User justin noel - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:13:21Z http://mathoverflow.net/feeds/user/8818 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89753/geometric-realization-of-hochschild-complex/89853#89853 Answer by Justin Noel for Geometric realization of Hochschild complex Justin Noel 2012-02-29T10:31:39Z 2012-02-29T10:31:39Z <p>Sam Gunningham had the right idea, but I would like to fix it up for the construction of the homotopy pullback that I am aware of.</p> <p>Alternatively one can compute $HH(A;A)$ as $\mathrm{Tor}^{A^e}(A,A)$ where $A^e$ is the enveloping algebra of $A$, $A\otimes A^{op}$, since your $A$ is commutative $A^{e}\cong A\otimes A$. These Tor groups can be calculated via the simplicial complex which in degree $n$ is $A\otimes (A^{e})^{\otimes n}\otimes A\cong A^{\otimes {2n+2}}$. As a side note the geometric realization of this complex gives a topological commutative ring (Because $A$ is commutative) which models $THH^{HZ}(HA;HA)$ via the Eilenberg-Maclane functor. Applying your contravariant functor to spaces to the simplicial algebra we obtain a cosimplicial space modeling the homotopy pullback $Spec(A)^{top}\times_{Spec(A)^{top}\times Spec(A)^{top}}\times Spec(A)^{top}$ (see 3.3 of <a href="http://www.math.uiuc.edu/~reldred2/tot-primer.pdf" rel="nofollow">http://www.math.uiuc.edu/~reldred2/tot-primer.pdf</a> ) which as Sam pointed out, models the free loop space.</p> http://mathoverflow.net/questions/89687/dualizable-classifying-spaces/89760#89760 Answer by Justin Noel for Dualizable classifying spaces Justin Noel 2012-02-28T13:17:30Z 2012-02-29T09:39:22Z <p>I'll answer a related question: in the $K(n)$-local stable category, $BG$ is dualizable for all finite groups $G$, moreover, each is self-dual. You can find this in Hovey and Strickland's 'Morava $K$-theories and localisation' Corollary 8.7.</p> <p>More precisely, the result states that if $G$ is finite, then the $K(n)$ localized norm map: $$L_{K(n)}\Sigma^\infty_+ BG\rightarrow F(L_{K(n)}\Sigma^\infty_+ BG, L_{K(n)} S)$$ is a weak equivalence. This is because Tate cohomology lowers chromatic complexity (Hovey-Strickland 96, Kuhn 2004), so the cofiber of the localized norm map $L_{K(n)}t_G(L_{K(n)} S)^{G}$ is weakly contractible.</p> http://mathoverflow.net/questions/86985/if-a-in-mboxrings-subset-e-infty-mbox-rings-what-is-the-equivalence-betw/87058#87058 Answer by Justin Noel for If $A\in \mbox{Rings}\subset E_\infty\mbox{-rings}$, what is the equivalence between objects of $\mathcal{D}(\mbox{Mod}_A)$ and $A$-module spectra? Justin Noel 2012-01-30T19:48:56Z 2012-01-30T20:33:36Z <p>This question already has been answered in the comments. </p> <p>(Tilson) We regard a commutative ring as an $E_\infty$ spectrum via the EM functor $H$. This is definitely what Jacob is doing. One could also use associative rings and $A_\infty$ spectra for what follows.</p> <p>(Wilson) Many of the correspondences between algebra and stable homotopy theory are described in Chapter 7 in Lurie's higher algebra book. </p> <p>(Muro) The correspondence between algebras/modules and the associated EM-spectra is laid out in math.uic.edu/~bshipley/zdga17.pdf (Cor 2.15) which depends on her paper with Stefan Schwede "Equivalences of monoidal model categories."</p> <p>It is a bit technical, but is easier to work out the correspondence if you restrict to non-negatively graded $\mathbb{Z}$-chain complexes and connective $H\mathbb{Z}$-modules. The correspondence can be spread into two stages: 1) Use the Dold-Kan correspondence to move between chain complexes and simplicial abelian groups. 2) Take the geometric realization of your simplicial abelian group which is a topological abelian group and hence an infinite loop space, so we can take its associated connective spectrum (by repeatedly applying the bar construction). The fact that geometric realization preserves products can then be used to see that this spectrum is an $H\mathbb{Z}$-module.</p> <p>Now given an $H\mathbb{Z}$-module $M$ we can form the associated simplicial abelian group $H\mathbb{Z}-mod(H\mathbb{Z}\wedge\Sigma^\infty_+ \Delta^i, M)$ to go back. </p> <p>This equivalence induces an equivalence their associated stable infinity categories.</p> http://mathoverflow.net/questions/84944/question-about-formal-group-schemes/85050#85050 Answer by Justin Noel for Question about formal group schemes Justin Noel 2012-01-06T12:27:15Z 2012-01-06T12:27:15Z <p>Show your product on $G$ restricts to a product on formal neighborhoods of the identity (via the Hopf algebra correspondence you mentioned these are coalgebra structures on the quotients of the powers of the maximal ideal of $\mathcal O_{G,e}$), these small group schemes form a directed system, then take the associated Ind-scheme. This is $\widehat G$ and the directed system corresponds to the inverse system defining $\widehat{\mathcal{O}}_{G,e}$. Since products commute with Ind systems, the multiplications on the formal neighborhoods define a multiplication on $\widehat{G}$. </p> <p>Dually we obtain comultiplications on each of our quotient rings, which defines a comultiplication on the pro/topological-ring $\widehat{\mathcal{O}}_{G,e}$. I don't know a particular reference for this. I learned about formal groups through Hazewinkel's book. Formal schemes you can learn about in EGA.I.10.</p> http://mathoverflow.net/questions/84500/the-norm-map-in-group-cohomology-via-classifying-spaces/84506#84506 Answer by Justin Noel for The Norm Map in (group) cohomology via classifying spaces Justin Noel 2011-12-29T11:55:20Z 2012-01-03T08:27:13Z <p>First let me describe the norm in degree 0. Let $k$ be a commutative ring, given $\alpha\in H^0(BH,k)$, which corresponds to a an $H$ equivariant homotopy class of a map $EH_+\rightarrow k$ where $k$ has the trivial $H$ action. W can take the $n$th smash power of this map to obtain \begin{equation*} (EH_+)^{\wedge n}\cong E(H^n)_+\rightarrow k^{\wedge n}. \end{equation*} This map is naturally $\Sigma_n \wr H$ equivariant. Using the (equivariant) multiplication in $k$ we can make this map land in $k$ again. Now use the monomial representation above $G\rightarrow S_{G/H}\wr H\cong \Sigma_n \wr H$, we can regard this $\Sigma_n \wr H$ equivariant map as a $G$ map and $G$ acts freely on the source. Taking orbits we obtain an ordinary map $[BG_+, k]$ as desired. </p> <p>To move up to higher degrees, I find it helpful to use equivariant spectra. So now our map $\alpha$ in degree $i$ will corresponde to a map $\Sigma^\infty_+ EH \rightarrow S^i \wedge Hk$. We proceed exactly as above but with one little snag. After taking smash powers the $S^i$ becomes $S^{i\rho}$ where $\rho$ is the standard $n$ dimensional representation of $\Sigma_n$ ($H$ acted trivially on $S^i$). When $i$ is even and $k$ is a field of characteristic $p$ (all $i$ in characteristic $2$) we use that $S^{i\rho}$ is $k$-orientable to land us in the correct degree.</p> <p>For more details on the norm in spectra consult the article by May and Greenlees or Schwede's <a href="http://www.math.uni-bonn.de/~schwede/equivariant.pdf" rel="nofollow">course notes</a> especially chapter 7.</p> http://mathoverflow.net/questions/84381/computations-in-infty-categories/84389#84389 Answer by Justin Noel for Computations in $\infty$-categories Justin Noel 2011-12-27T15:45:17Z 2011-12-27T15:45:17Z <p>I believe that many computations could be done in the context of $\infty$-categories given enough outside input. For example, I would expect that most of the spectral sequences arising in topology could be constructed in the context of $\infty$-categories (indeed there are some basic spectral sequences constructed in Lurie's Higher Algebra). However, knowing that they converge to the right thing, might require you to know very specific details about limits and colimits in your $\infty$-category. Since the existence of these limits and colimits is typically deduced from the existence of homotopy colimits and limits in some model category we are often forced to consider the constructions in some model. For the issues of convergence, I am thinking of the Eilenberg-Moore and unstable Adams spectral sequences.</p> <p>Even if we can overcome these problems, we usually need a small stable of existing computations to get us started (e.g. $\pi_1 S^1\cong \mathbb{Z}$, the ring structure on $K(S^2)$, the group cohomology of cyclic groups,...). The $\infty$-category context is an unnatural place to do such computations. It is more natural to work in an actual category with actual point-set models for our objects that let us compute something from first principles. </p> <p>When using $\infty$-categories that arise from taking the nerves of the subcategories of bifibrant objects in some model categories, we find, after unwinding the definitions, that many of our $\infty$-category constructions involve taking lots of cofibrant/fibrant replacements. Sometimes we don't need to take a bifibrant model for a computation and we could get by with a cofibrant model. Sometimes we have no concrete idea what the fibrant replacement does to our objects. This is particularly true when we have taken a left Bousfield localization of a model category (in the combinatorial case, this corresponds to presentable $\infty$-categories). In such a case, the $\infty$-category context (unnecessarily) forces us to lose control of our point-set models.</p> http://mathoverflow.net/questions/84309/why-are-monads-useful/84314#84314 Answer by Justin Noel for Why are monads useful? Justin Noel 2011-12-26T13:32:36Z 2011-12-26T13:32:36Z <p>I would say first that monads are useful in that they offer a level of abstraction that can be used to describe lots of different algebraic phenomena. That is, categories of groups, abelian groups, rings, commutative rings,... can each be described equivalently as categories of algebras over some monad acting on sets (see <a href="http://ncatlab.org/nlab/show/Lawvere+theory%20" rel="nofollow">Lawvere Theory</a> for more examples). There are monads that don't arise in this way, but typically most examples are pretty close to this. So in one sense the theorems that are true in all of these situations are most naturally proven in the language of monads and their categories of algebras.</p> <p>One of the other ways they show up in 'real math' is via the following: there is always a forgetful functor from the category of $T$-algebras in $\scr{C}$ to $\scr{C}$ with a left adjoint, the free $T$-algebra functor. This forgetful functor must also preserve and reflect certain coequalizer diagrams. In a precise sense (see <a href="http://ncatlab.org/nlab/show/monadicity+theorem" rel="nofollow">Monadicity theorem</a>) we can identify whether or not a functor is a forgetful functor from a category of algebras over some monad by these properties. This gives a useful criterion for seeing whether or not something is in the image of such a functor and what are the maps between such objects. This (or its dual formulation) gives descent type theorems, which is probably what algebraic geometers care most about. These theorems will generally tell you something like when is the category of sheaves of some type equivalent to a category of descent data of a very particular form.</p> <p>I would recommend looking at Borceux's 'Handbook of Categorical Algebra' Vol. 2 Ch. 4 for more details. I'm sorry that I don't have a more geometric reference at hand.</p> http://mathoverflow.net/questions/84161/hopf-algebras-examples/84164#84164 Answer by Justin Noel for Hopf algebras examples Justin Noel 2011-12-23T14:17:04Z 2011-12-23T14:23:05Z <p>I will elaborate on Bruce's examples and add a couple of my own.</p> <p>Of course, the homology and cohomology of topological groups over a field are good examples. </p> <p>For each prime prime $p$ the Steenrod algebra $\mathcal{A}_p$ which is the algebra of endomorphisms of the cohomology theory $H^*(-;\mathbb{F}_p)$. The cohomology of this Hopf algebra is the $E_2$ term of a spectral sequence, due to Adams, converging to the $p$-completed stable homotopy groups of spheres. </p> <p>The functions on any affine algebraic groups over a field are another family of examples. </p> <p>Formal group laws over a field $k$. You can read about these in Husemoller's book.</p> <p>The rational homotopy groups of connected topological group or more generally an $H$-space is a Lie algebra. A nice result of Milnor-Moore shows that the universal enveloping algebra of this Lie algebra is isomorphic as Hopf algebras to the rational homology of the space.</p> http://mathoverflow.net/questions/84122/is-a-coflat-over-a-b/84158#84158 Answer by Justin Noel for Is $A$ coflat over $A//B$? Justin Noel 2011-12-23T12:48:24Z 2011-12-23T12:48:24Z <p>I'm going to assume that your Hopf algebras are connected in which case this follows from Theorem 4.10 of Milnor-Moore (On the structure of Hopf-algebras). That result shows that $A\cong B\otimes A//B$ as a left $B$-module and right $A//B$-comodule. I should point out that this result is remarkably useful.</p> <p>This means $A$ is an extended $A//B$-comodule over a field and hence it is injective in the category of $A//B$-comodules. The fact that extended coalgebras are injective (when working over a field) is an easy exercise, but you can also find the result in the context of Hopf-algebroids as A1.2.2 'in Ravenel's Complex Cobordism and Stable Homotopy.'</p> <p>Since $A$ is injective the functor $-\square _{A//B} A$ is exact. </p> http://mathoverflow.net/questions/83925/bousfield-kan-cosimplicial-replacement-of-a-fibration-of-diagrams-is-a-fibration/83935#83935 Answer by Justin Noel for Bousfield-Kan: Cosimplicial Replacement of a fibration of diagrams is a fibration? Justin Noel 2011-12-20T11:48:19Z 2011-12-20T11:48:19Z <p>First I would like to refer you Goerss and Jardine's book "Simplicial homotopy theory" as an alternative resource, with modern typesetting. In particular, Example VII.4.2 gives a construction of the cosimplicial replacement from Bousfield-Kan and there is a very helpful discussion in VIII.2. </p> <p>The cosimplicial replacement of a diagram $F:J\rightarrow \mathcal C$ produces a codegeneracy free cosimplicial object $X^\bullet$. In particular, $X^{n}=Z^nX\times M^{n-1}X$ where $$Z^nX(j)=\prod_{f\in J_n} F(\mathrm{target}f)$$ where $J_n$ is the set of all $n+1$ composable arrows in $J$, with the first arrow originating at $j$ and none of which are the identity. The matching object can be identified with those arrows that have at least one identity morphism. </p> <p>Now switching to your notation we are trying to see if the map: $$X^{n+1}\cong Z^{n+1}X \times M^n X\rightarrow Y^{n+1}\times_{M^n Y} M^n X\cong Z^{n+1}Y\times M^n X$$ is a fibration. The map is the identity on $M^n X$ and the map $Z^{n+1} X\rightarrow Z^{n+1} Y$ is just the product, indexed as above, of levelwise fibrations, hence a fibration. Note it is also easy to see that this construction takes levelwise weak equivalences to weak equivalences.</p> http://mathoverflow.net/questions/83406/weak-operad-and-deloopings/83416#83416 Answer by Justin Noel for Weak operad and deloopings Justin Noel 2011-12-14T12:20:05Z 2011-12-14T12:31:18Z <p>Some clarifications: 1) You need that $X$ is grouplike (so the induced multiplication makes $\pi_0 X$ a group). This condition is always satisfied for a loop space, but not satisfied by the discrete $E_\infty$ space $\mathbb{N}$.</p> <p>2) In order for $\mathcal{L}$ to be an $E_\infty$ operad we require more than $\mathcal{L}(n)\simeq *$, otherwise we could have just used the commutative operad and all grouplike $E_\infty$ spaces would be topological abelian groups. We also require that $\mathcal{L}(n)$ is a cofibrant $\Sigma_n$-space, so that it has a free $\Sigma_n$ action and consequently homotopy equivalent to $E\Sigma_n$. </p> <p>In May's 'Geometry of iterated loop spaces' especially Ch. 3 <a href="http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf" rel="nofollow">http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf</a> it is shown that the delooping machine does not depend on the choice of $E_\infty$ operad.</p> http://mathoverflow.net/questions/82051/somewhat-general-question-that-includes-do-quasi-isomorphic-cdgas-have-quasi-is/82415#82415 Answer by Justin Noel for Somewhat general question that includes: "Do quasi-isomorphic cdgas have quasi-isomorphic spaces of derivations?" Justin Noel 2011-12-01T23:17:18Z 2011-12-04T11:32:59Z <p>If you have two cdgas which are cofibrant (so built out of free cdgas and their cones via an iterated sequence of pushouts) and quasi-isomorphic then their homological invariants agree (one of the properties of cofibrant models is that any quasi-isomorphism between them admits a homotopy inverse). Vitali's example illustrates the problem in general: the minimal model for $H^* (S^2)$ is cofibrant, but $H^* (S^2)$ with trivial differential is not. </p> <p>To clarify things a bit further, as Tom pointed out in the comments above, when we talk about derivations we usually have a map $A\rightarrow B$ of simplicial commutative rings (in characteristic zero we can use commutative dgas) and a $B$-module $M$. Since you didn't mention this data I assumed we were taking $A$ to be the unnamed field of characteristic zero and $B$ to be augmented over $A$ so that we could take the module $M$ to be $A$ (I think this is a common situation). Now $Der_A (B;A)$ is contravariantly functorial in $B$ as an augmented $A$ algebra and takes homotopic maps to the same map. So if I have maps of augmented $A$ algebras $B\rightarrow C\rightarrow B\rightarrow C$ such that the composite of each two maps is homotopic to the identity (such as when we have a quasi-isomorphism between two cofibrant $A$-algebras) I can apply $Der_A (-,A)$ to the sequence and obtain an isomorphism between $Der_A(B,A)$ and $Der_A(C,A)$. </p> <p>You might find it helpful to read Quillen's 1970 paper: On the (co-)homology of commutative rings. These ideas are explained and generalized there.</p> http://mathoverflow.net/questions/65180/why-are-equivariant-homotopy-groups-not-rog-graded/69281#69281 Answer by Justin Noel for Why are equivariant homotopy groups not RO(G)-graded? Justin Noel 2011-07-01T17:56:29Z 2011-07-01T18:06:17Z <p>I would like to add a few points:</p> <ol> <li><p>You can define $RO(G)$-graded homotopy groups of $G$-spectra, see for example <a href="http://www.math.uni-bonn.de/~schwede/equivariant.pdf" rel="nofollow">Stefan Schwede's course notes on equivariant homotopy theory</a>. </p></li> <li><p>These groups are interesting. For example the inclusion of the 0-skeleton $S^0\rightarrow S^{\sigma}$ of the sign representation stabilizes to a non-trivial (and non-nilpotent!) element in $\pi_0^{C_2} S^{\sigma}\cong \pi_{-\sigma}^{C_2} S^0.$ This can be proved by showing its Hurewicz image in Bredon homology is non-trivial.</p></li> <li><p>At least for finite groups $G$-representation spheres can be triangulated and constructed out of the standard $G$-cells ($G/H_+\wedge S^n,$ $G/H_+\wedge D^n$). So ordinary equivariant weak equivalences between cell complexes induce $RO(G)$-graded equivalences. </p></li> <li><p>One reason to put representation spheres into your theory is so that you have a reasonable form of equivariant Spanier-Whitehead duality. At the very least we want finite G-sets to be dualizable, so we want them to equivariantly embed into a G-sphere, which should be one of the spaces in the sphere spectrum. This will not work if we only allow trivial spheres. </p></li> </ol> http://mathoverflow.net/questions/96393/coproducts-compute-with-finite-limits-in-a-reflective-subcategoroy/96398#96398 Comment by Justin Noel Justin Noel 2012-05-09T19:04:32Z 2012-05-09T19:04:32Z @David: The tensor product with a fixed commutative algebra preserves finite products of modules since they are also finite coproducts, so it preserves finite products of commutative algebras which can be calculated as products in modules. Now we just need to check that it preserves equalizers which can be calculated as the kernel of the difference of the maps, since our fixed commutative algebra is \emph{flat} (since we are working over the reals) tensoring with our algebra preserves kernels. http://mathoverflow.net/questions/90379/equivalence-between-e-infty-spaces-and-connective-spectra Comment by Justin Noel Justin Noel 2012-03-07T17:43:53Z 2012-03-07T17:43:53Z @Thomas: I don't localize at the inclusion $S^0\rightarrow QS^0$ but the map from $\mathbb{P}S^0\rightarrow QS^0$ adjoint to that map in the category of $E^\infty$ spaces. Here $\mathbb{P}S^0$ is the free $E_\infty$ space on the pointed space $S^0$. http://mathoverflow.net/questions/90291/does-the-bordism-homology-theory-satisfy-the-weak-equivalence-axiom Comment by Justin Noel Justin Noel 2012-03-07T09:19:22Z 2012-03-07T09:19:22Z @Ricardo: Knowing that the boundary of the manifold is collared tells you that the inclusion is a Hurewicz cofibration, but do you know if it is a Serre cofibration? http://mathoverflow.net/questions/90379/equivalence-between-e-infty-spaces-and-connective-spectra Comment by Justin Noel Justin Noel 2012-03-07T09:07:09Z 2012-03-07T09:07:09Z One way to make sense of 'restricting to an equivalence' is to consider the group-like $E_\infty$ spaces as a left Bousfield localization of the category of $E_\infty$ spaces. I think you just need to invert the map from the free $E_\infty$ space on $S^0$ to $QS^0$. The fibrant objects in this category will be group-like and fibrant replacement will be group completion. To do such a construction I would want the category of $E_\infty$ spaces to be left proper. This should follow from the $E_\infty$ operad being cofibrant, by Spitzweck's thesis. http://mathoverflow.net/questions/89753/geometric-realization-of-hochschild-complex/89769#89769 Comment by Justin Noel Justin Noel 2012-02-29T09:56:58Z 2012-02-29T09:56:58Z The standard cosimplicial space modeling the homotopy fiber product $X\times_Z Y$ in degree $n$ is $X\times Z^{n}\times Y$; you appear to be missing some terms. http://mathoverflow.net/questions/89687/dualizable-classifying-spaces/89760#89760 Comment by Justin Noel Justin Noel 2012-02-29T09:42:49Z 2012-02-29T09:42:49Z I expanded the answer to clarify what I meant. If the new answer does not allay your concerns, perhaps you can tell me what you mean by a twist? http://mathoverflow.net/questions/88986/are-all-manifolds-affine Comment by Justin Noel Justin Noel 2012-02-20T12:20:17Z 2012-02-20T12:20:17Z You can do this without explicitly using Whitney's embedding theorem, if you are willing to use $C^\infty$-rings instead of $\mathbb{R}$-algebras, but the argument I know requires partitions of unity and the Hausdorff condition in an essential way. http://mathoverflow.net/questions/88448/mapping-spaces-of-diagrams/88461#88461 Comment by Justin Noel Justin Noel 2012-02-15T17:27:53Z 2012-02-15T17:27:53Z @Cary: You're right, your category is not Reedy, but it is generalized Reedy: <a href="http://arxiv.org/pdf/0809.3341v1.pdf" rel="nofollow">arxiv.org/pdf/0809.3341v1.pdf</a>. Also Quillen equivalences will induce homotopy equivalences between Dwyer-Kan function complexes. The DK complex from $X$ to $Y$ will have the same homotopy type as the mapping space from a cofibrant replacement of $X$ to a fibrant replacement for $Y$ in any of the model structures if the model structure makes the category a simplicial/topological model category. http://mathoverflow.net/questions/88354/monadicity-theorem-in-homotopy-theory Comment by Justin Noel Justin Noel 2012-02-13T22:32:07Z 2012-02-13T22:32:07Z As Mike already said, $Ho(C^T)$ is almost never monadic over $Ho(C)$. However if you are willing to use a homotopy coherent version of a monad and an algebra, Jacob Lurie has an $\infty$-category version of the Barr-Beck/monadicity theorem in his book on higher algebra. http://mathoverflow.net/questions/88354/monadicity-theorem-in-homotopy-theory/88375#88375 Comment by Justin Noel Justin Noel 2012-02-13T22:22:42Z 2012-02-13T22:22:42Z @Mike: what you said is true if $T$ is the monad associated to an $E_\infty$ operad. In the $A_\infty$ case $Ho(T)$-algebras are homotopy associative H-spaces. Niles Johnson and I have worked out the obstruction theory in lifting a map of $Ho(T)$-algebras to a homotopy class of $T$-algebra maps (under appropriate restrictions) in a forthcoming paper. http://mathoverflow.net/questions/87658/computing-homotopy-colimits-in-a-nice-simplicial-model-category/87684#87684 Comment by Justin Noel Justin Noel 2012-02-07T10:47:58Z 2012-02-07T10:47:58Z I would like to add Dugger's primer on homotopy colimits <a href="http://pages.uoregon.edu/ddugger/hocolim.pdf" rel="nofollow">pages.uoregon.edu/ddugger/hocolim.pdf</a> to this extended bibliography. Although incomplete it does compare many notions of homotopy colimit in a single place. http://mathoverflow.net/questions/86805/coproducts-of-modules-over-an-algebraic-monad Comment by Justin Noel Justin Noel 2012-01-27T14:54:01Z 2012-01-27T14:54:01Z Could he just mean that there is a surjection $\Sigma(M\coprod N)\rightarrow M \oplus N$? Here $\Sigma$ is the free $\Sigma$-module functor. That seems reasonable. http://mathoverflow.net/questions/84944/question-about-formal-group-schemes/85050#85050 Comment by Justin Noel Justin Noel 2012-01-06T22:20:02Z 2012-01-06T22:20:02Z I agree that it is extremely helpful to work out a few explicit examples by hand and I like your elliptic curve example. I think it is also handy to have an abstract viewpoint available. An even more elementary example is to work this out in an affine case such as $\mathbb{G}_m$ to obtain $\hat\mathbb{G}_m$. http://mathoverflow.net/questions/84850/cohomology-of-a-space-with-coefficients-in-a-lie-group Comment by Justin Noel Justin Noel 2012-01-04T19:56:53Z 2012-01-04T19:56:53Z If I try to form the analogue of the normalized cochain complex using a cosimplicial group I don't see why the iterated boundary map should be trivial. http://mathoverflow.net/questions/84500/the-norm-map-in-group-cohomology-via-classifying-spaces/84506#84506 Comment by Justin Noel Justin Noel 2012-01-03T08:28:25Z 2012-01-03T08:28:25Z @Chris: Sorry about that. I'm stilling learning how to use this system. It's fixed now.