Tyler Lawson
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Registered User
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Interested in algebraic topology, K-theory, arithmetic topics. I am, in theory, taking a break to do some writing. |
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May 1 |
awarded | ● homotopy-theory |
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Apr 28 |
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Can one make the category of pairs of topological spaces a model category? Are you restricting $A$ to be a subspace? If you don't, then you can use a model structure on the category of functors $\{\cdot \to \cdot\} \to X$, such as the projective or injective model structure. |
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Apr 10 |
awarded | ● Nice Question |
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Mar 28 |
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Units of MO and MU One comment is that if $S$ is the sphere spectrum there is a map $U\to GL_1(S)$ (resp. $O \t GL_1(S)$) and the description of MU explicitly makes the composite $U\to GL_1(S) \to GL_1(MU)$ nullhomotopic as a map of infinite loop spaces (similarly for $MO$). Another is that even though $MO$ is equivalent to a product of Eilenberg-Mac Lane objects, $GL_1(MO)$ is (as an infinite loop space) not. $GL_1$ is tough, and a good chunk of literature in orientation theory (much by Peter May) is devoted to determining information about it. |
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Mar 28 |
awarded | ● Nice Answer |
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Mar 24 |
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cohomology of infinite product of EM spaces Eric, you're correct; I was too fast, and interpreted the question as about the cohomology of the infinite product rather than the honest mapping space. |
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Mar 23 |
answered | cohomology of infinite product of EM spaces |
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Mar 16 |
awarded | ● Famous Question |
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Mar 12 |
awarded | ● Good Answer |
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Feb 8 |
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Tools for collaborative paper-writing as an aside, there is a really nice tutorial on how to use git online now: try.github.com |
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Jan 30 |
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Is the derived category of abelian groups a subcategory of the stable homotopy category? Indeed, if a general categorical argument worked then you could replace $S \to HZ$ by $HZ \to HZ/p^2$ and "prove" that the derived category of $Z/p^2$ embeds into the derived category of $HZ$. |
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Jan 26 |
awarded | ● Nice Answer |
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Jan 26 |
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A canonical way to kill a subset of cohomology in a dg-algebra: via $A_\infty$-algebras? References? If you are in a good setting (maybe you have something commutative enough, with an ideal given by a regular sequence) then you can often construct a quotient that looks nice on the cohomology level, but there may multiple inequivalent ways to do it. |
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Jan 26 |
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A canonical way to kill a subset of cohomology in a dg-algebra: via $A_\infty$-algebras? References? Mikhail: The shortest quotient would be to take some kind of homotopy pushout; the category of modules would then be some kind of category of dg $A$-modules $M$ equipped with, for each $s \in S$, a chain homotopy from multiplication-by-$s$ to zero. But this only has a "versal" property, and it depends on the actual set of generators (e.g. if you use different generators or include redundant generators in $S$) you get a different and likely inequivalent algebra. This has kind of lousy properties. My own feeling is that quotients are just hard in the derived setting. |
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Jan 26 |
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A canonical way to kill a subset of cohomology in a dg-algebra: via $A_\infty$-algebras? References? Are you looking for the resulting map on cohomology to be some kind of quotient map? Are you looking for the quotient to have some kind of universal property? Are you looking for something else? For many definitions of "quotient" there is almost always a nontrivial moduli space of them. |
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Jan 24 |
awarded | ● Enlightened |
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Jan 24 |
awarded | ● Nice Answer |
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Jan 23 |
accepted | Computation of [ HZ/4, HZ/4] |
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Jan 23 |
answered | Computation of [ HZ/4, HZ/4] |
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Jan 21 |
awarded | ● Nice Question |
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Jan 18 |
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Extensions of discrete groups by spectra A possible short + lazy answer by turning a classification into a definition: An extension is classified by an element of $H^2(G,A)$. You can give $K(A,2)$ a $G$-action, form the associated bundle $K(A,2) \to Y \to BG$, and then this $H^2$ can be identified with the set of space of sections of $Y \to BG$. Now replace $K(A,2)$ with $B^2 X$ for $X$ an infinite loop space. (A more modern thinker than me might say something about extensions of the $\infty$-groupoid $BG$ by a symmetric monoidal $\infty$-groupoid.) |
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Jan 16 |
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Mod 3 Moore spectrum That paper of Toda's is a little concentrated. The nature of the answer to this question will probably depends pretty heavily on whether you know what a Massey product is. You can show that there must be a map from $M$ to $H\mathbb{Z}/3$ which preserves the unit and multiplication. The resulting map $H_* M \to H_* H\mathbb{Z}/3$ has as image a square-zero class in the dual Steenrod algebra ($\tau_0$) whose triple Massey product $\langle \tau_0, \tau_0, \tau_0 \rangle$ is not in the image. |
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Jan 12 |
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What is a simplicial commutative ring from the point of view of homotopy theory? I think I agree with your approach, but I suspect that asking for the target category to be "$\Sigma_n$-objects" may be underemphasizing the difference between nonequivariant and equivariant homotopy theory. It seems to me that the target category itself should probably be part of the data. (Side question that I've asked several people: what's the most refined structure that you can put on a smash power over $R$, where $R$ is some commutative object?) |
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Jan 12 |
awarded | ● Nice Answer |
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Jan 11 |
awarded | ● Nice Answer |
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Jan 10 |
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What is a simplicial commutative ring from the point of view of homotopy theory? Such a construction can't be too general, because it relies on the existence of "symmetric power" functors on simplicial $k$-modules that refine the homotopical symmetric powers (and those don't exist in general). The monad you'd want to write down is basically built out of these symmetric power functors, and I've heard it said that a refinement to a simplicial commutative ring is equivalent to being equipped with these symmetric powers. I don't know the argument. |
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Jan 8 |
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Counterexamples to Smallness of Harmonic Spectra @Fernando: In the maps on homotopy groups $\oplus [S^n,X_i] \to [S^n, \oplus X_i] \to [S^n, L(\oplus X_i)]$, the first map is always an isomorphism but the second can't be because the map isn't a weak equivalence. |
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Dec 29 |
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Hochschild homology and change of non-ground ring @Sasha: I'm afraid that I don't have a ready reference. You can prove it by constructing a double complex whose rows are the Hochschild complexes of $A \otimes R^{\otimes q} \otimes T$, with vertical differentials coming from the bar construction. (It's essentially the Hochschild complex of the DGA $A \otimes^{\mathbb L}_R T$.) |
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Dec 29 |
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Hochschild homology and change of non-ground ring If A is flat over R, or more generally $Tor^R(A,T) = 0$ in positive degrees, then there is a spectral sequence starting with $Tor^{HH_* R}(HH_* A, HH_* T)$ converging to your desired Hochschild term. Does that work in your situation? |
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Dec 18 |
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Connection of X(n) spectra to formal group laws It's known that the $X(n)$ spectra don't admit $E_4$ structures. To see this, there would have to be an $E_4$-map $X(n)\to H\mathbb{Z}/2$. Look at the associated map on mod-2 homology. The element $\xi^2_1$ is in the image, but it generates the subalgebra $\mathbb{Z}/2[\xi^2_i]$ under the $E_4$-Dyer-Lashof operations. The homology of X(n) is $\mathbb{Z}/2[b_1,\ldots,b_n]$, which is not big enough to surject onto it. |
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Dec 14 |
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Compact MU or BP Modules My suspicion is that there are almost no finite spectra which admit MU-module structures. |
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Dec 9 |
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Homotopy $\pi_4(SU(2))=Z_2$ This question is somewhat terse -- it's not clear whether you want to know a generating class (as in the Hopf map comment), or why this is the only class (a nontrivial calculation), or how to detect whether a map is nullhomotopic (along the lines checking the framing of a preimage of a point, as in solbap's comment). Some clarification of what you'd really like to know would help us understand how to help. |
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Dec 8 |
awarded | ● Nice Answer |
