User gregory arone - MathOverflow

Can one compare monads arising from homotopy equivalent adjunctions?

2013-03-18T13:04:06Z

Suppose we have a Quillen adjunction $L\colon {\mathcal C} \leftrightarrow {\mathcal D}: R$. For convenience let us assume that all objects of $\mathcal C$ are cofibrant and all objects of $\mathcal D$ are fibrant, so that $L$ and $R$ are homotopy functors. Let $(L', R')$ be another Quillen adjunction. Let's suppose that we have a natural transformation $\alpha_L\colon L \to L'$. It induces a natural transformation $\alpha_R\colon R' \to R$. Let us assume that $\alpha_L$ is a weak equivalence. If I am not mistaken, it follows that $\alpha_R$ is a weak equivalence.

Let $T$ and $T'$ be the monads $RL$ and $R'L'$ respectively. As far as I see, in general there is no map of monads between $T$ and $T'$.

Question : Are the homotopy categories of $T$ algebras and $T'$-algebras equivalent?

Remark : The adjunction $(L, R)$ induces a Quillen adjunction between $T$-algebras and $\mathcal D$. This adjunction is sometimes (quite often?) a Quillen equivalence. In this case one may say that $T$ satisfies homotopy descent. It seems that if $T$ satisfies homotopy descent, then so does $T'$, and in this case Ho$(T-\mathrm{alg})$ and Ho$(T'-\mathrm{alg})$ are equivalent, since both are equivalent to Ho$(\mathcal D)$. I want to know if there is a direct way to compare algebras over $T$ and $T'$, without using descent.

Answer by Gregory Arone for Is the derived category of abelian groups a subcategory of the stable homotopy category?

2013-01-29T21:24:56Z

I think the answer to Question 1 is positive. Think of $SH$ as the homotopy category of modules over the sphere spectrum $S$. The category $D(Ab)$ is equivalent to the homotopy category of modules over the Eilenberg-Mac Lane spectrum $HZ$. Your adjunction is equivalent to the adjunction between $S$-modules and $HZ$ modules, where the right adjoint is pullback along the natural map of ring spectra $S\to HZ$, and the left adjoint is the functor $M\mapsto HZ\wedge M$.

Your question is equivalent to this: given $HZ$-modules $M, N$, is the map

$ [M, N]_{HZ}\to [M, N]_S$

injective? By adjunction

$[M,N] = [HZ \wedge M, N]_{HZ}$

and the map * is induced by the map $HZ\wedge M \to M$. I claim that the last map is a split surjection in the homotopy category of $HZ$-modules.

Edited to account for Fernando's comment.

Since every $HZ$-module splits (non-naturally) as a wedge sum of Eilenberg Maclane modules It is enough to check this claim when $M=HA$, in which case it is an easy calculation. The homotopy groups of $HZ\wedge HA$ are isomorphic to the homology groups of $HA$. By Huriewicz theorem, this is $A$ in dimension zero. Using the general splitting result again, it follows that $HA$ is a summand of $HZ\wedge HA$ in the category of $HZ$-modules.

Therefore * is injective.

Answer by Gregory Arone for Cohomology of Complements by an analytic subset?

2013-01-22T06:38:20Z

[ Edited , following Serge Lvovsky's comment]

When $\Omega$ is a ball $H^1(\Omega\setminus S)\cong H_{2n-2}(S^c)$ by Alexander duality. Here $S^c$ is the one-point compactification of $S$.

Answer by Gregory Arone for Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?

2013-01-21T20:35:01Z

Just in the case $n=3$, it seems that the natural fibration $F_4S^2\to F_3S^2$ (forget the last point) has a section. The section $F_3S^2 \to F_4 S^2$ is defined by the formula $$(v_1, v_2, v_3)\mapsto (v_1, v_2, v_3, \frac{(v_2-v_1)\times (v_3 - v_1)}{||(v_2-v_1)\times (v_3 - v_1)||}).$$ Here $v_1, v_2, v_3$ are three distinct unit vectors in ${\mathbb R}^3$, representing a point in $F_3S^2$. The idea is that you consider the line through the origin perpendicular to the plane containing the points $v_1, v_2, v_3$, and take the intersection of that line with $S^2$ to be your fourth point. Use the ordering of the points to decide which of the two points to take.

Such a section defines a split injection from the third pure spherical braid group to the fourth.

Edit: Ricardo's answer shows a simpler way to add a fourth point, and also generalizes the argument to produce a section of the fibration $F_mS^2 \to F_nS^2$ for all $m>n \ge 3$.

Answer by Gregory Arone for Example: a pair of nonisomorphic parallel morphisms with isomorphic cones

2013-01-21T18:21:46Z

Let $C$ be a non-contractible complex. Let $X$ be a direct sum of a countably infinite number of copies of $C$ plus a countable infinite number of copies of $\Sigma C$. Then the inclusion of $C$ into $X$ as a direct summand, and the null map from $C$ to $X$, are non-isomorphic maps with isomorphic mapping cones.

Even if it works, this example feels like a swindle. Is there one with finitely generated modules?

Answer by Gregory Arone for Borel localization with Mayer-Vietoris sequence

2013-01-19T15:44:33Z

Regarding your last question: torsion is essential whenever the map $M^G \to M$ (inclusion of fixed points) fails to induce an isomorphism on equivariant cohomology (as opposed to isomorphism mod torsion). Probably the simplest example is given by the standard action of $S^1$ on $S^2$ with two fixed points. The inclusion of fixed points is the map $S^0 \to S^2$. The induced map on Borel constructions is equivalent to the map ${\mathbb C}P^\infty \coprod {\mathbb C}P^\infty \to {\mathbb C}P^\infty \vee {\mathbb C}P^\infty$, which is not a rational cohomology isomorphism.

Regarding your first question, Meyer-Vietoris sequence plays an important role in the proof, but it is not, I think, the entire proof. Basically, it is used to reduce the theorem to the following assertion: For a proper subgroup $H$ of $G$, the cohomology of $BH$ is a torsion module over $H^*(BG)$.

Answer by Gregory Arone for A version of the group ring using direct product rather than direct sum?

2013-01-19T13:02:30Z

OK, I will make my comment into an answer. The multiplication is not well defined, because of infinite sums :-)

Answer by Gregory Arone for Functorial properties of blow-up

2013-01-16T16:08:10Z

This does not answer your specific question, but here is a fact about the functoriality of blow-up.

Let $f\colon X \to Y$ be a smooth map between smooth varieties. Suppose that $X_0\subset X$ and $Y_0\subset Y$ are smooth subvarieties, such that $X_0=f^{-1}(Y_0)$. If $f$ induces a fiber-wise injection from the normal bundle of $X_0$ to the normal bundle of $Y_0$, then $f$ induces a map from the blow-up of $X$ and $X_0$ to the blow-up of $Y$ at $Y_0$.

This does not apply to your example, where $X_0$ consists of isolated singularities. But I would think that in any situation where you can sensibly say that ``$f$ induces a monomorphism of normal bundles'', $f$ will induce a map of blow-ups.

Answer by Gregory Arone for Is the dual notion of a presheaf useful?

2010-07-16T01:39:11Z

Covariant functors from the category of pointed sets to the category of pointed topological spaces are sometimes called $\Gamma$-spaces, and they have been important in algebraic topology. One reason is that $\Gamma$-spaces model infinite loop spaces (and therefore connective spectra) and are very helpful for understanding stable homotopy theory.

$\Gamma$-spaces also serve as a model for particularly well-behaved covariant functors from the category of pointed topological spaces to itself. Of course, these functors play an important role in topology as well. I like to think of Goodwillie's Calculus of Homotopy Functors (and also of Michael Weiss's Orthogonal Calculus) as a kind of "sheaf theory for covariant functors". In these theories, covariant functors are analogous to presheaves and linear functors are analogous to sheaves (The definition of a linear functor is essentially a homotopy-invariant version of the definition of a sheaf). The process of approximating a general functor by a linear one is analogous to sheafification, and so forth. These theories provide methods for studying certain types of functors, but of course they also tell you something about the category of spaces itself.

Answer by Gregory Arone for Are there homotopy equivalences that are not weak homotopy equivalences?

2010-07-15T19:34:49Z

f is always a weak homotopy equivalence. This is related to the following assertion: Let $g\colon X\to X$ be homotopic to the identity. Then for any $x_0\in X$. $g$ induces an isomorphism $g_*\colon\pi_*(X,x_0)\to \pi_*(X, g(x_0))$ . This is so because $g_*$ is the same as conjugation by the path $H(x_0\times I)$, where $H:X\times I \to X$ is the homotopy of $g$ with the identity. Conjugation by a path induces an isomorphism.

Now let $h$ be a homotopy inverse of $f$ and consider the homomorphisms

$$\pi_*(X, x_0)\to \pi_*(Y, f(x_0))\to \pi_*(X, hf(x_0)) \to \pi_*(Y, fhf(x_0))$$

the first and the third being induced by $f$ and the second by $h$. The composition of the first two is an isomorphism and so is the composition of the last two. It follows that all three homomorphisms are isomorphisms.

Answer by Gregory Arone for Question about the fundamental group and homotopy equivalence

2010-07-11T14:36:38Z

I think Y is homotopy equivalent to $S^2\vee S^1\vee S^1$. Proof: $Y$ is homotopy equivalent to the homotopy cofiber or the map from $3$ point to $S^2$. This map is null-homotopic, so $Y$ is equivalent to the wedge sum of $S^2$ with the suspension of three points. Suspension of three points is equivalent to $S^1\vee S^1$.

It is easy enough to construct the homotopy equivalence directly.

In conclusion: both (1) and (2) are false.

Answer by Gregory Arone for Obstruction theory for non-simple spaces

2010-07-09T03:12:39Z

Paul Olum developed some obstruction theory for maps into non-simple spaces back in the 1940-ies and 50-ies. You may want to check out his paper "Obstructions to extensions and homotopies", Annals of Mathematics, Vol 52, 1950, pp 1-50, if you have not looked at it yet.

Answer by Gregory Arone for Do we have a pullback operation on simplicial chains?

2010-06-24T03:25:17Z

For a general map, there is no such pullback operation, but there are things you can do in special cases. For example, if $f\colon X\to Y$ is a finite cover, there is a chain homomorphism $C(Y)\to C(X)$ that sends a singular simplex in $Y$ to the sum of its lifts in $X$. This induces the transfer homomorphism in homology.

There are more general versions of the transfer that can be realized on chain level. See for example this paper of H. Munkholm

Answer by Gregory Arone for Checking whether the image of a smooth map is a manifold

2010-06-15T15:46:36Z

To elaborate on Diego Matessi's answer (and Boyarsky's comment which I only saw after writing this)... the map $F$ is symmetric, so it factors as a composite map $$C^2\longrightarrow (C^2)_{\Sigma_2}\longrightarrow C^2.$$ Here$\Sigma_2$ is the group with $2$ elements acting on $C^2$ by switching coordinates, the first map is the quotient map, and the second map is well-known to be a homeomorphism. It can be thought of as the map that sends an unordered pair $[z_1, z_2]$ to the unique monic complex polynomial whose roots are $-z_1$ and $-z_2$. So the map $S^3\to F(S^3)$ is not injective. Nevertheless, it is an easy exercise to show that $F(S^3)$ is homeomorphic to $S^3$. But then again, it {\em seems} to me that $F(S^3)$ should not be a smooth submanifold of ${\mathbb C}^2$. $F(S^3)$ can be identified with the space of monic complex quadratic polynomials $z^2+bz+c$ whose pair of roots gives a unit complex vector. So it is the space of pairs of complex numbers $(b, c)$ that solve the equation $$|\frac{-b+\mbox{first square root of } b^2-4c}{2}|^2+|\frac{-b+\mbox{second square root of } b^2-4c}{2}|^2=1.$$ It seems that the space of solutions of this equation should not be smooth, and that a singularity should occur where the discriminant $b^2-4c$ is zero - which corresponds to the diagonal in the original $C^2$, but I could be mistaken.

Edit Let me point out that the equation can be simplified. Let $\Delta=b^2-4c$. Then the equation is in fact equivalent to $$(*) \,\, |b|^2+|\Delta|=2.$$ Since the map $(b,c)\to (b, b^2-4c)$ is a diffeomorphism, the question of whether $F(M)$ is a smooth submanifold is equivalent to the question whether the space of solutions of the equation ($\ast$) is a smooth submanifold of $C^2$. I just wanted to point this out because equation ($\ast$) looks simpler than the one I originally wrote. In the meantime, Sergei Ivanov gave an argument showing that it is not smooth.

Answer by Gregory Arone for Pointed vs. unpointed homotopy colimits

2010-06-08T16:02:22Z

I think the answer is yes. Here is an attempt at an argument.

Let $SS_*$ and $SS$ be the categories of pointed and unpointed simplicial sets. Let $[C, SS]$ be the category of all functors from $C$ to $SS$ and let $[C, SS_*]_*$ be the category of all pointed functors from $C$ to $SS_*$ . Define similarly the functor categories $[C,Sets]$ and $[C, Sets_*]_*$ . Consider $[C, Sets]$ and $[C,Sets_*]_*$ to be subcategories of $[C, SS]$ and $[C, SS_*]_*$ respectively.

The functor categories $[C, SS]$ and $[C, SS_*]_*$ have well-known model structures where weak equivalences and fibrations are defined pointwise. It is not difficult to describe the cofibrations explicitly. The cofibration in $[C, SS]$ are generated by maps of the form $$I\times \hom(x_0, -) \longrightarrow J\times \hom(x_0, -)$$ where $I, J$ are simplicial sets, $x_0$ is an object of $C$, $\hom(x_0, x)$ denotes the (pointed) set of morphisms in $C$, and the map is induced from a cofibration of simplicial sets $I\hookrightarrow J$. Similarly, the cofibrations in $[C, SS_*]_*$ are generated by maps of the form $$I_+\wedge \hom(x_0, -) \longrightarrow J_+\wedge \hom(x_0, -).$$

One can define homotopy tensor product using cofibrant replacement in this model structure. Namely, if F and G are two functors (either pointed or unpointed), then $B(G, C, F)\simeq cG \otimes cF$, where $c$ denotes a cofibrant replacement in the appropriate functor category. In fact, it is enough to take a cofibrant replacement of either $F$ or $G$. That is, $cG\otimes F\simeq G\otimes cF\simeq cG\otimes cF$.

There is an obvious forgetful functor that I will denote by $R$. $$R\colon [C, SS_*]_* \longrightarrow [C, SS].$$ Your question is equivalent to the following: does $R$ preserve homotopy coends? You only ask the question for set-valued functors, but I think the answer is yes in general. Let me formulate it a little more precisely. Let $F\colon C\to SS_*$ and $G\colon C^{op}\to SS_*$ be pointed functors. There is an evident natural map from the (unpointed) homotopy coend $RG\otimes^h RF$ to the pointed homotopy coend $G\otimes^h F$ . We want to show that this map is an equivalence. Let us first check it when $F$ has the form $F(-)=I_+\wedge \hom(x_0, -)$ for some simplicial set $I$ and object $x_0$ of $C$. In this case, $F$ is cofibrant in $[C, SS_*]_*$ , so the pointed homotopy coend of $F$ and $G$ is equivalent to the pointed strict coend which, by Yoneda Lemma, is equivalent to $I_+\wedge G(x_0)$ . Now let us consider $RF$ and $RG$. It is not immediately obvious whether $RF$ is cofibrant in $ [C, SS]$. On the other hand, $RF$ is objectwise equivalent to the following homotopy pushout $$*\times \hom(0, -)\longleftarrow I\times \hom(0, -) \longrightarrow I\times \hom(x_0, -) .$$ Taking homotopy coend with $RG$ preserves objectwise homotopy pushouts. It follows that $RF \otimes^h RG$ is equivalent to the following homotopy pushout $$*\times \hom(0, -)\otimes^h RG\longleftarrow I\times \hom(0, -)\otimes^h RG\longrightarrow I\times\hom(x_0, -) \otimes^h RG.$$ Which, again using Yoneda Lemma, together with the fact that $RG(0)=*$, implies that $RF \otimes^h RG$ is equivalent to $I_+\wedge G(x_0)$ . So we obtain that $RF\otimes^h RG$ is equivalent to $F\otimes^h G$ . With a little more careful diagram-chasing it should not be hard to see that the canonical map $RF\otimes^h RG\longrightarrow F\otimes^h G$ induces this equivalence.

For a general pointed $F$, one can present $F$ as a homotopy colimit along generating cofibrations in $[C, SS_*]_*$ (take a cofibrant replacement of $F$), and one obtains the result using a similar calculation plus induction.

Edit: When I was writing this post, I actually changed my mind in the middle about how I wanted to do this, so I think it came out a bit unfocused. The idea is straightforward. Fix a contravariant pointed functor $G$. We want to check that the natural map from $F\otimes^h G$ (unpointed derived tensor product) to $F\otimes^h_* G$ (pointed derived tensor product) is an equivalence for functors $F$ of the form $F=I_+\wedge\hom(x_0, -)$. This is good enough, because all other homotopy types of pointed functors can be built as repeated homotopy pushouts of functors of this type. So, I need to calculate both the pointed and unpointed derived tensor products of $F$ and $G$ for this type of $F$. The pointed tensor product is easy, because $F$ is cofibrant in the pointed models structure, so the derived product is equivalent to the strict product, which can be calculated using the YL. The unpointed tensor product is slightly less obvious, because it is not clear that $F$ is cofibrant in the unpointed model structure, and this is why the derived case does not follow immediately from the strict case. But, $F$ can be presented as a homotopy pushout of free (in the unpointed sense!) functors, and an elementary little calculation shows that the unpointed derived tensor product agrees with the pointed one.

This is a proof by calculation. Since the "calculation" is extremely easy, I feel it is not too bad. But it would be nice to see a conceptual reason why it ought to be true. I believe such a reason exists, but I have not been able to nail it down.

Answer by Gregory Arone for What are normalized singular chains good for?

2010-06-10T19:15:08Z

I don't know Costello's reasons, but for example I find it convenient that the normalized chains on the one-point space are concentrated in degree zero (so the normalized chains functor takes the unit objet to unit object). Generally, when two construction have all the same good properties one tends to prefer the smaller one...

Comment by Gregory Arone

2013-03-19T07:15:50Z

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.

Comment by Gregory Arone

2013-03-18T19:23:56Z

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"?

Comment by Gregory Arone

2013-01-29T23:28:14Z

+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.

Comment by Gregory Arone

2013-01-25T21:14:36Z

@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 $SO(4)$.

Comment by Gregory Arone

2013-01-22T08:32:17Z

@Serge: You are right. I edited.

Comment by Gregory Arone

2013-01-22T00:41:05Z

@Ricardo - OK, I think I got it, thanks.

Comment by Gregory Arone

2013-01-22T00:03:44Z

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 $S^2$.

Comment by Gregory Arone

2013-01-21T22:11:39Z

$X$ 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.

Comment by Gregory Arone

2013-01-21T16:54:59Z

It seems to me that $f$ and $f'$ are isomorphic, via multiplication by $-1$. Notice that isomorphic in the homotopy category does not mean the same in the homotopy category.

Comment by Gregory Arone

2013-01-20T07:24:11Z

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 $M$.

Comment by Gregory Arone

2013-01-19T20:36:32Z

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...

Comment by Gregory Arone

2013-01-18T12:50:25Z

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.

Comment by Gregory Arone

2013-01-18T12:19:03Z

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.

Comment by Gregory Arone

2010-07-13T18:30:23Z

I also wonder if we are supposed to think of a discrete category or a topological one. If each point in each open set in the chart corresponds to an object, then the set of objects should have a natural topology. Do we take it into account?

Comment by Gregory Arone

2010-07-13T18:20:52Z

Are you sure that all the morphisms in your category are invertible? IF you want to encode a specific manifold as a category, then I would guess that you want to include morphisms that correspond to inclusions of opens subsets of $M$ into each other. My feeling is that the classifying space of this category would be equivalent to $M$. I could be totally off though.