Gregory Arone
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Registered User
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Mar 19 |
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Can one compare monads arising from homotopy equivalent adjunctions? Hi Peter. This is a helpful answer, thanks. It is interesting that the assumption that the map $L\to L'$ is an equivalence is not directly relevant to your argument. What matters is that $R$ and $R'$ can be moved past the bar constructions. The assumptions end up being equivalent to a form of Barr-Beck criterion for monadic descent. They guarantee that $T$ and $T'$-algebras embed into $\mathcal D$. You show that the images are the same. I had wondered whether the assumption that the maps $\alpha_L, \alpha_R$ are equivalences would give a more direct way to compare the monads. Perhaps not. |
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Mar 18 |
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Can one compare monads arising from homotopy equivalent adjunctions? Thanks. I would be happy to see a little more details of the argument that the map that you constructed is multiplicative. Do you have a reference to the statement that "adjoint pairs are unique"? |
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Mar 18 |
awarded | ● Student |
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Mar 18 |
asked | Can one compare monads arising from homotopy equivalent adjunctions? |
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Jan 29 |
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Is the derived category of abelian groups a subcategory of the stable homotopy category? added 423 characters in body |
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Jan 29 |
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Is the derived category of abelian groups a subcategory of the stable homotopy category? +1 You are right. I was originally going to say that it is enough to check it for $M$ an Eilenberg-Maclane spectrum, using the same splitting argument as you did (honest). Then somehow convinced myself in a hurry that I could get away with a categorical argument. But it does not work. The map $HZ \wedge M \to M$ splits, but not naturally. I will edit. |
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Jan 29 |
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Is the derived category of abelian groups a subcategory of the stable homotopy category? added 68 characters in body; deleted 9 characters in body |
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Jan 29 |
answered | Is the derived category of abelian groups a subcategory of the stable homotopy category? |
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Jan 25 |
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Nontrivial copies of SO(r) in SO(n) @mariano: do you mean topologically isomorphic, or abstractly isomorphic? If a subgroup of $SO(n)$ is topologically isomorphic to $SO(r)$, then it is compact, and therefore closed. But if you are thinking about an abstract isomorphism, then for example $SO(2)$ is abstractly isomorphic to a direct sum of $Q/Z$ and an uncountable number of $Q$s. I think it is not hard to embed this group for example into $S^1\times S^1$ and therefore into $SO(4)$. |
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Jan 22 |
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Cohomology of Complements by an analytic subset? @Serge: You are right. I edited. |
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Jan 22 |
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Cohomology of Complements by an analytic subset? added 93 characters in body |
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Jan 22 |
answered | Cohomology of Complements by an analytic subset? |
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Jan 22 |
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Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group? added 63 characters in body; added 2 characters in body |
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Jan 22 |
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Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group? @Ricardo - OK, I think I got it, thanks. |
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Jan 22 |
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Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group? Are you sure that the generalization works? The homeomorphism of $S^2\setminus x_1$ with ${\mathbb R}^2$ has to depend continuously on $x_1$. It seems that constructing such a homeomorphism is equivalent to trivializing the tangent bundle of $S^2$. |
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Jan 21 |
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Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group? added 15 characters in body |
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Jan 21 |
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Example: a pair of nonisomorphic parallel morphisms with isomorphic cones $X$ is a direct sum of infinitely many copies of $C$ and infinitely many copies of $\Sigma C$. Consider the map $C\to X$ that is inclusion into the first summand. The cone of this map is obtained by removing one copy of $C$ from $X$. This is isomorphic to $X$, because $\infty -1=\infty$. Now consider the null map from $C$ to $X$. The cone of this map is $X\oplus \Sigma C$. This, again, is isomorphic to $X$, because $\infty +1 =\infty$. Thus the two cones are isomorphic to each other. |
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Jan 21 |
answered | Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group? |
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Jan 21 |
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Example: a pair of nonisomorphic parallel morphisms with isomorphic cones deleted 8 characters in body |
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Jan 21 |
answered | Example: a pair of nonisomorphic parallel morphisms with isomorphic cones |
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Jan 20 |
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Tangent space of the moduli stack of Drinfeld modules Googling "Hochschild cohomology and derivations" will bring up a few references. It also is easy to figure it out from scratch by considering the Hochschild complex $M\stackrel{d^0}{\to}\hom(R, M) \stackrel{d_1}{\to} \hom(R\otimes R, M)\to \cdots$. $d_0$ sends $m$ to the inner derivation $f(r)=rm-mr$. So if $R$ is commutative, $d^0=0$. $d^1$ sends a homomorphism $f$ to the homomorphism $f(r_1\otimes r_2)-r_1f(r_2)-f(r_1)r_2$. So the kernel of $d^1$ consists exactly of derivations from $R$ to $M$. |
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Jan 19 |
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Tangent space of the moduli stack of Drinfeld modules It is well-known that for a commutative $k$-algebra $R$ and and $R$-module $M$, $HH^1(R, M)\cong Der(R, M) \cong \hom_R(\Omega_{R/k}, M)$. Could it be just this? I am not sure if the dimension shift signifies something important or is just a matter of grading... |
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Jan 19 |
answered | Borel localization with Mayer-Vietoris sequence |
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Jan 19 |
accepted | A version of the group ring using direct product rather than direct sum? |
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Jan 19 |
answered | A version of the group ring using direct product rather than direct sum? |
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Jan 18 |
awarded | ● Enlightened |
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Jan 18 |
awarded | ● Nice Answer |
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Jan 18 |
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A version of the group ring using direct product rather than direct sum? It looks like you will have trouble with infinite sums on the right hand side. For example, consider $x=\Sigma g$, the sum of all elements of $G$. What is the coefficient of the identity element in the expansion of $x^2$? Looks like it is the order of $G$, which is infinite. |
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Jan 18 |
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Extensions of discrete groups by spectra John, it might help if you try to indicate what kind of answer you are hoping for. My first thought is something along the lines of `a fiber bundle over $BG$ with fibers $A$'' (a bundle of spectra). But it is a variation of the same `short and lazy'' answer. |
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Jan 16 |
answered | Functorial properties of blow-up |
