation of differentials except in trivial cases, I think). </p> <p>For the moment I'm going to leave this not community wikified, since I think the request for an answer is specific and non-subjective enough that a person who gives a good answer deserves higher reputation for it. If anyone with the power to thinks otherwise, then feel free to hit it with the hammer.</p> http://mathoverflow.net/questions/102437/are-reflective-subcategories-of-complete-infinity-categories-complete/132656#132656 Answer by Dylan Wilson for Are reflective subcategories of complete infinity categories complete? Dylan Wilson 2013-06-03T16:39:49Z 2013-06-03T16:39:49Z <p>Here's a proof which is certainly overkill, but it has the merit of using references so you can read the proofs in detail.</p> <p>We have $i: \mathcal{C} \subset \mathcal{D}$ a fully faithful subcategory with $r$ a reflector.</p> <p>Step 1. <em>The inclusion $i$ is monadic</em>. Proof: It is clearly conservative, and it preserves and reflects $i$-split simplicial objects since $i$ is fully faithful so we can realize the splitting already in $\mathcal{C}$. By Barr-Beck (HA.6.2.2.5) the functor $i$ is monadic.</p> <p>Step 2. <em>Monadic functors 'create' limits.</em> Proof: This is the statement of HA.4.2.3.3. where the '$\mathcal{C}$' in that corollary corresponds to $\text{End}(\mathcal{C})$ here, the $\mathcal{M}$ corresponds to our $\mathcal{C}$, and the algebra $A$ corresponds to the monad $i \circ r$. </p> http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Answer by Dylan Wilson for how to make the category of chain complexes into an $(\infty,1)$-category Dylan Wilson 2013-06-02T00:00:44Z 2013-06-02T03:22:09Z <p>I don't know what 'explicit' means, but this is all covered in Higher Algebra very well.</p> <p>If you want a simplicial category at the end of the day, you can either...</p> <ol> <li>Take your favorite $\infty$-category (quasi-category) presentation, and straighten it to a simplicial category. (But no one wants to do this... it doesn't sound like fun.)</li> <li>Take the category of chain complexes and view at as a category enriched over itself via the internal hom. Now truncate the Hom-complexes and use Dold-Kan to get a (fibrant) simplicial set. This makes a fibrant simplicial category, which is equivalent to all the other things you might want. (For example, the homotopy category is correct, and the corresponding $\infty$-category is equivalent to one you might make from the dg-category of chain complexes, both of which agree with the $\infty$-category underlying the model category of chain complexes for a Grothendieck abelian category when this makes sense. All of your dreams come true!)</li> </ol> <p>See $\S$1.3 of Higher Algebra for all the details.</p> http://mathoverflow.net/questions/131386/equivariant-versus-retractive-spaces-a-reference-request/131397#131397 Answer by Dylan Wilson for Equivariant versus retractive spaces: a reference request Dylan Wilson 2013-05-21T23:13:38Z 2013-05-21T23:13:38Z <p>Appendix A in this paper seems to do this, unless I've misunderstood:</p> <p><a href="http://arxiv.org/abs/0810.4535" rel="nofollow">http://arxiv.org/abs/0810.4535</a></p> http://mathoverflow.net/questions/131033/directed-colimits-of-maps-in-a-combinatorial-model-category/131041#131041 Answer by Dylan Wilson for Directed colimits of maps in a combinatorial model category Dylan Wilson 2013-05-18T12:24:18Z 2013-05-18T12:24:18Z <p>The answer is 'no' (sorry I didn't realize this before!)</p> <p>I remembered someone showing me a weird counterexample in model categories a couple weeks ago (elaborated on from this paper by Rosicky: <a href="http://www.math.muni.cz/~rosicky/papers/comb2.pdf" rel="nofollow">http://www.math.muni.cz/~rosicky/papers/comb2.pdf</a>)</p> <p>and it seems to do the trick here:</p> <p>Consider the category ${\bf Pos}$ of posets. This is certainly presentable. Let $C$ denote the collection of split monomorphisms (which is weakly saturated and generated by split monomorphisms between finite posets). Then we can define a combinatorial model structure on ${\bf Pos}$ by taking <em>all</em> morphisms as weak equivalences (this satisfies the conditions of, say, Higher Topos Theory A.2.6.7. or one can see this directly). However, $C$ is not closed under sequential colimits because, for example, the non-split monomorphism arrow $\omega \rightarrow \omega \cup \infty$ is the sequential limit of the arrows $[n] \hookrightarrow [n] \cup \infty$, each of which is split.</p> <p>P.S. What is your distinction between directed colimits and filtered colimits (this terminology has always confused me...)? The way you're using the word makes it seem like "directed" is the same as "finitely filtered". </p> http://mathoverflow.net/questions/129074/cofibrant-replacements-of-a-given-object-in-a-combinatorial-model-category/129177#129177 Answer by Dylan Wilson for Cofibrant replacements of a given object in a combinatorial model category Dylan Wilson 2013-04-30T04:10:34Z 2013-04-30T16:47:41Z <p>I believe the answer is yes. Here is a reduction to a more basic question...</p> <p>First note it suffices to show that the subcategory of cofibrant objects is accessible. Indeed, if we have this then we can prove your result by forming the homotopy pullback of accessible categories and accessible functors:</p> <p><code>$$W \times_{\mathcal{C}} \mathcal{C}^{[1]}\times_{\mathcal{C}} \mathcal{C}_{X/}\times_{\mathcal{C}} \mathcal{C}^{cof}$$</code></p> <p>This is the category you're interested in, and replacing $W$ by $F \cap W$ we get the other possible category where objects are $(X, Y \rightarrow X$) with the morphism a trivial fibration.</p> <p>So we just need that the category of cofibrant objects is accessible... and I don't actually see how to prove this at the moment!</p> http://mathoverflow.net/questions/127507/question-about-topological-monoid-maps/127545#127545 Answer by Dylan Wilson for Question about topological monoid maps Dylan Wilson 2013-04-14T16:36:10Z 2013-04-15T01:53:56Z <p>EDIT 2: Sorry about the confusion, I will try to be careful now, and I'll put some comments at the end to clear up the business about adjoints etc. Here's what we're going to do.</p> <p>We'd like to show that <code>$$\text{Map}_{\text{Mon}}(X, Y) \rightarrow \text{Map}_{Spaces_*}(BX, BY)$$</code> is an equivalence when $X$ is cofibrant and $Y$ is grouplike. The strategy will be to consider the string <code>$$\text{Map}_{\text{Mon}}(X,Y) \rightarrow \text{Map}_{\text{Spaces}_*}(BX,BY) \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow$$</code></p> <p><code>$$\text{Map}_{\text{Mon}}(X, \Omega'BY) \cong \text{Map}_{\text{Mon}}(X,Y)$$</code> where $\Omega'$ denotes the Moore loop space (strictly associative multiplication), and show that the composite is homotopic to the identity. This clearly reduces down to showing that the third map is a weak equivalence, since I may as well have started the string with the inverse of the last equivalence ($\Omega' BY$ and $Y$ are interchangeable).</p> <p>So we want to show that <code>$$\text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY)$$</code> is a weak equivalence when $X$ is cofibrant ($Y$ can be arbitrary).</p> <p>We will prove this by showing that these two spaces represent the same functor on the homotopy category of spaces, the fundamental fact being that the natural map $\Omega'B Y \rightarrow \Omega'B\Omega'B Y$ is a weak equivalence.</p> <p>Given a map $K \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY)$ we get a map <code>$K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'B\Omega'BY)$</code>, and composing with the natural weak equivalence gives us a map <code>$K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)$</code>. One can check that composition gives back the original map, up to homotopy. This proves surjectivity of</p> <p><code>$$\text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)) \rightarrow \text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(X, \Omega'BY))$$</code> to get injectivity just note that we can recover the original value by applying $\Omega B'$ and using the weak equivalence again.</p> <p>This completes the proof.</p> <hr> <p>In my "second edition" I erroneously stated that there was a left adjoint to the inclusion of group-like topological monoids into monoids. Ricardo gracefully explains why this can't be true. However, it is a consequence of the above argument that there is a homotopical left adjoint, namely $\Omega'B$. (We've shown that the image of $\Omega'B$ is a homotopically reflective subcategory, and so all that's left is to note that the essential image is all group-like topological monoids.)</p> <p>I've also erased the original answer, since it seems silly now that we have this one. </p> <p>A reference for the argument I gave that basically showed that $\Omega'B$ was a localization functor can be found in <em>Higher Topos Theory</em> Proposition 5.2.7.4. Lots more love can be found throughout <em>Higher Algebra</em>; one can, for example, deduce a similar statement about $\mathbb{E}_1$-spaces in general, which actually implies this result since the inclusion of monoids into $\mathbb{E}_1$-spaces is an equivalence of $\infty$-categories.</p> <p>Hopefully this is all correct now! Let me know if there's still errors. </p> http://mathoverflow.net/questions/2917/where-does-a-math-person-go-to-learn-quantum-mechanics/125383#125383 Answer by Dylan Wilson for Where does a math person go to learn quantum mechanics? Dylan Wilson 2013-03-23T15:20:14Z 2013-03-23T15:20:14Z <p>Dan Dugger has wonderful mostly-finished notes on this here:</p> <p><a href="http://pages.uoregon.edu/ddugger/qftbook.pdf" rel="nofollow">http://pages.uoregon.edu/ddugger/qftbook.pdf</a></p> <p>It's written for topologists, very clearly, and he does a great job of giving both physical and "mathematical" explanations/intuitions.</p> <p>Really well done.</p> http://mathoverflow.net/questions/124855/can-one-compare-monads-arising-from-homotopy-equivalent-adjunctions/124861#124861 Answer by Dylan Wilson for Can one compare monads arising from homotopy equivalent adjunctions? Dylan Wilson 2013-03-18T14:26:40Z 2013-03-18T14:26:40Z <p>Unless I'm mistaken (very possible), the answer is "yes".</p> <p>Here's the general idea: Every algebra may be obtained as the geometric realization of its bar complex. So we may reduce to the case of a free algebra, i.e. we need only construct some sort of natural weak equivalence $TX \rightarrow T'X$ preserving the multiplication structure. I don't know how to make precise this choice in a model category setting without using an ugly zig-zag and convoluted argument (but that is more likely due to ignorance than anything else.)</p> <p>If we replace all model categories in sight with their underlying $\infty$-category (I'm ignoring some set-theoretic issues here, or assuming that the categories are simplicially enriched), then the above becomes an actual proof: The Quillen adjunction induces an adjunction of $\infty$-categories, and in this case a natural isomorphism of functors admits an $\infty$-categorical inverse. So we can define the map $$T \rightarrow T'$$ as the composition $$\alpha_R^{-1}L' \circ R\alpha_L: RL \rightarrow RL' \rightarrow R'L'$$</p> <p>(Here is where, in the model category case, I would have a zig-zag, and would have to do some sort of work... I don't actually know how the argument would go off the top of my head.)</p> <p>This map respects the multiplication (the only proof I can think of uses that adjoint pairs are unique up to unique isomorphism, where "unique" in this setting means "parameterized by a trivial Kan complex". But it's possible this is more obvious than I think.) </p> <p>It is a natural isomorphism, and so we're good.</p> <p>I don't think we can get away with any other type of proof: you need some sort of "multiplication preserving zig-zag of weak equivalences of functors" argument to prove this at all, because if you provide some natural zig-zag of weak equivalences between $T$-algebras and $T'$-algebras then, in particular, you need to do it for <em>free</em> $T$-algebras and $T'$-algebras, and multiplication preserving weak-equivalences like this can only come in the form of some zig-zag of natural transformations of $T$ with $T'$ since that's where the algebra structure on free algebras comes from. </p> http://mathoverflow.net/questions/124687/simplicial-space-whose-all-face-degeneracy-maps-are-homotopy-equivalences/124690#124690 Answer by Dylan Wilson for Simplicial space whose all face/degeneracy maps are homotopy equivalences Dylan Wilson 2013-03-16T13:37:53Z 2013-03-16T13:37:53Z <p>An easy spectral sequence argument tells us that the natural map induces an isomorphism for every generalized cohomology theory... so that deals with everything modulo fundamental groups. In general maybe you have to ask for the degeneracies to be cofibrations, then this might follow from a model structure argument (the geometric realization would be a hocolim.) (The spectral sequence mentioned above is in, for example, Segal's paper "Classifying Spaces and Spectral Sequences.")</p> http://mathoverflow.net/questions/123278/mayer-vietoris-sequence-for-arbitrary-bicartesian-square-of-spectra/123294#123294 Answer by Dylan Wilson for Mayer-Vietoris Sequence for Arbitrary Bicartesian Square of Spectra Dylan Wilson 2013-03-01T04:20:32Z 2013-03-01T04:20:32Z <p>Ok, I'll state this a bit more confidently... (but I'm still worried I'm missing something).</p> <p>Any time we're in, say, a stable model category where every object is cofibrant (or stable $\infty$-category), then given an object $X$ and a homotopy push-pull diagram (which I won't draw), involving $A \rightarrow B$, $A \rightarrow C$ and $C, B \rightarrow D$ we get a long exact sequence like $$\cdots \rightarrow [D, X] \rightarrow [B, X]\oplus [C, X] \rightarrow [A, X] \rightarrow [\Sigma^{-1} D, X] \rightarrow\cdots$$ Indeed, we have a homotopy push-pull as above precisely if we have a (co)fiber sequence $$A \rightarrow B \oplus C \rightarrow D$$ where the first map is the difference of the two obvious ones. There's probably a good way to do this without being fancy, but the easiest way I see to do this is in the setting of $\infty$-categories: the pushout and cofiber displayed above both have manifestly the same universal property. (I think I'm using cofibrancy here to say that the coproduct and homotopy coproduct should agree... maybe I don't need this- I'm bad with model categories, someone should correct me.)</p> <p>Anyway, this <em>must</em> be in one of the obvious references. Adams? Neeman's book on triangulated categories? Something like that.</p> <p>It's not immediately obvious to me that this specializes to Mayer-Vietoris, but that's me revealing too much of my ignorance. Inclusions of open subsets don't seem to be cofibrations, so why should the usual square we write down be a homotopy pullback/pushout (after taking suspension spectra)?</p> http://mathoverflow.net/questions/99614/contacting-an-eminent-mathematician/122370#122370 Answer by Dylan Wilson for Contacting an eminent mathematician Dylan Wilson 2013-02-20T02:42:07Z 2013-02-20T02:42:07Z <p>The answers to this question are probably useful:</p> <p><a href="http://mathoverflow.net/questions/34540/when-and-how-is-it-appropriate-for-an-undergraduate-to-email-a-professor-out-of-t" rel="nofollow">http://mathoverflow.net/questions/34540/when-and-how-is-it-appropriate-for-an-undergraduate-to-email-a-professor-out-of-t</a></p> http://mathoverflow.net/questions/121746/serre-spectral-sequence-of-representations/121756#121756 Answer by Dylan Wilson for Serre Spectral Sequence of Representations Dylan Wilson 2013-02-13T23:31:40Z 2013-02-16T03:10:36Z <p>: This is wrong!! See Peter's answer below.</p> <p>Unless I'm missing something, this follows from the naturality of the Serre spectral sequence. That is, each element of $g$ gives a map of fiber sequences, whence a map of Serre spectral sequences converging to the map on $H_*(E;\mathbb{Z})$.</p> http://mathoverflow.net/questions/121916/using-schemes-to-prove-things-about-rings Using schemes to prove things about rings Dylan Wilson 2013-02-15T16:20:56Z 2013-02-16T01:02:08Z <p>I apologize for asking a big list question, I've tried to avoid doing so for a while. I'll give my justification in a moment.</p> <p>The question is as follows:</p> <blockquote> <p>What are examples of strict applications of the language of schemes/stacks/algebraic geometry to commutative rings?</p> </blockquote> <p>Here a "strict" application means that the statement of the problem can be formulated without using any algebro-geometric language (stick to rings and modules and complexes, etc.) but a solution either requires or is very naturally obtained by using algebro-geometric language.</p> <p>I don't know examples of this phenomenon off the top of my head, but here are two examples from algebraic topology:</p> <ol> <li><p>Work on exotic spheres via homotopy theory (an example where this is the only known method to produce the results.)</p></li> <li><p>(one of) Quillen's proof(s) of the Atiyah-Swan conjecture. While there is a purely algebraic, group-cohomological proof, it turns out to be very natural to prove this theorem using spaces with an action, as opposed to specializing to when the space is a point.</p></li> </ol> <p><strong>Motivation</strong> Thanks to work of (insert all the usual suspects here), we now have a very strong theory of <em>spectral</em> algebraic geometry, i.e. algebraic geometry done with commutative ring spectra as opposed to commutative rings. While I don't know of any (hence this question), I am positive there exist strict applications of algebraic geometry to ring theory. It would be very neat if we could transplant these into strict applications of spectral algebraic geometry to the theory of ring spectra. Obviously I don't expect this to be straightforward, or literally possible, but I maintain that answers to this question would provide a useful insight in how to think about the relationship between non-affine and affine phenomena.</p> http://mathoverflow.net/questions/119869/model-category-structures-on-dgas-in-a-ringed-topos/121369#121369 Answer by Dylan Wilson for Model category structures on dga's in a ringed topos Dylan Wilson 2013-02-10T01:01:12Z 2013-02-10T01:01:12Z <p>What you suggest should work. We're going to transport the model structure on the category of chain complexes via the adjunction using the criteria given in this answer: <a href="http://mathoverflow.net/questions/92538/transporting-model-structures-via-adjunctions" rel="nofollow">http://mathoverflow.net/questions/92538/transporting-model-structures-via-adjunctions</a></p> <p>(A proof of this transport theorem for our case, where adjunction functor is monadic, can be found, for example, as Lemma 2.3 in this paper of Schwede and Shipley, <a href="http://homepages.math.uic.edu/~bshipley/monoidal.pdf" rel="nofollow">http://homepages.math.uic.edu/~bshipley/monoidal.pdf</a>)</p> <p>We have an adjunction between the functors </p> <p>$$\text{Free}: \mathbf{Ch}_{\mathcal{O}} \rightarrow \mathcal{O}\text{-}\mathbf{dga}$$ and</p> <p>$$\text{Forget}: \mathcal{O}\text{-}\mathbf{dga} \rightarrow \mathbf{Ch}_{\mathcal{O}}$$</p> <p>where the first is left adjoint to the second. Since $\mathcal{O}$-modules form a Grothendieck abelian category, there is a combinatorial model structure on $\mathbf{Ch}_{\mathcal{O}}$ with the fibrations and weak equivalences you described. Every object in both categories is small, since they are presentable.</p> <p>Now I need to show that everything that can be obtained by sequential limits from cobase changing the arrows $\text{Free}(g)$,where $g$ is a (generating) acyclic cofibration in $\mathbf{Ch}_{\mathcal{O}}$, is a quasi-isomorphism. I think this is true, but I'm not sure so I'll include my argument in case there's something wrong with it.</p> <p>First I claim that the relative tensor product $\text{Free}(D) \otimes_{\text{Free}(C)} (-)$ is exact on chain complexes. Indeed, given an exact sequence of chain complexes $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ I can form the diagram</p> <p><img src="http://i.imgur.com/jwY8UsU.png" alt="alt text"></p> <p>All three columns and the top two rows are exact, since free algebras are free as graded modules and the exactness diagrams of chain complexes is determined by exactness as graded objects. Therefore the bottom row is exact (diagram chase or spectral sequence argument.) (The unadorned tensors are over $\mathcal{O}$). </p> <p>Now I claim that cobase changing a morphism $\text{Free}(C) \rightarrow \text{Free}(D)$ where $C \rightarrow D$ is a monomorphism that's a quasi-isomorphism, gives a quasi-isomorphism. Indeed, we have a convergent spectral sequence $$\text{Tor}_{p,q}^{H^*F(C)} (H^*F(D), H^*A) \Rightarrow H(F(D) \otimes^{\mathbb{L}}A)$$ But since $F(D)$ is a flat $F(C)$ module this converges to the cohomology of $F(D) \otimes_{F(C)} A$. On the other hand, $$H^*F(C) \rightarrow H^*F(D)$$ is an isomorphism so the $E_2$-term collapses to an edge, and moreover the edge is just $H^*A$. The edge homomorphism is then an isomorphism, but the edge homomorphism is precisely the map induced by $A {\rightarrow} F(D) \otimes_{F(C)} A$, whence this map is a quasi-isomorphism, which was to be shown.</p> <p>Since sequential colimits in the category of algebras are the same as those in the category of chain complexes, we already know that sequential colimits of quasi-isomorphisms are quasi-isomorphisms. The result follows. </p> <p>(I didn't use that the arrow $F(C) \rightarrow F(D)$ was monic... so that worries me.)</p> http://mathoverflow.net/questions/121253/skeleton-category-of-the-category-of-skeleton-categories/121279#121279 Answer by Dylan Wilson for Skeleton category of the category of skeleton categories? Dylan Wilson 2013-02-09T07:41:53Z 2013-02-09T07:41:53Z <p>I'm not entirely sure what you're looking for in an answer, but maybe I'll flesh out my comment.</p> <p>It looks like what you're describing is equivalent to the homotopy category associated to the model structure on Cat where the weak equivalences are equivalences of categories. (I can say "the" because there is only one such, as pointed out in the comments. The cofibrations are functors injective on objects, and the fibrations are "isofibrations".)</p> <p>I would say that in this context your category has been much studied. In particular, it is interesting to ask questions about homotopy limits and colimits in this category because many useful constructions arise in this way. (Homotopy (co)limits with this model structure are the same as "2-(co)limits" which is the name appearing in most of the literature, especially older literature.)</p> <p>An example application of this language is the following theorem: The subcategory of presentable (resp. accessible) categories is closed under homotopy limits.</p> <p>Using this one can prove that most of your favorite things are presentable (resp. accessible). For example, the category of modules over a monad arises via a homotopy limit construction, and this takes care of most things of interest.</p> <p>Here's a neat application of this (which is the ordinary category version of a result that can be found, for example, in Lurie's HTT, 5.5.4.16.). </p> <p>Say you want to localize a category $\mathcal{C}$ with respect to some collection of morphisms, $S$. Usually $S$ will not be given as a set, but if $\mathcal{C}$ is presentable you're usually okay if $S$ is <em>generated</em> by a set. Well, it turns out that if $F: \mathcal{C} \rightarrow \mathcal{D}$ is a colimit preserving functor between presentable categories, and $S$ is a (strongly saturated) collection of morphisms in $\mathcal{D}$ that is generated by a set, then $f^{-1}S$ is a (strongly saturated) collection of morphisms generated by a set. The argument goes by way of showing that the subcategory of the category of morphisms generated by $f^{-1}S$ is presentable, using a homotopy pullback square.</p> <p>Adapting this to the model category or $\infty$-category setting, one sees immediately that localizing with respect to homology theories is totally okay and follows formally from this type of argument. (Basically, after fiddling around with cells to prove the category of spectra is presentable, you don't have to fiddle any more to get localizations. This is in contrast to the usual argument found in Bousfield's paper. You've moved the cardinality bookkeeping into a general argument about homotopy limits of presentable categories.)</p> <p>Anyway, apologies for the very idiosyncratic application of this language; these things have been on my mind recently. I'm sure there are much more elementary reasons why one would care about using the model category structure on Cat.</p> http://mathoverflow.net/questions/116504/integral-cohomology-operations-related-to-landweber-novikov Integral cohomology operations related to Landweber-Novikov Dylan Wilson 2012-12-16T02:19:32Z 2012-12-17T19:36:09Z <p>Let $U^* \rightarrow H^*$ be the homomorphism describing the complex orientation of $H^*$ from complex cobordism. Let $t_1, t_2, ...$ be indeterminates. </p> <p>My question is: Does there exist an integral cohomology operation $H^* (X) \rightarrow H^*(X)[\mathbf{t}]$ that makes the following diagram commute for any $X$? If so can we describe it in a way that doesn't involve complex cobordism?</p> <p><img src="http://i.stack.imgur.com/hTavM.png" alt="alt text"></p> <p>Here $s_{\mathbf{t}}$ is the total Landweber-Novikov operation.</p> <p>In particular, since the Landweber-Novikov operation satisfies the Riemann-Roch type formula for proper, complex-oriented maps $$s_{\mathbf{t}} f_{*}x = f_{*}(c_{\mathbf{t}}(\nu_{f}) \cdot s_{\mathbf{t}}x)$$ (where $\nu_{f}$ is the virtual class $1-\nu_i$ and $\nu_i$ is the stable normal bundle of the proper, complex-oriented map $f: Z \rightarrow X$) we would expect ths operations to satisfy something similar. </p> <p>EDIT: Whoops, original title didn't have to do with the question :) But it did have to do with the motivation behind asking it! Maybe another time...</p> http://mathoverflow.net/questions/114251/stable-infinity-categories-vs-dg-categories/114315#114315 Answer by Dylan Wilson for Stable infinity categories vs dg-categories Dylan Wilson 2012-11-24T06:19:26Z 2012-11-24T06:19:26Z <p>Here are a few observations...</p> <ol> <li>I think there exist stable infinity categories that are not the dg-nerve (resp. $A_\infty$-nerve) of a dg-category (resp. $A_\infty$ category). In particular, the category of spectra should not arise in this way. I think Keller has a paper on differential graded categories that answers this question; he notes at some point that the homotopy category of spectra is not "algebraic" but that homotopy categories of differential graded categories are (and in fact sort of encompass all such algebraic categories.) Basically it comes down to something like the existence of Hopf maps. Now- could one define somehow the "closest dg-category approximation" to a given stable infty category? Probably. I don't know how. Or maybe I could come up with how, but I'm not sure how useful this would be if the functor wasn't an equivalence? </li> <li>To answer Fernando's question, see DAG X.5 or DAG VII.6.2. That is, a stable $\infty$-category over a field $k$ is a presentable, stable $\infty$-category "equipped with an action of the monoidal $\infty$-category of $k$-module spectra". Unless I'm mistaken I think this basically implies that it is enriched and tensored over k-module spectra.</li> <li>Here would be a precise formulation of the statement about categories over a field of characteristic zero: The dg-nerve functor induces an equivalence of $\infty$-categories between the $\infty$-category underlying the model category of dg-categories over k and the $\infty$-category of stable, k-linear $\infty$-categories. (I don't mean to overwhelm with "infinities", I would state this in the perhaps friendlier world of model categories, but I'm not sure what precisely the model category is that corresponds to stable, k-linear $\infty$-categories.) I don't know of a reference for a proof, though Lurie alludes to this a lot. It would be great if someone wrote this down!</li> </ol> http://mathoverflow.net/questions/112451/spectral-sequence-for-cobordism-without-leaving-smooth-category spectral sequence for cobordism without leaving smooth category Dylan Wilson 2012-11-15T05:03:11Z 2012-11-15T06:05:57Z <p>In Bott &amp; Tu's marvelous book there is a derivation of the spectral sequence for a (smooth) fiber bundle for deRham cohomology done entirely in the realm of the smooth category. Unfortunately, as it's written there it is not clear how to generalize to other cohomology theories (they write down a double complex, and they use very much the fact that cohomology is given as the cohomology of this chain complex.)</p> <p>Now, in Quillen's paper computing the complex cobordism ring, he introduces a geometric description of complex cobordism as a cohomology theory on smooth manifolds. I've been attempting, to no avail, to come up with a derivation of the Atiyah-Hirzebruch spectral sequence for a smooth fibration without using CW complex-esque techniques. My question is:</p> <blockquote> <p>Does anyone know of a derivation of the Atiyah-Hirzebruch spectral sequence of a smooth fiber bundle for a generalized cohomology theory that does not leave the realm of manifolds?</p> </blockquote> <p>So far, the most promising bet I have found is Segal's approach to this spectral sequence in the paper "Classifying spaces and spectral sequences." The trouble appears, however, in his use of a complex $BX_U$ for a covering $U$ of $X$. This is most definitely not a manifold, though it is homotopy equivalent to one for numerable covers. The question in this context, however, reduces to:</p> <blockquote> <p>Does the natural filtration on $BX_U$ induce a filtration on $X$ that gives rise to the spectral sequence for a covering? If so, is there a nice description of this filtration using, say, just the data of $X$ and the numerable cover?</p> </blockquote> <p>For this question I should be more specific about $BX_U$. This is defined as the geometric realization of the nerve of the topological category $X_U$ whose points are pairs $(x, U_{\sigma})$ where $U_\sigma$ is a finite intersection of elements of $U$ and $x \in U_\sigma$. Morphisms are inclusions $U_\sigma \subset U_\tau$. The filtration on $BX_U$ is given by looking at the images of $(NX_U)_n \times \Delta^n$ in $BX_U$, where $NX_U$ is the nerve. </p> http://mathoverflow.net/questions/111582/proper-maps-and-transversality Proper maps and transversality Dylan Wilson 2012-11-05T21:14:24Z 2012-11-06T09:13:14Z <p>I'll begin with the question, which is intrinsically interesting:</p> <blockquote> <p>Let <em>M</em> be a manifold with some submanifold <em>Y</em>. Suppose that $W \rightarrow M$ is a smooth, proper map. Does there exist another map $W \rightarrow M$ homotopic to the original that is ALSO proper and transverse to the submanifold <em>Y</em>?</p> </blockquote> <p>Let me note that I am definitely not assuming the manifolds are compact. </p> <p>Why I care: This question came up while thinking about the geometric description of complex cobordism given by Quillen in "Elementary proofs of some results of cobordism theory using Steenrod operations." I have a geometric description of the coboundary map in the Mayer-Vietoris sequence but as of now it relies on the answer to the above question being "yes."</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/110458#110458 Answer by Dylan Wilson for Slick ways to make annoying verifications Dylan Wilson 2012-10-23T18:52:51Z 2012-10-23T18:52:51Z <p><strong>To show that an object S has property P, first show that the collection of all objects satisfying P is closed under a bunch of operations, prove that certain very simple objects have property P, and show that S can be "decomposed" or "filtered" or somehow unscrewed into these simple objects using the operations mentioned above.</strong></p> <p>This is a sort of induction, and it is used all the time to turn annoying verifications into verifying that something is true for like... a point. Maybe a shorthand for this "slick method" would be "think like Grothendieck."</p> <p>For lots of examples of this see any proof in Higher Topos Theory or Higher Algebra by Lurie.</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/33054#33054 Answer by Dylan Wilson for Slick ways to make annoying verifications Dylan Wilson 2010-07-23T06:23:31Z 2012-10-23T18:47:42Z <p>Riffing off of Nate's answer, another result in functional analysis (often proven in the same breath as the others) is the Uniform Boundedness Principle. This is exceedingly useful in getting uniform estimates from pointwise ones... and it's pretty magical. It says that if we have a collection $\mathcal{F}$ of continuous linear operators from a Banach space to a normed vector space, then these are uniformly bounded (i.e. $\sup_{T \in \mathcal{F}} \Vert T \Vert &lt; \infty$) if they are pointwise bounded (i.e., for every $x$ we have $\sup_{T \in\mathcal{F}} \Vert T(x)\Vert &lt; \infty$). </p> http://mathoverflow.net/questions/104750/about-a-letter-by-richard-palais-of-1965/104782#104782 Answer by Dylan Wilson for About a letter by Richard Palais of 1965. Dylan Wilson 2012-08-15T18:48:01Z 2012-08-16T10:14:23Z <blockquote> <p>I would like to know if, in the meanwhile, this letter was made available</p> </blockquote> <p>Yes! (See <a href="http://mathoverflow.net/questions/104750/about-a-letter-by-richard-palais-of-1965/104781#104781" rel="nofollow">here</a>).</p> http://mathoverflow.net/questions/99546/proof-for-which-primes-hg-has-torsion Proof for which primes H*G has torsion Dylan Wilson 2012-06-14T02:16:35Z 2012-06-15T17:14:35Z <p>In 1960 Borel proved a beautiful result:</p> <p><Blockquote> <strong>Theorem</strong>. Let G be a simple, simply connected Lie group. Suppose that <em>p</em> is a prime that does not divide any of the coefficients of the highest root (expressed as a linear combination of the simple roots). Then $H^*G$ has no <em>p</em>-torsion. </Blockquote></p> <p>Here $H^*G$ refers to the integral cohomology of $G$ as a space.</p> <p>(Interestingly, the converse of this statement does not hold. For example $Sp(n)$ does not have $2$-torsion, though the coefficients of the highest weight are (1,2). This counterexample is given by Borel).</p> <p>You can check out his proof <a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.tmj/1178244298" rel="nofollow">here</a> if you're interested, but basically it seems like what he does is this: he computes the cohomology of each of the simple, simply connected Lie groups with coefficients that have the relevant primes inverted and proves the result by observation.</p> <p>Since the statement of the theorem itself has little to do with the actual classification of simple Lie groups, and morally should only rely on the fact that we can recover a simple, simply-connected Lie algebra from its root system, it seems natural to ask:</p> <p><Blockquote> <strong>Question:</strong> In the intervening 50+ years since Borel proved this theorem, do we know of a direct proof of the theorem that does not use the classification of simple Lie groups? </Blockquote></p> <p>I really have no idea how to go about doing this... could one use Lie algebra cohomology? I can only seem to find a link between Lie algebra cohomology and deRham cohomology of Lie groups, which is of no help when asking questions of torsion. Maybe there's an integral version? </p> http://mathoverflow.net/questions/98982/non-noetherian-stable-homotopy/99017#99017 Answer by Dylan Wilson for Non-Noetherian Stable Homotopy Dylan Wilson 2012-06-07T08:27:15Z 2012-06-07T08:27:15Z <p>Alright, various things have been said in the comments, and I'd like to say something both coherent and correct since many of my comments above have been neither.</p> <p>Here's what I understand about all of this.</p> <p>First and foremost, unless I'm missing something big, it is absolutely the case that $D(R)$ is a stable homotopy category for <em>any</em> ring $R$. Here is a silly way-too-much-machinery reason why, with the benefit that it has references that I know of off the top of my head: The category of dg-modules over R (equiv. chain complexes) is Quillen equivalent to the category of HR-module spectra (http://homepages.math.uic.edu/~bshipley/zdga17.pdf). Thanks to, for example, May (pick a paper of his at random and it will probably contain this result), this has the structure of a symmetric monoidal stable model category, which turns its homotopy category into a stable homotopy category in the sense of HPS.</p> <p>Second, if you're asking if people have a "general theory" for stable homotopy categories with a non-Noetherian endomorphism ring, the answer is absolutely yes, in many different guises. One can study symmetric monoidal stable model categories (of which there are many, most of which do not have a Noetherian endomorphism ring for generators). One can study symmetric monoidal stable $\infty$-categories which are basically the same. One can try to prove results similar to the nilpotence and classification theorem in these settings, this has been done for: (i) D(R) where R is <em>any</em> commutative ring [Thomason], (ii) D(R) where R is <em>any</em> epsilon-commutative, G-graded ring [Dell'Ambroglio, Stevenson], (iii) stmod(kG) where G is a finite group scheme [Friedlander-Pevtsova, Benson-Carlson-Rickard], (iv) D(X) where X is a quasi-compact, quasi-separated scheme [Thomason], and (v) $\mathcal{S}$ the category of finite spectra [Devinatz-Hopkins-Smith]. There are some others but I'm less familiar with them... </p> <p>The thing that absolutely does <em>not</em> work, fails miserably actually, for non-Noetherian situations is an attempt to classify the localizing subcategories. Luke Wolcott knows a lot about how bad this can get (his very recent thesis was about it). I'm pretty sure Balmer has written some things about what one can say generally if you <em>are</em> in the Noetherian case. The point is that it's not even clear what a "general theory" would look like in the non-Noetherian case for stuff like localizing subcategories... again, Luke knows much more about this than me, so you should ask him. Fernando sums it up nicely in his original comment.</p> <p>Finally, let me try to clear up two things I said in the comments (someone should correct me if I'm wrong):</p> <ol> <li><p>The category of chain complexes on a Grothendieck abelian category can be given the structure of a stable model category in which the weak equivalences are the quasi-isos.</p></li> <li><p>The category of chain complexes of $\mathcal{O}$-modules on any ringed space admits a symmetric monoidal model structure, which means that the unbounded derived category is at the very least a tensor-triangulated category (it's not immediately obvious that the tensor structure plays nice with the triangular structure, but it would be very strange to me if this wasn't true or obvious to someone else?) </p></li> </ol> http://mathoverflow.net/questions/93572/applications-of-descent/93579#93579 Answer by Dylan Wilson for Applications of Descent? Dylan Wilson 2012-04-09T16:00:02Z 2012-04-09T16:00:02Z <p>Probably not quite what you're looking for, but it certainly involves modular representation theory, and it's really neat! A baby version of descent is used in a proof of Quillen's stratification theorem. </p> <p>Suppose we have a compact Lie group, $G$, and a smooth $G$-manifold $X$ (if you'd like to feel modular, take $G$ to be finite and $X$ to be a point). We'd like to show that the map $H_G^*X \rightarrow \lim H_G^*A$ is an F-isomorphism, where cohomology has coefficients in some fixed prime, the limit is taken over a certain category involving components of fixed point subspaces and the elementary abelian subgroups of $G$, and an "F-isomorphism" is a map of rings such that all elements of the kernel are nilpotent and every element, $s$, in the codomain satisfies $s^{p^n} \in \text{image}$ for some $n$. (This implies, in particular, that the Krull dimensions of each ring are the same.) In the case of the point, the right hand side is the limit over elementary abelian subgroups of $G$.</p> <p>Here's one of the ways that Quillen does this. First he constructs a space $\mathcal{F}$ as follows: Choose a faithful unitary representation of $G$, say on $U(n)$, and let $\mathcal{F} = U(n)/S$ where $S$ is the subgroup of the maximal torus consisting of elements of order $p$ (i.e. matrices with $p$-roots of unity on the diagonal and zeros elsewhere). This space has all sorts of nice properties as a $G$-manifold (in particular, all of the isotropy groups are elementary abelian subgroups.)</p> <p>Now, Quillen proves the theorem for the $G$-manifold $X \times \mathcal{F}$ (notice that even if all you cared about was when $X$ is a point, you'd still need the general statement of the theorem) and then uses a fun argument using a baby version of faithfully flat descent to deduce the theorem for $X$! </p> <p>I won't give the argument here, but basically he uses the sequence $X \times \mathcal{F} \times \mathcal{F} \Rightarrow X \times \mathcal{F} \rightarrow X$ (I don't know how to make a double arrow here...) and shows that applying H^* gives an equalizer sequence (this is the "descent" part), and applying the functor appearing on the right hand side of the theorem also gives an exact sequence. Then one does a 3-lemma esque argument to conclude the result. It's really neat! For a much more detailed and entertaining account, see</p> <p><a href="http://www.math.washington.edu/~mitchell/Quillen/qnew.pdf" rel="nofollow">http://www.math.washington.edu/~mitchell/Quillen/qnew.pdf</a></p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89370#89370 Answer by Dylan Wilson for Example of a manifold which is not a homogeneous space of any Lie group Dylan Wilson 2012-02-24T06:54:00Z 2012-02-24T06:54:00Z <p>It is a result of Mostov that any compact homogeneous manifold must have nonnegative Euler characteristic:</p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=2174096" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2174096</a></p> <p>That should provide plenty of counterexamples. :)</p> http://mathoverflow.net/questions/87382/detecting-equivalences-of-infinity-categories-by-nerves/87400#87400 Answer by Dylan Wilson for Detecting equivalences of (infinity) categories by nerves Dylan Wilson 2012-02-03T05:24:48Z 2012-02-03T05:24:48Z <p>Just a few ideas/observations...</p> <p>For 1:</p> <p>a. A functor is an <em>isomorphism</em> if and only if the induced map is an isomorphism of simplicial sets.</p> <p>b. So a functor is an equivalence if and only if the induced map on the nerves of skeleta is an isomorphism. (This isn't very helpful though...)</p> <p>c. Any map of simplicial sets $NC \rightarrow ND$ gives us a functor $C \rightarrow D$ (just look at the map on 0-simplices, 1-simplices, and 2-simplices to see what to do). So a map $NC \times \Delta^1 \rightarrow ND$ corresponds to functor $F: C \times [1] \rightarrow D$ (since $N[1] = \Delta^1$), but this is precisely a natural transformation of functors $C \rightarrow D$. Thus, if we have $NC \rightarrow ND$ and $ND \rightarrow NC$ such that the composites are "homotopic" to the identity (where I mean, use $\Delta^1$), then the original functor is an equivalence.</p> <p>d. None of what we have said so far is any easier than just proving your original map is an equivalence. One thing you could do, that allows for homotopy theory, is associate to $C$ a stronger invariant. For example, take the <em>bisimplicial set</em> call it $\mathfrak{N}C$ given by: $\mathfrak{N}C_k = N(\text{iso }C^{[k]})$ (where "iso C" means "maximal groupoid" or "just take isomorphisms as your morphisms." This is a special case of a construction of Rezk's). It turns out that a functor is an equivalence if and only if the induced map of these special nerves is a weak equivalence. </p> <p>e. I don't think that checking it's a weak equivalence works for the usual nerve, since (for example) any category with a final object has a nerve that's weakly equivalent to a point. <em>However</em> it is probably the case that if the induced map on nerves is a categorical equivalence then the original categories are equivalent. This brings us to point 2...</p> <p>For 2</p> <p>a. I'm not sure if there is any way of showing that a functor between $\infty$-categories is an equivalence (in general) that does not amount to showing the induced map on mapping spaces is an equivalence. However, you can do all of this without mentioning simplicial categories. Given an $\infty$-category, $\mathcal{C}$, we can define something (or several things) equivalent to the mapping space between two objects in $\mathfrak{C}[\mathcal{C}]$; let Hom_C^R(x,y) be the simplicial set defined by requiring that a map $\Delta^n \rightarrow \text{Hom}_{\mathcal{C}}^R(x,y)$ be a map $z: \Delta^{n+1} \rightarrow \mathcal{C}$ such that $d_0z$ is the constant diagram on $y$ and $z$ evaluated on the initial vertex is $x$. (Actually this might be $\text{Hom}^L$, I can't remember off the top of my head). Anyway, this turns out to be a Kan complex (i.e. a space), and Lurie/Joyal show that it's the same as what you get after doing $\mathfrak{C}$. See Lurie for more.</p> <p>b. Given the description above, we can say a functor is an equivalence if it induces a weak equivalence of Kan complexes for all mapping spaces as we defined above (i.e. fully faithful), and essentially surjective (i.e. every object is equivalent to something in the image). This isn't easy to do but...</p> <p>c. In general, an equivalence of $\infty$-categories is kind of a big deal. In particular, it should be at least as hard as writing down various Quillen equivalences of model categories, right? Maybe slightly easier, but still quite hard. </p> <p>Anyway, I realize this probably isn't what you want, but it's the best I can do at the moment! I'll let you know if I think of anything more helpful (but probably someone brighter will end up posting something more helpful soon...)</p> http://mathoverflow.net/questions/84161/hopf-algebras-examples/84190#84190 Answer by Dylan Wilson for Hopf algebras examples Dylan Wilson 2011-12-23T20:53:26Z 2011-12-23T20:53:26Z <p>A couple people mentioned the Steenrod algebra briefly, but you can do a few more topologically-related things:</p> <ul> <li><p>The subalgebras $\mathcal{A}(n)$ of the Steenrod algebra generated by $Sq^1, ..., Sq^n$ are neat. In particular, it is a good exercise in cohomology to compute $Ext_{\mathcal{A}(n)}(k,k)$. (One can do a minimal resolution and try to look for a pattern, and then prove that it works using a spectral sequence.)</p></li> <li><p>You can show that the Hopf algebras given by $\mathbb{Z}[c_1, c_2, ...]$ ($c_i$ living in degree 2i) and $\mathbb{Z}/2 [w_1, w_2, ...]$ ($w_i$ living in degree i) and comultiplications given by $y_n \mapsto \sum y_i \otimes y_j$ on the generators, are self-dual Hopf algebras and explicitly describe the relationship between itself and the dual. This is neat in and of itself, but then you can mention that these results lead to quick calculations of $H_*MU$ and $H_*MO$ as comodules over the dual of the Steenrod algebra, and thus allow for computations of cobordism groups via the Adams spectral sequence. This self-duality can also be used for a quick proof of the Bott periodicity theorem, though the only reference I know of for this is not yet published (by May), though it will be soon. </p></li> <li><p>If you're doing cohomology, it's always nice to do the cohomology of an exterior algebra; i.e. a Hopf algebra that's an exterior algebra on primitive generators. It's a very easy result, but you can use it to compute other things via spectral sequences.</p></li> </ul> <p>I know I've forgotten several things I wanted to mention... if I remember them, I'll edit them in.</p> http://mathoverflow.net/questions/134217/strict-applications-of-deformation-theory-in-which-to-dip-ones-toe Comment by Dylan Wilson Dylan Wilson 2013-06-20T06:34:43Z 2013-06-20T06:34:43Z Silly of me to forget to mention Goerss-Hopkins obstruction theory!! And of course I'm omitting the whole derived story- that's what I really want to get at, but I was hoping to get some of the more classical or algebraic story for the purposes of this question. In any event some of the other references I hadn't heard of- looks like lots of fun! Thanks Sean :) http://mathoverflow.net/questions/133614/krull-dimension-in-equivariant-cohomology Comment by Dylan Wilson Dylan Wilson 2013-06-13T14:48:43Z 2013-06-13T14:48:43Z (Sorry, the above comment is for the case $X = pt$!) http://mathoverflow.net/questions/133614/krull-dimension-in-equivariant-cohomology Comment by Dylan Wilson Dylan Wilson 2013-06-13T13:51:37Z 2013-06-13T13:51:37Z With rational coefficients the cohomology should be invariants of the cohomology of the torus by the Weyl group, which shouldn't change the Krull dimension. http://mathoverflow.net/questions/133513/equivariant-euler-class Comment by Dylan Wilson Dylan Wilson 2013-06-12T22:01:51Z 2013-06-12T22:01:51Z Transversality is a dangerous and mostly nonexistent game, equivariantly. http://mathoverflow.net/questions/133051/a-category-with-weak-equivalences-which-is-not-a-model-category/133062#133062 Comment by Dylan Wilson Dylan Wilson 2013-06-07T13:19:30Z 2013-06-07T13:19:30Z Can you give a little indication of the argument for why this does not have a model structure? (This is a neat example!) http://mathoverflow.net/questions/102437/are-reflective-subcategories-of-complete-infinity-categories-complete Comment by Dylan Wilson Dylan Wilson 2013-06-03T17:14:20Z 2013-06-03T17:14:20Z @David: Right, I was being silly :) Proof below... http://mathoverflow.net/questions/102437/are-reflective-subcategories-of-complete-infinity-categories-complete/132656#132656 Comment by Dylan Wilson Dylan Wilson 2013-06-03T17:13:44Z 2013-06-03T17:13:44Z (For a more direct proof one could use the characterization of a limit diagram in HTT.4.2.4.3 to explicitly construct an inverse to the counit applied to the limit of the diagram in the big category, but then you have to be slightly fussy about checking that actual maps are equivalences, but it's not difficult.) http://mathoverflow.net/questions/102437/are-reflective-subcategories-of-complete-infinity-categories-complete Comment by Dylan Wilson Dylan Wilson 2013-06-03T07:54:39Z 2013-06-03T07:54:39Z Why doesn't this follow from HTT 5.2.3.5? http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Comment by Dylan Wilson Dylan Wilson 2013-06-02T18:19:06Z 2013-06-02T18:19:06Z @Hiro: &quot;...it may happen that nonequivalent objects become equivalent...&quot; I don't see how this can happen: the homotopy category is the same as the one underlying the relative category $(Ch, Quasi-isos)$. (I'm assuming the complexes are bounded from one side, but this should also be true without that assumption if we have a Grothendieck abelian category, yeah?) http://mathoverflow.net/questions/132561/construction-of-the-spectral-sequence-of-katz-oda Comment by Dylan Wilson Dylan Wilson 2013-06-02T09:18:08Z 2013-06-02T09:18:08Z How is this different than the Leray spectral sequence? http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Comment by Dylan Wilson Dylan Wilson 2013-06-02T09:14:02Z 2013-06-02T09:14:02Z @Fernando: What do you mean? http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Comment by Dylan Wilson Dylan Wilson 2013-06-02T03:21:43Z 2013-06-02T03:21:43Z @Nevermind: Lurie constructs it in several ways (as I mentioned), and one way he does it is via the Dold-Kan correspondence to get a simplicial category and then takes the coherent nerve. http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Comment by Dylan Wilson Dylan Wilson 2013-06-02T00:13:26Z 2013-06-02T00:13:26Z (If it's a simplicial model category then you can just look at the category of cofibrant-fibrant objects, but some part of me feels like chain complexes don't always form a simplicial model category...) http://mathoverflow.net/questions/132547/how-to-make-the-category-of-chain-complexes-into-an-infty-1-category/132549#132549 Comment by Dylan Wilson Dylan Wilson 2013-06-02T00:12:44Z 2013-06-02T00:12:44Z Also, more generally, if you want to get a simplicial category from a model category, you can always use one of the many variations of Dwyer-Kan localizations. http://mathoverflow.net/questions/132347/homotopy-pullback-pushout Comment by Dylan Wilson Dylan Wilson 2013-05-30T16:19:02Z 2013-05-30T16:19:02Z Yes. This is a consequence of the Blakers-Massey theorem.